my math isnt too great but im trying to learn though..
What im trying to do is give my games missiles a helix rocket effect, but i dont know how to work the Sin and Cos to make the helix play out in the right direction..
This is a 3D game by the way:
The problem is, depending on which direction the missile faces, the helix looks warped or flat..
Whats the best way to mathematically calculate a helix based on the missiles X,Y,Z/direction?, ive been trying to figure it out for a long time :/
Thanks alot!
double x = radius * Math.cos(theta);
double y = radius * Math.sin(theta);
double z = radius * Math.cos(theta);
location.add(x,y,z);
missile.shootFlame(location,2);
location.subtract(x,y,z);
Basis vectors
you need the overall direction of missile as 3D vector let call it W. From it you need to get 2 another vectors U,V which are all perpendicular to each other. To get them you can exploit cross product. So:
make W unit vector
Just W = W/|W| so:
l=sqrt(Wx*Wx+Wy*Wy+Wz*Wz);
Wx/=l;
Wy/=l;
Wz/=l;
choose U as any direction non parallel to W
so start with U=(1.0,0.0,0.0) and if U==W choose U=(0.0,1.0,0.0). If you got anything to lock to use that as U direction so the coordinate system will not rotate with time (like Up,North,Sun ...)
U should be unit so if not normalize it just like in #1
compute V
It should be perpendicular to U,W so use cross product:
V = W x U
Cross product of unit vectors is also unit vector so no need to normalize.
recompute U
As we choose the U manually we need to make sure it is also perpendicular to V,W so:
U = V x W
Now we have 3 perpendicular basis vectors where U,V lies in plane of the helix screws and W is the overall movement direction.
If you do not know how to compute the cross product see:
Understanding 4x4 homogenous transform matrices look for [edit2].
Now the Helix is easy:
Defines
so we have U,V,W on top of that we need radius r [units], movement speed v [units/s], angular speed o [rad/s] time t>=0.0 [s] and start position P0.
Helix equation
What we need is equation returning actual position for time so:
ang = o*t;
P(t) = P0 + U*r*cos(ang) + V*r*sin(ang) + W*v*t;
rewritten to scalars:
ang = o*t;
x = P0x + Ux*r*cos(ang) + Vx*r*sin(ang) + Wx*v*t;
y = P0y + Uy*r*cos(ang) + Vy*r*sin(ang) + Wy*v*t;
z = P0z + Uz*r*cos(ang) + Vz*r*sin(ang) + Wz*v*t;
[edit1] as you are incompetent to copy paste and or changing my code correctly...
Vector w = loc.getDirection();
double wX = w.getX();
double wY = w.getY();
double wZ = w.getZ();
double l = Math.sqrt((wX * wX) + (wY * wY) + (wZ * wZ));
wX = wX / l;
wY = wY / l;
wZ = wZ / l;
w = new Vector(wX,wY,wZ); // you forget to change W and use it latter ...
Vector u = new Vector(0, 1.0, 0);
if (Math.abs(wX)<1e-3) // if U and W are the same chose different U
if (Math.abs(wZ)<1e-3)
u = new Vector(1.0, 0.0, 0);
Vector v = w.crossProduct(u);
u = v.crossProduct(w);
double radius = 10; // helix radius [units]
double speed = 2.00; // movement speed [unit/s]
double omega = 0.628; // angular speed [rad/s]
//double omega = 36.0; // angular speed [deg/s]
for (double i = 0; i < 100; i += 1.0) // time [s]
{
double angle = omega * i; // actual screw position [rad] or [deg]
double x = u.getX() * radius * Math.cos(angle) + v.getX() * radius * Math.sin(angle) + wX * speed * i;
double y = u.getY() * radius * Math.cos(angle) + v.getY() * radius * Math.sin(angle) + wY * speed * i;
double z = u.getZ() * radius * Math.cos(angle) + v.getZ() * radius * Math.sin(angle) + wZ * speed * i;
loc.add(x,y,z); // What is this? should not you set the x,y,z instead of adding?
