this is basic question. im having touble setting the camra not how to set it but what values should i set it to. is there any app that can help with setting camra like you set it and it gives you the values or can you explain what the values stand for and how to are they scaled.
D3DXMatrixLookAtLH(&matView,
&D3DXVECTOR3 (value, value, value), // the camera position
&D3DXVECTOR3 (value, value, value), // the look-at position
&D3DXVECTOR3 (value, value, value));
The D3DXMatrixLookAtLH function is generating a camera matrix, that gets stored in your matView.
After that, the function gets three vectors:
position
look-at
up
Those three vectors stand for:
the position of your camera - where your camera is. It could be (0,0,0) for example. (Those are your x, y and z coordinates in the world.)
the look-at point - this is where your camera looks at. It consists of your position + your (usually) normalized view direction. So when you stand at (0,0,0) and want to look down the negative z-axis, your look-at point is (0,0,-1). If you stand at (1,2,3) and look down the negative x-axis, it's (0,2,3)
the up-vector points up - normally this is (0,1,0).
If you move around now, without looking around, the position and the look-at vectors change to reflect your new position.
If you stand still and look around, only the look-at point changes.
The up-vector only changes if you roll the camera.
There are plenty of nice camera tutorials out there, that show you how to change those three vectors when looking around with the camera - for example this one.
Related
As shown in the image below.
The user moves the ball by changing x,y,z coordinates which correspond to right,left, up, down, near, far movements respectively. But when we change the camera from position A to position B things look weird. Right doesn't look right any more, that because the ball still moves in previous coordinate frame shown by previous z in the image. How can I make the ball move in such a away that changing camera doesn't affect they way its displacement looks.
simple example: if we place the camera such that it looking from positive X axis, the change in the values of z coordinate now, will look like right and left movements. However in reality changing z should be near and far always.
Thought i will answer it here:
I solved it by simply multiplying the cam model view matrix to the balls coordinates.
Here is the code:
glGetDoublev( GL_MODELVIEW_MATRIX, cam );
matsumX=cam[0]*tx+cam[1]*ty+cam[2]*tz+cam[3];
matsumY=cam[4]*tx+cam[5]*ty+cam[6]*tz+cam[7];
matsumZ=cam[8]*tx+cam[9]*ty+cam[10]*tz+cam[11];
where tx,ty,tz are ball's original coordinates and matsumX, matsumY, matsumZ are the new coordinates which change according the camera.
I want to test if any given point in the world is on a quad/plane? The quad/plane can be translated/rotated/scaled by any values but it still should be able to detect if the given point is on it. I also need to get the location where the point should have been, if the quad was not applied any rotation/scale/translation.
For example, consider a quad at 0, 0, 0 with size 100x100, rotated at an angle of 45 degrees along z axis. If my mouse location in the world is at ( x, y, 0, ), I need to know if that point falls on that quad in its current transformation? If yes, then I need to know if no transformations were applied to the quad, where that point would have been on it? Any code sample would be of great help
A ray-casting approach is probably simplest:
Use gluUnProject() to get the world-space direction of the ray to cast into the scene. The ray's origin is the camera position.
Put this ray into object space by transforming it by the inverse of your rectangle's transform. Note that you need to transform both the ray's origin point and direction vector.
Compute the intersection point between this ray and the XY plane with a standard ray-plane intersection test.
Check that the intersection point's x and y values are within your rectangle's bounds, if they are then that's your desired result.
A math library such as GLM will be very helpful if you aren't confident about some of the math involved here, it has corresponding functions such as glm::unProject() as well as functions to invert matrices and do all the other transformations you'd need.
I am currently struggling in finding a formula to rotate my OpenGL "Camera" (I tried do do it via a scene rotation, but have the same issue).
Basically my Camera is at a given position, looking a given point (all indicated to gluLookAt) and I would like to rotate the camera upwards for example, and still looking at the same point.
What should be the right process ?
