Solving for velocity in the x/y/z axes?

Question

In a 3D environment I have an object with a displacement x that I know I need to traverse in a given time interval. I have the object's heading and elevation and I want to figure out the distance the object needs to travel on the x/y/z axis to move x units in the given direction. How can I calculate this? Are there two distinct ways to calculate it (using trigonometry or linear algebra)? Which way is the best in what scenarios?

EDIT: To clarify, I will make an analogy to a simple 2D scenario that is similar to the 3D scenario I'm working with. Let's say the object has an angle of 35 degrees from the x-axis. Making a right triangle, let's say we know that the hypotenuse (or displacement of the object) is 10 units. Now to figure out the x and y deltas to get from the object's current position to the object's new position 10 units away at a 35 degree angle, you simply do:

x = 10 * cos(35) 
y = 10 * sin(35)

I simply need to solve the same problem except in 3 dimensions.

Answer 1

Ok, I think I get what you're asking now...

If you have the 3D position of the target object (B) and your own position (A), the vector B - A is the 3D displacement to the target.

If you only have a bearing and an elevation to the target, you can get a vector that points at it but you can't determine how far away it is with only those angles. The components for that vector would be: x = cos(bearing) * cos(elevation); y = sin(bearing) * cos(elevation); z = sin(elevation); You may need to convert between degrees and radians and may need to add a constant to your bearing and/or elevation to properly line them up with the world axes.

If you have the bearing, elevation, and distance, multiply the vector components above by that distance.

Answer 2

accoring to your changes I can assume you have two angles, one that I call theta which is just like the one you have in 2d plane and has some value between -pi -> +pi. and you have a second angle called (let's call this one phi) and it has some value between -pi/2 -> pi/2. you can see in the picture what are these angles exactly specifying.

using these 2 angles you can compute normal 3d vector using this formula :

 x = cos(theta) * cos(phi);
 y = sin(theta) * cos(phi);
 z = sin(phi);

Answer 3

Things simplify considerably if you're willing to use quaternions.

In this case you have to simply store a "direction" unit vector(D) and an "upside" unit vector(U). Optionally you can store the "right" unit vector that points to the right direction (R) or you can compute doing the cross product (D x U).

When you want to change your pitch or bearing or orientation you simply have to construct a quaternion using the right vector and multiply the other two for that quaternion.

With this strategy you can store a "direction" vector that even points where you are going and rotate without the problem you may encounter if you touch some nasty rotation pole.

[EDIT]

Quaternions are 4-tuple of real numbers that represent a real part and the coefficients for tree orthogonal imaginary parts i, j, k. A quaterion may be write as (s,v) where s is the real part and v is the imaginary 3D vector.

A point in 3D (p) may be represented as the quaternion (0,p) and the rotation of θ rads about the r unit vector is: q = (s,v) where s = cos(θ/2) and v is r·sin(θ/2).

now p* = q · p · q' is the rotated point where q' is the inverse of q.

The inversion of a unit quaternion equals its conjugate so q' = (s, -v)

The quaternion multiplication shows a little complexity: while the a quaternion can be written as q = w + x i +y j +z k thus p · q is a simple polinomial multiplication, there are the relations so:

we can say that a multiplication between quaternion can be written as:

this lead us to the conclusion that if q1 is the vector Q = [W1, X1, Y1, Z1] then

so q1 · q2 = Q · A in term of matrix multiplications.

You can see that a rotation using a quaternion involves 32 multiplications and 24 sums Vs the 9 multiplications and 6 sums for the rotation using the standard matrix rotations.

This difference is due to the fact that a fourth dimension is used to avoid the pole problem that often makes the matrix rotation unviable: the euclideans matrix rotations is not faster, it simply compute less things and you pay this if you pass through a pole.

Answer 4

If you don't want do deal with some nasty trig functions, try parametrizing the movements on each axis. For example, imagine an object moving in a vector (variably scaled) of <1,2,3> for 2 seconds at 3 meters per second. Define the units of each axis as one meter. x=t y=2t z=3t Continue time to find where the object would be at t seconds. To find velocity, differentiate each parametric for a velocity vector. For a projectile, z would usually be something like -9.8t^2+t+20 and that projectile's velocity vector would be scalars for x and y and a linear with t for z. I personally like this better because differentiation is more intuitive, but 3D polars are good for some things too