//do work
loc.subtract(x,y,z); // what is this?
}
This should provide you with helix points with traveled linear distance
speed*imax = 2.0*100.0 = 200.0 units
And screws:
omega*imax/(2*Pi) ~= 0.628*100.0/6.28 ~= 10 screws // in case of sin,cos want [rad]
omega*imax/360.0 = 36.0*100.0/360 = 10.0 screws // in case of sin,cos want [deg]
Do not forget to rem in/out the correct omega line (I choose [rad] as that is what I am used that my math libs use). Not sure If I translated to your environment correctly there may be bugs like abs has different name or u = new Vector(1.0, 0.0, 0); can be done on intermediate or declaration of variable only etc which I do not know as I do not code in it.
Related
The issue I have is that I'm attempting to add drag to an object in this basic physics simulation (Java [Processing]), but once I add the appropriate formula, it causes the objects velocity to increase drastically in the opposite direction. Of course the problem is that drag for some reason is being calculated too high but I'm not sure why thats happening as I'm using the real world equation.
void setup(){size(1280,720);}
class Circle{
float x,y,r,m,dx,dy,ax,ay,fx,fy;
Circle(float xPos, float yPos, float Radius, float Mass){
x = xPos;
y = yPos;
r = Radius;
m = Mass;
}
void ADD_DRAG(){
fx -= 0.5 * 1.225 * dx * dx * 0.5 * r * PI;
fy -= 0.5 * 1.225 * dy * dy * 0.5 * r * PI;
}
void update(){
ADD_DRAG();
ax = fx / m;
ay = fy / m;
dx += ax / frameRate;
dy += ay / frameRate;
x += dx / frameRate;
y += dy / frameRate;
}
}
Circle[] SceneObjects = {new Circle(50,50,20,20000),new Circle(50,50,2,20)};
void draw(){
background(51);
for (Circle c : SceneObjects){
c.update();
circle(c.x * 3,c.y * 3,c.r * 3);
}
}
void mouseClicked(){
if(SceneObjects[1].fx != 2000)
SceneObjects[1].fx = 2000;
else
SceneObjects[1].fx = 0;
}
This is the code, essentially there is a Circle class which stores the objects properties and then the forces applies are updated each draw loop. The mouseClicked void is just for testing by adding a force to the objects. All and any help is appreciated, thanks!
Maths I am Using:
Rearranged F=ma for ax = fx / m;
Acceleration * time = Speed for dx += ax / frameRate; (frameRate is 1/time)
Distance = Speed * time = for x += dx / frameRate; (same as above)
For drag im using this equation https://www.grc.nasa.gov/WWW/K-12/rocket/drageq.html with the constants eg air density etc added as seen.
There are a couple of things wrong here.
You haven't given us numbers (or a minimal complete example), but the vector algebra is off.
Yes, the acceleration is f = -kv2, and |v|2 = vx2 + vy2, but that doesn't mean that you can decompose f into fx=kvx2 and fy=kvy2. Not only is your magnitude off, but your acceleration is now not (in general) aligned with the motion; the path of your projectile will tend to curve toward a diagonal between the axes (e.g. x=y).
Also, your code always gives acceleration in the negative x and negative y directions. If your projectile happens to start out going that way, your version of air resistance will speed it up.
Finally, your time interval may simply be too large.
There is a better way. The differential equation is v' = -k v|v|, and the exact solution is v = (1/kt) z, (with appropriate choice of the starting time) where z is the unit direction vector. (I don't know how to put a caret over a letter.) This leads to v(t) = (1/t)v(t=1.0)
So you can either work out a fictional time t0 and calculate each new velocity using 1/(kt), or you can calculate the new velocity from the previous velocity: vn+1 =vn/(kd vn + 1), where d is the time interval. (And then of course you have to decompose v into vx and vy properly.)
If you're not familiar with vector algebra, this may seem confusing, but you can't get an air-resistance sim to work without learning the basics.
