I've been banging my head on the keyboard for about one week now and I can't figure a proper solution for my problem. I think it's more Math related than HTML Canvas... hopefully someone can point me into the right direction.
I'm having an HTML Canvas where users can draw lines using they mouse and the very simple moveTo() and lineTo() functions. When the user is done I save the coords in a MongoDB. When the user hits the page later again I want to display his drawing BUT I don't want to load the entire picture with all stored coordinates at once, I want to return it in tiles (for better performance by caching each tile).
The tiles are 200x200 pixels (fixed offsets and width, starting at 0 -> 200-> 400 ->...).
Now, when the user draws a line from let's say 50,50(x/y) to 250,250(x/y) there's only one dot in each bounding box (tile). I need to split the lines and calculate the start and ending points of each line in each bounding box (tile). Otherwise I can't draw the image partially (in tiles). It get's even more complicated when a single line crosses multiple bounding boxes (tiles). For instance: 100,100 (x/y) -> -1234,-300 (x/y).
The lines can go from any point (+/-) to ANY direction for ANY distance.
Of course I looked at Bresenham's good old algorithm and it worked - partially, but it seems like the longest and most resource-hungry solution to me.
So, the reason I'm here is that I hope someone can point me into the right direction with (perhaps) another approach of calculating the start/ending points of my lines for each bounding box.
Code examples are very welcome in JavaScript or PHP.
Thank you for reading and thinking about it :)
tl;dr: Use planes, maths explained below. There's a canvas example at the bottom.
Given that all of your cells are axis-aligned bounding boxes, you could use the plane equation to find the intersection of your line with the edges.
Planes
You can think of your box as a set of four geometric planes. Each plane has a normal, or a vector of length one, indicating which direction is the "front" of the plane. The normals for the planes that make up your cell's sides would be:
top = {x: 0, y: -1};
bottom = {x: 0, y: 1};
left = {x: -1, y: 0};
right = {x: 1, y: 0};
Given a point on the plane, the plane has the equation:
distance = (normal.x * point.x) + (normal.y * point.y)
You can use this equation to calculate the distance of the plane. In this case, you know the top-left corner of your box (let's say x is 10 and y is 100) is on the top plane, so you can do:
distance = (0 * 10) + (-1 * 100)
distance = -100
Checking a point against a plane
Once you have the distance, you can reuse the equation to check where any point is, relative to the plane. For a random point p (where x is -50 and y is 90), you can do:
result = (normal.x * p.x) + (normal.y * p.y) - distance
result = (0 * -50) + (-1 * 90) - (-100)
result = 0 + (-90) - (-100)
result = -90 + 100
result = 10
There are two possible results:
if (result >= 0) {
// point is in front of the plane, or coplanar.
// zero means it is coplanar, but we don't need to distinguish.
} else {
// point is behind the plane
}
Checking a line against a plane
You can check both endpoints of a line from a to b in this way:
result1 = (normal.x * a.x) + (normal.y * a.y) - distance
result2 = (normal.x * b.x) + (normal.y * b.y) - distance
There are four possible results:
if (result1 >= 0 && result2 >= 0) {
// the line is completely in front of the plane
} else if (result1 < 0 && result2 < 0) {
// the line is completely behind the plane
} else if (result1 >= 0 && result2 < 0) {
// a is in front, but b is behind, line is entering the plane
} else if (result1 < 0 && result2 >= 0) {
// a is behind, but b is in front, line is exiting the plane
}
When the line intersects the plane, you want to find the point of intersection. It helps to think of a line in vector terms:
a + t * (b - a)
If t == 0, you are at the start of the line, and t == 1 is the end of the line. In this context, you can calculate the point of intersection as:
time = result1 / (result1 - result2)
And the point of intersection as:
hit.x = a.x + (b.x - a.x) * time
hit.y = a.y + (b.y - a.y) * time
Checking a line against the box
With that math, you can figure out the lines of intersection with your box. You just need to test the endpoints of your line against each plane, and find the minimum and maximum values of time.
Because your box is a convex polygon, there is an early out in this check: if the line is completely in front of any one plane in your box, it cannot intersect with your box. You can skip checking the rest of the planes.
In JavaScript, your result might look something like this:
/**
* Find the points where a line intersects a box.
*
* #param a Start point for the line.
* #param b End point for the line.
* #param tl Top left of the box.
* #param br Bottom right of the box.
* #return Object {nearTime, farTime, nearHit, farHit}, or false.
*/
function intersectLineBox(a, b, tl, br) {
var nearestTime = -Infinity;
var furthestTime = Infinity;
var planes = [
{nx: 0, ny: -1, dist: -tl.y}, // top
{nx: 0, ny: 1, dist: br.y}, // bottom
{nx: -1, ny: 0, dist: -tl.x}, // left
{nx: 1, ny: 0, dist: br.x} // right
];
for (var i = 0; i < 4; ++i) {
var plane = planes[i];
var nearDist = (plane.nx * a.x + plane.ny * a.y) - plane.dist;
var farDist = (plane.nx * b.x + plane.ny * b.y) - plane.dist;
if (nearDist >= 0 && farDist >= 0) {
// both are in front of the plane, line doesn't hit box
return false;
} else if (nearDist < 0 && farDist < 0) {
// both are behind the plane
continue;
} else {
var time = nearDist / (nearDist - farDist);
if (nearDist >= farDist) {
// entering the plane
if (time > nearestTime) {
nearestTime = time;
}
} else {
// exiting the plane
if (time < furthestTime) {
furthestTime = time;
}
}
}
}
if (furthestTime < nearestTime) {
return false;
}
return {
nearTime: nearestTime,
farTime: furthestTime,
nearHit: {
x: a.x + (b.x - a.x) * nearestTime,
y: a.y + (b.y - a.y) * nearestTime
},
farHit: {
x: a.x + (b.x - a.x) * furthestTime,
y: a.y + (b.y - a.y) * furthestTime
}
};
}
If this is still too slow, you can also do broadphase culling by dividing the world up into big rects, and assigning lines to those rects. If your line and cell aren't in the same rect, they don't collide.
I've uploaded a canvas example of this.
This looks like you'd have to figure out at what point each line intersects with the bounds of each tile.
Check out the answer to this question: Is there an easy way to detect line segment intersections?
The answers don't provide code, but it shouldn't be too hard to convert the equations to PHP or Javascript...
EDIT:
Why, exactly, do you want to split the lines? I understand you don't want to load all the lines at once, since that could take a while. But what's wrong with just loading and drawing the first few lines, and drawing the remainder later on?
Methinks that would be a lot simpler than having to cut up each line to fit in a specific tile. Tiling is a nice way of optimizing bitmap loading; I don't think it's very appropriate for vector-based drawings.
You could also consider sending an Ajax request, and start drawing the whole thing whenever it comes in; this would not interfere with the loading of the page.
