Trig in 3D Rotations?
Say I was going to rotate a 3D field...like a fighting arena, on its y axis. I know that the matrix for y rotations involves trigonomic functions...but for that to happen, you need a right triangle and a reference angle.
My question is: if it's not clearly a triangle...and it's something like a character, or an entire level....how you do determine its reference angle, and the size and location of the triangle?
Actually you've got a little mis conception, and the source of the error is considering things on too grand a scheme. You see you are thinking you are rotating a triangle, or a character--rest assured you are not! Actually you are rotating the VERTICES of the triangle of the character. Rotations all come down to the rotation of a point in space (2d, in cartesian xy; 3d in cartesian xyz). Which also means infact, you do have the trig identies. Because of the way the points of space are depicted. X and Y (possibly Z). The x y and z components of a point are what form a right angle with the axis of reference and presumably the origin. I find it hard to explain with out a chalk board, or white board (which makes me kinda laugh when I think about how theres always a chalk/white board in math).
Any ways the key is: You dont move/rotate/scale an object, you move/rotate/scale verticies that make up an object.
HOpe that helps.
c++->visualc++->directx->opengl->c++;
(it should be realized my posts are all in a light hearted manner. And should not be taken offense to.)
Look into matrix math and matrix concatenation. You use 4x4 homogneous matrices to represent the rotations either in euler angles or in quaternion form. Impossible to explain it all here.
Get a book.
