Learning Experiences KNOWLEDGE IN ACTION
Put your curiosity to the test.
Change the settings yourself and see the principles in action.
What will you discover today?
Many fields, small discoveries
Choose what sparks your curiosity today.
Experiences are shown in random order.
01 / IT · The Path to an Answer
Search one step at a time,
or narrow the search by half.
Find a target among numbers sorted in order starting from 1.
Follow both methods and see how many numbers each checks.
Changing the target or number of values restarts the search comparison.
The count is the number of values checked for a match with the target. While searches are still running, the final difference between the two methods may change.
Change the target and search to discover today's mission.
After searching for yourself
Checking the middle number lets you rule out the half without the answer in one step.
Binary search can be used when the data is sorted. This experience uses numbers that are already sorted, so the cost of sorting is excluded from the comparison. If the target is at the very beginning, linear search may find it faster.
Read more about the principle · NIST: Binary Search
02 / IT · Waiting and Delivery
Even on a fast connection,
there is still a wait.
Change the file size, connection speed, and round-trip latency.
Examine the wait for a response separately from the time spent transferring the file.
1 byte equals 8 bits. Here, 1 MB = 100 ten thousand bytes, and 1 Mbps = 100 ten thousand bits per second.
Change the size, speed, and latency to discover today's mission.
After breaking down the wait
File transfer speed and the time spent waiting for a response are separate factors.
Total time = one round-trip latency + file size × 8 ÷ connection speed. This simplified model assumes a constant transfer speed with no connection setup or congestion. Actual downloads vary with repeated communications, server processing, congestion, and transmission loss.
Read more about the principle · Cloudflare: Network Latency
03 / Chemistry · Reading Differences with Logarithms
When pH differs by one unit,
concentration differs tenfold.
Change the pH of two virtual solutions.
Compare A and B using approximate hydrogen ion concentrations.
The solution colors are conceptual aids to distinguish the solutions. They do not represent actual indicator colors or solution safety.
Change the pH of both solutions to discover today's mission.
After comparing for yourself
The lower the pH, the higher the approximate hydrogen ion concentration.
Here, we calculate [H₃O⁺] ≈ 10−pH mol/L. Strictly, pH is defined using activity rather than concentration. A neutral pH of 7 applies at 25°C, and pH alone cannot determine safety or the strength of the acid itself. 0–14 is the exploration range for this experience; not all solutions have a pH within it.
Read more about the principle · OpenStax: pH and pOH
04 / Chemistry · The Reactant That Runs Out First
Two kinds of molecules—
can they all react?
Pair virtual molecules according to their reaction ratio.
Find how much water can form according to 2H₂ + O₂ → 2H₂O.
This virtual grouping model pairs 2 hydrogen molecules with 1 oxygen molecule per group. The water shown is the maximum number of molecules this model can produce.
Change the molecule counts to discover today's mission.
After pairing for yourself
The molecules that can be paired in the required ratio determine how many water molecules form.
Each group requires 2 H₂ molecules and 1 O₂ molecule. The number of possible groups is the smaller of the integer part of 'hydrogen count ÷ 2' and the oxygen count. An odd hydrogen count leaves 1 hydrogen molecule unpaired, so both kinds may remain. This is not an experiment reproducing actual reaction rates, conditions, or yields.
Read more about the principle · OpenStax: Limiting Reactants and Reaction Yields
05 / Physics · Three Values for Understanding Electricity
When voltage and resistance change,
what happens to current?
This virtual circuit applies a DC voltage across a single resistor.
Compare the current flowing and the power consumed.
Relative power is compared against 0.36 W at 6 V and 100 Ω. 1 A equals 1,000 mA.
Change the voltage and resistance to discover today's mission.
After changing the circuit values
At constant resistance, current is proportional to voltage, and power is proportional to voltage squared.
Calculations use Ohm's law, I = V ÷ R, and power P = V² ÷ R. This is a DC model of a single resistor with constant resistance. In real materials, light bulbs, and electronic devices, resistance may change with temperature and operating conditions. This on-screen experience explains the relationships and does not provide instructions for actual wiring.
Read more about the principle · OpenStax: Ohm's Law · Power and Electrical Energy
06 / Physics · Energy Changing Form
As height decreases,
motion increases.
A virtual object falls from rest.
As potential and kinetic energy change, does their sum stay the same?
This indicates how far the object has fallen from its initial height, not the fraction of falling time elapsed.
Change the height, mass, and falling position to discover today's mission.
After changing the position
Kinetic energy increases by the amount that potential energy decreases.
This model uses gravitational acceleration of 9.8 m/s² with no friction or air resistance. If the initial height is h and the fraction fallen is p, potential energy is mgh(1 − p), kinetic energy is mghp, and speed is √(2ghp). Potential energy is measured relative to the ground, and 100% means just before impact. Transformations into deformation, heat, and sound after impact are excluded.
Read more about the principle · OpenStax: Gravitational Potential Energy · NASA: Conservation of Energy
07 / What's 'Typical' in the Numbers
If just one person changes,
does the mean follow?
These are the monthly incomes of six fictional people.
Change the income of the person on the far right and compare.
Bar heights are always compared on a scale of 0–1,000 ten thousand won.
Move the slider or use the arrow keys to adjust.
Change the numbers to discover today's mission.
After making changes yourself
When incomes are similar, the mean and median are close.
The mean is the sum of all values divided by the number of people. The median is the middle value when the values are ordered by size. For six people, it is the mean of the two middle values. The appropriate measure of central tendency depends on the distribution and the question.
08 / Scenes Shaped by Color
The same poster
changes through color alone.
Choose background, text, and accent colors.
Explore changes in mood and text readability together.
Small discoveries,
great delight.
Where curiosity lingers, Japhak Library
Small discoveries,
great delight.
Where curiosity lingers, Japhak Library
Change the colors to discover today's mission.
After making changes yourself
Changing colors gives the same layout a different impression.
The background color changes the impression of large areas, while the accent color changes where the eye is drawn. The brightness difference between text and background affects readability. The displayed value is the contrast ratio calculated from the relative luminance of the two colors.
09 / Rules for Finding a Book's Place
Six books—
what order should they go in?
Select books to shelve them in order from the left.
Arrange them from the lowest classification number to the highest.
Books still waiting for a place
Fictional books · Select a book to move it to the shelf.Select a shelved book again to remove it. Use Tab and Enter on a keyboard.
Help the six scattered books find their places.
After shelving for yourself
The numbers group books by subject and indicate where to find them on the shelf.
KDC is a system for classifying books by subject. Here, you practice ordering books using classification numbers for broad subjects. In actual libraries, books with the same classification number are ordered using other parts of the call number, such as author marks, and library rules.
Economics and Society · Time and Compound Interest
When interest earns interest,
time makes a difference.
Compare simple and compound interest using the same principal and interest rate.
How does the gap between the two amounts change over a longer period?
Assume the same interest rate each year, with no additional deposits or withdrawals.
Change the values to discover today's mission.
After making changes yourself
With compound interest, future interest is earned on both the principal and the interest already accumulated.
Simple-interest amount = principal × (1 + annual interest rate × period); annually compounded amount = principal × (1 + annual interest rate)period. For calculations, divide the percentage interest rate by 100. This educational calculation excludes taxes, fees, and changes in prices, and does not predict returns on actual financial products.
Read more about the principle · Investor.gov: How Compound Interest Works
Math and Statistics · Base Rates and Probability
Even accurate alerts
can be wrong. Here's why.
Turn on spam alerts for 10,000 virtual messages.
Change the spam rate and detection performance to compare how trustworthy the alerts are.
Counts are expected values calculated from probabilities. No actual messages are checked or alerts sent.
Change the values to discover today's mission.
After making changes yourself
When spam is rare, even a low false-positive rate can apply to many legitimate messages.
Correct alerts = 10,000 × spam rate × detection rate; false alerts = 10,000 × legitimate-message rate × false-positive rate. The share of spam among alerts is correct alerts ÷ all alerts. Even with the same detection performance, results depend on how common spam is to begin with.
Read more about the principle · OpenStax: Probability and Conditional Probability
Life and Environment · Populations and Growth
After growing rapidly,
why does growth slow?
Explore a virtual population with limited food and space.
Change the initial population, carrying capacity, and time to compare growth curves.
Population size represents an average trend, so it may be displayed as a decimal. A step is this model's unit of time.
Change the values to discover today's mission.
After making changes yourself
As the population approaches the environment's limit, there is less room for further growth.
The logistic model uses dN/dt = rN(1 − N/K). N is population size, K is the fixed environmental carrying capacity, and r is the growth coefficient. This experience displays the equation's continuous-time solution. This simplified model excludes migration, seasons, and changes in food supply, and does not predict actual ecosystems.
Read more about the principle · OpenStax: The Logistic Equation
Math and Statistics · Optimization Under the Same Constraints
The same length of fencing,
but different areas.
Build a rectangle with a fixed length of fencing.
The balance between width and height changes the area you can enclose.
The rectangle is fenced on all four sides. Cases using a wall or leaving an entrance are excluded.
Change the values to discover today's mission.
After making changes yourself
As width and height become more equal, the same fencing encloses more space.
Width + height = perimeter ÷ 2, and area = width × height. Among rectangles with a fixed perimeter, a square with equal width and height has the greatest area. Compare using maximum area = (perimeter ÷ 4)2.
Read more about the principle · OpenStax: Comparing Areas with the Same Perimeter
Earth and Space · Map Scale
A short distance on a map—
how far is it in reality?
Choose a scale and change the length on the map.
Explore both the actual distance and the area of a square region.
Map lengths are virtual values, not lengths measured with a ruler on the screen.
Change the values to discover today's mission.
After making changes yourself
A larger scale denominator means the same map length represents a larger real-world space.
At a scale of 1 : S, actual distance (km) = map length (cm) × S ÷ 100,000. The actual area of a square is its actual side length (km) squared. Map projection distortion and terrain slopes are excluded. Zooming in on a map on-screen does not automatically change the numbers in its printed scale label.
Read more about the principle · USGS: Guide to Maps and Scale
Earth and Space · Seasons and the Sun
The seasons change
the angle of midday sunlight.
Change the latitude and the Sun's seasonal position.
The Sun's noon altitude and the shadow of a 1 m stick change together.
Positive values indicate north latitude; negative values indicate south latitude. 0° is the equator.
Noon means solar noon, when the Sun is highest at that location, rather than 12 o'clock. Earth's axial tilt is held fixed at approximately 23.4°.
Change the values to discover today's mission.
After making changes yourself
Even at the same seasonal position, sunlight angles differ between the Northern and Southern Hemispheres.
Solar-noon altitude ≈ 90° − |latitude − solar declination|. For a 1 m stick standing perpendicular to a horizontal surface, shadow length is calculated as 1 ÷ tan(altitude). At an altitude of 90°, the shadow length is 0. Atmospheric refraction and terrain are excluded, and representative seasonal declinations are used.
Read more about the principle · NOAA: Solar Position Equations
IT · The Data Behind Images
As pixels increase,
so does the data to store.
Change the image width, height, and bits per pixel.
Compare the storage occupied by pixel data before compression.
The grid is a scaled-down model for explaining pixels. It does not represent screen size or print size.
Change the values to discover today's mission.
After making changes yourself
Doubling both width and height quadruples the pixel count and data size.
Pixel data (bytes) = width in pixels × height in pixels × bits per pixel ÷ 8. Here, 1 MB is 1,000,000 bytes. File headers, palettes, end-of-row padding, and compression are excluded, so actual image file sizes differ. 8 bits is one way to represent data such as grayscale brightness; the extra 8 bits in 32 bits represent transparency in this example.
Read more about the principle · Microsoft: Bitmap Pixel Data and Storage Structure
Chemistry · Dilution and Amount of Solute
Adding water dilutes the solution,
but the dissolved amount stays the same.
Change the virtual solution's concentration and volume, and the amount of water added.
Explore how concentration changes while the amount of solute stays constant.
g/L expresses the mass of solute dissolved in 1 L of solution. The added water contains no solute.
Change the values to discover today's mission.
After making changes yourself
Doubling the solution volume halves the concentration, but the amount of solute stays the same.
Initial concentration × initial volume = final concentration × final volume. Use the same units for both volumes. This model assumes that the solute neither reacts nor evaporates, mixes completely, and that the volumes of the solution and water can be added directly.
Read more about the principle · OpenStax: Solution Concentration and Dilution
Physics · Decibels and Combining Sounds
If sound intensity doubles,
do decibels double too?
Change the decibel level of one sound and the number of sounds at that level.
Compare how intensity and decibels add at one point.
No actual sound is played. The intensity multiplier differs from the multiplier of perceived loudness.
Change the values to discover today's mission.
After making changes yourself
Adding the intensities of two sounds at the same level raises the level by about 3 dB.
Combined level of N sounds = level of one sound + 10 × log₁₀(N). This assumes sounds with equal levels at the same observation point that are independent and maintain no fixed phase relationship. This model does not calculate the sum of phase-aligned waves, perceived loudness, or hearing recommendations.
Read more about the principle · University of Manchester: Adding the Intensities of Different Sounds
Economics & Society · Prices and Purchasing Power
Is your money growing
faster than prices?
Change the growth rate of the amount and the inflation rate separately.
Compare future money by converting it to prices at the starting point.
This assumes that the price of the same basket of goods rises at the set rate. No measured prices or financial product returns are used.
Change the values to discover today's mission.
After making changes yourself
Even if your money grows, its purchasing power falls if prices rise faster.
Nominal amount = initial amount × (1 + amount growth rate)period, amount at initial prices = nominal amount ÷ (1 + inflation rate)period. The inflation-adjusted annual growth rate is calculated as (1 + amount growth rate) ÷ (1 + inflation rate) − 1. For calculations, divide percentage rates by 100. This educational model uses constant hypothetical rates and excludes taxes, fees, and differences in individual spending.
Read more about the principle · European Central Bank: Inflation and Purchasing Power
Life & Environment · Probabilities of Inheritance
One allele from each parent
determines the offspring's combination.
Explore one gene that determines pea seed shape.
Change both parents' genotypes and compare the possible combinations.
A and a are two alleles of the same gene. In this example, at least one A produces round seeds.
Change the values to discover today's mission.
After making changes yourself
Crossing Aa with Aa gives aa in one of the four combinations. The theoretical probability of wrinkled seeds is 25%.
This is a single-gene model of complete dominance in which each parent passes on either allele with equal probability. The four cells show equally likely combinations; they do not mean that every four actual offspring must have a 3 : 1 ratio. Each fertilization is a separate probability event. Not all traits in all organisms follow this rule, and dominant does not mean better or more common.
Read more about the principle · OpenStax: Mendel's Laws of Inheritance and Punnett Squares
Physics · Force and Distance
Using less force means
pulling more rope.
Change the object's mass, the number of supporting rope segments, and the lifting height.
Compare the required force and pulling distance together.
Count the rope segments supporting the moving object. This differs from the number of ropes pulled by hand.
Mass is shown in kg and force in N. More supporting rope segments mean less force carried by each segment.
Change the values to discover today's mission.
After making changes yourself
Doubling the number of supporting rope segments halves the force and doubles the pulling distance. The work needed to lift the same object to the same height stays the same.
Pulling force = mass × gravitational acceleration ÷ number of supporting rope segments; pulling distance = number of supporting rope segments × height; work done = force × distance pulled. This ideal model uses gravitational acceleration of 9.8 m/s², ignores rope and pulley mass and friction, and lifts the object at constant speed. The illustration explains how force is shared and does not provide instructions for rigging actual ropes.
Read more about the principle · OpenStax: Pulleys and Mechanical Advantage
Earth and Space · Moon Phases
From which direction
do we see the Moon's lit half?
Change the Moon's position in its orbit around Earth.
Compare the fraction and shape of the illuminated portion seen from Earth.
Moon shape at this position · First quarter
The first quarter at 90° and last quarter at 270° are both half illuminated, but on opposite sides. This illustration uses a north-up reference view.
Change the values to discover today's mission.
After making changes yourself
Half the Moon is always illuminated by sunlight, but its apparent shape changes as the direction from which Earth views that lit half changes.
Lunar phases are not caused by Earth's shadow. Earth's shadow falling on the Moon causes a lunar eclipse. This model assumes uniform motion in a circular orbit and an average synodic month of 29.53 days. The illuminated percentage is calculated as 50 × (1 − cos position angle), and elapsed days as 29.53 × position angle ÷ 360. This is not a calendar for predicting the Moon's actual position, lunar calendar dates, lunar eclipses, or observation times.
Read more about the principle · NASA: Moon Phases and the Illuminated Side
Mathematics & Statistics · Samples and Error
If we ask more people,
how much does error shrink?
Change the survey sample size and the proportion in favor.
Compare changes in the margin of error for an approximate 95% confidence interval.
50% ±9.8%p means a range from 40.2% to 59.8%. The symbol %p means percentage points, which express the difference between percentages.
Change the values to discover today's mission.
After making changes yourself
At the same proportion, quadrupling the sample size halves the margin of error. The margin does not shrink in direct proportion to the sample size increase.
Margin of error (%p) = 1.96 × √(p × (1 − p) ÷ sample size) × 100, where p is the percentage in favor divided by 100. This is a normal approximation for independent random sampling from a very large population. It is less accurate for small samples or proportions close to 0% or 100%. If surveys were repeated using the same method to construct intervals, about 95% of those intervals would contain the true proportion. It does not mean that this particular interval has a 95% probability of containing the true proportion. Increasing the sample size does not resolve response bias or the exclusion of groups from the survey.
Read more about the principle · NIST: Confidence Intervals for Proportions and the Normal Approximation
Everyday Knowledge · Photographic Angle of View
From the same spot,
how wide a view can you capture?
Change the sensor width and lens focal length.
Compare the horizontal angle of view and the width captured at the subject distance.
The angle of view is the angular span captured in the frame. The captured width is the horizontal span on a plane at the selected distance.
Change the values to discover today's mission.
After making changes yourself
With the same camera position and sensor, a longer focal length captures a narrower view. A smaller sensor width also narrows the angle of view.
Horizontal angle of view = 2 × arctan(sensor width ÷ (2 × focal length)); captured width = subject distance × sensor width ÷ focal length. Enter sensor width and focal length in the same units; the angle of view is converted to degrees for display. This ideal rectilinear pinhole model excludes distortion and changes in angle of view caused by focusing. Subject distance is measured from the model's center of projection, not the front of an actual lens barrel. Changing lenses from the same position changes the angle of view; perspective changes that alter the relative sizes of near and far objects occur when the camera position moves.
Read more about the principle · Edmund Optics: Focal Length and Angle of View
IT · Logic for Combining Conditions
When two conditions combine,
how does the result change?
If only one input is on, will the output be on too?
Change two conditions and the rule for combining them to explore how notifications, searches, and programs make decisions.
0 = Off · False / 1 = On · True
OR is true even when both inputs are true. NOT only reverses A, so it does not use B.
Change the values to discover today's mission.
After trying the combinations
Even with the same conditions, the result depends on whether you use AND, OR, XOR, or NOT.
This logic model uses only two values, 0 and 1. AND returns 1 when both A and B have that value; OR returns 1 when at least one has that value; XOR returns 1 when they differ. NOT returns 1 when A is 0, and 0 when A is 1. It shows the structure of conditional decisions without reproducing actual circuit voltages or timing.
Read more about the principle · OpenStax: Conditions and Logical Operations · Analog Devices: XOR Logic
Chemistry · Equilibrium in Motion
If nothing changes,
has the reaction stopped too?
Set how quickly A and B convert into each other, then move through time.
Check whether conversion continues in both directions even when the amount of each substance stops changing.
Initially, A is 80 and B is 20, and their total is always 100. Here, no change does not mean equal amounts of A and B; it means that each amount stays constant over time.
Change the values to discover today's mission.
After moving through time
At equilibrium, the amounts of A and B need not be equal. Their conversion rates are equal in both directions, so each amount stops changing.
