I am writing a program that will draw a solid along the curve of a spline. I am using visual studio 2005, and writing in C++ for OpenGL. I'm using FLTK to open my windows (fast and light toolkit).
I currently have an algorithm that will draw a Cardinal Cubic Spline, given a set of control points, by breaking the intervals between the points up into subintervals and drawing linesegments between these sub points. The number of subintervals is variable.
The line drawing code works wonderfully, and basically works as follows: I generate a set of points along the spline curve using the spline equation and store them in an array (as a special datastructure called Pnt3f, where the coordinates are 3 floats and there are some handy functions such as distance, length, dot and crossproduct). Then i have a single loop that iterates through the array of points and draws them as so:
glBegin(GL_LINE_STRIP);
for(pt = 0; pt<=numsubsegements ; ++pt) {
glVertex3fv(pt.v());
}
glEnd();
As stated, this code works great. Now what i want to do is, instead of drawing a line, I want to extrude a solid. My current exploration is using a 'cylinder' quadric to create a tube along the line. This is a bit trickier, as I have to orient openGL in the direction i want to draw the cylinder. My idea is to do this:
Psuedocode:
Push the current matrix,
translate to the first control point
rotate to face the next point
draw a cylinder (length = distance between the points)
Pop the matrix
repeat
My problem is getting the angles between the points. I only need yaw and pitch, roll isnt important. I know take the arc-cosine of the dot product of the two points divided by the magnitude of both points, will return the angle between them, but this is not something i can feed to OpenGL to rotate with. I've tried doing this in 2d, using the XZ plane to get x rotation, and making the points vectors from the origin, but it does not return the correct angle.
My current approach is much simpler. For each plane of rotation (X and Y), find the angle by:
arc-cosine( (difference in 'x' values)/distance between the points)
the 'x' value depends on how your set your plane up, though for my calculations I always use world x.
Barring a few issues of it making it draw in the correct quadrant that I havent worked out yet, I want to get advice to see if this was a good implementation, or to see if someone knew a better way.
You are correct in forming two vectors from the three points in two adjacent line segments and then using the arccosine of the dot product to get the angle between them. To make use of this angle you need to determine the axis around which the rotation should occur. Take the cross product of the same two vectors to get this axis. You can then build a transformation matrix using this axis-angle or pass it as parameters to glRotate.
A few notes:
first of all, this:
for(pt = 0; pt<=numsubsegements ; ++pt) {
glBegin(GL_LINE_STRIP);
glVertex3fv(pt.v());
}
glEnd();
is not a good way to draw anything. You MUST have one glEnd() for every single glBegin(). you probably want to get the glBegin() out of the loop. the fact that this works is pure luck.
second thing
My current exploration is using a
'cylinder' quadric to create a tube
along the line
This will not work as you expect. the 'cylinder' quadric has a flat top base and a flat bottom base. Even if you success in making the correct rotations according to the spline the edges of the flat tops are going to pop out of the volume of your intended tube and it will not be smooth. You can try it in 2D with just a pen and a paper. Try to draw a smooth tube using only shorter tubes with a flat bases. This is impossible.
Third, to your actual question, The definitive tool for such rotations are quaternions. Its a bit complex to explain in this scope but you can find plentyful information anywhere you look.
If you'd have used QT instead of FLTK you could have also used libQGLViewer. It has an integrated Quaternion class which would save you the implementation. If you still have a choice I strongly recommend moving to QT.
Have you considered gluLookAt? Put your control point as the eye point, the next point as the reference point, and make the up vector perpendicular to the difference between the two.
Related
I am using OpenTK(OpenGL) and a general hint will be helpful.
I have a 3d terrain. I have one point on this terrain O(x,y,z) and two perpendicular lines passing through this point that will serve as my X and Y axes.
Now I have a set of 2d points with are in polar coordinates (range,theta). I need to find which points on the terrain correspond to these points. I am not sure what is the best way to do it. I can think of two ideas:
Lets say I am drawing A(x1,y1).
Find the intersection of plane passing through O and A which is perpendicular to the XY plane. This will give me a polyline (semantics may be off). Now on this line, I find a point that is visible from O and is at a distance of the range.
Create a circle which is perpendicular to the XY plane with radius "range", find intersection points on the terrain, find which ones are visible from O and drop rest.
I understand I can find several points which satisfy the conditions, so I will do further check based on topography, but for now I need to get a smaller set which satisfy this condition.
I am new to opengl, but I get geometry pretty well. I am wondering if something like this exists in opengl since it is a standard problem with ground measuring systems.
As you say, both of the options you present will give you more than the one point you need. As I understand your problem, you need only to perform a change of bases from polar coordinates (r, angle) to cartesian coordinates (x,y).
