I have been attempting to get matrix rotation around an arbitrary axis working, I think I'm close but I have a bug. I am relatively new to 3D rotations and have a basic understanding of what is going on.
public static Matrix4D Rotate(Vector3D u, Vector3D v)
{
double angle = Acos(u.Dot(v));
Vector3D axis = u.Cross(v);
double c = Cos(angle);
double s = Sin(angle);
double t = 1 - c;
return new Matrix4D(new double [,]
{
{ c + Pow(axis.X, 2) * t, axis.X * axis.Y * t -axis.Z * s, axis.X * axis.Z * t + axis.Y * s, 0 },
{ axis.Y * axis.X * t + axis.Z * s, c + Pow(axis.Y, 2) * t, axis.Y * axis.Z * t - axis.X * s, 0 },
{ axis.Z * axis.X * t - axis.Y * s, axis.Z * axis.Y * t + axis.X * s, c + Pow(axis.Z, 2) * t, 0 },
{ 0, 0, 0, 1 }
});
}
The above code is the algorithm for the matrix rotation. When I test the algorithm with unit vectors I get the following:
Matrix4D rotationMatrix = Matrix4D.Rotate(new Vector3D(1, 0, 0), new Vector3D(0, 0, 1));
Vector4D vectorToRotate = new Vector4D(1,0,0,0);
Vector4D result = rotationMatrix * vectorToRotate;
//Result
X = 0.0000000000000000612;
Y = 0;
Z = 1;
Length = 1;
With a 90 degree rotation I find it works almost perfectly. Now lets look at a 45 degree rotation:
Matrix4D rotationMatrix = Matrix4D.Rotate(new Vector3D(1, 0, 0), new Vector3D(1, 0, 1).Normalize());
Vector4D vectorToRotate = new Vector4D(1,0,0,0);
Vector4D result = rotationMatrix * vectorToRotate;
//Result
X = .70710678118654746;
Y = 0;
Z = .5;
Length = 0.8660254037844386;
When we take the atan(.5/.707) we find that we have a 35.28 degree rotation instead of a 45 degree rotation. Also the length of the vector is changing from 1 to .866. Does anyone have any tips on what I am doing wrong?
Your matrix code looks correct, but you need to normalize the axis too (just because u and v are normalized doesn't mean u cross v also is).
(I would also recommend using a simple multiply instead of Pow, for performance and accuracy reasons; but this is minor and not the source of your problem)
Just for posterity, this is the code that I use for this exact thing. I alway recommend using Atan2(dy,dx) instead of Acos(dx) because it is more stable numerically at angles near 90°.
public static class Rotations
{
public static Matrix4x4 FromTwoVectors(Vector3 u, Vector3 v)
{
Vector3 n = Vector3.Cross(u, v);
double sin = n.Length(); // use |u×v| = |u||v| SIN(θ)
double cos = Vector3.Dot(u, v); // use u·v = |u||v| COS(θ)
// what if u×v=0, or u·v=0. The full quadrant arctan below is fine with it.
double angle = Atan2(sin, cos);
n = Vector3.Normalize(n);
return FromAxisAngle(n, angle);
}
public static Matrix4x4 FromAxisAngle(Vector3 n, double angle)
{
// Asssume `n` is normalized
double x = n.X, y = n.Y, z = n.Z;
double sin = Sin(angle);
double vcos = 1.0-Cos(angle);
return new Matrix4x4(
1.0-vcos*(y*y+z*z), vcos*x*y-sin*z, vcos*x*z+sin*y, 0,
vcos*x*y+sin*z, 1.0-vcos*(x*x+z*z), vcos*y*z-sin*x, 0,
vcos*x*z-sin*y, vcos*y*z+sin*x, 1.0-vcos*(x*x+y*y), 0,
0, 0, 0, 1.0);
}
}
Code is C#
Related
I am trying to place evenly-spaced markers/dots on a quadratic curve drawn with HTML Canvas API. Found a nice article explaining how the paths are calculated in the first place, at determining coordinates on canvas curve.
