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I want to find the center point of an arc.
I have start point, end point, and offset of an arc.
I have tried the below code. With this code, I got the center point of arc in a 90% case. But for certain data its fail to find center point.
I have this definition of arc drawing
X169290Y2681101I90207J39371*
X267716Y2779527J98426*
X169290Y2877952I98425*
X79082Y2818897J98425*
The answer for all above four arc center coordinate is (4.299940599999999, 70.5999858)
This is an arc image, This arc is made from four single-quadrant arcs.
private Coordinate findCenter(Coordinate start, Coordinate end, Coordinate offset) {
double twoPi = 2 * Math.PI;
if (quadrantMode.equals("single-quadrant")) {
// The Gerber spec says single quadrant only has one possible center,
// and you can detect it based on the angle. But for real files, this
// seems to work better - there is usually only one option that makes
// sense for the center (since the distance should be the same
// from start and end). We select the center with the least error in
// radius from all the options with a valid sweep angle.
double sqdistDiffMin = Double.MAX_VALUE;
Coordinate center = null;
List<Coordinate> qFactors = new ArrayList<Coordinate>();
qFactors.add(new Coordinate(1, 1));
qFactors.add(new Coordinate(1, -1));
qFactors.add(new Coordinate(-1, 1));
qFactors.add(new Coordinate(-1, -1));
for (Coordinate factors : qFactors) {
Coordinate testCenter = new Coordinate(start.getX() + offset.getX() * factors.getX(),
start.getY() + offset.getY() * factors.getY());
// Find angle from center to start and end points
double startAngleX = start.getX() - testCenter.getX();
double startAngleY = start.getY() - testCenter.getY();
double startAngle = Math.atan2(startAngleY, startAngleX);
double endAngleX = end.getX() - testCenter.getX();
double endAngleY = end.getY() - testCenter.getY();
double endAngle = Math.atan2(endAngleY, endAngleX);
// # Clamp angles to 0, 2pi
double theta0 = (startAngle + twoPi) % twoPi;
double theta1 = (endAngle + twoPi) % twoPi;
// # Determine sweep angle in the current arc direction
double sweepAngle;
if (direction.equals("counterclockwise") ) {
theta1 += twoPi;
// sweepAngle = Math.abs(theta1 - theta0);
sweepAngle = Math.abs(theta1 - theta0) % twoPi;
} else {
theta0 += twoPi;
sweepAngle = Math.abs(theta0 - theta1) % twoPi;
}
// # Calculate the radius error
double sqdistStart = sqDistance(start, testCenter);
double sqdistEnd = sqDistance(end, testCenter);
double sqdistDiff = Math.abs(sqdistStart - sqdistEnd);
// Take the option with the lowest radius error from the set of
// options with a valid sweep angle
// In some rare cases, the sweep angle is numerically (10**-14) above pi/2
// So it is safer to compare the angles with some tolerance
boolean isLowestRadiusError = sqdistDiff < sqdistDiffMin;
boolean isValidSweepAngle = sweepAngle >= 0 && sweepAngle <= Math.PI / 2.0 + 1e-6;
if (isLowestRadiusError && isValidSweepAngle) {
center = testCenter;
sqdistDiffMin = sqdistDiff;
}
}
return center;
}
else {
return new Coordinate(start.getX() + offset.getX(), start.getY() + offset.getY());
}
}
public static double sqDistance(Coordinate point1, Coordinate point2) {
double diff1 = point1.getX() - point2.getX();
double diff2 = point1.getY() - point2.getY();
return diff1 * diff1 + diff2 * diff2;
}
Explaination graph for arc drawing
You are dealing with an accuracy problem, caused by the value 1e-6. I'd advise you to do two things:
Check what happens when you replace this value by zero (any side-effects which might cause problems?)
In case zero value does not work, try to replace by a value, smaller than 1e-6, like 1e-9, 1e-12, ...
But most of all: you need to do some exhaustive testing while modifying that value (do you have a testlist that needs to be covered + invent some realistic testing cases).
