If I have 4 points
var x1;
var y1;
var x2;
var y2;
var x3;
var y3;
var x4;
var y4;
that make up a box. So
(x1,y1) is top left
(x2,y2) is top right
(x3,y3) is bottom left
(x4,y4) is bottom right
And then each point has a weight ranging from 0-522. How can I calculate a coordinate (tx,ty) that lies inside the box, where the point is closer to the the place that has the least weight (but taking all weights into account). So for example. if (x3,y3) has weight 0, and the others have weight 522, the (tx,ty) should be (x3,y3). If then (x2,y2) had weight like 400, then (tx,ty) should be move a little closer towards (x2,y2) from (x3,y3).
Does anyone know if there is a formula for this?
Thanks
Creating a minimum, complete, verifiable exmample
You have a little bit of a tricky problem here, but it's really quite fun. There might be better ways to solve it, but I found it most reliable to use Point and Vector data abstractions to model the problem better
I'll start with a really simple data set – the data below can be read (eg) Point D is at cartesian coordinates (1,1) with a weight of 100.
|
|
| B(0,1) #10 D(1,1) #100
|
|
| ? solve weighted average
|
|
| A(0,0) #20 C(1,0) #40
+----------------------------------
Here's how we'll do it
find the unweighted midpoint, m
convert each Point to a Vector of Vector(degrees, magnitude) using m as the origin
add all the Vectors together, vectorSum
divide vectorSum's magnitude by the total magnitude
convert the vector to a point, p
offset p by unweighted midpoint m
Possible JavaScript implementation
I'll go thru the pieces one at a time then there will be a complete runnable example at the bottom.
The Math.atan2, Math.cos, and Math.sin functions we'll be using return answers in radians. That's kind of a bother, so there's a couple helpers in place to work in degrees.
// math
const pythag = (a,b) => Math.sqrt(a * a + b * b)
const rad2deg = rad => rad * 180 / Math.PI
const deg2rad = deg => deg * Math.PI / 180
const atan2 = (y,x) => rad2deg(Math.atan2(y,x))
const cos = x => Math.cos(deg2rad(x))
const sin = x => Math.sin(deg2rad(x))
Now we'll need a way to represent our Point and Point-related functions
// Point
const Point = (x,y) => ({
x,
y,
add: ({x: x2, y: y2}) =>
Point(x + x2, y + y2),
sub: ({x: x2, y: y2}) =>
Point(x - x2, y - y2),
bind: f =>
f(x,y),
inspect: () =>
`Point(${x}, ${y})`
})
Point.origin = Point(0,0)
Point.fromVector = ({a,m}) => Point(m * cos(a), m * sin(a))
And of course the same goes for Vector – strangely enough adding Vectors together is actually easier when you convert them back to their x and y cartesian coordinates. other than that, this code is pretty straightforward
// Vector
const Vector = (a,m) => ({
a,
m,
scale: x =>
Vector(a, m*x),
add: v =>
Vector.fromPoint(Point.fromVector(Vector(a,m)).add(Point.fromVector(v))),
inspect: () =>
`Vector(${a}, ${m})`
})
Vector.zero = Vector(0,0)
Vector.fromPoint = ({x,y}) => Vector(atan2(y,x), pythag(x,y))
Lastly we'll need to represent our data above in JavaScript and create a function which calculates the weighted point. With Point and Vector by our side, this will be a piece of cake
// data
const data = [
[Point(0,0), 20],
[Point(0,1), 10],
[Point(1,1), 100],
[Point(1,0), 40],
]
// calc weighted point
const calcWeightedMidpoint = points => {
let midpoint = calcMidpoint(points)
let totalWeight = points.reduce((acc, [_, weight]) => acc + weight, 0)
let vectorSum = points.reduce((acc, [point, weight]) =>
acc.add(Vector.fromPoint(point.sub(midpoint)).scale(weight/totalWeight)), Vector.zero)
return Point.fromVector(vectorSum).add(midpoint)
}
console.log(calcWeightedMidpoint(data))
// Point(0.9575396819442366, 0.7079725827019256)
Runnable script
// math
