- #1

darrencubitt

- 3

- 0

## Homework Statement

This isn't exactly homework, but this seemed like the right place for this question. I'm working on an add-on for Orbiter - the space flight simulator - and would like to be able to determine the equation of motion for an object traveling at an initial velocity (

**v**), with a linear acceleration term (

_{0}**a**), a velocity-squared drag proportionality constant (

**k**), and elapsed time (

**t**).

## Homework Equations

v = v + a * t - k * v^2 * t

## The Attempt at a Solution

Being of limited mathematical ability, I scoured the web for some solutions. I found many dealing with velocity-proportional drag, and also some velocity-squared functions which, unfortunately, assume v0=0. Reluctantly, I tried my own hand and I managed to hammer out an equation which can deal with velocity and drag but, alas, no acceleration.

Looking back at how I came to this solution, I am sure that my logic is utterly, utterly wrong. This is almost certainly a case of being right for the wrong reasons. For your entertainment (and also in the hope of getting an answer), my working follows:

I started by writing the basic equation:

1. v

1. v

_{t}= v_{0}- k * v_{0}^2 * tObviously, since velocity is constantly changing through time, the (k * v

_{0}^2) term will only be valid at t=0, so I divided both sides by v^2 to remove it from this part of the equation:

**2. 1/v**

_{t}= (1/v_{0}- k * t)I then reasoned that since I am now dealing with the reciprocal of the velocity, I should add the drag term rather than subtract it.

**3. 1/v**

_{t}= (1/v_{0}+ k * t)Finally, I multiplied the reciprocals out:

**4. v**

_{t}= v_{0}/ (1 + k * t * v)And, realising velocity may be negative, used the magnitude of the velocity on the right hand side.

5. v

5. v

_{t}= v_{0}/ (1 + k * t * |v|)After studying the relationships between the initial and final velocity, I concluded that displacement could be found by:

**6. LOG(1 + k * t * |v|) / k**(for k != 0)

I guess you could call eqn. 6 "solving by imperical observation".

I am sure that my workings are hilariously wrong but, to my surprise, these equations actually work. I compared the equation to the output of a numerical approximation and it agrees almost 100%, with some slight deviations which I put down to the latter being an approximation.

I tried repeating this trick when including a linear acceleration term (a) but always end up with a nasty (a/v^2) term stuck inside the equation (replacing the k*v^2 problem I was initially trying to solve).

Can anyone provide me with any hints on how to perform this calculation with linear acceleration included? I have seen some solutions bordering on what I am looking for, but either assume v

_{0}=0 or break down in situations with negative acceleration, etc. Is what I am looking for even possible?

Thanks for any help.