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Draw an arc in opengl GL10

I want to draw an arc using center point,starting point,ending point on opengl surfaceview.I have tried this given below code so far. This function draws the expected arc if we give the value for start_line_angle and end_line_angle manually (like start_line_angle=0 and end_line_angle=90) in degree.
But I need to draw an arc with the given co-ordinates(center point,starting point,ending point) and calculating the start_line_angle and end_line_angle programatically.
This given function draws an arc with the given parameters but not giving the desire result. I've wasted my 2 days for this. Thanks in advance.
private void drawArc(GL10 gl, float radius, float cx, float cy, float start_point_x, float start_point_y, float end_point_x, float end_point_y) {
gl.glLineWidth(1);
int start_line_angle;
double sLine = Math.toDegrees(Math.atan((cy - start_point_y) / (cx - start_point_x))); //normal trigonometry slope = tan^-1(y2-y1)/(x2-x1) for line first
double eLine = Math.toDegrees(Math.atan((cy - end_point_y) / (cx - end_point_x))); //normal trigonometry slope = tan^-1(y2-y1)/(x2-x1) for line second
//cast from double to int after round
int start_line_Slope = (int) (sLine + 0.5);
/**
* mapping the tiriogonometric angle system to glsurfaceview angle system
* since angle system in trigonometric system starts in anti clockwise
* but in opengl glsurfaceview angle system starts in clock wise and the starting angle is 90 degree of general trigonometric angle system
**/
if (start_line_Slope <= 90) {
start_line_angle = 90 - start_line_Slope;
} else {
start_line_angle = 360 - start_line_Slope + 90;
}
// int start_line_angle = 270;
// int end_line_angle = 36;
//casting from double to int
int end_line_angle = (int) (eLine + 0.5);
if (start_line_angle > end_line_angle) {
start_line_angle = start_line_angle - 360;
}
int nCount = 0;
float[] stVertexArray = new float[2 * (end_line_angle - start_line_angle)];
float[] newStVertextArray;
FloatBuffer sampleBuffer;
// stVertexArray[0] = cx;
// stVertexArray[1] = cy;
for (int nR = start_line_angle; nR < end_line_angle; nR++) {
float fX = (float) (cx + radius * Math.sin((float) nR * (1 * (Math.PI / 180))));
float fY = (float) (cy + radius * Math.cos((float) nR * (1 * (Math.PI / 180))));
stVertexArray[nCount * 2] = fX;
stVertexArray[nCount * 2 + 1] = fY;
nCount++;
}
//taking making the stVertextArray's data in reverse order
reverseArray = new float[stVertexArray.length];//-2 so that no repeatation occurs of first value and end value
int count = 0;
for (int i = (stVertexArray.length) / 2; i > 0; i--) {
reverseArray[count] = stVertexArray[(i - 1) * 2 + 0];
count++;
reverseArray[count] = stVertexArray[(i - 1) * 2 + 1];
count++;
}
//reseting the counter to initial value
count = 0;
int finalArraySize = stVertexArray.length + reverseArray.length;
newStVertextArray = new float[finalArraySize];
/**Now adding all the values to the single newStVertextArray to draw an arc**/
//adding stVertextArray to newStVertextArray
for (float d : stVertexArray) {
newStVertextArray[count++] = d;
}
//adding reverseArray to newStVertextArray
for (float d : reverseArray) {
newStVertextArray[count++] = d;
}
Log.d("stArray", stVertexArray.length + "");
Log.d("reverseArray", reverseArray.length + "");
Log.d("newStArray", newStVertextArray.length + "");
ByteBuffer bBuff = ByteBuffer.allocateDirect(newStVertextArray.length * 4);
bBuff.order(ByteOrder.nativeOrder());
sampleBuffer = bBuff.asFloatBuffer();
sampleBuffer.put(newStVertextArray);
sampleBuffer.position(0);
gl.glEnableClientState(GL10.GL_VERTEX_ARRAY);
gl.glVertexPointer(2, GL10.GL_FLOAT, 0, sampleBuffer);
gl.glDrawArrays(GL10.GL_LINE_LOOP, 0, nCount * 2);
gl.glLineWidth(1);
}

