Missing triangles when drawing a sphere - c++

I'm trying to draw a sphere with openGL, but I can't find my mistake...
Only half of the triangles are drawn, like here in the picture.
Here is my algorithm so far:
// The angle step used in iteration
float a = (2.0f*M_PI)/8.0;
float c = 0.0f;
for (float theta = 0.0f; theta < 2.0f*M_PI; theta += a, c += a/2.0f)
for (float phi = 0.0f; phi < 2.0f*M_PI; phi += a) {
// Here something is missing...
glBegin(GL_TRIANGLES);
float p_1[3] = {sin(theta)*cos(phi+c),
sin(theta)*sin(phi+c),
cos(theta)};
glVertex3f(p_1[0], p_1[1], p_1[2]);
float p_3[3] = {sin(theta+a)*cos(phi+c+a/2.0f),
sin(theta+a)*sin(phi+c+a/2.0f),
cos(theta+a)};
glVertex3f(p_3[0], p_3[1], p_3[2]);
float p_2[3] = {sin(theta)*cos(phi+c+a),
sin(theta)*sin(phi+c+a),
cos(theta)};
glVertex3f(p_2[0], p_2[1], p_2[2]);
glEnd();
}

Could it be that half your triangles have the wrong orientation?
http://math.hws.edu/graphicsnotes/c3/s2.html

Several problems with this code:
Using floating point tests for loop condition. Keep in mind that float calculations are not exact. You have about a 50% chance of going through the loop one more time than you intended.
With standard spherical coordinates, the range of theta should be from 0..pi, not 0..2*pi.
You need to generate a quad inside the loop, not a triangle. Picture a map. If you look at the area between two lines of longitude and two lines of latitude, it has 4 corners.
I don't understand what your value c is doing. Not sure if this is a problem, or if I'm just not getting it.
Not a correctness problem, but you can place your glBegin and glEnd calls outside the loop.

Related

Why do I have to divide by Z?