I wrote a code that should turn a point around another point counterclockwise. But it does not work correctly.
public boolean contains(double x, double y) {
double ox = this.x.get() + (this.width.get()/2);
double oy = this.y.get() + (this.height.get()/2);
double theta = rotate.get() - (rotate.get() * 2);
double px1 = Math.cos(theta) * (x-ox) - Math.sin(theta) * (y-oy) + ox;
double py1 = Math.sin(theta) * (x-ox) + Math.cos(theta) * (y-oy) + oy;
return shape.contains(px1, py1);
}
x, y - are the coordinates of the point to be rotated.
ox,oy - is the coordinates of the point around which you want to rotate.
rotate.get() - angle to rotate
Update: Changes in the code that solved the problem, who can come in handy:
double px1 = Math.cos(Math.toRadians(theta)) * (x-ox) - Math.sin(Math.toRadians(theta)) * (y-oy) + ox;
double py1 = Math.sin(Math.toRadians(theta)) * (x-ox) + Math.cos(Math.toRadians(theta)) * (y-oy) + oy;
Please check, if your rotate.get() will give you a degrees value (e.g. 45°) or a radians value (e.g. 0.5*pi). Math.sin() and Math.cos() will only accept radians.
To convert them you could use something like angle = Math.toRadians(45)
Although this is answered, another simple way to get this done is using the built-in method of Rotate class. This way you dont need to worry about the Math stuff ;)
Rotate r = new Rotate();
r.setPivotX(ox);
r.setPivotY(oy);
r.setAngle(angleInDegrees);
Point2D point = r.transform(new Point2D(x, y));
I am trying to generate random points on a sphere that is filled with a cube.
Because I had no idea how to do that i started with 2d. (A circle filled with a quadrat.)
What I am trying to do: Generating random points inside the outer circle, but outside the green square.
Basically in the blue areas.
The square is located at (-1|-1),(1|-1),(1|1),(-1|1).
The circle has a radius of r = sqrt(2) and is centered at (0|0).
I already have scripts to:
generate a random point on a circle (uniformly):
float a = 2 * MathUtils.PI * MathUtils.random(1f); // angle between 0 and 2pi
float r = radius * Math.sqrt(MathUtils.random(0, 1f)
float x = r * MathUtils.cos(a);
float y = r * MathUtils.sin(a);
calculating the radius for a given angle to form a square:
float r = (1/Math.sqrt(2)) / MathUtils.cos(((a+45)%90-45)/180*MathUtils.PI);
with (1/Math.sqrt(2)) being half the side length of the square
Before anyone asks:
I know that I could just re-generate points which are inside the green square until I get one that is outside, but I don't want to do it this way.
I appreciate any help. Thank you :)
It is rather hard to generate points only in region of sphere outside the cube (caps and wedges), so rejecting method looks reasonable.
But you can diminish the number of useless points, generating points in the ring only in 2D case and in spherical shell in 3D case.
So pseudocode might look as
//2d
SquaredR = RandomUniformInRange(0.5, 1)
R = Sqrt(SquaredR)
//3d
CubedR = RandomUniformInRange(Pow(3, -3/2), 1)
R = Pow(CubedR, 1/3)
//generate point on the circle or on the sphere with radius R
if Abs(x) > Sqrt(2)/2 or Sqrt(3)/3 and so on - reject
Having R, you can generate point on the sphere using any approach from here
Here is rough sketch of the idea. You select one quadrant to sample, say, one on the right.
