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noisy line between two specific points P5.js

I'm trying to draw a noisy line (using perlin noise) between two specific points.
for example A(100, 200) and B(400,600).
The line could be a points series.
Drawing random noisy line is so clear but I dont know how can I calculate distance specific points.
working of P5.js.
I don't have any code written yet to upload.
Please can anyone help me?

I tried to add sufficient comments that you would be able to learn how such a thing is done. There are a number of things that you should make yourself aware of if you aren't already, and it's hard to say which if these you're missing:
for loops
drawing lines using beginShape()/vertex()/endShape()
Trigonometry (in this case sin/cos/atan2) which make it possible to find angles and determine 2d offsets in X and Y components at a given angle
p5.Vector() and its dist() function.
// The level of detail in the line in number of pixels between each point.
const pixelsPerSegment = 10;
const noiseScale = 120;
const noiseFrequency = 0.01;
const noiseSpeed = 0.1;
let start;
let end;
function setup() {
createCanvas(400, 400);
noFill();
start = createVector(10, 10);
end = createVector(380, 380);
}
function draw() {
background(255);
let lineLength = start.dist(end);
// Determine the number of segments, and make sure there is at least one.
let segments = max(1, round(lineLength / pixelsPerSegment));
// Determine the number of points, which is the number of segments + 1
let points = 1 + segments;
// We need to know the angle of the line so that we can determine the x
// and y position for each point along the line, and when we offset based
// on noise we do so perpendicular to the line.
let angle = atan2(end.y - start.y, end.x - start.x);
let xInterval = pixelsPerSegment * cos(angle);
let yInterval = pixelsPerSegment * sin(angle);
beginShape();
// Always start with the start point
vertex(start.x, start.y);
// for each point that is neither the start nor end point
for (let i = 1; i < points - 1; i++) {
// determine the x and y positions along the straight line
let x = start.x + xInterval * i;
let y = start.y + yInterval * i;
// calculate the offset distance using noice
let offset =
// The bigger this number is the greater the range of offsets will be
noiseScale *
(noise(
// The bigger the value of noiseFrequency, the more erretically
// the offset will change from point to point.
i * pixelsPerSegment * noiseFrequency,
// The bigger the value of noiseSpeed, the more quickly the curve
// fluxuations will change over time.
(millis() / 1000) * noiseSpeed
) - 0.5);
// Translate offset into x and y components based on angle - 90°
// (or in this case, PI / 2 radians, which is equivalent)
let xOffset = offset * cos(angle - PI / 2);
let yOffset = offset * sin(angle - PI / 2);
vertex(x + xOffset, y + yOffset);
}
vertex(end.x, end.y);
endShape();
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/p5.js/1.4.0/p5.js"></script>
This code makes jaggy lines, but they could be smoothed using curveVertex(). Also, making the line pass through the start and end points exactly is a little tricky because the very next point may be offset by a large amount. You could fix this by making noiseScale very depending on how far from an endpoint the current point is. This could be done by multiplying noiseScale by sin(i / points.length * PI) for example.

Find the Points of Intersection of a Circle with a Line in Javascript

I'm trying to animate a given element to go around a pre-defined radius and I'm having trouble getting the position of the element at a Y point given.
I'm trying to find each point with the circle equation, but I can only get one point out of the two possible ones.
In Javascript, I use Math.sqrt( Math.pow(radius, 2) - Math.pow(y, 2) , 2) to get the point. assuming the center of the of the circle is 0,0.
but then I need to translate it to pixels on the screen since there are no negative pixels in positions on the browser.
All the sizing is relative to the window. so the radius, for example, is 80% of the height of the window in my tests.
Also, I'm trying to calculate what the distance of the element between each frame should be for the duration, but I'm not using it yet because I try to fix the issue above first.
This is what I have(a cleaned up version):
let height = window.innerHeight * 0.8,
radius = height / 2,
circumferance = (radius * 2) * Math.PI,
container = document.getElementById('container'),
rotating = document.querySelector('.rotating'),
centerX = radius - (rotating.offsetWidth / 2),
centerY = radius - (rotating.offsetHeight / 2),
duration = 10,
stepDistance = circumferance / 16;
// Setting the dimensions of the container element.
container.style.height = height + 'px';
container.style.width = height + 'px';
// return positive X of any given Y.
function getXOffset(y) {
return Math.sqrt( Math.pow(radius, 2) - Math.pow(y, 2) , 2);
}
// Setting the position of the rotating element to the start.
rotating.style.top = 0 + 'px';
rotating.style.left = centerX + 'px';
setInterval(() => {
let top = parseInt(rotating.style.top),
y = radius - top;
rotating.style.top = (top + 1) + 'px';
rotating.style.left = (centerX + getXOffset(y)) + 'px';
}, 16);
Here is a fiddle with a bit more code for trying to get the right amount of distance between points for a smoother animation(currently needs fixing, but it doesn't bother me yet.)
https://jsfiddle.net/shock/1qcfvr4y/
Last note: I know that there might be other ways to do this with CSS, but I chose to use javascript for learning purposes.

