Help me please to get camera following target point. I can't do it for 2 days already for the new prototype project in my company. Currently i have:
void Camera::followTargetPoint(CIwVec3 target, int32 distance, int32 height)
{
CIwVec3 viewSpaceLocation = (CIwVec3)IwGxGetModelViewMatrix().TransformVec(target);
IwTrace(SCOOTER, ("viewSpaceLocation: x: %d, y: %d, z: %d", viewSpaceLocation.x, viewSpaceLocation.y, viewSpaceLocation.z));
// set the viewspace origin to the camera location in viewspace
//IwGxSetViewSpaceOrg(&viewSpaceLocation);
CIwVec3 pos = CIwVec3(viewSpaceLocation.x, viewMatrix.t.y, viewSpaceLocation.z);
viewMatrix.SetTrans(pos);
//viewMatrix.LookAt(CIwVec3(0,0,-100), target, CIwVec3(0,-100,0));
IwGxSetViewMatrix(&viewMatrix);
}
this method called in my renderer like:
p->Render();
cam->followTargetPoint(p->getModelMatrix().GetTrans(), 3, 100);
What i need is a third person view camera that is located at a distance and height specified by followTargetPoint method. I can only get a first person view with this code cuz when i trying to subtract a distance from z coordinate of target vector i get a flickering model and i know the reason but the question i how to get rid of it?
You need to know which direction your point is travelling in, in order to be able to follow it. Unless you can make assumptions about this, e.g. is it a top-down 2d game?
Anyway, given a direction, your camera position in world space would be something like:
CIwVec3 camPos = ( target - ( targetDirection * distance ) );
camPos.z += height; // assuming Z is up
You then need to build a matrix using this position, but you'll also need 3 axes. The forward axis is:
( target - camPos ).Normalized();
Then you can use some cross products assuming a rough up vector of 0,0,1, to construct the other 2 axes. You should be able to figure out/google for how to build that matrix given a point and a forward vector. I can't just spoon-feed you the whole thing ;)
Then you can set the inverse of your newly constructed matrix as your new view matrix:
viewMatrix = viewMatrix.Inverse();
IwGxSetViewMatrix(&viewMatrix);
Sorry I'm not familar with the IwGx API so I probably typed some function names incorrectly.
Related
I think this is ultimately a pretty simple question, but it's hard to describe, thus, I provide a working example here (in the sample press 'z' to see rotation with unwanted translation and 'x' keys to rotate with a compensating re-position).
Basically, I am trying to rotate an object (a thumbstick) about the z-axis of a complex model loaded via gltf (a model of the oculus rift touch controller). It's easy to rotate about the x-axis because it's 90 deg. orthogonal to the x-axis. About the z-axis, it's harder because the plane the thumbstick is attached to is angled at 30 deg. I realize that if the thumbstick were using local coordinates, this wouldn't be a problem, but 'thumb.rotation.z' does not seem to be using local coordinates and is rotating about the model's (as a whole), or maybe even the scene's global y and z (?). Anyway, after a bunch of futzing around, I was able to get things to work by doing the following:
// occulus plane is angle at 30 deg, which corresponds to
// 5 units forward to 3 units down.
var axis = new THREE.Vector3(0, 5, -3).normalize();
factory.thumbstick.geometry.center();
var dir = (evt.key === 'x' ? 1 : -1);
thumb.rotateOnAxis(axis, factory.ONE_DEG * 5.0 * dir);
Basically, I'm rotating about a "tilted" axis, and then calling 'center' to make thumbstick centered on the pivot point, so it rotates about the pivot point, rather than around the pivot point (like the earth orbiting the sun).
Only problem is that when you call 'geometry.center()' and then call 'rotateOnAxis', it translates the thumbstick to the pivot point:
Note: the position on the thumbstick object is (0,0,0) before and after the calls.
