how do I get mouse world position. X Y plane only in unity . ScreenToWorldPosition isn't working. I think I need to cast a ray to mouse but not sure.
This is what I am using. doesnt seem to give the correct coordinates or right plane. need for targeting and raycasting.
private void Get3dMousePoint()
{
var screenPosition = Input.mousePosition;
screenPosition.z = 1;
worldPosition = mainCamera.ScreenToWorldPoint(screenPosition);
worldPosition.z = 0;
}
Just need XY coords.
I tried with ScreenToWorldPoint () and it works.
The key I think is in understanding the z coordinate of the position.
Geometrically, in 3D space we need 3 coordinates to define a point. With only 2 coordinates we have a straight line with variable z parameter. To obtain a point from that line, we must choose at what distance (i.e. set z) we want the point sought to be.
Obviously, since the camera is perspective, the coordinates you have at z = 1 are different from those at z = 100, differently from the 2D plane.
If you can figure out how far away, that is, to set the z correctly, you can find the point you want.
Just remember that the z must be greater than the minimum rendering distance of the chamber. I set that very value in the script.
Also remember that the resulting vector will have the z equal to the z position of the camera + the z value of the vector used in ScreenToWorldPoint.
void Get3dMousePoint()
{
Vector3 worldPosition = Camera.main.ScreenToWorldPoint(new Vector3(Input.mousePosition.x, Input.mousePosition.y, Camera.main.nearClipPlane));
print(worldPosition);
}
if you think my answer helped you, you can mark it as accepted and vote positively. I would very much appreciate it :)
Related
This is a question for Unity people or Math geniuses.
I'm making a game where I have a circle object that I can move, but I don't want it to intersect or go into other (static) circles in the world (Physics system isn't good enough in Unity to simply use that, btw).
It's in 3D world, but the circles only ever move on 2 axis.
I was able to get this working perfectly if circle hits only 1 other circle, but not 2 or more.
FYI: All circles are the same size.
Here's my working formula for 1 circle to move it to the edge of the colliding circle if intersecting:
newPosition = PositionOfStaticCircleThatWasJustIntersected + ((positionCircleWasMovedTo - PositionOfStaticCircleThatWasJustIntersected).normalized * circleSize);
But I can't figure out a formula if the moving circle hits 2 (or more) static circles at the same time.
One of the things that confuse me the most is the direction issue depending on how all the circles are positioned and what direction the moving circle is coming from.
Here's an example image of what I'm trying to do.
Since we're operating in a 2D space, let's approach this with some geometry. Taking a close look at your desired outcome, a particular shape become apparent:
There's a triangle here! And since all circles are the same radius, we know even more: this is an isosceles triangle, where two sides are the same length. With that information in hand, the problem basically boils down to:
We know what d is, since it's the distance between the two circles being collided with. And we know what a is, since it's the radius of all the circles. With that information, we can figure out where to place the moved circle. We need to move it d/2 between the two circles (since the point will be equidistant between them), and h away from them.
Calculating the height h is straightforward, since this is a right-angle triangle. According to the Pythagorean theorem:
// a^2 + b^2 = c^2, or rewritten as:
// a = root(c^2 - b^2)
float h = Mathf.Sqrt(Mathf.Pow(2 * a, 2) - Mathf.Pow(d / 2, 2))
Now need to turn these scalar quantities into vectors within our game space. For the vector between the two circles, that's easy:
Vector3 betweenVector = circle2Position - circle1Position
But what about the height vector along the h direction? Well, since all movement is on 2D space, find a direction that your circles don't move along and use it to get the cross product (the perpendicular vector) with the betweenVector using Vector3.Cross(). For
example, if the circles only move laterally:
Vector3 heightVector = Vector3.Cross(betweenVector, Vector3.up)
Bringing this all together, you might have a method like:
Vector3 GetNewPosition(Vector3 movingCirclePosition, Vector3 circle1Position,
Vector3 circle2Position, float radius)
{
float halfDistance = Vector3.Distance(circle1Position, circle2Position) / 2;
float height = Mathf.Sqrt(Mathf.Pow(2 * radius, 2) - Mathf.Pow(halfDistance, 2));
Vector3 betweenVector = circle2Position - circle1Position;
Vector3 heightVector = Vector3.Cross(betweenVector, Vector3.up);
// Two possible positions, on either side of betweenVector
Vector3 candidatePosition1 = circle1Position
+ betweenVector.normalized * halfDistance
+ heightVector.normalized * height;
Vector3 candidatePosition2 = circle1Position
+ betweenVector.normalized * halfDistance
- heightVector.normalized * height;
// Absent any other information, the closer position will be assumed as correct
float distToCandidate1 = Vector3.Distance(movingCirclePosition, candidatePosition1);
float distToCandidate2 = Vector3.Distance(movingCirclePosition, candidatePosition2);
if (distToCandidate1 < distToCandidate2){
return candidatePosition1;
}
else{
return candidatePosition2;
}
}
I want a determined angle in a local rotated axis system. Basically I want to achieve the angle in a plane of a determined rotated axis system. The best way to explain it is graphically.