Math.sqrt would only return the positive root. You'll have to account for the negative value based on the application. In this case, you need the positive x value during the 1st half of the cycle and negative during the 2nd half.
To do that, you should implement a method to track the progress and reverse the sign accordingly.
Here is a sample. I modified upon yours.
edit:
Instead of Math.sqrt( Math.pow(radius, 2) - Math.pow(y, 2) , 2) You can use the full formula to get x if you do not want to assume origin as center, which in this case is Math.sqrt( Math.pow(radius, 2) - Math.pow((actualY - centerY), 2) , 2)
explanation:
The original equation (x-a)² + (y'-b)² = r²
becomes x = √(r² - (y'-b)²) + a
Assuming .rotating box have 0 width and height.
The variable equivalents in your code are centerX = a, centerY = b.
By assuming origin as center you're basically doing a pre-calculation so that your y value becomes the equivalent of (y'-b). Hence x = √(r² - y²) + a is valid.
At initial state top = 0
i.e (y'-b) => height - centerY.
In your code y = radius => height/2.
Now (height - centerY) being equal to (height/2) is a side effect of your circle being bound by a square container whose height determines the y value.
In other words, when you use origin as center, you are taking the center offsets outside of circle equation and handling it separately. You could do the same thing by using the whole formula, that is, x = √(r² - (y'-b)²) + a

Rotating One Object Around Another In createJS/easelJS

In easelJS, what is the best way to rotate an object around another? What I'm trying to accomplish is a method to rotate the crosshair around the circle pictured below, just like a planet orbits the sun:
I've been able to rotate objects around their own center point, but am having a difficult time devising a way to rotate one object around the center point of a second object. Any ideas?

Might make sense to wrap content in a Container. Translate the coordinates so the center point is where you want it, and then rotate the container.

To build on what Lanny is suggesting, there may be cases where you don't want to rotate the entire container. An alternative would be to use trigonometric functions and an incrementing angle to calculate the x/y position of the crosshair. You can find the x/y by using an angle (converted to radians) and Math.cos(angleInRadians) for x and Math.sin(angleInRadians) for y, the multiply by the radius of the orbit.
See this working example for reference.
Here's a complete snippet.
var stage = new createjs.Stage("stage");
var angle = 0;
var circle = new createjs.Shape();
circle.graphics.beginFill("#FF0000").drawEllipse(-25, -25, 50, 50).endFill();
circle.x = 100;
circle.y = 100;
var crosshair = new createjs.Shape();
crosshair.graphics.setStrokeStyle(2).beginStroke("#FF0000").moveTo(5, 0).lineTo(5, 10).moveTo(0, 5).lineTo(10, 5).endStroke();
stage.addChild(circle);
stage.addChild(crosshair);
createjs.Ticker.addEventListener("tick", function(){
angle++;
if(angle > 360)
angle = 1;
var rads = angle * Math.PI / 180;
var x = 100 * Math.cos(rads);
var y = 100 * Math.sin(rads);
crosshair.x = x + 100;
crosshair.y = y + 100;
stage.update();
});

Put another point respect to origin point with the same direction
var one_meter = 1 / map_resolution;
// get one meter distance from pointed points
var extra_x = one_meter * Math.cos(temp_rotation);
var extra_y = one_meter * Math.sin(-temp_rotation);
var new_x = mapXY.x + extra_x;
var new_y = mapXY.y + extra_y;
var home_point = new createjs.Shape().set({ x: new_x, y: new_y });
home_point.graphics.beginFill("Blue").drawCircle(0, 0, 10);
stage.addChild(home_point);
stage.update();

For the point inside circle, find in which quarter it is?

