I am experiment kinect on winrt for metro app.
I am trying to obtain angle at the elbow.
normally i will do the following
Vector3D handLeftVector = new Vector3D(HandLeftX, HandLeftY, HandLeftZ);
handLeftVector.Normalize();
Vector3D ElbowLeftEVector = new Vector3D(ElbowLeftX, ElbowLeftY, ElbowLeftZ);
ElbowLeftEVector.Normalize();
Vector3D ShoulderLeftVector = new Vector3D(ShoulderLeftX, ShoulderLeftY, ShoulderLeftZ);
ShoulderLeftVector.Normalize();
Vector3D leftElbowV1 = ShoulderLeftVector - ElbowLeftEVector;
Vector3D leftElbowV2 = handLeftVector - ElbowLeftEVector;
double leftElbowAngle = Vector3D.AngleBetween(leftElbowV1, leftElbowV2);
However Vector3D object isn't available in winrt.
I had decided to replicate the Vector3D method as below. However the result doesn't seem to be as expected. Did I make a mistake anywhere?
double leftElbowV1X = ShoulderLeftX - ElbowLeftX;
double leftElbowV1Y = ShoulderLeftY - ElbowLeftY;
double leftElbowV1Z = ShoulderLeftZ - ElbowLeftZ;
double leftElbowV2X = handLeftX - ElbowLeftX;
double leftElbowV2Y = handLeftY - ElbowLeftY;
double leftElbowV2Z = handLeftZ - ElbowLeftZ;
double product = leftElbowV1X * leftElbowV2X + leftElbowV1Y * leftElbowV2Y + leftElbowV1Z * leftElbowV2Z;
double magnitudeA = Math.Sqrt(Math.Pow(leftElbowV1X, 2) + Math.Pow(leftElbowV1Y, 2) + Math.Pow(leftElbowV1Z, 2));
double magnitudeB = Math.Sqrt(Math.Pow(leftElbowV2X, 2) + Math.Pow(leftElbowV2Y, 2) + Math.Pow(leftElbowV2Z, 2));
magnitudeA = Math.Abs(magnitudeA);
magnitudeB = Math.Abs(magnitudeB);
double cosDelta = product / (magnitudeA * magnitudeB);
double angle = Math.Acos(cosDelta) *180.0 / Math.P;
And is there a need to normalize it?
i had managed to resolve it, however i am thinking if there is a more efficient way of doing.
Not sure if this helps but this is some old angle code I use, return in degrees:
float AngleBetween(Vector3 from, Vector3 dest) {
float len = from.magnitude * dest.magnitude;
if(len < Mathf.Epsilon) len = Mathf.Epsilon;
float f = Vector3.Dot(from,dest) / len;
if(f>1.0f)f=1.0f;
else if ( f < -1.0f) f = -1.0f;
return Mathf.Acos(f) * 180.0f / (float)Math.PI;
}
It's obviously using API specific syntax but I think the method is clear.
Related
I recently found out about the Mandelbrot set and now I am trying to generate a Mandelbrot set in Processing 3.0. I found a tutorial on youtube about programming it and tried to impelement it in Progressing.
background(255);
size(1280, 720);
}
void draw(){
int maxIteration = 100;
double reMid = -0.75;
double iMid = 0;
double rangeR = 3.5;
double rangeI = 2;
double xPixels = 1280;
double yPixels = 720;
for(int x = 0; x < xPixels; x++){
for(int y = 0; y < yPixels; y++){
double xP = (double)x / xPixels;
double yP = (double)y / yPixels;
double cReal = xP * rangeR + reMid - rangeR / 2;
double cIm = yP * rangeI + iMid - rangeI / 2;
double zReal = 0;
double zIm = 0;
int iteration = 0;
while(iteration < maxIteration && zReal * zReal + zIm * zIm <= 4) {
double temp = zReal * zReal - cIm * cIm + cReal;
zIm = 2 * zReal * zIm + cIm;
zReal = temp;
iteration++;
}
if(iteration >= maxIteration - 1){
stroke(0);
}
else{
stroke(255);
}
point(x, y);
}
}
}
But when i generated the Mandelbrot set, it looked different than it should:
I have already compared my code with the one in the video, but i did not find a mistake in my code.