I have empirically determined that if I alter the position of the thumbstick after the translation like so:
// magic numbers compensating position
var zDisp = 0.0475;
var yDisp = zDisp / 6.0
thumb.position.x = 0.001;
thumb.position.y = -yDisp;
thumb.position.z = zDisp;
Then it (almost) returns back to it's original position:
Problem is these numbers were just determined by interactively and repeatedly trying to re-position the thumbstick i.e. empirically. I simply cannot find a programmatic, analytical, api kind of way to restore the original position. Note: saving the original position doesn't work, because it's zero before and after the translation. Some of the things I tried were taking the difference between the bounding spheres of the global object and the thumbstick object, trying to come up with some 'sin x- cos x' relation on one distance etc. but nothing works.
My question is, how can I progammatically reverse the offset due to calling 'geometry.center()' and rotateOnAxis (which translates to the pivot point), without having to resort to hacked, empircal "magic" numbers, that could conceivably change if the gltf model changes.
Of course, if someone can also come up with a better way to achieve this rotation, that would be great too.
What's throwing me is the (peceived?) complexity of the gltf model itself. It's confusing because I have a hard time interpreting it and it's various parts: I'm really not sure where the "center" is, and in certain cases, it appears with the 'THREE.AxesHelper' I'm attaching that what it shows as 'y' is actually 'z' and sometimes 'up' is really 'down' etc, and it gets confusing fast.
Any help would be appreciated.
The breakthrough for me on this was to re-frame the problem as how do I change the pivot point for the thumbstick, rather than how do I move the thumbstick to the (default and pre-existing) pivot point. To paraphrase JFK, "ask not how you can move to the pivot, but ask how the pivot can move to you" :-)
After changing my angle of attack, I pretty quickly found the aforementioned link, which yielded my solution.
I posted an updated glitch here, so now pressing z works as I expected. Here is the relevant code portion:
factory.onModelLoaded = function(evt) {
console.log(`onModelLoaded: entered`);
factory.thumbstick = this.scene.children[1].children[2]
let thumb = factory.thumbstick;
// make the thumb red so it's easier to see
thumb.material = (new THREE.MeshBasicMaterial({color: 0xFF7777}));
// use method from https://stackoverflow.com/questions/28848863/threejs-how-to-rotate-around-objects-own-center-instead-of-world-center/28860849#28860849
// to translate the pivot point of the thumbstick to the the thumbstick center
factory.thumbParent = thumb.parent;
let thumbParent = factory.thumbParent;
thumbParent.remove(thumb);
var box = new THREE.Box3().setFromObject( thumb );
box.getCenter( thumb.position ); // this basically yields my prev. "magic numbers"
// thumb.position.multiplyScalar( - 1 );
var pivot = new THREE.Group();
thumbParent.add( pivot );
pivot.add( thumb );
thumb.geometry.center();
// add axeshelp after centering, otherwise the axes help, as a child of thumb,
// will increase the bounding box of thumb, and positioning will be wrong.
axesHelper = new THREE.AxesHelper();
thumb.add(axesHelper);
}
Which allows my "z" handler to just rotate without having to do translation:
case 'z':
case 'Z':
var axis = new THREE.Vector3(0, 5, -3).normalize();
var dir = (evt.key === 'z' ? 1 : -1);
thumb.rotateOnAxis(axis, factory.ONE_DEG * 5.0 * dir);
break;
Interestingly, it's the call to box.getCenter() that generates numbers very close to my "magic numbers":
box.getCenter()
Vector3 {x: 0.001487499801442027, y: -0.007357006114165027, z: 0.04779449797522323}
My empirical guess was {x: 0.001, y: -0.00791666666, z: 0.0475} which is %error {x: 32.7%, y: 7.6%, z: 0.61%}, so I was pretty close esp. on the z component, but still not the "perfect" numbers of box.getCenter().
I'm using the blender game engine and python I made a script that makes an empty follow my cursor in 3D space. (I use the keyboard for height for now).
Now I wanted to implement a LookAt function for a general object rather than a camera, using python. I want the object to look exactly at the point I'm hovering (the empty position) at the screen. For now I'm using a cube so basically one face of the cube should always face the empty.
So, I thought of using matrices or quaternions but the problem is that All I have is a direction vector and I chose the x axis for the local look direction. So either way I need to calculate the euler angles and convert them to axis-rotation angles. (theta*[axis^]).
