3D Vector Components using Angle of Projection

11-26-2007, 10:05 PM	#1
xombie	By angle of projection, I mean the 3-dimensional angle that the vector "follows", if that is clear at all. I know that to find the angle of projection given 3 coordinates (3D vector) the equation is: Atan2(dz,SquareRoot(dxdx+dydy+0.1)) However, say you have this angle, the question would be how do you find 'dx', 'dy', and 'dz'?

11-26-2007, 10:17 PM

HINDYhat

Quote:

Originally Posted by iNfraNe

you must transform your usage of spherical/polar(in case of 2d) coordinates (explained here) to real coordinates:

angle a1 = angle from xyplane to z axis (angle of attack)
angle a2 = angle in the xy plane (direction)
speed = total speed

speedx = CosBJ(a1)*CosBJ(a2)*speed
speedy = CosBJ(a1)*SinBJ(a2)*speed
speedz = SinBJ(a1)*speed

You're welcome.