I can do that projecting the direction from origin to target in my plane, and then use Vector3.Angle(origin forward dir, Projected direction in plane).
Is there is a way to obtain this in a similar way like Quaternion.FromToRotation(from, to).eulerAngles; but, with the Euler angles, with respect to a coordinate system that is not the world's one, but the local rotated one (the one represented by the rotated plane in the picture above)?
So that the desired angle would be, for the rotation in the local y axis: Quaternion.FromToRotation(from, to).localEulerAngles.y, as the locan Euler angles would be (0, -desiredAngle, 0), based on this approach.
Or is there a more direct way than the way I achieved it?
If I understand you correct there are probably many possible ways to go.
I think you could e.g. use Quaternion.ToAngleAxis which returns an Angle and the axis aroun and which was rotated. This axis you can then convert into the local space of your object
public Vector3 GetLocalEulerAngles(Transform obj, Vector3 vector)
{
// As you had it already, still in worldspace
var rotation = Quaternion.FromToRotation(obj.forward, vector);
rotation.ToAngleAxis(out var angle, out var axis);
// Now convert the axis from currently world space into the local space
// Afaik localAxis should already be normalized
var localAxis = obj.InverseTransformDirection(axis);
// Or make it float and only return the angle if you don't need the rest anyway
return localAxis * angle;
}
As alternative as mentioned I guess yes, you could also simply convert the other vector into local space first, then Quaternion.FromToRotation should already be in local space
public Vector3 GetLocalEulerAngles(Transform obj, Vector3 vector)
{
var localVector = obj.InverseTransformDirection(vector);
// Now this already is a rotation in local space
var rotation = Quaternion.FromToRotation(Vector3.forward, localVector);
return rotation.eulerAngles;
}
I am trying to simulate liquid conformity in a container. The container is a Unity cylinder and so is the liquid. I track current volume and max volume and use them to determine the coordinates of the center of where the surface should be. When the container is tilted, each vertex in the upper ring of the cylinder should maintain it's current local x and z values but have a new local y value that is the same height in the global space as the surface center.
In my closest attempt, the surface is flat relative to the world space but the liquid does not touch the walls of the container.
Vector3 v = verts[i];
Vector3 newV = new Vector3(v.x, globalSurfaceCenter.y, v.z);
verts[i] = transform.InverseTransformPoint(newV);
(I understand that inversing the point after using v.x and v.z changes them, but if I change them after the fact the surface is no longer flat...)
I have tried many different approaches and I always end up at this same point or a stranger one.
Also, I'm not looking for any fundamentally different approach to the problem. It's important that I alter the vertices of a cylinder.
EDIT
Thank you, everyone, for your feedback. It helped me make progress with this problem but I've reached another roadblock. I made my code more presentable and took some screenshots of some results as well as a graph model to help you visualize what's happening and give variable names to refer to.
In the following images, colored cubes are instantiated and given the coordinates of some of the different vectors I am using to get my results.
F(red) A(green) B(blue)
H(green) E(blue)
Graphed Model
NOTE: when I refer to capital A and B, I'm referring to the Vector3's in my code.
The cylinders in the images have the following rotations (left to right):
(0,0,45) (45,45,0) (45,0,20)
As you can see from image 1, F is correct when only one dimension of rotation is applied. When two or more are applied, the surface is flat, but not oriented correctly.
If I adjust the rotation of the cylinder after generating these results, I can get the orientation of the surface to make sense, but the number are not what you might expect.
For example: cylinder 3 (on the right side), adjusted to have a surface flat to the world space, would need a rotation of about (42.2, 0, 27.8).
Not sure if that's helpful but it is something that increases my confusion.