This is a closed hypothetical model in which A ↔ B converts in a 1 : 1 ratio, and each conversion rate equals the amount of the substance × its rate coefficient. The coefficients are constant, and the total remains 100. At equilibrium, amount of B = 100 × forward coefficient ÷ (forward coefficient + reverse coefficient). Conversion continues in both directions, but the net change approaches 0. Complex rates involving actual substances, temperatures, and reaction steps are not included.
Read more about the principle · OpenStax: Chemical Equilibrium and Dynamic Equilibrium
Life & Environment · Energy in Food Chains
As you move up a food chain,
where does the energy go?
How much energy stored in producers reaches the next level?
Change the transfer percentage and trophic level to compare energy pyramids.
The illustration compares energy, not organism numbers or body mass. It assumes the same transfer percentage at every level; this is not always fixed at 10% in real ecosystems.
Change the values to discover today's mission.
After moving up the levels
If 10% is transferred at each level, 1% of the initial energy reaches two levels up.
Energy at the selected level = producer energy × (transfer percentage ÷ 100)number of levels. Energy not transferred is converted to heat through life processes or moves along other pathways through uneaten parts, waste, and dead organisms. It has not disappeared. Nutrients cycle through ecosystems, but energy enters, moves through them, and ultimately disperses into the surroundings as heat.
Read more about the principle · OpenStax: Energy Flow in Ecosystems
Economics & Society · Division of Labor and Comparative Advantage
Even if one person is faster at both,
can dividing the work help?
Two people each spend 8 hours making hypothetical products X and Y.
Change how each person allocates time to see whether they can produce more than when splitting it equally.
| Person | X | Y |
|---|---|---|
| A | 4 units | 2 units |
| B | 1 unit | 1 unit |
The remaining time is spent making Y. The comparison baseline stays fixed at the initial allocation, with each person spending 50% of their time on each product (X: 20, Y: 12).
Change the values to discover today's mission.
After dividing the work
A makes both products faster, but gives up less Y when making X.
To make 1 unit of X, A gives up 0.5 units of Y, while B gives up 1 unit. The person with this lower opportunity cost has a comparative advantage in X. If A spends 75% of the time on X and B spends 0%, output is X: 24 and Y: 12, producing 4 more units of X than initially. This hypothetical example assumes output is proportional to working time and excludes quality, travel, and coordination costs.
Read more about the principle · OpenStax: Absolute and Comparative Advantage
Life and Environment · Natural Selection
Even in the same environment,
trait proportions can change.
Compare two types, A and B, that reproduce their own traits.
Change their contributions to the next generation and the number of generations to explore changes in their proportions.
The rest are type B. The proportions of A and B always add up to 100%.
The coefficient represents contribution to the next generation through survival and reproduction in the given environment. With 1.1 and 1, A contributes 10% more than B. This does not mean the trait itself is always better.
Change the values to discover today's mission.
After moving through generations
The type that contributes more to the next generation makes up a larger proportion of the population over successive generations.
With the initial proportion of A as p₀ (0–1), contribution coefficients wA·wB, and generation count g, the proportion of A is p₀ × wAg ÷ [p₀ × wAg + (1 − p₀) × wBg]. This educational model assumes that both types reproduce asexually, pass on their traits unchanged, and have a constant environment and coefficients. It differs from a model of genotype crosses and excludes mutation, migration, and random genetic drift. It compares proportions without predicting actual population counts. Evolution has no predetermined goal or endpoint.
Read more about the principle · UC Berkeley: A Proportion Model of Natural Selection (PDF) · OpenStax: Environment and Relative Fitness
Economics and Society · Price Elasticity
Even after a price cut,
revenue can fall.
Change a hypothetical product's price and the responsiveness of demand.
How do sales quantity and revenue change when the price falls by 5 units?
P is price and Q is sales quantity. Both demand equations are hypothetical relationships.
The discount is always 5 price units. Revenue equals price × sales quantity and differs from profit after costs.
Change the values to discover today's mission.
After trying a discount
A price cut reduces revenue if sales quantity does not increase enough.
Arc elasticity is the percentage change in sales quantity divided by the absolute percentage change in price, with each percentage change based on the average of the values before and after the discount. Below 1, revenue falls after the discount; at 1, it stays the same; above 1, it rises. Choosing price 27.5 and Q = 100 − 2P gives equal revenue of 1,237.5 before and after the discount. This hypothetical linear demand model excludes costs, competition, seasons, and differences among consumers. It compares the interval before and after a discount without predicting demand or returns for actual products.
Read more about the principle · OpenStax: The Midpoint Method and Price Elasticity · The Relationship Between Elasticity and Revenue
Earth & Space · The Weight of Air
The higher you go,
the less air remains above you.
Assume a uniform temperature throughout the atmosphere and change the altitude.
With sea-level pressure set to 100%, how does pressure change at higher altitudes?
This is the model temperature applied throughout the atmosphere, not the actual temperature at a particular altitude.
Atmospheric pressure is related to the weight of the air above. A warmer atmosphere spreads out more, so pressure decreases more slowly with altitude.
Change the values to discover today's mission.
After changing the altitude
As altitude increases, less air weight remains above, so atmospheric pressure falls.
This is an ideal isothermal atmosphere model with pressure P = P₀ × exp[−Mgh ÷ (RT)]. Sea-level pressure P₀ is 1,013.25 hPa, air molar mass M is 0.0289644 kg/mol, gravitational acceleration g is 9.80665 m/s², and the gas constant R is approximately 8.31446 J/(mol·K). Here, h is altitude (m), and T is temperature in Celsius + 273.15 (K). Temperature, composition, and gravity are constant throughout the atmosphere; water vapor and weather changes are excluded. This differs from standard-atmosphere calculations in which temperature decreases with altitude, and cannot be used for actual pressure forecasts or altitude measurements.
Read more about the principle · NASA: Ideal Gases Under Gravity and the Isothermal Approximation
Everyday Knowledge · Music and Vibration
Going up an octave
doubles the vibration frequency.
Choose how many semitones to move up or down from the reference note.
Compare the number of waves drawn over the same time and the frequencies of the two notes.
Press to hear the reference note briefly, followed by the selected note.
Hz is the number of vibrations repeated in 1 second. The waves are drawn with equal amplitude. Pitch and loudness are different properties.
Change the values to discover today's mission.
After changing the note
One octave contains 12 semitone steps. Moving up 12 steps doubles the frequency; moving down halves it.
In 12-tone equal temperament, frequency = reference frequency × 2(semitones ÷ 12). Each semitone step multiplies frequency by approximately 1.05946, so the difference in Hz for the same number of steps depends on the reference note. The two sine-wave tones play at equal amplitude only when the button is pressed. Actual instrument timbres, harmonics, and tuning differences are excluded. Doubling the frequency ratio does not mean doubling perceived loudness.
Read more about the principle · UNSW: Note Frequencies and 12-Tone Equal Temperament
IT · Lossless Compression
If we store repeated runs together,
is the result always smaller?
Consecutive identical characters are stored together as “count + character.”
Change the character pattern and repetition count to compare the original with the compressed result.
Each original character uses 1 byte. Each compressed run uses 1 byte for the count + 1 byte for the character. Negative savings mean the compressed version is larger.
Change the values to discover today's mission.
After grouping repeated characters
Longer runs of identical characters let you store the original in fewer groups.
This method is run-length encoding (RLE). For example, AAAAAAAA can be stored as “8, A” and restored by writing A 8 times. If A and B alternate every time, each character uses 2 bytes, doubling the size. Counts are displayed as readable numbers, but the storage model counts each number as 1 byte. This educational model uses only A and B and excludes headers; it does not reproduce the actual BMP file format or a complete general-purpose compression program. A compression method's effectiveness depends on the data pattern.
Read more about the principle · Microsoft: RLE Storage Using Run Counts and Color Values
Physics · Density and Buoyancy
Even if an object gets heavier,
its floating fraction can stay the same.
Change the object's density and volume and the type of liquid.
Compare the downward weight with the liquid's upward force.
If the object is denser than the liquid, a net downward force remains even when it is fully submerged. Equal densities give neutral buoyancy: weight and buoyant force balance when fully submerged.
Change the values to discover today's mission.
After floating the object
Increasing volume increases both weight and maximum buoyant force. The ratio of the two densities determines the floating fraction.
By Archimedes' principle, buoyant force equals the weight of the displaced liquid. When an object floats, buoyant force = weight, so submerged volume fraction = object density ÷ liquid density. If this ratio exceeds 1, the object cannot float at the surface, and the submerged fraction is shown as 100%. The object is assumed to have uniform density and be incompressible; the liquid is still, and surface tension, fluid drag, and bottom contact are excluded. The downward force is weight minus buoyant force after full submersion, not sinking speed. N is a unit of force, and gravitational acceleration is 9.80665 m/s².
Read more about the principle · OpenStax: Archimedes' Principle
Math and Statistics · The Birthday Problem
With just 23 people,
what is the chance of a shared birthday?
Increase the group size and count the pairs you can compare.
Compare any shared birthday with someone having a birthday on a specific date.
The “specific date” is chosen before looking at people's birthdays. Every date has the same probability in this model.
Change the values to discover today's mission.
After increasing the group size
Among 23 people, there are 253 pairs to compare. The chance of any shared birthday is much higher than the chance of a birthday on a specific date.
The probability of a shared birthday is 1 minus the probability that everyone's birthday is different. With n people, there are n(n − 1) ÷ 2 pairs to compare. Shared-birthday events for all pairs cannot be treated as mutually independent, and multiplying the number of pairs by 1/365 does not give the overall probability. The model assumes birthdays are uniformly distributed over 365 days and independent between people, excluding leap day and actual seasonal birth patterns. It means the probability becomes high even in a small group, not that a shared birthday is guaranteed.
Read more about the principle · UC Berkeley: The Birthday Problem and Complementary Events (PDF)
Earth and Space · Humidity and Dew Point
When air cools,
how does humidity change?
Set the initial temperature and relative humidity, then cool the air.
Humidity rises even without added water vapor, and droplets form below the dew point.
Air pressure stays constant, and no water vapor is added. Before saturation, vapor pressure is treated as constant for comparison. This model explores how air cools around a cold cup.
Change the values to discover today's mission.
After cooling the air
Cooling lowers saturation vapor pressure, raising relative humidity. Below the dew point, some water vapor condenses into liquid water.
Relative humidity = current vapor pressure ÷ saturation vapor pressure at that temperature × 100%. Saturation vapor pressure over liquid water is calculated using the Magnus approximation: eₛ(T) = 6.112 × exp[17.62T ÷ (243.12 + T)] hPa. The dew point is the temperature at which saturation vapor pressure equals the initial air's vapor pressure. Below that target temperature, excess vapor condenses, reducing vapor pressure to saturation, and relative humidity is shown as 100%. “Excess above saturation” is the difference between vapor pressure before condensation and saturation vapor pressure at the target temperature, not droplet mass or condensation rate. Dew points below freezing are also shown relative to liquid water and differ from frost points. This equilibrium model excludes supersaturation, pressure changes, and heat transfer rates.
Read more about the principle · NPL: Dew Point and Relative Humidity
Chemistry · Molecular Shape and Polarity
Even polar bonds can cancel
depending on their directions.
Change the angle between two identical bonds connected to a central atom.
What remains when you add the arrows showing each bond's polarity?
This describes how unevenly electrons are distributed. It is not the energy needed to break a bond or the bond strength.
The arrows have equal magnitudes and point from the center toward the outer atoms. At 180°, they point in opposite directions and cancel.
Change the values to discover today's mission.
After adding the arrows yourself
Even with polar bonds, the sum can be 0 if they point symmetrically in opposite directions.
If two bond dipoles of equal magnitude m form an angle θ, the resultant vector has magnitude 2m × cos(θ/2). Understanding molecular polarity requires considering both bond angles and directions. This activity is a geometric model that adds only the vectors of two identical bonds; it does not calculate a real molecule's full electron distribution or dipole moment. The angle 104.5° is an example for comparing bent shapes, and the relative units shown are neither measured values for water nor Debye units. Actual bond angles are determined by factors such as electron-pair arrangements and cannot change freely like the slider.
Read more about the principle · OpenStax: Molecular Structure and Polarity
Life & Environment · Enzyme Work Sites
Even if you keep adding material,
the rate has a limit.
Increase the concentration of substrate, the material an enzyme processes.
If you also change the amount of enzyme, how do the maximum rate and the concentration needed for half that rate change?
mM means mmol/L, the amount of substrate per 1L of solution. We compare rates just after the reaction starts, using one enzyme type at the same temperature and pH.
Change the values to discover today's mission.
After adding substrate yourself
When most enzymes are processing substrate, adding more substrate increases the rate only slightly. Increasing the enzyme amount also increases the maximum rate.
This uses the Michaelis–Menten initial-rate model v = Vₘₐₓ × [S] ÷ (Kₘ + [S]). Substrate concentration [S] and Kₘ are in mM, with Kₘ fixed at 2mM. The maximum rate Vₘₐₓ is set to the relative enzyme amount × 10 in hypothetical rate units. When the substrate concentration equals Kₘ, the rate is half the maximum regardless of enzyme amount. The model assumes substrate is in sufficient excess over enzyme and follows simple saturation kinetics. It excludes inhibition, cooperativity, substrate depletion over time, and enzyme denaturation. It does not predict total product yield or the actual rate of a particular experiment.
Read more about the principle · MIT: Enzyme Catalysis and Reaction Rates (PDF)
Economics and Society · Choice and Cooperation
The best choice for each person
may not be best for both.
Two people each choose once, without knowing the other's choice.
Change the combinations of cooperation and defection to compare each person's score and their total.
A higher score is better for you. If both cooperate, each gets 3 points. If both defect, each gets 1 point. If only one defects, the defector gets 5 points and the cooperator gets 0 points.
Change the values to discover today's mission.
After changing each person's choice yourself
Whatever the other person chooses, defection gives you a higher individual score. But the combined score when both defect is lower than when both cooperate.
This payoff structure is called the prisoner's dilemma. Both decide simultaneously just once, with no communication or binding agreements before choosing. If each considers only their own score, mutual defection is the Nash equilibrium. Equilibrium means neither can benefit by changing their choice alone; it does not mean the outcome is best for everyone. The scores are hypothetical rewards, not prison terms. Results may change when trust, retaliation, negotiation in repeated relationships, or concern for others is included.
Read more about the principle · MIT: The Prisoner's Dilemma and Nash Equilibrium (PDF)
Everyday Life and Culture · Queues
Just a little busier,
but a much longer wait.
Change the customer arrival rate and service rate at a single counter.
How long does the average wait become when the counter is almost never idle?
In this model, intervals between arrivals and individual service times vary. A service rate of 6 people/min means an average service time of 10 seconds. An average of 0.5 people does not mean half a person; it is the number waiting averaged across different times.
Change the values to discover today's mission.
After changing the queue conditions yourself
As the arrival rate approaches the service rate, there is less spare capacity to clear the backlog, so waiting time rises sharply.
This is a steady-state M/M/1 model with one counter serving people in arrival order. It assumes an independent Poisson arrival process, independent exponentially distributed service times, and sufficient waiting space. With average arrival rate λ and average service rate μ in people/min, waiting time = 60λ ÷ [μ(μ − λ)] seconds, time in the system = 60 ÷ (μ − λ) seconds, and number waiting = λ² ÷ [μ(μ − λ)] people. Finite steady-state averages exist only when λ < μ, and λ/μ is then the counter's utilization. When λ ≥ μ, λ/μ is the ratio of requested load to service capacity, not actual utilization. This means stable averages cannot be shown, not that every customer necessarily waits forever. These values do not predict instantaneous queue lengths or crowding in real stores with reservations, multiple counters, or departures during service.
Read more about the principle · MIT: The M/M/1 Queueing Model (PDF)
IT · Error Detection
With one check bit,
can you detect an error?
One check bit is appended to 4 data bits before transmission.
If bits flip during transmission, can the receiver detect an error?
Flipping a bit changes 0 to 1 and 1 to 0. Flipping the same position twice restores its original value. The check uses only the five received bits.
Change the values to discover today's mission.
After flipping bits yourself
Even parity adds a check bit so the number of 1s is even. An error is detected if the received number of 1s is odd.
The four data bits represent values of 8, 4, 2, and 1 from left to right. The rightmost check bit is not included in the data value. An odd number of ultimately changed bits among the five transmitted bits is detected, but an even number can go undetected. 'Not detected' does not guarantee correct data. Changing only the check bit also triggers an error, while the data value can remain unchanged. This method alone cannot identify the incorrect bit or restore the original data. This activity compares at most two bit flips.
Read more about the principle · MIT: Digital Information and Parity Checks
Life and Environment · Osmosis
Across a water-permeable membrane,
what if concentrations differ?
Change solute concentrations on either side of a membrane that allows only water through.
At the initial moment, with equal hydrostatic pressure on both sides, which way does water move overall?
mM is concentration in mmol per 1L of solution. The model's solute neither dissociates into ions in water nor passes through the membrane. Water moves in both directions; the arrow shows the net movement combining both.
Change the values to discover today's mission.
After changing both sides of the membrane yourself
With equal hydrostatic pressure on both sides, water moves overall from lower to higher solute concentration. Equal concentrations produce no net movement due to this difference.
The osmotic pressure of a dilute ideal nonelectrolyte solution is π = CRT. C is in mol/m³, R is approximately 8.31446J/(mol·K), and T is the Celsius temperature + 273.15 in K. Numerically, 1mM = 1mol/m³, so pressure in kPa is the displayed concentration × R × T ÷ 1,000. The inside − outside osmotic pressure difference is the inside − outside hydrostatic pressure difference needed to prevent net water movement; it does not mean this hydrostatic pressure difference is actually present initially. Only initial concentrations are compared; changes in volume or concentration over time and movement rates are not calculated. This model does not predict complex solutes and membrane transport in real cells or responses of particular organisms.
Read more about the principle · OpenStax: Osmotic Pressure of Solutions
Math and Statistics · Groups and the Whole
Ahead in every group,
yet behind overall.
Assume methods A and B are each used to solve 100 problems.
Will overall success rates stay the same if only the proportions of easy and difficult problems change?
Easy-problem success rates are fixed at A 90% and B 80%; difficult-problem rates are A 40% and B 30%. The remaining problems are difficult. %p means the difference between two percentages.
Change the values to discover today's mission.
After changing the problem mix yourself
The overall success rate is an average of group success rates weighted by problem counts. With different proportions of easy problems, A can lead in each group yet trail overall.
This reversal is called Simpson's paradox. If a and b are the proportions of easy problems, ranging from 0 to 1, A overall = 90a + 40(1 − a), and B overall = 80b + 30(1 − b). With equal proportions, A always leads by 10%p, but at A 10% and B 90%, overall success rates become A 45% and B 75%. Differences within difficulty levels and the overall difference answer different questions. These are hypothetical data assigning different mixes to the two methods; they alone cannot establish which method actually causes success. Real comparisons also require examining how groups were defined and why their compositions differ.
Read more about the principle · Williams College: Weighted Averages and Simpson's Paradox (PDF)
Everyday Life and Culture · Heat Pumps
When electricity moves heat,
more heat arrives than electricity goes in.
This is an ideal heat pump moving heat between two places at constant temperatures.
Change both temperatures and the energy input to compare transferred heat and the energy sum.
The heating coefficient of performance is heat delivered ÷ energy input; the cooling coefficient of performance is heat removed ÷ energy input. kWh is an amount of energy, without separately specifying power or operating time.
Change the values to discover today's mission.
After following the heat transfer yourself
Delivered heat includes both the input energy and heat taken from the colder side. Delivering more heat than the electrical energy supplied does not create new energy.
The ideal limit for a reversible Carnot heat pump is heating COP = Thot ÷ (Thot − Tcold), cooling COP = Tcold ÷ (Thot − Tcold). Temperatures must be calculated as absolute temperatures in K by adding 273.15 to Celsius temperatures. Heat delivered to the hotter side = heat taken from the colder side + work input, and the two COPs always differ by 1. Both sides are heat reservoirs whose temperatures remain constant despite heat exchange, and all electrical energy is assumed to become work. These are theoretical upper limits excluding friction, compression losses, and temperature differences needed for heat exchange in real equipment. The displayed values cannot predict the performance or running costs of real air conditioners or heaters.