The resources I have in the Blender Game Engine is: mathutils (provide quarternions, euler based rotations (via axis-angles), matrices) - though it doesn't have any updated documentation which is just annyoingly horrible! I have to print help to get some sort of info!
Now I've been able to make the object look at the empty when I rotate only the Z axis. I used a little trick that handles the angle sign for me using simple trigonometry, so sign is handled and I don't need any matrix trickery or quarternions. The problem begins when I try to rotate once again - I want to rotate the Y axis for the up-down look (as known in 3D we need two sorts of rotations to face someone, the third is just for rotating the view upside-down - "rolling the camrea") since this rotation axis is the look direction vector.
Here's my script:
import bge
from mathutils import Vector, Matrix
import math
# Basic stuff
cont = bge.logic.getCurrentController()
own = cont.owner
scene = bge.logic.getCurrentScene()
c = scene.objects["Cube"]
e = scene.objects["Empty"]
# axises (we're using localOrientation)
x = Vector((1.0,0.0,0.0))
y = Vector((0.0,1.0,0.0))
z = Vector((0.0,0.0,1.0))
vec = Vector(e.worldPosition - c.worldPosition) # direction vector
# Converting direction vector into euler angles
# Using trigonometry we get: tan(psi) = cos(phi2)/cos(phi1)
# Where phi1 is the angle between x axises (euler angle)
# and phi2 is the euler of the y axises.
# psi is the z rotation angle.
# get cos(euler_angle)
phi1 = vec.dot(x)/vec.length # = cos p1
phi2 = vec.dot(y)/vec.length # = cos p2
phi3 = vec.dot(z)/vec.length # = cos p3
# get the rotation/steer angles
zAngle = math.atan(phi2/phi1)
yAngle = math.atan2(phi3,phi1)
xAngle = math.atan(phi2/phi3)
# use only 2 as the third must adapt (also: view concept - x is the looking direction, rotating it would make rolling)
r = c.localOrientation.to_euler()
r.z = zAngle
r.y = -yAngle
#r.x = xAngle
c.localOrientation = r
Seperately each axis works perfectly, but when combined, there are little jump glitches when I get through the global Y axis.
Also, it seems that the "local" orientation in blender is just the same as the "worldOrientation" which is also annoying cause I'm not sure anymore in what frame of reference I'm working anymore. If anyone knows, please help !
Edit 1:
Appearantely there's a built in logic block that handles this for me and when I press "3D" it tracks AND succeeds on rotating BOTH axises. Though, I still want to know what's the problem with my script! What did the 3D button do that I didn't?
Edit 2:
I tried stop making trigo tricks and found out that when I use local orientation I ALWAYS get a gimbal lock in one axis. That's probably what happened behind the scenes. Thanks for anyone interested, if you have any good trick I'd still be glad to hear =]!
I have a youtube tutorial on how to make the camera look at specific objects. It may help.
https://www.youtube.com/watch?v=hwbObDkiJrE
But the concept, when using the gui, is to open the object->relations panel and for the object you want to be doing the LookAt, you make it the child of the object you want it to follow (the parent). You then select 'Vertex' as the relationship. This will then affect the rotation angles of the child object only.
Try this,
bpy.data.objects['child'].parent = bpy.data.objects['parent']
bpy.data.objects['child'].parent_type = 'VERTEX'
and actually there is more info here
https://blender.stackexchange.com/questions/26108/how-do-i-parent-objects
So according to the Unity documentation RectTransform.anchoredPosition will return the screen coordinates of a UI element if the anchors are touching at the pivot point of the RectTransform. However, if they are separated (in my case positioned at the corners of the rect) they will give you the position of the anchors relative to the pivot point. This is wonderful unless you want to keep appropriate dimensions of a UI object through multiple resolutions and position a different object based on that position at the same time.
Let's break this down. I have object1 and object2. object1 is positioned at (322.5, -600) and when the anchor points meet at the center (pivot) of the object anchoredPosition returns just that and object2 is positioned just fine. On the other hand once I have placed the anchors at the 4 corners of object1 anchoredPosition returns (45.6, -21). Thats just no good. I've even tried using Transform.position and then Camera.WorldToScreenPoint(), but that does just about as much to getting me to my goal.
