I couldn't find the correct and understandable expression of picking in 3D with method of ray-tracing. Has anyone implemented this algorithm in any language? Share directly working code, because since pseudocodes can not be compiled, they are genereally written with lacking parts.
What you have is a position in 2D on the screen. The first thing to do is convert that point from pixels to normalized device coordinates — -1 to 1. Then you need to find the line in 3D space that the point represents. For this, you need the transformation matrix/ces that your 3D app uses to create a projection and camera.
Typically you have 3 matrics: projection, view and model. When you specify vertices for an object, they're in "object space". Multiplying by the model matrix gives the vertices in "world space". Multiplying again by the view matrix gives "eye/camera space". Multiplying again by the projection gives "clip space". Clip space has non-linear depth. Adding a Z component to your mouse coordinates puts them in clip space. You can perform the line/object intersection tests in any linear space, so you must at least move the mouse coordinates to eye space, but it's more convenient to perform the intersection tests in world space (or object space depending on your scene graph).
To move the mouse coordinates from clip space to world space, add a Z-component and multiply by the inverse projection matrix and then the inverse camera/view matrix. To create a line, two points along Z will be computed — from and to.
In the following example, I have a list of objects, each with a position and bounding radius. The intersections of course never match perfectly but it works well enough for now. This isn't pseudocode, but it uses my own vector/matrix library. You'll have to substitute your own in places.
vec2f mouse = (vec2f(mousePosition) / vec2f(windowSize)) * 2.0f - 1.0f;
mouse.y = -mouse.y; //origin is top-left and +y mouse is down
mat44 toWorld = (camera.projection * camera.transform).inverse();
//equivalent to camera.transform.inverse() * camera.projection.inverse() but faster
vec4f from = toWorld * vec4f(mouse, -1.0f, 1.0f);
vec4f to = toWorld * vec4f(mouse, 1.0f, 1.0f);
from /= from.w; //perspective divide ("normalize" homogeneous coordinates)
to /= to.w;
int clickedObject = -1;
float minDist = 99999.0f;
for (size_t i = 0; i < objects.size(); ++i)
{
float t1, t2;
vec3f direction = to.xyz() - from.xyz();
if (intersectSphere(from.xyz(), direction, objects[i].position, objects[i].radius, t1, t2))
{
//object i has been clicked. probably best to find the minimum t1 (front-most object)
if (t1 < minDist)
{
minDist = t1;
clickedObject = (int)i;
}
}
}
//clicked object is objects[clickedObject]
Instead of intersectSphere, you could use a bounding box or other implicit geometry, or intersect a mesh's triangles (this may require building a kd-tree for performance reasons).
[EDIT]
Here's an implementation of the line/sphere intersect (based off the link above). It assumes the sphere is at the origin, so instead of passing from.xyz() as p, give from.xyz() - objects[i].position.
//ray at position p with direction d intersects sphere at (0,0,0) with radius r. returns intersection times along ray t1 and t2
bool intersectSphere(const vec3f& p, const vec3f& d, float r, float& t1, float& t2)
{
//http://wiki.cgsociety.org/index.php/Ray_Sphere_Intersection
float A = d.dot(d);
float B = 2.0f * d.dot(p);
float C = p.dot(p) - r * r;
float dis = B * B - 4.0f * A * C;
if (dis < 0.0f)
return false;
float S = sqrt(dis);
t1 = (-B - S) / (2.0f * A);
t2 = (-B + S) / (2.0f * A);
return true;
}
vec4f from = toWorld * vec4f(mouse, -1.0f, 1.0f);
vec4f to = toWorld * vec4f(mouse, 1.0f, 1.0f);
I'm assuming that 'from' is the position of the mouse cursor? If so then why is its z negative one, if we are assuming openGL coordinates.
Also in this way do we assume that the depth at this time is -1 to +1 right? Rather than the depth of our frustrum.
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 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
I am trying to determine a players depth position on a plane, which defines the walkable ground in a 2D brawler game. The problem is depictured in the following drawing:
C represents the players current position. I need to find the magnitude of vector V. Since I am not strong on linear algebra, the one thing I can think of is: determining the intersection point P of L1 and L2, and then take the magnitude from AP. However, I get the feeling there must be an easier way to find V, since I already know the angle the vector should have, given by vector from AB.
Any input would be appreciated, since I am looking forward to step up my linear algebra game.
Edit: As it is unclear thanks to my lack of drawing skills: the geometry depicted above is a parallelogram. The vector V I am looking for is parallel to the left and right side of the parallelogram. Depth does not mean, that I am looking for the vector perpendicular to the top side, but it refers to the fake depth of a purely 2D game. The parallelogram is therefore used as a means for creating the feeling of walking along a z axis.
The depth of your player (length of V) as measured from the top line in your drawing, is just the difference between A.y and C.y. This is seperate from the slant in the parralelogram, as we're just looking at depth.
example:
float v;
Vector2 a = new Vector2(100, 100); //The point you're measuring from
Vector2 c = new Vector2(150, 150); //Your character position
v = c.y - a.y; // This is the length of V.
//In numbers: 50 = 150 - 100
Illustrated: image not to scale
This works for any coördinate in your plane.
Now if you'd want to get the length of AC is when you'd need to apply some pythagoras, which is a² + b² = c². In the example that would mean in code:
Vector2 a = new Vector2(100, 100);
Vector2 c = new Vector2(150, 150);
float ac1 = Mathf.Sqrt(Mathf.Pow(c.x - a.x, 2) + Mathf.Pow(c.y - a.y, 2));
Now that is quite a sore to have to type out every time, and looks quite scary. But Unity has you covered! There is a Vector method called Distance
float ac2 = Vector2.Distance(a, c);
Which both return 70.71068 which is the length of AC.