Read more about the principle · OpenStax: Key Equations for Thermodynamics and Heat Pumps
Chemistry · Cells and Electrodes
Change the two metals,
and voltage and flow change too.
Choose two electrodes, each metal immersed in a solution of its own ions.
The difference in standard reduction potentials produces the cell voltage.
Comparisons use the reaction M²⁺ + 2e⁻ → M, in which a metal ion gains 2 electrons. Swapping left and right preserves the voltage magnitude but reverses electron flow.
Change the values to discover today's mission.
After choosing electrodes yourself
Metal releases electrons on the side with the lower standard reduction potential, while metal ions gain electrons on the higher-potential side. Electrons in the external circuit move from anode to cathode.
The standard voltage of an ideal galvanic cell is the cathode's standard reduction potential − the anode's standard reduction potential. The voltage magnitude is shown for the spontaneous reaction direction. Identical metals at identical ion activities have no potential difference. Comparisons assume pure solid metals and an activity of 1 for each M²⁺ ion at 25°C, referenced to the standard hydrogen electrode. Table potentials are rounded to three decimal places. The magnitude of maximum electrical work is nFE, displayed in kJ/mol using n = 2mol electrons/mol reaction and F = 96,485.33212C/mol. Concentration changes, internal resistance, overpotential, and side reactions in real cells are excluded. Voltage differs from capacity, the amount of stored charge; this value alone cannot determine actual power output or operating time.
Read more about the principle · OpenStax: Electrode Potentials and Cell Voltage · Table of Standard Half-Cell Potentials
Physics · Refraction and Total Internal Reflection
When light reaches a boundary,
does it bend toward the normal?
Choose the materials light travels from and into, then change the angle.
When traveling toward a lower refractive index, light sometimes cannot pass through the boundary.
The normal is an imaginary line perpendicular to the boundary. Angles are measured from the normal, not the boundary. When reflection occurs, the reflection angle equals the incidence angle.
Change the values to discover today's mission.
After changing the light's direction yourself
Light bends toward the normal when entering a material with a higher refractive index. It bends away from the normal when entering one with a lower index, and undergoes total internal reflection above the critical angle.
The refraction angle is calculated using Snell's law, n₁sinθ₁ = n₂sinθ₂. A critical angle θc = asin(n₂/n₁) exists only when n₁ is greater than n₂. At the critical angle, the refraction angle is 90°, the limiting direction along the boundary; total internal reflection occurs only above the critical angle. 'Conditions not met' means the two refractive indices do not allow total internal reflection. The model assumes a flat boundary between two transparent, uniform materials with fixed refractive indices, excluding real wavelength and temperature dependence. This activity compares light directions and does not calculate the intensity of partially reflected light, polarization, or absorption. With equal refractive indices, light does not change direction optically, and no boundary reflection is shown.
Read more about the principle · OpenStax: Refraction and Snell's Law
Economics and Society · Progressive Taxation
A higher tax rate applies
only to income in that bracket.
Gradually increase income in an educational, hypothetical tax schedule.
Crossing a bracket boundary does not apply the new rate to all your income.
The first 200 units are taxed at 0%; the portion above 200 through 400 at 10%; above 400 through 800 at 20%; and above 800 at 30%. No real country's tax rates or currency are used.
Change the values to discover today's mission.
After crossing income brackets yourself
The average tax rate is the share of total income paid in tax; the marginal tax rate applies to the next additional income. These figures answer different questions.
Bracket-based progressive taxation adds each bracket's income portion × its tax rate. At an income of 400 units, tax on the first 200 is 0 and tax on the next 200 is 20, totaling 20 units. Further income is taxed at 20%, so the marginal rate at a boundary is shown for the 'next income increment.' All bracket rates in this model are below 100%, so after-tax income increases with income and does not drop when crossing a boundary. Only hypothetical taxable income and tax rates are used; deductions, tax credits, welfare benefits, and other taxes are excluded. This activity compares the principle of bracket taxation rather than calculating actual taxes or take-home income.
Read more about the principle · IRS: How Tax Applies to Income Within Each Bracket
Earth and Space · Seismic Waves
The gap between P- and S-wave arrivals
can help estimate distance.
Change the time gap between the earlier P wave and the later S wave.
This models two waves traveling the same path at different speeds.
With both speeds fixed, a larger time gap means the station is farther from the hypocenter. One station's time gap alone cannot reveal the hypocenter's direction.
Change the values to discover today's mission.
After comparing the two waves yourself
The arrival-time gap between the faster P wave and slower S wave grows with distance. Finding the distance requires applying both speeds to this gap.
For the same distance d, the arrival-time difference is Δt = d/vS − d/vP, so d = Δt ÷ (1/vS − 1/vP). This activity assumes both waves travel along the same straight path at constant speeds through a uniform solid. Hypocentral distance is the distance between the station and the underground earthquake source, not the surface distance to the epicenter on a map. P- and S-wave times are model travel times with the earthquake's start set to 0. 'Distance for a time gap of 1 second' is a distance conversion factor, not the speed of either wave. The model excludes speed changes between Earth's layers, refraction, and differing paths. Determining the hypocenter's direction and location requires records from multiple stations and a more detailed model. This activity does not predict earthquakes.
Read more about the principle · USGS: Locating Earthquakes Using Seismic Waves and Arrival Times
IT · Cache Replacement
Can adding storage slots
increase source reads?
A small storage area called a cache handles the same twelve requests.
Change the number of slots and the replacement policy to compare how often missing data must be read from the source.
The request sequence is fixed: 1 → 2 → 3 → 4 → 1 → 2 → 5 → 1 → 2 → 3 → 4 → 5. The cache starts empty. Missing data is read from source storage into the cache, including initial reads in the count. Data already cached does not require a source read.
Change the values to discover today's mission.
After changing the storage policy yourself
Cache effectiveness depends on replacement policy and request order as well as slot count. Follow how the same requests are stored step by step.
FIFO preserves insertion order, even on cache hits. LRU moves a hit item to the most recently used position. When an item is missing and all slots are full, each policy replaces its oldest item. For this sequence, FIFO misses increase from 9 with 3 slots to 10 with 4 slots, a phenomenon called Belady's anomaly. LRU with the same 4 slots has 8 misses. This result alone does not mean FIFO is always bad or LRU is better for every sequence. Hit rate is hits ÷ processed requests and is undefined for 0 requests. This replacement-policy model treats each number as one equal-sized data item. It does not calculate real browser cache expiration, communication time, capacity, or write costs.
Read more about the principle · MIT: FIFO, LRU, and Memory Replacement
Math and Statistics · Models and Errors
Even with a perfect fit to observations,
what happens at other points?
Build a polynomial model from six observation points with small added perturbations.
Change complexity and perturbation size, and compare errors at twenty reference points not used for fitting.
Observation points are used to fit the model; reference points are used only to compare errors. RMSE is the square root of the average squared error. Smaller values mean a closer fit to those points.
Change the values to discover today's mission.
After changing model complexity yourself
A model passing through every observation can still deviate more at other reference points. Examine errors at both the fitted points and the separate comparison points.
Observed x values are −1, −0.6, −0.2, 0.2, 0.6, and 1. To y = 3 + 2x, we add the perturbation multiplier × [0.2, −0.2, 0.3, −0.3, 0.2, −0.2]. The polynomial is fitted by minimizing the sum of squared errors at these six points. Reference x values are twenty points from −0.95 to 0.95 at intervals of 0.1; reference y is the unperturbed 3 + 2x. This activity compares reference errors for hypothetical data with a known generating process; the reference points are not independently collected observations. A degree-5 polynomial can pass through all six perturbed observations but may curve sharply between them, increasing reference error. This does not mean complex models are always bad or simple models always good. With perturbation set to 0, all three models fit the same straight line. These values do not guarantee errors for future outcomes that are unknown in practice.
Read more about the principle · scikit-learn: Underfitting and Overfitting
Life and Environment · Biodiversity
Even with the same species count,
different proportions make a difference.
Change the number of species and the proportion belonging to the first species.
Compare distributions in which the other species share the remaining proportion equally.
The values show each species' proportion of the total, not individual counts. The first species is not always the most abundant. These are hypothetical distributions without species names or traits.
Change the values to discover today's mission.
After changing species proportions yourself
Even with the same species count, diversity decreases when a few species dominate the distribution. Species count and evenness of proportions are different characteristics.
For species proportions p, Shannon diversity is H = −Σp ln(p), using natural logarithms. exp(H) is the effective number of species in an even distribution with the same H; it does not mean a fractional number of species actually exists. Evenness is displayed as H ÷ ln(species count) × 100 and equals 100% when all species have equal proportions. With 4 species, a first-species proportion of 25% makes all proportions 25%; at 70%, the others are each 10%. The slider moves in 5% increments, so the exactly even proportion ⅓ for 3 species cannot be selected. This compares ideal distributions with all proportions positive, without addressing sample size, observation error, or relationships among species. A higher index does not always mean a healthy ecosystem; species identity, rarity, and conservation value require separate consideration.
Read more about the principle · Hokkaido University: Diversity Indices
Everyday Life and Culture · Task Order
If one task gets faster,
does total time stay the same?
After preparation, two teams work simultaneously along the upper and lower paths.
Change task durations and find the completion time that must wait for both paths.
Preparation A is fixed at 2 minutes and finishing task F at 1 minute. D starts only after B finishes, E only after C finishes, and F only after both D and E finish.
Change the values to discover today's mission.
After changing the task durations yourself
The longest task path that determines the completion time is called the critical path. Shortening only tasks on a shorter path may leave the overall completion time unchanged.
The total completion time is 2 + max(B + D, C + E) + 1. The “sum of task durations” adds the durations of all six tasks and differs from actual elapsed time because the two paths run simultaneously. Each path’s slack is the longer path’s duration minus its own duration; it can be delayed by that amount without delaying overall completion. With the defaults, the lower path takes 8 minutes and the upper path takes 7 minutes, leaving 1 minute of slack on the upper path. When the paths have equal durations, both are critical paths; shortening just one still leaves you waiting for the other. Shortening tasks can also change the critical path itself. This dependency model assumes separate teams share no resources and each task duration is fixed. It gives the earliest completion time, excluding staffing conflicts, interruptions, and uncertainty in durations.
Read more about the principle · MIT: Critical Path Method (PDF)
Chemistry · Saturation and Solubility
Even if you add plenty of solid,
it may not all dissolve.
Add a hypothetical solid whose solubility varies with temperature to water.
Change the temperature and the amounts of water and solid, then compare how much dissolves and how much remains.
The hypothetical solid’s solubility is set to (20 + temperature) g per 100 g of water. At 40 °C, up to 60 g dissolves in 100 g of water. This is not an actual solubility table for a specific substance.
Change the values to discover today's mission.
After changing the dissolution limit yourself
There is a limit to how much can dissolve under given conditions. If you add more, the excess remains undissolved. If cooling lowers the limit, some dissolved material comes out as solid again.
Maximum soluble amount = solubility × water mass ÷ 100; dissolved amount = min(amount added, maximum soluble amount); remaining solid = amount added − dissolved amount. The displayed ratio is dissolved-solid mass ÷ water mass × 100, which differs from mass percent based on the total solution mass. We assume enough time has passed to reach dissolution equilibrium under each condition. A solution is saturated when the maximum soluble amount is dissolved; undissolved solid does not have to be present. Water evaporation, supersaturation, dissolution rate, and heat of dissolution are excluded. This hypothetical substance is defined to become more soluble as temperature rises; not all solids follow this relationship.
Read more about the principle · OpenStax: Solubility and Saturated Solutions
Physics · Electromagnetic Induction
What produces voltage
is a changing magnetic field.
Change the magnetic field passing through a coil over a set time.
Change the field change, elapsed time, number of turns, and angle to compare average induced voltages.
The area of each coil turn is fixed at 0.01 m². Only the magnetic field changes; the area and angle remain constant. A change of 0 T means the magnetic field is unchanged, not absent.
Change the values to discover today's mission.
After changing the rate of change yourself
The same magnetic field change produces a larger average induced voltage magnitude when it happens over a shorter time. Reversing the direction of change also reverses the voltage’s sign.
Magnetic flux measures the magnetic field passing through the coil plane, accounting for area and direction. In a uniform magnetic field, Φ = BAcosθ. Here, θ is the angle between the normal perpendicular to the coil plane and the magnetic field. Faraday’s law gives the average induced electromotive force as ε = −NΔΦ ÷ Δt, where N is the number of turns. 1 mWb equals 0.001 Wb. The sign is shown relative to the chosen normal and winding direction; a negative value does not mean a smaller voltage magnitude. The minus sign in Lenz’s law indicates a direction that opposes the change in magnetic flux. We assume every coil turn has the same area and magnetic field, excluding detailed time waveforms, coil resistance, and self-induction. This compares average voltage over a set time interval without calculating actual current or power.
Read more about the principle · OpenStax: Magnetic Flux · Faraday’s and Lenz’s Laws
Economics and Society · External Costs
The cost producers pay
differs from the cost society bears.
Change the cost of making one additional item.
Compare production when only producer costs count and when harm to 3rd parties is also included.
At output q, the benefit of 1 additional item is (120 − q) cost units/item. Producer and external costs remain constant regardless of output. The figures describe a hypothetical market.
Change the values to discover today's mission.
After adding the omitted costs yourself
When harm creates costs that the parties to a transaction do not pay, market output can differ from the output that maximizes society’s total net benefit.
External costs are costs borne by 3rd parties who do not directly participate in a transaction. Social marginal cost = private marginal cost + marginal external cost. In a competitive market, output reflecting only private costs is q = 120 − private cost; output maximizing total net benefit including external costs is q = 120 − private cost − external cost. Between these output levels, social cost exceeds additional benefit. The lost net benefit is the triangular area: ½ × output difference × marginal external cost. This loss differs from the total external harm from all production. This continuous quantity model includes only negative externalities from production, excluding estimates of actual pollution damage, market power, policy enforcement costs, and positive externalities. It compares hypothetical products and cost units and does not predict the effects of actual regulations or taxes.
Read more about the principle · OpenStax: The Economics of Pollution and External Costs
Earth and Space · Reflection and Energy
Reflecting more sunlight
reduces the energy absorbed.
Change the fraction of sunlight reflected by a hypothetical planet with no atmosphere.
Explore the relationship between planet-wide average energy and radiative equilibrium temperature.
Radiation received by a cross section perpendicular to sunlight is fixed at 1,361 W/m². A sphere’s cross-sectional area is ¼ of its total surface area, so incoming energy averaged over the planet’s entire surface is 340.25 W/m².
Change the values to discover today's mission.
After changing reflectance yourself
Reflecting more reduces absorbed energy and lowers the equilibrium temperature at which the same amount is radiated away. Celsius temperature is not proportional to absorption.
The fraction of sunlight reflected across all wavelengths and the entire planet is called the Bond albedo a. Average absorption over the entire surface = S(1 − a) ÷ 4, and blackbody emission = σT⁴. The temperature at which they are equal is T = [S(1 − a) ÷ (4σ)]¼. We use S = 1,361 W/m² and σ = 5.670374419 × 10⁻⁸ W/(m²·K⁴); Celsius temperature is T − 273.15. This simple model treats an entire spherical planet as having a single temperature, absorbing only solar radiation and having a thermal emissivity of 1. It excludes the atmosphere and greenhouse effect, temperature differences between day and night or across latitudes, ice and cloud feedback, and the time needed to reach equilibrium. The displayed temperature is not Earth’s actual surface temperature or a climate prediction for any location.
Read more about the principle · NASA: Earth’s Energy Budget · Solar Radiation and Earth Data
IT · Characters and Bytes
Even when it looks like one cluster,
its stored size can differ.
Change the character example and repetition count.
Compare visible character clusters, code points, UTF-16 units, and UTF-8 bytes.
For these six examples, each repetition is one extended grapheme cluster. Whether a cluster appears as a single shape depends on the font and device.
Change the values to discover today's mission.
After choosing characters yourself
The Hangul character ‘가’ is one code point but takes 3 bytes in UTF-8. Counting characters and measuring storage size use different criteria.
A Unicode code point is a number that identifies a character, and UTF-8 is an encoding that stores each Unicode scalar value in 1–4 bytes. JavaScript string length counts UTF-16 code units; one smiling emoji 😊 uses 2 code units. Extended grapheme clusters use Unicode boundary rules to group multiple code points together, as with combining accents and joined emoji. Although é (U+00E9) and e + combining accent (U+0065 U+0301) look similar, their stored code points differ, and their UTF-8 sizes are 2 and 3 bytes respectively. This activity compares six fixed examples; it does not normalize input or segment arbitrary strings. The number of shapes drawn on screen does not always equal the number of Unicode clusters. File headers, encoding markers, and compressed sizes are excluded.
Read more about the principle · Python Documentation: Unicode and UTF-8 · Unicode: Grapheme Cluster Boundaries
Math and Statistics · Paths and Connections
Can you traverse every path
without using any path twice?
Remove or add paths in the Seven Bridges of Königsberg.
The number of paths meeting at each point and overall connectivity affect whether a complete traversal is possible.
You may visit points multiple times, but you must traverse each path exactly once. Parallel paths connecting the same two points count as distinct paths.
Change the values to discover today's mission.
After changing the connections yourself
At intermediate points, entering and leaving paths form pairs. To traverse every path exactly once, consider both connectivity and the number of odd-degree points.
The number of paths meeting at a point is its degree in the graph. When all points with paths are connected, 0 odd-degree points allow an Euler circuit that starts and ends at the same point; 2 odd-degree points allow an open Euler trail starting and ending at those two points. With more odd-degree points, traversing every path exactly once is impossible. If the points with paths are disconnected, a single complete traversal is impossible even when all degrees are even. Isolated points with no paths are excluded from the connectivity condition. This activity uses hypothetical graphs without directions or travel restrictions. It does not calculate drawn path lengths, distances, or optimal routes on actual roads.
Read more about the principle · MIT: Graphs and Euler Trails (PDF)
Life and Environment · Estimating Population Size
Use the proportion seen again
to estimate the total population.
Change how many individuals were marked in the first observation and how many were encountered in the second.
Estimate the total population from how many individuals in the second observation were previously marked.
R counts individuals included in both observations, so it cannot exceed the smaller of M and C. Both observations concern the same hypothetical population.
Change the values to discover today's mission.
After changing the overlap yourself
If marked individuals have mixed sufficiently throughout the population, their proportion in the second observation is expected to resemble their proportion in the whole population.
The Lincoln–Petersen point estimate of total population is N̂ = M × C ÷ R. It follows from assuming that the population’s marked proportion M/N is similar to the second observation’s marked proportion R/C. The number of distinct individuals directly observed is M + C − R, which differs from the total population estimate. If R is 0, this formula cannot produce an estimate. When R is small, a difference of just a few overlapping individuals can greatly change the estimate. Assumptions include a closed population with no births, deaths, or migration between observations; sufficient mixing; retained marks and accurate identification; equal capture probability for every individual; and no behavioral changes after marking. Small-sample bias corrections and confidence intervals are not calculated. This activity explores relationships among hypothetical observation counts; it does not provide instructions for capturing or marking actual animals or estimating a real population.
Read more about the principle · University of Wyoming: Mark–Recapture and Population Estimation
Everyday Life and Culture · Perspective in Photography
As the camera moves closer,
the size difference between near and far objects grows.
This pinhole camera views two objects of equal actual width placed at different distances.
Change the camera position and focal length to distinguish projected sizes from their ratio.
Each object’s actual width is fixed at 0.3 m. Only their depth positions differ; the displayed mm values are widths projected onto the image plane.
Change the values to discover today's mission.