I researched google but couldn't find the keywords for search. So I ask here if my algorithm and code is efficient?
http://sketchtoy.com/66429941 (algorithm)
The algoritm is: I have four points which are: north, east, south and west of circle. I check 4 distances (distanceToNorth, distanceToEast, distanceToSouth, distanceToWest). And I find minimum of them so that is the quarter.
Here is the code but it does not seem efficient for me.
(firstQuarter is North, secondQuarter is East and so on..
note: assume that mousemove is inside the circle.
var firstQuarterX = centerX;
var firstQuarterY = centerY - radius;
var secondQuarterX = centerX + radius;
var secondQuarterY = centerY;
var thirdQuarterX = centerX;
var thirdQuarterY = centerY + radius;
var fourthQuarterX = centerX - radius;
var fourthQuarterY = centerY;
var distanceToFirst = Math.sqrt(Math.pow(x-firstQuarterX, 2) + Math.pow(y-firstQuarterY, 2));
var distanceToSecond = Math.sqrt(Math.pow(x-secondQuarterX, 2) + Math.pow(y-secondQuarterY, 2));
var distanceToThird = Math.sqrt(Math.pow(x-thirdQuarterX, 2) + Math.pow(y-thirdQuarterY, 2));
var distanceToFourth = Math.sqrt(Math.pow(x-fourthQuarterX, 2) + Math.pow(y-fourthQuarterY, 2));
var min = Math.min(distanceToFirst, distanceToSecond, distanceToThird, distanceToFourth);
var numbers = [distanceToFirst, distanceToSecond, distanceToThird, distanceToFourth];
var index = numbers.indexOf(min); // it will give 0 or 1 or 2 or 3
var quarter = index + 1;

Observe that the boundaries between your quarters lie along the lines with equations y = x and y = -x, relative to an origin at the center of the circle. You can use those to evaluate which quarter each point falls in.
If your point is (x, y), then its coordinates relative to the center of the circle are xRelative = x - centerX and yRelative = y - centerY. Then
your point is in the first (south in your code) quarter if yRelative < 0 and Math.abs(xRelative) < -yRelative
your point is in the second (east) quarter if xRelative > 0 and Math.abs(yRelative) < xRelative
your point is in the third (north) quarter if yRelative > 0 and Math.abs(xRelative) < yRelative
your point is in the fourth (west) quarter if xRelative < 0 and Math.abs(yRelative) < -xRelative
I leave it to you to determine to which quarter to assign points that fall exactly on a boundary. Also, you can implement a little decision tree based on those criteria if you prefer; that should be a little more efficient then testing each criterion in turn.

Not so sure but I think this might work. Math.atan2(CenterY - y, CenterX - x) * 180 / Math.PI gives the apparent angle between the points. Do the remaining math to figure out the quarter.

What about something like:
return x>centerX?(y>centerY?"Quad 2":"Quad 1"):(y>centerY?"Quad 3":"Quad 4");
Less graceful, more slim.

For more efficient algorithm, you can compute the quadrant just by analyzing the signs of dx + dy and dx - dy quantities (dx, dy being x, y minus centerX, centerY respectively) (I presume that as your animation shows, your quadrants are rotated by 45 degrees against 'standard' quadrants.

Find column, row on 2D isometric grid from x,y screen space coords (Convert equation to function)

I'm trying to find the row, column in a 2d isometric grid of a screen space point (x, y)
Now I pretty much know what I need to do which is find the length of the vectors in red in the pictures above and then compare it to the length of the vector that represent the bounds of the grid (which is represented by the black vectors)
Now I asked for help over at mathematics stack exchange to get the equation for figuring out what the parallel vectors are of a point x,y compared to the black boundary vectors. Link here Length of Perpendicular/Parallel Vectors
but im having trouble converting this to a function
Ideally i need enough of a function to get the length of both red vectors from three sets of points, the x,y of the end of the 2 black vectors and the point at the end of the red vectors.
Any language is fine but ideally javascript