This is fairly straight forward to do. Assuming that the two coordinate spaces share the origin O and that the angle is measured from the x-axis, then point (r_i, angle_i) maps to x_i = r_i*cos(angle_i) and y_i = r_i*sin(angle_i). If those assumptions aren't correct (i.e. if the origins aren't coincident or the angle is not measured from a radii parallel to the x-axis), then the transformation is a bit more complicated but can still be done.
If your terrain is represented as a height map, or 2D array of heights (e.g. Terrain[x][y] = z), once you have the point in cartesian coordinates (x_i,y_i) you can find the height at that point. Of course (x_i, y_i) might not be exactly one of the [x] or [y] indices of the height map.
In that case, I think you have a few options:
Choose the closest (x,y) point and take that height; or
Interpolate the height at (x_i,y_i) based on the surrounding points in the height map.
Unfortunately I am also learning OpenGL and can not provide any specific insights there, but I hope this helps solve your problem.
Reading your description I see a bit of confusion... maybe.
You have defined point O(x,y,z). Fine, this is your pole for the 3D coordinate system. Then you want to find a point defined by polar coordinates. That's fine also - it gives you 2D location. Basically all you need to do is to pinpoint the location in 3D A'(x,y,0), because we are assuming you know the elevation of the A at (r,t), which you of course do from the terrain there.
Angle (t) can be measured only from one axis. Choose which axis will be your polar north and stick to. Then you measure r you have and - voila! - you have your location. What's the point of having 2D coordinate set if you don't use it? Instead, you're adding visibility to the mix - I assume it is important, but highest terrain point on azimuth (t) NOT NECESSARILY will be in the range (r).
You have specific coordinates. Just like RonL suggest, convert to (x,y), find (z) from actual terrain and be done with it.
Unless that's not what you need. But in that case a different question is in order: what do you look for?
- SOLVED -
Warning : I'm not native English speaker
Hi,
I'm currently trying to make a 3D camera, surely because of some mistakes or math basics that I don't have, anyway, I think I will definitely become insane if I don't ask for someone help.
OK lets go.
First, I've a custom game engine that allow to deal with camera only by setting up:
the projection parameters (according to an orthographic or perspective mode)
the view: with a vector 3 for the position and a quaternion for orientation
(and no, we will not discuss about this design right now)
Now I'm writing a camera in my gameplay code (which use the functionalities of the previous engine)
My camera's environment has the following specs:
up_vector = (0, 1, 0)
forward_vector = (0, 0, 1)
angles are in degrees
glm as math lib
In my camera code I handle the player input, convert them into data that I send to my engine.
In the engine I only do:
glm::mat4 projection_view = glm::perspective(...parameters...) * glm::inverse(view_matrix)
And voila I have my matrix for the rendering step.
And now a little scenario with simple geometry.
In a 3D space we have 7 circles, drawn from z = -300 to 300.
The circle at z = -300 is red and the one at 300 is blue,
There are decorative shapes (triangles/box), they are there to facilitate the identification of up and right
When I run the scenario I have got the following disco result !! Thing that I don't want.
As you can see on my exemple of colorful potatoid above, the blue circle is the bigger but is setup to be the farest on z. According to the perspective it should be the smaller. What happened ?
On the other hand, when I use an orthographic camera everything works well.
Any ideas ?
About the Perspective matrix
I generate my perspective matrix with the function glm::perspective(), After a quick check , I have confirmed that my parameters' values are always good, so I can easily imagine that my issue doesn't come from there.
About the View matrix
First, I think my problem must be around here, maybe ... So, I have a vector3 for the position of the camera and 3 float for describing its rotation on each axes.
And here is the experimental part where I don't know what I'm doing !
I copy the previous three float in a vector 3 that I use as an Euleur angles and use a glm quaternion constructor that can create a quat from Euler angles, like that :
glm::quat q(glm::radians(euler_angles));
Finally I send the quaternion like that into the engine, without having use my up and forward vector (anyway I do not see now how to use them)
I work on it for too long and I think my head will explode, the saddest is I think I'm really close.
PS0: Those who help me have my eternal gratitude
PS1: Please do not give me some theory links : I no longer have any neuron, and have already read two interesting and helpless books. Maybe because I have not understood everything yet.
(3D Math Primer for Graphics and Game Development / Mathematics for 3D Game Programming and Computer Graphics, Third Edition)
SOLUTION
It was a dumb mistake ... at the very end of my rendering pipeline, I forget to sort the graphical objects on them "z" according to the camera orientation.
You said:
In my camera code I handle the player input, convert them into data
that I send to my engine. In the engine I only do:
glm::mat4 projection_view = glm::perspective(...parameters...) *
glm::inverse(view_matrix)
And voila I have my matrix for the rendering step.