I needed to implement 'choosing an object' in a 3D environment. So instead of going with robust, accurate approach, such as raycasting, I decided to take the easy way out. First, I transform the objects world position onto screen coordinates:
glm::mat4 modelView, projection, accum;
glGetFloatv(GL_PROJECTION_MATRIX, (GLfloat*)&projection);
glGetFloatv(GL_MODELVIEW_MATRIX, (GLfloat*)&modelView);
accum = projection * modelView;
glm::mat4 transformed = accum * glm::vec4(objectLocation, 1);
Followed by some trivial code to transform the opengl coordinate system to normal window coordinates, and do a simple distance from the mouse check. BUT that doesn't quite work. In order to translate from world space to screen space, I need one more calculation added on to the end of the function shown above:
transformed.x /= transformed.z;
transformed.y /= transformed.z;
I don't understand why I have to do this. I was under the impression that, once one multiplied your vertex by the accumulated modelViewProjection matrix, you had your screen coordinates. But I have to divide by Z to get it to work properly. In my openGL 3.3 shaders, I never have to divide by Z. Why is this?
EDIT: The code to transform from from opengl coordinate system to screen coordinates is this:
int screenX = (int)((trans.x + 1.f)*640.f); //640 = 1280/2
int screenY = (int)((-trans.y + 1.f)*360.f); //360 = 720/2
And then I test if the mouse is near that point by doing:
float length = glm::distance(glm::vec2(screenX, screenY), glm::vec2(mouseX, mouseY));
if(length < 50) {//you can guess the rest
EDIT #2
This method is called upon a mouse click event:
glm::mat4 modelView;
glm::mat4 projection;
glm::mat4 accum;
glGetFloatv(GL_PROJECTION_MATRIX, (GLfloat*)&projection);
glGetFloatv(GL_MODELVIEW_MATRIX, (GLfloat*)&modelView);
accum = projection * modelView;
float nearestDistance = 1000.f;
gameObject* nearest = NULL;
for(uint i = 0; i < objects.size(); i++) {
gameObject* o = objects[i];
o->selected = false;
glm::vec4 trans = accum * glm::vec4(o->location,1);
trans.x /= trans.z;
trans.y /= trans.z;
int clipX = (int)((trans.x+1.f)*640.f);
int clipY = (int)((-trans.y+1.f)*360.f);
float length = glm::distance(glm::vec2(clipX,clipY), glm::vec2(mouseX, mouseY));
if(length<50) {
nearestDistance = trans.z;
nearest = o;
}
}
if(nearest) {
nearest->selected = true;
}
mouseRightPressed = true;
The code as a whole is incomplete, but the parts relevant to my question works fine. The 'objects' vector contains only one element for my tests, so the loop doesn't get in the way at all.
I've figured it out. As Mr David Lively pointed out,
Typically in this case you'd divide by .w instead of .z to get something useful, though.
My .w values were very close to my .z values, so in my code I change the statement:
transformed.x /= transformed.z;
transformed.y /= transformed.z;
to:
transformed.x /= transformed.w;
transformed.y /= transformed.w;
And it still worked just as before.
https://stackoverflow.com/a/10354368/2159051 explains that division by w will be done later in the pipeline. Obviously, because my code simply multiplies the matrices together, there is no 'later pipeline'. I was just getting lucky in a sense, because my .z value was so close to my .w value, there was the illusion that it was working.