After moving the camera yourself
The ratio of the objects’ distances from the camera determines their projected size ratio. Changing only focal length at the same position changes both widths together while preserving their ratio.
In ideal pinhole projection, image-plane width w′ = f × w ÷ z. Here, f is focal length, w is actual width, and z is distance from the camera to the object. For two objects of equal actual width, the projected width ratio is far-object distance ÷ near-object distance = (d + separation) ÷ d; focal length cancels out. As the camera moves back, this ratio approaches 1. If you change focal length and also move the camera to maintain the same framing, the position change alters perspective. The layout above shows the distance relationships between the camera and objects. The bars’ 36 mm scale provides a fixed reference for comparing projected widths. This geometric model assumes parallel object and image planes and does not calculate lens distortion, defocus blur, depth of field, or changes in facial shape.
Read more about the principle · MIT Course Text: Images and Pinhole Projection · Columbia: Geometry of Perspective Projection
Chemistry · Light Absorption
As light travels a longer path,
how does the remaining light change?
Light of a single color passes through a uniform hypothetical solution.
Change concentration, path length, and incoming light to distinguish absorbance from transmittance.
This hypothetical solute has an absorption coefficient ε set to 1 L/(mmol·cm). The brightness units and coefficient are values for comparing principles, not measurements of an actual substance.
Change the values to discover today's mission.
After changing the light’s travel conditions yourself
Concentration multiplied by path length determines absorbance. Doubling absorbance does not halve the fraction of light transmitted.
The Beer–Lambert law gives absorbance A = ε × concentration c × optical path length l = −log₁₀(transmitted light ÷ incoming light). This activity calculates transmittance as 100 × 10⁻ᴬ % and transmitted light as incoming light × 10⁻ᴬ. Absorbance is logarithmic and differs from the percentage of light absorbed. When A increases from 0.5 to 1, transmittance falls from about 31.623% to 10%. At the same concentration and path length, halving only incoming light halves both transmitted and absorbed light, while transmittance and absorbance remain unchanged. At a concentration of 0, all light passes through in this model. This ideal linear absorption model assumes monochromatic light passing through a uniform solution without reflection, scattering, or light emission, and excludes absorption by the container and solvent. It is not a tool for measuring actual substance concentrations or predicting a specific liquid’s transparency.
Read more about the principle · IUPAC: Beer–Lambert Law
Physics · Collisions and Momentum
Although velocities change after a collision,
something stays the same when added together.
Object 1 on the left and object 2 on the right move along a straight line.
Change masses, velocities, and bounciness to compare momentum and kinetic energy before and after collision.
For velocity, + means rightward and − means leftward. A collision is calculated only when the left object’s velocity exceeds the right object’s, narrowing the gap. Otherwise, velocities remain unchanged.
Change the values to discover today's mission.
After changing collision conditions yourself
If the external impulse on the two objects is 0, total momentum is the same before and after collision. Whether kinetic energy is conserved must be examined separately.
Total momentum is p = m₁v₁ + m₂v₂, and kinetic energy is ½m₁v₁² + ½m₂v₂². When the objects approach each other, their post-collision velocities are v₁′ = [p − m₂e(v₁ − v₂)] ÷ (m₁ + m₂) and v₂′ = [p + m₁e(v₁ − v₂)] ÷ (m₁ + m₂). Here, e is relative separation speed after collision ÷ relative approach speed before collision. When e = 1, kinetic energy is conserved; when e = 0, both objects move at the same velocity p ÷ (m₁ + m₂). Moving together does not always mean stopping. The decrease in kinetic energy is the portion converted into other forms such as deformation, heat, and sound; it does not mean total energy disappears. This is an ideal collision along a straight line with no external impulse. Friction, rotation, collision duration, and the deformation process are not calculated. No collision event is applied when the gap stays constant or increases.
Read more about the principle · University of Florida: Collisions Between Two Objects (PDF)
Economics and Society · Voting Methods
With the same preferences,
can the counting method change the result?
Three groups rank A, B, and C in different preference orders.
Change group sizes and voting methods to compare results from the same preferences.
Candidates earlier in the arrow sequence are preferred. The total population changes with your settings and is not always 100 people. Everyone ranks all three candidates.
Change the values to discover today's mission.
After changing how votes are counted yourself
Plurality considers only first choices, while Borda awards points for every rank. Even with the same preferences, the winner can change depending on what is added up.
Plurality gives one vote to each person’s rank-1 candidate. Borda awards 2 points for rank 1, 1 point for rank 2, and 0 points for rank 3, giving A = 2a + c, B = a + 2b, and C = b + 2c. The top-two runoff first compares rank-1 votes. A candidate with a majority—more than half the total—wins immediately. If nobody has a majority and the top two can be identified, each person votes for the one they prefer among those remaining candidates. If a tie for second place prevents selecting two candidates, “Additional rule needed” is displayed rather than choosing arbitrarily. Equal highest totals in plurality, Borda, and the final runoff are shown as ties. With the default preferences, the winners differ: A with 40 votes under plurality, B with 110 points under Borda, and A with 65 votes in the runoff. All preferences are hypothetical and complete; actual election rules, strategic voting, abstentions, and incomplete rankings are excluded. This does not mean one method is best in every situation.
Read more about the principle · OpenStax: Voting Methods
Earth and Space · Groundwater Flow
Even water flowing through pores
needs a driving difference.
Imagine a uniform porous layer completely filled with water.
Change hydraulic conductivity, the head difference between the ends, and flow distance to compare water flow.
Cross-sectional area is fixed at 5 m² and effective porosity at 0.3. Hydraulic head expresses water’s energy due to elevation and pressure as a height; it does not mean only the difference in ground-surface elevation.
Change the values to discover today's mission.
After changing water-flow conditions yourself
With the same head difference, a more conductive layer carries more water, while a longer flow distance reduces the difference per unit distance.
Darcy’s law gives hydraulic gradient i = Δh ÷ L, velocity based on the full cross section q = K × i, and discharge Q = q × cross-sectional area. The value q uses the entire cross section, including solids, rather than only the pores where water actually flows. With effective porosity nₑ = 0.3, average pore-water velocity is q ÷ nₑ, about 3.333 times the Darcy velocity here. When the head difference is 0, flow in this model is 0. Doubling only K also doubles discharge. Holding the head difference fixed and doubling distance halves both hydraulic gradient and discharge. This is steady flow through a uniform, fully water-saturated layer, with water properties, cross-sectional area, and porosity fixed. K uses hypothetical values chosen for comparison, not measurements of specific soils or rocks. The model does not predict rainfall infiltration into unsaturated soil, actual groundwater arrival times, contaminant transport, or ground stability.
Read more about the principle · USGS: Groundwater Hydraulics (PDF) · USGS: Pores and Groundwater Flow
IT · Concurrent Execution
Two participants each add once,
but only one addition may take effect.
A and B each add 1 to the shared number 10.
Change the read/write order and locking setting, then examine execution step by step.
A read retrieves the shared value and prepares a value with 1 added; a write puts that prepared value into the shared value. With locking enabled, whoever reads first finishes writing before the other starts.
Change the values to discover today's mission.
After changing the execution order yourself
Even when the code adds once for each participant, one addition can be lost depending on the order of reads and writes to the shared value.
If A and B both read 10, each prepares 11. Even if they write 11 in turn, the final value is 11, so one update is lost. This is called a lost update. Protecting the entire read–calculate–write sequence with one lock lets the first person finish before the next reads the new value, making the final value 12. Locking only the write does not fix the earlier reading of an outdated value. The target 12 is the value after both additions finish, not the correct value at an intermediate step. When both have only finished reading, the shared value has not changed and no update has been lost yet. This activity models the order of operations without running actual browser threads. It does not cover synchronization across multiple computers or deadlocks.
Read more about the principle · OSTEP: Threads and Shared Values (PDF) · OSTEP: Locks (PDF)
Math and Statistics · Multiple Testing
If you look for enough differences,
even chance can look like a difference.
Run multiple independent tests under conditions with no actual differences.
Change the number of tests and the error threshold to compare the probability of at least one false positive.
A false positive means concluding that a difference exists when there is no actual difference. We assume all null hypotheses are true and the tests are independent. An average of 0.05 cases means the average over many repetitions under the same conditions, not that a single result is split into fractions.
Change the values to discover today's mission.
After changing the number of tests yourself
Even with a small error threshold for each test, the probability of at least one false positive can grow when many tests are performed.
If p is the false-positive probability per test and m is the number of tests, the probability of at least one false positive among independent tests with all null hypotheses true is 1 − (1 − p)ᵐ. The average false-positive count is mp. Applying an unadjusted threshold of α = 5% to 20 tests gives an overall probability of about 64.15%. Bonferroni adjustment lowers each threshold to α ÷ m to keep the overall error rate at or below α. This upper-bound guarantee does not require independent tests, but the exact probability shown here assumes independence. The overall probability and the average count are different quantities. This activity does not generate test results or actual p-values. A p-value is not the probability that the null hypothesis is true or that an effect exists. Stricter thresholds can make small differences harder to detect, and actual research must also consider how hypotheses are grouped and how tests are related.
Read more about the principle · MIT: Error Probabilities in Multiple Testing (PDF) · NIST: The Bonferroni Principle
Life and Environment · The Genetic Code
Three letters form a group,
but what if one changes?
Change one letter in GAA, the second codon of a hypothetical RNA.
Compare outcomes that produce the same amino acid, a different amino acid, or an early stop.
RNA reads AUG · GAA · UUC · UAA in the 5′ → 3′ direction. Translation starts with Met (methionine) at AUG; GAA specifies Glu (glutamic acid), and UUC specifies Phe (phenylalanine). A stop is a signal to end translation, not an amino acid.
Change the values to discover today's mission.
After changing one RNA letter yourself
Multiple codons can specify the same amino acid, so whether a single-letter change alters the amino acid depends on the codon.
A codon consists of three consecutive RNA bases. In the standard genetic code (table 1), GAA and GAG both specify Glu, so this is a synonymous change that leaves the amino acid sequence unchanged. GUA specifies Val (valine), making it a missense change, while a change to UAA is a nonsense change that brings the stop signal forward. If the second codon is a stop, translation ends after producing one Met, and the following UUC is not translated. Entering the original letter is classified as no change. The starting position and reading frame of three-letter groups are fixed; insertions and deletions are not covered. Genetic codes have exceptions across organisms and organelles, and synonymous changes are not always without effects in actual organisms. This model illustrates translation principles using a short synthetic RNA, not the effects of actual genes or diseases.
Read more about the principle · NCBI: The Standard Genetic Code · OpenStax: Codons and the Genetic Code
Everyday Life and Culture · Photographic Exposure
Even at the same brightness,
light quantity and motion trails can differ.
Change the aperture, shutter time, and ISO for the same scene.
Compare light received by the sensor, model brightness, and motion trails separately.
The reference is f/4 · 1/200 s · ISO 100. Scene lighting, focal length, camera position, and subject speed remain the same. ISO amplifies brightness in this model without increasing the number of photons reaching the sensor.
Change the values to discover today's mission.
After changing the photo settings yourself
Aperture and shutter time change the incoming light, while ISO changes the model's brightness for that light. Motion trails can differ even at the same brightness.
Under otherwise identical conditions, light quantity is proportional to shutter time and inversely proportional to the square of the f-number. Opening from f/4 to f/2 quadruples the light, while shortening the time from 1/200 s to 1/800 s reduces it to ¼, so changing both gives the same quantity as the reference. The shorter time makes the motion trail ¼ as long. Using f/8 with ISO 400 gives ¼ of the received light but the same model brightness as the reference. A brightness difference of +1 stop means twice as bright, and −1 stop means half as bright. Model brightness uses linear amplification proportional to ISO. It does not reproduce highlight clipping, actual photo tone processing, metering, or noise. The motion trail is the relative length left by a point moving at constant speed, not a measure of camera shake or overall sharpness. Differences in lens light transmission, depth of field, and automatic exposure are also excluded.
Read more about the principle · Nikon: Aperture, Shutter, and Exposure Basics · Nikon: ISO Sensitivity
Chemistry · Buffer Solutions
Even when acid is added,
what if a partner can accept it?
Change the amounts of weak acid HA and conjugate base A−, then add a little strong acid.
Check the remaining conjugate base, the pH change, and the range where the approximation applies.
Volume is fixed at 1 L and does not change when acid is added. A hypothetical weak acid with pKa = 5 is used. These are not measured values for an actual substance or instructions for preparing a solution.
Change the values to discover today's mission.
After changing the acid amount yourself
When the conjugate base accepts incoming H+ and becomes weak acid, it can reduce the pH change. But the amount available to accept it is limited.
First, calculate the stoichiometric reaction A− + H+ → HA. A− decreases and HA increases by the amount accepted by the conjugate base. The amounts in the diagram are calculated from neutralization, not exact ion amounts accounting for slight acid dissociation and water equilibrium. The Henderson–Hasselbalch approximation is pH ≈ pKa + log₁₀(A− ÷ HA). Two solutions with the same ratio have the same approximate initial pH, but the one containing more weak acid and conjugate base can accept the same amount of strong acid with a smaller pH change. This activity displays the approximation only when at least 10 mmol of A− remains after calculation and A− ÷ HA is within 0.1–10. Outside this range, it shows “Outside approximation range,” which does not mean the actual pH is nonexistent or 0. Once all the prepared conjugate base is used up, the remaining strong acid is shown, but this equation is no longer used to calculate pH. This is an ideal-solution approximation applied after the stoichiometric reaction of the weak acid and conjugate base. Exact acid dissociation equilibrium, ion activities, and volume changes are not calculated.
Read more about the principle · OpenStax: Buffers and Buffer Capacity
Physics · Torque
If you push farther out,
will it turn more easily?
Change the magnitude, position, and direction of the force on the right side of the bar.
Compare its torque with that of the fixed force on the left to find rotational balance.
On the left, a downward force of 20 N acts 1 m from the pivot. The right-side angle is measured downward from the bar's rightward direction; 90° means a downward force.
Change the values to discover today's mission.
After changing where you apply the force
Torque depends on the force magnitude, its distance from the pivot, and its direction.
Torque magnitude is τ = rF sin θ. The lever arm r sin θ is the perpendicular distance from the rotation axis to the force's line of action. Applying the same force perpendicularly at twice the distance doubles the torque. At an angle of 30°, the torque is halved for the same distance and force. At 0° or 180°, parallel to the bar, torque is 0, so the right-side force alone cannot cancel the left-side torque. The “balancing force” is the calculated force required and may exceed the adjustable range of 10–40 N. A + net torque initially tends to turn the bar counterclockwise; − tends to turn it clockwise. 0 indicates rotational balance in this model, not that all forces on a free object are balanced. The pivot is fixed, and its reaction force constrains the bar's translation. The model assumes a massless rigid bar initially at rest with no rotational friction; motion over time is not calculated. Torque in N·m is not displayed as J, the unit of energy.
Read more about the principle · OpenStax: Torque and Lever Arms · OpenStax: Force and Torque Balance
Economics and Society · Auction Formats
Even if you bid highest,
must you pay your own bid?
A, B, and C submit sealed bids for one hypothetical item.
Keep the bids unchanged and change only the winner's payment rule.
A's valuation of this item is fixed at 80 points. A bid and a valuation are different. If the highest bids tie, an additional rule is needed to select the winner.
Change the values to discover today's mission.
After changing the payment rule yourself
Even when the same person wins, the payment depends on whether they pay their own bid or the next-highest bid.
If there is a unique highest bid, that bidder wins under both rules. In a sealed-bid auction with price rank 1, the winner pays their own bid; with price rank 2, they pay the highest bid among the other participants. Nonwinners pay 0 points. A's net benefit is 80 − payment if A wins, and 0 otherwise. If A bids 90 points and wins at price rank 1, the net benefit can be −10 points. If the highest bids tie, the model does not arbitrarily select a winner; it displays payment and net benefit as “Undetermined.” The model uses one item, one round of bidding, and hypothetical valuations and points, with no reserve price, entry fees, or commissions. It does not calculate participants' strategic choices, information differences, or the best strategy across all auctions.
Read more about the principle · MIT: Auction Payment Rules and Strategies (PDF)
Earth and Space · Orbital Periods
A longer path around,
but does it take longer?
Change the radius of a hypothetical circular orbit and the mass of its central star.
Compare orbital period and speed, then adjust the time to check the current position.
For a circular orbit of 1 AU around 1 solar mass, the period is set to 1 model year and the speed to 1×. A model year of 1 is a reference chosen to approximate Earth's orbital period, not an exact prediction of an actual celestial body's position.
Change the values to discover today's mission.
After changing the orbital conditions yourself
For the same central mass, more distant circular orbits have lower speeds and take longer to complete one revolution.
When gravity provides the centripetal force for a circular orbit, the period is T ∝ √(r³ ÷ M) and the speed is v ∝ √(M ÷ r). Here, radius r is in AU and M is in solar masses, normalized so that T = √(r³ ÷ M) model years and speed relative to the reference = √(M ÷ r)×. At the same mass, doubling the radius multiplies the period by √8 and the speed by 1 ÷ √2. At the same radius, doubling the central mass multiplies the period by 1 ÷ √2 and the speed by √2. Cumulative revolutions equal time ÷ period, and the current angle represents the position within one revolution. The model assumes the central star is much more massive than the orbiting object and the orbit is circular. Each setting compares a different possible orbit, rather than reproducing an instantaneous orbital change by a spacecraft. The star's drawn size is fixed and does not imply that its actual radius is calculated from its mass. Elliptical orbits, interactions among multiple bodies, and actual planetary positions are not calculated.
Read more about the principle · OpenStax: Gravity and Orbital Periods
IT · Table Joins
Joined on the same key,
yet rows multiply or disappear.
Match A, B, and C in the left table with keys in the right table.
Change the right-hand data and join type, then inspect each resulting row.
The left table is fixed at 3 rows. NULL indicates that no corresponding right-hand row exists; it is neither the number 0 nor an empty string.
Change the values to discover today's mission.
After changing the table matches yourself
A join forms every pair of rows with the same key. Duplicate right-hand keys can turn one left-hand row into multiple result rows.
INNER JOIN includes every pair of rows with matching keys. Unmatched rows from either table are omitted. LEFT JOIN also includes every matching pair, but adds one row with NULL right-hand values for each left-hand row with no match at all. LEFT JOIN therefore preserves the left-hand data, but the result does not necessarily have the same 3 rows as the left table. If the right table has two A rows, the left A matches twice. The different keys D and E do not match A, B, or C on the left. This screen compares small fixed tables without connecting to an actual database. For illustration, rows follow a fixed left-table and right-table order; actual SQL requires ORDER BY to guarantee ordering.
Read more about the principle · PostgreSQL: Joining Tables · PostgreSQL: SELECT and Joined Tables
Math and Statistics · Area Under a Curve
Add up small areas,
and approach the curve.
Divide the area under y = x² from x = 0 to 2.
Change the number of pieces and the method for choosing heights, then compare with the exact area.
The function is fixed at x² and the interval at [0, 2]. The error's sign distinguishes overestimation from underestimation; its magnitude without the sign is the absolute error.
Change the values to discover today's mission.
After dividing the area yourself
The area under a curve can be approximated by adding areas over small intervals. Even with the same number of pieces, the choice of heights makes a difference.
With n pieces, the width is Δx = 2 ÷ n. Rectangles sum x² × Δx at the selected positions; trapezoids sum the average of the endpoint heights × Δx. The exact area under this curve is ∫₀² x² dx = 8 ÷ 3. Because x² is increasing and convex upward here, left-endpoint and midpoint sums underestimate, while right-endpoint and trapezoidal sums overestimate. These relationships do not apply unchanged to every function. Displayed values are rounded to three decimal places; calculations use unrounded values. No actual physical unit is assigned to y, and the area under x² on [0, 2] is not interpreted as a probability. A probability density function requires separate checks, including that its total area is 1.
Read more about the principle · OpenStax: Approximating Area with Small Sums · OpenStax: Midpoint and Trapezoidal Integration
Life and Environment · Surface Area and Volume
The inside grows quickly,
but is there enough surface?