Are you using the projection matrix when you render the coloured circles?
Should you be using an identity matrix to draw the circle, the model is then viewed according to the view/perspective matrices ?
The triangles and squares look correct - do you have a different transform in effect when you render the circles ?
Hi TonyWilk and thanks
Are you using the projection matrix when you render the coloured circles?
Yes, I generate my projection matrix from the glm::perspective() function and after use my projection_view matrix on my vertices when rendering, as indicated in the first block of code.
Should you be using an identity matrix to draw the circle, the model is then viewed according to the view/perspective matrices ?
I don't know if I have correctly understood this question, but here is an answer.
Theoretically, I do not apply directly the perspective matrix into vertices. I use, in pseudo code:
rendering_matrix = projection_matrix * inverse_camera_view_matrix
The triangles and squares look correct - do you have a different transform in effect when you render the circles ?
At the end, I always use the same matrix. And if the triangles and squares seem to be good, that is only due to an "optical effect". The biggest box is actually associated to the blue circle, and the smaller one to the red
I'm trying to export (3D) bezier curves from Blender to my C++ program. I asked a related question a while back, where I was successfully directed to use De Casteljau's Algorithm to evaluate points (and tangents to these points) along a bezier curve. This works well. In fact, perfectly. I can export the curves and evaluate points along the curve, as well as the tangent to these points, all within my program using De Casteljau's Algorithm.
However, in 3D space a point along a bezier curve and the tangent to this point is not enough to define a "frame" that a camera can lock into, if that makes sense. To put it another way, there is no "up vector" which is required for a camera's orientation to be properly specified at any point along the curve. Mathematically speaking, there are an infinite amount of normal vectors at any point along a 3D bezier curve.
I've noticed when constructing curves in Blender that they aren't merely infinitely thin lines, they actually appear to have a proper 3D orientation defined at any point along them (as shown by the offshooting "arrow lines" in the screenshot below). I'd like to replicate what blender does here as closely as possible in my program. That is, I'd like to be able to form a matrix that represents an orientation at any point along a 3D bezier curve (almost exactly as it would in Blender itself).
Can anyone lend further guidance here, perhaps someone with an intimate knowledge of Blender's source code? (But any advice is welcome, Blender background or not.) I know it's open source, but I'm having a lot of troubles isolating the code responsible for these curve calculations due to the vastness of the program.
Some weeks ago, I have found a solution to this problem. I post it here, in case someone else would need it :
1) For a given point P0, calculate the tangent vector T0.
One simple, easy way, is to take next point on the curve, subtract current point, then normalize result :
T0 = normalize(P1 - P0)
Another, more precise way, to get tangent is to calculate the derivative of your bezier curve function.
Then, pick an arbitrary vector V (for example, you can use (0, 0, 1))
Make N0 = crossproduct(T0, V) and B0 = crossproduct(T0, N0) (dont forget to normalize result vectors after each operation)
You now have a starting set of coordinates ( P0, B0, T0, N0)
This is the initial camera orientation.
2) Then, to calculate next points and their orientation :
Calculate T1 using same method as T0
Here is the trick, new reference frame is calculated from previous frame :
N1 = crossproduct(B0, T1)
B1 = crossproduct(T1, N1)
Proceed using same method for other points. It will results of having camera slightly rotating around tangent vector depending on how curve change its direction. Loopings will be handled correctly (camera wont twist like in my previous answer)
You can watch a live example here (not from me) : http://jabtunes.com/labs/3d/webgl_geometry_extrude_splines.html
Primarily, we know, that the normal vector you're searching for lies on the plane "locally perpendicular" to the curve on the specific point. So the real problem is to choose a single vector on this plane.
I've made an empty object to track the curve and noticed, that it behave similarly to the cart of a rollercoaster: its "up" vector was correlated to the centrifugal force while it was moving along the curve. This one can be uniquely evaluated from the local shape of the curve.
I'm not very good at physics, but I would try to estimate that vector by evaluating two planes: the first is previously mentioned perpendicular plane and the second is a plane made of three neighboring points of a curve segment (if the curve is not straight, these will form a triangle, which describes exactly one plane). Intersection of these two planes will give you an axis and you'll only have to choose a direction of such calculated normal vector.
If I understand you question correcly, what you want is to get 3 orientation vectors (left, front, up) for any point of the curve.
Here is a simple method ( there is a limitation, (*) see below ) :
1) Front vector :
Calculate a 3d point on bezier curve for a given position (t). This is the point for which we will calculate front, left, up vectors. We will call it current_point.
Calculate another 3d point on the curve, next to first one (t + 0.01), let's call it next_point.