The divide-by-Z step effectively applies the perspective transformation. Without it, you'd have an iso view. Imagine two view-space vertices: A(-1,0,1) and B(-1,0,100).
Without the divide by Z step, the screen coordinates are equal (-1,0).
With the divide-by-Z, they are different: A(-1,0) and B(-0.01,0). So, things farther away from the view-space origin (camera) are smaller in screen space than things that are closer. IE, perspective.
That said: if your projection matrix (and matrix multiplication code) is correct, this should already be happening, as the projection matrix will contain 1/Z scaling components which do this. So, some questions:
Are you really using the output of a projection transform, or just the view transform?
Are you doing this in a pixel/fragment shader? Screen coordinates there are normalized (-1,-1) to (+1,+1), not pixel coordinates, with the origin at the middle of the viewport. Typically in this case you'd divide by .w instead of .z to get something useful, though.
If you're doing this on the CPU, how are you getting this information back to the host?
I guess it is because you are going from 3 dimensions to 2 dimensions, so you are normalizing the 3 dimension world to a 2 dimensional coordinates.
P = (X,Y,Z) in 3D will be q = (x,y) in 2D where x=X/Z and y = Y/Z
So a circle in 3D will not be circle in 2D.
You can check this video out:
https://www.youtube.com/watch?v=fVJeJMWZcq8
I hope I understand your question correctly.

Get 3D model coordinate with 2D screen coordinates gluUnproject

I try to get the 3D coordinates of my OpenGL model. I found this code in the forum, but I donĀ“t understand how the collision is detected.
-(void)receivePoint:(CGPoint)loke
{
GLfloat projectionF[16];
GLfloat modelViewF[16];
GLint viewportI[4];
glGetFloatv(GL_MODELVIEW_MATRIX, modelViewF);
glGetFloatv(GL_PROJECTION_MATRIX, projectionF);
glGetIntegerv(GL_VIEWPORT, viewportI);
loke.y = (float) viewportI[3] - loke.y;
float nearPlanex, nearPlaney, nearPlanez, farPlanex, farPlaney, farPlanez;
gluUnProject(loke.x, loke.y, 0, modelViewF, projectionF, viewportI, &nearPlanex, &nearPlaney, &nearPlanez);
gluUnProject(loke.x, loke.y, 1, modelViewF, projectionF, viewportI, &farPlanex, &farPlaney, &farPlanez);
float rayx = farPlanex - nearPlanex;
float rayy = farPlaney - nearPlaney;
float rayz = farPlanez - nearPlanez;
float rayLength = sqrtf((rayx*rayx)+(rayy*rayy)+(rayz*rayz));
//normalizing rayVector
rayx /= rayLength;
rayy /= rayLength;
rayz /= rayLength;
float collisionPointx, collisionPointy, collisionPointz;
for (int i = 0; i < 50; i++)
{
collisionPointx = rayx * rayLength/i*50;
collisionPointy = rayy * rayLength/i*50;
collisionPointz = rayz * rayLength/i*50;
}
}
In my opinion there a break condition missing. When do I find the collisionPoint?
Another question is:
How do I manipulate the texture at these collision point? I think that I need the corresponding vertex!?
best regards
That code takes the ray from your near clipping place to your far at the position of your loke then partitions it in 50 and interpolates all the possible location of your point in 3D along this ray. At the exit of the loop, in the original code you posted, collisionPointx, y and z is the value of the far most point. There is no "collision" test in that code. you actually need to test your 3D coordinates against a 3D object you want to collide with.