There is a formula, at the end, to calculate the angle:
function getQuadraticAngle(t, sx, sy, cp1x, cp1y, ex, ey) {
var dx = 2*(1-t)*(cp1x-sx) + 2*t*(ex-cp1x);
var dy = 2*(1-t)*(cp1y-sy) + 2*t*(ey-cp1y);
return Math.PI / 2 - Math.atan2(dx, dy);
}
The x/y pairs that we pass, are the current position, the control point and the end position of the curve - exactly what is needed to pass to the canvas context, and t is a value from 0 to 1. Unless I somehow misunderstood the referenced article.
I want to do something very similar - place my markers over the distance specified s, rather than use t. This means, unless I am mistaken, that I need to calculate the length of the "curved path" and from there, I could probably use the above formula.
I found a solution for the length in JavaScript at length of quadratic curve. The formula is similar to:
.
And added the below function:
function quadraticBezierLength(x1, y1, x2, y2, x3, y3) {
let a, b, c, d, e, u, a1, e1, c1, d1, u1, v1x, v1y;
v1x = x2 * 2;
v1y = y2 * 2;
d = x1 - v1x + x3;
d1 = y1 - v1y + y3;
e = v1x - 2 * x1;
e1 = v1y - 2 * y1;
c1 = a = 4 * (d * d + d1 * d1);
c1 += b = 4 * (d * e + d1 * e1);
c1 += c = e * e + e1 * e1;
c1 = 2 * Math.sqrt(c1);
a1 = 2 * a * (u = Math.sqrt(a));
u1 = b / u;
a = 4 * c * a - b * b;
c = 2 * Math.sqrt(c);
return (
(a1 * c1 + u * b * (c1 - c) + a * Math.log((2 * u + u1 + c1) / (u1 + c))) /
(4 * a1)
);
}
Now, I am trying to space markers evenly. I thought that making "entropy" smooth - dividing the total length by the step length would result in the n markers, so going using the 1/nth step over t would do the trick. However, this does not work. The correlation between t and distance on the curve is not linear.
How do I solve the equation "backwards" - knowing the control point, the start, and the length of the curved path, calculate the end-point?
Not sure I fully understand what you mean by "space markers evenly" but I do have some code that I did with curves and markers that maybe can help you ...
Run the code below, it should output a canvas like this:
function drawBezierCurve(p0, p1, p2, p3) {
distance = 0
last = null
for (let t = 0; t <= 1; t += 0.0001) {
const x = Math.pow(1 - t, 3) * p0[0] + 3 * Math.pow(1 - t, 2) * t * p1[0] + 3 * (1 - t) * Math.pow(t, 2) * p2[0] + Math.pow(t, 3) * p3[0];
const y = Math.pow(1 - t, 3) * p0[1] + 3 * Math.pow(1 - t, 2) * t * p1[1] + 3 * (1 - t) * Math.pow(t, 2) * p2[1] + Math.pow(t, 3) * p3[1];
ctx.lineTo(x, y);
if (last) {
distance += Math.sqrt((x - last[0]) ** 2 + (y - last[1]) ** 2)
if (distance >= 30) {
ctx.rect(x - 1, y - 1, 2, 2);
distance = 0
}
}
last = [x, y]
}
}
const canvas = document.getElementById('canvas');
const ctx = canvas.getContext('2d');
ctx.beginPath();
drawBezierCurve([0, 0], [40, 300], [200, -90], [300, 150]);
ctx.stroke();
<canvas id="canvas" width=300 height=150></canvas>
I created the drawBezierCurve function, there I'm using a parametric equation of a bezier curve and then I use lineTo to draw between points, and also we get a distance between points, the points are very close so my thinking is OK to use the Pythagorean theorem to calculate the distance, and the markers are just little rectangles.
Now, I know similar questions have been asked. But none of the answers has helped me to find the result I need.
Following situation:
We have a line with a point-of-origin (PO), given as lx, ly. We also have an angle for the line in that it exits PO, where 0° means horizontally to the right, positive degrees mean clockwise. The angle is in [0;360[. Additionally we have the length of the line, since it is not infinitely long, as len.
