← All tools

Flywheel Design Studio

Surface speed, gearing, compression and stored energy together. Starts with the thing most flywheel maths gets wrong: a single wheel launches at half its surface speed, not all of it.

Inputs

Layout

in
%

Only used for two-wheel setups. Below 100 puts backspin on the object.

Drive

The object

in
oz
in

Flywheel surface to the opposite surface.

%

Of the theoretical figure. Slip always loses some; measure a real shot and put the ratio here.

Stored energy

oz

Everything that spins: wheels, shaft, gears.

in

Where the mass effectively sits. About 0.7 of the wheel radius for a solid wheel, closer to the rim for a spoked one.

Results

Exit speed the model givespredicted

ft/s

Geometry only, and there is no target to miss here. A single wheel against a plate launching at half its surface speed is the expected physics, not a shortfall.

Exit speed after your measured slipmeasured

ft/s

What you actually get. The gap between these two is the slip figure you entered, and it is the honest part of this page.

Object leaves at

ft/s

Flywheel surface speed

ft/s

Of surface speed

%

Before slip.

Flywheel speed

RPM

Backspin on the object

rpm

Compression

%

Energy stored at speed

J

Energy per shot, as a share

%

How this is calculated

A flywheel does not throw an object at its surface speed. What it does is drag one side of the object along while something else holds the other side back, and the object's centre ends up moving at the average of the two surfaces.

surface speed = π × diameter × RPM ÷ 720 (ft/s) exit speed = (near surface + far surface) ÷ 2

The factor of two

With one flywheel and a fixed plate, the far surface is stationary. The average of the wheel speed and zero is half the wheel speed, so the object leaves at roughly half what the rim is doing. It also picks up heavy backspin, because the two sides are moving at very different speeds.

With two flywheels at the same speed, both surfaces move together, so the object leaves at roughly the full surface speed and with almost no spin. Same wheels, same RPM, twice the exit speed.

This is the single most common flywheel design error: sizing a single flywheel from a surface speed calculation and finding the shot lands half as far as expected. The arrangement matters more than the wheel.

Backspin is a lever, not a side effect

spin rate = (near surface − far surface) ÷ (π × object diameter)

Backspin makes an object fly flatter and further than plain projectile maths predicts, through the same effect that makes a golf ball carry. It also makes shots bounce out of goals differently.

Running two wheels at different speeds is how you dial it in: equal speeds give a fast flat shot with no spin, and slowing one wheel trades exit speed for backspin. That is the real reason to build a dual flywheel with independent control.

Stored energy decides the dip, not the motors

inertia I = m × k² k is the radius of gyration energy E = ½ × I × ω²

Every shot takes energy out of the wheel and the speed drops. How far it drops is set by how much energy the wheel was holding compared with what the shot takes, so a heavier or faster wheel dips less. The motors barely affect the dip at all; they only affect how fast it climbs back.

So a large dip wants more spinning mass, and a slow recovery wants more motor. Adding motors to fix a big dip is the usual wasted change. Therecovery analyzer measures what actually happens on your robot.

Compression, and why the exit speed figure exists

Too little compression and the wheel slips across the object instead of carrying it, so real exit speed falls well short. Too much and the object jams or the wheel stalls. There is no published figure for what a given game object wants, so the honest approach is to fire a real shot, measure the exit speed or the range, and put the ratio into the field above.

Sources & assumptions

Cartridge speeds and gear tooth counts are VEX published figures. Surface speed, the averaging that gives exit speed, backspin and rotational energy are all standard mechanics.

The half-speed result for a single flywheel assumes the object rolls without slipping against both surfaces, which is the ideal case. Real slip only reduces it further, which is what the measured exit speed field is for.

Radius of gyration is yours to estimate and no default would be right: it depends on whether your wheels are solid, spoked or weighted at the rim. The energy figure scales with its square, so a rough guess there makes the energy figure rough too. Everything else on the page is unaffected by it.

No aim assist from spin is modelled, and no air resistance. Thelaunch studio covers where the shot actually goes once it leaves.

Save this run, and compare

Keeps what is on screen so you can change something and see both sides of the change. Saved in this browser only, never uploaded.

Save this as evidence

Collects what you entered, what came out, how it was worked out, and anything the tool flagged, with a timestamp and a version so someone else can reproduce it.

This is evidence, not a notebook entry. It deliberately does not write your problem statement, your reasoning, or your conclusion, because under RECF rules an Engineering Notebook has to be the students' own work and no tool may generate or organise its content. Take the numbers, decide what matters, and write it yourself.