First, sample angles from -pi/4 to pi/4
float a = -MathUtils.PI/4.0f + MathUtils.PI/2.0 * MathUtils.random(0.f,1.f);
float c = MathUtils.cos(a);
float s = MathUtils.sin(a);
Second, find minimum radius. With ray going from (0,0) at angle a will intersect quadrant line at x=1 at minimum
float rmin = 1.0f / c;
float rmax = Math.sqrt(2.0f);
Sample from rmin to rmax = sqrt(2), taking into account that for plane you sample squared radius and then use sqrt(), and for 3d space you sample cubed radius and then use cbrt().
float r2 = rmin*rmin + (rmax*rmax-rmin*rmin)*MathUtils.random(0.f,1.f);
float r = Math.sqrt(r);
float x = r * c;
float y = r * s;
Now, we constructed (x,y) is a such way it is guaranteed to be in the right quadrant below circle and on the right of the x=1 line.
To cover all four quadrants just sample to which quadrant you will move the point
float q = MathUtils.random(0.f,1.f);
if (q < 0.25f) // top quadrant
return (y, x);
if (q < 0.5f) // left quadrant
return (-x, y);
if (q < 0.75f) // bottom quadrant
return (y, -x);
return (x,y); // right quadrant
Please bear with me - my Java is quite rusty, and I have no ways to test the code.
In 3D case you'll have to deal with two angles, cubed radius, eight octants instead of four quadrants, but general logic is the same
UPDATE
I was wrong, sampling like I propose would lead to non-uniform point distribution.
From PDF:
PDF(phi, r) = S_(-\pi/4)^\phi d\phi S_1/cos(\phi)^\sqrt(2) r dr
One could get that we have to make \phi sampling non-uniform. Unfortunately, from
U(0,1) to get to sampled \phi requires solving non-linear equation
\pi/2 (0.5*(\phi/\pi/4 + 1) - U(0,1)) = 0.5*(tan(phi) + 1) - U(0,1)
So algorithm would be:
Sample U(0,1)
Find appropriate \phi by solving equation above
Find lower R boundary
Sample R
Quick code (in Python, sorry) to plot this non-linear function
import numpy as np
import matplotlib.pyplot as plt
def f(phi, ksi):
c = 0.5 * np.pi
c_2 = 0.5 * c
left = c * (0.5 * (phi/c_2 + 1.0) - ksi)
rght = (0.5 * (np.tan(phi) + 1.0) - ksi)
return left - rght
nof_points = 41
phi = np.linspace(-0.25*np.pi, 0.25*np.pi, nof_points)
y0_00 = f(phi, 0.00)
y0_25 = f(phi, 0.25)
y0_50 = f(phi, 0.50)
y0_75 = f(phi, 0.75)
y0_99 = f(phi, 1.00)
plt.plot(phi, y0_00, 'ro', phi, y0_25, 'b+', phi, y0_50, 'gx', phi, y0_75, 'm.', phi, y0_99, 'y^')
plt.show()
and plotted functions for five values of U(0,1) (ksi in the code)
Sampling could be rearranged such that r sampling is non-linear, but it exhibits the same problem - need to solve non-linear equation with polynomial and trigonometric parts
UPDATE II
And just for the record, if you want to sample r first, then it has to be sampled from the solution of the non-linear equation:
r2 sec-1(r) - sqrt(r2 - 1) = U(0,1)*(\pi/2 - 1)
in the interval [1...sqrt(2)]
After solving it and finding sampled r, \phi could be sampled uniformly in the interval allowed by r: [-cos-1(1/r) ... +cos-1(1/r)]
I'm programming a software renderer in Java, and am trying to use Z-buffering for the depth calculation of each pixel. However, it appears to work inconsistently. For example, with the Utah teapot example model, the handle will draw perhaps half depending on how I rotate it.
