So I am creating a simulation of a bouncing ball, and the user can place lines the ball can collide with on the canvas by dragging from one point to another. There are essentially four lines that can be created:
So the object that stores a line is defined as such:
export interface pathSection {
xfrom: number;
yfrom: number;
xto: number;
yto: number;
length: number;
}
The first and third lines in the image for example dont give the same value from
Math.atan2(yto - yfrom, xto - from);
So given the (relative) complexity of the surfaces, I need to find the angle between a moving object and that surface at the point of collision:
The ball strikes the surface at an angle a, which is what I want!
However I am having trouble finding the angle between the two vectors. This is what I understood would work:
var dx = this.path[index_for_path_section].xfrom - this.path[index_for_path_section].xto;
var dy = this.path[index_for_path_section].yfrom - this.path[index_for_path_section].yto;
var posX = this.particle.pos.x;
var posY = this.particle.pos.y;
var posNextX = posX + this.particle.v.x;
var posNextY = posY + this.particle.v.y;
var angleOfRamp = Math.atan2(dy, dx);
var angleOfvelocity = Math.atan2(posNextY - posY, posNextX - posX);
var angleBetween = angleOfRamp - angleOfvelocity;
This is then used to calculate the speed of the object after the collision:
var spd = Math.sqrt(this.particle.v.x * this.particle.v.x + this.particle.v.y * this.particle.v.y);
var restitution = this.elasticity / 100;
this.particle.v.x = restitution * spd * Math.cos(angleBetween);
this.particle.v.y = restitution * spd * Math.sin(angleBetween);
However the angle calculated is around -4.5 Pi, about -90 degrees for the object directly down and the surface at what looks to be around 45-60 degrees…
The red arrow shows the path of the object moving through the surface - the white dots show where a collision has been detected between the surface and the object.
Any help on how to get the correct and usable angle between the two velocity and the line would be appreciated!
Note I have tried utilizing this solution, but have struggled to adapt it to my own work.
So it took me some time, and I am not 100% sure still of why it works because I think im finding the JavaScript angles system a bit tricky, but:
var dx = this.path[collided].xfrom - this.path[collided].xto;
var dy = this.path[collided].yfrom - this.path[collided].yto;
var spd = Math.sqrt(this.particle.v.x * this.particle.v.x + this.particle.v.y * this.particle.v.y);
var angleOfRamp = Math.atan2(dy, dx);
var angleOfvelocity = Math.atan2(this.particle.v.y, this.particle.v.x);
var angleBetween = angleOfRamp * 2 - angleOfvelocity; // not sure why :)
if (angleBetween < 0) { angleBetween += 2*Math.PI; } // not sure why :)
const restitution = this.elasticity / 100;
this.particle.v.x = restitution * spd * Math.cos(angleBetween);
this.particle.v.y = restitution * spd * Math.sin(angleBetween);
Thanks to all who looked :)
I have been trying to convert the 360 degree camera, single fish eye image, to equirectangular viewer in node js for the past two days. In stackoverflow, the same question is asked and answered in pseudo code. I have been trying to convert pseudo code to node js and cleared some errors. Now the project runs without error but the output image is blank.
From that pseudo, I dont know the polar_w, polar_h and geo_w, geo_h, geo and polar value, so, it gave static value to show the output. Here is a link which i followed to convert pseudo code to node js.
How to convert spherical coordinates to equirectangular projection coordinates?.