Related
I'm attempting to write code that will generate fractals according to the Chaos game
In particular, I'm trying to debug the faulty generation/rendering of this fractal:
I'm doing this with Javascript in a Canvas element. The relevant Javascript is below:
canvas = document.getElementById('myCanvas');
context = canvas.getContext('2d');
//constants
border = 10 //cardinal distance between vertices and nearest edge(s)
class Point{
constructor(_x, _y){
this.x = _x;
this.y = _y;
}
}
vertices = []
secondLastVertex = 0;
lastVertex = 0;
//vertices in clockwise order (for ease of checking adjacency)
vertices.push(new Point(canvas.width / 2, border)); //top
vertices.push(new Point(canvas.width - border, canvas.height * Math.tan(36 * Math.PI / 180) / 2)); //upper right
vertices.push(new Point(canvas.width * Math.cos(36 * Math.PI / 180), canvas.height - border)); //lower right
vertices.push(new Point(canvas.width * (1 - (Math.cos(36 * Math.PI / 180))), canvas.height - border)); //lower left
vertices.push(new Point(border, canvas.height * Math.tan(36 * Math.PI / 180) / 2)); //upper left
//move half distance towards random vertex but it can't neighbor the last one IF the last two were the same
function updatePoint(){
//pick a random vertex
v = Math.floor(Math.random() * vertices.length);
if(lastVertex == secondLastVertex)
//while randomly selected vertex is adjacent to the last approached vertex
while((v == (lastVertex - 1) % 5) || (v == (lastVertex + 1) % 5))
//pick another random vertex
v = Math.floor(Math.random() * vertices.length);
//cycle the last two vertices
secondLastVertex = lastVertex;
lastVertex = v;
//move half way towards the chosen vertex
point.x = (vertices[v].x + point.x) / 2;
point.y = (vertices[v].y + point.y) / 2;
}
//starting point (doesn't matter where)
point = new Point(canvas.width / 2, canvas.height / 2);
for (var i = 0; i < 1000000; i++){
//get point's next location
updatePoint();
//draw the point
context.fillRect(Math.round(point.x), Math.round(point.y), 1, 1);
}
The rendering that is produced looks like this:
So far I haven't been able to determine what is causing the rendering to be so skewed and wrong. One possibility is that I've misunderstood the rules that generate this fractal (i.e. "move half the distance from the current position towards a random vertex that is not adjacent to the last vertex IF the last two vertices were the same")
Another is that I have some bug in how I'm drawing fractals. But the same code with rule/starting-vertex modifications is able to draw things like the Sierpinkski triangle/carpet and even other pentagonal fractals apparently perfectly. Though one other pentagonal fractal ended up with some weird skewing and "lower right fourth of each self-similar substructure" weirdness.
I tried making some slight modifications to how I interpreted the rules (e.g. "next vertex can't be adjacent OR EQUAL TO previous vertex if last two vertices were the same") but nothing like that helped. I also tried not rounding the coordinates of the target point before plotting it, but though this slightly changed the character/sharpness of the details, it didn't change any larger scale features of the plot.
My issue as kindly pointed out by ggorlen, was that I wasn't comparing vertices for adjacency correctly. I mistakenly thought Javascript evaluated something like (-1 % 5) as 4, rather than -1.
To fix this, I add 4 to the index instead of subtracting 1, before modding it against 5 (the number of vertices)
This completely fixed the render. (in not just this case but other cases I'd been testing with different fractals)
Summary
This question is in JavaScript, but an answer in any language, pseudo-code, or just the maths would be great!
I have been trying to implement the Separating-Axis-Theorem to accomplish the following:
Detecting an intersection between a convex polygon and a circle.
Finding out a translation that can be applied to the circle to resolve the intersection, so that the circle is barely touching the polygon but no longer inside.
Determining the axis of the collision (details at the end of the question).
I have successfully completed the first bullet point and you can see my javascript code at the end of the question. I am having difficulties with the other parts.
Resolving the intersection
There are plenty of examples online on how to resolve the intersection in the direction with the smallest/shortest overlap of the circle. You can see in my code at the end that I already have this calculated.
However this does not suit my needs. I must resolve the collision in the opposite direction of the circle's trajectory (assume I already have the circle's trajectory and would like to pass it into my function as a unit-vector or angle, whichever suits).
You can see the difference between the shortest resolution and the intended resolution in the below image:
How can I calculate the minimum translation vector for resolving the intersection inside my test_CIRCLE_POLY function, but that is to be applied in a specific direction, the opposite of the circle's trajectory?
My ideas/attempts:
My first idea was to add an additional axis to the axes that must be tested in the SAT algorithm, and this axis would be perpendicular to the circle's trajectory. I would then resolve based on the overlap when projecting onto this axis. This would sort of work, but would resolve way to far in most situations. It won't result in the minimum translation. So this won't be satisfactory.
My second idea was to continue to use magnitude of the shortest overlap, but change the direction to be the opposite of the circle's trajectory. This looks promising, but there are probably many edge-cases that I haven't accounted for. Maybe this is a nice place to start.
Determining side/axis of collision
I've figured out a way to determine which sides of the polygon the circle is colliding with. For each tested axis of the polygon, I would simply check for overlap. If there is overlap, that side is colliding.
This solution will not be acceptable once again, as I would like to figure out only one side depending on the circle's trajectory.
My intended solution would tell me, in the example image below, that axis A is the axis of collision, and not axis B. This is because once the intersection is resolved, axis A is the axis corresponding to the side of the polygon that is just barely touching the circle.
My ideas/attempts:
Currently I assume the axis of collision is that perpendicular to the MTV (minimum translation vector). This is currently incorrect, but should be the correct axis once I've updated the intersection resolution process in the first half of the question. So that part should be tackled first.
Alternatively I've considered creating a line from the circle's previous position and their current position + radius, and checking which sides intersect with this line. However, there's still ambiguity, because on occasion there will be more than one side intersecting with the line.