My z-buffer algorithm:
for(int i = 0; i < m_triangles.size(); i++)
{
if(triangleIsBackfacing(m_triangles.get(i))) continue; //Backface culling
for(int y = minY(m_triangles.get(i)); y < maxY(m_triangles.get(i)); y++)
{
if((y + getHeight()/2 < 0) || (y + getHeight()/2 >= getHeight())) continue; //getHeight/2 and getWidth/2 is for moving the model to the centre of the screen
for(int x = minX(m_triangles.get(i)); x < maxX(m_triangles.get(i)); x++)
{
if((x + getWidth()/2 < 0) || (x + getWidth()/2 >= getWidth())) continue;
rayOrigin = new Point2D(x, y);
if(pointWithinTriangle(m_triangles.get(i), rayOrigin))
{
zDepth = zValueOfPoint(m_triangles.get(i), rayOrigin);
if(zDepth > zbuffer[x + getWidth()/2][y + getHeight()/2])
{
zbuffer[x + getWidth()/2][y + getHeight()/2] = zDepth;
colour[x + getWidth()/2][y + getHeight()/2] = m_triangles.get(i).getColour();
g2.setColor(m_triangles.get(i).getColour());
drawDot(g2, rayOrigin);
}
}
}
}
}
Method for calculating the z value of a point, given a triangle and the ray origin:
private double zValueOfPoint(Triangle triangle, Point2D rayOrigin)
{
Vector3D surfaceNormal = getNormal(triangle);
double A = surfaceNormal.x;
double B = surfaceNormal.y;
double C = surfaceNormal.z;
double d = -(A * triangle.getV1().x + B * triangle.getV1().y + C * triangle.getV1().z);
double rayZ = -(A * rayOrigin.x + B * rayOrigin.y + d) / C;
return rayZ;
}
Method for calculating if the ray origin is within a projected triangle:
private boolean pointWithinTriangle(Triangle triangle, Point2D rayOrigin)
{
Vector2D v0 = new Vector2D(triangle.getV3().projectPoint(modelViewer), triangle.getV1().projectPoint(modelViewer));
Vector2D v1 = new Vector2D(triangle.getV2().projectPoint(modelViewer), triangle.getV1().projectPoint(modelViewer));
Vector2D v2 = new Vector2D(rayOrigin, triangle.getV1().projectPoint(modelViewer));
double d00 = v0.dotProduct(v0);
double d01 = v0.dotProduct(v1);
double d02 = v0.dotProduct(v2);
double d11 = v1.dotProduct(v1);
double d12 = v1.dotProduct(v2);
double invDenom = 1.0 / (d00 * d11 - d01 * d01);
double u = (d11 * d02 - d01 * d12) * invDenom;
double v = (d00 * d12 - d01 * d02) * invDenom;
// Check if point is in triangle
if((u >= 0) && (v >= 0) && ((u + v) <= 1))
{
return true;
}
return false;
}
Method for calculating surface normal of a triangle:
private Vector3D getNormal(Triangle triangle)
{
Vector3D v1 = new Vector3D(triangle.getV1(), triangle.getV2());
Vector3D v2 = new Vector3D(triangle.getV3(), triangle.getV2());
return v1.crossProduct(v2);
}
Example of the incorrectly drawn teapot:
What am I doing wrong? I feel like it must be some small thing. Given that the triangles draw at all, I doubt it's the pointWithinTriangle method. Backface culling also appears to work correctly, so I doubt it's that. The most likely culprit to me is the zValueOfPoint method, but I don't know enough to know what's wrong with it.
My zValueOfPoint method was not working correctly. I'm unsure why :( however, I changed to a slightly different method of calculating the value of a point in a plane, found here: http://forum.devmaster.net/t/interpolation-on-a-3d-triangle-using-normals/20610/5
To make the answer here complete, we have the equation of a plane:
A * x + B * y + C * z + D = 0
Where A, B, and C are the surface normal x/y/z values, and D is -(Ax0 + By0 + Cz0).
x0, y0, and z0 are taken from one of the vertices of the triangle. x, y, and z are the coordinates of the point where the ray intersects the plane. x and y are known values (rayOrigin.x, rayOrigin.y) but z is the depth which we need to calculate. From the above equation we derive:
z = -A / C * x - B / C * y - D
Then, copied from the above link, we do:
"Note that for every step in the x-direction, z increments by -A / C, and likewise it increments by -B / C for every step in the y-direction.
