I am trying to implement skeletal animation in OpenGL. At this stage, I am trying to have my bone system deform a mesh. I believe the problem is with how I create the matrix Bone::getMatrix().
Background:
Each bone has an offset to it's parent, and if a bone does not have a parent the offset will instead be the world position of the entire skeleton. A collection of bones comprise a skeleton. Currently, for testing purposes, I have 3 bones in a skeleton deforming a skinned mesh.
Again, the issue is that when deforming the mesh, the visual result does not produce the desired effect.
Code:
glm::mat4 Bone::getMatrix(){
Bone* parent = getParent();
glm::mat4 rot = glm::mat4(1.0f);
glm::mat4 trans= glm::mat4(1.0f);
glm::vec3 orient = getOrientation();
//Create Rotation Matrix;
rot *= glm::rotate(orient.x,1.0f,0.0f,0.0f);
rot *= glm::rotate(orient.y,0.0f,1.0f,0.0f);
rot *= glm::rotate(orient.z,0.0f,0.0f,1.0f);
//Create Translation Matrix
if(parent != nullptr){
glm::vec3 OFF = offset;
trans = glm::translate(OFF.x,OFF.y,OFF.z);
}else{
glm::vec3 pos = getPosition();
//trans = glm::translate(pos.x,pos.y,pos.z);
}
glm::mat4 PM = glm::mat4(1.0f);
if(parent != nullptr){
PM *= glm::inverse( parent->getMatrix() );
}else{
}
return PM * trans * rot ;
}
Deforming the mesh:
void Armature::draw(const float& dt){
if(mesh == nullptr){
postMsg("reference copy to mesh is nullptr","draw",34);
return;
}
ArrayVertex3* mesh_Vertex = mesh->getVertex3();
ArrayVertex3 render_vertex;
if(mesh_Vertex == nullptr){
postMsg("mesh vertex structure is null during draw","draw",30);
}else{
render_vertex.reserve(mesh_Vertex->size());
}
int i=0,j=0,k=0;
glPushMatrix();
glTranslatef(10,0,0);
glBegin(GL_POINTS);
for( i =0; i<this->getBoneCount(); i++){
glVertex3fv( glm::value_ptr( glm::vec3( glm::vec4() * getBone(i)->getMatrix() ) ));
}
glEnd();
glBegin(GL_LINES);
for( i =1; i<this->getBoneCount(); i++){
// glVertex3fv( glm::value_ptr(getBone(i)->getTranslationMatrix() * glm::mat3(getBone(i)->getRotationMatrix()) ));
// glVertex3fv( glm::value_ptr(getBone(i-1)->getTranslationMatrix() * glm::mat3(getBone(i-1)->getRotationMatrix()) ));
}
glEnd();
glPopMatrix();
//Loop all vertices
for( i = 0 ; i < int(mesh->getNumberVertexes()); i ++){
render_vertex[i].vtx = glm::vec3(0.0f, 0.0f, 0.0f);
int inf_count = (*mesh_Vertex)[i].numInfluences;
//Loop all this vertices influences
for( j=0; j < inf_count; j ++){
Bone *bone = getBone( (*mesh_Vertex)[i].boneIDs[j] );
float weight = (*mesh_Vertex)[i].boneWeights[j];
glm::vec4 original_vert = glm::vec4((*mesh_Vertex)[i].vtx,1.0f);
glm::vec4 original_norm = glm::vec4((*mesh_Vertex)[i].norm,0.0f);
glm::vec3 T = bone->getTranslationMatrix();
render_vertex[i].vtx += glm::vec3( bone->getMatrix() * original_vert ) * (weight) ;
//Transform normal
render_vertex[i].norm = glm::vec3( bone->getRotationMatrix() * original_norm ) ;
}
(*mesh_Vertex)[i].vtx = render_vertex[i].vtx;
(*mesh_Vertex)[i].norm = glm::normalize(render_vertex[i].norm);
}
}
Result
The rotation is not done according to its joint location
I fount your bone matrix function is a bit chaotic. From my understanding, all the vertexes need to be transformed by global transformation matrix. A global transformation matrix is the concatenation of all the local transformation matrix. The following code is used to calculate global transformation. This article is useful.