const pythag = (a,b) => Math.sqrt(a * a + b * b)
const rad2deg = rad => rad * 180 / Math.PI
const deg2rad = deg => deg * Math.PI / 180
const atan2 = (y,x) => rad2deg(Math.atan2(y,x))
const cos = x => Math.cos(deg2rad(x))
const sin = x => Math.sin(deg2rad(x))
// Point
const Point = (x,y) => ({
x,
y,
add: ({x: x2, y: y2}) =>
Point(x + x2, y + y2),
sub: ({x: x2, y: y2}) =>
Point(x - x2, y - y2),
bind: f =>
f(x,y),
inspect: () =>
`Point(${x}, ${y})`
})
Point.origin = Point(0,0)
Point.fromVector = ({a,m}) => Point(m * cos(a), m * sin(a))
// Vector
const Vector = (a,m) => ({
a,
m,
scale: x =>
Vector(a, m*x),
add: v =>
Vector.fromPoint(Point.fromVector(Vector(a,m)).add(Point.fromVector(v))),
inspect: () =>
`Vector(${a}, ${m})`
})
Vector.zero = Vector(0,0)
Vector.unitFromPoint = ({x,y}) => Vector(atan2(y,x), 1)
Vector.fromPoint = ({x,y}) => Vector(atan2(y,x), pythag(x,y))
// data
const data = [
[Point(0,0), 20],
[Point(0,1), 10],
[Point(1,1), 100],
[Point(1,0), 40],
]
// calc unweighted midpoint
const calcMidpoint = points => {
let count = points.length;
let midpoint = points.reduce((acc, [point, _]) => acc.add(point), Point.origin)
return midpoint.bind((x,y) => Point(x/count, y/count))
}
// calc weighted point
const calcWeightedMidpoint = points => {
let midpoint = calcMidpoint(points)
let totalWeight = points.reduce((acc, [_, weight]) => acc + weight, 0)
let vectorSum = points.reduce((acc, [point, weight]) =>
acc.add(Vector.fromPoint(point.sub(midpoint)).scale(weight/totalWeight)), Vector.zero)
return Point.fromVector(vectorSum).add(midpoint)
}
console.log(calcWeightedMidpoint(data))
// Point(0.9575396819442366, 0.7079725827019256)
Going back to our original visualization, everything looks right!
|
|
| B(0,1) #10 D(1,1) #100
|
|
| * <-- about right here
|
|
|
| A(0,0) #20 C(1,0) #40
+----------------------------------
Checking our work
Using a set of points with equal weighting, we know what the weighted midpoint should be. Let's verify that our two primary functions calcMidpoint and calcWeightedMidpoint are working correctly
const data = [
[Point(0,0), 5],
[Point(0,1), 5],
[Point(1,1), 5],
[Point(1,0), 5],
]
calcMidpoint(data)
// => Point(0.5, 0.5)
calcWeightedMidpoint(data)
// => Point(0.5, 0.5)
Great! Now we'll test to see how some other weights work too. First let's just try all the points but one with a zero weight
const data = [
[Point(0,0), 0],
[Point(0,1), 0],
[Point(1,1), 0],
[Point(1,0), 1],
]
calcWeightedMidpoint(data)
// => Point(1, 0)
Notice if we change that weight to some ridiculous number, it won't matter. Scaling of the vector is based on the point's percentage of weight. If it gets 100% of the weight, it (the point) will not pull the weighted midpoint past (the point) itself
const data = [
[Point(0,0), 0],
[Point(0,1), 0],
[Point(1,1), 0],
[Point(1,0), 1000],
]
calcWeightedMidpoint(data)
// => Point(1, 0)
Lastly, we'll verify one more set to ensure weighting is working correctly – this time we'll have two pairs of points that are equally weighted. The output is exactly what we're expecting
const data = [
[Point(0,0), 0],
[Point(0,1), 0],
[Point(1,1), 500],
[Point(1,0), 500],
]
calcWeightedMidpoint(data)
// => Point(1, 0.5)
Millions of points
Here we will create a huge point cloud of random coordinates with random weights. If points are random and things are working correctly with our function, the answer should be pretty close to Point(0,0)
const RandomWeightedPoint = () => [
Point(Math.random() * 1000 - 500, Math.random() * 1000 - 500),
Math.random() * 1000
]
let data = []
for (let i = 0; i < 1e6; i++)
data[i] = RandomWeightedPoint()
calcWeightedMidpoint(data)
// => Point(0.008690554978970092, -0.08307212085822799)
A++
Assume w1, w2, w3, w4 are the weights.