Here is the code I tried for converting spherical image to equirectangular viewer:
exports.sphereImage=(request, response)=>{
var Jimp = require('jimp');
// Photo resolution
var img_w_px = 1280;
var img_h_px = 720;
var polar_w = 1280;
var polar_h = 720;
var geo_w = 1280;
var geo_h = 720;
var img_h_deg = 70;
var img_w_deg = 30;
// Camera field-of-view angles
var img_ha_deg = 70;
var img_va_deg = 40;
// Camera rotation angles
var hcam_deg = 230;
var vcam_deg = 60;
// Camera rotation angles in radians
var hcam_rad = hcam_deg/180.0*Math.PI;
var vcam_rad = vcam_rad/180.0*Math.PI;
// Rotation around y-axis for vertical rotation of camera
var rot_y = [
[Math.cos(vcam_rad), 0, Math.sin(vcam_rad)],
[0, 1, 0],
[-Math.sin(vcam_rad), 0, Math.cos(vcam_rad)]
];
// Rotation around z-axis for horizontal rotation of camera
var rot_z = [
[Math.cos(hcam_rad), -Math.sin(hcam_rad), 0],
[Math.sin(hcam_rad), Math.cos(hcam_rad), 0],
[0, 0, 1]
];
Jimp.read('./public/images/4-18-2-42.jpg', (err, lenna) => {
polar = new Jimp(img_w_px, img_h_px);
geo = new Jimp(img_w_px, img_h_px);
for(var i=0; i<img_h_px; ++i)
{
for(var j=0; j<img_w_px; ++j)
{
// var p = img.getPixelAt(i, j);
var p = lenna.getPixelColor(i, j)
// var p = getPixels(img, { x: i, y: j })
// Calculate relative position to center in degrees
var p_theta = (j - img_w_px / 2.0) / img_w_px * img_w_deg / 180.0 * Math.PI;
var p_phi = -(i - img_h_px / 2.0) / img_h_px * img_h_deg / 180.0 *Math. PI;
// Transform into cartesian coordinates
var p_x = Math.cos(p_phi) * Math.cos(p_theta);
var p_y = Math.cos(p_phi) * Math.sin(p_theta);
var p_z = Math.sin(p_phi);
var p0 = {p_x, p_y, p_z};
// Apply rotation matrices (note, z-axis is the vertical one)
// First vertically
var p1 = rot_y[1][2][3] * p0;
var p2 = rot_z[1][2][3] * p1;
// Transform back into spherical coordinates
var theta = Math.atan2(p2[1], p2[0]);
var phi = Math.asin(p2[2]);
// Retrieve longitude,latitude
var longitude = theta / Math.PI * 180.0;
var latitude = phi / Math.PI * 180.0;
// Now we can use longitude,latitude coordinates in many different
projections, such as:
// Polar projection
{
var polar_x_px = (0.5*Math.PI + phi)*0.5 * Math.cos(theta)
/Math.PI*180.0 * polar_w;
var polar_y_px = (0.5*Math.PI + phi)*0.5 * Math.sin(theta)
/Math.PI*180.0 * polar_h;
polar.setPixelColor(p, polar_x_px, polar_y_px);
}
// Geographical (=equirectangular) projection
{
var geo_x_px = (longitude + 180) * geo_w;
var geo_y_px = (latitude + 90) * geo_h;
// geo.setPixel(geo_x_px, geo_y_px, p.getRGB());
geo.setPixelColor(p, geo_x_px, geo_y_px);
}
// ...
}
}
geo.write('./public/images/4-18-2-42-00001.jpg');
polar.write('./public/images/4-18-2-42-00002.jpg');
});
}
And tried another method by slicing image into four parts to detect car. Sliced image into four parts using image-slice module and to read and write jimp module is used. But unfortunately cars not detected properly.
Here is the code i used for slicing image:
exports.sliceImage=(request, response)=>{
var imageToSlices = require('image-to-slices');
var lineXArray = [540, 540];
var lineYArray = [960, 960];
var source = './public/images/4-18-2-42.jpg'; // width: 300, height: 300
imageToSlices(source, lineXArray, lineYArray, {
saveToDir: './public/images/',
clipperOptions: {
canvas: require('canvas')
}
}, function() {
console.log('the source image has been sliced into 9 sections!');
});
}//sliceImage
And for detect car from image i used opencv4nodejs. Cars are not detected properly. here is the code i used for detect car:
function runDetectCarExample(img=null){
if(img==null){
img = cv.imread('./public/images/section-1.jpg');
}else
{
img=cv.imread(img);
}
const minConfidence = 0.06;
const predictions = classifyImg(img).filter(res => res.confidence > minConfidence && res.className=='car');
const drawClassDetections = makeDrawClassDetections(predictions);
const getRandomColor = () => new cv.Vec(Math.random() * 255, Math.random() * 255, 255);
drawClassDetections(img, 'car', getRandomColor);
cv.imwrite('./public/images/section-'+Math.random()+'.jpg', img);
var name="distanceFromCamera";
var focalLen= 1.6 ;//Focal length in mm
var realObjHeight=254 ;//Real Height of Object in mm
var cameraFrameHeight=960;//Height of Image in pxl
var imgHeight=960;//Image Height in pxl
var sensorHeight=10;//Sensor height in mm
var R = 6378.1 //#Radius of the Earth
var brng = 1.57 //#Bearing is 90 degrees converted to radians.
