Develop Reference

Javascript canvas draw line strange behavior using algorithm

There are plenty of examples on how to draw lines on canvas, in js.
But for only educational purposes i want to draw line using algorithm. basically method gets two Vector2 points, from them it finds middle point, then it continues like that recursively until minimum distance of 2 pixels is reached.
I have DrawPoint method to basically draw 1 point on canvas, and DrawLine method that does all the job.
For now I have 2 problems:
1: points are not colored red, as they should be.
2:
It doesnt look like a line.
For Vector2 i used "Victor.js" plugin, and it seems to be working well.
this is code i have:
JS:
var point2 = new Victor(100, 100);
var point3 = new Victor(150, 150);
DrawLine(point2, point3);
function DrawLine(vec0, vec1)
{
var point0 = new Victor(vec0.x, vec0.y);
var point1 = new Victor(vec1.x, vec1.y);
var dist = point1.distance(point0);
if (dist < 2)
return;
//this is how it should look like in c# var middlePoint = point0 + (point1 - point0)/2; But looks like i cant just divide by 2 using victor js because i can only divide vector by vector.
var middlePoint = point0.add(point1.subtract(point0).divide(new Victor(2,2)));
DrawPoint(middlePoint);
DrawLine(point0, middlePoint);
DrawLine(middlePoint, point1);
}
function DrawPoint(point){
var c = document.getElementById("screen");
var ctx = c.getContext("2d");
ctx.fillStyle = "FF0000";
ctx.fillRect(point.x, point.y, 3,1);
}
I really appreciate any help you can provide.

The victor.js documentation shows that most functions of Victors do not return new Victors, but operate on the current instance. In a way, v1.add(v2) is semantically more like v1 += v2 and not v1 + v2.
The problem is with calculating the midpoint. You could use the mix() method, which blends two vectors with a weight. You must clone() the Victor first, otherwise point0will be midofied:
var middlePoint = point0.clone().mix(point1, 0.5);
If you don't change the original Vectors, you don't need to create new instances of Victors from the arguments, you can use the arguments directly:
function DrawLine(point0, point1)
{
var dist = point1.distance(point0);
if (dist < 2) return;
var middlePoint = point0.clone().mix(point1, 0.5);
DrawPoint(middlePoint);
DrawLine(point0, middlePoint);
DrawLine(middlePoint, point1);
}
Finally, as Sven the Surfer has already said in a comment, "FF0000" isn't a valid colour. Use "#FF0000", note the hash mark, or one of the named web colours such as "crimson".

canvas draw - change rotation axis without clearing old state

I'm using html5 canvas to create a simple 3d polygon program. the program allows to change the rotation of each axis - x,y,z. on the event of change x/y/z angle, a corresponding call to the drawing function is done. the problem is every time I make a new call to the draw function it clears the older position and the result is it jumps. basically each function works on its own but they do not work together.
example code:
var Perspective = function(rotate){
var canvas = document.getElementById('myCanvas');
var context = canvas.getContext('2d');
//clear
...
switch(rotate){
case "x" : {
var transform = Mat3.rotationX(Math.radians(rotateX.getValue())); //rotation x
break;
}
case "y" : {
var transform = Mat3.rotationY(Math.radians(rotateY.getValue())); //rotation y
break;
}
case "z" : {
var transform = Mat3.rotationZ(Math.radians(rotateZ.getValue())); //rotation Z
}
...
draw(...settings..)
}
set a listener to each slider change event (also for Y and Z). AngleX makes the call to Perspective, passing the string "x" as rotate param.
var angleX = $('#AngleX').slider()
.on('slide change', AngleX)
.data('slider');
how can i make the changes if different axis more dynamic ?

anyway i'v solved it no thanks to you QBM5 .. the switch case was a mistake, i had to multiply all axis transformation matrixs like this :
var transform = Mat3.rotationX(-Math.radians(rotateY.getValue())) .multiply(Mat3.rotationY(Math.radians(rotateX.getValue())));

How to connect objects with Bezier-like curves using Paper.js

I have a web app prototype where nodes similar to Blender shader editor are connected to each other. I am using Paper.js framework
I want them to be connected using those smooth Bezier-like curves. So I have 2 shapes and I can connect them by making a straight line, but now I want to have handles at the endpoints that smooth these objects out, kinda like this:
So 2 handles on endpoints, pointing horizontally for half the bounding box of the path.
The problem is I can't figure out how to add and edit those handles using Paper.js
The code I have is this:
function makeRectangle(topLeft, size, cornerSize, colour){
var rectangle = new Rectangle(topLeft, size);
var cornerSize = cornerSize;
var path = new Path.RoundRectangle(rectangle, cornerSize);
path.fillColor = colour;
return path;
}
var xy1 = new Point(50,50); //Position of 1st rectangle.
var size = new Size(100, 80); //Size
var c = new Size(8,8); //Corner radius
var col = "#167ee5"; //Colour
var r1 = makeRectangle(xy1, size, c, col); //Make first rectangle
var xy2 = new Point(467,310); //Position of second rectangle
var size2 = new Size(115, 70); //Size of second rectangle
var r2 = makeRectangle(xy2, size2, c, col); //Make secont rectangle
var r1cent = r1.bounds.center; //Get the center points, they will be used as endpoints for the curve.
var r2cent = r2.bounds.center;
var connector = new Path(r1cent, r2cent); //Ok so I made this path... Now what? How do access and edit the handlers at endpoints like in the image?
connector.strokeColor = 'black'; //Give it some colour so we can see it.
You can paste all this code here without any setup, it's a good way to test the framework.