To begin with the trigonometry you may not simply use the atan to find degrees of the angle. You need to check what quadrant the vector is in and increase or decrease the result you get from atan. Better yet use atan2 which should include both dx and dy and do the job for you.
You seem to create the buffer so that a point is created per degree. This is not the best solution as for large radius that might be too small and for small radius this is way too much. Tessellation should include the radius as well such that number of points N is N = abs((int)(deltaAngle*radius*tessellationFactor)) then use angleFragment = deltaAngle/N but make sure that N is greater then 0 (N = N?N:1). The buffer size is then 2*(N+1) of floats and the iteration if for(int i=0; i<=N; i++) angle = startAngle + angleFragment*i;.
As already pointed out you need to define the radius of the arc. It is quite normal to use an outside source the way you do and simply force it to that value but use the 3 points for center and the two borders. Some other options that usually make sense are:
getting the radius from the start line
getting the radius from the shorter of the two lines
getting the average of the two
interpolate the two to get an elliptic curve (explained below)
To interpolate the radius you need to get the two radiuses startRadius and endRadius. Then you need to find the overall radius which was already used as deltaAngle above (watch out when computing this one, it is more complicated as it seems, for instance drawing from 320 degrees to 10 degrees results in deltaAngle = 50). Anyway the radius for a specific point is then simply radius = startRadius + (endRadius-startRadius)*abs((angleFragment*i)/deltaAngle). This represents a simple linear interpolation in polar coordinate system which is usually used to interpolate vector in matrices and is the core functionality to get nice animations.
There are some other ways of getting the arc points which may be better performance wise but I would not suggest them unless and until you need to optimize your code which should be very late in production. You may simply keep stepping toward the next point and correcting the radius (this is only a concept):
vec2 start, end, center; // input values
float radius; // input value
// making the start and end relative to center
start -= center;
end -= center;
vec2 current = start/length(start) * radius; // current position starts in first vector
vec2 target = end/length(end) * radius; // should be the last point
outputBuffer[0] = current+center; // insert the first point
for(int i=1;; i++) { // "break" will need to exit the loop, we need index only for the buffer
vec2 step = vec2(current.y, -(current.x)); // a tangential vector from current start point according to center
step = step/length(step) / tessellationScale; // normalize and apply tessellation
vec2 next = current + step; // move tangentially
next = next/length(next) * radius; // normalize and set the
if(dot(current-target, next-target) > .0) { // when we passed the target vector
current = next; // set the current point
outputBuffer[i] = current+center; // insert into buffer
}
else {
current = target; // simply use the target now
outputBuffer[i] = current+center; // insert into buffer
break; // exit
}
}

strange thing with cylinder algorithm

I want to render a cylinder in Opengl. For that i wrote an simple algorithm, that
generates me the points mesh by the parameters radius, height, xSubDivisions and ySubDivisions:
(Java)
for(int yDivision = 0; yDivision < yDivisionCount; yDivision++){
for(int xDivision = 0; xDivision < xDivisionCount; xDivision++){
float line[] = getVboLine(xDivision, yDivision, radius, height, xDivisionCount, yDivisionCount);
string.append(line[0] + ", " + line[1] + ", " + line[2] + ", " + line[3] + ", " + line[4] + ", ");
}
}
public float[] getVboLine(int xDivision, int yDivision, float radius, float height, int xDivisionCount, int yDivisionCount){
float xDegrees = 360.0f / xDivisionCount * xDivision;
float xRadian = (float) Math.toRadians(xDegrees);
float x = (float) Math.sin(xRadian) * radius;
float z = (float) Math.cos(xRadian) * radius;
float y = (float) yDivision * (height / (yDivisionCount - 1));
float s = xDegrees * (1.0f / 360.0f);
float t = yDivision * (1.0f / (yDivisionCount - 1));
return new float[]{
x, y, z, s, t
};
}
The result is actually an cylinder, (i created an IBO to render this points) but sometimes, with different inputs for x and yDivisions there is a strange gap in it.
I couldn't find a rule, but the values i found this bug with were 200, 100.
To debug i rendered only the points. The result was:
How is this possible? One points is just missing (where i added the reed circle with paint).
Where is the problem with my algorithm?