What input data should I take to decide the amount of movement ? 2D mouse coordinates evolution or 3D unprojected mouse coordinates evolution ?
The trick is to see that a camera-rotation is the same as a scene rotation if you do it at the correct position. Move the camera into the point around which you want to rotate, then rotate the camera, then move back out by the same distance you moved in.
The amount by which you rotate depends on your application. Take G-Earth as an example: if you are close to the surface the rotation is (absolute) small, if you are far from the surface it is large.
If you're creating orbiting(oribitng around LookAt) camera for openGL I sugest you make it with these data:
LookAtPosition- 3D vector
CamUp - 3D unit vector
RelativeCamPosition - 3D unit vector
CamDistance - decimal number
LookAtPosition is a point on which you'll be looking. CamUp is vector that points up from camera, you can see it on this image. It's best to initialize camera at no rotation, so that CamUp = [0,1,0]. Note that it's unit vector so it's magnitude/size/length is always 1. RelativeCamPosition is again unit vector. You get it by taking LookAt to Camera
vector and dividing by it's magnitude, which you'll save in CamDistance. In intialized state it might look as this:
LookAtPosition = [0,0,0]
CamUp = [0,1,0]
RelativeCamPosition = [1,0,0]
CamDistance = 10
You can now get camera position by
CamPosition = LookAtPosition + RelativeCamPosition * CamDistance
But you need to rotate that camera arround right? Well there's a reason for unit vectors - they are easy to use in calculations. I believe you use angles for rotating so you need to use only sine and cosine. Rotate function might look like this:
Rotate(angleX, angleY){
RelativeCamPosition.x = sin(angleX)*cos(angleY);
RelativeCamPosition.z = cos(angleX)*cos(angleY);
RelativeCamPosition.y = sin(angleY);
}
where angleX and angleY are absolute (NOT RELATIVE) rotations in horizontal and vertical direction. You should always use absolute roations because there can be floating point errors while adding. Anyway I just made those calculations on scrap of paper so I hope they're allright.
Edit: I've just noticed that this will work just if your intiial state is like I wrote RelativeCamPosition = [1,0,0]. However it shouldn't be hard to edit them so it works for arbirtary initial state.
Using OpenGL I'm attempting to draw a primitive map of my campus.
Can anyone explain to me how panning, zooming and rotating is usually implemented?
For example, with panning and zooming, is that simply me adjusting my viewport? So I plot and draw all my lines that compose my map, and then as the user clicks and drags it adjusts my viewport?
For panning, does it shift the x/y values of my viewport and for zooming does it increase/decrease my viewport by some amount? What about for rotation?
For rotation, do I have to do affine transforms for each polyline that represents my campus map? Won't this be expensive to do on the fly on a decent sized map?
Or, is the viewport left the same and panning/zooming/rotation is done in some otherway?
For example, if you go to this link you'll see him describe panning and zooming exactly how I have above, by modifying the viewport.
Is this not correct?
They're achieved by applying a series of glTranslate, glRotate commands (that represent camera position and orientation) before drawing the scene. (technically, you're rotating the whole scene!)
There are utility functions like gluLookAt which sorta abstract some details about this.
To simplyify things, assume you have two vectors representing your camera: position and direction.
gluLookAt takes the position, destination, and up vector.
If you implement a vector class, destinaion = position + direction should give you a destination point.
Again to make things simple, you can assume the up vector to always be (0,1,0)
Then, before rendering anything in your scene, load the identity matrix and call gluLookAt
glMatrixMode(GL_MODELVIEW);
glLoadIdentity();
gluLookAt( source.x, source.y, source.z, destination.x, destination.y, destination.z, 0, 1, 0 );
Then start drawing your objects
You can let the user span by changing the position slightly to the right or to the left. Rotation is a bit more complicated as you have to rotate the direction vector. Assuming that what you're rotating is the camera, not some object in the scene.
One problem is, if you only have a direction vector "forward" how do you move it? where is the right and left?
