Need help with some maths.

[mortiz-Redux]

I thought I'd post this here because that's how I roll.

I want to physically simulate some trank tracks (in 2D) that envolopes two wheels, what I need to know is how to work out how many tracks I'm going to need to cover the two wheels:

I know the following:

1. The radius / circumference of the front and rear wheels.
2. The distance the two wheels are apart.
3. The length of each track.
4. The distance each track is apart from one another.

Obviously i'll have a fraction of a track left over but I can always round up. Any math's guru's around that can lend a man a hand with the mathematic equation to solve this?

 

[theotherguy]

You can represent it as domino tiling in one dimension.

The total distance that needs to be covered is half of the (circumference of each wheel + distance between each wheel)*2

This distance must be evenly divisible by (number of tracks*length of each track+(number of tracks-1)*(distance between each track))

So I think what you'd need is:
Let C be the cirucumference of a wheel.
Let D be the distance beetween the centers of the wheel.
Let T be the length of a track.
Let X be the distance between each track.

We aim to find N, the number of tracks required.

Then let Y, the total distance=(C+D)*2

So, ((C+D)*2)/(N*T+(N-1)(X)) is an integer. In particular, it must equal one.

1=((C+D)*2)/(N*T+(N-1)(X))

=>1/(N*T+(N-1)(X))=((C+D)*2)
=>(N*T)+(N-1)(X)=1/((C+D)*2)
=>(N)(T)+NX-X=1/((C+D)*2)
=>N(T+X)=1/((C+D)*2)+X
=>N=(1/((C+D)*2)+X)/(T+X)

QED

(This may not be possible if N is not a natural number)

 

[Kinslayer]

I disagree.

Wouldn't it be one total circumference (1/2 + 1/2) and then two total distances? What you're saying is that you have one total circumference but then you're only going one total distance between them. Unless I'm reading what you wrote wrong.

You could also more simply get the size of one "domino tile" by adding the distance between the tracks to the size of one track, getting you the size of one track + distance. Then it's simple division.

So you would have:

N = (C + 2D) / (T + X)

Edit: Marvelous picture here to illustrate my reasoning

http://img510.imageshack.us/img510/8019/14859775.jpg

 

[theotherguy]

Yes, my mistake. I forgot about the track on the other side.

Also, I failed to take into account that the track is a loop, rather than a straight line.

 

[Dan]

Just order a thousand tracks and ditch the spare ones in Iraq when you leave. Suspension and tension rollers can make up for rounding leftovers in the number of tracks so that you are left with a whole number.

 

[Teta_Bonita]

If the wheels have radii r1 and r2, and are a distance d apart, then the total length track needed to cover them is:

I tested it for a few simple values and it works. Hope that helps, and thanks for giving me a math problem to sink my nerdy teeth in. ^_^