 Space Calc
Doing the math, so you don't have to.

# Artificial Gravity

## Introduction

To get an acceleration relative to the floor in space (where the floor does not produce a significant acceleration due to gravity), one can spin the floor in circles, so that you are thrown outward against it. The effect is not unlike the acceleration due to real gravity.

This calculator calculates how this works out for a given situation.

Motion within such frames is typically unintuitive, especially when the object is given an initial velocity. To help you out, this calculator will also plot trajectories inside a stationary and a spinning reference frame.

## Calculator

Quantity Value Solve For
(diameter)
Spin: (angular velocity)
(tangential velocity)
(rotational period)
Artificial Gravity:

## Interactive Trajectory Plots

### Impact

 Impact Time: Co-rotated net relative speed:

Click and drag to move the launch point up or down or to change the launch angle or speed. After you release the vector, the scales will renormalize so that they fit nicely (at-least better) on the screen. Note therefore that the vectors are not scaled with respect to the dimensions of the ring. You can hold SHIFT to constrain the angle to the nearest 15°. You can also click+drag to zoom.

## Coriolis and Spin-Gravity Trajectories

The Coriolis effect (named for some guy) is the phenomenon that in a spinning reference frame, objects' motions appear to be deflected in weird ways. Of course, this is caused by the observer spinning, not the laws of Physics being broken. Indeed, we can construct Newton's Laws in a rotating frame (Obligatory XKCD), and we'll see the appropriate correction terms emerge.

In a rotating frame, Newton's Second Law (you-know, $F = m a$?) looks like:

$\mathbf{F} ~~{\color{Magenta} - m \left( \frac{d \mathbf{\Omega}}{d t} \times \mathbf{r} \right) } ~~{\color{Blue} - 2 m \left( \mathbf{\Omega} \times \mathbf{v} \right) } ~~{\color{DarkGreen} - m \left( \mathbf{\Omega} \times \left( \mathbf{\Omega} \times \mathbf{r} \right) \right) } = m \mathbf{a}$

Here, bolded text represents vectors; $\mathbf{r}$ is position, $\mathbf{v}$ is velocity, and $\mathbf{\Omega}$ is the rotation vector (i.e., the rotation axis, a unit vector, scaled by the angular speed (in radians per second)).

As we can see, being in a spinning reference frame adds three perturbing factors, which I've colored (and are in left-to right order):

• Magenta is the Euler Force, which relates to the rotation rate speeding-up or slowing down. For-example, if you're on a playground merry-go-round, and someone stops it, you'll be thrown spinward. If the rotation rate is constant (as it is here), this term is zero.
• The blue term is the Coriolis force. It tends to twist movements.
• The green term is centifugal force. Intuitively, it throws you outward; this is the spin gravity.

One thing to note is that the two cross products in the centrifugal force term make the centrifugal force proportional to the distance to the center and proportional to the square of the angular speed. Spin three times as fast, and you're thrown outward nine times as hard. Move inward, and you'll feel less, until at the center you feel no outward force at-all (just dizzy).

Another important thing is that the Coriolis force only depends on the rotation and velocity. It does not depend on position. Being closer to the spin axis will not make the Coriolis force "stronger". Of-course, as you move inward, as-above the centrifugal force becomes weaker, so Coriolis is relatively stronger. That is, toward the center, Coriolis is more-significant, and toward the rim, centrifugal force becomes more-significant by virtue of becoming stronger.