http://graphics.ucsd.edu/courses/cse169_w05/2-Skeleton.htm
Some of my previous work using GLSL + assimp + glm
http://www.youtube.com/watch?v=bXBfVl-msYw&list=HL1385555185&feature=mh_lolz
glm::mat4 T = glm::translate(bone->offset);
glm::mat4 R = glm::toMat4(bone->rotate);
glm::mat4 S = glm::scale(bone->scale);
glm::mat4 localTransform = T * R * S;
if(bone->getID()!=0)
{
bone->globalTransform = bone->getParent()->globalTransform * localTransform;
}
else
{
bone->globalTransform = localTransform;
}
for (uint i = 0 ; i < bone->getChildren().size() ; i++) {
updateTransform(bone->getChildren()[i]);
}
I am writing a ray tracer program for my computer graphics class. So far I only have spheres implemented and a shadow ray. The current problem is that when i move my sphere off center it stretches. here is the code that i use to calculate if a ray is intersecting a sphere:
bool Sphere::onSphere(Ray r)
{
float b = (r.dir*2).innerProduct(r.pos + centre*-1);
float c = (r.pos + centre*-1).innerProduct(r.pos + centre*-1) - radius*radius;
return b*b - 4*c >= 0;
}
here is the code that i use to spawn each ray:
for(int i = -cam.width/2; i < cam.width/2; i++)
{
for(int j = -cam.height/2; j < cam.height/2; j++)
{
float normi = (float)i;
float normj = (float)j;
Vector pixlePos = cam.right*normi + cam.up*normj + cam.forward*cam.dist + cam.pos*1;
Vector direction = pixlePos + cam.pos*-1;
direction.normalize();
Vector colour = recursiveRayTrace(Ray(pixlePos, direction), 30, 1, 0);
float red = colour.getX()/255;
float green = colour.getY()/255;
float blue = colour.getZ()/255;
fwrite (&red, sizeof(float), 1, myFile);
fwrite (&green, sizeof(float), 1, myFile);
fwrite (&blue, sizeof(float), 1, myFile);
}
}
recursiveRayTrace:
Vector Scene::recursiveRayTrace(Ray r, float maxDist, int maxBounces, int bounces)
{
if(maxBounces < bounces)
return Vector(0,0,0);
int count = 0;
for(int i = 0; i < spheres.size(); i++)
{
if(spheres.at(i).onSphere(r))
{
Vector colour(ambiant.colour);
for(int j = 0; j < lights.size(); j++)
{
Vector intersection(r.pos + r.dir*spheres.at(i).getT(r));
Ray nRay(intersection, lights.at(i).centre + intersection*-1);
colour = colour + lights.at(i).colour;
}
return colour;
}
}
return Vector(0,0,0);
}
What i get is an sphere that is stretched in the direction of the vector from the center to the center of the circle. I'm not looking for anyone to do my homework. I am just having a really hard time debugging this on. Any hints are appreciated :) Thanks!
Edit: cam.dist is the distance from the camera to the view plane
The stretching is actually a natural consequence of perspective viewing and it is exaggerated if you have a very wide field of view. In other words moving the camera back from your image plane should make it seem more natural.
I'm trying to draw the following shape using OpenGL ES 1.1. And well, I'm stuck, I don't really know how to go about it.
My game currently uses Android's Canvas API, which isn't hardware accelerated, so I'm rewriting it with OpenGL ES. The Canvas class has a method called drawArc which makes drawing this shape very very easy; Canvas.drawArc
Any advice/hints on doing the same with OpenGL ES?
Thank you for reading.
void gltDrawArc(unsigned int const segments, float angle_start, float angle_stop)
{
int i;
float const angle_step = (angle_stop - angle_start)/segments;
GLfloat *arc_vertices;
arc_vertices = malloc(2*sizeof(GLfloat) * (segments+2));
arc_vertices[0] = arc_vertices[1] = 0.
for(i=0; i<segments+1; i++) {
arc_vertices[2 + 2*i ] = cos(angle_start + i*angle_step);
arc_vertices[2 + 2*i + 1] = sin(angle_start + i*angle_step);
}
glVertexPointer(2, GL_FLOAT, 0, arc_vertices);
glEnableClientState(GL_VERTEX_ARRAY);
glDrawArrays(GL_TRIANGLE_FAN, 0, segments+2);
free(arc_vertices);
}
What about just sampling the circle at discrete angles and drawing a GL_TRIANGLE_FAN?