My code so far
function test_CIRCLE_POLY(circle, poly, circleTrajectory) {
// circleTrajectory is currently not being used
let axesToTest = [];
let shortestOverlap = +Infinity;
let shortestOverlapAxis;
// Figure out polygon axes that must be checked
for (let i = 0; i < poly.vertices.length; i++) {
let vertex1 = poly.vertices[i];
let vertex2 = poly.vertices[i + 1] || poly.vertices[0]; // neighbouring vertex
let axis = vertex1.sub(vertex2).perp_norm();
axesToTest.push(axis);
}
// Figure out circle axis that must be checked
let closestVertex;
let closestVertexDistSqr = +Infinity;
for (let vertex of poly.vertices) {
let distSqr = circle.center.sub(vertex).magSqr();
if (distSqr < closestVertexDistSqr) {
closestVertexDistSqr = distSqr;
closestVertex = vertex;
}
}
let axis = closestVertex.sub(circle.center).norm();
axesToTest.push(axis);
// Test for overlap
for (let axis of axesToTest) {
let circleProj = proj_CIRCLE(circle, axis);
let polyProj = proj_POLY(poly, axis);
let overlap = getLineOverlap(circleProj.min, circleProj.max, polyProj.min, polyProj.max);
if (overlap === 0) {
// guaranteed no intersection
return { intersecting: false };
}
if (Math.abs(overlap) < Math.abs(shortestOverlap)) {
shortestOverlap = overlap;
shortestOverlapAxis = axis;
}
}
return {
intersecting: true,
resolutionVector: shortestOverlapAxis.mul(-shortestOverlap),
// this resolution vector is not satisfactory, I need the shortest resolution with a given direction, which would be an angle passed into this function from the trajectory of the circle
collisionAxis: shortestOverlapAxis.perp(),
// this axis is incorrect, I need the axis to be based on the trajectory of the circle which I would pass into this function as an angle
};
}
function proj_POLY(poly, axis) {
let min = +Infinity;
let max = -Infinity;
for (let vertex of poly.vertices) {
let proj = vertex.projNorm_mag(axis);
min = Math.min(proj, min);
max = Math.max(proj, max);
}
return { min, max };
}
function proj_CIRCLE(circle, axis) {
let proj = circle.center.projNorm_mag(axis);
let min = proj - circle.radius;
let max = proj + circle.radius;
return { min, max };
}
// Check for overlap of two 1 dimensional lines
function getLineOverlap(min1, max1, min2, max2) {
let min = Math.max(min1, min2);
let max = Math.min(max1, max2);
// if negative, no overlap
let result = Math.max(max - min, 0);
// add positive/negative sign depending on direction of overlap
return result * ((min1 < min2) ? 1 : -1);
};
I am assuming the polygon is convex and that the circle is moving along a straight line (at least for a some possibly small interval of time) and is not following some curved trajectory. If it is following a curved trajectory, then things get harder. In the case of curved trajectories, the basic ideas could be kept, but the actual point of collision (the point of collision resolution for the circle) might be harder to calculate. Still, I am outlining an idea, which could be extended to that case too. Plus, it could be adopted as a main approach for collision detection between a circle and a convex polygon.
I have not considered all possible cases, which may include special or extreme situations, but at least it gives you a direction to explore.
Transform in your mind the collision between the circle and the polygon into a collision between the center of the circle (a point) and a version of the polygon thickened by the circle's radius r, i.e. (i) each edge of the polygon is offset (translated) outwards by radius r along a vector perpendicular to it and pointing outside of the polygon, (ii) the vertices become circular arcs of radius r, centered at the polygons vertices and connecting the endpoints of the appropriate neighboring offset edges (basically, put circles of radius r at the vertices of the polygon and take their convex hull).
Now, the current position of the circle's center is C = [ C[0], C[1] ] and it has been moving along a straight line with direction vector V = [ V[0], V[1] ] pointing along the direction of motion (or if you prefer, think of V as the velocity of the circle at the moment when you have detected the collision). Then, there is an axis (or let's say a ray - a directed half-line) defined by the vector equation X = C - t * V, where t >= 0 (this axis is pointing to the past trajectory). Basically, this is the half-line that passes through the center point C and is aligned with/parallel to the vector V. Now, the point of resolution, i.e. the point where you want to move your circle to is the point where the axis X = C - t * V intersects the boundary of the thickened polygon.
So you have to check (1) first axis intersection for edges and then (2) axis intersection with circular arcs pertaining to the vertices of the original polygon.
Assume the polygon is given by an array of vertices P = [ P[0], P[1], ..., P[N], P[0] ] oriented counterclockwise.
(1) For each edge P[i-1]P[i] of the original polygon, relevant to your collision (these could be the two neighboring edges meeting at the vertex based on which the collision is detected, or it could be actually all edges in the case of the circle moving with very high speed and you have detected the collision very late, so that the actual collision did not even happen there, I leave this up to you, because you know better the details of your situation) do the following. You have as input data:
C = [ C[0], C[1] ]
V = [ V[0], V[1] ]
P[i-1] = [ P[i-1][0], P[i-1][1] ]
P[i] = [ P[i][0], P[i][1] ]
Do:
Normal = [ P[i-1][1] - P[i][1], P[i][0] - P[i-1][0] ];
Normal = Normal / sqrt((P[i-1][1] - P[i][1])^2 + ( P[i][0] - P[i-1][0] )^2);
// you may have calculated these already
Q_0[0] = P[i-1][0] + r*Normal[0];
Q_0[1] = P[i-1][1] + r*Normal[1];
Q_1[0] = P[i][0]+ r*Normal[0];
Q_1[1] = P[i][1]+ r*Normal[1];
Solve for s, t the linear system of equations (the equation for intersecting ):
Q_0[0] + s*(Q_1[0] - Q_0[0]) = C[0] - t*V[0];
Q_0[1] + s*(Q_1[1] - Q_0[1]) = C[1] - t*V[1];
if 0<= s <= 1 and t >= 0, you are done, and your point of resolution is
R[0] = C[0] - t*V[0];
R[1] = C[1] - t*V[1];
else
(2) For the each vertex P[i] relevant to your collision, do the following:
solve for t the quadratic equation (there is an explicit formula)
norm(P[i] - C + t*V )^2 = r^2
or expanded:
(V[0]^2 + V[1]^2) * t^2 + 2 * ( (P[i][0] - C[0])*V[0] + (P[i][1] - C[1])*V[1] )*t + ( P[i][0] - C[0])^2 + (P[i][1] - C[1])^2 ) - r^2 = 0
or if you prefer in a more code-like way:
a = V[0]^2 + V[1]^2;
b = (P[i][0] - C[0])*V[0] + (P[i][1] - C[1])*V[1];
c = (P[i][0] - C[0])^2 + (P[i][1] - C[1])^2 - r^2;
D = b^2 - a*c;
if D < 0 there is no collision with the vertex
i.e. no intersection between the line X = C - t*V
and the circle of radius r centered at P[i]
else
D = sqrt(D);
t1 = ( - b - D) / a;
t2 = ( - b + D) / a;
where t2 >= t1
Then your point of resolution is
R[0] = C[0] - t2*V[0];
R[1] = C[1] - t2*V[1];
Circle polygon intercept
If the ball is moving and if you can ensure that the ball always starts outside the polygon then the solution is rather simple.
We will call the ball and its movement the ball line. It starts at the ball's current location and end at the position the ball will be at the next frame.
To solve you find the nearest intercept to the start of the ball line.
There are two types of intercept.
