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