EDIT: Something like this will just draw a sector of a unit circle around the origin in 2D:
glBegin(GL_TRIANGLE_FAN);
glVertex2f(0.0f, 0.0f);
for(angle=startAngle; angle<=endAngle; ++angle)
glVertex2f(cos(angle), sin(angle));
glEnd();
Actually take this more as pseudocode, as sin and cos usually work on radians and I'm using degrees, but you should get the point.
I am new to android programming so I am sure there is probably a better way to do this. But I was following the OpenGL ES 1.0 tutorial on the android developers site http://developer.android.com/resources/tutorials/opengl/opengl-es10.html which walks you through drawing a green triangle. You can follow the link and you will see most of the code I used there. I wanted to draw a circle on the triangle. The code I added is based on the above example posted by datenwolf. And is shown in snippets below:
public class HelloOpenGLES10Renderer implements GLSurfaceView.Renderer {
// the number small triangles used to make a circle
public int segments = 100;
public float mAngle;
private FloatBuffer triangleVB;
// array to hold the FloatBuffer for the small triangles
private FloatBuffer [] segmentsArray = new FloatBuffer[segments];
private void initShapes(){
.
.
.
// stuff to draw holes in the board
int i = 0;
float angle_start = 0.0f;
float angle_stop = 2.0f * (float) java.lang.Math.PI;
float angle_step = (angle_stop - angle_start)/segments;
for(i=0; i<segments; i++) {
float[] holeCoords;
FloatBuffer holeVB;
holeCoords = new float [ 9 ];
// initialize vertex Buffer for triangle
// (# of coordinate values * 4 bytes per float)
ByteBuffer vbb2 = ByteBuffer.allocateDirect(holeCoords.length * 4);
vbb2.order(ByteOrder.nativeOrder());// use the device hardware's native byte order
holeVB = vbb2.asFloatBuffer(); // create a floating point buffer from the ByteBuffer
float x1 = 0.05f * (float) java.lang.Math.cos(angle_start + i*angle_step);
float y1 = 0.05f * (float) java.lang.Math.sin(angle_start + i*angle_step);
float z1 = 0.1f;
float x2 = 0.05f * (float) java.lang.Math.cos(angle_start + i+1*angle_step);
float y2 = 0.05f * (float) java.lang.Math.sin(angle_start + i+1*angle_step);
float z2 = 0.1f;
holeCoords[0] = 0.0f;
holeCoords[1] = 0.0f;
holeCoords[2] = 0.1f;
holeCoords[3] = x1;
holeCoords[4] = y1;
holeCoords[5] = z1;
holeCoords[6] = x2;
holeCoords[7] = y2;
holeCoords[8] = z2;
holeVB.put(holeCoords); // add the coordinates to the FloatBuffer
holeVB.position(0); // set the buffer to read the first coordinate
segmentsArray[i] = holeVB;
}
}
.
.
.
public void onDrawFrame(GL10 gl) {
.
.
.
// Draw hole
gl.glColor4f( 1.0f - 0.63671875f, 1.0f - 0.76953125f, 1.0f - 0.22265625f, 0.0f);
for ( int i=0; i<segments; i++ ) {
gl.glVertexPointer(3, GL10.GL_FLOAT, 0, segmentsArray[i]);
gl.glDrawArrays(GL10.GL_TRIANGLES, 0, 3);
}
}
first post here, and probably an easy one.
I've got the code from Processing's reference site:
float a = 0.0;
float inc = TWO_PI/25.0;
for(int i=0; i<100; i=i+4) {
line(i, 50, i, 50+sin(a)*40.0);
a = a + inc;
}
http://processing.org/reference/sin_.html
However, what I need is a line that follows the curve of a Sin wave, not lines representing points along the curve and ending at the 0 axis. So basically I need to draw an "S" shape with a sin wave equation.
Can someone run me through how to do this?
Thank you in advance,
-Askee
To draw a curve you need to store the previous point's position.
float a = 0.0;
float inc = TWO_PI/25.0;
float prev_x = 0, prev_y = 50, x, y;
for(int i=0; i<100; i=i+4) {
x = i;
y = 50 + sin(a) * 40.0;
line(prev_x, prev_y, x, y);
prev_x = x;
prev_y = y;
a = a + inc;
}
I have an application that defines a real world rectangle on top of an image/photograph, of course in 2D it may not be a rectangle because you are looking at it from an angle.