I am working on the problem of dividing an ellipse into equal sized segments. This question has been asked but the answers suggested numerical integration so that I what I'm attempting. This code short-circuits the sectors so the integration itself should never cover more than 90 degrees. The integration itself is being done by totaling the area of intermediate triangles. Below is the code I have tried, but it is sweeping more than 90 degrees in some cases.
public class EllipseModel {
protected double r_x;
protected double r_y;
private double a,a2;
private double b,b2;
boolean flip;
double area;
double sector_area;
double radstep;
double rot;
int xp,yp;
double deviation;
public EllipseModel(double r_x, double r_y, double deviation)
{
this.r_x = r_x;
this.r_y = r_y;
this.deviation = deviation;
if (r_x < r_y) {
flip = true;
a = r_y;
b = r_x;
xp = 1;
yp = 0;
rot = Math.PI/2d;
} else {
flip = false;
xp = 0;
yp = 1;
a = r_x;
b = r_y;
rot = 0d;
}
a2 = a * a;
b2 = b * b;
area = Math.PI * r_x * r_y;
sector_area = area / 4d;
radstep = (2d * deviation) / a;
}
public double getArea() {
return area;
}
public double[] getSweep(double sweep_area)
{
System.out.println(String.format("getSweep(%f) a = %f b = %f deviation = %f",sweep_area,a,b,deviation));
double[] ret = new double[2];
double[] next = new double[2];
double t_base, t_height, swept,x_mid,y_mid;
double t_area;
sweep_area = sweep_area % area;
if (sweep_area < 0d) {
sweep_area = area + sweep_area;
}
if (sweep_area == 0d) {
ret[0] = r_x;
ret[1] = 0d;
return ret;
}
double sector = Math.floor(sweep_area/sector_area);
double theta = Math.PI * sector/2d;
double theta_last = theta;
System.out.println(String.format("- Theta start = %f",Math.toDegrees(theta)));
ret[xp] = a * Math.cos(theta + rot);
ret[yp] = (1 + (((theta / Math.PI) % 2d) * -2d)) * Math.sqrt((1 - ( (ret[xp] * ret[xp])/a2)) * b2);
next[0] = ret[0];
next[1] = ret[1];
swept = sector * sector_area;
System.out.println(String.format("- Sweeping for %f sector_area=%f",sweep_area-swept,sector_area));
int c = 0;
while(swept < sweep_area) {
c++;
ret[0] = next[0];
ret[1] = next[1];
theta_last = theta;
theta += radstep;
// calculate next point
next[xp] = a * Math.cos(theta + rot);
next[yp] = (1 + (((theta / Math.PI) % 2d) * -2d)) * // selects +/- sqrt
Math.sqrt((1 - ( (ret[xp] * ret[xp])/a2)) * b2);
// calculate midpoint
x_mid = (ret[xp] + next[xp]) / 2d;
y_mid = (ret[yp] + next[yp]) / 2d;
// calculate triangle metrics
t_base = Math.sqrt( ( (ret[0] - next[0]) * (ret[0] - next[0]) ) + ( (ret[1] - next[1]) * (ret[1] - next[1])));
t_height = Math.sqrt((x_mid * x_mid) + (y_mid * y_mid));
// add triangle area to swept
t_area = 0.5d * t_base * t_height;
swept += t_area;
}
System.out.println(String.format("- Theta end = %f (%d)",Math.toDegrees(theta_last),c));
return ret;
}
}
In the output I see the following case where it sweeps over 116 degrees.
getSweep(40840.704497) a = 325.000000 b = 200.000000 deviation = 0.166667
- Theta start = 0.000000
- Sweeping for 40840.704497 sector_area=51050.880621
- Theta end = 116.354506 (1981)
Is there any way to fix the integration formula to create a function that returns the point on an ellipse that has swept a given area? The application that is using this code divides the total area by the number of segments needed, and then uses this code to determine the angle where each segment starts and ends. Unfortunately it doesn't work as intended.
* edit *
I believe the above integration failed because the base and height formula's aren't correct.