What you need is a base transformation:
Suppose the coordinates of the first black vector are (x1, x2) and the coordinates of the second vector are (y1, y2).
Therefore, finding the red vectors that get at a point (z1, z2) is equivalent to solving the following linear system:
x1*r1 + y1*r2 = z1
x2*r1 + y2*r2 = z2
or in matrix form:
A x = b
/x1 y1\ |r1| = |z1|
\x2 y2/ |r2| |z2|
x = inverse(A)*b
For example, lets have the black vector be (2, 1) and (2, -1). The corresponding matrix A will be
2 2
1 -1
and its inverse will be
1/4 1/2
1/4 -1/2
So a point (x, y) in the original coordinates will be able to be represened in the alternate base, bia the following formula:
(x, y) = (1/4 * x + 1/2 * y)*(2,1) + (1/4 * x -1/2 * y)*(2, -1)

What exactly is the point of doing it like this? Any isometric grid you display usually contains cells of equal size, so you can skip all the vector math and simply do something like:
var xStep = 50,
yStep = 30, // roughly matches your image
pointX = 2*xStep,
pointY = 0;
Basically the points on any isometric grid fall onto the intersections of a non-isometric grid. Isometric grid controller:
screenPositionToIsoXY : function(o, w, h){
var sX = ((((o.x - this.canvas.xPosition) - this.screenOffsetX) / this.unitWidth ) * 2) >> 0,
sY = ((((o.y - this.canvas.yPosition) - this.screenOffsetY) / this.unitHeight) * 2) >> 0,
isoX = ((sX + sY - this.cols) / 2) >> 0,
isoY = (((-1 + this.cols) - (sX - sY)) / 2) >> 0;
// isoX = ((sX + sY) / isoGrid.width) - 1
// isoY = ((-2 + isoGrid.width) - sX - sY) / 2
return $.extend(o, {
isoX : Math.constrain(isoX, 0, this.cols - (w||0)),
isoY : Math.constrain(isoY, 0, this.rows - (h||0))
});
},
// ...
isoToUnitGrid : function(isoX, isoY){
var offset = this.grid.offset(),
isoX = $.uD(isoX) ? this.isoX : isoX,
isoY = $.uD(isoY) ? this.isoY : isoY;
return {
x : (offset.x + (this.grid.unitWidth / 2) * (this.grid.rows - this.isoWidth + isoX - isoY)) >> 0,
y : (offset.y + (this.grid.unitHeight / 2) * (isoX + isoY)) >> 0
};
},

Okay so with the help of other answers (sorry guys neither quite provided the answer i was after)
I present my function for finding the grid position on an iso 2d grid using a world x,y coordinate where the world x,y is an offset screen space coord.
WorldPosToGridPos: function(iPosX, iPosY){
var d = (this.mcBoundaryVectors.upper.x * this.mcBoundaryVectors.lower.y) - (this.mcBoundaryVectors.upper.y * this.mcBoundaryVectors.lower.x);
var a = ((iPosX * this.mcBoundaryVectors.lower.y) - (this.mcBoundaryVectors.lower.x * iPosY)) / d;
var b = ((this.mcBoundaryVectors.upper.x * iPosY) - (iPosX * this.mcBoundaryVectors.upper.y)) / d;
var cParaUpperVec = new Vector2(a * this.mcBoundaryVectors.upper.x, a * this.mcBoundaryVectors.upper.y);
var cParaLowerVec = new Vector2(b * this.mcBoundaryVectors.lower.x, b * this.mcBoundaryVectors.lower.y);
var iGridWidth = 40;
var iGridHeight = 40;
var iGridX = Math.floor((cParaLowerVec.length() / this.mcBoundaryVectors.lower.length()) * iGridWidth);
var iGridY = Math.floor((cParaUpperVec.length() / this.mcBoundaryVectors.upper.length()) * iGridHeight);
return {gridX: iGridX, gridY: iGridY};
},
The first line is best done once in an init function or similar to save doing the same calculation over and over, I just included it for completeness.
The mcBoundaryVectors are two vectors defining the outer limits of the x and y axis of the isometric grid (The black vectors shown in the picture above).
Hope this helps anyone else in the future