I am not JAVA coder but you are mixing int and float together
for example:
xDegrees = 360.0f / xDivisionCount * xDivision
[float] [float] [int] [int]
I would rather use this:
xDegrees = float(360*xDivision)/float(xDivisionCount)
multiplication should go always first (if operands are >= 1)
and division after that to preserve accuracy
some weird rounding could cause your problem but it would be more noticeable for lower xDivisionCount not bigger one
Bug breakpoint
add to your code last generated point
after new point computation compute the distance from last point
add if (|distance-some_avg_distance|>1e-10)
and add breakpoint inside
some_avg_distance set by distance that should be there (get it from trace)
this way you can breakpoint the point causing problems (or the next point to it)
so you can actually see what is wrong
my bet is that by rounding you get the same angle as prev/next point
and therefore you do not have missing point but some duplicate instead
you can check that also by Blending

How to fix circle and rectangle overlap in collision response?

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.

Moving an object towards a point android

I have this in the initialization of a bullet object:
x = startX;
y = startY;
double distance = Math.sqrt(((endX - x) ^ 2) + ((endY - y) ^ 2));
speedX = (6 * (endX - x)) / distance;
speedY = (6 * (endY - y)) / distance;
It goes to where I touch on the screen, but the further away I touch, the faster it goes. This works fine on paper, I've tried it with different lengths and it should work, but bullets need to move 6 pixels on the line from the player to the point touched every step. And its update method moves of course. But why do bullets move at different speeds?

If I remember my Java operators...
Replace
double distance = Math.sqrt(((endX - x) ^ 2) + ((endY - y) ^ 2));
with
double distance = Math.sqrt(Math.pow(endX - x, 2) + Math.pow(endY - y, 2));

Assuming that all measurements are in pixels and you want the speed to be 6 pixels per step, then you can calculate the velocity by using a little bit of trig:
double theta = Math.atan2(endY - startY, endX - startX);
velX = 6 * Math.cos(theta);
velY = 6 * Math.sin(theta);
Note that I am using the terms "speed" and "velocity" as a physicist would; speed is a scalar value and velocity is a vector with magnitude and direction.

How to draw a smooth line through a set of points using Bezier curves?