No transformation needed use parametric equations for ellipse ...
x=x0+rx*cos(a)
y=y0+ry*sin(a)
where a = < 0 , 2.0*M_PI >
if you divide ellipse by lines from center to x,y from above equation
and angle a is evenly encreased
then the segments will have the same size
btw. if you apply affine transform you will get the same result (even the same equation)
This code will divide ellipse to evenly sized chunks:
double a,da,x,y,x0=0,y0=0,rx=50,ry=20; // ellipse x0,y0,rx,ry
int i,N=32; // divided to N = segments
da=2.0*M_PI/double(N);
for (a=0.0,i=0;i<N;i++,a+=da)
{
x=x0+(rx*cos(a));
y=y0+(ry*sin(a));
// draw_line(x0,y0,x,y);
}
This is what it looks like for N=5
[edit1]
I do not understood from your comment what exactly you want to achieve now
sorry but my English skills are horrible
ok I assume these two possibilities (if you need something different please specify closer)
0.but first some global or member stuff needed
double x0,y0,rx,ry; // ellipse parameters
// [Edit2] sorry forgot to add these constants but they are I thin straight forward
const double pi=M_PI;
const double pi2=2.0*M_PI;
// [/Edit2]
double atanxy(double x,double y) // atan2 return < 0 , 2.0*M_PI >
{
int sx,sy;
double a;
const double _zero=1.0e-30;
sx=0; if (x<-_zero) sx=-1; if (x>+_zero) sx=+1;
sy=0; if (y<-_zero) sy=-1; if (y>+_zero) sy=+1;
if ((sy==0)&&(sx==0)) return 0;
if ((sx==0)&&(sy> 0)) return 0.5*pi;
if ((sx==0)&&(sy< 0)) return 1.5*pi;
if ((sy==0)&&(sx> 0)) return 0;
if ((sy==0)&&(sx< 0)) return pi;
a=y/x; if (a<0) a=-a;
a=atan(a);
if ((x>0)&&(y>0)) a=a;
if ((x<0)&&(y>0)) a=pi-a;
if ((x<0)&&(y<0)) a=pi+a;
if ((x>0)&&(y<0)) a=pi2-a;
return a;
}
1.is point inside segment ?
bool is_pnt_in_segment(double x,double y,int segment,int segments)
{
double a;
a=atanxy(x-x0,y-y0); // get sweep angle
a/=2.0*M_PI; // convert angle to a = <0,1>
if (a>=1.0) a=0.0; // handle extreme case where a was = 2 Pi
a*=segments; // convert to segment index a = <0,segments)
a-=double(segment );
// return floor(a); // this is how to change this function to return points segment id
// of course header should be slightly different: int get_pnt_segment_id(double x,double y,int segments)
if (a< 0.0) return false; // is lower then segment
if (a>=1.0) return false; // is higher then segment
return true;
}
2.get edge point of segment area
void get_edge_pnt(double &x,double &y,int segment,int segments)
{
double a;
a=2.0*M_PI/double(segments);
a*=double(segment); // this is segments start edge point
//a*=double(segment+1); // this is segments end edge point
x=x0+(rx*cos(a));
y=y0+(ry*sin(a));
}
for booth:
x,y is point
segments number of division segments.
segment is sweep-ed area < 0,segments )
Apply an affine transformation to turn your ellipse into a circle, preferrably the unit circle. Then split that into equal sized segments, before you apply the inverse transform. The transformation will scale all areas (as opposed to lengths) by the same factor, so equal area translates to equal area.
Since in the digital world a real collision almost never happens, we will always have a situation where the "colliding" circle overlaps the rectangle.
How to put back the circle in the situation where it collides perfectly with the rectangle without overlap?
Suppose that the rectangle is stopped (null velocity) and axis-aligned.
I would solve this problem with a posteriori approach (in two dimensions).
In short I have to solve this equation for t:
Where:
is a number that answers to the question: how many frames ago did the
collision happen perfectly?
is the radius of the circle.
is the center of the circle
is its velocity.
and are functions that return the x and y coordinates of
the point where the circle and the rectangle collide (when the circle is
at position, that is in the position in which perfectly collide with the rectangle).
Recently I solved a similar problem for collisions between circles, but now I don't know the law of the functions A and B.