So these are the gradients we're looking for to perform linear interpolation. In the plane equation (A, B, C) is the normal vector of the plane.
It can easily be computed with a cross product.
Now that we have the gradients, let's call them dz/dx (which is -A / C) and dz/dy (which is -B / C), we can easily compute z everywhere on the triangle.
We know the z value in all three vertex positions.
Let's call the one of the first vertex z0, and it's position coordinates (x0, y0). Then a generic z value of a point (x, y) can be computed as:"
z = z0 + dz/dx * (x - x0) + dz/dy * (y - y0)
This found the Z value correctly and fixed my code. The new zValueOfPoint method is:
private double zValueOfPoint(Triangle triangle, Point2D rayOrigin)
{
Vector3D surfaceNormal = getNormal(triangle);
double A = surfaceNormal.x;
double B = surfaceNormal.y;
double C = surfaceNormal.z;
double dzdx = -A / C;
double dzdy = -B / C;
double rayZ = triangle.getV1().z * modelViewer.getModelScale() + dzdx * (rayOrigin.x - triangle.getV1().projectPoint(modelViewer).x) + dzdy * (rayOrigin.y - triangle.getV1().projectPoint(modelViewer).y);
return rayZ;
}
We can optimize this by only calculating most of it once, and then adding dz/dx to get the z value for the next pixel, or dz/dy for the pixel below (with the y-axis going down). This means that we cut down on calculations per polygon significantly.
this must be really slow
so much redundant computations per iteration/pixel just to iterate its coordinates. You should compute the 3 projected vertexes and iterate between them instead look here:
triangle/convex polygon rasterization
I dislike your zValueOfPoint function
can not find any use of x,y coordinates from the main loops in it so how it can compute the Z value correctly ?
Or it just computes the average Z value per whole triangle ? or am I missing something? (not a JAVA coder myself) in anyway it seems that this is your main problem.
if you Z-value is wrongly computed then Z-Buffer can not work properly. To test that look at the depth buffer as image after rendering if it is not shaded teapot but some incoherent or constant mess instead then it is clear ...
Z buffer implementation
That looks OK
[Hints]
You have too much times terms like x + getWidth()/2 why not compute them just once to some variable? I know modern compilers should do it anyway but the code would be also more readable and shorter... at least for me
I'm sorry if this question was asked before, I did search, and I did not find an answer.
My problem is, that I'd like to make movement on all 3 axes with the X and Y rotation of the camera being relevant.
This is what I did:
private static void fly(int addX, int addY){ //parameters are the direction change relative to the current rotation
float angleX = rotation.x + addX; //angle is basically the direction, into which we will be moving(when moving forward this is always the same as our actual rotation, therefore addX and addY would be 0, 0)
float angleY = rotation.y + addY;
float speed = (moveSpeed * 0.0002f) * delta;
float hypotenuse = speed; //the length that is SUPPOSED TO BE moved overall on all 3 axes
/* Y-Z side*/
//Hypotenuse, Adjacent and Opposite side lengths of a triangle on the Y-Z side
//The point where the Hypotenuse and the Adjacent meet is where the player currently is.
//OppYZ is the opposite of this triangle, which is the ammount that should be moved on the Y axis.
//the Adjacent is not used, don't get confused by it. I just put it there, so it looks nicer.
float HypYZ = speed;
float AdjYZ = (float) (HypYZ * Math.cos(Math.toRadians(angleX))); //adjacent is on the Z axis
float OppYZ = (float) (HypYZ * Math.sin(Math.toRadians(angleX))); //opposite is on the Y axis
/* X-Z side*/
//Side lengths of a triangle on the Y-Z side
//The point where the Hypotenuse and the Adjacent meet is where the player currently is.
float HypXZ = speed;
float AdjXZ = (float) (HypXZ * Math.cos(Math.toRadians(angleY))); //on X
float OppXZ = (float) (HypXZ * Math.sin(Math.toRadians(angleY))); //on Z
position.x += AdjXZ;
position.y += OppYZ;
position.z += OppXZ;
}
I only implement this method when moving forwards(parameters: 0, 90) or backwards(params: 180, 270), since movement can't happen on the Y axis while going sideways, since you don't rotate on the Z axis. ( the method for going sideways(strafing) works just fine, so I won't add that.)
the problem is that when I look 90 degrees up or -90 down and then move forward I should be moving only on the Y axis(vertically) but for some reason I also move forwards(which means on the Z axis, as the X axis is the strafing).