The problem is, say that the rectangle needs to have grid lines drawn on it, for example if it is 3x5 so I need to draw 2 lines from side 1 to side 3, and 4 lines from side 2 to side 4.
As of right now I am breaking up each line into equidistant parts, to get the start and end point of all the grid lines. However the more of an angle the rectangle is on, the more "incorrect" these lines become, as horizontal lines further from you should be closer together.
Does anyone know the name of the algorithm that I should be searching for?
Yes I know you can do this in 3D, however I am limited to 2D for this particular application.
Here's the solution.
The basic idea is you can find the perspective correct "center" of your rectangle by connecting the corners diagonally. The intersection of the two resulting lines is your perspective correct center. From there you subdivide your rectangle into four smaller rectangles, and you repeat the process. The number of times depends on how accurate you want it. You can subdivide to just below the size of a pixel for effectively perfect perspective.
Then in your subrectangles you just apply your standard uncorrected "textured" triangles, or rectangles or whatever.
You can perform this algorithm without going to the complex trouble of building a 'real' 3d world. it's also good for if you do have a real 3d world modeled, but your textriangles are not perspective corrected in hardware, or you need a performant way to get perspective correct planes without per pixel rendering trickery.
Image: Example of Bilinear & Perspective Transform (Note: The height of top & bottom horizontal grid lines is actually half of the rest lines height, on both drawings)
========================================
I know this is an old question, but I have a generic solution so I decided to publish it hopping it will be useful to the future readers.
The code bellow can draw an arbitrary perspective grid without the need of repetitive computations.
I begin actually with a similar problem: to draw a 2D perspective Grid and then transform the underline image to restore the perspective.
I started to read here:
http://www.imagemagick.org/Usage/distorts/#bilinear_forward
and then here (the Leptonica Library):
http://www.leptonica.com/affine.html
were I found this:
When you look at an object in a plane from some arbitrary direction at
a finite distance, you get an additional "keystone" distortion in the
image. This is a projective transform, which keeps straight lines
straight but does not preserve the angles between lines. This warping
cannot be described by a linear affine transformation, and in fact
differs by x- and y-dependent terms in the denominator.
The transformation is not linear, as many people already pointed out in this thread. It involves solving a linear system of 8 equations (once) to compute the 8 required coefficients and then you can use them to transform as many points as you want.
To avoid including all Leptonica library in my project, I took some pieces of code from it, I removed all special Leptonica data-types & macros, I fixed some memory leaks and I converted it to a C++ class (mostly for encapsulation reasons) which does just one thing:
It maps a (Qt) QPointF float (x,y) coordinate to the corresponding Perspective Coordinate.
If you want to adapt the code to another C++ library, the only thing to redefine/substitute is the QPointF coordinate class.
I hope some future readers would find it useful.
The code bellow is divided into 3 parts:
A. An example on how to use the genImageProjective C++ class to draw a 2D perspective Grid
B. genImageProjective.h file
C. genImageProjective.cpp file
//============================================================
// C++ Code Example on how to use the
// genImageProjective class to draw a perspective 2D Grid
//============================================================
#include "genImageProjective.h"
// Input: 4 Perspective-Tranformed points:
// perspPoints[0] = top-left
// perspPoints[1] = top-right
// perspPoints[2] = bottom-right
// perspPoints[3] = bottom-left
void drawGrid(QPointF *perspPoints)
{
(...)
// Setup a non-transformed area rectangle
// I use a simple square rectangle here because in this case we are not interested in the source-rectangle,
// (we want to just draw a grid on the perspPoints[] area)
// but you can use any arbitrary rectangle to perform a real mapping to the perspPoints[] area
QPointF topLeft = QPointF(0,0);
QPointF topRight = QPointF(1000,0);
QPointF bottomRight = QPointF(1000,1000);
QPointF bottomLeft = QPointF(0,1000);
float width = topRight.x() - topLeft.x();
float height = bottomLeft.y() - topLeft.y();
// Setup Projective trasform object
genImageProjective imageProjective;
imageProjective.sourceArea[0] = topLeft;
imageProjective.sourceArea[1] = topRight;
imageProjective.sourceArea[2] = bottomRight;
imageProjective.sourceArea[3] = bottomLeft;
imageProjective.destArea[0] = perspPoints[0];
imageProjective.destArea[1] = perspPoints[1];
imageProjective.destArea[2] = perspPoints[2];
imageProjective.destArea[3] = perspPoints[3];
// Compute projective transform coefficients
if (imageProjective.computeCoeefficients() != 0)
return; // This can actually fail if any 3 points of Source or Dest are colinear
// Initialize Grid parameters (without transform)
float gridFirstLine = 0.1f; // The normalized position of first Grid Line (0.0 to 1.0)
float gridStep = 0.1f; // The normalized Grd size (=distance between grid lines: 0.0 to 1.0)
// Draw Horizonal Grid lines
QPointF lineStart, lineEnd, tempPnt;
for (float pos = gridFirstLine; pos <= 1.0f; pos += gridStep)
{
// Compute Grid Line Start
tempPnt = QPointF(topLeft.x(), topLeft.y() + pos*width);
imageProjective.mapSourceToDestPoint(tempPnt, lineStart);
// Compute Grid Line End
tempPnt = QPointF(topRight.x(), topLeft.y() + pos*width);
imageProjective.mapSourceToDestPoint(tempPnt, lineEnd);
// Draw Horizontal Line (use your prefered method to draw the line)
(...)