Precision issue - viewpoint far from origin - OpenGL C++

I have a camera class for controlling the camera, with the main function:
void PNDCAMERA::renderMatrix()
{
float dttime=getElapsedSeconds();
GetCursorPos(&cmc.p_cursorPos);
ScreenToClient(hWnd, &cmc.p_cursorPos);
double d_horangle=((double)cmc.p_cursorPos.x-(double)cmc.p_origin.x)/(double)screenWidth*PI;
double d_verangle=((double)cmc.p_cursorPos.y-(double)cmc.p_origin.y)/(double)screenHeight*PI;
cmc.horizontalAngle=d_horangle+cmc.d_horangle_prev;
cmc.verticalAngle=d_verangle+cmc.d_verangle_prev;
if(cmc.verticalAngle>PI/2) cmc.verticalAngle=PI/2;
if(cmc.verticalAngle<-PI/2) cmc.verticalAngle=-PI/2;
changevAngle(cmc.verticalAngle);
changehAngle(cmc.horizontalAngle);
rightVector=glm::vec3(sin(horizontalAngle - PI/2.0f),0,cos(horizontalAngle - PI/2.0f));
directionVector=glm::vec3(cos(verticalAngle) * sin(horizontalAngle), sin(verticalAngle), cos(verticalAngle) * cos(horizontalAngle));
upVector=glm::vec3(glm::cross(rightVector,directionVector));
glm::normalize(upVector);
glm::normalize(directionVector);
glm::normalize(rightVector);
if(moveForw==true)
{
cameraPosition=cameraPosition+directionVector*(float)C_SPEED*dttime;
}
if(moveBack==true)
{
cameraPosition=cameraPosition-directionVector*(float)C_SPEED*dttime;
}
if(moveRight==true)
{
cameraPosition=cameraPosition+rightVector*(float)C_SPEED*dttime;
}
if(moveLeft==true)
{
cameraPosition=cameraPosition-rightVector*(float)C_SPEED*dttime;
}
glViewport(0,0,screenWidth,screenHeight);
glScissor(0,0,screenWidth,screenHeight);
projection_matrix=glm::perspective(60.0f, float(screenWidth) / float(screenHeight), 1.0f, 40000.0f);
view_matrix = glm::lookAt(
cameraPosition,
cameraPosition+directionVector,
upVector);
gShader->bindShader();
gShader->sendUniform4x4("model_matrix",glm::value_ptr(model_matrix));
gShader->sendUniform4x4("view_matrix",glm::value_ptr(view_matrix));
gShader->sendUniform4x4("projection_matrix",glm::value_ptr(projection_matrix));
gShader->sendUniform("camera_position",cameraPosition.x,cameraPosition.y,cameraPosition.z);
gShader->sendUniform("screen_size",(GLfloat)screenWidth,(GLfloat)screenHeight);
};
It runs smooth, I can control the angle with my mouse in X and Y directions, but not around the Z axis (the Y is the "up" in world space).
In my rendering method I render the terrain grid with one VAO call. The grid itself is a quad as the center (highes lod), and the others are L shaped grids scaled by powers of 2. It is always repositioned before the camera, scaled into world space, and displaced by a heightmap.
rcampos.x = round((camera_position.x)/(pow(2,6)*gridscale))*(pow(2,6)*gridscale);
rcampos.y = 0;
rcampos.z = round((camera_position.z)/(pow(2,6)*gridscale))*(pow(2,6)*gridscale);
vPos = vec3(uv.x,0,uv.y)*pow(2,LOD)*gridscale + rcampos;
vPos.y = texture(hmap,vPos.xz/horizontal_scale).r*vertical_scale;
The problem:
The camera starts at the origin, at (0,0,0). When I move it far away from that point, it causes the rotation along the X axis discontinuous. It feels like the mouse cursor was aligned with a grid in screen space, and only the position at grid points were recorded as the cursor movement.
I've also recorded the camera position when it gets pretty noticeable, it's about at 1,000,000 from the origin in X or Z directions. I've noticed that this 'lag' increases linearly with distance, (from the origin).
There is also a little Z-fighting at this point(or similar effect), even if I use a single plane with no displacement, and no planes can overlap. (I use tessellation shaders and render patches.) Black spots appear on the patches. May be caused by fog:
float fc = (view_matrix*vec4(Pos,1)).z/(view_matrix*vec4(Pos,1)).w;
float fResult = exp(-pow(0.00005f*fc, 2.0));
fResult = clamp(fResult, 0.0, 1.0);
gl_FragColor = vec4(mix(vec4(0.0,0.0,0.0,0),vec4(n,1),fResult));
Another strange behavior is the little rotation by the Z axis, this increases with distance too, but I don't use this kind of rotation.
Variable formats:
The vertices are unsigned short format, the indexes are in unsigned int format.
The cmc struct is the camera/cursor struct with double variables.
PI and C_SPEED are #define constants.
Additional information:
The grid is created with the above mentioned ushort array, with the spacing of 1. In the shader I scale it with a constant, then use tessellation to achieve the best performance and the largest view distance.
The final position of a vertex is calculated in the tessellation evaluation shader.
mat4 MVP = projection_matrix*view_matrix*model_matrix;
As you could see I send my matrices to the shader with the glm library.
+Q:
How could the length of a float (or any other format) cause this kind of 'precision loss', or whatever causes the problem. The view_matrix could be a cause of this, but I still cannot output it on the screen at runtime.
PS: I don't know If this helps, but the view matrix at about the 'lag start location' is
-0.49662 -0.49662 0.863129 0
0.00514956 0.994097 0.108373 0
-0.867953 0.0582648 -0.493217 0
1.62681e+006 16383.3 -290126 1
EDIT
Comparing the camera position and view matrix:
view matrix = 0.967928 0.967928 0.248814 0
-0.00387854 0.988207 0.153079 0
-0.251198 -0.149134 0.956378 0
-2.88212e+006 89517.1 -694945 1
position = 2.9657e+006, 6741.52, -46002
It's a long post so I might not answer everything.
I think it is most likely precision issue. Lets start with the camera rotation problem. I think the main problem is here
view_matrix = glm::lookAt(
cameraPosition,
cameraPosition+directionVector,
upVector);
As you said, position is quite a big number like 2.9657e+006 - and look what glm does in glm::lookAt:
GLM_FUNC_QUALIFIER detail::tmat4x4<T> lookAt
(
detail::tvec3<T> const & eye,
detail::tvec3<T> const & center,
detail::tvec3<T> const & up
)
{
detail::tvec3<T> f = normalize(center - eye);
detail::tvec3<T> u = normalize(up);
detail::tvec3<T> s = normalize(cross(f, u));
u = cross(s, f);
In your case, eye and center are these big (very similar) numbers and then glm subtracts them to compute f. This is bad, because if you subtract two almost equal floats, the most significant digits are set to zero, which leaves you with the insignificant (most erroneous) digits. And you use this for further computations, which only emphasizes the error. Check this link for some details.
The z-fighting is similar issue. Z-buffer is not linear, it has the best resolution near the camera because of the perspective divide. The z-buffer range is set according to your near and far clipping plane values. You always want to have the smallest possible ration between far and near values (generally far/near should not be greater than 30000). There is a very good explanation of this on the openGL wiki, I suggest you read it :)
Back to the camera issue - first, I would consider if you really need such a huge scene. I don't think so, but if yes, you could try computing your view matrix differently, compute rotation and translation separately, which could help your case. The way I usually handle camera:
glm::vec3 cameraPos;
glm::vec3 cameraRot;
glm::vec3 cameraPosLag;
glm::vec3 cameraRotLag;
int ox, oy;
const float inertia = 0.08f; //mouse inertia
const float rotateSpeed = 0.2f; //mouse rotate speed (sensitivity)
const float walkSpeed = 0.25f; //walking speed (wasd)
void updateCameraViewMatrix() {
//camera inertia
cameraPosLag += (cameraPos - cameraPosLag) * inertia;
cameraRotLag += (cameraRot - cameraRotLag) * inertia;
// view transform
g_CameraViewMatrix = glm::rotate(glm::mat4(1.0f), cameraRotLag[0], glm::vec3(1.0, 0.0, 0.0));
g_CameraViewMatrix = glm::rotate(g_CameraViewMatrix, cameraRotLag[1], glm::vec3(0.0, 1.0, 0.0));
g_CameraViewMatrix = glm::translate(g_CameraViewMatrix, cameraPosLag);
}
void mousePositionChanged(int x, int y) {
float dx, dy;
dx = (float) (x - ox);
dy = (float) (y - oy);
ox = x;
oy = y;
if (mouseRotationEnabled) {
cameraRot[0] += dy * rotateSpeed;
cameraRot[1] += dx * rotateSpeed;
}
}
void keyboardAction(int key, int action) {
switch (key) {
case 'S':// backwards
cameraPos[0] -= g_CameraViewMatrix[0][2] * walkSpeed;
cameraPos[1] -= g_CameraViewMatrix[1][2] * walkSpeed;
cameraPos[2] -= g_CameraViewMatrix[2][2] * walkSpeed;
break;
...
}
}
This way, the position would not affect your rotation. I should add that I adapted this code from NVIDIA CUDA samples v5.0 (Smoke Particles), I really like it :)
Hope at least some of this helps.

