Date: Thu, 16 Feb 95 10:27:46 CST
From: rusin (Dave Rusin)
To: sulis@uxa.cso.uiuc.edu
Subject: Re: Probably an easy Circular motion question - hints?
Newsgroups: sci.physics,sci.math,comp.graphics,comp.graphics.algorithms

In article <3huogg$fgb@vixen.cso.uiuc.edu> you write:
>I tried my hardest to work with some formulas and drew out some ideas, but
>it's been awhile since I've really dug into any serious math. I came up
>with this - each point has: x position, y position, dx (change in x pos),
>dy (change in y pos), and K (curvature).
>
>So I just loop over and over:
>    tempdx=dx  (these just so the data doesn't change in the middle of the
>	tempdy=dy   dx and dy calculations)
>
>	    dx = tempdx + K (tempdy)
>		dy = tempdy + K (tempdx)
>		    x=x+dx
>			y=y+dy

I'd suggest you change the coefficients of one of the  K's  to a minus sign.
That way your increment in  (dx, dy) will be proportional to (dy, -dx).
Viewing p=(x,y) as position, you have arranged it so  (dx,dy) = velocity  v
and acceleration = change in velocity is  K(dy, dx). If as I suggest you
make acceleration be  K(dy, -dx), then you'll have acceleration perpendicular
to velocity, so that the dot product  v.v  will have a derivative of zero,
i.e., the particle will travel at constant speed.

You have then set up the general equation of motion for a particle at
constant speed. You get various kinds of motion depending on what you
specify for  K -- if it's not constant, it might depend on time elapsed,
current position, and/or current velocity. It's easy to see that
|K| is || acceleration || / || velocity ||, so now that we've set
speed constant,  having a constant  K  is equivalent to having an
acceleration of constant magnitude. This should indeed give circular motion.
Moreover, you should be able to illustrate, say, gravitational attraction
by having  K  be proportional to   (distance to some fixed point) ^ (-2).

Of course, this is all idealized, that is, I am replacing your discrete
motion by continuous, differentiable motion in order to perform this
analysis; this and round-off error may well allow a supposedly closed orbit
to spiral in or out, for example.

dave