}
// Draw Vertical Grid lines
for (float pos = gridFirstLine; pos <= 1.0f; pos += gridStep)
{
// Compute Grid Line Start
tempPnt = QPointF(topLeft.x() + pos*height, topLeft.y());
imageProjective.mapSourceToDestPoint(tempPnt, lineStart);
// Compute Grid Line End
tempPnt = QPointF(topLeft.x() + pos*height, bottomLeft.y());
imageProjective.mapSourceToDestPoint(tempPnt, lineEnd);
// Draw Vertical Line (use your prefered method to draw the line)
(...)
}
(...)
}
==========================================
//========================================
//C++ Header File: genImageProjective.h
//========================================
#ifndef GENIMAGE_H
#define GENIMAGE_H
#include <QPointF>
// Class to transform an Image Point using Perspective transformation
class genImageProjective
{
public:
genImageProjective();
int computeCoeefficients(void);
int mapSourceToDestPoint(QPointF& sourcePoint, QPointF& destPoint);
public:
QPointF sourceArea[4]; // Source Image area limits (Rectangular)
QPointF destArea[4]; // Destination Image area limits (Perspectivelly Transformed)
private:
static int gaussjordan(float **a, float *b, int n);
bool coefficientsComputed;
float vc[8]; // Vector of Transform Coefficients
};
#endif // GENIMAGE_H
//========================================
//========================================
//C++ CPP File: genImageProjective.cpp
//========================================
#include <math.h>
#include "genImageProjective.h"
// ----------------------------------------------------
// class genImageProjective
// ----------------------------------------------------
genImageProjective::genImageProjective()
{
sourceArea[0] = sourceArea[1] = sourceArea[2] = sourceArea[3] = QPointF(0,0);
destArea[0] = destArea[1] = destArea[2] = destArea[3] = QPointF(0,0);
coefficientsComputed = false;
}
// --------------------------------------------------------------
// Compute projective transform coeeeficients
// RetValue: 0: Success, !=0: Error
/*-------------------------------------------------------------*
* Projective coordinate transformation *
*-------------------------------------------------------------*/
/*!