There is also a circle with the given center-point (CP), given as cx, cy. The radius is given as cr.
I now need a function that takes these numbers as parameters and returns the distance of the closest intersection between line and circle to the PO, or -1 if no intersection occures.
My current approach is a follows:
float getDistance(float lx, float ly, float angle, float len, float cx, float cy, float cr) {
float nlx = lx - cx;
float nly = ly - cy;
float m = tan(angle);
float b = (-lx) * m;
// a = m^2 + 1
// b = 2 * m * b
// c = b^2 - cr^2
float[] x_12 = quadraticFormula(sq(m) + 1, 2*m*b, sq(b) - sq(cr));
// if no intersections
if (Float.isNaN(x_12[0]) && Float.isNaN(x_12[1]))
return -1;
float distance;
if (Float.isNaN(x_12[0])) {
distance = (x_12[1] - nlx) / cos(angle);
} else {
distance = (x_12[0] - nlx) / cos(angle);
}
if (distance <= len) {
return distance;
}
return -1;
}
// solves for x
float[] quadraticFormula(float a, float b, float c) {
float[] results = new float[2];
results[0] = (-b + sqrt(sq(b) - 4 * a * c)) / (2*a);
results[1] = (-b - sqrt(sq(b) - 4 * a * c)) / (2*a);
return results;
}
But the result is not as wished. Sometimes I do get a distance returned, but that is rarely correct, there often isn't even an intersection occuring. Most of the time no intersection is returned though, although there should be one.
Any help would be much appreciated.
EDIT:
I managed to find the solution thanks to MBo's answer. Here is the content of my finished getDistance(...)-function - maybe somebody can be helped by it:
float nlx = lx - cx;
float nly = ly - cy;
float dx = cos(angle);
float dy = sin(angle);
float[] results = quadraticFormula(1, 2*(nlx*dx + nly*dy), sq(nlx)+sq(nly)-sq(cr));
float dist = -1;
if (results[0] >= 0 && results[0] <= len)
dist = results[0];
if (results[1] >= 0 && results[1] <= len && results[1] < results[0])
dist = results[1];
return dist;
Using your nlx, nly, we can build parametric equation of line segment
dx = Cos(angle)
dy = Sin(Angle)
x = nlx + t * dx
y = nly + t * dy
Condition of intersection with circumference:
(nlx + t * dx)^2 + (nly + t * dy)^2 = cr^2
t^2 * (dx^2 + dy^2) + t * (2*nlx*dx + 2*nly*dy) + nlx^2+nly^2-cr^2 = 0
so we have quadratic equation for unknown parameter t with
a = 1
b = 2*(nlx*dx + nly*dy)
c = nlx^2+nly^2-cr^2
solve quadratic equation, find whether t lies in range 0..len.
