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View Full Version : Simple trig problem but always comes up with error

Schlogenburg
09-05-2000, 03:12 PM
I have a vector with magnitude of 50.
I have an angle on the x axis of 45, and angle on y axis of 132.
(don't tell me specifically the answer, give me how you got it)

and I have the coordinate in 3d space(xyz) that I want the vector to go from. So this vector would point out from this coordinate at the given angles and I want the coordinate it ends at.

it should be something like this?
nx=cos()*vectormagnitude
ny=sin()*vectormagnitude
nz=tan()*vectormagnitude

but the way I have it the Y get's all screwed up, way out of what it should be.

john
09-05-2000, 04:10 PM
In other words, you want to rotate a point in three dimensions. Then you need to solve this:

v'=ZYXv

where v is the start position of your point (your vector with a magnitiude of 50 at xyz), and X, Y & Z are your matricies to rotate a point laong the X, Y and Z axis respectivly.

the eqns you'll get if you expand the matricies out have a few more terms than you'd care to poke a stick at =)

that is, if I understood your point correctly...

if you want the rotation matricies, check out the web or the back of the red book. they rotate homogeneous coords; in practice, you probably don't want to compte the w coordinate.

cheers
john

09-06-2000, 02:46 AM
For a vector with angle PHI to the y-axis and THETA to the x-axis (assuming a left-handed coord system):

mag = magnitude(vector); // in your case 50

vy = mag*cos(PHI);
vtemp = mag*sin(PHI);
vx = vtemp*cos(THETA);
vz = vtemp*sin(THETA);

-or-
vy = mag*cos(PHI);
vx = mag*sin(PHI)*cos(THETA);
vz = mag*sin(PHI)*sin(THETA);

in either case, add those three elements to the coordinates of the point from which you want it to originate and you'll get the final coords.

09-06-2000, 02:56 AM
Uh.. actually the code above assumes (perhaps erroneously) that the angle to the x-axis lies in the x-z plane. That is, I assumed that the projection of your vector onto the x-z plane had an angle of THETA with the x-axis.

Sorry.

09-06-2000, 03:15 AM
I'm interested in the answer to this problem. I think this question is similar to the one I posted a few questions ago. (however, he must have asked it much more elegantly, for I haven't got any replies yet... http://www.opengl.org/discussion_boards/ubb/smile.gif

Schlogenburg
09-06-2000, 06:15 PM
Can you tell me the difference between Theta and Phi, and how to acquire those?

steveo
09-07-2000, 09:45 AM
If 45 is the angle between the vector and the x axis ... ( as distinct from the angle projected onto a plane see prev. post) then

nx=50*cos(45)
ny=50*cos(132)

nz=square root of (50*50-nx*nx-ny*ny)

BTW if you know what the angle between the vector and the z axis is then
nz=50*cos(the angle)

Steve

09-11-2000, 07:16 PM
Don't forget, by the way, that the math.h routines for C/C++ work on radians, not degrees.