* computeCoeefficients()
*
* Input: this->sourceArea[4]: (source 4 points; unprimed)
* this->destArea[4]: (transformed 4 points; primed)
* this->vc (computed vector of transform coefficients)
* Return: 0 if OK; <0 on error
*
* We have a set of 8 equations, describing the projective
* transformation that takes 4 points (sourceArea) into 4 other
* points (destArea). These equations are:
*
* x1' = (c[0]*x1 + c[1]*y1 + c[2]) / (c[6]*x1 + c[7]*y1 + 1)
* y1' = (c[3]*x1 + c[4]*y1 + c[5]) / (c[6]*x1 + c[7]*y1 + 1)
* x2' = (c[0]*x2 + c[1]*y2 + c[2]) / (c[6]*x2 + c[7]*y2 + 1)
* y2' = (c[3]*x2 + c[4]*y2 + c[5]) / (c[6]*x2 + c[7]*y2 + 1)
* x3' = (c[0]*x3 + c[1]*y3 + c[2]) / (c[6]*x3 + c[7]*y3 + 1)
* y3' = (c[3]*x3 + c[4]*y3 + c[5]) / (c[6]*x3 + c[7]*y3 + 1)
* x4' = (c[0]*x4 + c[1]*y4 + c[2]) / (c[6]*x4 + c[7]*y4 + 1)
* y4' = (c[3]*x4 + c[4]*y4 + c[5]) / (c[6]*x4 + c[7]*y4 + 1)
*
* Multiplying both sides of each eqn by the denominator, we get
*
* AC = B
*
* where B and C are column vectors
*
* B = [ x1' y1' x2' y2' x3' y3' x4' y4' ]
* C = [ c[0] c[1] c[2] c[3] c[4] c[5] c[6] c[7] ]
*
* and A is the 8x8 matrix
*
* x1 y1 1 0 0 0 -x1*x1' -y1*x1'
* 0 0 0 x1 y1 1 -x1*y1' -y1*y1'
* x2 y2 1 0 0 0 -x2*x2' -y2*x2'
* 0 0 0 x2 y2 1 -x2*y2' -y2*y2'
* x3 y3 1 0 0 0 -x3*x3' -y3*x3'
* 0 0 0 x3 y3 1 -x3*y3' -y3*y3'
* x4 y4 1 0 0 0 -x4*x4' -y4*x4'
* 0 0 0 x4 y4 1 -x4*y4' -y4*y4'
*
* These eight equations are solved here for the coefficients C.
*
* These eight coefficients can then be used to find the mapping
* (x,y) --> (x',y'):
*
* x' = (c[0]x + c[1]y + c[2]) / (c[6]x + c[7]y + 1)
* y' = (c[3]x + c[4]y + c[5]) / (c[6]x + c[7]y + 1)
*
*/
int genImageProjective::computeCoeefficients(void)
{
int retValue = 0;
int i;
float *a[8]; /* 8x8 matrix A */
float *b = this->vc; /* rhs vector of primed coords X'; coeffs returned in vc[] */
b[0] = destArea[0].x();
b[1] = destArea[0].y();
b[2] = destArea[1].x();
b[3] = destArea[1].y();
b[4] = destArea[2].x();
b[5] = destArea[2].y();
b[6] = destArea[3].x();
b[7] = destArea[3].y();
for (i = 0; i < 8; i++)
a[i] = NULL;
for (i = 0; i < 8; i++)
{
if ((a[i] = (float *)calloc(8, sizeof(float))) == NULL)
{
retValue = -100; // ERROR_INT("a[i] not made", procName, 1);
goto Terminate;
}
}
a[0][0] = sourceArea[0].x();
a[0][1] = sourceArea[0].y();
a[0][2] = 1.;
a[0][6] = -sourceArea[0].x() * b[0];
a[0][7] = -sourceArea[0].y() * b[0];
a[1][3] = sourceArea[0].x();
a[1][4] = sourceArea[0].y();
a[1][5] = 1;
a[1][6] = -sourceArea[0].x() * b[1];
a[1][7] = -sourceArea[0].y() * b[1];
a[2][0] = sourceArea[1].x();
a[2][1] = sourceArea[1].y();
a[2][2] = 1.;
a[2][6] = -sourceArea[1].x() * b[2];
a[2][7] = -sourceArea[1].y() * b[2];
a[3][3] = sourceArea[1].x();
a[3][4] = sourceArea[1].y();
a[3][5] = 1;
a[3][6] = -sourceArea[1].x() * b[3];
a[3][7] = -sourceArea[1].y() * b[3];
a[4][0] = sourceArea[2].x();
a[4][1] = sourceArea[2].y();
a[4][2] = 1.;
a[4][6] = -sourceArea[2].x() * b[4];
a[4][7] = -sourceArea[2].y() * b[4];
a[5][3] = sourceArea[2].x();
a[5][4] = sourceArea[2].y();
a[5][5] = 1;
a[5][6] = -sourceArea[2].x() * b[5];
a[5][7] = -sourceArea[2].y() * b[5];
a[6][0] = sourceArea[3].x();
a[6][1] = sourceArea[3].y();
a[6][2] = 1.;
a[6][6] = -sourceArea[3].x() * b[6];
a[6][7] = -sourceArea[3].y() * b[6];
a[7][3] = sourceArea[3].x();
a[7][4] = sourceArea[3].y();
a[7][5] = 1;
a[7][6] = -sourceArea[3].x() * b[7];