My code: (refer to graph model for variable names)
Vector3 v = verts[iter];
Vector3 D = globalSurfaceCenter;
Vector3 E = transform.TransformPoint(new Vector3(v.x, surfaceHeight, v.z));
Vector3 H = new Vector3(gsc.x, E.y, gsc.z);
float a = Vector3.Distance(H, D);
float b = Vector3.Distance(H, E);
float i = (a / b) * a;
Vector3 A = H - D;
Vector3 B = H - E;
Vector3 F = ((A + B)) + ((A + B) * i);
Instantiate(greenPrefab, transform).transform.position = H;
Instantiate(bluePrefab, transform).transform.position = E;
//Instantiate(redPrefab, transform).transform.position = transform.TransformPoint(F);
//Instantiate(greenPrefab, transform).transform.position = transform.TransformPoint(A);
//Instantiate(bluePrefab, transform).transform.position = transform.TransformPoint(B);
Some of the variables in my code and in the graphed model may not be necessary in the end, but my hope is it gives you more to work with.
Bear in mind that I am less than proficient in geometry and math in general. Please use Laymans's terms. Thank you!
And thanks again for taking the time to help me.
As a first step, we can calculate the normal of the upper cylinder surface in the cylinder's local coordinate system. Given the world transform of your cylinder transform, this is simply:
localNormal = inverse(transform) * (0, 1, 0, 0)
Using this normal and the cylinder height h, we can define the plane of the upper cylinder in normal form as
dot(localNormal, (x, y, z) - (0, h / 2, 0)) = 0
I am assuming that your cylinder is centered around the origin.
Using this, we can calculate the y-coordinate for any x/z pair as
y = h / 2 - (localNormal.x * x + localNormal.z * z) / localNormal.y
In Unity, say you have a 3D object,
Of course, it's trivial to get the AABB, Unity has direct functions for that,
(You might have to "add up all the bounding boxes of the renderers" in the usual way, no issue.)
So Unity does indeed have a direct function to give you the 3D AABB box instantly, out of the internal mesh/render pipeline every frame.
Now, for the Camera in question, as positioned, that AABB indeed covers a certain 2D bounding box ...
In fact ... is there some sort of built-in direct way to find that orange 2D box in Unity??
Question - does Unity have a function which immediately gives that 2D frustrum box from the pipeline?
(Note that to do it manually you just make rays (or use world to screen space as Draco mentions, same) for the 8 points of the AABB; encapsulate those in 2D to make the orange box.)
I don't need a manual solution, I'm asking if the engine gives this somehow from the pipeline every frame?
Is there a call?
(Indeed, it would be even better to have this ...)
My feeling is that one or all of the
occlusion system in particular
the shaders
the renderer
would surely know the orange box, and perhaps even the blue box inside the pipeline, right off the graphics card, just as it knows the AABB for a given mesh.
We know that Unity lets you tap the AABB 3D box instantly every frame for a given mesh: In fact does Unity give the "2D frustrum bound" as shown here?
As far as I am aware, there is no built in for this.
However, finding the extremes yourself is really pretty easy. Getting the mesh's bounding box (the cuboid shown in the screenshot) is just how this is done, you're just doing it in a transformed space.
Loop through all the verticies of the mesh, doing the following:
Transform the point from local to world space (this handles dealing with scale and rotation)
Transform the point from world space to screen space
Determine if the new point's X and Y are above/below the stored min/max values, if so, update the stored min/max with the new value
After looping over all vertices, you'll have 4 values: min-X, min-Y, max-X, and max-Y. Now you can construct your bounding rectangle
You may also wish to first perform a Gift Wrapping of the model first, and only deal with the resulting convex hull (as no points not part of the convex hull will ever be outside the bounds of the convex hull). If you intend to draw this screen space rectangle while the model moves, scales, or rotates on screen, and have to recompute the bounding box, then you'll want to do this and cache the result.
Note that this does not work if the model animates (e.g. if your humanoid stands up and does jumping jacks). Solving for the animated case is much more difficult, as you would have to treat every frame of every animation as part of the original mesh for the purposes of the convex hull solving (to insure that none of your animations ever move a part of the mesh outside the convex hull), increasing the complexity by a power.
3D bounding box
Get given GameObject 3D bounding box's center and size
Compute 8 corners
Transform positions to GUI space (screen space)
Function GUI3dRectWithObject will return the 3D bounding box of given GameObject on screen.
2D bounding box
Iterate through every vertex in a given GameObject
Transform every vertex's position to world space, and transform to GUI space (screen space)
Find 4 corner value: x1, x2, y1, y2
Function GUI2dRectWithObject will return the 2D bounding box of given GameObject on screen.