Line segment (ball line) with Line segment (polygon edge)
Line segment (ball line) with circle (circle at each (convex only) polygon corner)
The example code has a Lines2 object that contains the two relevant intercept functions. The intercepts are returned as a Vec2containing two unit distances. The intercept functions are for lines (infinite length) not line sgements. If there is no intercept then the return is undefined.
For the line intercepts Line2.unitInterceptsLine(line, result = new Vec2()) the unit values (in result) are the unit distance along each line from the start. negative values are behind the start.
To take in account of the ball radius each polygon edge is offset the ball radius along its normal. It is important that the polygon edges have a consistent direction. In the example the normal is to the right of the line and the polygon points are in a clockwise direction.
For the line segment / circle intercepts Line2.unitInterceptsCircle(center, radius, result = new Vec2()) the unit values (in result) are the unit distance along the line where it intercepts the circle. result.x will always contain the closest intercept (assuming you start outside the circle). If there is an intercept there ways always be two, even if they are at the same point.
Example
The example contains all that is needed
The objects of interest are ball and poly
ball defines the ball and its movement. There is also code to draw it for the example
poly holds the points of the polygon. Converts the points to offset lines depending on the ball radius. It is optimized to that it only calculates the lines if the ball radius changes.
The function poly.movingBallIntercept is the function that does all the work. It take a ball object and an optional results vector.
It returns the position as a Vec2 of the ball if it contacts the polygon.
It does this by finding the smallest unit distance to the offset lines, and point (as circle) and uses that unit distance to position the result.
Note that if the ball is inside the polygon the intercepts with the corners is reversed. The function Line2.unitInterceptsCircle does provide 2 unit distance where the line enters and exits the circle. However you need to know if you are inside or outside to know which one to use. The example assumes you are outside the polygon.
Instructions
Move the mouse to change the balls path.
Click to set the balls starting position.
Math.EPSILON = 1e-6;
Math.isSmall = val => Math.abs(val) < Math.EPSILON;
Math.isUnit = u => !(u < 0 || u > 1);
Math.TAU = Math.PI * 2;
/* export {Vec2, Line2} */ // this should be a module
var temp;
function Vec2(x = 0, y = (temp = x, x === 0 ? (x = 0 , 0) : (x = x.x, temp.y))) {
this.x = x;
this.y = y;
}
Vec2.prototype = {
init(x, y = (temp = x, x = x.x, temp.y)) { this.x = x; this.y = y; return this }, // assumes x is a Vec2 if y is undefined
copy() { return new Vec2(this) },
equal(v) { return (this.x - v.x) === 0 && (this.y - v.y) === 0 },
isUnits() { return Math.isUnit(this.x) && Math.isUnit(this.y) },
add(v, res = this) { res.x = this.x + v.x; res.y = this.y + v.y; return res },
sub(v, res = this) { res.x = this.x - v.x; res.y = this.y - v.y; return res },
scale(val, res = this) { res.x = this.x * val; res.y = this.y * val; return res },
invScale(val, res = this) { res.x = this.x / val; res.y = this.y / val; return res },
dot(v) { return this.x * v.x + this.y * v.y },
uDot(v, div) { return (this.x * v.x + this.y * v.y) / div },
cross(v) { return this.x * v.y - this.y * v.x },
uCross(v, div) { return (this.x * v.y - this.y * v.x) / div },
get length() { return this.lengthSqr ** 0.5 },
set length(l) { this.scale(l / this.length) },
get lengthSqr() { return this.x * this.x + this.y * this.y },
rot90CW(res = this) {
const y = this.x;
res.x = -this.y;
res.y = y;
return res;
},
};
const wV1 = new Vec2(), wV2 = new Vec2(), wV3 = new Vec2(); // pre allocated work vectors used by Line2 functions
function Line2(p1 = new Vec2(), p2 = (temp = p1, p1 = p1.p1 ? p1.p1 : p1, temp.p2 ? temp.p2 : new Vec2())) {
this.p1 = p1;
this.p2 = p2;
}
Line2.prototype = {
init(p1, p2 = (temp = p1, p1 = p1.p1, temp.p2)) { this.p1.init(p1); this.p2.init(p2) },
copy() { return new Line2(this) },
asVec(res = new Vec2()) { return this.p2.sub(this.p1, res) },
unitDistOn(u, res = new Vec2()) { return this.p2.sub(this.p1, res).scale(u).add(this.p1) },
translate(vec, res = this) {
this.p1.add(vec, res.p1);
this.p2.add(vec, res.p2);
return res;
},
translateNormal(amount, res = this) {
this.asVec(wV1).rot90CW().length = -amount;
this.translate(wV1, res);
return res;
},
unitInterceptsLine(line, res = new Vec2()) { // segments
this.asVec(wV1);
line.asVec(wV2);
const c = wV1.cross(wV2);
if (Math.isSmall(c)) { return }
wV3.init(this.p1).sub(line.p1);
res.init(wV1.uCross(wV3, c), wV2.uCross(wV3, c));
return res;
},
unitInterceptsCircle(point, radius, res = new Vec2()) {
this.asVec(wV1);
var b = -2 * this.p1.sub(point, wV2).dot(wV1);
const c = 2 * wV1.lengthSqr;
const d = (b * b - 2 * c * (wV2.lengthSqr - radius * radius)) ** 0.5
if (isNaN(d)) { return }
return res.init((b - d) / c, (b + d) / c);
},
};
/* END of file */ // Vec2 and Line2 module
/* import {vec2, Line2} from "whateverfilename.jsm" */ // Should import vec2 and line2
const POLY_SCALE = 0.5;
const ball = {
pos: new Vec2(-150,0),
delta: new Vec2(10, 10),
radius: 20,
drawPath(ctx) {
ctx.beginPath();
ctx.arc(this.pos.x, this.pos.y, this.radius, 0, Math.TAU);
ctx.stroke();
},
}
const poly = {
bRadius: 0,
lines: [],
set ballRadius(radius) {
const len = this.points.length
this.bRadius = ball.radius;
i = 0;
while (i < len) {
let line = this.lines[i];
if (line) { line.init(this.points[i], this.points[(i + 1) % len]) }
else { line = new Line2(new Vec2(this.points[i]), new Vec2(this.points[(i + 1) % len])) }
this.lines[i++] = line.translateNormal(radius);
}
this.lines.length = i;
},
points: [
new Vec2(-200, -150).scale(POLY_SCALE),
new Vec2(200, -100).scale(POLY_SCALE),
new Vec2(100, 0).scale(POLY_SCALE),
new Vec2(200, 100).scale(POLY_SCALE),
new Vec2(-200, 75).scale(POLY_SCALE),
new Vec2(-150, -50).scale(POLY_SCALE),
],
drawBallLines(ctx) {
if (this.lines.length) {
const r = this.bRadius;
ctx.beginPath();
for (const l of this.lines) {
ctx.moveTo(l.p1.x, l.p1.y);
ctx.lineTo(l.p2.x, l.p2.y);
}