a[7][7] = -sourceArea[3].y() * b[7];
retValue = gaussjordan(a, b, 8);
Terminate:
// Clean up
for (i = 0; i < 8; i++)
{
if (a[i])
free(a[i]);
}
this->coefficientsComputed = (retValue == 0);
return retValue;
}
/*-------------------------------------------------------------*
* Gauss-jordan linear equation solver *
*-------------------------------------------------------------*/
/*
* gaussjordan()
*
* Input: a (n x n matrix)
* b (rhs column vector)
* n (dimension)
* Return: 0 if ok, 1 on error
*
* Note side effects:
* (1) the matrix a is transformed to its inverse
* (2) the vector b is transformed to the solution X to the
* linear equation AX = B
*
* Adapted from "Numerical Recipes in C, Second Edition", 1992
* pp. 36-41 (gauss-jordan elimination)
*/
#define SWAP(a,b) {temp = (a); (a) = (b); (b) = temp;}
int genImageProjective::gaussjordan(float **a, float *b, int n)
{
int retValue = 0;
int i, icol=0, irow=0, j, k, l, ll;
int *indexc = NULL, *indexr = NULL, *ipiv = NULL;
float big, dum, pivinv, temp;
if (!a)
{
retValue = -1; // ERROR_INT("a not defined", procName, 1);
goto Terminate;
}
if (!b)
{
retValue = -2; // ERROR_INT("b not defined", procName, 1);
goto Terminate;
}
if ((indexc = (int *)calloc(n, sizeof(int))) == NULL)
{
retValue = -3; // ERROR_INT("indexc not made", procName, 1);
goto Terminate;
}
if ((indexr = (int *)calloc(n, sizeof(int))) == NULL)
{
retValue = -4; // ERROR_INT("indexr not made", procName, 1);
goto Terminate;
}
if ((ipiv = (int *)calloc(n, sizeof(int))) == NULL)
{
retValue = -5; // ERROR_INT("ipiv not made", procName, 1);
goto Terminate;
}
for (i = 0; i < n; i++)
{
big = 0.0;
for (j = 0; j < n; j++)
{
if (ipiv[j] != 1)
{
for (k = 0; k < n; k++)
{
if (ipiv[k] == 0)
{
if (fabs(a[j][k]) >= big)
{
big = fabs(a[j][k]);
irow = j;
icol = k;
}
}
else if (ipiv[k] > 1)
{
retValue = -6; // ERROR_INT("singular matrix", procName, 1);
goto Terminate;
}
}
}
}
++(ipiv[icol]);
if (irow != icol)
{
for (l = 0; l < n; l++)
SWAP(a[irow][l], a[icol][l]);
SWAP(b[irow], b[icol]);
}
indexr[i] = irow;
indexc[i] = icol;
if (a[icol][icol] == 0.0)
{
retValue = -7; // ERROR_INT("singular matrix", procName, 1);
goto Terminate;
}
pivinv = 1.0 / a[icol][icol];
a[icol][icol] = 1.0;
for (l = 0; l < n; l++)
a[icol][l] *= pivinv;
b[icol] *= pivinv;
for (ll = 0; ll < n; ll++)
{
if (ll != icol)
{
dum = a[ll][icol];
a[ll][icol] = 0.0;
for (l = 0; l < n; l++)
a[ll][l] -= a[icol][l] * dum;
b[ll] -= b[icol] * dum;
}
}
}
for (l = n - 1; l >= 0; l--)
{
if (indexr[l] != indexc[l])
{
for (k = 0; k < n; k++)
SWAP(a[k][indexr[l]], a[k][indexc[l]]);
}
}
Terminate:
if (indexr)
free(indexr);
if (indexc)
free(indexc);
if (ipiv)
free(ipiv);
return retValue;
}
// --------------------------------------------------------------
// Map a source point to destination using projective transform
// --------------------------------------------------------------
// Params:
// sourcePoint: initial point
// destPoint: transformed point
// RetValue: 0: Success, !=0: Error
// --------------------------------------------------------------
// Notes:
// 1. You must call once computeCoeefficients() to compute
// the this->vc[] vector of 8 coefficients, before you call
// mapSourceToDestPoint().
// 2. If there was an error or the 8 coefficients were not computed,
// a -1 is returned and destPoint is just set to sourcePoint value.