How to properly move the camera in the direction it's facing

I'm trying to figure out how to make the camera in directx move based on the direction it's facing.
Right now the way I move the camera is by passing the camera's current position and rotation to a class called PositionClass. PositionClass takes keyboard input from another class called InputClass and then updates the position and rotation values for the camera, which is then passed back to the camera class.
I've written some code that seems to work great for me, using the cameras pitch and yaw I'm able to get it to go in the direction I've pointed the camera.
However, when the camera is looking straight up (pitch=90) or straight down (pitch=-90), it still changes the cameras X and Z position (depending on the yaw).
The expected behavior is while looking straight up or down it will only move along the Y axis, not along the X or Z axis.
Here's the code that calculates the new camera position
void PositionClass::MoveForward(bool keydown)
{
float radiansY, radiansX;
// Update the forward speed movement based on the frame time
// and whether the user is holding the key down or not.
if(keydown)
{
m_forwardSpeed += m_frameTime * m_acceleration;
if(m_forwardSpeed > (m_frameTime * m_maxSpeed))
{
m_forwardSpeed = m_frameTime * m_maxSpeed;
}
}
else
{
m_forwardSpeed -= m_frameTime * m_friction;
if(m_forwardSpeed < 0.0f)
{
m_forwardSpeed = 0.0f;
}
}
// ToRadians() just multiplies degrees by 0.0174532925f
radiansY = ToRadians(m_rotationY); //yaw
radiansX = ToRadians(m_rotationX); //pitch
// Update the position.
m_positionX += sinf(radiansY) * m_forwardSpeed;
m_positionY += -sinf(radiansX) * m_forwardSpeed;
m_positionZ += cosf(radiansY) * m_forwardSpeed;
return;
}
The significant portion is where the position is updated at the end.
So far I've only been able to deduce that I have horrible math skills.
So, can anyone help me with this dilemma? I've created a fiddle to help test out the math.
Edit: The fiddle uses the same math I used in my MoveForward function, if you set pitch to 90 you can see that the Z axis is still being modified
Thanks to Chaosed0's answer, I was able to figure out the correct formula to calculate movement in a specific direction.
The fixed code below is basically the same as above but now simplified and expanded to make it easier to understand.
First we determine the amount by which the camera will move, in my case this was m_forwardSpeed, but here I will define it as offset.
float offset = 1.0f;
Next you will need to get the camera's X and Y rotation values (in degrees!)
float pitch = camera_rotationX;
float yaw = camera_rotationY;
Then we convert those values into radians
float pitchRadian = pitch * (PI / 180); // X rotation
float yawRadian = yaw * (PI / 180); // Y rotation
Now here is where we determine the new position:
float newPosX = offset * sinf( yawRadian ) * cosf( pitchRadian );
float newPosY = offset * -sinf( pitchRadian );
float newPosZ = offset * cosf( yawRadian ) * cosf( pitchRadian );
Notice that we only multiply the X and Z positions by the cosine of pitchRadian, this is to negate the direction and offset of your camera's yaw when it's looking straight up (90) or straight down (-90).
And finally, you need to tell your camera the new position, which I won't cover because it largely depends on how you've implemented your camera. Apparently doing it this way is out of the norm, and possibly inefficient. However, as Chaosed0 said, it's what makes the most sense to me!
To be honest, I'm not entirely sure I understand your code, so let me try to provide a different perspective.
The way I like to think about this problem is in spherical coordinates, basically just polar in 3D. Spherical coordinates are defined by three numbers: a radius and two angles. One of the angles is yaw, and the other should be pitch, assuming you have no roll (I believe there's a way to get phi if you have roll, but I can't think of how currently). In conventional mathematics notation, theta is your yaw and phi is your pitch, with radius being your move speed, as shown below.
Note that phi and theta are defined differently, depending on where you look.
Basically, the problem is to obtain a point m_forwardSpeed away from your camera, with the right pitch and yaw. To do this, we set the "origin" to your camera position, obtain a spherical coordinate, convert it to cartesian, and then add it to your camera position:
float radius = m_forwardSpeed;
float theta = m_rotationY;
float phi = m_rotationX
//These equations are from the wikipedia page, linked above
float xMove = radius*sinf(phi)*cosf(theta);
float yMove = radius*sinf(phi)*sinf(theta);
float zMove = radius*cosf(phi);
m_positionX += xMove;
m_positionY += yMove;
m_positionZ += zMove;
Of course, you can condense a lot of this code, but I expanded it for clarity.
You can think about this like drawing a sphere around your camera. Each of the points on the sphere is a potential position in the next timestep, depending on the camera's rotation.
This is probably not the most efficient way to do it, but in my opinion it's certainly the easiest way to think about it. It actually looks like this is nearly exactly what you're trying to do in your code, but the operations on the angles are just a little bit off.

Add sin wave to triangle mesh

can someone help me addd a sin wave onto my triangle mesh to help me get a wave effect.
for(int i = 0; i<150; i++){
for(int j = 0; j<150; j++){
grid[i][j] = 0;
glBegin(GL_LINE_LOOP);
glVertex3f(i*3,grid[i][j],j*3);
glVertex3f(i*3,grid[i][j],j*3+3);
glVertex3f(i*3+3,grid[i][j],j*3);
glEnd();
glBegin(GL_LINE_LOOP);
glVertex3f(i*3,grid[i][j],j*3+3);
glVertex3f(i*3+3,grid[i][j],j*3+3);
glVertex3f(i*3+3,grid[i][j],j*3);
glEnd();
}
}
If i've got it right, all i should need to do is add a sin value to grid[i][j], am i right?
Are all the y values to be set to the same grid[i][j]?
It really depends on what you are trying to accomplish.
Are you trying to set up a surface that when looked on edge it looks like a sine wave?
If that is the case then assuming that you are modulating the y-axis and the z-axis plays no effect then you need to determine the frequency you want to use.
i.e y = A * sine (w * x + p) where A is amplitude, w is angular frequency, and p is phase.
You will also have to take into account the number of sample points on the x-axis so that it doesn't look to aliased. Sine is a continuous function but you are take only 150 samples.
Also you may want to reconsider how to calculate and draw your final triangle mesh. Your current code is not the most efficient because you are recalculating your mesh every frame.
You may want to consider initializing grid and then drawing triangle strips, etc. There is a lot online that discusses that.