var hc=(200/100);//Camera height in m
predictions
.forEach((data)=> {
// imgHeight=img.rows;//Image Height in pxl
// realObjHeight=data.rect.height;
// data.rect[name]=((focalLen)*(realObjHeight)*
(cameraFrameHeight))/((imgHeight)*(sensorHeight));
var dc=(((data.rect.width * focalLen) / img.cols)*2.54)*100; // meters
console.log(Math.floor(parseInt(data.rect.width)));
// var dc=((Math.floor(parseInt(data.rect.width)* 0.264583) * focalLen) / img.cols); // mm
var lat1=13.0002855;//13.000356;
var lon1=80.2046441;//80.204632;
// Gate 13.0002855,80.2046441
// Brazil Polsec : -19.860566, -43.969436
// var d=Math.sqrt((dc*dc)+(hc*hc));
// d=(data.rect[name])/1000;
data.rect[name]=d=dc/1000;
lat1 =toRadians(lat1);
lon1 = toRadians(lon1);
brng =toRadians(90);
// lat2 = Math.asin( Math.sin(lat1)*Math.cos(d/R) +
// Math.cos(lat1)*Math.sin(d/R)*Math.cos(brng));
// lon2 = lon1 +
Math.atan2(Math.sin(brng)*Math.sin(d/R)*Math.cos(lat1),
// Math.cos(d/R)-Math.sin(lat1)*Math.sin(lat2));
var lat2 = Math.asin(Math.sin(lat1) * Math.cos(d/6371) +
Math.cos(lat1) * Math.sin(d/6371) * Math.cos(brng));
var lon2 = lon1 + Math.atan2(Math.sin(brng) * Math.sin(d/6371) * Math.cos(lat1),
Math.cos(d/6371) - Math.sin(lat1) * Math.sin(lat2));
lat2 = toDegrees(lat2);
lon2 = toDegrees(lon2);
data.rect['latLong']=lat2+','+lon2;
// console.log(brng);
});
response.send(predictions);
cv.imshowWait('img', img);
};
here is the fish eye image which need to be converted to equirectangular.
Any help much appreciated pls....
You are asking how to convert a 360deg fish-eye projection to an equirectangular projection.
In order to do this, for every pixel on the fish-eye image you need to know where to place in onto the output image.
Your input image is 1920x1080, let us assume you want to output it to an equirectangular projection of the same size.
The input circle mapping is defined as:
cx = 960; // center of circle on X-axis
cy = 540; // center of circle on Y-axis
radius = 540; // radius of circle
If you have a pixel at (x,y) in the input image, then we can calculate the spherical coordinates using:
dx = (x - cx) * 1.0 / radius;
dy = (y - cy) * 1.0 / radius;
theta_deg = atan2(dy, dx) / MATH_PI * 180;
phi_deg = acos(sqrt(dx*dx + dy*dy)) / MATH_PI * 180;
outputx = (theta_deg + 180) / 360.0 * outputwidth_px;
outputy = (phi_deg + 90) / 180.0 * outputheight_px;
So there we translated (x,y) from the fish-eye image to the (outputx,outputy) in the equirectangular image. In order to not leave the implementation as the dreaded "exercise to the reader", here is some sample Javascript-code using the Jimp-library as used by the OP:
var jimp = require('jimp');
var inputfile = 'input.png';
jimp.read(inputfile, function(err, inputimage)
{
var cx = 960;
var cy = 540;
var radius = 540;
var inputwidth = 1920;
var inputheight = 1080;
var outputwidth = 1920;
var outputheight = 1080;
new jimp(outputwidth, outputheight, 0x000000ff, function(err, outputimage)
{
for(var y=0;y<inputheight;++y)
{
for(var x=0;x<inputwidth;++x)
{
var color = inputimage.getPixelColor(x, y);
var dx = (x - cx) * 1.0 / radius;
var dy = (y - cy) * 1.0 / radius;
var theta_deg = Math.atan2(dy, dx) / Math.PI * 180;
var phi_deg = Math.acos(Math.sqrt(dx*dx + dy*dy)) / Math.PI * 180;
var outputx = Math.round((theta_deg + 180) / 360.0 * outputwidth);
var outputy = Math.round((phi_deg + 90) / 180.0 * outputheight);
outputimage.setPixelColor(color, outputx, outputy);
}
}
outputimage.write('output.png');
});
});
Note that you will still need to do blending of the pixel with neighbouring pixels (for the same reason as when you're resizing the image).
Additionally, in your case, you only have half of the sphere (you can't see the sun in the sky). So you would need to use var outputy = Math.round(phi_deg / 90.0 * outputheight). In order to keep the right aspect ratio, you might want to change the height to 540.
