Develop Reference

Directional helix for game missiles

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.

Calculate angle between 3 points with vector, more precision needed (java)

I need to calculate the angle between 3 points. I've done that using vectors, it looks like it's working, but sometimes I get NaN as a result. To calculate the angle I used the arcos(dot(v1,v2)/(length(v1)*length(v2))) formula. Here's the code:
private static double angleBetween(Point previous, Point center, Point next) {
Vector2D vCenter = new Vector2D(center.x, center.y );
Vector2D vPrev = new Vector2D(previous.x, previous.y );
Vector2D vNext = new Vector2D(next.x, next.y );
Vector2D a = vCenter.minus(vPrev);
Vector2D b = vCenter.minus(vNext);
double dot = Vector2D.dot(a, b);
double la = a.length();
double lb = b.length();
double l = la*lb;
double acos = Math.acos(dot/l);
double deg = Math.toDegrees(acos);
return deg;
}
Vector2D class:
public double length() {
return Math.sqrt(x * x + y * y);
};
public static double dot(Vector2D v1, Vector2D v2) {
return v1.x * v2.x + v1.y * v2.y;
};
public Vector2D minus(Vector2D v) {
return new Vector2D(x - v.x, y - v.y);
};
Debugging the program I've discovered why this happens. for example let be:
center = (127,356)
previous = (117,358)
next = (137,354)
//calculating the vectors
a = (-10,2) //center - prev
b = (10,-2) //center - next
dot = -10*10 + 2*-2 = 104
la = sqrt(-10*-10 + 2*2) = sqrt(104) = see attachment
lb = sqrt(10*10 + -2*-2) = sqrt(104) = see attachment
l = la * lb = see attachment
acos = NaN because dot/l>1 because I lost precision because of sqrt() which didn't give me the exact value therefore la*lb isn't 104.
now as far as I know double is the most precise number type in java. How can I solve this problem?
PS It may looks like a very rare situation, but I'm experiencing quite a lot of them, so I can't just ignore it.

The best way to solve this problem is to use an appropriate data type like java.math.BigDecimal and define a precision for you computation using an instance of type java.math.MathContext. For instance:
double l = la*lb;
BigDecimal lWrapper = new BigDecimal(l, new MathContext(5));
l = lWrapper.doubleValue();
There is an other way to work around this problem. Use the following formula:
angle = atan(length(crossProduct(a, b)) / dotProduct(a, b)) // Because the domain of definition of the tan function is R
Derivation of the formula:
cos(angle) = dotProduct(a, b) / (length(a) * length(b)) and
sin(angle) = length(crossProduct(a, b)) / (length(a) * length(b))
One has
tan(angle) = sin(angle) / cos(angle)
so
tan(angle) = length(crossProduct(a, b)) / (length(a) * length(b)) / dotProduct(a, b) / (length(a) * length(b))
tan(angle) = length(crossProduct(a, b)) / dotProduct(a, b)
Applying the invert function of tan:
angle = atan(length(crossProduct(a, b)) / dotProduct(a, b))
The cross product of A, B ∈ ℜ2:
|| A x B || = det(A,B) = ((A.x * B.y) - (A.y * B.x))
Remarks:
||x|| is the length of the vector x ⇔ length(a)
∀ A, B ∈ ℜ2: || A x B || equal the determinant of A, B
You can use the sign(|| A x B ||) to find out the orientation

The angle between vectors that are colinear is 180 degrees. Using arccos won't work very well because arccos uses pythagoras' theorem to derive the angle and pythagoras' theorem only works with triangles. Two colinear vectors do not make a triangle.
The simplest answer is to check for Nan and special case it.
double deg = Math.toDegrees(Double.isNaN(acos) ? Math.PI : acos);

