3d trig

primalvisions

I have a rectangle i am rotating in 3d on multiple axis. i am trying to calculate the 2d height of the rectangle and im quite lost.

For example, if i rotate the rectangle on the x axis so the top of the rectangle comes forward and the bottom moves back. i can calculate the height of this fine.

But then i rotate it on the y axis so the left of the rect moves to the right and the right hand side moves left.

Whats the total height from the bottom left corner to the top right corner extent when rotated on 2 axis?

Its for an orthographic view of the object where you show the front, top and side views of the 3d object.

Any clue how i can calculate this?

Cheers =)

Office_Shredder

Can you give it a better description as to where the rectangle is (centered on the x-y plane is what it sounds like)? And what is your definition of height here? You mean the distance when you project the rotated rectangle back onto x-y plane? Or the distance when projected onto the z-axis?

primalvisions

yes it is centered on the x-y plane and yes the height when projected back onto the x-y plane.

the total distance from the bottom to the top in a 2d view when rotated in 3d about the x-axis then the y-axis.

=)

Office_Shredder

Fortunately the axes of rotation are perpendicular, so you have an easy time doing this.

If you start off with a point at (x,y,0) and you rotate about the x axis by an angle t, the new point you get is

(x,ycos(t),ysin(t)).

Since you know the x coordinate of the point is unchanged, you can see this by just considering the rotation of the point (y,0) in the y-z plane around the origin by an angle of t)

Now for a caveat. When I decided I was going around by an angle t, that's the angle you see when you look at the y-z plane such that y is horizontal, and z is the vertical axis (which is how you defined it).

Next we look at the x-z axis. The point (x,ysin(t)) is rotated by an angle of r - we omit the y coordinate since it's fixed when rotating about the y axis. Now we hit a problem here, since the left of the rectangle moves to the right regardless of which direction you rotate it in. I'll assume the left of the rectangle is being rotated upwards, if it's moving down just use -r here. Then the new point is
(xcos(r)-ysin(t)cos(r),ysin(t)sin(r) + xsin(r)). so going back into three dimensions we get

(xcos(r)-ysin(t)cos(r),ycos(t),ysin(t)sin(r)+xsin(r))