Change the size of one sphere and the number of smaller spheres sharing the same total volume.
Explore the relationship between exchange through a surface and internal volume using a geometric model.
The circles are cross sections of spheres; the values describe each whole sphere's surface and volume. The original and smaller spheres use the same length scale.
Change the values to discover today's mission.
After changing the size and division yourself
As a sphere grows, its volume increases with the cube of its radius, while its surface area increases with the square. Dividing the same total volume into smaller spheres creates more surface.
A sphere's surface area is 4πr², its volume is 4πr³ ÷ 3, and its surface-area-to-volume ratio is 3 ÷ r. Dividing the volume of a sphere of radius R into N equal spheres gives a smaller radius of R ÷ ∛N and a total surface area ∛N times the original. With 8 spheres, the surface area is 2 times as large; with 64, it is 4 times as large. Division preserves only total volume, creates new surfaces, and assumes each sphere's surface is independently exposed. The diagram does not show smaller spheres packed without gaps inside the original, nor does it reproduce actual cell division. This model explains the relationship between surface area available for exchange and internal demand proportional to volume. Actual cells differ in shape, membrane folds, transport systems, and metabolic rates, so this ratio alone cannot predict exchange rates or viable sizes. Displayed decimals are rounded; calculations use unrounded values.
Read more about the principle · OpenStax: Cell Surface Area and Volume
Everyday Life and Culture · Line Breaks and Spacing
Fill the first lines,
but will the total cost fall?
Keep words in order while breaking them into lines of a fixed width.
Compare filling earlier lines first with a layout that minimizes the total whitespace cost.
Each character and each space between words occupies 1 unit. Words are not split, and the last line's cost is set to 0.
Change the values to discover today's mission.
After changing the line-breaking rules yourself
Filling each earlier line as much as possible may not minimize the total cost. You also need to consider the whitespace on later lines.
The greedy layout fills each line while the next word still fits, then starts a new line. The minimum-cost layout compares possible line breaks to find the smallest sum of squared or cubed unused widths. In this activity, the last line's cost is set to 0, and words on a line are separated by a space of 1 unit. Word order is preserved; omission, duplication, and splitting are not allowed. At width 6, starting with “one of” leaves 4 units of whitespace on the next “us” line, giving a squared cost of 16. Breaking into “one” and “of us” reduces the cost to 9 + 1 = 10. If minimum costs tie, the layout with fewer lines is preferred; if line counts also tie, the layout with more words on earlier lines is displayed. This may show one of several optimal layouts. This hypothetical cost model uses short English words with equal character widths and does not judge actual writing's beauty or readability. Proportional fonts, Korean punctuation, language-specific line breaking, and hyphenation are not calculated.
Read more about the principle · MIT: Line Breaking by Comparing Total Costs
Chemistry · Gas States
If you reduce the space,
what happens to pressure?
Change the temperature and volume of a fixed amount of ideal gas.
Compare with the reference state to explore what affects pressure.
The amount of gas is fixed at 0.1 mol. The reference is 300 K · 5 L. The 20 dots represent the same amount of gas, not the actual number of molecules.
Change the values to discover today's mission.
After changing temperature and space yourself
For a fixed amount of ideal gas, pressure rises with higher absolute temperature or smaller volume. What you hold constant matters.
The ideal gas equation is PV = nRT. Here, n = 0.1 mol and R = 8.314462618 J/(mol·K). With volume in L and pressure in kPa, kPa × L = J. Pressure is calculated as nRT ÷ V, always using absolute temperature in K. Celsius is displayed as K − 273.15 but is not used in ratio calculations. Pressure × volume remains constant only at constant temperature; raising temperature increases this product too. Both gas diagrams use the same volume scale and the same number of representative dots. The dots do not reproduce actual particle counts or collisions. This ideal gas model ignores intermolecular forces and particle volume, so it does not predict real gases at high pressure, liquefaction, or phase changes. Displayed values are rounded; calculations use unrounded values.
Read more about the principle · OpenStax: Ideal Gas Pressure, Volume, and Temperature · OpenStax: The Kinetic Molecular Model of Gases
Physics · Angular Momentum
Closer to the rotation axis,
how much faster will it spin?
Move both masses inward or outward by the same distance relative to the rotation axis.
Compare how angular momentum and rotational energy each change.
Both initial distances are 1 m. This ideal rotation model fixes the central rotor's rotational inertia at 1 kg·m² and uses two point masses of 1 kg each, one on either side.
Change the values to discover today's mission.
After changing the masses' distance yourself
Moving the masses closer to the axis reduces rotational inertia and increases angular velocity. Equal angular momentum does not necessarily mean equal rotational energy.
Angular momentum is L = Iω, and rotational kinetic energy is K = Iω² ÷ 2. If each 1kg point mass is at distance r from the axis, the total rotational inertia is I = 1kg·m² + 2 × 1kg × r². Initially, r = 1m, so I = 3kg·m². With no external torque about the rotation axis, angular momentum remains unchanged, and the new angular velocity is L ÷ I. The net internal work done to rearrange the masses equals the change in rotational energy: W = K_after − K_before. Moving them inward requires positive work; negative work when moving them outward represents ideal energy recovery. The radial velocity of the masses is assumed to be 0 at the moments compared. Friction, air resistance, and heat loss are excluded. This model does not calculate the motion of human arms or an actual body. Both arrangements use the same distance scale.
Read more about the principle · OpenStax: Conservation of Angular Momentum · OpenStax: Calculating Rotational Inertia
Economics and Society · Sunk Costs
Costs already incurred,
and the choice ahead?
Adjust irreversible past costs and future costs and benefits.
Compare continuing and stopping using hypothetical points.
If you stop, future costs and benefits are both 0. Past costs remain the same for both choices, and their effect in lowering the total scores stays visible.
Change the values to discover today's mission.
After adjusting costs and benefits yourself
Costs already incurred remain on record. Because both choices share them, they cancel when comparing future differences.
The total score for continuing is −past costs −future costs +future benefits; for stopping, it is −past costs. Subtracting gives continue −stop = future benefits −future costs, so past costs cancel. If the difference is positive, continuing scores higher; if negative, stopping scores higher; if 0, the choices are equal. Even when both total scores are negative, continuing may mean a smaller loss if future net benefit is positive. This hypothetical points model assumes no refunds, resale proceeds, or cancellation penalties when stopping, certain future values, and past costs that cannot be recovered under either choice. Recoverable costs and outcomes that differ between choices must be included separately in the future comparison.
Read more about the principle · OpenStax: Relevant Costs and Sunk Costs
Earth and Space · Plate Boundaries
Beyond individual motion,
differences in direction
Adjust the horizontal and vertical velocities of the plates on either side of the boundary.
Resolve relative motion into components to explore convergence, divergence, sliding, and oblique motion.
The boundary is a fixed vertical line. Right is +horizontal, up is +vertical, and relative velocity is B − A.
Change the values to discover today's mission.
After adjusting both plates' velocities yourself
Changes at the boundary between two plates depend on their relative velocity. Examine motion perpendicular and parallel to the boundary separately.
The perpendicular component is B's horizontal velocity −A's horizontal velocity; the parallel component is B's vertical velocity −A's vertical velocity. A negative perpendicular component means convergence, and a positive one means divergence. If the perpendicular component is 0 and only the parallel component remains, the motion is sliding at a transform boundary. When both components are present, oblique motion is also shown as convergence +sliding or divergence +sliding. Plates with equal velocities are at relative rest even if both are moving. Relative speed is the square root of the sum of the squared components, distinct from the directional components. This planar schematic compares velocity vectors relative to a fixed boundary; it does not reproduce actual ground movement or collisions. Convergence does not always cause subduction; this depends on plate types and geological conditions. The values in this activity do not predict earthquake occurrence, magnitude, timing, or actual fault activity.
Read more about the principle · USGS: Three Types of Plate Boundaries and Relative Motion
IT · Signal Sampling
The same sample points,
different waveforms
Adjust the signal frequency and the number of samples taken in 1 second.
Compare changes hidden between samples and different waveforms passing through the same points.
At 20 samples/second, 1 second is divided into 20 intervals, giving 21 points including both endpoints. Both waveforms start at time 0 seconds with value 1; the curves show model values between points.
Change the values to discover today's mission.
After adjusting the sampling interval yourself
A fast waveform can produce the same values as a slow one when sampled sparsely. The sample points alone cannot distinguish them.
The original waveform is cos(2πft), sampled at t = n ÷ fₛ. Here, f is the signal frequency, and fₛ is the number of samples taken in 1 second. If r is the remainder when f is divided by fₛ, the folded frequency is min(r, fₛ − r), between 0 and fₛ ÷ 2. This activity fixes both cosine waves' initial phase at 0, so their values match at every sample point. If the folded frequency is 0, the original signal still oscillates between samples even though all sampled values are 1. Including both 0 and 1 seconds gives fₛ + 1 points. Lines connecting points do not reconstruct the original signal. To distinguish ideal band-limited signals, the sampling frequency must exceed twice the highest frequency. At exactly twice that frequency, information may be lost depending on the initial phase, so this boundary does not provide a general reconstruction guarantee. This is a sampling model for a single waveform; it does not calculate hardware audio or video quality, noise, or filters.
Read more about the principle · MIT: Sampling and Aliasing (PDF)
Math and Statistics · Instantaneous Rate of Change
Change between two points,
rate of change at one point
Adjust the starting position on the curve and the spacing between two points.
See how the average rate of change approaches the instantaneous rate of change.
The function is fixed at y = x². The second point is at x + h, and h is always positive, so 0 is never actually substituted.
Change the values to discover today's mission.
After bringing the points closer yourself
The average rate of change is the change in y between two points divided by their spacing in x. The instantaneous rate is defined as the limit as this spacing approaches 0.
For y = x², the average rate of change is [(x + h)² − x²] ÷ h = 2x + h, and the instantaneous rate at the starting point is 2x. Their difference is h, so it shrinks as the spacing decreases. This result applies to the quadratic function used here; the error is not h for every curve. While the spacing remains positive, the average rate is not exactly equal to the instantaneous rate. The secant connects the two points, and the short tangent segment shows the instantaneous slope at the starting point. A tangent is not defined as a line that meets a curve only once. The change in y and the rate of change, in y units/x units, also have different units. A negative rate means y decreases as x increases. This function model uses no time or distance data, so its values do not predict actual speed or future changes.
Read more about the principle · OpenStax: Average Rate of Change and the Definition of the Derivative
Life and Environment · Genetic Drift
Even without an advantage,
frequencies change.
Sample gene copies for the next generation from the same starting frequencies.
Adjust copy count, generation count, and replicate runs to explore the effects of chance.
Initially, A is 50% and B is 50%. N counts gene copies, not individuals carrying two copies of the gene.
Change the values to discover today's mission.
After sampling successive generations yourself
Even when neither type has an advantage, random sampling of the next generation can change their frequencies. Such chance changes in frequency are called genetic drift.
This neutral model independently samples N copies with replacement from the previous generation each generation. The probability of drawing A equals its frequency in the previous generation. Neither A nor B has a reproductive advantage, and there is no mutation or migration. The initial and final diagrams each show all N copies. Once the frequency reaches 0 or 100%, it stays there because no new copies enter this model. From a starting frequency of 0.5, the theoretical standard deviation of frequency after g generations is √{0.25 × [1 − (1 − 1/N)ᵍ]}. This describes the entire distribution of ideal replicate runs. It is neither the standard deviation calculated from the 8 runs on screen nor an error range for the current result. Smaller copy counts give a larger theoretical standard deviation, but do not guarantee a larger change in every individual run. % expresses frequency; %p expresses the difference between two percentages. The replicate number selects a model run that can be revisited with the same settings. Changing the copy count produces different model experiments; these do not predict the evolution or health of an actual population.
Read more about the principle · OpenStax: Population Genetics and Genetic Drift · Simons Institute: Neutral Copy-Sampling Model (PDF)
Everyday Life and Culture · White Balance
Light changes colors,
but what is the reference?
Adjust the light illuminating the same materials and the correction reference.
Compare using gray as the reference with choosing a colored material.
Channel differences are calculated for the actual gray material, rather than the selected reference color. The materials stay fixed; only lighting and correction change.
Change the values to discover today's mission.
After changing the correction reference yourself
To remove a lighting color cast, choose an actually neutral white or gray reference. Treating a colored material as neutral can distort other colors.
Observed linear RGB is calculated by multiplying each material's fixed reflectance by the lighting value channel by channel. The mean of the selected reference's three channels is the target, and each correction factor is target ÷ channel value. At 0% strength, the factor is 1; at 100%, the full calculated factor is applied. This arithmetic mean is the model's brightness reference, not physical luminance or the maximum brightness of white. Neutrality is compared using the actual gray material's maximum channel value −minimum channel value. A colored reference can make the actual gray change even when the reference's own channels become equal. Calculations use linear RGB; corrected colors are converted to sRGB for display. Values are clipped to 0–1 only during conversion, and the number of channels exceeding 1 is reported separately. Statistics use values before display clipping, so clipped screen colors and calculated values are distinguished. This simple per-channel lighting and correction model does not reproduce actual light spectra, color temperatures, camera calibration, or precise human color perception.
Read more about the principle · Adobe: Neutral References for White Balance · W3C: Linear RGB and sRGB Conversion
Chemistry · Partitioning Between Two Phases
Even with the same start,
retention times differ.
Two components spend time partitioned between the mobile and stationary phases.
Adjust partitioning and flow rate to compare which component emerges first.
This ideal model has a length of 10cm and mobile- and stationary-phase volumes of 1mL each. K is stationary-phase concentration ÷ mobile-phase concentration and has no units.
Change the values to discover today's mission.
After adjusting the retention fractions yourself
A component that spends more time in the stationary phase spends less time moving with the mobile phase and emerges later. Different partitioning makes components arrive at different times even when they start together.
Rapid linear partition equilibrium between the two phases is assumed. Because the stationary and mobile phases have equal volumes, the mobile-phase fraction is 1 ÷ (1 + K), and the stationary-phase fraction is K ÷ (1 + K). This differs from a reaction that converts two substances into each other or dilution by adding solvent. In general, the retention factor is k′ = K × stationary-phase volume ÷ mobile-phase volume, and retention time is tᵣ = tₘ × (1 + k′). In this model, tₘ is mobile-phase volume ÷ flow rate. Doubling the flow rate halves both retention times and their difference. A larger time difference does not necessarily mean better actual resolution. The markers show arrival times of component centers; band width, diffusion, column efficiency, and sample overload are not calculated. Equal retention times mean the centers of both components arrive together in this model.
Read more about the principle · Agilent: Partitioning Between Two Phases and Retention (PDF) · Waters: Retention Factor and Time
Physics · Polarization and Transmitted Intensity
Between filters blocking light,
what if you add another?
Keep the first filter's axis fixed and rotate the final filter.
How do polarization direction and transmitted intensity change when you insert a middle filter?
The original light is unpolarized, and the first filter's axis is at 0°. The middle filter's angle is unused when it is absent. Angles are measured relative to the first filter's axis.
Change the values to discover today's mission.
After adjusting the polarization axes in sequence yourself
Light transmission depends on the angle between its preceding polarization direction and the next filter's axis. The middle filter changes polarization direction and intensity without creating light.
An ideal first polarizer transmits half the intensity of unpolarized light. Each subsequent filter's transmission fraction follows Malus's law: cos²(angle difference from the preceding polarization direction). With only two filters at 0° and 90°, transmitted intensity is 0. Inserting a filter at 45° between them transmits 50% × cos²45° × cos²45° = 12.5% of the original light. The final filter's transmission fraction is relative to the light reaching it, while final transmitted intensity is relative to the original light, so their denominators differ. Even if incident intensity is 0, this fraction is a theoretical coefficient determined by the axes, not a measured ratio of 0 divided by 0. Rotating any filter's axis by 180° gives the same axis; this activity compares angles from 0° to 90°. The process selects the electric field's oscillation direction perpendicular to light propagation, rather than bending the propagation direction as refraction does. This ideal polarizer model does not calculate absorption, scattering, or leakage in actual materials.
Read more about the principle · OpenStax: Polarization and Malus's Law
The article below provides background on polarization and optical activity. Rotation of the polarization plane by a chiral solution differs from the absorptive polarizers in this activity.
Economics and Society · Risk Sharing and Correlation
When sharing together,
some variation remains.
Increase the number of people sharing hypothetical losses equally.
Adjust how closely losses move together and compare variation in each person's share.
Each person has the same expected loss of 20 points and standard deviation of 10 points. This hypothetical model assigns correlation ρ to every pair. The bar shows variance (points²), not standard deviation.
Change the values to discover today's mission.
After increasing common shocks yourself
Pooling independent losses reduces variation in each person's share. Shocks everyone experiences together do not disappear through sharing alone, even in larger groups.
The variance of average loss for N people is 100 × [ρ + (1 − ρ) ÷ N] points². Add the common-shock variance 100ρ to the shared individual-shock variance 100(1 − ρ) ÷ N, then take the square root to obtain the standard deviation of each person's share. With correlation 0, it is 10 ÷ √N points; with correlation 1, it is 10 points regardless of group size. The standard deviation of total loss is N × the standard deviation of each share, so variation in the total may increase even as variation in each share decreases. Expected loss per person remains 20 points; this activity does not eliminate losses themselves. Standard deviation measures variation, not loss probability, maximum loss, or an interval with a specified probability. The model consists of independent common and individual shocks; in general, correlation 0 does not necessarily imply independence. It compares hypothetical loss sharing under equal conditions, not actual insurance premiums or investment risk.
Read more about the principle · MIT: Covariance and Correlation (PDF)
The article on the law of large numbers below provides background on pooling independent losses. Conclusions for independent repetitions cannot be applied unchanged when common shocks are present.
Earth and Space · Parallax and Distance
Angles from two positions,
a clue to distant stars
Adjust the spacing between two observation points and the target's distance.
Compare how little the direction to a distant target changes on a common angular scale.
The target lies on the perpendicular through the baseline's midpoint. The upper schematic shows relative positions, not actual scale. The angles below are enlarged for visibility, using the same scale for both observations.
Change the values to discover today's mission.
After adjusting the observation-point spacing yourself
With the same baseline, more distant targets show a smaller difference in viewing direction. Converting this angle to distance requires knowing the spacing between observation points as well.
The half-angle p is calculated as arctan[B ÷ (2D)], with B and D converted to the same length unit. Here, D is the perpendicular distance from the baseline's midpoint to the target. 1pc is exactly (648000 ÷ π)AU, 1 arcsecond is 1/3600 degree, and 1mas is 1/1000 arcsecond. This ideal geometry places the baseline perpendicular to the line of sight and the target on the perpendicular through its midpoint. The full difference between viewing directions is 2p. With a baseline of 2AU, as between opposite ends of Earth's orbit, annual stellar parallax is p, half the full change. The small-angle distance formula is D(pc) ≈ [B(AU) ÷ 2] ÷ p(arcseconds), while 1 ÷ p assumes B = 2AU. Do not substitute a value in mas directly. If the baseline is 4AU but you assume 2AU, you will incorrectly infer half the distance from the same angle. Exact arctan angles and approximate small-angle distances are calculated separately, then rounded for display. Actual observational noise, proper motion, and changes in baseline direction are not calculated.
Read more about the principle · OpenStax: Measuring Stellar Distances with Parallax · IAU: Exact Definition of the Parsec (PDF)
The astronomy section of the unit-conversion article below provides background on light-years and parsecs. It is not a dedicated explanation of parallax measurement.
IT · Deadlock and Resource Ordering
Holding one resource each,
how do they get stuck?
Change the order in which two tasks acquire resources X and Y.
Advance one attempt at a time and follow ownership and request relationships.
Solid arrows run from resource → owning task; dashed arrows run from task → resource it is waiting for. Changing the order restarts the same execution rules under the new conditions.
Change the values to discover today's mission.