I do realize that movement speed this way is not constant. If you have a solution for that, I'd gladly accept it as well.
I think your error lies in the fact that you don't fully project your distance (your quantity of movement hypothenuse) on your horizontal plane and vertical one.
In other words, whatever the chosen direction, what you are doing right now is moving your point of hypothenuse in the horizontal plane X-Z, even though you already move it of a portion of hypothenuse in the vertical direction Y.
What you probably want to do is moving your point of a hypothenuse quantity as a total.
So you have to evaluate how much of the movement takes place in the horizontal plane and how much in the vertical axis. Your direction gives you the answer.
Now, it is not clear to me right now what your 2 angles represent. I highly recommend you to use Tait–Bryan angles in this situation (using only yawn and pitch, since you don't seem to need the rolling - what you call the Z-rotation), to simplify the calculations.
In this configuration, the yawn angle would be apparently similar to your definition of your angleY, while the pitch angle would be the angle between the horizontal plane and your hypothenuse vector (and not the angle of the projection in the plane Y-Z).
A schema to clarify:
With :
s your quantity of movement from your initial position P_0 to P_1 (hypothenuse)
a_y the yawn angle and a_p the pitch one
D_x, D_y, D_z the displacements for each axis (to be added to position, ie AdjXZ, OppYZ and OppXZ)
So if you look at this representation, you can see that your triangle in X-Z doesn't have s as hypotenuse but its projection s_xz. The evaluation of this distance is quite straightforward: if you place yourself in the triangle P_0 P_1 P_1xz, you can see that s_xz = s * cos(a_p). Which gives you:
float HypXZ = speed * Math.cos(Math.toRadians(angleP))); // s_xz
float AdjXZ = (float) (HypXZ * Math.cos(Math.toRadians(angleY))); // D_x
float OppXZ = (float) (HypXZ * Math.sin(Math.toRadians(angleY))); // D_z
As for D_y ie OppYZ, place yourself in the triangle P_0 P_1 P_1xz again, and you'll obtain:
float OppYZ = (float) (speed * Math.sin(Math.toRadians(angleP))); // D_y
Now, if by angleX you actually meant the angle of elevation as I suppose you did, then angleP = angleX and HypXZ = AdjYZ in your code.
With this correction, if angleX = 90 or angleX = -90, then
HypXZ = speed * cos(angleX) = speed * cos(90deg) = speed * 0;
... and thus AdjXZ = 0 and OppXZ = 0. No movement in the horizontal plane.
Note:
To check if your calculations are correct, you can verify if you actually move your point of the wanted quantity of movement (hypothenuse ie speed ie s). Using Pythagorean theorem:
s² = s_xz² + D_z² // Applied in the triangle P_0 P_1 P_1xz
= D_x² + D_y² + D_z² // Applied in the triangle P_0 P_1x P_1xz
With the definitions of the displacements given above:
D_x² + D_y² + D_z²
= (s * cos(a_p) * cos(a_y))² + (s * cos(a_p) * sin(a_y))² + (s * sin(a_p))²
= s² * (cos(a_p)² * cos(a_y)² + cos(a_p)² * sin(a_y)² + sin(a_p)²)
= s² * (cos(a_p)² * (cos(a_y)² + sin(a_y)²) + sin(a_p)²)
= s² * (cos(a_p)² * 1 + sin(a_p)²)
= s² * (cos(a_p)² + sin(a_p)²)
= s² * 1 // Correct
Hope it helped... Bye!