I have an instance of Object3D and visually it's at (0, 0, 0) but its actual position is shifted (and positions of this object's internals are shifted as well). So, basically geometry of this object is shifted in such a way that it compensates for position shift. So, I want to keep it visuallyy at (0, 0, 0) and change its position to (0, 0, 0).
In simple words, I want to shift object's position vector to this object's geometry position.
How can I do it?
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().
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)
This is most likely a math issue. I have a line geometry that feeds its vertices positions to an arrow helper. The arrow helper runs normalize() on its direction vector to render its output. If I lay them over each other in the view the arrow vector shoots upward while the line is perfectly drawing to the right of the screen. I am expecting the arrow helper to be in sync with the line geometry.
var origin = new THREE.Vector3(0, 10, 0);
var direction = new THREE.Vector3(80, 10, 0);
direction.normalize();
/*
direction after its normalized
Object
x: 0.9922778767136676
y: 0.12403473458920845
z: 0
*/
http://jsbin.com/nazapabaxaco/1/edit?html,js,output
Mathematically this is actually correct for the arrow helper vector. But its not the expected result. If I increase the distance X to say 500 the two converge again but not completely, they are still out of sync and just shooting apart at a really long distance.
The vector in use would be coming from a moving object and not the camera's unproject method so the points need to remain dynamically fed into.
You compute the direction vector by subtracting one line segment endpoint from the other, and then normalizing.
Ok, I am using structs to handle my animations. Right now I am unable to animate both my projectile and my character. As soon as I shoot my projectile my character won't move anymore.
I am using the same variables for character starting position and projectile starting position because I need the projectile to start at the character.
I am wondering on how to use the same variables, but separate them, so when the projectile's coordinates are altered... the characters are not.
I would recommend using separate variables to store the position of your character and your projectile, and initially setting them to the same value.
The following should be something like what you have now:
pos = (0, 0)
def animate():
pos = add(pos, (1, 1))
And the following is an example of using separate variables to store positions; it allows the character to take a path different from the projectile's.
characterposition = (0, 0)
projectileposition = (0, 0)
def animate():
characterposition = add(characterposition, (1, 0))
projectileposition = add(projectileposition, (1, 1))
If you absolutely must use the same variable for both of their positions, you will need to incorporate character and projectile position into that variable separately (i.e. pos = [[0, 0], [0, 0]]).
I'm trying to achieve something simple: Set a translation on X axis, and rotate the object around it's center by a fixed angle.
To achieve this, as far my current knowledge, it's necessary to move the object to the center, rotate, and move back to the original position. Okay. The problem I get although, is that it looks like the object rotate it's local axis and do the last translation along these axis, so it ends in a wrong position.
This is my code:
public void draw(GL10 gl) {
gl.glLoadIdentity();
GLU.gluLookAt(gl, 0, 0, 5, 0, 0, 0, 0, 1, 0);
gl.glTranslatef(x, 0, 0);
gl.glTranslatef(-x, 0, 0);
gl.glRotatef(-80, 0, 1, 0);
gl.glTranslatef(x, 0, 0);
gl.glBindTexture(GL10.GL_TEXTURE_2D, textureId);
gl.glEnableClientState(GL10.GL_VERTEX_ARRAY);
gl.glEnableClientState(GL10.GL_TEXTURE_COORD_ARRAY);
gl.glFrontFace(GL10.GL_CW);
gl.glVertexPointer(3, GL10.GL_FLOAT, 0, verticesBuffer);
gl.glTexCoordPointer(2, GL10.GL_FLOAT, 0, textureBuffer);
gl.glDrawElements(GLES20.GL_TRIANGLES, indices.length, GLES10.GL_UNSIGNED_SHORT, indicesBuffer);
}
Before the rotation the object should be at 0,0,0. It rotates correctly. But then it comes near to the screen as if the x axis would be pointing to me (80°).
Note: I let only "opengl" as tag, since this is a general OpenGL question, the answer should not be Android related.
This is the deprecated way of doing this, but I guess that is no excuse for not answering the question.
OpenGL performs matrices multiplications in reverse order if multiple transforms are applied to a vertex. For example, If a vertex is transformed by MA first, and transformed by MB second, then OpenGL performs MB x MA first before multiplying the vertex. So, the last transform comes first and the first transform occurs last in your code.
gl.glPushMatrix();
gl.glTranslatef(globalX, 0, 0);
gl.glTranslatef(localX, 0 ,0);
gl.glRotatef(-80, 0, 1, 0);
gl.glTranslatef(-globalX, 0, 0);
gl.glPopMatrix();
First move from where you are in a hierarchy of transforms to the origin.
Then rotate around that origin.
Apply some local movement along any axis.
Move the object back to its global positioning.
Use glPushMatrix() and glPopMatrix() to undo changes for elements in the same level of relative positioning, this is having the same parent element to which they are relatively positioned.
The push preserves translations from previous (parent) objects that OpenGL applies after operations in the local code above, as it is the order of a common stack (LIFO), in this case the matrix stack.