for (const p of this.points) {
ctx.moveTo(p.x + r, p.y);
ctx.arc(p.x, p.y, r, 0, Math.TAU);
}
ctx.stroke()
}
},
drawPath(ctx) {
ctx.beginPath();
for (const p of this.points) { ctx.lineTo(p.x, p.y) }
ctx.closePath();
ctx.stroke();
},
movingBallIntercept(ball, res = new Vec2()) {
if (this.bRadius !== ball.radius) { this.ballRadius = ball.radius }
var i = 0, nearest = Infinity, nearestGeom, units = new Vec2();
const ballT = new Line2(ball.pos, ball.pos.add(ball.delta, new Vec2()));
for (const p of this.points) {
const res = ballT.unitInterceptsCircle(p, ball.radius, units);
if (res && units.x < nearest && Math.isUnit(units.x)) { // assumes ball started outside poly so only need first point
nearest = units.x;
nearestGeom = ballT;
}
}
for (const line of this.lines) {
const res = line.unitInterceptsLine(ballT, units);
if (res && units.x < nearest && units.isUnits()) { // first unit.x is for unit dist on line
nearest = units.x;
nearestGeom = ballT;
}
}
if (nearestGeom) { return ballT.unitDistOn(nearest, res) }
return;
},
}
const ctx = canvas.getContext("2d");
var w = canvas.width, cw = w / 2;
var h = canvas.height, ch = h / 2
requestAnimationFrame(mainLoop);
// line and point for displaying mouse interaction. point holds the result if any
const line = new Line2(ball.pos, ball.pos.add(ball.delta, new Vec2())), point = new Vec2();
function mainLoop() {
ctx.setTransform(1,0,0,1,0,0); // reset transform
if(w !== innerWidth || h !== innerHeight){
cw = (w = canvas.width = innerWidth) / 2;
ch = (h = canvas.height = innerHeight) / 2;
}else{
ctx.clearRect(0,0,w,h);
}
ctx.setTransform(1,0,0,1,cw,ch); // center to canvas
if (mouse.button) { ball.pos.init(mouse.x - cw, mouse.y - ch) }
line.p2.init(mouse.x - cw, mouse.y - ch);
line.p2.sub(line.p1, ball.delta);
ctx.lineWidth = 1;
ctx.strokeStyle = "#000"
poly.drawPath(ctx)
ctx.strokeStyle = "#F804"
poly.drawBallLines(ctx);
ctx.strokeStyle = "#F00"
ctx.beginPath();
ctx.arc(ball.pos.x, ball.pos.y, ball.radius, 0, Math.TAU);
ctx.moveTo(line.p1.x, line.p1.y);
ctx.lineTo(line.p2.x, line.p2.y);
ctx.stroke();
ctx.strokeStyle = "#00f"
ctx.lineWidth = 2;
ctx.beginPath();
if (poly.movingBallIntercept(ball, point)) {
ctx.arc(point.x, point.y, ball.radius, 0, Math.TAU);
} else {
ctx.arc(line.p2.x, line.p2.y, ball.radius, 0, Math.TAU);
}
ctx.stroke();
requestAnimationFrame(mainLoop);
}
const mouse = {x:0, y:0, button: false};
function mouseEvents(e) {
const bounds = canvas.getBoundingClientRect();
mouse.x = e.pageX - bounds.left - scrollX;
mouse.y = e.pageY - bounds.top - scrollY;
mouse.button = e.type === "mousedown" ? true : e.type === "mouseup" ? false : mouse.button;
}
["mousedown","mouseup","mousemove"].forEach(name => document.addEventListener(name,mouseEvents));
#canvas {
position: absolute;
top: 0px;
left: 0px;
}
<canvas id="canvas"></canvas>
Click to position ball. Move mouse to test trajectory
Vec2 and Line2
To make it easier a vector library will help. For the example I wrote a quick Vec2 and Line2 object (Note only functions used in the example have been tested, Note The object are designed for performance, inexperienced coders should avoid using these objects and opt for a more standard vector and line library)
It's probably not what you're looking for, but here's a way to do it (if you're not looking for perfect precision) :
You can try to approximate the position instead of calculating it.
The way you set up your code has a big advantage : You have the last position of the circle before the collision. Thanks to that, you can just "iterate" through the trajectory and try to find a position that is closest to the intersection position.
I'll assume you already have a function that tells you if a circle is intersecting with the polygon.
Code (C++) :
// What we need :
Vector startPos; // Last position of the circle before the collision
Vector currentPos; // Current, unwanted position
Vector dir; // Direction (a unit vector) of the circle's velocity
float distance = compute_distance(startPos, currentPos); // The distance from startPos to currentPos.
Polygon polygon; // The polygon
Circle circle; // The circle.
unsigned int iterations_count = 10; // The number of iterations that will be done. The higher this number, the more precise the resolution.
// The algorithm :
float currentDistance = distance / 2.f; // We start at the half of the distance.
Circle temp_copy; // A copy of the real circle to "play" with.
for (int i = 0; i < iterations_count; ++i) {
temp_copy.pos = startPos + currentDistance * dir;
if (checkForCollision(temp_copy, polygon)) {
currentDistance -= currentDistance / 2.f; // We go towards startPos by the half of the current distance.
}
else {
currentDistance += currentDistance / 2.f; // We go towards currentPos by the half of the current distance.
}
}
// currentDistance now contains the distance between startPos and the intersection point
// And this is where you should place your circle :
Vector intersectionPoint = startPos + currentDistance * dir;
I haven't tested this code so I hope there's no big mistake in there. It's also not optimized and there are a few problems with this approach (the intersection point could end up inside the polygon) so it still needs to be improved but I think you get the idea.
The other (big, depending on what you're doing) problem with this is that it's an approximation and not a perfect answer.
Hope this helps !
I'm not sure if I understood the scenario correctly, but an efficient straight forward use case would be, to only use a square bounding box of your circle first, calculating intersection of that square with your polygone is extremely fast, much much faster, than using the circle. Once you detect an intersection of that square and the polygone, you need to think or to write which precision is mostly suitable for your scenarion. If you need a better precision, than at this state, you can go on as this from here:
From the 90° angle of your sqare intersection, you draw a 45° degree line until it touches your circle, at this point, where it touches, you draw a new square, but this time, the square is embedded into the circle, let it run now, until this new square intersects the polygon, once it intersects, you have a guaranteed circle intersection. Depending on your needed precision, you can simply play around like this.