Does anyone know, what I did wrong?
I zoomed it out a bit, and the long tails keep extending to infinity. Since all |C| > 2 should diverge, this makes it easy to find a specific case that fails, such as cReal = 2; cImg = -1.5;
Your code says it converges, but doing it by hand shows it diverges very quickly:
Z0 = 0 + 0i
Z1 = (0 + 0i)^2 + 2 - 1.5i = 2 - 1.5i
Z2 = 2*2 - 2*2*1.5i - 1.5^2 = 1.75 - 6i
Stepping through your code gives zReal, zImg
-1.5, -0.25
-0.75, -0.1875
-1.21875, -0.21484375
-0.976318359375, -0.2038421630859375
-1.1019703075289726, -0.20844837254844606
[...]
In other words, your loop is wrong. The immediately suspect line of code is this:
double temp = zReal * zReal - cIm * cIm + cReal;
It's doing cIm*cIm, but there's not supposed to be any multiplication of any components of C: it's simply added at the end.
So what's happened is that you accidentally switched zIm for cIm.
Switch them back and you should get a better result:
double temp = zReal * zReal - zIm * zIm + cReal;
I am trying to figure out where a bunch of line-segments clip into a window around them. I saw the Liang–Barsky algorithm, but that seems to assume the segments already clip the edges of the window, which these do not.
Say I have a window from (0,0) to (26,16), and the following segments:
(7,6) - (16,3)
(10,6) - (19,6)
(13,10) - (21,3)
(16,12) - (19,14)
Illustration:
I imagine I need to extend the segments to a certain X or Y point, till they hit the edge of the window, but I don't know how.
How would I find the points where these segments (converted to lines?) clip into the edge of the window? I will be implementing this in C#, but this is pretty language-agnostic.
If you have two line segments P and Q with points
P0 - P1
Q0 - Q1
The line equations are
P = P0 + t(P1 - P0)
Q = Q0 + r(Q1 - Q0)
then to find out where they intersect after extension you need to solve the following equation for t and r
P0 + t(P1 - P0) = Q0 + r(Q1 - Q0)
The following code can do this. ( Extracted from my own code base )
public static (double t, double r )? SolveIntersect(this Segment2D P, Segment2D Q)
{
// a-d are the entries of a 2x2 matrix
var a = P.P1.X - P.P0.X;
var b = -Q.P1.X + Q.P0.X;
var c = P.P1.Y - P.P0.Y;
var d = -Q.P1.Y + Q.P0.Y;
var det = a*d - b*c;
if (Math.Abs( det ) < Utility.ZERO_TOLERANCE)
return null;
var x = Q.P0.X - P.P0.X;
var y = Q.P0.Y - P.P0.Y;
var t = 1/det*(d*x - b*y);
var r = 1/det*(-c*x + a*y);
return (t, r);
}
If null is returned from the function then it means the lines are parallel and cannot intersect. If a result is returned then you can do.
var result = SolveIntersect( P, Q );
if (result != null)
{
var ( t, r) = result.Value;
var p = P.P0 + t * (P.P1 - P.P0);
var q = Q.P0 + t * (Q.P1 - Q.P0);
// p and q are the same point of course
}
The extended line segments will generally intersect more than one box edge but only one of those intersections will be inside the box. You can check this easily.
bool IsInBox(Point corner0, Point corner1, Point test) =>
(test.X > corner0.X && test.X < corner1.X && test.Y > corner0.Y && test.Y < corner1.Y ;
That should give you all you need to extend you lines to the edge of your box.
I managed to figure this out.
I can extend my lines to the edge of the box by first finding the equations of my lines, then solving for the X and Y of each of the sides to get their corresponding point. This requires passing the max and min Y and the max and min X into the following functions, returning 4 values. If the point is outside the bounds of the box, it can be ignored.
My code is in C#, and is making extension methods for EMGU's LineSegment2D. This is a .NET wrapper for OpenCv.