Note : i don't write formula here, because i believe you already how to
do that.
Then, to calculate front vector, just substract the two points calculated previously :
vector front = next_point - current_point
Don't forget to normalize the result.
2) Left vector
Define a temporary "up" vector
vector up = vector(0.0f, 1.0f, 0.0f);
Now you can calculate left easily, using front and up :
vector left = CrossProduct(front, up);
3) Up vector
vector up = CrossProduct(left, front);
Using this method you can always calculate a front, left, up for any point along the curve.
(*) NOTE : this wont work in all cases. Imagine you have a loop in you curve, just like a rollercoaster loop. On the top of the loop your calculated up vector will be (0, 1, 0), while you maybe want it to be (0, -1, 0). Only way to solve that is to have two curves : one for points and one for up vectors (from which left and front can be calculated easily).
I'm developing a game that basically has its entire terrain made out of AABB boxes. I know the verticies, minimum, and maximum of each box. I also set up my camera like this:
glMatrixMode(GL_MODELVIEW);
glLoadIdentity();
glRotatef(Camera.rotx,1,0,0);
glRotatef(Camera.roty,0,1,0);
glRotatef(Camera.rotz,0,0,1);
glTranslatef(-Camera.x,-Camera.y,-Camera.z);
What I'm trying to do is basically find the cube the mouse is on. I thought about giving the mouse position a forward directional vector and simply iterating through until the 'mouse bullet' hits something. However this envolves interating through all objects several times. Is there a way I could do it by only iterating through all the objects once?
Thanks
This is usually referred to as 'picking' This here looks like a good gl based link
If that is tldr, then a basic algorithm you could use
sort objects by z (or keep them sorted by z, or depth buffer tricks etc)
iterate and do a bounds test, stopping when you hit the first one.
This is called Ray Tracing (oops, my mistake, it's actually Ray Casting). Every Physics engine has this functionality. You can look at one of the simplest - ODE, or it's derivative - Bullet. They are open-source so you can take out what you don't need. They both have a handy math library that handles all oftenly needed matrix and vertex operations.
They all have demos on how to do exactly this task.
I suggest you consider looking at this issue from a bigger perspective.
The boxes are just points at a lower resolution. The trick is to reduce the resolution of the mouse to figure out which box it is on.
You may have to perform a 2d to 3d conversion (or vice versa). In most games, the mouse lives in a 2d coordinate world. The stuff "under" the mouse is a 2d projection of a 3d universe.
You want to use a 3D picking algorithm. The idea is that you draw a ray from the user's position in the virtual world in the direction of the click. This blog post explains very clearly how to implement such an algorithm. Essentially your screen coordinates need to be transformed from the screen space to the virtual world space. There's a website that has a very good description about the various transformations involved and I can't post the link due to my rank. Search for book of hook's mouse picking algorithm [I do not own the site and I haven't authored the document].
Once you get a ray in the desired direction, you need to perform tests for intersection with the geometries in the real world. Since you have AABB boxes entirely, you can use simple vector equations to check which geometry intersects the ray. I would say that approximating your boxes as a sphere would make life very easy since there is a very simple sphere-ray intersection test. So, your ray would be described by what you obtain from the first step (the ray drawn in the first step) and then you would need to use an intersection test. If you're ok with using spheres, the center of the sphere would be the point you draw your box and the diameter would be the width of your box.
Good Luck!
i have an object in 3d space that i want to align according to a vector.
i already got the Y-rotation out by doing an atan2 on the x and z component of the vector. but i would also like to have an X-rotation to make the object look downwards or upwards.
imagine a plane that does it's pitch yaw roll, just without the roll.
i am using openGL to set the rotations so i will need an Y-angle and an X-angle.
I would not use Euler angles, but rather a Euler axis/angle. For that matter, this is what Opengl glRotate uses as input.
If all you want is to map a vector to another vector, there are an infinite number of rotations to do that. For the shortest one, (the one with the smallest angle of rotation), you can use the vector found by the cross product of your from and to unit vectors.
axis = from X to
from there, the angle of rotation can be found from from.to = cos(theta) (assuming unit vectors)
theta = arccos(from.to)
glRotate(axis, theta) will then transform from to to.
But as I said, this is only one of many rotations that can do the job. You need a full referencial to define better how you want the transform done.
You should use some form of quaternion interpolation (Spherical Linear Interpolation) to animate your object going from its current orientation to this new orientation.
If you store the orientations using Quaternions (vector space math), then you can get the shortest path between two orientations very easily. For a great article, please read Understanding Slerp, Then Not Using It.
If you use Euler angles, you will be subject to gimbal lock and some really weird edge cases.
Actually...take a look at this article. It describes Euler Angles which I believe is what you want here.