I'm not sure what your next problem is from here? If it has to be only the inverse of the circles trajectory, than it simply must be that reverse, I'm really not sure what I'm missing here.
function distance(x1, y1, x2, y2) {
var x = x1 - x2;
var y = y1 - y2;
return(Math.sqrt((x*x) + (y*y)))
};
function collisionCirc(circ1, circ2) {
var d = distance(circ1.x, circ1.y, circ2.x, circ2.y);
var r = circ1.radius + circ2.radius;
return(r > d);
};
function collisionCircPoint(circ1, circ2) {
var cx = ((circ1.x * circ2.radius) + (circ2.x * circ1.radius)) / (circ1.radius + circ2.radius);
var cy = ((circ1.y * circ2.radius) + (circ2.y * circ1.radius)) / (circ1.radius + circ2.radius);
var p = [cx, cy];
return p;
};
function angleDegrees(x1, y1, x2, y2) {
return (Math.atan2(y2 - y1, x2 - x1) * 180 / Math.PI) + 180;
};
function updateCollisions() {
var a;
var p;
Player.hitArea = new PIXI.Circle(Player.sprite.x, Player.sprite.y, 20);
MapObjects.chest.hitArea = new PIXI.Circle(MapObjects.chest.x, MapObjects.chest.y, 20);
if (collisionCirc(Player.hitArea, MapObjects.chest.hitArea)) {
a = angleDegrees(Player.sprite.x, Player.sprite.y, MapObjects.chest.x, MapObjects.chest.y);
p = collisionCircPoint(Player.hitArea, MapObjects.chest.hitArea);
Player.sprite.x = p[0];
Player.sprite.y = p[1];
};
};
I have 2 sprites on the map and each has a circle hitArea defined. I am trying to make a smooth circular collision that the player cannot pass through. I thought I could just set the Player.sprite's coordinates to the point of collision but it just warps him to the MapObjects.chest's coordinates, even though the point of collision is correct and is 20 pixels from the MapObject.chest's center. What am I doing wrong or what more information is needed to create a collision much like the JavaScript physics libraries where I can circle around a circle object?
The collision point is between the player and the obstacle. If you move the player towards the collision point, you are actually moving the player closer. For example, if there's exactly 40 px (r1+r2) between the player and the obstacle, the collision point is between them, at only 20 px from the obstacle!
When you have multiple objects, getting it right when the collision has already happened is difficult. If there is only one obstacle nearby, you can simply move the player directly away from the obstacle. However, this way the player might actually end up inside another obstacle.
Another solution is to go back to the start and try smaller movements, until there is no collision. This way you would eventually get it right, but this might also be slow.
The mathematically correct solution is to calculate the maximum distance to move before the collision happens. This is done by solving the following vector equation:
# p = player position before moving
# o = obstacle position
# u = player direction (unit vector)
# d = distance to move
distance(o, p + d * u) = o.radius + p.radius
That's mathematics, you may solve it by yourself or using a tool like Wolfram Alpha.
Solving this equation will give you zero, one or two possible values for the distance. Negative values you can dismiss, as they mean that the player is already past the obstacle. If you get only one value, it means that the player would merely brush the obstacle, which you can also dismiss. Two values mean that the collision happens between these distances; the smaller value is where the collision starts, and the larger value is where the player would already be through the obstacle. Also, if one value is positive and the other is negative, it means that the player is already inside the obstacle, which should never happen.
You should run this check for all nearby obstacles and then move the player according to the smallest non-negative result (that is, zero or positive), or less, if the player can't move that fast.
Finally, to circle around a round object, you can move the player a little bit in a perpendicular direction (either left or right, depending on which side of the obstacle the player will be passing) after a collision, if this doesn't cause any new collisions.
There are many other possible implementations.
Player.hitArea = new PIXI.Circle(Player.sprite.x, Player.sprite.y, 20);
MapObjects.chest.hitArea = new PIXI.Circle(MapObjects.chest.x, MapObjects.chest.y, 20);
if (collisionCirc(Player.hitArea, MapObjects.chest.hitArea)) {
p = collisionCircPoint(Player.hitArea, MapObjects.chest.hitArea);
a = angleDegrees(Player.sprite.x, Player.sprite.y, MapObjects.chest.x, MapObjects.chest.y);
if (Player.sprite.x - MapObjects.chest.x > 0) {
Player.sprite.x += 1;
} else if (Player.sprite.x + MapObjects.chest.x > 0) {
Player.sprite.x -= 1;
};
if (Player.sprite.y - MapObjects.chest.y > 0) {
Player.sprite.y += 1;
} else if (Player.sprite.y + MapObjects.chest.y > 0) {
Player.sprite.y -= 1;
};
};
};
I added that and it actually works well enough minus the player speed being slightly too fast when running into the MapObjects.chest's hitArea at certain angles. Work on that later.
So, i'm trying to implement hough transform, this version is 1-dimensional (its for all dims reduced to 1 dim optimization) version based on the minor properties.
Enclosed is my code, with a sample image... input and output.
Obvious question is what am i doing wrong. I've tripled check my logic and code and it looks good also my parameters. But obviously i'm missing on something.
Notice that the red pixels are supposed to be ellipses centers , while the blue pixels are edges to be removed (belong to the ellipse that conform to the mathematical equations).
also, i'm not interested in openCV / matlab / ocatve / etc.. usage (nothing against them).
Thank you very much!
var fs = require("fs"),
Canvas = require("canvas"),
Image = Canvas.Image;
var LEAST_REQUIRED_DISTANCE = 40, // LEAST required distance between 2 points , lets say smallest ellipse minor
LEAST_REQUIRED_ELLIPSES = 6, // number of found ellipse
arr_accum = [],
arr_edges = [],
edges_canvas,
xy,
x1y1,
x2y2,
x0,
y0,
a,
alpha,
d,
b,
max_votes,
cos_tau,
sin_tau_sqr,
f,
new_x0,
new_y0,
any_minor_dist,
max_minor,
i,
found_minor_in_accum,
arr_edges_len,
hough_file = 'sample_me2.jpg',
edges_canvas = drawImgToCanvasSync(hough_file); // make sure everything is black and white!