My Code:
public static float GetYIntersection(this LineSegment2D line, float x)
{
Point p1 = line.P1;
Point p2 = line.P2;
float dx = p2.X - p1.X;
if(dx == 0)
{
return float.NaN;
}
float m = (p2.Y - p1.Y) / dx; //Slope
float b = p1.Y - (m * p1.X); //Y-Intercept
return m * x + b;
}
public static float GetXIntersection(this LineSegment2D line, float y)
{
Point p1 = line.P1;
Point p2 = line.P2;
float dx = p2.X - p1.X;
if (dx == 0)
{
return float.NaN;
}
float m = (p2.Y - p1.Y) / dx; //Slope
float b = p1.Y - (m * p1.X); //Y-Intercept
return (y - b) / m;
}
I can then take these points, check if they are in the bounds of the box, discard the ones that are not, remove duplicate points (line goes directly into corner). This will leave me with one x and one y value, which I can then pair to the corresponding min or max Y or X values I passed into the functions to make 2 points. I can then make my new segment with the two points.
Wiki description of Liang-Barsky algorithm is not bad, but code is flaw.
Note: this algorithm intended to throw out lines without intersection as soon as possible. If most of lines intersect the rectangle, then approach from your answer might be rather effective, otherwise L-B algorithm wins.
This page describes approach in details and contains concise effective code:
// Liang-Barsky function by Daniel White # http://www.skytopia.com/project/articles/compsci/clipping.html
// This function inputs 8 numbers, and outputs 4 new numbers (plus a boolean value to say whether the clipped line is drawn at all).
//
bool LiangBarsky (double edgeLeft, double edgeRight, double edgeBottom, double edgeTop, // Define the x/y clipping values for the border.
double x0src, double y0src, double x1src, double y1src, // Define the start and end points of the line.
double &x0clip, double &y0clip, double &x1clip, double &y1clip) // The output values, so declare these outside.
{
double t0 = 0.0; double t1 = 1.0;
double xdelta = x1src-x0src;
double ydelta = y1src-y0src;
double p,q,r;
for(int edge=0; edge<4; edge++) { // Traverse through left, right, bottom, top edges.
if (edge==0) { p = -xdelta; q = -(edgeLeft-x0src); }
if (edge==1) { p = xdelta; q = (edgeRight-x0src); }
if (edge==2) { p = -ydelta; q = -(edgeBottom-y0src);}
if (edge==3) { p = ydelta; q = (edgeTop-y0src); }
if(p==0 && q<0) return false; // Don't draw line at all. (parallel line outside)
r = q/p;
if(p<0) {
if(r>t1) return false; // Don't draw line at all.
else if(r>t0) t0=r; // Line is clipped!
} else if(p>0) {
if(r<t0) return false; // Don't draw line at all.
else if(r<t1) t1=r; // Line is clipped!
}
}
x0clip = x0src + t0*xdelta;
y0clip = y0src + t0*ydelta;
x1clip = x0src + t1*xdelta;
y1clip = y0src + t1*ydelta;
return true; // (clipped) line is drawn
}
I am trying to implement an angular constraint into a simple verlet integration based 2D physic engine. This is the code I am currently using:
int indexA = physicAngularConstraint[i].indexA;
int indexB = physicAngularConstraint[i].indexB;
int indexC = physicAngularConstraint[i].indexC;
CGPoint e = CGPointSubtract(physicParticle[indexB].pos,physicParticle[indexA].pos);
CGPoint f = CGPointSubtract(physicParticle[indexC].pos,physicParticle[indexB].pos);
float dot = CGPointDot(e, f);
float cross = CGPointCross(e, f);
float angle = atan2f(cross, dot);
float da = (angle < physicAngularConstraint[i].minAngle)? angle - physicAngularConstraint[i].minAngle : (angle > physicAngularConstraint[i].maxAngle)? angle - physicAngularConstraint[i].maxAngle : 0.0f;
if (da != 0.0f)
{
physicParticle[indexA].pos = CGPointRotate(physicParticle[indexA].pos,
physicParticle[indexB].pos, da);
physicParticle[indexC].pos = CGPointRotate(physicParticle[indexC].pos,
physicParticle[indexB].pos, -da);
}
The CGPointRotate function looks like this:
CGPoint CGPointRotate(CGPoint pt, CGPoint center, float angle)
{
CGPoint ret;
pt = CGPointSubtract(pt, center);
float co = cosf(angle);
float si = sinf(angle);
ret.x = pt.x * co - pt.y * si;
ret.y = pt.x * si + pt.y * co;
ret = CGPointAdd(ret, center);
return ret;
}
I am testing this implementation with a row of particles which are connected through distant constraints. Without the angular constraints they act like a rope. I am trying to give the "rope" some stiffness through the angular constraints. But my implementation above is gaining energy and blowing up the system after some millisecs. Why does this constrain implementation is gaining energie?
