This is something I can't seem to figure out:
[attachment=5990:problem.png]
I have two 2D points P and Z, and two 2D vectors V and D. I need to find a rotation angle a such that Z lies on a line with the origin P+V[sup]R[/sup] and direction D[sup]R [/sup]where V[sup]R [/sup]is V rotated by a and the same for D.
Note that D doesn't have to be a vector, it can be an angle if that makes it easier.
I'd appreciate any help on this issue.
Rotating an offset vector to point at a position
This is something I can't seem to figure out:
[attachment=5990:problem.png]
I have two 2D points P and Z, and two 2D vectors V and D. I need to find a rotation angle a such that Z lies on a line with the origin P+V[sup]R[/sup] and direction D[sup]R [/sup]where V[sup]R [/sup]is V rotated by a and the same for D.
Note that D doesn't have to be a vector, it can be an angle if that makes it easier.
I'd appreciate any help on this issue.
Here's a crack at it, untested and could be completely wrong
theta = angle between V, D (given)
||P - Z|| / sin(theta) = ||V - P|| / sin(phi)
phi = arcsin(sin(theta) * ||V - P|| / ||P - Z||)
beta = arctan((Zy - Py)/(Zx - Px))
omega = 2 * pi - theta - phi - beta
Vr.x = ||V - P|| * cos(omega)
Vr.y = ||V - P|| * sin(omega)
Author
I tried out your solution pantaloons and it seems to work properly, but only when V.y is less than zero and when D is pointing directly to the right.
The restriction on D isn't really an issue, because V and D can always be rotated beforehand to make D point to the right, but the restriction on V is a bigger problem.
Still, thanks a lot!
Edit: Figured out how to make it work when V.y > 0. Kind of an unintuitive solution, but omega = beta - (PI - theta - phi), and Z.x should be mirrored relative to P.x when V.y > 0
The restriction on D isn't really an issue, because V and D can always be rotated beforehand to make D point to the right, but the restriction on V is a bigger problem.
Still, thanks a lot!
Edit: Figured out how to make it work when V.y > 0. Kind of an unintuitive solution, but omega = beta - (PI - theta - phi), and Z.x should be mirrored relative to P.x when V.y > 0
It should have been:
[color="#1C2837"]omega = pi - theta - phi - beta
[color="#1C2837"]If you didn't catch that. Although you are right I think quadrant issues get in the way with the right triangle approach to computing Vr. We can make it easier:
[color="#1C2837"]theta = angle between V, D (given)[color="#1C2837"]
[color="#1C2837"]
[color="#1C2837"]
[color="#1C2837"]
[color="#1C2837"]omega = pi - theta - phi - beta
[color="#1C2837"]If you didn't catch that. Although you are right I think quadrant issues get in the way with the right triangle approach to computing Vr. We can make it easier:
[color="#1C2837"]theta = angle between V, D (given)[color="#1C2837"]
||P - Z|| / sin(theta) = ||V|| / sin(phi)
phi = arcsin(sin(theta) * ||V|| / ||P - Z||)
[color="#1C2837"]omega = pi - theta - phi[color="#1C2837"]
X = (Z - P)/||Z - P||
[color="#1C2837"]if(0 <= theta <= pi) [color="#1C2837"]Vr = ||V|| * R(omega) * X [color="#1C2837"]else Vr = ||V|| * R(-omega) * X[color="#1C2837"]
[color="#1C2837"]
Where R(omega) is a standard 2x2 rotation matrix through angle omega. Notice that in fact, discounting the direction of vector D, there are two solutions that are equally rotated away from the vector Z - P, however you want the solution where D points towards Z which necessitates a negative rotation if -pi < theta < 0 (or turn(P, V, D) == right)).
Author
After looking over the solution again, I found that I didn't actually need beta or Vr. Once I have omega, I can compare it to the angle between V and Z-P to figure out how much rotation is needed.
It now works for all the cases I'll need.
Really appreciate it.
It now works for all the cases I'll need.
Really appreciate it.
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