arr_edges = getEdgesArr(edges_canvas);
arr_edges_len = arr_edges.length;
var hough_canvas_img_data = edges_canvas.getContext('2d').getImageData(0, 0, edges_canvas.width,edges_canvas.height);
for(x1y1 = 0; x1y1 < arr_edges_len ; x1y1++){
if (arr_edges[x1y1].x === -1) { continue; }
for(x2y2 = 0 ; x2y2 < arr_edges_len; x2y2++){
if ((arr_edges[x2y2].x === -1) ||
(arr_edges[x2y2].x === arr_edges[x1y1].x && arr_edges[x2y2].y === arr_edges[x1y1].y)) { continue; }
if (distance(arr_edges[x1y1],arr_edges[x2y2]) > LEAST_REQUIRED_DISTANCE){
x0 = (arr_edges[x1y1].x + arr_edges[x2y2].x) / 2;
y0 = (arr_edges[x1y1].y + arr_edges[x2y2].y) / 2;
a = Math.sqrt((arr_edges[x1y1].x - arr_edges[x2y2].x) * (arr_edges[x1y1].x - arr_edges[x2y2].x) + (arr_edges[x1y1].y - arr_edges[x2y2].y) * (arr_edges[x1y1].y - arr_edges[x2y2].y)) / 2;
alpha = Math.atan((arr_edges[x2y2].y - arr_edges[x1y1].y) / (arr_edges[x2y2].x - arr_edges[x1y1].x));
for(xy = 0 ; xy < arr_edges_len; xy++){
if ((arr_edges[xy].x === -1) ||
(arr_edges[xy].x === arr_edges[x2y2].x && arr_edges[xy].y === arr_edges[x2y2].y) ||
(arr_edges[xy].x === arr_edges[x1y1].x && arr_edges[xy].y === arr_edges[x1y1].y)) { continue; }
d = distance({x: x0, y: y0},arr_edges[xy]);
if (d > LEAST_REQUIRED_DISTANCE){
f = distance(arr_edges[xy],arr_edges[x2y2]); // focus
cos_tau = (a * a + d * d - f * f) / (2 * a * d);
sin_tau_sqr = (1 - cos_tau * cos_tau);//Math.sqrt(1 - cos_tau * cos_tau); // getting sin out of cos
b = (a * a * d * d * sin_tau_sqr ) / (a * a - d * d * cos_tau * cos_tau);
b = Math.sqrt(b);
b = parseInt(b.toFixed(0));
d = parseInt(d.toFixed(0));
if (b > 0){
found_minor_in_accum = arr_accum.hasOwnProperty(b);
if (!found_minor_in_accum){
arr_accum[b] = {f: f, cos_tau: cos_tau, sin_tau_sqr: sin_tau_sqr, b: b, d: d, xy: xy, xy_point: JSON.stringify(arr_edges[xy]), x0: x0, y0: y0, accum: 0};
}
else{
arr_accum[b].accum++;
}
}// b
}// if2 - LEAST_REQUIRED_DISTANCE
}// for xy
max_votes = getMaxMinor(arr_accum);
// ONE ellipse has been detected
if (max_votes != null &&
(max_votes.max_votes > LEAST_REQUIRED_ELLIPSES)){
// output ellipse details
new_x0 = parseInt(arr_accum[max_votes.index].x0.toFixed(0)),
new_y0 = parseInt(arr_accum[max_votes.index].y0.toFixed(0));
setPixel(hough_canvas_img_data,new_x0,new_y0,255,0,0,255); // Red centers
// remove the pixels on the detected ellipse from edge pixel array
for (i=0; i < arr_edges.length; i++){
any_minor_dist = distance({x:new_x0, y: new_y0}, arr_edges[i]);
any_minor_dist = parseInt(any_minor_dist.toFixed(0));
max_minor = b;//Math.max(b,arr_accum[max_votes.index].d); // between the max and the min
// coloring in blue the edges we don't need
if (any_minor_dist <= max_minor){
setPixel(hough_canvas_img_data,arr_edges[i].x,arr_edges[i].y,0,0,255,255);
arr_edges[i] = {x: -1, y: -1};
}// if
}// for
}// if - LEAST_REQUIRED_ELLIPSES
// clear accumulated array
arr_accum = [];
}// if1 - LEAST_REQUIRED_DISTANCE
}// for x2y2
}// for xy
edges_canvas.getContext('2d').putImageData(hough_canvas_img_data, 0, 0);
writeCanvasToFile(edges_canvas, __dirname + '/hough.jpg', function() {
});
function getMaxMinor(accum_in){
var max_votes = -1,
max_votes_idx,
i,
accum_len = accum_in.length;
for(i in accum_in){
if (accum_in[i].accum > max_votes){
max_votes = accum_in[i].accum;
max_votes_idx = i;
} // if
}
if (max_votes > 0){
return {max_votes: max_votes, index: max_votes_idx};
}
return null;
}
function distance(point_a,point_b){
return Math.sqrt((point_a.x - point_b.x) * (point_a.x - point_b.x) + (point_a.y - point_b.y) * (point_a.y - point_b.y));
}
function getEdgesArr(canvas_in){
var x,
y,
width = canvas_in.width,
height = canvas_in.height,
pixel,
edges = [],
ctx = canvas_in.getContext('2d'),
img_data = ctx.getImageData(0, 0, width, height);
for(x = 0; x < width; x++){
for(y = 0; y < height; y++){
pixel = getPixel(img_data, x,y);
if (pixel.r !== 0 &&
pixel.g !== 0 &&
pixel.b !== 0 ){
edges.push({x: x, y: y});
}
} // for
}// for
return edges
} // getEdgesArr
function drawImgToCanvasSync(file) {
var data = fs.readFileSync(file)
var canvas = dataToCanvas(data);
return canvas;
}
function dataToCanvas(imagedata) {
img = new Canvas.Image();
img.src = new Buffer(imagedata, 'binary');
var canvas = new Canvas(img.width, img.height);
var ctx = canvas.getContext('2d');
ctx.patternQuality = "best";
ctx.drawImage(img, 0, 0, img.width, img.height,
0, 0, img.width, img.height);
return canvas;
}
function writeCanvasToFile(canvas, file, callback) {
var out = fs.createWriteStream(file)
var stream = canvas.createPNGStream();
stream.on('data', function(chunk) {
out.write(chunk);
});
stream.on('end', function() {
callback();
});
}
function setPixel(imageData, x, y, r, g, b, a) {
index = (x + y * imageData.width) * 4;
imageData.data[index+0] = r;
imageData.data[index+1] = g;
imageData.data[index+2] = b;
imageData.data[index+3] = a;
}
function getPixel(imageData, x, y) {
index = (x + y * imageData.width) * 4;
return {
r: imageData.data[index+0],
g: imageData.data[index+1],
b: imageData.data[index+2],
a: imageData.data[index+3]
}
}
It seems you try to implement the algorithm of Yonghong Xie; Qiang Ji (2002). A new efficient ellipse detection method 2. p. 957.
Ellipse removal suffers from several bugs
In your code, you perform the removal of found ellipse (step 12 of the original paper's algorithm) by resetting coordinates to {-1, -1}.
You need to add:
`if (arr_edges[x1y1].x === -1) break;`
at the end of the x2y2 block. Otherwise, the loop will consider -1, -1 as a white point.
More importantly, your algorithm consists in erasing every point which distance to the center is smaller than b. b supposedly is the minor axis half-length (per the original algorithm). But in your code, variable b actually is the latest (and not most frequent) half-length, and you erase points with a distance lower than b (instead of greater, since it's the minor axis). In other words, you clear all points inside a circle with a distance lower than latest computed axis.