Drawn a line from a point A to point B. Let d be offset. Let C be point to be tested.
I am going to do a kind of hit testing around the line with offset.
How can i do the hit testing around the line with the given offset.
Ex: A = (10,10), B (30,30), offset = 2. choose C as any point. Please Refer the image in the link please.
http://s10.postimg.org/6by2dzvax/reference.png
Please help me.
Thanks in advance.
Find offset for C.
e.g. dx1 and dy1. If dy1/dx1=dy/dx then your C hits the line.
For segment you should also check if whether dx1 < dx or dy1 < dy.
In other words, you want to check if that point C is inside a certain rectangle, with dimensions 2*d and |A-B|+2*d. You need to represent the line as u*x+v*y+w=0, this can be accomplished by
u = A.y-B.y
v = B.x-A.x
w = A.x*B.y - A.y * B.x
Then the (signed) distance of C from that line would be
d = (u*C.x + v*C.y +w) / sqrt( u*u+v*v)
You compare abs(d) to your offset.
The next step would be to check the position of C in the direction of the line. To that end you consider the orthogonal line u2*x+v2*y+w2=0 with
u2 = v
v2 = -u
w2 = -u2*(A.x+B.x)/2 - v2*(A.y+B.y)/2
and the distance
d2 = (u2 * C.x + v2 * C.y + w2 ) / sqrt( u2*u2+v2*v2 )
This distance must be compared to something like the length of the line+offset:
abs(d2) < |A-B| / 2 + offset
A convenient trick is to rotate and translate the plane in such a way that the segment AB maps to the segment (0, 0)-(0, L) (just like on the image), L being the segment length.
If you apply the same transform to C, then it a very simple matter to test inclusion in the rectangle.
That useful transform is given by:
x = ((X - XA).(XB - XA) + (Y - YA).(YB - YA)) / L
y = ((X - XA).(YB - YA) - (Y - YA).(XB - XA)) / L
maybe you can use this function to count the shortest distance of the point to the line. If the distance is <= offset, then that point is hitting the line.
private double pointDistanceToLine(PointF line1, PointF line2, PointF pt)
{
var isValid = false;
PointF r = new PointF();
if (line1.Y == line2.Y && line1.X == line2.X)
line1.Y -= 0.00001f;
double U = ((pt.Y - line1.Y ) * (line2.Y - line1.Y )) + ((pt.X - line1.X) * (line2.X - line1.X));
double Udenom = Math.Pow(line2.Y - line1.Y , 2) + Math.Pow(line2.X - line1.X, 2);
U /= Udenom;
r.Y = (float)(line1.Y + (U * (line2.Y - line1.Y ))); r.X = (float)(line1.X + (U * (line2.X - line1.X)));
double minX, maxX, minY , maxY ;
minX = Math.Min(line1.Y , line2.Y );
maxX = Math.Max(line1.Y , line2.Y );
minY = Math.Min(line1.X, line2.X);
maxY = Math.Max(line1.X, line2.X);
isValid = (r.Y >= minX && r.Y <= maxX) && (r.X >= minY && r.X <= maxY );
//return isValid ? r : null;
if (isValid)
{
double result = Math.Pow((pt.X - r.X), 2) + Math.Pow((pt.Y - r.Y), 2);
result = Math.Sqrt(result);
return result;
}
else {
double result1 = Math.Pow((pt.X - line1.X), 2) + Math.Pow((pt.Y - line1.Y), 2);
result1 = Math.Sqrt(result1);
double result2 = Math.Pow((pt.X - line2.X), 2) + Math.Pow((pt.Y - line2.Y), 2);
result2 = Math.Sqrt(result2);
return Math.Min(result1, result2);
}
}
I have been grappling with the Gribb/Hartmann method of extracting the Frustum planes for some time now, with little success. I want to build a camera view-frustum to cull my scene.