After years of staring at this problem, and never coming up with a perfect solution, I've finally done it!
It's pretty much a straight forward algorithm, no need for looping and approximations.
This is how it works at a higher level:
Calculate intersection times with each side's plane IF the path from current point to future point crosses that plane.
Check each side's quadrant for single-side intersection, return the intersection.
Determine the corner that the circle is colliding with.
Solve the triangle between the current point, the corner, and the intersecting center (radius away from the corner).
Calculate time, normal, and intersection center.
And now to the gory details!
The input to the function is bounds (which has a left, top, right, bottom) and a current point (start) and a future point (end).
The output is a class called Intersection which has x, y, time, nx, and ny.
{x, y} is the center of the circle at intersection time.
time is a value from 0 to 1 where 0 is at start and 1 is at end
{nx, ny} is the normal, used for reflecting the velocity to determine the new velocity of the circle
We start off with caching variables we use often:
float L = bounds.left;
float T = bounds.top;
float R = bounds.right;
float B = bounds.bottom;
float dx = end.x - start.x;
float dy = end.y - start.y;
And calculating intersection times with each side's plane (if the vector between start and end pass over that plane):
float ltime = Float.MAX_VALUE;
float rtime = Float.MAX_VALUE;
float ttime = Float.MAX_VALUE;
float btime = Float.MAX_VALUE;
if (start.x - radius < L && end.x + radius > L) {
ltime = ((L - radius) - start.x) / dx;
}
if (start.x + radius > R && end.x - radius < R) {
rtime = (start.x - (R + radius)) / -dx;
}
if (start.y - radius < T && end.y + radius > T) {
ttime = ((T - radius) - start.y) / dy;
}
if (start.y + radius > B && end.y - radius < B) {
btime = (start.y - (B + radius)) / -dy;
}
Now we try to see if it's strictly a side intersection (and not corner). If the point of collision lies on the side then return the intersection:
if (ltime >= 0.0f && ltime <= 1.0f) {
float ly = dy * ltime + start.y;
if (ly >= T && ly <= B) {
return new Intersection( dx * ltime + start.x, ly, ltime, -1, 0 );
}
}
else if (rtime >= 0.0f && rtime <= 1.0f) {
float ry = dy * rtime + start.y;
if (ry >= T && ry <= B) {
return new Intersection( dx * rtime + start.x, ry, rtime, 1, 0 );
}
}
if (ttime >= 0.0f && ttime <= 1.0f) {
float tx = dx * ttime + start.x;
if (tx >= L && tx <= R) {
return new Intersection( tx, dy * ttime + start.y, ttime, 0, -1 );
}
}
else if (btime >= 0.0f && btime <= 1.0f) {
float bx = dx * btime + start.x;
if (bx >= L && bx <= R) {
return new Intersection( bx, dy * btime + start.y, btime, 0, 1 );
}
}
We've gotten this far so we know either there's no intersection, or it's collided with a corner. We need to determine the corner:
float cornerX = Float.MAX_VALUE;
float cornerY = Float.MAX_VALUE;
if (ltime != Float.MAX_VALUE) {
cornerX = L;
} else if (rtime != Float.MAX_VALUE) {
cornerX = R;
}
if (ttime != Float.MAX_VALUE) {
cornerY = T;
} else if (btime != Float.MAX_VALUE) {
cornerY = B;
}
// Account for the times where we don't pass over a side but we do hit it's corner
if (cornerX != Float.MAX_VALUE && cornerY == Float.MAX_VALUE) {
cornerY = (dy > 0.0f ? B : T);
}
if (cornerY != Float.MAX_VALUE && cornerX == Float.MAX_VALUE) {
cornerX = (dx > 0.0f ? R : L);
}
Now we have enough information to solve for the triangle. This uses the distance formula, finding the angle between two vectors, and the law of sines (twice):
double inverseRadius = 1.0 / radius;
double lineLength = Math.sqrt( dx * dx + dy * dy );
double cornerdx = cornerX - start.x;
double cornerdy = cornerY - start.y;
double cornerdist = Math.sqrt( cornerdx * cornerdx + cornerdy * cornerdy );
double innerAngle = Math.acos( (cornerdx * dx + cornerdy * dy) / (lineLength * cornerdist) );
double innerAngleSin = Math.sin( innerAngle );
double angle1Sin = innerAngleSin * cornerdist * inverseRadius;
// The angle is too large, there cannot be an intersection
if (Math.abs( angle1Sin ) > 1.0f) {
return null;
}
double angle1 = Math.PI - Math.asin( angle1Sin );
double angle2 = Math.PI - innerAngle - angle1;
double intersectionDistance = radius * Math.sin( angle2 ) / innerAngleSin;
Now that we solved for all sides and angles, we can determine time and everything else:
// Solve for time
float time = (float)(intersectionDistance / lineLength);
// If time is outside the boundaries, return null. This algorithm can
// return a negative time which indicates the previous intersection.