Also note that the given implementation may not be efficient at all, it's better to use the buffer directly.
Anyway, without blending I came up with the result as demonstrated here:
So in order to do blending, you could use the simplest method which is the nearest neighbour approach. In that case, you should invert the formulas in the above example. Instead of moving the pixels from the input image to the right place in the output image, you can go through every pixel in the output image and ask which input pixel we can use for that. This will avoid the black pixels, but may still show artifacts:
var jimp = require('jimp');
var inputfile = 'input.png';
jimp.read(inputfile, function(err, inputimage)
{
var cx = 960;
var cy = 540;
var radius = 540;
var inputwidth = 1920;
var inputheight = 1080;
var outputwidth = 1920;
var outputheight = 1080/2;
var blendmap = {};
new jimp(outputwidth, outputheight, 0x000000ff, function(err, outputimage)
{
for(var y=0;y<outputheight;++y)
{
for(var x=0;x<outputwidth;++x)
{
var theta_deg = 360 - x * 360.0 / outputwidth - 180;
var phi_deg = 90 - y * 90.0 / outputheight;
var r = Math.sin(phi_deg * Math.PI / 180)
var dx = Math.cos(theta_deg * Math.PI / 180) * r;
var dy = Math.sin(theta_deg * Math.PI / 180) * r;
var inputx = Math.round(dx * radius + cx);
var inputy = Math.round(dy * radius + cy);
outputimage.setPixelColor(inputimage.getPixelColor(inputx, inputy), x, y);
}
}
outputimage.write('output.png');
});
});
For reference, in order to convert between Cartesian and Spherical coordinate systems. These are the formulas (taken from here). Note that the z is in your case just 1, a so-called "unit" sphere, so you can just leave it out of the equations. You should also understand that since the camera is actually taking a picture in three dimensions, you also need formulas to work in three dimensions.
Here is the generated output image:
Since I don't see your original input image in your question anymore, in order for anyone to test the code from this answer, you can use the following image:
Run the code with:
mkdir /tmp/test
cd /tmp/test
npm install --permanent jimp
cat <<EOF >/tmp/test/main.js
... paste the javascript code from above ...
EOF
curl https://i.stack.imgur.com/0zWt6.png > input.png
node main.js
Note: In order to further improve the blending, you should remove the Math.round. So for instance, if you need to grab a pixel at x is 0.75, and the pixel on the left at x = 0 is white, and the pixel on the right at x = 1 is black. Then you want to mix both colors into a dark grey color (using ratio 0.75). You would have to do this for both dimensions simultaneously, if you want a nice result. But this should really be in a new question imho.
I started a project using the raycasting technique GitHub Project
To find the length of the ray (distance from players pos to wall) I just increment by one. But there are several problems with that, its time consuming, inaccurate & will be difficult for texturing.
I tried to implement the daa algorithm, which doesnt just increments by 1 -> he goes through the grids and returns exact positions.
http://www.geeksforgeeks.org/dda-line-generation-algorithm-computer-graphics/
Has anyone experience with that or any tips?
No algorithm way:
for(let resolution = 0; resolution < display.width / 2; resolution++){ //every 2nd px gets scanned
let ray = this.pov + (-this.fov / 2 + this.fov / (display.width / 2) * resolution);
let distance = 0, hit = false;
/*ugly way of raycasting!*/
do{
let x = this.x + distance * Math.cos(ray * (Math.PI / 180));
let y = this.y + distance * Math.sin(ray * (Math.PI / 180));
if(map[Math.floor(x / block)][Math.floor(y / block)]){
distance = Math.sqrt(Math.pow(this.x - x, 2) + Math.pow(this.y - y, 2));
hit = true
}
distance += 1;
}while(!hit);
distance = convert / distance;
canvas.fillStyle = "#fff";
canvas.fillRect(resolution * 2, display.height / 2 - distance / 2, 2, distance);
}
You don't need DDA or Bresenham algorithm to find intersections of the ray with walls.
If you need one intersection with given border (or box edges) - just calculate it with ray equation and border position.
If you want to get intersections with grid cells - use voxelization algorithm like Amanatides-Woo
I've been trying to implement clicking in my webgl app for the last 6 hours and I can't find anything clear enough about this subject.
What I have came up with so far is, in pseudo code:
screenSpace = mousePosition;
normalizedScreenSpace = (screenSpace.x/screen.width, screenSpace.y/screen.height);
camSpace = invertProjectionMatrix * normalizedScreenSpace;
worldSpace = invertViewMatrix * camSpace;
Printing out the worldSpace coordinates, and it doesn't corresponds to other objects in the scene. What am I doing wrong?