Your sample image can actually be processed with a clearing of all points inside a circle with a distance lower than selected major axis with:
max_minor = arr_accum[max_votes.index].d;
Indeed, you don't have overlapping ellipses and they are spread enough. Please consider a better algorithm for overlapping or closer ellipses.
The algorithm mixes major and minor axes
Step 6 of the paper reads:
For each third pixel (x, y), if the distance between (x, y) and (x0,
y0) is greater than the required least distance for a pair of pixels
to be considered then carry out the following steps from (7) to (9).
This clearly is an approximation. If you do so, you will end up considering points further than the minor axis half length, and eventually on the major axis (with axes swapped). You should make sure the distance between the considered point and the tested ellipse center is smaller than currently considered major axis half-length (condition should be d <= a). This will help with the ellipse erasing part of the algorithm.
Also, if you also compare with the least distance for a pair of pixels, as per the original paper, 40 is too large for the smaller ellipse in your picture. The comment in your code is wrong, it should be at maximum half the smallest ellipse minor axis half-length.
LEAST_REQUIRED_ELLIPSES is too small
This parameter is also misnamed. It is the minimum number of votes an ellipse should get to be considered valid. Each vote corresponds to a pixel. So a value of 6 means that only 6+2 pixels make an ellipse. Since pixels coordinates are integers and you have more than 1 ellipse in your picture, the algorithm might detect ellipses that are not, and eventually clear edges (especially when combined with the buggy ellipse erasing algorithm). Based on tests, a value of 100 will find four of the five ellipses of your picture, while 80 will find them all. Smaller values will not find the proper centers of the ellipses.
Sample image is not black & white
Despite the comment, sample image is not exactly black and white. You should convert it or apply some threshold (e.g. RGB values greater than 10 instead of simply different form 0).
Diff of minimum changes to make it work is available here:
https://gist.github.com/pguyot/26149fec29ffa47f0cfb/revisions
Finally, please note that parseInt(x.toFixed(0)) could be rewritten Math.floor(x), and you probably want to not truncate all floats like this, but rather round them, and proceed where needed: the algorithm to erase the ellipse from the picture would benefit from non truncated values for the center coordinates. This code definitely could be improved further, for example it currently computes the distance between points x1y1 and x2y2 twice.
I asked this question earlier but didn't communicated it clearly, so forgive me for the duplicate. This should be better.
I need to figure the position of a coordinate, given three other coordinates and 2 slopes. Basically, the intersection point of two lines. However, I don't have all of the information normally available to solve this.
I have an arbitrary shape defined by a bunch of vertices. The user may drag a line between these vertices and the shape should react as the diagrams below show.
So in the first example, the user drags line EF from the position on the left to the position on the right (line E2F2). What needs to happen is that line EF grows/shrinks such that it's slope stays the same and that it's beginning and ending coordinates remain on the lines DE and AF respectively. This is shown as line E2F2.
This needs to be generic enough that it can handle any sort of strange or regular angles I throw at it. The second set of shapes shows a simpler approach. The user drags line CD to the position of C2D2. Notice how the slopes stay the same and D2 essentially slides down that diagonal line and B2C2 and C2D2 both extend in length. The result is that all 3 slopes stay the same but lines B2C2 and C2D2 grow in length to stay connected, while line D2E2 shrinks.
You'll need to understand that when dragging line EF, you're actually moving the coordinate "E". So, figuring the first coordinate is easy. And the previous and next one's never change. So I essentially have the slopes of the 3 relevant lines and 3 of the 4 necessary coordinates. I need the 4th, so in my example, F2 or D2.
This code is called on an event every time the coordinate moves. Lets say we're dragging line EF - the coordinate is E then.
var next = this.model.get("next"), // coordinate F
nextNext = next.get("next"), // coordinate A
nextDx = nextNext.get("x") - next.get("x"), // delta X of AF
nextDy = nextNext.get("y") - next.get("y"), // delta Y of AF
prev = this.model.get("prev"), // coordinate D
prevDx = prev.get("x") - this.model.get("x"), // delta X of DF
prevDy = prev.get("y") - this.model.get("y"), // delta Y of DF
selfDx = next.get("x") - this.model.get("x"), // delta X of EF
selfDy = next.get("y") - this.model.get("y"), // delta Y of EF
selfX = this.initialCoords.x + this.shape.getX(), // the new position of E
selfY = this.initialCoords.y + this.shape.getY(),
selfM, selfB, prevM, prevB, nextM, nextB, m, x, y, b;
// check for would-be infinities
if (selfDx == 0) {
// **** THIS WHOLE BLOCK IS CORRECT ****
// i'm vertical
// we can safely assume prev/next aren't also vertical. i think? right?
prevM = prev.get("slope");
prevB = prev.get("y") - prevM * prev.get("x");
var myX = selfX,
myY = prevM * myX + prevB;
this.model.set({
x: myX,
y: myY
});
nextM = next.get("slope");
nextB = next.get("y") - nextM * next.get("x");
var nextX = selfX,
nextY = nextM * nextX + nextB;
next.set({
x: nextX,
y: nextY
});
} else if (selfDy == 0) {
//***** THIS WHOLE BLOCK IS CORRECT **** //
// i'm horizontal
if (prevDx == 0) {
// prev is a vertical line
this.model.set({
y: selfY
});
} else {
prevM = prev.get("slope");
prevB = prev.get("y") - prevM * prev.get("x");
var myY = selfY,
myX = (selfY - prevB) / prevM;
this.model.set({
x: myX,
y: myY
});
}
if (nextDx == 0) {
// next is a vertical line
next.set({
y: selfY
});
} else {
nextM = next.get("slope");
nextB = next.get("y") - nextM * next.get("x");
var nextY = selfY,
nextX = (selfY - nextB) / nextM;
next.set({
x: nextX,
y: nextY
});
}
} else {
// HELP HERE - you've chosen to drag an arbitrarily angled line. Figure out the "next" coordinate given the "current" one.
selfM = this.model.get("slope");
selfB = this.model.get("y") - this.model.get("slope") * this.model.get("x");
if (selfM < 0) {
prevM = prev.get("slope");
prevB = prev.get("y") - prevM * prev.get("x");
var myY = selfY,
myX = (selfY - prevB) / prevM;
// CORRECT, but need "next" position based on this
this.model.set({
x: myX,
y: myY
});
} else {
// CORRECT but need "next" position based on this.
var myX = selfX;
this.model.set({
x: myX
});
}
}
I had a similar situation and had some success using this page as reference :
http://en.wikipedia.org/wiki/Line-line_intersection
You should be able to enumerate all your lines testing for any points where they cross your moving line. These will be the new coordinates.
The equations in the wiki article assume lines of infinite length which you should be aware of but should actually be what you want (I think - there are probably edge cases).