After following each task's requests yourself
If each task waits for a resource held by the other, neither may be able to proceed. Following the same global resource order prevents circular wait in this model.
X and Y are exclusive resources that only one task can hold at a time. A task acquires its first resource, then its second, and on its next attempt finishes its work and releases both together. Attempts alternate starting with A, skipping completed tasks. If a task cannot acquire a resource, it waits while retaining those already held. The model assumes mutual exclusion, hold and wait, and no preemption. Deadlock occurs when ownership and requests for the two resources form a cycle. Different orders do not always cause deadlock; this activity shows whether the fixed alternating execution history reaches that state. If both tasks follow the same global order, no cycle forms, and they retry and finish once resources are released. Selecting an order does not recover actual tasks already executed. Real database schedulers, timeouts, and rollback policies are not reproduced.
Read more about the principle · OSTEP: Concurrency and Deadlock (PDF)
The database article below provides background on units of work, transactions, and rollback. It is not a dedicated explanation of locking or deadlock.
Math and Statistics · Regression to the Mean
An unusually high value,
will the next be the same?
Adjust the first observation and the correlation between two observations.
Distinguish the next observation's expected value from the result of a fluctuation you select.
Both observations have an overall mean of 0 and standard deviation of 1. You select one example fluctuation; the slider range does not represent all possible outcomes or equal probabilities.
Change the values to discover today's mission.
After separating the expected value from one example yourself
After an extreme first observation, the next expected value may lie closer to the center. This does not mean every subsequent outcome must move toward the center.
Let X and the independent fluctuation E be mutually independent standard normal variables with mean 0 and variance 1, and define Y = ρX + √(1 − ρ²)E. In this bivariate normal model, when the first observation is x, the conditional expectation of the next observation is ρx and its conditional standard deviation is √(1 − ρ²). Knowing only the correlation coefficient does not make this formula universally valid for general data. For 0 ≤ ρ < 1, the expected next value for a group selected for extreme first values moves closer to 0. This differs from a change in the overall population mean of 0 and does not guarantee an individual's next result. If the correlation is 1, the next value equals the first, and the fluctuation is unused. A manually selected E is one example from the normal distribution; it neither randomly samples the full distribution nor limits the range of outcomes. The key is to avoid attributing changes that can occur without intervention to praise, punishment, treatment, or similar effects. This activity does not predict people's grades, abilities, or health.
Read more about the principle · MIT: Regression to the Mean (PDF) · Penn State: Bivariate Normal and Conditional Distributions
The section critiquing measurement methods in the Dunning–Kruger article below connects comparisons of extreme groups with cautions about regression to the mean. It is not material for evaluating individuals using this model.
Life and Environment · Predators and Prey
Each population changes
the other's direction of growth or decline
Adjust the initial prey and predator indices and the time to explore.
Track how the two populations change each other's rates of change on the same scale.
Populations are dimensionless indices, and time also uses model units. These calculations use a continuous idealized model and do not represent actual animal counts or calendar dates.
Change the values to discover today's mission.
After tracking the two populations yourself
More prey affects predator growth, while more predators suppress prey growth. Examine the signs of the current rates of change separately from the peaks of the curves.
The dimensionless Lotka–Volterra model is dx/dt = x(1 − y), dy/dt = y(x − 1). Here, x is the prey index and y is the predator index. Predators increase when prey exceeds 1, and prey decreases when predators exceed 1. The point where both equal 1 is an equilibrium. Other positive initial values follow neutrally stable closed orbits without damped convergence to that point. H = x − ln x + y − ln y is a conserved quantity in this idealized model. The displayed curves show the numerical calculation history up to the selected current time, without adding a future that has not yet elapsed. The time lag between peaks arises from the interaction, rather than a fixed external delay. Unlike single-population growth with a carrying capacity, the two populations change each other's rates of change, but the model excludes carrying capacity, migration, seasons, random extinction, and environmental shocks. It does not imply that real ecosystems must follow the same cycles and does not predict actual population counts, control measures' effects, or extinction risk. Displayed values are rounded; calculations use unrounded values.
Read more about the principle · MIT: Predator–Prey Models in Ecology (PDF) · MIT: Predator–Prey Equations
The red tide article below provides background on feeding relationships and ecosystem changes. It does not validate the cycles of the predator equations or the effects of actual control measures.
Everyday Life and Culture · Layered Insulation
Heat passing through three layers:
Where does the temperature drop?
Adjust the hypothetical middle layer's thickness and thermal conductivity, and the outer layer's thickness.
Compare heat flux and the temperature drop across each layer at the same temperature difference.
The inner layer is fixed at 1cm with k = 0.20, and the outer layer at k = 0.80 W/(m·K). These thermal conductivities are hypothetical values, not the performance of specific products.
Change the values to discover today's mission.
After tracking temperatures across the layers yourself
At the same temperature difference, greater total thermal resistance reduces conductive heat flow. Each layer's temperature drop is proportional to its thermal resistance, not just its thickness.
The thermal resistance per unit area of each of the three layers is Rᵢ = Lᵢ ÷ kᵢ, with L converted from cm to m. The total resistance of layers in series is ΣRᵢ, and heat flux is q = temperature difference ÷ ΣRᵢ. At steady state, the same heat flux passes through all layers, and each layer's temperature drop is qRᵢ. The diagram's horizontal axis uses the same physical thickness scale for all three layers, and the temperature profile connects boundary values at the corresponding depths. When the middle layer's thickness is 0, that layer disappears, and its selected thermal conductivity does not affect the calculation. This hypothetical example sets the reference outside temperature to 0°C and the inside temperature higher by the selected temperature difference. Temperature differences have the same numerical value in K and °C, but absolute temperature units are not used interchangeably. Only 1-dimensional steady-state heat conduction with perfect contact is calculated; convection, radiation, thermal bridges, moisture, and heat storage over time are excluded. This model does not predict actual building heating costs or savings from specific products.
Read more about the principle · OpenStax: Heat Conduction and Thermal Resistance of Multiple Layers
The eco-friendly polymer article below provides background on polymers used as building insulation. It does not certify this activity's hypothetical thermal conductivities or compare product performance.
IT · Public-Key Cryptography
A locking key anyone can use:
How can only I unlock it?
Follow encryption and decryption using very small, publicly known numbers.
Change the key pair and message to see whether the original number returns.
Do not enter real passwords or private keys. These very small public examples and unpadded RSA calculations are not secure practical encryption.
Change the values to discover today's mission.
After exploring public-key encryption yourself
Encryption and decryption use different exponents but share the same modulus n. Calculate the results for each message to examine the relationship between the two keys.
Using distinct primes p and q, form n = pq and φ = (p − 1)(q − 1). The public exponent e and decryption exponent d are paired so that ed leaves a remainder of 1 when divided by φ. The ciphertext is c = mᵉ mod n, and decryption is cᵈ mod n. Integer remainders are calculated repeatedly instead of computing the entire large power. Integer messages with 0 ≤ m < n are recovered even if they equal 0 or share a common factor with p·q. Changing the key pair changes the message slider's maximum to n − 1 and reduces any value exceeding the range. The p, q, and d shown here are all small, publicly known educational numbers. Real public-key encryption requires sufficiently large keys, secure padding, protocols, and more. Raw, unpadded RSA like this must not be used to encrypt actual secrets. This does not reproduce the full communication process used by HTTPS.
Read more about the principle · Original RSA Paper: Public Keys and Decryption (PDF)
The communications article below provides basic background on TLS, HTTPS, and the role of encryption. This small RSA model does not explain every encryption process in actual TLS.
Chemistry · Activation Energy
The hill a reaction must climb
changes with temperature and catalysts.
Adjust the temperature, activation barrier, and barrier reduction provided by a catalyst.
Explore rate ratios relative to the same reaction at 300K.
The same hypothetical frequency factor is used throughout. A rate ratio of 2 does not mean that every real reaction takes exactly half the time.
Change the values to discover today's mission.
After exploring activation energy yourself
Temperature changes the fraction of molecules able to cross the barrier, while a catalyst lowers the barrier through a different pathway. Both effects contribute to the rate ratio.
The Arrhenius equation k = A exp(−Eₐ/RT) is used. All pathways share the same hypothetical frequency factor A, and the gas constant is R = 8.31446261815324 J/(mol·K). Barrier energies are converted from kJ/mol to J/mol for calculation. The ratio from temperature alone is exp[(Eₐ/R)(1/300 − 1/T)], and the effect of a catalyst lowering the barrier by C at the same temperature is exp(C/RT). The combined ratio is the product of these ratios, relative to the uncatalyzed reaction at 300K with the same Eₐ. The graph's vertical axis uses a logarithmic scale for the rate ratio, so equal intervals represent equal multiplicative differences. The pathway diagram keeps the starting and ending energy levels the same with and without a catalyst. A catalyst changes the activation barrier and pathway, but not the reaction's energy difference ΔE or the equilibrium itself. This hypothetical comparison excludes reaction mechanisms, concentration changes, catalyst consumption, and other factors, and does not predict actual substances' reaction times or hazards.
Read more about the principle · OpenStax: Collision Theory and the Arrhenius Equation
The catalyst article below provides further reading on differences in activation energy and reaction pathways.
Physics · Standing Waves and Resonance
Even on a moving string,
some points always stay still.
Adjust the string length, vibration mode, and instantaneous phase.
Distinguish node positions from moments when the entire string is flat.
Phase is an angle selecting an instant in time, not elapsed seconds. Permanent nodes are marked separately even when every point is momentarily at 0.
Change the values to discover today's mission.
After exploring standing waves and resonance yourself
When waves reflected at both ends overlap, certain positions always remain still. Not every point on a momentarily flat string is a node.
For an ideal string fixed at both ends, the wave speed is fixed at v = 100m/s. The nth mode has wavelength λ = 2L/n and natural frequency f = nv/(2L). Normalized displacement is displayed as y(x) = sin(nπx/L) cos(phase). Nodes occur at x = jL/n, with n − 1 excluding the endpoints, and there are n antinodes. At 90° and 270°, the permanent nodes remain in place even though every point's instantaneous displacement is 0. The diagram's horizontal axis uses a common physical length scale of 0–2m for all strings, and its vertical axis shows normalized displacement from −1 to 1. Changing the string length also moves the right endpoint. Phase selects an instant in a still image; it does not represent actual elapsed time or the same time across all modes. Tension, linear density, damping, and combined vibrations are excluded, and sound does not play automatically.
Read more about the principle · OpenStax: Standing Waves and Resonance
The resonance article below connects string length and natural frequency with resonance in musical instruments.
Math and Statistics · Random Walks
Even with random steps,
the spread follows a pattern.
Change the number of steps and the example path.
Distinguish one path's final position from the probabilities across all possible paths.
Each independent step moves 1 space left or right, each with probability 1/2. Changing the example path does not change the exact probability distribution or theoretical values.
Change the values to discover today's mission.
After exploring random walks yourself
Even though each direction is random, the mean squared distance increases with the number of steps. One path's final distance differs from the average across all paths.
In an independent, symmetric 1-dimensional walk, taking k rightward steps out of n gives a final position X = 2k − n. Its probability is calculated exactly as the number of combinations C(n,k)/2ⁿ. The overall mean position is 0, the mean squared distance is n, and the RMS distance is √n. The mean absolute distance E|X| sums each position's absolute value multiplied by its probability, so it differs from RMS. Paths A, B, and C are individual examples generated from public fixed random seeds. Pressing a button neither draws a new random sample nor pools these paths to estimate theoretical values. The exact probability distribution is calculated separately from the example paths. The path graph shows step count on the horizontal axis and position on the vertical axis; the probability bars show final position on the horizontal axis and exact probability on the vertical axis. One path's actual final distance need not equal √n. This does not directly reproduce continuous Brownian motion or predict stocks or prices.
Read more about the principle · MIT: Random Walks and Diffusion
The Brownian motion article below explains the connection between irregular motion and random walks.
Life and Environment · Substance Accumulation
The amount taken in through food
and the amount leaving the body
Adjust the food concentration, elimination rate, stage, and time.
Compare the current concentration with the equilibrium concentration reached after sufficient time.
The food reservoir's concentration stays constant, and all consumers start at concentration 0. These are hypothetical concentrations and model time, not toxicity or food safety standards for actual substances.
Change the values to discover today's mission.
After exploring substance accumulation yourself
Whether concentrations rise at higher stages depends on the relationship between uptake and elimination rates. Over a short time, concentrations may not yet have reached equilibrium.
Let the constant food concentration be C₀, with concentrations at all four consumer stages starting at 0. The uptake coefficient at each stage is a = 0.5/model time, and the elimination rate k is the same at every stage. Uptake and elimination are calculated together as dCⱼ/dt = aCⱼ₋₁ − kCⱼ. Uptake at the first stage is aC₀. The solution is Cⱼ(t) = C₀(a/k)ʲ[1 − exp(−kt)Σᵢ₌₀ʲ⁻¹(kt)ⁱ/i!], and equilibrium is Cⱼ* = C₀(a/k)ʲ. Current values are distinguished from equilibrium values; at time 0, consumer concentrations are exactly 0. When a/k > 1, equilibrium concentrations increase at each stage. When a/k ≤ 1, however, concentrations do not automatically increase simply by moving up the food chain. The diagram compares substance concentrations and uptake and elimination rates, not energy transfer. This simple linear model excludes changes in food in real environments, growth dilution, substance-specific metabolism, direct uptake from water, and other factors. It is not a basis for assessing actual pollutant concentrations, exposure risks, toxicity, or food consumption.
Read more about the principle · US EPA: Uptake and Elimination Models for Substance Accumulation in Food Chains
The environmental pollution article below provides basic background on persistent substances and their connections to ecosystems. It does not provide the hypothetical coefficients used here or toxicity standards.
Economics and Society · Tax Incidence
The person who remits a tax
and the person who actually bears it
Adjust the price responsiveness of demand and supply and the tax per transaction.
See whether the outcome changes depending on who remits the same tax.
A steeper slope means quantity changes less for the same price change. Even at a tax of 0, the displayed shares represent the burden shares for a positive tax; the actual burden is 0.
Change the values to discover today's mission.
After exploring tax burdens yourself
Changing who legally remits the tax leaves prices and quantities unchanged in this competitive model. The side whose quantity responds less bears more of the burden through price changes.
Hypothetical demand gives the buyer price Pb = 60 − bD·Q, and supply gives seller receipts Ps = 20 + bS·Q. Before tax, quantity is Q₀ = 40/(bD + bS) and price is P₀ = 20 + bS·Q₀. With a per-unit tax t, Pb − Ps = t, giving Q = (40 − t)/(bD + bS). The buyer's per-unit burden is Pb − P₀ = bD·t/(bD + bS), and the seller's is P₀ − Ps = bS·t/(bD + bS); together they sum to t. For a positive tax, the buyer's share is bD/(bD + bS), and the seller's is bS/(bD + bS). If the tax is 0, actual burdens and tax revenue are also 0. Changing the legal taxpayer changes only who remits the tax, not this model's economic outcome. The graph uses actual coordinates to show the pretax point and the difference between buyer and seller prices at the same post-tax quantity. It assumes perfect competition, no tax avoidance, and no price controls, with the consumer and producer curves themselves held fixed. Unlike the existing progressive income tax activity, this addresses the market price burden of a transaction tax. It does not provide guidance on actual tax law, policy effects, or personal tax decisions.
Read more about the principle · OpenStax: Elasticity and Tax Incidence
The supply and demand article below provides basic background on market prices and price responsiveness. It does not explain this hypothetical transaction tax model or actual tax law.
Earth and Space · Radiometric Dating
Using the elements remaining and formed,
trace back the elapsed time.
Adjust the half-life, the number of elapsed half-lives, and the daughter element present initially.
Compare ages calculated with and without correcting for the initial daughter element.
The parent element starts with an expected amount of 64. Numerical results use expected amounts and are distinct from the single example in the individual-atom diagram below. These are not ages for actual isotopes.
Change the values to discover today's mission.
After exploring radiometric ages yourself
To determine elapsed time, you must distinguish daughter material produced by decay from daughter material present initially. Treating all initial material as decay products yields an older calculated age.
In a closed system, the initial expected parent amount is P₀ = 64, and the initial daughter amount is D₀ = 64 × the selected initial ratio. If s is the number of elapsed half-lives and h is the half-life, then P = 64 × 2⁻ˢ, the daughter amount produced by decay is 64 − P, and the total daughter amount is D = D₀ + 64 − P. Thus, the sum of parent and total daughter amounts is conserved at 64 + D₀. The age corrected for the initial daughter amount is h·log₂[1 + (D − D₀)/P], while the age ignoring it is h·log₂(1 + D/P). These calculations use expected amounts. The individual-atom diagram is one example with exponentially distributed lifetimes generated from a public fixed seed, not a rule under which exactly half decay every half-life. Each surviving atom has a constant probability of decay per unit time, and the initial daughter element is displayed separately from individual parent decays. Ideal conditions are assumed: no external influx or loss of parent or daughter elements, and a known initial daughter amount. For real rocks, one must examine whether the system has remained closed since mineral formation, along with initial composition and the isotope system. This hypothetical element is not used to assess actual rock ages or radiation risks.
Read more about the principle · OpenStax: Radioactive Decay and Dating
The nuclear fission and fusion article below provides basic background on nuclear decay and isotopes. It does not validate this hypothetical age calculation or actual rock ages.
Everyday Life and Culture · Feedback Control
What if your current control adjustment
follows a delayed temperature reading?
Adjust the control gain, measurement delay, target, and time.
Compare the temperature history so far with the heating or cooling command for the next instant.
The starting difference from ambient temperature is 0, and negative commands mean active cooling. This does not reproduce a typical heater's on/off switch or recommended settings for actual equipment.
Change the values to discover today's mission.
After exploring feedback control yourself
A strong response to delayed readings can keep heating even after the target is exceeded. With proportional control alone, heat lost to the surroundings leaves a gap from the target.
Calculations use the normalized difference x with ambient temperature set to 0 and x₀ = 0. The thermal model is x(k + 1) = 0.8x(k) + 0.2u(k), and control is u(k) = gain × [target − x(max(0, k − delay))]. Measurements before the start are set to 0. Positive commands mean heating, and negative commands mean active cooling. The command at the final step is intended for the next, not-yet-calculated step and is distinguished from the temperature history so far. At equilibrium, x* = gain × target/(1 + gain), revealing the proportional-control offset that leaves a steady-state gap from the target. This equilibrium is a fixed point of the equation; not every gain–delay combination is guaranteed to converge to it. The diagram shows only the history up to the selected current step, with the target and equilibrium marked separately on a common temperature-difference scale of 0–5. Even if high gain and delay produce overshoot or oscillations, a short observation window cannot prove long-term stability. This is a hypothetical active heating and cooling device; it does not predict actual heater switching, power use, or safe operation, or recommend settings.
Read more about the principle · MIT: Fundamentals of Feedback and Control · MIT: Discrete-Time Feedback and Oscillations
The smart heater article below provides basic background on the connection between temperature sensors and actuators. It does not recommend this activity's hypothetical control gains or delay values.
Chemistry · Heat Capacity and Thermal Equilibrium
Two objects at different temperatures:
At what temperature will they meet?
Adjust both objects' masses and initial temperatures, and A's hypothetical specific heat capacity.
Explore the final temperature together with the heat exchanged.
B's specific heat capacity is fixed at 1 kJ/(kg·K). If their heat capacities—the product of specific heat capacity and mass—are equal, the final temperature is midway between their initial temperatures, even if their masses differ.
Change the values to discover today's mission.
After changing heat capacity yourself
The temperature at which the objects equalize is not determined by mass alone. Also consider heat capacity: an object with greater heat capacity exchanges more heat for the same temperature change.