if (time > 1.0f || time < 0.0f) {
return null;
}
// Solve the intersection and normal
float ix = time * dx + start.x;
float iy = time * dy + start.y;
float nx = (float)((ix - cornerX) * inverseRadius);
float ny = (float)((iy - cornerY) * inverseRadius);
return new Intersection( ix, iy, time, nx, ny );
Woo! That was fun... this has plenty of room for improvements as far as efficiency goes. You could reorder the side intersection checking to escape as early as possible while making as few calculations as possible.
I was hoping there would be a way to do it without trigonometric functions, but I had to give in!
Here's an example of me calling it and using it to calculate the new position of the circle using the normal to reflect and the intersection time to calculate the magnitude of reflection:
Intersection inter = handleIntersection( bounds, start, end, radius );
if (inter != null)
{
// Project Future Position
float remainingTime = 1.0f - inter.time;
float dx = end.x - start.x;
float dy = end.y - start.y;
float dot = dx * inter.nx + dy * inter.ny;
float ndx = dx - 2 * dot * inter.nx;
float ndy = dy - 2 * dot * inter.ny;
float newx = inter.x + ndx * remainingTime;
float newy = inter.y + ndy * remainingTime;
// new circle position = {newx, newy}
}
And I've posted the full code on pastebin with a completely interactive example where you can plot the starting and ending points and it shows you the time and resulting bounce off of the rectangle.
If you want to get it running right away you'll have to download code from my blog, otherwise stick it in your own Java2D application.
EDIT:
I've modified the code in pastebin to also include the collision point, and also made some speed improvements.
EDIT:
You can modify this for a rotating rectangle by using that rectangle's angle to un-rotate the rectangle with the circle start and end points. You'll perform the intersection check and then rotate the resulting points and normals.
EDIT:
I modified the code on pastebin to exit early if the bounding volume of the path of the circle does not intersect with the rectangle.
Finding the moment of contact isn't too hard:
You need the position of the circle and rectangle at the timestep before the collision (B) and the timestep after (A). Calculate the distance from the center of the circle to the line of the rectangle it collides with at times A and B (ie, a common formula for a distance from a point to a line), and then the time of collision is:
tC = dt*(dB-R)/(dA+dB),
where tC is the time of collision, dt is the timestep, dB is the distance to line before the collision, dA is the distance after the collision, and R is the radius of the circle.
This assumes everything is locally linear, that is, that your timesteps are reasonably small, and so that the velocity, etc, don't change much in the timestep where you calculate the collision. This is, after all, the point of timesteps: in that with a small enough timestep, non-linear problems are locally linear. In the equation above I take advantage of that: dB-R is the distance from the circle to the line, and dA+dB is the total distance moved, so this question just equates the distance ratio to the time ratio assuming everything is approximately linear within the timestep. (Of course, at the moment of collision the linear approximation isn't its best, but to find the moment of collision, the question is whether it's linear within a timestep up to to moment of collision.)
It's a non-linear problem, right?
You take a time step and move the ball by its displacement calculated using velocity at the start of the step. If you find overlap, reduce the step size and recalculate til convergence.
Are you assuming that the balls and rectangles are both rigid, no deformation? Frictionless contact? How will you handle the motion of the ball after contact is made? Are you transforming to a coordinate system of the contact (normal + tangential), calculating, then transforming back?