You can use Segment objects when defining the connector rather than using Points (or you can set the handleIn and handleOut properties after creating the path).
The doc is here: Segment
And here is a sketch showing how to use handleIn and handleOut with your code:
sketch.paperjs.org solution

Trying to workout how to use quaternions to rotate a camera that is moving along a path to look in new direction vector

I am trying to rotate the camera smoothly and without altering the y-vector of the camera direction, i can use look at, and it changes the camera direction in a flash, but this is not working for me, I would like a smooth transition as the direction of the camera changes. I have been reading up, and not understanding everything, but it seems to me that quaternions are the solution to this problem.
I have this.object (my camera) moving along a set path (this.spline.points). The location of the camera at any one time is (thisx,thisy, thisz)
I have cc[i] the direction vector for the direction I would like the camera to face (formerly I was using lookat(cc[i]) which changes the direction correctly, but too quickly/instantaneously)
Using info I have read, I have tried this below, and it just resulted in the screen going black at the point when the camera is due to move.
Could anyone please explain if I am on the right track, how to correct my code.
Thanks
var thisx = this.object.matrixWorld.getPosition().x.toPrecision(3);
var thisy = this.object.matrixWorld.getPosition().y.toPrecision(3);
var thisz = this.object.matrixWorld.getPosition().z.toPrecision(3);
var i = 0;
do {
var pathx = this.spline.points[i].x.toPrecision(3);
var pathz = this.spline.points[i].z.toPrecision(3);
if (thisx == pathx && thisz == pathz){
this.object.useQuaternion = true;
this.object.quaternion = new THREE.Quaternion(thisx, thisy, thisz, 1);
var newvect;
newvect.useQuaternion = true;
newvect.quaternion = new THREE.Quaternion(thisx+cc[i].x, thisy+cc[i].y, thisz+cc[i].z, 1);
var newQuaternion = new THREE.Quaternion();
THREE.Quaternion.slerp(this.object.quaternion, newvect.quaternion, newQuaternion, 0.5);
this.object.quaternion = newQuaternion;
//this.object.lookAt( cc[i]);
i = cc.length;
} else i++;
} while(i < cc.length);

There is no need to call this.object.useQuaternion = true. That is default behavior.
Also, this.object.quaternion contains the current rotation, so no need to generate that either.
You might want to try a different approach - construct the rotation matrix from the spline position, lookAt and up vectors, creating a path of quaternions as a preprocessing step:
var eye = this.spline.points[i].clone().normalize();
var center = cc[i].normalize();
var up = this.object.up.normalize();
var rotMatrix = new THREE.Matrix4().lookAt(eye, center, up);
You could then create the quaternions from the rotation matrix:
var quaternionAtSplineCoordinates = [];
quaternionAtSplineCoordinates.push(new THREE.Quaternion().setFromRotationMatrix(rotMatrix));
Once you have that path, you could apply the quaternion to the camera in your animation loop - provided you have a large enough number of samples. Otherwise, you could consider using slerp to generate the intermediate points.

Get curved result set rather than angular in JavaScript (maths help needed)

I've got a script that creates a gradient by shading cells based on their distance from a set of coordinates. What I want to do is make the gradient circular rather than the diamond shape that it currently is. You can see an en example here: http://jsbin.com/uwivev/9/edit
var row = 5, col = 5, total_rows = 20, total_cols = 20;
$('table td').each(function(index, item) {
// Current row and column
var current_row = $(item).parent().index(),
current_col = $(item).index();
// Percentage based on location, always using positive numbers
var percentage_row = Math.abs(current_row-row)/total_rows;
var percentage_col = Math.abs(current_col-col)/total_cols;
// I'm thinking this is what I need to change to achieve the curve I'm after
var percentage = (percentage_col+percentage_row)/2;
$(this).find('div').fadeTo(0,percentage*3);
});
If you can give me hand with the right maths function to get the curve I'm after that would be great! Thanks!
Darren

// Current row and column
var current_row = $(item).parent().index(),
current_col = $(item).index();
// distance away from the bright pixel
var dist = Math.sqrt(Math.pow(current_row - row, 2) + Math.pow(current_col - col, 2))
// do something with dist, you might change this
var percentage = dist / total_cols;
$(this).find('div').fadeTo(0,percentage*3);

You can use the square of the distance formula:
((current_row - row)*(current_row - row) + (current_col - col)*(current_col - col))
then multiply it by whatever scale factor you need.

Here is a circle drawing procudure I wrote many moons ago in Pascal which you can use as pseudo code to understand how to color pixels at the radius from an (X,Y) and work your way in. Multiple shrinking circles should cover the entire area you need. The code also gives you the formula for accessing the radius.
PROCEDURE DrawCircle(X,Y,Radius:Integer);
VAR A,B,Z:LongInt;
BEGIN
Z:=Round(Sqrt(Sqr(LongInt(Radius))/2));
FOR A:=Z TO Radius DO
FOR B:=0 TO Z DO
IF Radius=Round(Sqrt(A*A+B*B)) THEN
BEGIN
PutPixel(X+A,Y+B,8);
PutPixel(X+A,Y-B,9);
PutPixel(X-A,Y+B,10);
PutPixel(X-A,Y-B,11);
PutPixel(X+B,Y+A,12);
PutPixel(X+B,Y-A,13);
PutPixel(X-B,Y+A,14);
PutPixel(X-B,Y-A,15);
END;
END;
NB: "Longint()" is a compiler typecast for larger numeric computations so don't let that worry you.
NB: Inner-most brackets are executed first.