My approach in this case is to just take the cross product of "direction" and (0,1,0).
Now you can move the camera to the left and to the right using something like:
position = position + right * amount; //amount < 0 moves to the left
You can move forward using the "direction vector", but IMO it's better to restrict movement to a horizontal plane, so get the forward vector the same way we got the right vector:
forward = cross( up, right )
To be honest, this is somewhat of a hackish approach.
The proper approach is to use a more "sophisticated" data structure to represent the "orientation" of the camera, not just the forward direction. However, since you're just starting out, it's good to take things one step at a time.
All of these "actions" can be achieved using model-view matrix transformation functions. You should read about glTranslatef (panning), glScalef (zoom), glRotatef (rotation). You also should need to read some basic tutorial about OpenGL, you might find this link useful.
Generally there are three steps that are applied whenever you reference any point in 3d space within opengl.
Given a Local point
Local -> World Transform
World -> Camera Transform
Camera -> Screen Transform (usually a projection. depends on if you're using perspective or orthogonal)
Each of these transforms is taking your 3d point, and multiplying by a matrix.
When you are rotating the camera, it is generally changing the world -> camera transform by multiplying the transform matrix by your rotation/pan/zoom affine transformation. Since all of your points are re-rendered each frame, the new matrix gets applied to your points, and it gives the appearance of a rotation.
Heyo,
I'm currently working on a project where I need to place the camera such that the full motion of a character would be viewable without moving the camera. I have the position where the character starts, as well as the maximum distance that the character will travel in all three directions (X,Y, & Z). I also have the field of view (which is 90 degrees).
Is there an equation that'll figure out where I need to place the camera so it won't have to move to see the full motion?
Note: this is using OpenGL.
Clarification: The camera should be "in front" of the character that's in the motion, not above.
It'll also be moving along a ground plane.
If you make a bounding sphere of the points, all you need to do is keep the camera at a distance greater than or equal to the radius of the bounding sphere / sin(FOV/2).
For example, if you have a bounding sphere with radius Radius, and a specified Field of View FOV, your camera just needs to be at a point "Dist" away, pointing towards the center of the bounding sphere.
The equation for calculating the distance is:
Dist = Radius / sin( FOV/2 );
This will work in 3D, for a camera at any orientation.
Simply having the maximum range of (X, Y, Z) is not on its own sufficient, because the viewing port is essentially pyramid shaped, with the apex of the pyramid being at the eye position.
For the sake of argument, let's assume that all movement is in the (X, Z) plane (i.e. the ground), and the eye is directly above the origin 10m along the Y axis.
Assuming a square viewport, with your 90˚ field of view you'd be able to see from ±10m along both the X and Z axis, but only for objects who are on the ground (Y = 0). As soon as they come off the ground your view is reduced. If it's 1m of the ground then your (X, Z) extent is only ±9m.
Clearly a real camera could be placed anyway in the scene, facing any direction. Even the "roll" angle of the camera could change how much is visible. There are actually infinitely many such camera points, so you will need to constrain your criteria somewhat.
Take the line segment from the startpoint to the endpoint. Construct a plane orthogonal to this line segment through the midpoint of the line segment. Then position the camera somewhere in this plane at an distance of more than the following from the intersection point of plane and line looking at the intersection point. The up vector of the camera must be in the plane and the horizontal field of view must be 90 degrees.
distance = sqrt(dx^2 + dy^2 + dz^2) / 2
This camera positions will all have the startpoint and the endpoint on the left or right border of the view port and verticaly centered.
Another solution might be to write a function that takes the startpoint, the endpoint, and the desired position of both points on the screen. Then just solve the projection equation for the camera transformation.
It depends, for example, if the object is gonna move in a plane, you can just place the camera outside a ball circumscribed its movement area (this depends on the fact that FOV is 90, which is a fortunate angle).
If the object is gonna move in 3D, it's much more difficult. It would help if you'd specify the region where the object moves (cube vs. ball...) and the direction you want to see it from.