Ah, thanks Celierra. 3 planes, yes that makes a lot more sense. Hmm, trying to step through your calculations. I think you have to subtract the angle difference rather than add it on. I just tried visually going through your calcs and it seems to hit right on your perfect value p3.
http://i48.tinypic.com/15x2wz8.png - Nexii On Sat, Feb 6, 2010 at 1:28 AM, Celierra Darling <celie...@gmail.com> wrote: > I'm not sure about this -- I tried this with a free-flight parabola and got > a nonsense value out. Calculations attached at the end. > What is the significance of the "anchor"? It looks like you're trying to > estimate the center of the path's curvature? If you were doing that, you > want to intersect using the lines through p and *perpendicular* to v (in > 2D), not intersect the lines created by v. In 3D, I think you'd want to > intersect 3 planes, the two planes through p perpendicular to their > respective v and also the plane that contains both points. > I'm not sure how to extend that to handle torques on the object.... > Have you tried estimating a = (v1 - v2)/t, and extrapolating using that > (p_next = p_cur + v_cur*dt + 1/2*a*dt^2)? It would also extend to torques > and angular velocity using the analogous equation. (I'm sure you could even > fit a line or spline to multiple previous values of v to get something even > higher-order, if you wanted, though with diminishing returns.) > > Celi > The aforementioned calculation: > > dt = 0.1 > a = g = [0,-2] > initial position and velocity: > p1 = [0, 0] > v1 = [1, 0] > next position and velocity > p2 = [0.1, -0.01] > v2 = [1, -0.2] > perfect value: p3 = [0.2, -0.04] > lines are (p1 + s*v1) and (p2 + t*v2) > intersect at s = 0.05, t = -0.05, position = [0.05, 0] > angle change = clockwise 3pi/4 (radians) > current angle = -pi/4 > angle after "twist" = pi/2 (upwards!?) > (it just gets worse from here - you estimate a distance upwards, and the > step forward using the current velocity doesn't get you back to y=0.)
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