Heat capacity C = mass × specific heat capacity. If the two objects have heat capacities C_A and C_B, the final temperature is T_f = (C_A·T_A + C_B·T_B)/(C_A + C_B). The heat gained by each object is Q_A = C_A(T_f − T_A) and Q_B = C_B(T_f − T_B), positive for heat gained and negative for heat lost. With no heat exchange with the surroundings, Q_A + Q_B = 0, and the final temperature lies between the two initial temperatures. Heat flows from the hotter object to the cooler one; if their initial temperatures are equal, the heat transferred is 0. A Celsius temperature difference of 1°C equals 1K. This assumes two hypothetical objects with constant specific heat capacities and perfect thermal contact, excluding phase changes, chemical reactions, and heat loss to the surroundings. It is not a recipe for calculating the results of experiments that mix real substances or of cooking.
Read more about the principle · OpenStax: Specific Heat and Calorimetry
The article below introduces energy conservation and energy balances in chemical processes. The two objects and hypothetical specific heat capacities in this activity do not represent mixtures of materials in real processes.
Physics · Diffraction Through a Circular Aperture
With a smaller aperture,
light spreads out more.
Change the circular aperture diameter, light wavelength, and distance to the screen.
Compare the first dark ring and the relative intensity away from the center.
The central intensity is normalized to 1 for each setting. Moving the screen farther away increases the ring's physical radius, but its angular width stays the same for the same aperture and wavelength.
Change the values to discover today's mission.
After changing the aperture and wavelength
A smaller aperture spreads light out more through diffraction rather than concentrating it into a point. A longer wavelength also produces a wider pattern through the same aperture.
This is an ideal Airy pattern formed by monochromatic light uniformly illuminating a circular aperture. Express diameter D, wavelength λ, screen distance z, and distance from the center r in m. With u = πDr/(λz), relative intensity is I = [2J₁(u)/u]². J₁ is the Bessel function of order 1, and at the center u = 0, I = 1. The first dark ring occurs at the first positive zero of J₁, u ≈ 3.831706. Its angle is approximately 1.21967λ/D, its screen radius is this value multiplied by z, and the area inside it is πr². Each curve is normalized to a central intensity of 1, so it does not compare absolute intensity or total incoming light. This uses the far-field, small-angle diffraction approximation for coherent monochromatic light and does not predict overall camera sharpness or other optical losses.
Read more about the principle · OpenStax: Circular Apertures and Diffraction Width · MIT: Relative Intensity of the Airy Pattern (PDF)
The article below offers a comparison with diffraction, where ocean waves spread around structures. This activity covers monochromatic light and a circular aperture; it is not a model for calculating ocean wave heights.
Economics and Society · Stable Matching
Match mutual preferences
by holding proposals tentatively.
Change both sides' preference rankings and which side proposes first.
Follow proposals step by step to see why matches stay or change.
Hypothetical A·B·C form one side, and X·Y·Z the other. Everyone ranks without ties, and blocking pairs are counted relative to the current matching, including unmatched participants.
Change the values to discover today's mission.
After following the proposals in order
You can hold a proposal tentatively and replace it when a preferred one arrives. At the end, no other pair mutually prefers each other to their current partners.
This is the Gale–Shapley deferred acceptance procedure. Select the first unmatched proposer in alphabetical order. They propose to the highest-ranked person they have not yet proposed to. Recipients hold their most preferred proposal and release the previous proposer if a better proposal arrives. Rejected or released proposers propose to their next choice, repeating until everyone is matched. Two participants who prefer each other to their current partners count as a blocking pair. Being unmatched ranks below every possible partner, so counts during the procedure indicate possible improvements to the incomplete matching. A completed matching is stable if it has no such pairs. Stability does not mean the greatest total satisfaction or the fairest outcome. With equal numbers on both sides and complete, strict rankings, as here, this procedure produces the stable matching most favorable to the proposing side. Changing the preference profile or proposing side shows a history recalculated from the start under those conditions. It does not evaluate real people's worth or compatibility.
Read more about the principle · MIT: Stable Matching and Deferred Acceptance (PDF)
The article below introduces game theory, where others' choices affect your outcomes. This activity separately covers the proposal procedure and ranking conditions for stable matching.
Earth and Space · Rotating Coordinates
The same straight-line motion
looks curved on a rotating screen
Change the rotation rate and the object's initial speed, direction, and elapsed time.
Compare the same motion in nonrotating and rotating coordinates.
The starting point is (1, 0), and an initial direction of 0° points right. Changing only the rotation rate shows the same actual straight-line path from a different perspective. Changing the object's initial speed or direction also changes its actual path.
Change the values to discover today's mission.
After rotating the viewing coordinates
Even when an object moves straight in inertial coordinates, it appears to curve in rotating coordinates. Explaining its acceleration in those coordinates requires both the Coriolis and centrifugal terms.
With no external force, an object moves at constant velocity in nonrotating inertial coordinates: r(t) = (1, 0) + vt. The rotation rate ω is constant, and rotating coordinates are calculated as r′ = R(−ωt)r. R is a planar rotation transformation. Velocity in rotating coordinates is v′ = R(−ωt)v − ω × r′, and acceleration is the sum of the Coriolis term −2ω × v′ and the centrifugal term −ω × (ω × r′). In planar components, these are (2ωv′y, −2ωv′x) and (ω²x′, ω²y′), respectively. Acceleration in inertial coordinates is 0. The two terms arise from expressing motion in rotating coordinates, rather than from newly applied physical forces. Both paths use the same length scale and are shown up to the selected time. This is a hypothetical coordinate model of a uniformly rotating flat platform; it does not predict actual winds, typhoons, or atmospheric motion in a particular hemisphere on Earth.
Read more about the principle · MIT: Rotating Reference Frames
The article below introduces inertia: maintaining constant straight-line velocity when no external force acts. This activity separately covers the path seen in rotating coordinates and the two inertial terms.
IT · Shortest Paths
Even with a cheap first step,
the total cost can differ.
Change the arrow costs and travel from S to T.
Compare choosing the cheapest next step with taking the path of minimum total cost.
The remaining costs are A → C = 8, B → C = 1, and B → T = 9. Drawn line lengths are unrelated to costs, and movement follows arrow directions only.
Change the values to discover today's mission.
After changing the path costs
A cheap first segment does not guarantee the cheapest complete path. Add the costs of the later segments to find the minimum cost from S to T.
A path's total cost is the sum of the costs of its arrows. This small directed network has four paths: S → A → T, S → A → C → T, S → B → T, and S → B → C → T. The local choice selects the cheapest next arrow from the current position, breaking ties alphabetically by destination vertex name. The minimum total cost is found by comparing all four complete paths. If paths tie for the minimum, their count is shown and one is drawn as a representative in alphabetical order. This local choice rule differs from Dijkstra's algorithm. This educational model uses positive hypothetical costs and does not predict actual communication network conditions or travel times.
Read more about the principle · MIT: Shortest Paths and Summed Costs (PDF)
The article below introduces nodes, links, routers, and alternative routes. This activity separately covers calculating the minimum total cost across the four paths.
Math and Statistics · Repetition and Measurement
Trace the same line more closely,
and longer bends emerge.
Repeatedly add small triangular bends to line segments.
Compare the length measured by skipping vertices with the length following every bend.
Instead of moving a ruler of fixed length, this connects vertices at a fixed vertex-count interval. Lower stages allow fewer vertex intervals.
Change the values to discover today's mission.
After measuring along the small bends
Measurement lines that skip widely miss small bends. Following the same curve's vertices more closely brings the measured length closer to the total length of its segments.
The Koch curve replaces each segment with four segments, each 1/3 of its original length. At iteration stage n, the segment count is 4ⁿ, each segment has length L/3ⁿ for initial length L, and the total length is L(4/3)ⁿ. Measurement lines connect vertices retained every 2ʳ vertices and sum their lengths, including the final endpoint. This rule differs from measuring a real coastline by moving a fixed-length ruler. The display shows finite polygonal lines at stages 0–4, whose own dimension is 1. log 4 / log 3 is the dimension of the ideal, infinitely repeated self-similar structure, not a dimension measured from the current display. Both lines use the same cm scale, and this does not predict the length or dimension of a real coastline.
Read more about the principle · Brown: Length and Growth of Repeated Structures · Wisconsin: Dimension of Self-Similar Structures
Life and Environment · Rhythm Synchronization
Can a clock with a different period
synchronize to a signal?
Change the internal clock's period and its coupling to an external signal.
Observe the phase difference between the two rhythms at the same time each day.
The external period is fixed at 24 hours. Phase hours express the angular difference between the rhythms as time; they are not actual bedtimes.
Change the values to discover today's mission.
After observing the difference between the clocks
Different free-running periods cause the phase difference to accumulate. A sufficiently strong signal creates a stable phase in the model, but that condition alone does not establish that the rhythms are already synchronized.
Let φ be the hypothetical oscillator's internal phase minus the external phase. The rate of phase change is dφ/dt = Δ − K sin φ, with the free-running period difference expressed as Δ = 2π(1/τ − 1/24). τ is the internal free-running period, and K is the model's coupling coefficient. A stable phase exists when K > |Δ|; the equality boundary is excluded from the strict stability condition. If K = 0 and τ = 24, every phase remains neutrally unchanged, unlike entrainment in which a signal restores a phase difference. Even with a stable point, transient responses take time, and a special trajectory starting exactly at the opposite unstable point can remain there. The current phase is wrapped into the interval from −12 inclusive to +12 exclusive, while cumulative change and the graph include full cycles. This simple phase model shows the state at the same external time each day and the past trajectory up to the selected day. K is neither light brightness nor food intake, and the model does not prescribe human sleep or meal times or assess health.
Read more about the principle · Original Study: Phase Entrainment of Biological Rhythms
The article below introduces the relationship between the body's central and peripheral clocks and environmental signals. Do not interpret the numbers in this hypothetical model as a meal or sleep schedule.
Everyday Life and Culture · Selective Attention
Set aside the word's meaning
and choose its ink color.
Compare conditions where the word and color match or differ.
Answer 9 questions to experience the conflict between reading and choosing colors.
Choose the word's ink color, rather than the color it names. Respond using the buttons or keyboard keys 1·2·3.
Choose a condition and start.
Completed results for each condition are compared only while you view this page. Changing conditions cancels the current question; changing the sequence or resetting also clears comparison records.
Congruent condition · no completed results yet.
Incongruent condition · no completed results yet.
Respond to discover today's mission.
After ignoring the words and choosing colors
Familiar reading can interfere with choosing an ink color. Compare correct response times and errors side by side for congruent and incongruent conditions.
Stroop's original study involved ignoring words and naming their ink colors. This screen is a short educational activity with button responses, unlike the original study's spoken-response test. Each sequence displays red, blue, and green ink three times each. After you press start and the stimulus appears, the device's clock measures time until your first response. After responding, you manually select the next question. Only correct response times are sorted to calculate the median; no time is shown if there are no correct answers. Errors are counted separately rather than converted to a time of 0 or penalty points. Leaving the tab or hiding the window cancels the current activity. Device and input delays, color vision, practice, and condition order affect results, so the activity does not force incongruent responses to be slower. Do not interpret results as personal intelligence or attention scores or clinical assessments. Individual response times and comparison records are neither transmitted nor saved to your device; only mission completion is added to the existing activity record.
Read more about the principle · Stroop: Original Study of Interference Between Reading and Color Responses
The article below introduces attention: selecting relevant information and excluding irrelevant information. Do not use this activity's response times as evidence of training effects or health status.
Chemistry · Charge and Metal Deposition
Even with the same charge,
the metal produced differs.
Change the current, supply duration, and metal ion.
Explore how the amount of electrons relates to the moles and mass of metal.
This ideal model assumes all supplied charge reduces the selected metal ions. Aluminum assumes suitable electrolysis conditions and is not an example of aqueous electroplating.
Change the values to discover today's mission.
After relating electron amount to metal amount
One copper ion accepts two electrons to become metal, while one silver ion accepts one. Even with the same charge, the mass produced depends on the required electron count and molar mass.
A constant current I supplied for t seconds delivers total charge Q = It. Dividing by the Faraday constant F ≈ 96485.33212331001 C/mol, the charge of 1mol of electrons, gives Q/F moles of electrons. If each metal ion requires n electrons, the amount of metal is Q/(nF) moles, and its mass is MQ/(nF), multiplying by molar mass M. Representative molar masses for copper, silver, and aluminum are 63.546, 107.8682, and 26.9815384 g/mol, requiring 2·1·3 electrons, respectively. F is the product of the elementary charge and the Avogadro constant. The three straight lines show ideal mass production for the same charge, and the selected point shows the current calculated result. If supplied charge is 0, the electron and metal amounts are both 0. Calculations assume sufficient metal ions and 100% current efficiency. In real electrolysis, other reactions, potentials, electrolytes, and surface conditions also affect results. This does not imply that aluminum can be electroplated from an aqueous solution.
Read more about the principle · OpenStax: Electrolysis and Amount of Substance Produced · NIST: SI Defining Constants · CIAAW: Standard Atomic Weights
The electrochemistry article below covers electrolytic cells, Faraday's quantitative relationships, and electroplating. Read it while distinguishing this activity's ideal yield from the results of real electroplating processes.
Physics · Static and Kinetic Friction
Friction's limit and actual value:
when do they match or differ?
Change the slope, friction coefficients, mass, and state of motion.
Compare an object at rest with one already sliding downhill under the same conditions.
Static friction acts only as strongly as needed and cannot exceed its limit. For an object already sliding, kinetic friction is used, and the current acceleration direction is shown.
Change the values to discover today's mission.
After examining actual friction
Friction on an object at rest is not always at its maximum. It is only strong enough to resist the downhill force; once the object is sliding, a different coefficient is used.
The tangent of incline angle θ is height/horizontal length. The downhill component of gravity is mg sin θ, and the normal force N is mg cos θ, using g = 9.80665 m/s². Maximum static friction is μsN. If the object is initially at rest and tan θ ≤ μs, actual friction equals mg sin θ, making the net force 0. Even exactly at the limit, this ideal model allows the object to remain at rest. On a steeper incline it starts sliding, and kinetic friction μkN is also used when it is already sliding downhill. μk equals μs multiplied by the selected coefficient ratio. Downhill acceleration is then g(sin θ − μk cos θ). A negative value means the downhill-moving object is slowing at that instant; it does not mean it has already stopped or is moving uphill. The diagram shows gravity, the normal force, and actual friction on the same scale. The downhill gravity component resolves gravity and does not add a separate fourth force. This is an instantaneous comparison using a simple contact-friction model; it does not calculate velocity or trajectory.
Read more about the principle · OpenStax: Static and Kinetic Friction
The classical mechanics article below introduces the relationship between net force, mass, and acceleration. For incline friction coefficients and the maximum static friction formula, continue with the formula reference above.
Economics and Society · Monopoly and Quantity Traded
With the same demand and costs,
selling rules change quantity traded.
Change the demand intercept price, slope, and marginal cost.
Compare quantity, price, and welfare under the competitive benchmark and a single-price monopoly.
This hypothetical market omits fixed costs, external costs, and price discrimination. When quantity traded is 0, there are no actual sales, so the transaction price is omitted and all surpluses and deadweight loss are shown as 0.
Change the values to discover today's mission.
After changing quantity sold and price together
With a single selling price, increasing sales also lowers the price of units already being sold. Marginal revenue is therefore below price, and the profit-maximizing quantity is smaller than under the competitive benchmark.
Hypothetical inverse demand is P = A − Bq, with constant marginal cost c. Total revenue is Pq = Aq − Bq², and marginal revenue, the extra revenue from selling one more unit, is MR = A − 2Bq. Under the competitive benchmark, P = c, giving qC = max(0, (A − c)/B). Under a single-price monopoly, MR = c, giving qM = qC/2. The monopoly maximizes profit, revenue minus cost cq, rather than revenue itself. Consumer surplus is the triangle between willingness to pay and price; producer surplus is the rectangle between price and cost. With no fixed costs, producer surplus equals profit in this model. Their sum is total surplus, and the reduction from the competitive benchmark's total surplus is deadweight loss. Deadweight loss is the net benefit of trades that disappear and is a different area from consumer surplus. If cost c is at least the intercept price A, neither rule produces any trade, so the actual selling price is omitted. The graph draws only the marginal revenue segment within the positive price scale and shows surplus and loss areas in actual quantity–price coordinates.
Read more about the principle · OpenStax: Monopoly Profit Maximization and Welfare
The microeconomics article below introduces how demand responds to price and how pricing power differs between competition and monopoly. For exact calculations of marginal revenue and deadweight loss, continue with the formula reference above.
Earth and Space · Differential Gravity
Even when all are pulled the same way,
their differences from the center vary.
Change the external body's gravitational strength, distance, and surface position.
Read gravity itself alongside differential gravity with the Earth's central value subtracted.
View the difference in external gravity between the surface and the center in a frame moving with Earth's center. The external body is a point mass; Earth's own gravity and ocean levels are not shown.
Change the values to discover today's mission.
Comparing with Earth's center
The near side is pulled more strongly than the center, and the far side less strongly. The difference relative to the center can point outward on both sides, but actual gravity has not reversed direction.
The model uses radius R = 1, external point-mass gravitational parameter μ, and center-to-center distance D. With the center at (0, 0), the external body at (D, 0), and the surface at p = (cos α, sin α), external gravitational acceleration is g(p) = μ(D − x, −y)/[(D − x)² + y²]³ᐟ². The center is pulled with g(0) = (μ/D², 0), and differential gravity between the surface and center is Δg = g(p) − g(0). The projection of Δg onto the outward surface direction indicates outward when positive and inward when negative. Even on the far side, external gravity itself points toward the body; the weaker pull relative to the center appears as an outward difference. The enlarged surface view compares twelve positions using the same arrow scale. When the body is sufficiently distant, the characteristic magnitude of differential gravity is approximately inversely proportional to the cube of distance, distinct from the inverse-square dependence of external gravity itself. Actual tides also depend on coastlines, the seafloor, rotation, other celestial bodies, and the ocean's response. This diagram does not predict ocean levels, high or low tide times, or local tide schedules.
Read more about the principle · NOAA: Differential Gravity That Produces Tides · OpenStax: Tidal Forces and Differences Relative to the Center
IT · Indivisible Choices
Even when you choose by value per unit weight,
the best combination may differ.
Change the bag's capacity, item list, and selection method.
Compare packing by value per unit weight with all possible combinations.
There is one of each item, and items cannot be split. Compare total values among combinations that stay within the bag's capacity.
Change the values to discover today's mission.
Comparing good items with good combinations
One item with a high value-to-weight ratio can leave awkward unused space. For indivisible items, the combination matters as well as each item's ranking.
The 0/1 knapsack problem involves choosing whether to pack each item whole or leave it out. The ratio method considers items in descending value/weight order and packs each if it fits the remaining capacity. Ties are considered alphabetically, starting with A. The overall optimization method compares all 64 combinations, including the empty bag, and chooses the highest value within capacity. If several combinations share the maximum value, one is displayed. The ratio method is not always wrong: with capacity 6 in list B, both methods give a value of 16. This model is distinct from the fractional knapsack problem, where items can be split.
Read more about the principle · MIT: The 0/1 Knapsack Problem and Combinations of Choices
Math and Statistics · Remainders and Cycles
For two repeating schedules,
starting offsets must align for them to coincide.
Change the pair of periods and their respective remainders.
Find when both conditions for the same time hold together.
Find times t whose remainders upon division by each period match the specified values. Even though a least common multiple exists, the schedules never coincide if their starting offsets are incompatible.
Change the values to discover today's mission.
Matching the remainders as well as the periods
For two coprime periods, a shared time exists for any choice of remainders. If the periods share a divisor, the difference between the remainders must be divisible by that divisor.
The conditions are t ≡ rA (mod m) and t ≡ rB (mod n). A solution exists if the greatest common divisor g = gcd(m, n) divides the remainder difference rB − rA, and coincidences recur every least common multiple mn/g after the first solution. Otherwise, no solution exists at any time. The displayed least common multiple describes the periods themselves; it does not imply an actual coincidence interval when no solution exists. The graph includes integer times 0–70 with both endpoints, and the first solution is sought from 0 inclusive to the least common multiple exclusive. This integer-period model omits real calendar holidays and irregular schedules.