// https://openprocessing.org/sketch/8009#
// by https://openprocessing.org/user/54?view=sketches
float circleX = 200;
float circleY = 200;
float circleRadius = 100;
float lineX1 = 350;
float lineY1 = 350;
float lineX2, lineY2;
void setup() {
size(400, 400);
ellipseMode(RADIUS);
smooth();
}
void draw() {
background(204);
lineX2 = mouseX;
lineY2 = mouseY;
if (circleLineIntersect(lineX1, lineY1, lineX2, lineY2, circleX, circleY, circleRadius) == true) {
noFill();
}
else {
fill(255);
}
ellipse(circleX, circleY, circleRadius, circleRadius);
line(lineX1, lineY1, lineX2, lineY2);
}
// Code adapted from Paul Bourke:
// http://local.wasp.uwa.edu.au/~pbourke/geometry/sphereline/raysphere.c
boolean circleLineIntersect(float x1, float y1, float x2, float y2, float cx, float cy, float cr ) {
float dx = x2 - x1;
float dy = y2 - y1;
float a = dx * dx + dy * dy;
float b = 2 * (dx * (x1 - cx) + dy * (y1 - cy));
float c = cx * cx + cy * cy;
c += x1 * x1 + y1 * y1;
c -= 2 * (cx * x1 + cy * y1);
c -= cr * cr;
float bb4ac = b * b - 4 * a * c;
//println(bb4ac);
if (bb4ac < 0) { // Not intersecting
return false;
}
else {
float mu = (-b + sqrt( b*b - 4*a*c )) / (2*a);
float ix1 = x1 + mu*(dx);
float iy1 = y1 + mu*(dy);
mu = (-b - sqrt(b*b - 4*a*c )) / (2*a);
float ix2 = x1 + mu*(dx);
float iy2 = y1 + mu*(dy);
// The intersection points
ellipse(ix1, iy1, 10, 10);
ellipse(ix2, iy2, 10, 10);
float testX;
float testY;
// Figure out which point is closer to the circle
if (dist(x1, y1, cx, cy) < dist(x2, y2, cx, cy)) {
testX = x2;
testY = y2;
} else {
testX = x1;
testY = y1;
}
if (dist(testX, testY, ix1, iy1) < dist(x1, y1, x2, y2) || dist(testX, testY, ix2, iy2) < dist(x1, y1, x2, y2)) {
return true;
} else {
return false;
}
}
}
I am working on a game that rotates an object on the z axis. I need to limit the total rotation to 80 degrees. I tried the following code, but it doesn't work. minAngle = -40.0f and maxAngle = 40.0f
Vector3 pos = transform.position;
pos.z = Mathf.Clamp(pos.z, minAngle, maxAngle);
transform.position = pos;
The code you posted clamps the z position. What you want is to use transform.rotation
void ClampRotation(float minAngle, float maxAngle, float clampAroundAngle = 0)
{
//clampAroundAngle is the angle you want the clamp to originate from
//For example a value of 90, with a min=-45 and max=45, will let the angle go 45 degrees away from 90
//Adjust to make 0 be right side up
clampAroundAngle += 180;
//Get the angle of the z axis and rotate it up side down
float z = transform.rotation.eulerAngles.z - clampAroundAngle;
z = WrapAngle(z);
//Move range to [-180, 180]
z -= 180;
//Clamp to desired range
z = Mathf.Clamp(z, minAngle, maxAngle);
//Move range back to [0, 360]
z += 180;
//Set the angle back to the transform and rotate it back to right side up
transform.rotation = Quaternion.Euler(transform.rotation.eulerAngles.x, transform.rotation.eulerAngles.y, z + clampAroundAngle);
}
//Make sure angle is within 0,360 range
float WrapAngle(float angle)
{
//If its negative rotate until its positive
while (angle < 0)
angle += 360;
//If its to positive rotate until within range
return Mathf.Repeat(angle, 360);
}
Here's a static version of the nice solution by Imapler that, instead of changing the angle itself, it returns the campled angle, so it can be used with any axis.
public static float ClampAngle(
float currentValue,
float minAngle,
float maxAngle,
float clampAroundAngle = 0
) {
return Mathf.Clamp(
WrapAngle(currentValue - (clampAroundAngle + 180)) - 180,
minAngle,
maxAngle
) + 360 + clampAroundAngle;
}
public static float WrapAngle(float angle)
{
while (angle < 0) {
angle += 360;
}
return Mathf.Repeat(angle, 360);
}
Or if you don't expect to use the WrapAngle method, here's an all-in-one version:
public static float ClampAngle(
float currentValue,
float minAngle,
float maxAngle,
float clampAroundAngle = 0
) {
float angle = currentValue - (clampAroundAngle + 180);
while (angle < 0) {
angle += 360;
}
angle = Mathf.Repeat(angle, 360);
return Mathf.Clamp(
angle - 180,
minAngle,
maxAngle
) + 360 + clampAroundAngle;
}
So now you can do:
transform.localEulerAngles.x = YourMathf.ClampAngle(
transform.localEulerAngles.x,
minX,
maxX
);
I am writing a simple ray shader and I am trying to prodcue a dice with a cube and a number of spheres representing the dots. The spheres are correct, but the sides of the cube are on the x, y and z axes. The cube is centred around 0, 0, 0.