Z-buffering algorithm not drawing 100% correctly

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

Triangle Coordinate Geometry

I have a triangle ABC inscribed in a circle. Point B is located in the centre of the circle, A and C are two points on the circle.
Given
Given AB (length)
Given coords of A and B
Given angle B (angle ABC)
Needed
Find coords of C
What I know
AB = BC, both are radius's
What I am using this for
I am making a basic 3D java game, for android. This will be used for looking left and right, so if you click on the right part of the screen the objects will move around you by adding one degree to angle B.
The code I tried for finding coords of C
rect.get(index)[5] = (int) ((di * Math.cos(Math.toRadians(angle-90)))+.5);
rect.get(index)[6] = (int) ((di * Math.sin(Math.toRadians(angle-90)))+.5);
rect.get(index)[5] = shapes x coord
rect.get(index)[6] = shapes y coord
di = radius length
angle = angle B
and I added the .5 so that when the coord is truncated it is rounded.
My complete code
double di = distance(playerx, playery, rect.get(index)[5], rect.get(index)[6]);
double side1 = di;
System.out.println("Side1: "+ side1);
double side2 = side1;
System.out.println("Side2: "+ side2);
double side3 = distance(rect.get(index)[5], rect.get(index)[6], playerx, playery+di);
System.out.println("Side3: "+ side3);
double angle = ((side1*side1)+(side2*side2)-(side3*side3));
angle = angle/(2*side1*side2);
angle = Math.acos(angle)*(180/Math.PI);
System.out.println("Angle: "+angle);
if(playerx > rect.get(index)[5]){
if(lookdirection.equals("left")){
angle += 5;
}
if(lookdirection.equals("right")){
angle -= 5;
}
}
else{
if(lookdirection.equals("left")){
angle -= 5;
}
if(lookdirection.equals("right")){
angle += 5;
}
}
System.out.println("Angle: "+angle);
rect.get(index)[5] = -(di * Math.cos(Math.toRadians(angle-90)));
rect.get(index)[6] = -(di * Math.sin(Math.toRadians(angle-90)));
di = distance(playerx, playery, rect.get(index)[5], rect.get(index)[6]);
side1 = di;
System.out.println("Side1: "+ side1);
side2 = side1;
System.out.println("Side2: "+ side2);
side3 = distance(rect.get(index)[5], rect.get(index)[6], playerx, playery+di);
System.out.println("Side3: "+ side3);
angle = ((side1*side1)+(side2*side2)-(side3*side3));
angle = angle/(2*side1*side2);
angle = Math.acos(angle)*(180/Math.PI);
System.out.println("Angle: "+angle);
repaint();
}
The angles are now working but the X and Y coords
but
rect.get(index)[5] = -(di * Math.cos(Math.toRadians(angle-90)));
rect.get(index)[6] = -(di * Math.sin(Math.toRadians(angle-90)));
are getting very large/ small values. They should stay 'di' distance away from point B.

UPDATED
It seems that this is a geometry problem!
I haven't tested your code but I guess what you want is not
rect.get(index)[6] = (int) ((di * Math.sin(Math.toRadians(angle-90)))+.5);
but
rect.get(index)[6] = (int) (-(di * Math.sin(Math.toRadians(angle-90)))+.5);
Try to visualise what you are doing and check the identities under symmetry, shifts, and periodicity here. The image is turning and reflecting along the x axis with what you are doing now. You have to take care of rounding issues as well. If coordinates are close to 0 the image might not move.
My suggestion is to keep your coordinates as doubles and only round them when you are going to render them into a pixel without overwriting their value

Better way to go up/down slope based on yaw?

Alright, so I got a bit of movement code and I'm thinking I'm going to need to manually input when to go up/down a slope. All I got to work with is the slope's normal, and vector, and My current and previous position, and my yaw.
Is there a better way to rotate whether I go up or down the slope based on my yaw?
Vector3f move = new Vector3f(0,0,0);
move.x = (float)-Math.cos(Math.toRadians(yaw));
move.z = (float)-Math.sin(Math.toRadians(yaw));
System.out.println("slopeNormal.z: " + slopeNormal.z + "move.z: " + move.z);
move.normalise();
float vx = (float) (Math.sqrt(Math.pow(move.y, 2) + Math.pow(move.z, 2)) * move.x);
float vy = (float) (Math.sqrt(Math.pow(move.x, 2) + Math.pow(move.z, 2)) * move.y);
float vz = - vx * slopeNormal.x - vy * slopeNormal.y;
move.scale(movementSpeed * delta);
if(vz < 0)
move.y -= slopeVec.y * 1.5f;
if(vz > 0)
move.y += slopeVec.y * 1.5f;
Vector3f.add(pos, move, pos);
Edit: updated code.

First off, the following is incorrect:
move.x = (float)-Math.toDegrees(Math.cos(Math.toRadians(yaw)));
move.z = (float)-Math.toDegrees(Math.sin(Math.toRadians(yaw)));
Math.toDegrees converts an angle in radians to one in degrees, but the results of Math.cos and Math.sin are not angles.
Assume zero yaw is in the positive x-direction... and define vx, vy, vz = rate of motion along 3 axes, s = speed, and slope normal = nx, ny, nz where nx^2 + ny^2 + nz^2 = 1. So nx = ny = 0, nz = 1 would be flat.
First, I define x', y' = axes relative to the flat ground (motion is constrained to ground). Then (the following is not valid Java, but I'm enclosing it in code format anyway):
vx' = cos(yaw) * s
vy' = sin(yaw) * s
Then I need to rotate from x', y' coordinates to real-world coordinates. That is done using the slope normal:
vx = sqrt(vy^2 + vz^2) vx'
vy = sqrt(vx^2 + vz^2) vy'
vz = - vx' nx - vy' ny
A check on this transformation: vx^2 + vy^2 + vz^2 must equal vx'^2 + vy'^2 = s^2. I think this works out.
So to answer your question: up or down? vz > 0 is up, vz < 0 is down.