I need to draw a smooth line through a set of vertices. The set of vertices is compiled by a user dragging their finger across a touch screen, the set tends to be fairly large and the distance between the vertices is fairly small. However, if I simply connect each vertex with a straight line, the result is very rough (not-smooth).
I found solutions to this which use spline interpolation (and/or other things I don't understand) to smooth the line by adding a bunch of additional vertices. These work nicely, but because the list of vertices is already fairly large, increasing it by 10x or so has significant performance implications.
It seems like the smoothing should be accomplishable by using Bezier curves without adding additional vertices.
Below is some code based on the solution here:
http://www.antigrain.com/research/bezier_interpolation/
It works well when the distance between the vertices is large, but doesn't work very well when the vertices are close together.
Any suggestions for a better way to draw a smooth curve through a large set of vertices, without adding additional vertices?
Vector<PointF> gesture;
protected void onDraw(Canvas canvas)
{
if(gesture.size() > 4 )
{
Path gesturePath = new Path();
gesturePath.moveTo(gesture.get(0).x, gesture.get(0).y);
gesturePath.lineTo(gesture.get(1).x, gesture.get(1).y);
for (int i = 2; i < gesture.size() - 1; i++)
{
float[] ctrl = getControlPoint(gesture.get(i), gesture.get(i - 1), gesture.get(i), gesture.get(i + 1));
gesturePath.cubicTo(ctrl[0], ctrl[1], ctrl[2], ctrl[3], gesture.get(i).x, gesture.get(i).y);
}
gesturePath.lineTo(gesture.get(gesture.size() - 1).x, gesture.get(gesture.size() - 1).y);
canvas.drawPath(gesturePath, mPaint);
}
}
}
private float[] getControlPoint(PointF p0, PointF p1, PointF p2, PointF p3)
{
float x0 = p0.x;
float x1 = p1.x;
float x2 = p2.x;
float x3 = p3.x;
float y0 = p0.y;
float y1 = p1.y;
float y2 = p2.y;
float y3 = p3.y;
double xc1 = (x0 + x1) / 2.0;
double yc1 = (y0 + y1) / 2.0;
double xc2 = (x1 + x2) / 2.0;
double yc2 = (y1 + y2) / 2.0;
double xc3 = (x2 + x3) / 2.0;
double yc3 = (y2 + y3) / 2.0;
double len1 = Math.sqrt((x1-x0) * (x1-x0) + (y1-y0) * (y1-y0));
double len2 = Math.sqrt((x2-x1) * (x2-x1) + (y2-y1) * (y2-y1));
double len3 = Math.sqrt((x3-x2) * (x3-x2) + (y3-y2) * (y3-y2));
double k1 = len1 / (len1 + len2);
double k2 = len2 / (len2 + len3);
double xm1 = xc1 + (xc2 - xc1) * k1;
double ym1 = yc1 + (yc2 - yc1) * k1;
double xm2 = xc2 + (xc3 - xc2) * k2;
double ym2 = yc2 + (yc3 - yc2) * k2;
// Resulting control points. Here smooth_value is mentioned
// above coefficient K whose value should be in range [0...1].
double k = .1;
float ctrl1_x = (float) (xm1 + (xc2 - xm1) * k + x1 - xm1);
float ctrl1_y = (float) (ym1 + (yc2 - ym1) * k + y1 - ym1);
float ctrl2_x = (float) (xm2 + (xc2 - xm2) * k + x2 - xm2);
float ctrl2_y = (float) (ym2 + (yc2 - ym2) * k + y2 - ym2);
return new float[]{ctrl1_x, ctrl1_y, ctrl2_x, ctrl2_y};
}

Bezier Curves are not designed to go through the provided points! They are designed to shape a smooth curve influenced by the control points.
Further you don't want to have your smooth curve going through all data points!
Instead of interpolating you should consider filtering your data set:
Filtering
For that case you need a sequence of your data, as array of points, in the order the finger has drawn the gesture:
You should look in wiki for "sliding average".
You should use a small averaging window. (try 5 - 10 points). This works as follows: (look for wiki for a more detailed description)
I use here an average window of 10 points:
start by calculation of the average of points 0 - 9, and output the result as result point 0
then calculate the average of point 1 - 10 and output, result 1
And so on.
to calculate the average between N points:
avgX = (x0+ x1 .... xn) / N
avgY = (y0+ y1 .... yn) / N
Finally you connect the resulting points with lines.
If you still need to interpolate between missing points, you should then use piece - wise cubic splines.
One cubic spline goes through all 3 provided points.
You would need to calculate a series of them.
But first try the sliding average. This is very easy.

Nice question. Your (wrong) result is obvious, but you can try to apply it to a much smaller dataset, maybe by replacing groups of close points with an average point. The appropriate distance in this case to tell if two or more points belong to the same group may be expressed in time, not space, so you'll need to store the whole touch event (x, y and timestamp). I was thinking of this because I need a way to let users draw geometric primitives (rectangles, lines and simple curves) by touch

What is this for? Why do you need to be so accurate? I would assume you only need something around 4 vertices stored for every inch the user drags his finger. With that in mind:
Try using one vertex out of every X to actually draw between, with the middle vertex used for specifying the weighted point of the curve.
int interval = 10; //how many points to skip
gesture.moveTo(gesture.get(0).x, gesture.get(0).y);
for(int i =0; i +interval/2 < gesture.size(); i+=interval)
{
Gesture ngp = gesture.get(i+interval/2);
gesturePath.quadTo(ngp.x,ngp.y, gp.x,gp.y);
}
You'll need to adjust this to actually work but the idea is there.