I was hoping that you might be able to help me find a way to get the actual screen coordinates of this object. If anyone has any insight into this subject it would be greatly appreciated.
Notes: I've already attempted to use RectTranfrom.rect.center and it returned (0, 0)
I've also looked into RectTransformUtility and those helper functions have done all of squat.
anchoredPosition returns "The position of the pivot of this RectTransform relative to the anchor reference point." It has nothing to do with screen coordinates or world space.
If you're looking for the screen coordinates of a UI element in Unity, you can either use rectTransform.TransformPoint or rectTransform.GetWorldCorners to get any of the Vector3s you'd need in world space. Which ever you decide to go with, you can then pass them into Camera.WorldToScreenPoint()
Here's a glimpse on how finding world space coordinates of UI elements works if your stuck and need to roll your own transformations from view-space to world-space.
This may be beneficial if say you need something more than rectTransform.TransformPoint or want to know how this works.
Ok, so you want to do a transformation from normalised UI coordinates in the range [-1, 1], and de-project them back into world space coordinates.
To do this you could use something like Camera.main.ScreenToWorldPoint or Camera.main.ViewportToWorldPoint, or even rectTransform.position if your a lacker.
This is how to do it with just the camera's projection matrix.
/// <summary>
/// Get the world position of an anchor/normalised device coordinate in the range [-1, 1]
/// </summary>
private Vector3 GetAnchor(Vector2 ndcSpace)
{
Vector3 worldPosition;
Vector4 viewSpace = new Vector4(ndcSpace.x, ndcSpace.y, 1.0f, 1.0f);
// Transform to projection coordinate.
Vector4 projectionToWorld = (_mainCamera.projectionMatrix.inverse * viewSpace);
// Perspective divide.
projectionToWorld /= projectionToWorld.w;
// Z-component is backwards in Unity.
projectionToWorld.z = -projectionToWorld.z;
// Transform from camera space to world space.
worldPosition = _mainCamera.transform.position + _mainCamera.transform.TransformVector(projectionToWorld);
return worldPosition;
}
I've found out that you can multiply your coordinate by the 2 times the camera size and divide it to screen height.
I have a panel placed at (0, 1080) on a fullHD screen (1920 x 1080), camera size is 7. So the Y coordinate in world space will be 1080 * 7 * 2 / 1080 = 14 -> (0, 14).
ScreenToWorldPoint convert canvas position to world position :
Camera.main.ScreenToWorldPoint(transform.position)
I'm trying to set the ProjectionMatrix of a Three.js Perspective Camera to match a projection Matrix I calculated with a different program.
So I set the camera's position and rotation like this:
self.camera.position.x = 0;
self.camera.position.y = 0;
self.camera.position.z = 142 ;
self.camera.rotation.x = 0.0;// -0.032
self.camera.rotation.y = 0.0;
self.camera.rotation.z = 0;
Next I created a 4x4 Matrix (called Matrix4 in Three.js) like this:
var projectionMatrix = new THREE.Matrix4(-1426.149, -145.7176, -523.0170, 225.07519, -42.40711, -1463.2367, -23.6839, 524.3322, -0.0174, -0.11928, -0.99270, 0.43826, 0, 0, 0, 1);
and changed the camera's projection Matrix entries like this:
for ( var i = 0; i < 16; i++) {
self.camera.projectionMatrix.elements[i] = projectionMatrix.elements[i];
}
when I now render the scene I just get a black screen and can't see any of the objects I inserted. Turning the angle of the Camera doesn't help either. I still can't see any objects.
If I insert a
self.camera.updateProjectionMatrix();
after setting the camera's projection Matrix to the values of my projectionMatrix the camera is set back to the original Position (x=0,y=0,z=142 and looking at the origin where I created some objects) and the values I set in the camera's matrix seem to have been overwritten. I checked that by printing the cameras projection Matrix to the console. If I do not call the updateProjectionMatrix() function the values stay as I set them.
Does somebody have an idea how to solve this problem?