It's not a trivial problem.
Maybe you should look into a physics engine, like Box2D, rather than coding it yourself.
Given two bearing, how do I find the smallest angle between them?
So for example if 1 heading is 340 degrees and the second is 10 degrees the smallest angle will be 30 degrees.
I've attached a picture to show what I mean. I've tried subtracting one from the other but that didn't work because of the wrap around effect of a circle. I've also tried using negative degrees (180 - 359 being -180 to 0) but that got messed up when trying to calculate the angle between positive and negative number.
I'm sure there must be an easier way that having lots of if statements.
Thank for your help.
Adam
BTW. This is a navigation question so the radius of the circle is unknown.
I ended up using the following formula found on this message board because I needed the result to be signed based on the direction (clockwise or counterclockwise). It has a good explanation of exactly what's going on.
((((bearing - heading) % 360) + 540) % 360) - 180
float getDifference(float a1, float a2) {
return Math.min((a1-a2)<0?a1-a2+360:a1-a2, (a2-a1)<0?a2-a1+360:a2-a1)
}
What about:
angle = Math.abs(a1-a2);
if (angle > 180)
angle = 360 - angle;
You mention an issue regarding positive and negative numbers, so perhaps there is something I'm not considering here...
For navigation, you might want to know if b1 is left or right of b2, so here it is in a nice function. (Assumes exactly 0 is not a use case)
function isBearing1LeftOrRightOfBearing2 (b1, b2) {
if (Math.sign(((b1 - b2 + 540) % 360) - 180) > 0) {
return 'left'
} else {
return 'right'
}
}
If angle direction is needed, then this will work:
int maxBearing = Math.max(bearing0, bearing1);
int minBearing = Math.min(bearing0, bearing1);
int firstDir = maxBearing - minBearing;
int secondDir = minBearing + 360 - maxBearing;
int diff = Math.min(firstDir, secondDir);
boolean anticlock_dir = false;
int anticlock = bearing1 + diff;
if (anticlock >= 360)
anticlock = anticlock - 360;
if (anticlock == bearing0)
anticlock_dir = true;
You need to consider the difference in both directions.
public static double bearingDiff(double a, double b) {
double maxBearing = Math.max(a, b);
double minBearing = Math.min(a, b);
double antiClockwiseDiff = maxBearing - minBearing;
double clockwiseDiff = minBearing + 360 - maxBearing;
return Math.min(antiClockwiseDiff, clockwiseDiff);
}
I have a problem that I can't seem to get a working algorithm for, I've been trying to days and get so close but yet so far.
I want to draw a triangle defined by 3 points (p0, p1, p2). This triangle can be any shape, size, and orientation. The triangle must also be filled inside.
Here's a few things I've tried and why they've failed:
1
Drawing lines along the triangle from side to side
Failed because the triangle would have holes and would not be flat due to the awkwardness of drawing lines across the angled surface with changing locations
2
Iterate for an area and test if the point falls past the plane parallel to the triangle and 3 other planes projected onto the XY, ZY, and XZ plane that cover the area of the triangle
Failed because for certain triangles (that have very close sides) there would be unpredictable results, e.g. voxels floating around not connected to anything
3
Iterate for an area along the sides of the triangle (line algorithm) and test to see if a point goes past a parallel plane
Failed because drawing a line from p0 to p1 is not the same as a line from p1 to p0 and any attempt to rearrange either doesn't help, or causes more problems. Asymmetry is the problem with this one.
This is all with the intent of making polygons and flat surfaces. 3 has given me the most success and makes accurate triangles, but when I try to connect these together everything falls apart and I get issues with things not connecting, asymmetry, etc. I believe 3 will work with some tweaking but I'm just worn out from trying to make this work for so long and need help.
There's a lot of small details in my algorithms that aren't really relevant so I left them out. For number 3 it might be a problem with my implementation and not the algorithm itself. If you want code I'll try and clean it up enough to be understandable, it will take me a few minutes though. But I'm looking for algorithms that are known to work. I can't seem to find any voxel shape making algorithms anywhere, I've been doing everything from scratch.