I am working with column-major matrices in a right-handed coordinate system. (OpenGL style - I'm using C# and Playstation Mobile, but the math should be the same)
I want to get my planes in World-Space, so I build my frustum from the View-Projection Matrix (that's projectionMatrix * viewMatrix). The view Matrix is the inverse of the camera's World-Transform.
The problem is; regardless of what I tweak, I can't seem to get a correct frustum. I think that I may be missing something obvious.
If I "strafe" my camera left or right while still looking down the z-axis, the normals of my planes change so that they are always pointing at the origin of the scene - which makes me think that they are not in world-space...
The planes from a projection matrix can be extracted using the Gribb/Hartmann method as follows, (column major):
void extract_planes_from_projmat(
const float mat[4][4],
float left[4], float right[4],
float bottom[4], float top[4],
float near[4], float far[4])
{
for (int i = 4; i--; ) left[i] = mat[i][3] + mat[i][0];
for (int i = 4; i--; ) right[i] = mat[i][3] - mat[i][0];
for (int i = 4; i--; ) bottom[i] = mat[i][3] + mat[i][1];
for (int i = 4; i--; ) top[i] = mat[i][3] - mat[i][1];
for (int i = 4; i--; ) near[i] = mat[i][3] + mat[i][2];
for (int i = 4; i--; ) far[i] = mat[i][3] - mat[i][2];
}
Where mat4 is the product of the projection matrix and the model-view matrix.
See:
https://fgiesen.wordpress.com/2012/08/31/frustum-planes-from-the-projection-matrix/
http://www8.cs.umu.se/kurser/5DV051/HT12/lab/plane_extraction.pdf
Note: if the matrix components aren't normalized and you require a Hessian Normal Form plane, then you will need to normalize the resulting planes.
The missing part:
comboMatrix = projection_matrix * Matrix4_Transpose(modelview_matrix)
Then the world-space frustum plane extraction for OpenGL is exactly as mentioned in the Gribb/Hartmann method:
p_planes[0].a = comboMatrix._41 + comboMatrix._11;
p_planes[0].b = comboMatrix._42 + comboMatrix._12;
p_planes[0].c = comboMatrix._43 + comboMatrix._13;
p_planes[0].d = comboMatrix._44 + comboMatrix._14;
// Right clipping plane
p_planes[1].a = comboMatrix._41 - comboMatrix._11;
p_planes[1].b = comboMatrix._42 - comboMatrix._12;
p_planes[1].c = comboMatrix._43 - comboMatrix._13;
p_planes[1].d = comboMatrix._44 - comboMatrix._14;
// Top clipping plane
p_planes[2].a = comboMatrix._41 - comboMatrix._21;
p_planes[2].b = comboMatrix._42 - comboMatrix._22;
p_planes[2].c = comboMatrix._43 - comboMatrix._23;
p_planes[2].d = comboMatrix._44 - comboMatrix._24;
// Bottom clipping plane
p_planes[3].a = comboMatrix._41 + comboMatrix._21;
p_planes[3].b = comboMatrix._42 + comboMatrix._22;
p_planes[3].c = comboMatrix._43 + comboMatrix._23;
p_planes[3].d = comboMatrix._44 + comboMatrix._24;
// Near clipping plane
p_planes[4].a = comboMatrix._41 + comboMatrix._31;
p_planes[4].b = comboMatrix._42 + comboMatrix._32;
p_planes[4].c = comboMatrix._43 + comboMatrix._33;
p_planes[4].d = comboMatrix._44 + comboMatrix._34;
// Far clipping plane
p_planes[5].a = comboMatrix._41 - comboMatrix._31;
p_planes[5].b = comboMatrix._42 - comboMatrix._32;
p_planes[5].c = comboMatrix._43 - comboMatrix._33;
p_planes[5].d = comboMatrix._44 - comboMatrix._34;
These planes now are in world-space and can be used to frustum cull world-space objects.
for(int i = 0; i < 6; i++)
{
var dist = dot3(world_space_point.xyz, p_planes[i].xyz) + p_planes[i].d + sphere_radius;
if(dist < 0) return false; // sphere culled
}