If I do not call the updateProjectionMatrix() function the values stay as I set them.
Correct, updateProjectionMatrix() calculates those 16 numbers you pasted in your projection matrix based on a bunch of parameters. Those parameters are, the position and rotation you set above, plus the parameters you passed (or default) for the camera. (these actually make the matrixWorld and its inverse.
In case of a perspective camera, you don't have much - near, far, fov and aspect. Left,right,top,bottom are derived from these, with an orthographic camera you set them directly. These are then used to compose the projection matrix.
Scratch a pixel has a REALLY good tutorial on this subject. The next lesson on the openGL projection matrix is actually more relevant to WebGL. left right top and bottom are made from your FOV and your aspect ratio. Add near and far and you've got yourself a projection matrix.
Now, in order for this thing to work, you either have to know what you're doing, or get really lucky. Pasting these numbers from somewhere else and getting it to work is short of winning the lottery. Best case scenario, you can have your scale all wrong and clipping your scene. Worst case, you've mixed a completely different matrix, different XYZ convention, and there's no way you'll get it to work, or at least make sense.
Out of curiosity, what are you trying to do? Are you trying to match your camera to a camera from somewhere else?
I've searched far and wide, so if there's a similar question please forgive me but I just couldn't find it.
To put what I'm trying to do in context: I want to create an infinitely-generated field of stars that disappear as they go offscreen and reappear at the edge of the screen where the camera is moving. I'm working with a top-down view, so it must be pretty simple to achieve this, but alas I haven't a clue.
I'm using the following code to determine whether a star has gone off-screen and then replace it:
//update camera frustum
camera.projScreenMatrix.multiplyMatrices(
camera.projectionMatrix,
camera.matrixWorldInverse
);
camera.frustum.setFromMatrix(camera.projScreenMatrix);
//loop through stars
var stars=scene.stars.geometry.vertices;
for(var i=0;i<stars.length;i++) {
if(!camera.frustum.containsPoint(stars[i])) {
stars[i]=new THREE.Vector3(
// fill in the blank
);
scene.stars.geometry.verticesNeedUpdate=true;
}
}
Since I'm using a perspective camera, I know I'll need to somehow factor in camera.fov and other perspective elements, but as you can tell I'm no expert on the third dimension.
Assuming I have an angle or normalized vector telling me the direction the view is panning, how would I go about creating a vertex along the edge of the screen regardless of its Z position?
If I'm not clear enough, I'll be happy to clarify. Thanks.
I know this is an old question, but I came across it while looking for an answer and found a simple, trigonometry reliant method to get the left edge of the camera frustum, and I'm sharing it in case someone else might find it useful:
// Get half of the cameras field of view angle in radians
var fov = camera.fov / 180 * Math.PI / 2;
// Get the adjacent to calculate the opposite
// This assumes you are looking at the scene
var adjacent = camera.position.distanceTo( scene.position );
// Use trig to get the leftmost point (tangent = o / a)
var left = Math.tan( fov ) * adjacent * camera.aspect;
Basically, this gets the leftmost point, but if you don't multiply by the aspect ratio you should get a point in a circle around your camera frustum, so you could translate a point any direction away from the cameras focus and it would always be outside the frustum.
It works by assuming that the imaginary plane that is the camera is perpendicular to the line connecting the camera and its focus, so there is a straight angle. This should work if you want objects further away as well (so if you want them at a further point from the camera you just need to increase the distance between the focus and the camera).
Well, countless headaches and another question later, I've come up with a fairly makeshift answer. Just in case by some unlikely chance someone else has the same question, the following function plots a point on the scene relative to the camera's current view with whatever Z specified:
//only needs to be defined once
var projector=new THREE.Projector();
//input THREE.Vector3
function(vector) {
var z=vector.z;
vector.z=0;
projector.unprojectVector(vector,camera);
return camera.position.clone().add(
vector
.sub(camera.position)
.normalize()
.multiplyScalar(
-(camera.position.z-z)/vector.z
)
);
The x and y, in this case, both range from -1 to 1 for bottom-left to top-right. You can use position/window.Width and position/window.Height for extra precision (using mouse coordinates or what have you).