EDIT:
Here's the third attempt. It's a mess, but I tried to clean it up.
// Point3i is a class I made, however the Vector3fs you'll see are from lwjgl
public void drawTriangle (Point3i r0, Point3i r1, Point3i r2)
{
// Util is a class I made with some useful stuff inside
// Starting values for iteration
int sx = (int) Util.min(r0.x, r1.x, r2.x);
int sy = (int) Util.min(r0.y, r1.y, r2.y);
int sz = (int) Util.min(r0.z, r1.z, r2.z);
// Ending values for iteration
int ex = (int) Util.max(r0.x, r1.x, r2.x);
int ey = (int) Util.max(r0.y, r1.y, r2.y);
int ez = (int) Util.max(r0.z, r1.z, r2.z);
// Side lengths
float l0 = Util.distance(r0.x, r1.x, r0.y, r1.y, r0.z, r1.z);
float l1 = Util.distance(r2.x, r1.x, r2.y, r1.y, r2.z, r1.z);
float l2 = Util.distance(r0.x, r2.x, r0.y, r2.y, r0.z, r2.z);
// Calculate the normal vector
Vector3f nn = new Vector3f(r1.x - r0.x, r1.y - r0.y, r1.z - r0.z);
Vector3f n = new Vector3f(r2.x - r0.x, r2.y - r0.y, r2.z - r0.z);
Vector3f.cross(nn, n, n);
// Determines which direction we increment for
int iz = n.z >= 0 ? 1 : -1;
int iy = n.y >= 0 ? 1 : -1;
int ix = n.x >= 0 ? 1 : -1;
// Reorganize for the direction of iteration
if (iz < 0) {
int tmp = sz;
sz = ez;
ez = tmp;
}
if (iy < 0) {
int tmp = sy;
sy = ey;
ey = tmp;
}
if (ix < 0) {
int tmp = sx;
sx = ex;
ex = tmp;
}
// We're we want to iterate over the end vars so we change the value
// by their incrementors/decrementors
ex += ix;
ey += iy;
ez += iz;
// Maximum length
float lmax = Util.max(l0, l1, l2);
// This is a class I made which manually iterates over a line, I already
// know that this class is working
GeneratorLine3d g0, g1, g2;
// This is a vector for the longest side
Vector3f v = new Vector3f();
// make the generators
if (lmax == l0) {
v.x = r1.x - r0.x;
v.y = r1.y - r0.y;
v.z = r1.z - r0.z;
g0 = new GeneratorLine3d(r0, r1);
g1 = new GeneratorLine3d(r0, r2);
g2 = new GeneratorLine3d(r2, r1);
}
else if (lmax == l1) {
v.x = r1.x - r2.x;
v.y = r1.y - r2.y;
v.z = r1.z - r2.z;
g0 = new GeneratorLine3d(r2, r1);
g1 = new GeneratorLine3d(r2, r0);
g2 = new GeneratorLine3d(r0, r1);
}
else {
v.x = r2.x - r0.x;
v.y = r2.y - r0.y;
v.z = r2.z - r0.z;
g0 = new GeneratorLine3d(r0, r2);
g1 = new GeneratorLine3d(r0, r1);
g2 = new GeneratorLine3d(r1, r2);
}
// Absolute values for the normal
float anx = Math.abs(n.x);
float any = Math.abs(n.y);
float anz = Math.abs(n.z);
int i, o;
int si, so;
int ii, io;
int ei, eo;
boolean maxx, maxy, maxz,
midy, midz, midx,
minx, miny, minz;
maxx = maxy = maxz =
midy = midz = midx =
minx = miny = minz = false;
// Absolute values for the longest side vector
float rnx = Math.abs(v.x);
float rny = Math.abs(v.y);
float rnz = Math.abs(v.z);
int rmid = Util.max(rnx, rny, rnz);
if (rmid == rnz) midz = true;
else if (rmid == rny) midy = true;
midx = !midz && !midy;
// Determine the inner and outer loop directions
if (midz) {
if (any > anx)
{
maxy = true;
si = sy;
ii = iy;
ei = ey;
}
else {
maxx = true;
si = sx;
ii = ix;
ei = ex;
}
}
else {
if (anz > anx) {
maxz = true;
si = sz;
ii = iz;
ei = ez;
}
else {
maxx = true;
si = sx;
ii = ix;
ei = ex;
}
}
if (!midz && !maxz) {
minz = true;
so = sz;
eo = ez;
}
else if (!midy && !maxy) {
miny = true;
so = sy;
eo = ey;
}
else {
minx = true;
so = sx;
eo = ex;
}
// GeneratorLine3d is iterable
Point3i p1;
for (Point3i p0 : g0) {
// Make sure the two 'mid' coordinate correspond for the area inside the triangle
if (midz)
do p1 = g1.hasNext() ? g1.next() : g2.next();
