# Calculate velocity components x,y and z

1. Dec 19, 2011

### Redweasel

Hello,

I want to calculate the velocity components x, y and z based on a velocity value and two angles (yaw along the z-axis and pitch).

I know how to calculate the x and y components in an 2D-environment:
initial velocity x = initial velocity * cos(theta)
initial velocity y = initial velocity * sin(theta)

But I can't figure out to apply this to an 3D-environment. Can anybody give me some good resources for learning this concept or provide some examples?

Redweasel
P.S.: Sorry for my bad english

2. Dec 19, 2011

### rcgldr

If the yaw is relative to z axis, then which axis is pitch relative to? This sound similar to spherical coordinates, with r equal to the magnitude of velocity. Note that there are conflicting usages of the symbols used to represent the angles. Link to mathworld article:

http://mathworld.wolfram.com/SphericalCoordinates.html

3. Dec 19, 2011

### Redweasel

The pitch should be related to the x axis.

The purpose for this is java programming. I want to create a little program where you enter the initial velocity, yaw and pitch and then throw a ball. It then displays the time when it hits the floor (time = zInitialVelocity / g * 2) and the x and y coordinates. Is there a way without the spherical coordinate system?

4. Dec 19, 2011

### rcgldr

Related to the x-axis in which direction, towards the y axis or towards the z axis? You didn's specify the relationship bettween x, y, z axis and the directions left-right, forward-back, up-down. If z-axis is vertical, then pitch could be the angle from horizontal (the x-y plane) (which could be restated as π/2 - angle from z-axis), and yaw would be the angle from x or y axis along the x-y plane.

A velocity, yaw, and pitch implies a spherical like coordinate (magnitude and two angles). You can always translate the initial condition into components of velocity vx, vy, vz. Acceleration can also be split up into ax, ay, az components and position would be x, y, z.

Last edited: Dec 19, 2011
5. Dec 20, 2011

### Redweasel

I guess I have to learn this spherical coordinate stuff then. Looks quite confusing, but I think I'll find a way. Thanks for your answer