Code
public static Rect GUI3dRectWithObject(GameObject go)
{
Vector3 cen = go.GetComponent<Renderer>().bounds.center;
Vector3 ext = go.GetComponent<Renderer>().bounds.extents;
Vector2[] extentPoints = new Vector2[8]
{
WorldToGUIPoint(new Vector3(cen.x-ext.x, cen.y-ext.y, cen.z-ext.z)),
WorldToGUIPoint(new Vector3(cen.x+ext.x, cen.y-ext.y, cen.z-ext.z)),
WorldToGUIPoint(new Vector3(cen.x-ext.x, cen.y-ext.y, cen.z+ext.z)),
WorldToGUIPoint(new Vector3(cen.x+ext.x, cen.y-ext.y, cen.z+ext.z)),
WorldToGUIPoint(new Vector3(cen.x-ext.x, cen.y+ext.y, cen.z-ext.z)),
WorldToGUIPoint(new Vector3(cen.x+ext.x, cen.y+ext.y, cen.z-ext.z)),
WorldToGUIPoint(new Vector3(cen.x-ext.x, cen.y+ext.y, cen.z+ext.z)),
WorldToGUIPoint(new Vector3(cen.x+ext.x, cen.y+ext.y, cen.z+ext.z))
};
Vector2 min = extentPoints[0];
Vector2 max = extentPoints[0];
foreach (Vector2 v in extentPoints)
{
min = Vector2.Min(min, v);
max = Vector2.Max(max, v);
}
return new Rect(min.x, min.y, max.x - min.x, max.y - min.y);
}
public static Rect GUI2dRectWithObject(GameObject go)
{
Vector3[] vertices = go.GetComponent<MeshFilter>().mesh.vertices;
float x1 = float.MaxValue, y1 = float.MaxValue, x2 = 0.0f, y2 = 0.0f;
foreach (Vector3 vert in vertices)
{
Vector2 tmp = WorldToGUIPoint(go.transform.TransformPoint(vert));
if (tmp.x < x1) x1 = tmp.x;
if (tmp.x > x2) x2 = tmp.x;
if (tmp.y < y1) y1 = tmp.y;
if (tmp.y > y2) y2 = tmp.y;
}
Rect bbox = new Rect(x1, y1, x2 - x1, y2 - y1);
Debug.Log(bbox);
return bbox;
}
public static Vector2 WorldToGUIPoint(Vector3 world)
{
Vector2 screenPoint = Camera.main.WorldToScreenPoint(world);
screenPoint.y = (float)Screen.height - screenPoint.y;
return screenPoint;
}
Reference: Is there an easy way to get on-screen render size (bounds)?
refer to this
It needs the game object with skinnedMeshRenderer.
Camera camera = GetComponent();
SkinnedMeshRenderer skinnedMeshRenderer = target.GetComponent();
// Get the real time vertices
Mesh mesh = new Mesh();
skinnedMeshRenderer.BakeMesh(mesh);
Vector3[] vertices = mesh.vertices;
for (int i = 0; i < vertices.Length; i++)
{
// World space
vertices[i] = target.transform.TransformPoint(vertices[i]);
// GUI space
vertices[i] = camera.WorldToScreenPoint(vertices[i]);
vertices[i].y = Screen.height - vertices[i].y;
}
Vector3 min = vertices[0];
Vector3 max = vertices[0];
for (int i = 1; i < vertices.Length; i++)
{
min = Vector3.Min(min, vertices[i]);
max = Vector3.Max(max, vertices[i]);
}
Destroy(mesh);
// Construct a rect of the min and max positions
Rect r = Rect.MinMaxRect(min.x, min.y, max.x, max.y);
GUI.Box(r, "");
I am working on a game in sprite kit and have been trying to get a point in front of a node. I've been reading up on trigonometry but have not been able to do it.
The problem: Get a CGPoint x units in front of an SKSpriteNode, relative to zRotation. See the illustration here: http://i.stack.imgur.com/TGZ51.png
I have understood that i can use the adjacent and opposite lengths in the triangle to calculate the distance of the hypotenuse (and that the hypotenuse is a vector?). However, i've failed to understand how to get this vector relative to current zPosition and how to get a point from the vector.
I would be grateful if anyone can provide some sample code or point me in a direction where i can find more info.
Thanks a lot!
I solved it after trying some more and here's how i did it:
- (CGVector)convertAngleToVector:(CGFloat)radians {
CGVector vector;
vector.dx = cos(radians) * 40;
vector.dy = sin(radians) * 40;
return vector;
}
I call the method with the sprites zRotation which gives me a vector. The number 40 decides how long the vector is. Then i just added the vector to the current position.