// --------------------------------------------------------------
int genImageProjective::mapSourceToDestPoint(QPointF& sourcePoint, QPointF& destPoint)
{
if (coefficientsComputed)
{
float factor = 1.0f / (vc[6] * sourcePoint.x() + vc[7] * sourcePoint.y() + 1.);
destPoint.setX( factor * (vc[0] * sourcePoint.x() + vc[1] * sourcePoint.y() + vc[2]) );
destPoint.setY( factor * (vc[3] * sourcePoint.x() + vc[4] * sourcePoint.y() + vc[5]) );
return 0;
}
else // There was an error while computing coefficients
{
destPoint = sourcePoint; // just copy the source to destination...
return -1; // ...and return an error
}
}
//========================================
Using Breton's subdivision method (which is related to Mongo's extension method), will get you accurate arbitrary power-of-two divisions. To split into non-power-of-two divisions using those methods you will have to subdivide to sub-pixel spacing, which can be computationally expensive.
However, I believe you may be able to apply a variation of Haga's Theorem (which is used in origami to divide a side into Nths given a side divided into (N-1)ths) to the perspective-square subdivisions to produce arbitrary divisions from the closest power of 2 without having to continue subdividing.
The most elegant and fastest solution would be to find the homography matrix, which maps rectangle coordinates to photo coordinates.
With a decent matrix library it should not be a difficult task, as long as you know your math.
Keywords: Collineation, Homography, Direct Linear Transformation
However, the recursive algorithm above should work, but probably if your resources are limited, projective geometry is the only way to go.
I think the selected answer is not the best solution available. A better solution is to apply perspective (projective) transformation of a rectangle to simple grid as following Matlab script and image show. You can implement this algorithm with C++ and OpenCV as well.
function drawpersgrid
sz = [ 24, 16 ]; % [x y]
srcpt = [ 0 0; sz(1) 0; 0 sz(2); sz(1) sz(2)];
destpt = [ 20 50; 100 60; 0 150; 200 200;];
% make rectangular grid
[X,Y] = meshgrid(0:sz(1),0:sz(2));
% find projective transform matching corner points
tform = maketform('projective',srcpt,destpt);
% apply the projective transform to the grid
[X1,Y1] = tformfwd(tform,X,Y);
hold on;
%% find grid
for i=1:sz(2)
for j=1:sz(1)
x = [ X1(i,j);X1(i,j+1);X1(i+1,j+1);X1(i+1,j);X1(i,j)];
y = [ Y1(i,j);Y1(i,j+1);Y1(i+1,j+1);Y1(i+1,j);Y1(i,j)];
plot(x,y,'b');
end
end
hold off;
In the special case when you look perpendicular to sides 1 and 3, you can divide those sides in equal parts. Then draw a diagonal, and draw parallels to side 1 through each intersection of the diagonal and the dividing lines drawn earlier.
This a geometric solution I thought out. I do not know whether the 'algorithm' has a name.
Say you want to start by dividing the 'rectangle' into n pieces with vertical lines first.
The goal is to place points P1..Pn-1 on the top line which we can use to draw lines through them to the points where the left and right line meet or parallel to them when such point does not exist.
If the top and bottom line are parallel to each other just place thoose points to split the top line between the corners equidistantly.
Else place n points Q1..Qn on the left line so that theese and the top-left corner are equidistant and i < j => Qi is closer to the top-left cornern than Qj.
In order to map the Q-points to the top line find the intersection S of the line from Qn through the top-right corner and the parallel to the left line through the intersection of top and bottom line. Now connect S with Q1..Qn-1. The intersection of the new lines with the top line are the wanted P-points.
Do this analog for the horizontal lines.
Given a rotation around the y axis, especially if rotation surfaces are planar, the perspective is generated by vertical gradients. These get progressively closer in perspective. Instead of using diagonals to define four rectangles, which can work given powers of two... define two rectangles, left and right. They'll be higher than wide, eventually, if one continues to divide the surface into narrower vertical segments. This can accommodate surfaces that are not square. If a rotation is around the x axis, then horizontal gradients are needed.
What you need to do is represent it in 3D (world) and then project it down to 2D (screen).
This will require you to use a 4D transformation matrix which does the projection on a 4D homogeneous down to a 3D homogeneous vector, which you can then convert down to a 2D screen space vector.
I couldn't find it in Google either, but a good computer graphics books will have the details.
Keywords are projection matrix, projection transformation, affine transformation, homogeneous vector, world space, screen space, perspective transformation, 3D transformation
And by the way, this usually takes a few lectures to explain all of that. So good luck.