Read more about the principle · MIT: Chinese Remainder Theorem
Life and Environment · Spread in a Closed Population
Even with the same transmission rate,
a shrinking susceptible group changes the outcome.
Change the transmission and removal rates and the initial amount transmitting.
Explore how the three groups change as the susceptible group shrinks.
A hypothetical population of 100 units with no outside inflow. Decimal values represent continuous model quantities, not predictions of actual headcounts, dates, or infection risk.
Change the values to discover today's mission.
Watching the pool of susceptible recipients shrink
The transmission coefficient alone does not determine growth or decline. As the susceptible amount S decreases, the current effective reproduction ratio also falls, potentially creating a period when the transmitting amount decreases.
The closed SIR model links three groups through dS/dt = −βSI/100, dI/dt = βSI/100 − γI, and dR/dt = γI. Their sum is always 100, with no new arrivals, births, or retransmission. R is the group removed from transmission, not necessarily restricted to those who have recovered. β/γ is the baseline ratio, and the current effective ratio is (β/γ)S/100. When I is positive, it increases at that instant if this ratio exceeds 1 and decreases if it is smaller. If it is exactly 1, the instantaneous rate of change is merely 0; this does not mean it remains constant. If initially I = 0, no transmission occurs even with a large ratio. Model time and coefficients are not clinical data for a real disease and do not estimate individual infection risk or outbreak size.
Read more about the principle · Duke: SIR Assumptions and Differential Equations · Duke: Growth Conditions and Model Analysis
Everyday Life and Culture · The Eye's Blind Spot
Between the screen you view and the image in your eye,
there is a small gap.
Observe while changing the eye used and the target spacing.
Explore distance and left–right image relationships in a separate projection model.
Gently cover the other eye without pressing on it, keep looking at the cross, and slowly change your actual viewing distance. Stop if uncomfortable, and use both eyes if you prefer. Whether the target disappears does not affect your mission score.
Change the eye direction and model values to discover today's mission.
Distinguishing the eye's blind spot from projection
The optic disc, where the optic nerve leaves the retina, has no light-sensitive cells, creating a blind spot. Target visibility varies with the person, screen, and viewing distance.
The right visual field projects onto the left retina, and the left visual field onto the right retina. The ideal pinhole projection model below sets the projection-plane distance to 1 arbitrary unit and shows the direction angle arctan(a/d) and image position −a/d for lateral spacing a and forward distance d. It does not reproduce the eye's focusing, curved retina, or individual blind spot position and size. Its angles and lengths are not measurements of the actual screen's visual angle or retinal position, and it is not a test of vision or eye health. You can complete the mission through the controls regardless of your observations.
Read more about the principle · OpenStax: Retina and Optic Disc · Exploratorium: How to Observe Your Blind Spot
Chemistry · Electron Energy Levels and Light
What matters is less the starting level
than the energy difference in the drop.
Change the spacing and reference level of two energy levels in a hypothetical atom.
Compare the wavelength of one emitted photon with the total energy of multiple photons.
This hypothetical model allows one downward transition between two levels. Repeating the same transition produces photons of the same energy.
Change the values to discover today's mission.
Connecting energy levels to a single photon
Raising both levels equally leaves the energy drop unchanged. More photons increase the total energy, but the wavelength of each photon stays the same.
For an allowed transition, one photon's energy is ΔE = Eupper − Elower = hf = hc/λ. For N photons of the same light, the total energy is NΔE. We use Planck's constant h = 6.62607015 × 10⁻³⁴ J·s, the speed of light c = 299792458 m/s, and 1eV = 1.602176634 × 10⁻¹⁹ J. A larger energy gap means higher frequency and shorter wavelength. The diagram shows two levels and one type of allowed transition. The visible range is marked approximately 380–780nm, with boundaries that may vary by person. Real atoms have multiple levels and selection rules, so not all transitions are allowed. This model does not determine actual flame colors, element identities, relative emission-line intensities, or perceived colors.
Read more about the principle · OpenStax: Atomic Spectra and Energy-Level Transitions · NIST: SI Physical Constants
The flame-test article below offers further reading on excited-electron energy transitions, emission lines, and the background to elemental analysis.
Physics · Wavelength and Scattering
Even with the same tiny particles,
scattering varies with light's wavelength.
Change the wavelengths of two lights and the observation direction.
Separate the effects of wavelength and direction using scattering-intensity curves on the same scale.
This is elastic single scattering of unpolarized light by identical particles much smaller than its wavelength. Incident intensity, particle count, observation distance, and polarizability are held equal.
Change the values to discover today's mission.
Changing wavelength and direction separately
Shorter wavelengths scatter more strongly. The intensity ratio of two lights compared in the same direction is determined by their wavelengths, while each intensity varies with observation direction.
Under these conditions, scattering intensity is proportional to (1 + cos²θ)/λ⁴. Setting forward scattering of 550nm light to 1 gives relative intensity (550/λ)⁴ × (1 + cos²θ)/2. θ is the angle between the incident light's direction and the observation direction. Intensities at forward 0° and backward 180° are equal; at sideways 90°, each is halved. Dividing A by B in the same direction cancels the angular factor, giving the intensity ratio (λB/λA)⁴. The diagram's radius represents relative intensity, not photon travel distance or probability. Values may exceed 1 when stronger than the reference. Actual sky color also depends on the solar spectrum, atmospheric path, absorption, multiple scattering, and eye sensitivity, so this curve alone does not predict actual colors.
Read more about the principle · University of Texas: Light Scattering and Angular Distribution · NOAA/NWS: Why the Sky Is Blue
The Rayleigh scattering article below offers further reading on short-wavelength scattering and the background to blue skies and red sunsets.
Economics and Society · Orders and Inventory
Ordering less often
means more inventory waiting in storage.
Change annual demand, cost per order, and holding cost.
See how the sum of the two costs changes with the quantity ordered each time.
This repeating model assumes constant demand and immediate replenishment when inventory reaches 0, with no stockouts or discounts. The purchase price per item is constant and excluded from cost comparisons across order quantities.
Change the values to discover today's mission.
Breaking down recurring order costs
Larger orders reduce the number of orders and save ordering costs. They also increase average inventory and holding costs. The total is lowest where the two costs are equal.
For annual demand D, fixed cost per order K, annual holding cost per item H, and order quantity Q, the average annual order frequency is D/Q and average inventory is Q/2. Thus, annual relevant cost is DK/Q + HQ/2, the continuous model's optimal quantity is Q* = √(2DK/H), and minimum cost is √(2DKH). Annual ordering and holding costs are equal at the optimum. Ordering half or twice the optimal quantity makes their sum 1.25 times the minimum. The slider moves in steps of 20 items, but the continuous optimum may lie between them. A decimal D/Q represents average annual frequency over repeated long-term operation, not an integer order schedule for a particular year. The cost graph compares all three curves on the shared annual-cost scale displayed each time. This educational model omits demand fluctuations, lead times, space constraints, and discounts; it does not recommend order quantities for an actual business.
Read more about the principle · MIT: Inventory Management and Economic Order Quantity
The economic order quantity article below directly explains the relationship between ordering and holding costs and the same optimal-quantity formula.
Earth and Space · Water Wave Dispersion
Wave crests and wave groups—
do they travel at the same speed?
Change wavelength and water depth, then view the distances traveled after the same time.
Compare the phase speed of a traveling crest with the group speed of a narrowband wave group.
These are small-amplitude linear surface gravity waves in water of uniform depth. Currents, surface tension, and viscosity are omitted. Group speed is not the overall travel speed of water particles.
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Comparing the travel of crests and wave groups
When water is shallow relative to wavelength, the two speeds are similar. In relatively deep water, the group speed of a narrowband wave group approaches half the crest's phase speed.
For wavenumber k = 2π/λ, water depth h, and gravitational acceleration g = 9.80665 m/s², the dispersion relation is ω² = gk tanh(kh). Phase speed is cp = ω/k, and group speed is cg = dω/dk = cp[1 + 2kh/sinh(2kh)]/2. When kh is very small, both speeds approach √(gh); when kh is very large, group speed approaches half the phase speed. Here, shallow and deep describe the relationship between wavelength and depth, rather than depth alone. Distances traveled after the same time t are cp·t and cg·t; at 0 seconds, both distances are 0. Changing time alone does not change the speeds themselves. Group speed describes narrowband wave groups and energy transfer. This small-amplitude, uniform-depth model does not predict actual tsunami heights, arrival times, flooding, or navigation hazards.
Read more about the principle · MIT: Surface Gravity Wave Dispersion and Group Speed · NOAA: Tsunami and Long-Wave Basics
The tsunami article below provides background on the √(gh) relationship when wavelength is much greater than depth, and on the effects of depth. See the MIT resource above for the general dispersion relation and derivation of group speed.
IT · Parallel Processing and Sequential Tasks
Even when work is shared,
some tasks must still be done in sequence.
Change the serial fraction and core count for the same workload.
Compare the time reduced by adding cores with the time that cannot be reduced.
The total workload is fixed. Only the parallel portion is divided evenly. This is an ideal upper bound that omits communication and distribution costs.
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The serial portion sets the speedup limit
Adding cores does not shorten serial work. As the parallel portion shrinks, the remaining serial portion has a greater effect.
Setting single-core execution time to 1, serial fraction to f, and core count to N gives execution time T = f + (1−f)/N and speedup 1/T. Efficiency per core is speedup/N. If f exceeds 0, speedup only approaches 1/f even as core count grows without bound. The ∞ symbol at f=0 means there is no finite ideal upper bound; the speedup with the current N cores is N. If N=1 or all work is serial, speedup is 1. Total active work is always 1, while N×T is reserved core time including idle cores, distinct from CPU work actually performed. Real computers incur distribution, communication, and memory costs, so this curve does not guarantee hardware performance.
Read more about the principle · LLNL: Parallel Computing and Amdahl's Law
Math and Statistics · Reveal Rules and Probability
You saw a goat—
also consider how the doors were opened.
You initially choose door 1. The host opens the others, leaving only one other door closed.
Compare the reveal rules of a host who knows the answer and a random host.
Cases where the random host reveals the prize remain gray and are excluded from win-probability calculations. No additional condition specifies which numbered doors were opened.
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Consider how the revealed information was produced
If an informed host always reveals only goats, every initially wrong choice becomes a win for the switching strategy. With an uninformed host, calculations must be restricted to cases where only goats were revealed.
A prize is hidden uniformly among N doors, and you initially choose door 1. The host leaves just one other door closed and opens N−2 doors. An informed host never reveals the prize and, if your initial choice is correct, chooses uniformly which other door to leave closed. The stay win probability is then 1/N and the switch win probability is (N−1)/N. An uninformed host randomly chooses which door to leave closed and may reveal the prize. The fraction of cases where only goats are revealed is 2/N; under that condition, staying and switching each have win probability 1/2. Each cell shows an equally likely combination of N prize locations and N−1 host draws. When the informed host is forced to leave a particular door closed, the draw is unused, so the same outcome appears in multiple cells. This model counts every case exactly; it is not a short winning streak from random trials.
Read more about the principle · Jeffrey S. Rosenthal: Monty Hall Versus Random Reveals
Everyday Life and Culture · Image Compositing and Transparency
When you remove a green background,
will similar-colored clothing disappear too?
Change the color-removal tolerance and edge softness.
Compare the original, opacity matte, and composite on a new background side by side.
Black in the matte means alpha 0; white means alpha 1. Because clothing and background are distinguished by color alone, clothing matching the key color is also removed.
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Selecting colors and compositing backgrounds
Widening the tolerance removes more green background but may also remove similar colors in the subject. The alpha matte determines how much of each area remains; the new background is composited afterward.
The key color is RGB(0,180,0). Color distance d is the largest R, G, or B difference between each pixel and the key color. If d is at most tolerance t, alpha α=0. Outside that range, if soft-edge width s is 0, α=1; if s is positive, α=min(1,(d−t)/s). Compositing onto an opaque new background uses α×original+(1−α)×new background for each channel. Alpha 0 shows only the new background, 1 only the original, and intermediate values give a weighted sum of the two numbers. This educational model directly mixes 8-bit RGB channel values and rounds at the end. It does not reproduce perceptual color differences, actual light quantities, or commercial keying algorithms. Complex real-world edges such as hair and color-spill correction are omitted.
Read more about the principle · Adobe: Purpose and Controls of Keying Effects · W3C: Alpha Compositing Calculations
Physics · Flow and Pressure
When flow speeds up in a narrow section,
how does static pressure change?
Change the outlet area and elevation, and the inlet flow speed.
Compare pressure, speed, and elevation on the same energy scale.
This is steady, incompressible flow along the same streamline, with no friction, pumps, or turbines. Ideal inlet conditions are set to area 4cm², absolute pressure 300kPa, and elevation 2m.
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Where does the energy to speed up the flow come from?
At the same elevation, a narrower outlet increases flow speed by continuity and lowers static pressure. If elevation also changes, the change in potential energy must be included.
We use an ideal fluid with density ρ=1000kg/m³ and g=9.80665m/s². From the continuity equation Q=A₁v₁=A₂v₂, v₂=v₁/(A₂/A₁). Bernoulli’s equation along the same streamline, P/(ρg)+v²/(2g)+h=constant, means that the sum of pressure head, velocity head, and elevation head is conserved. Thus, P₂=P₁+ρ(v₁²−v₂²)/2−ρg(h₂−h₁). Each circle’s radius is proportional to the square root of its cross-sectional area, and all head bars use the same 0–36m scale. Head does not refer only to the pipe’s physical elevation. Each change is calculated under ideal boundary conditions that maintain the specified inlet velocity and pressure; this does not mean a pump’s flow rate stays unchanged when a real pipe’s cross-sectional area changes. Friction losses, compression, and external work are omitted, so this is not a calculator for actual piping design or for predicting pressures in two unrelated flows.
Read more about the principle · OpenStax: Bernoulli’s equation and the elevation term · NASA: Flow velocity and static pressure · NASA: Cross-sectional area and conservation of mass
Chemistry · Distillation and Component Balances
Even when both evaporate,
their proportions in the vapor change.
Separate 100mol of a hypothetical mixture in a single equilibrium stage.
Change the feed composition, relative volatility, and fraction separated as vapor, then compare purity and recovery.
This is ideal binary equilibrium with constant relative volatility. The vapor fraction is a specified separation condition, not a control that sets actual heat input, pressure, or temperature.
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Purity and amount recovered are different quantities
The more volatile component is enriched in the vapor. But as more is separated as vapor, its proportion in the vapor and the amount recovered from the feed change differently.
Let z be the feed mole fraction of component A, x its liquid mole fraction, y its vapor mole fraction, and β the vapor fraction of the total. At constant relative volatility α, y=αx/[1+(α−1)x], and the component balance is z=(1−β)x+βy. The total 100mol separates into liquid L=100(1−β) and vapor V=100β, and the amount of A fed, 100z, equals Lx+Vy. Recovery is Vy/(100z), so it differs from vapor purity y. When α=1, x=y=z and no separation occurs. The curve uses a composition scale, while the bars below share an amount scale of 0–100mol; the two phases are not stretched to equal lengths. This ideal single-stage equilibrium model does not predict actual substances’ boiling points, heating methods, azeotropes, reflux, or distillation column performance.
Read more about the principle · IChemE: Vapor–liquid equilibrium and total and component balances · HKUST: Relative volatility and distillation basics
Read the distillation article below for more on composition differences when two components evaporate together and the limits of actual separation.
Life and Environment · Genetic Linkage and Recombination
When are combinations of two genes
determined independently?
Explore two hypothetical gene loci together.
Change the recombination fraction and parental combinations, then compare the four expected proportions in a testcross.
A testcross between AaBb and ab/ab is assumed. There are no mutations, selection, or sex differences, and the numbers are expected values rather than random experimental results.
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Look at the combinations inherited together, beyond individual genes
When recombination is low, parental combinations are transmitted more often. Changing the parental combinations changes which gametes are more common, but the total probability of parental types stays the same.
An AaBb individual with two loci produces AB, Ab, aB, and ab gametes. In the AB/ab coupling phase, AB and ab are parental types; in the Ab/aB repulsion phase, Ab and aB are parental types. At recombination fraction r, each parental type has probability (1−r)/2, each recombinant has probability r/2, and the four probabilities sum to 1. The testcross partner ab/ab always contributes an ab gamete, so the corresponding offspring are AaBb, Aabb, aaBb, and aabb, with each expected count equal to the number of offspring × probability. At r=50%, each combination occurs at 25%, giving the same distribution as independent assortment. This does not necessarily mean the loci are on different chromosomes or physically unlinked. Even on the same chromosome, distant loci can have an observed recombination fraction close to 50%. These proportions are not converted into actual DNA distances, familial genetic risks, or disease predictions.
Read more about the principle · OpenStax: Genetic linkage, recombination, and genetic maps
Economics and Society · Shared Resources and Rules of Use
How long will
shared stock last?
Four people take turns using a shared stock of 24 units.
Change A’s and B’s requests, the number of periods, and the shared cap to see actual amounts received and remaining stock.
Each period follows the order A→B→C→D. C and D each request 2 units. Stock is neither replenished nor returned, and each person receives only what remains.
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Distinguishing requests from amounts actually received
Even small individual withdrawals add up and reduce shared stock. Requests reduced by the cap and amounts not received because stock ran out have different causes.
The initial stock is 24 units, and four people make requests in the order A→B→C→D each period. With a shared cap, each request is first reduced to at most 2 units, then compared with the remaining stock to determine the amount actually received, which is deducted immediately. Cumulative amounts received + remaining stock = 24 always holds for each period. At period 0, no one has received anything yet, so 24 remains. All unmet requests equal the sum of amounts reduced by the shared cap and amounts not received due to insufficient stock. This hypothetical model has no replenishment, so even a shared cap does not guarantee indefinite use, and stock can run out. Arrival order and request sizes also affect distribution. The model does not establish that a cap is always fairest or most efficient, nor does it predict actual policy effects or ecological resource recovery.
Read more about the principle · Official Nobel Prize scientific background: Common resources and usage rules
Read the commons article below for more on rivalry, overuse, and management rules established by users.
Earth and Space · Air Parcels and Atmospheric Stability
Will air moved upward
become less dense than its surroundings?
Compare the temperatures of a dry air parcel and the surrounding air separately.
Change the surface temperature, environmental lapse rate, and displacement height to see the direction of buoyancy at the same height.
The dry air parcel is assumed to follow a linear temperature profile, cooling by about 9.8℃ per 1km. Only buoyancy due to temperature differences is diagnosed, using the same ideal gas at equal pressure.
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The difference at the same height matters more than the starting temperature
If a lifted air parcel is colder than its surroundings, buoyancy pushes it back toward its starting position. A temperature inversion is stable, but dry air can be stable even without an inversion.
A dry air parcel moved from surface temperature T₀ to height z(km) is compared with the environment using the specified linear profiles Tₚ=T₀−9.8z and Tₑ=T₀−Γz. For ideal gases at the same pressure and composition, density is inversely proportional to absolute temperature, so buoyancy acceleration per unit parcel mass is b=g(TₚK/TₑK−1). Here g=9.80665m/s², and K is absolute temperature, found by adding 273.15 to ℃. A temperature difference of 1K equals a difference of 1℃. This dry model is stable when the environmental lapse rate Γ is below 9.8℃/km, neutral when equal, and unstable when greater. Thus, a dry parcel is stable even at 6.5℃/km, while an inversion with negative Γ is an example of strong stability in which the environment warms with height. At 0m, the temperature difference and buoyancy are 0, but the stability classification based on lapse rate remains unchanged. This is an educational comparison with fixed lapse rates, not a calculation integrating actual adiabatic or hydrostatic differential equations or an ascent trajectory. Condensation, latent heat, mixing, wind, and drag are omitted, so it does not predict actual weather or pollutant concentrations.
Read more about the principle · NWS: Comparing environmental and air parcel stability · NWS: Dry adiabatic lapse rate · NWS: Temperature inversions and stable layers
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