while (p1.z != p0.z);
else if (midy)
do p1 = g1.hasNext() ? g1.next() : g2.next();
while (p1.y != p0.y);
else
do p1 = g1.hasNext() ? g1.next() : g2.next();
while (p1.x != p0.x);
eo = (minx ? p0.x : miny ? p0.y : p0.z);
so = (minx ? p1.x : miny ? p1.y : p1.z);
io = eo - so >= 0 ? 1 : -1;
for (o = so; o != eo; o += io) {
for (i = si; i != ei; i += ii) {
int x = maxx ? i : midx ? p0.x : o;
int y = maxy ? i : midy ? p0.y : o;
int z = maxz ? i : midz ? p0.z : o;
// isPassing tests to see if a point goes past a plane
// I know it's working, so no code
// voxels is a member that is an arraylist of Point3i
if (isPassing(x, y, z, r0, n.x, n.y, n.z)) {
voxels.add(new Point3i(x, y, z));
break;
}
}
}
}
}
You could use something like Besenham's line algorithm, but extended into three dimensions. The two main ideas we want to take from it are:
rotate the initial line so its slope isn't too steep.
for any given x value, find an integer value that is closest to the ideal y value.
Just as Bresenham's algorithm prevents gaps by performing an initial rotation, we'll avoid holes by performing two initial rotations.
Get the normal vector and point that represent the plane your triangle lies on. Hint: use the cross product of (line from p0 to p1) and (line from p0 to p2) for the vector, and use any of your corner points for the point.
You want the plane to be sufficiently not-steep, to avoid holes. You must satisfy these conditions:
-1 >= norm.x / norm.y >= 1
-1 >= norm.z / norm.y >= 1
Rotate your normal vector and initial points 90 degrees about the x axis and 90 degrees about the z axis until these conditions are satisfied. I'm not sure how to do this in the fewest number of rotations, but I'm fairly sure you can satisfy these conditions for any plane.
Create a function f(x,z) which represents the plane your rotated triangle now lies on. It should return the Y value of any pair of X and Z values.
Project your triangle onto the XZ plane (i.e., set all the y values to 0), and use your favorite 2d triangle drawing algorithm to get a collection of x-and-z coordinates.
For each pixel value from step 4, pass the x and z values into your function f(x,z) from step 3. Round the result to the nearest integer, and store the x, y, and z values as a voxel somewhere.
If you performed any rotations in step 2, perform the opposite of those rotations in reverse order on your voxel collection.
Start with a function that checks for triangle/voxel intersection. Now you can scan a volume and find the voxels that intersect the triangle - these are the ones you're interested in. This is a lousy algorithm but is also a regression test for anything else you try. This test is easy to implement using SAT (separating axis theorem) and considering the triangle a degenerate volume (1 face, 3 edges) and considering the voxels symmetry (only 3 face normals).
I use octtrees, so my preferred method is to test a triangle against a large voxel and figure out which of the 8 child octants it intersects. Then use recursion on the intersected children until the desired level of subdivision is attained. Hint: at most 6 of the children can be intersected by the triangle and often fewer than that. This is tricky but will produce the same results as the first method but much quicker.
Rasterization in 3d is probably fastest, but IMHO is even harder to guarantee no holes in all cases. Again, use the first method for comparison.