I have checked that the coordinate of the vertices are correct. I am assuming that my ray calculation is correct as the spheres are in the correct positions.
Here is the code for the ray calculation
Ray Image::RayThruPixel(float i, float j)
{
float alpha = m_tanFOVx * ((j - m_halfWidth) / m_halfWidth);
float beta = m_tanFOVy * ((m_halfHeight - i) / m_halfHeight);
vec3 *coordFrame = m_camera.CoordFrame();
vec3 p1 = (coordFrame[U_VEC] * alpha) + (coordFrame[V_VEC] * beta) - coordFrame[W_VEC];
return Ray(m_camera.Eye(), p1);
}
where m_tanFOVx is tan(FOVx / 2) and m_tanFOVy is tan(FOVy / 2) FOVx and FOVy are in radians.
To find the intersection of the ray and triangle my code is as follows:
bool Triangle::Intersection(Ray ray, float &fDistance)
{
static float epsilon = 0.000001;
bool bHit = false;
float fMinDist(10000000);
float divisor = glm::dot(ray.p1, normal);
// if divisor == 0 then the ray is parallel with the triangle
if(divisor > -epsilon && divisor < epsilon)
{
bHit = false;
}
else
{
float t = (glm::dot(v0, normal) - glm::dot(ray.p0, normal)) / divisor;
if(t > 0)
{
vec3 P = ray.p0 + (ray.p1 * t);
vec3 v2 = P - m_vertexA;
v0 = m_vertexB - m_vertexA;
v1 = m_vertexC - m_vertexA;
normal = glm::normalize(glm::cross(v0, v1));
d00 = glm::dot(v0, v0);
d01 = glm::dot(v0, v1);
d11 = glm::dot(v1, v1);
denom = d00 * d11 - d01 * d01;
float d20 = glm::dot(v2, v0);
float d21 = glm::dot(v2, v1);
float alpha = (d11 * d20 - d01 * d21) / denom;
float beta = (d00 * d21 - d01 * d20) / denom;
float gamma = 1.0 - alpha - beta;
vec3 testP = alpha * m_vertexA + beta * m_vertexB + gamma * m_vertexC;
if((alpha >= 0 ) &&
(beta >= 0) &&
(alpha + beta <= 1))
{
bHit = true;
fDistance = t;
}
}
}
return bHit;
}
I am creating a map, i.e. bitmap where each pixel have some real value - height. I want to fill each pixel according to height if this pixel. I want to get something like this, but in 2D.
What is the function that maps height to (R,G,B) that I have such kind of pictures?
Edit: I code in python.
It's just a hue, starting at 360° and going backwards to 0°. You haven't specified a language, but here's how to do the conversion from HSL to RGB in JavaScript, for example:
/**
* Converts an HSL color value to RGB. Conversion formula
* adapted from http://en.wikipedia.org/wiki/HSL_color_space.
* Assumes h, s, and l are contained in the set [0, 1] and
* returns r, g, and b in the set [0, 255].
*
* #param Number h The hue
* #param Number s The saturation
* #param Number l The lightness
* #return Array The RGB representation
*/
function hslToRgb(h, s, l) {
var r, g, b;
if(s == 0) {
r = g = b = l; // achromatic
} else {
function hue2rgb(p, q, t){
if(t < 0) t += 1;
if(t > 1) t -= 1;
if(t < 1/6) return p + (q - p) * 6 * t;
if(t < 1/2) return q;
if(t < 2/3) return p + (q - p) * (2/3 - t) * 6;
return p;
}
var q = l < 0.5 ? l * (1 + s) : l + s - l * s;
var p = 2 * l - q;
r = hue2rgb(p, q, h + 1/3);
g = hue2rgb(p, q, h);
b = hue2rgb(p, q, h - 1/3);
}
return [r * 255, g * 255, b * 255];
}
You'll want a fixed saturation of 100% and a lightness of 50%.