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Sorry if this has been asked before but my maths days are long behind me.

What I want to know is how to find out the coordinates along the circumference of an ellipse.

So supposing I am at point X,Y which lies on the circumference, and I am traveling at velocity Z. How can i work out after given time t what the new X,Y values would be.

seems so simple but just cant think where to start.

Thanks for the help

Aaron

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is the speed constant? –  Santosh Linkha Jul 25 '12 at 22:06
    
yes it is and its z –  Jorge Fernández Jul 25 '12 at 22:08
    
youtube.com/watch?v=iUfu9RK6w44 –  Will Jagy Jul 25 '12 at 22:35
    
@WillJagy are you aware this link is to a song? –  Jorge Fernández Jul 25 '12 at 22:48
    
@ChuckFernández, yes, DevilWAH wrote "my maths days are long behind me" in a rather forlorn manner. The song says no, not really. Here is a different one: youtube.com/watch?v=KHVYBiVKldU –  Will Jagy Jul 25 '12 at 23:06

1 Answer 1

up vote 3 down vote accepted

I don't think this has a nice elementary solution -- since the period of such a solution would be the circumference of the ellipse, something that it hard enough to compute that it has its own class of special functions associated with it.

So in practice you need to do it numerically, and then it's easy enough to program. You know (I hope) how to generate points on the ellipse without the constant-velocity requirement:

$$ u \mapsto (a\cos u, b\sin u)$$ where $u$ is a parameter and $2a$ and $2b$ are the lengths of the ellipse's major and minor axes. Your only problem is then to find the next $u$ value at each step. If you do it with small enough steps you can simply differentiate this expression to find out how far a small change in $u$ makes the point move. You get something like $$ u(t+\Delta t) = u(t) + \frac{z\Delta t}{\sqrt{a^2\sin^2(u(t))+b^2\cos^2(u(t))}}$$ where you choose the time step $\Delta t$ by trial and error to be small enough that the speed ends up being close enough to constant for your needs.

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Cool cheers for that, seems like this is a little more complex than I had hopped. This is some thing I will defiantly be coming back to soon. I am currently moving the point using the first equation you gave, and for now this will do. Once I have the rest of the application in place, I want to come back to this to fine tune it, however for now I think my current method will do fine. Thanks for the solution though. –  DevilWAH Jul 28 '12 at 23:44
    
@Devil, spelling tip: You mean "definitely", not "defiantly" -- the latter means "while openly refusing to obey". –  Henning Makholm Jul 29 '12 at 6:04
    
:) yer that would be the one. but honestly to my mind those two words look identical, takes a bit of looking for me to see the difference. Thats dyslexics for you. –  DevilWAH Jul 29 '12 at 15:23

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