This works because for any point c in your area you can draw a right angled triangle from a to c.
Edit as per comment:
If you want your "depth" vector to be parallel with the sides of the paralellogram we can just create a triangle in the parallelogram of which we calculate the hypotenuse.
Since we want the new hypotenuse of our triangle to be parallel to the parallelogram we can use the same angle θ as point B has in your drawing (indicated by pink in mine), of which I understood you know the value.
We also know the length of the adjacent (indicated in blue) side of this new triangle, as that is the height we calculated earlier (c.y - a.y).
Using these two values we can use cosine to find the length of hypotenuse (indicated in red) of the triangle, which is equal to the vector V, in parallel with the parallelogram.
the formula for that is: hypotenuse = adjacent/cos(θ)
Now if we were to put some numbers in this, and for my example I took 55 for the angle θ. It would look like this
float v = 50/(cos(55));
image not to scale
Let's call the lower right vertex of the parallelogram D.
If the long sides of the parallelogram are horizontal, you can find magnitude of V vector by:
V.magnitude = (c.y - a.y) / sin(BAD)
Or if you prefer:
V.magnitude = AB.magnitude * (c.y - a.y)/(b.y - a.y)
In a surface shader, given the world's up axis (and the others too), a world space position and a normal in world space, how can we rotate the worldspace position into the space of the normal?
That is, given a up vector and a non-orthogonal target-up vector, how can we transform the position by rotating its up vector?
I need this so I can get the vertex position only affected by the object's rotation matrix, which I don't have access to.
Here's a graphical visualization of what I want to do:
Up is the world up vector
Target is the world space normal
Pos is arbitrary
The diagram is bidimensional, but I need to solve this for a 3D space.
Looks like you're trying to rotate pos by the same rotation that would transform up to new_up.
Using the rotation matrix found here, we can rotate pos using the following code. This will work either in the surface function or a supplementary vertex function, depending on your application:
// Our 3 vectors
float3 pos;
float3 new_up;
float3 up = float3(0,1,0);
// Build the rotation matrix using notation from the link above
float3 v = cross(up, new_up);
float s = length(v); // Sine of the angle
float c = dot(up, new_up); // Cosine of the angle
float3x3 VX = float3x3(
0, -1 * v.z, v.y,
v.z, 0, -1 * v.x,
-1 * v.y, v.x, 0
); // This is the skew-symmetric cross-product matrix of v
float3x3 I = float3x3(
1, 0, 0,
0, 1, 0,
0, 0, 1
); // The identity matrix
float3x3 R = I + VX + mul(VX, VX) * (1 - c)/pow(s,2) // The rotation matrix! YAY!
// Finally we rotate
float3 new_pos = mul(R, pos);
This is assuming that new_up is normalized.
If the "target up normal" is a constant, the calculation of R could (and should) only happen once per frame. I'd recommend doing it on the CPU side and passing it into the shader as a variable. Calculating it for every vertex/fragment is costly, consider what it is you actually need.
If your pos is a vector-4, just do the above with the first three elements, the fourth element can remain unchanged (it doesn't really mean anything in this context anyway).
I'm away from a machine where I can run shader code, so if I made any syntactical mistakes in the above, please forgive me.
Not tested, but should be able to input a starting point and an axis. Then all you do is change procession which is a normalized (0-1) float along the circumference and your point will update accordingly.
using UnityEngine;
using System.Collections;
public class Follower : MonoBehaviour {
Vector3 point;
Vector3 origin = Vector3.zero;
Vector3 axis = Vector3.forward;
float distance;
Vector3 direction;
float procession = 0f; // < normalized
void Update() {
Vector3 offset = point - origin;
distance = offset.magnitude;
direction = offset.normalized;
float circumference = 2 * Mathf.PI * distance;
angle = (procession % 1f) * circumference;
direction *= Quaternion.AngleAxis(Mathf.Rad2Deg * angle, axis);
Ray ray = new Ray(origin, direction);
point = ray.GetPoint(distance);
}
}
Let's see if I can explain myself.
When you set up the glFrustum view it will give the perspective effect. Near things near & big... far things far & small. Everything looks like it shrinks along its Z axis to create this effect.
Is there a way to make it NOT shrink that much?
To approach perspective view to an orthographic view.... but not that much to lose perspective completely?
Thanks
The angle is conformed by two parameters: heigth of the nearest clipping plane (determined by top and bottom parameters), and the distance of the nearest clipping plane (determined by zNearest).
To make a perspective matrix such that it doesn't shrink the image too much, you can set a smaller height or a further nearest clipping plane.
The thing is to understand that orthographic view is a view with a FOV of zero and a camera position at infinity. So there is a way to approach orthographic view by reducing FOV and moving the camera far away.
I can suggest the following code that computes a near-orthographic projection camera from a given theta FOV value. I use it in a personal project, though it uses custom matrix classes rather than glOrtho and glFrustum, so it might be incorrect. I hope it gives a good general idea though.
void SetFov(int width, int height, float theta)
{
float near = -(width + height);
float far = width + height;
/* Set the projection matrix */
if (theta < 1e-4f)
{
/* The easy way: purely orthogonal projection. */
glOrtho(0, width, 0, height, near, far);
return;
}
/* Compute a view that approximates the glOrtho view when theta
* approaches zero. This view ensures that the z=0 plane fills
* the screen. */
float t1 = tanf(theta / 2);
float t2 = t1 * width / height;
float dist = width / (2.0f * t1);
near += dist;
far += dist;
if (near <= 0.0f)
{
far -= (near - 1.0f);
near = 1.0f;
}
glTranslate3f(-0.5f * width, -0.5f * height, -dist);
glFrustum(-near * t1, near * t1, -near * t2, near * t2, near, far);
}