comparing angles from two trajectories

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laurie
laurie el 10 de Abr. de 2012
Hello there,
I am trying to compare the angles from two trajectories, (0,0) being at the center of the screen and the trajectories in the whole screen.
What I did was to use Arctan() to determine the polar coordinates rho(t),theta(t) from x(t),y(t) for both trajectories. To avoid problems of angles "jumping" between -pi and +pi I introduced a condition that minimalizes the gap between two successive angles, choosing between the angle returned by arctan() and the angle returned by arctan() +/- 2pi. I get less "jumps" in each angular trajectory but I am still having some important problems :
  • sometimes the trajectory crosses the origin (0,0) so that the angle cannot be calculated and I get "jumps" between the point after and the point before that.
  • this is bad because then it allows the two angular trajectories to differ more than they would differ without the "jump" (angle from the one is angle from the other +/- 2pi)
so i am looking for a way to compute a continuous angular trajectory even when the trajectory hits the origin (0,0), so that i can compare angular trajectories correctly. i could just set the origin elsewhere than in the center of the screen but the problem is that what i am really interested in is the angle with the center of the screen as a reference.
thank you, sorry i don't think this was clear at all
  3 comentarios
laurie
laurie el 13 de Abr. de 2012
wow sorry i did not see your answer !!!
ok sorry atan()
atan2 might well be what i need :)
from what i understand, atan2(y,x) gives the angle formed by a point of coordinates (x,y) in an orthonormal four-quadrants, and it gives the one measure that is between pi and -pi ?
what is good is that atan2(0,0) is 0 which is what i need. :)
i am still getting some problem with the following :
i have a point that moves from (100,0.1) to (100,-0.1) so those two positions are quite close.
atan2(0.1,100)= 3.1406
atan2(-0.1,100)=-3.1406
so the distance between those two measures is huge compared to the actual distance (you have to add +/-2pi to get them closer..)
anyway thank you maybe atan2 will be the answer i need :)
laurie
laurie el 13 de Abr. de 2012
AWESOME :) i made some minor arrangements to account for the "continuity" problem each time the point crosses the line between -pi and pi and it is now doing exactly what i want !! thank you !!

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