The viewProjection matrix brings a vec3 from world space to clip space and so its inverse does the reverse, clip space to world space. Whats missing is the perspective divide that gpu handles for you behind the hood so you have to account for that as well. Add in the screen width and height and you have your screen to world:
screenToWorld: function(invViewProjection, screenWidth, screenHeight){
// expects this[2] (z value) to be -1 if want position at zNear and +1 at zFar
var x = 2*this[0]/screenWidth - 1.0;
var y = 1.0 - (2*this[1]/screenHeight); // note: Y axis oriented top -> down in screen space
var z = this[2];
this.setXYZ(x,y,z);
this.applyMat4(invViewProjection);
var m = invViewProjection;
var w = m[3] * x + m[7] * y + m[11] * z + m[15]; // required for perspective divide
if (w !== 0){
var invW = 1.0/w;
this[0] *= invW;
this[1] *= invW;
this[2] *= invW;
}
return this;
},
And the reverse calculation:
worldToScreen: function(viewProjectionMatrix, screenWidth, screenHeight){
var m = viewProjectionMatrix;
var w = m[3] * this[0] + m[7] * this[1] + m[11] * this[2] + m[15]; // required for perspective divide
this.applyMat4(viewProjectionMatrix);
if (w!==0){ // do perspective divide and NDC -> screen conversion
var invW = 1.0/w;
this[0] = (this[0]*invW + 1) / 2 * screenWidth;
this[1] = (1-this[1]*invW) / 2 * screenHeight; // screen space Y goes from top to bottom
this[2] *= invW;
}
return this;
},
I have a function which gets the mouse position in world space, then checks to see if the mouse is over or near to the circle's line.
The added complication how ever is the circle is transformed at an angle so it's more of an ellipse. I can't see to get the code to detect that the mouse is near the border of circle and am unsure where I am going wrong.
This is my code:
function check(evt){
var x = (evt.offsetX - element.width/2) + camera.x; // world space
var y = (evt.offsetY - element.height/2) + camera.y; // world space
var threshold = 20/scale; //margin to edge of circle
for(var i = 0; i < obj.length;i++){
// var mainAngle is related to the transform
var x1 = Math.pow((x - obj[i].originX), 2) / Math.pow((obj[i].radius + threshold) * 1,2);
var y1 = Math.pow((y - obj[i].originY),2) / Math.pow((obj[i].radius + threshold) * mainAngle,2);
var x0 = Math.pow((x - obj[i].originX),2) / Math.pow((obj[i].radius - threshold) * 1, 2);
var y0 = Math.pow((y - obj[i].originY),2) / Math.pow((obj[i].radius - threshold) * mainAngle, 2);
if(x1 + y1 <= 1 && x0 + y0 >= 1){
output.innerHTML += '<br/>Over';
return false;
}
}
output.innerHTML += '<br/>out';
}
To understand it better, I have a fiddle here: http://jsfiddle.net/nczbmbxm/ you can move the mouse over the circle, it should say "Over" when you are within the threshold of being near the circle's perimeter. Currently it does not seem to work. And I can't work out what the maths needs to be check for this.
There is a typo on line 34 with orignX
var x1 = Math.pow((x - obj[i].orignX), 2) / Math.pow((obj[i].radius + threshold) * 1,2);
should be
var x1 = Math.pow((x - obj[i].originX), 2) / Math.pow((obj[i].radius + threshold) * 1,2);
now you're good to go!
EDIT: In regards to the scaling of the image and further rotation of the circle, I would set up variables for rotation about the x-axis and y-axis, such as
var xAngle;
var yAngle;
then as an ellipse can be written in the form
x^2 / a^2 + y^2 / b^2 = 1
such as in Euclidean Geometry,
then the semi-major and semi-minor axes would be determined by the rotation angles. If radius is the circles actual radius. then
var semiMajor = radius * cos( xAngle );
var semiMinor = radius;
or
var semiMajor = radius;
var semiMinor = radius * cos( yAngle );
you would still need to do some more transformations if you wanted an x and y angle.
so if (xMouseC, yMouseC) are the mouse coordinates relative to the circles centre, all you must do is check if that point satisfies the equation of the ellipse to within a certain tolerance, i.e. plug in
a = semiMajor;
b = semiMinor;
x = xMouseC;
y = yMouseC;
and see if it is sufficiently close to 1.
Hope that helps!