You can start with this (pseudocode):
M = 522
a = 1
b = 1 / ( (1 - w1/M)^a + (1 - w2/M)^a + (1 - w3/M)^a + (1 - w4/M)^a )
tx = b * (x1*(1-w1/M)^a + x2*(1-w2/M)^a + x3*(1-w3/M)^a + x4*(1-w4/M)^a)
ty = b * (y1*(1-w1/M)^a + y2*(1-w2/M)^a + y3*(1-w3/M)^a + y4*(1-w4/M)^a)
This should approximate the behavior you want to accomplish. For the simplest case set a=1 and your formula will be simpler. You can adjust behavior by changing a.
Make sure you use Math.pow instead of ^ if you use Javascript.
A very simple approach is this:
Convert each point's weight to 522 minus the actual weight.
Multiply each x/y co-ordinate by its adjusted weight.
Sum all multiplied x/y co-ordinates together, and --
Divide by the total adjusted weight of all points to get your adjusted average position.
That should produce a point with a position that is biased proportionally towards the "lightest" points, as described. Assuming that weights are prefixed w, a quick snippet (followed by JSFiddle example) is:
var tx = ((522-w1)*x1 + (522-w2)*x2 + (522-w3)*x3 + (522-w4)*x4) / (2088-(w1+w2+w3+w4));
var ty = ((522-w1)*y1 + (522-w2)*y2 + (522-w3)*y3 + (522-w4)*y4) / (2088-(w1+w2+w3+w4));
JSFiddle example of this
Even though this has already been answered, I feel the one, short code snippet that shows the simplicity of calculating a weighted-average is missing:
function weightedAverage(v1, w1, v2, w2) {
if (w1 === 0) return v2;
if (w2 === 0) return v1;
return ((v1 * w1) + (v2 * w2)) / (w1 + w2);
}
Now, to make this specific to your problem, you have to apply this to your points via a reducer. The reducer makes it a moving average: the value it returns represents the weights of the points it merged.
// point: { x: xCoordinate, y: yCoordinate, w: weight }
function avgPoint(p1, p2) {
return {
x: weightedAverage(p1.x, p1.w, p2.x, p2.w),
x: weightedAverage(p1.x, p1.w, p2.x, p2.w),
w: p1.w + pw.2,
}
}
Now, you can reduce any list of points to get an average coordinate and the weight it represents:
[ /* points */ ].reduce(avgPoint, { x: 0, y: 0, w: 0 })
I hope user naomik doesn't mind, but I used some of their test cases in this runnable example:
function weightedAverage(v1, w1, v2, w2) {
if (w1 === 0) return v2;
if (w2 === 0) return v1;
return ((v1 * w1) + (v2 * w2)) / (w1 + w2);
}
function avgPoint(p1, p2) {
return {
x: weightedAverage(p1.x, p1.w, p2.x, p2.w),
y: weightedAverage(p1.y, p1.w, p2.y, p2.w),
w: p1.w + p2.w,
}
}
function getAvgPoint(arr) {
return arr.reduce(avgPoint, {
x: 0,
y: 0,
w: 0
});
}
const testCases = [
{
data: [
{ x: 0, y: 0, w: 1 },
{ x: 0, y: 1, w: 1 },
{ x: 1, y: 1, w: 1 },
{ x: 1, y: 0, w: 1 },
],
result: { x: 0.5, y: 0.5 }
},
{
data: [
{ x: 0, y: 0, w: 0 },
{ x: 0, y: 1, w: 0 },
{ x: 1, y: 1, w: 500 },
{ x: 1, y: 0, w: 500 },
],
result: { x: 1, y: 0.5 }
}
];
testCases.forEach(c => {
var expected = c.result;
var outcome = getAvgPoint(c.data);
console.log("Expected:", expected.x, ",", expected.y);
console.log("Returned:", outcome.x, ",", outcome.y);
console.log("----");
});
const rndTest = (function() {
const randomWeightedPoint = function() {
return {
x: Math.random() * 1000 - 500,
y: Math.random() * 1000 - 500,
w: Math.random() * 1000
};
};
let data = []
for (let i = 0; i < 1e6; i++)
data[i] = randomWeightedPoint()
return getAvgPoint(data);
}());
console.log("Expected: ~0 , ~0, 500000000")
console.log("Returned:", rndTest.x, ",", rndTest.y, ",", rndTest.w);
.as-console-wrapper {
min-height: 100%;
}
Related
I want to create the array below with a for loop as its large
var centres = {
1979: { x: width * 1 / 41, y: height / 2 },
1980: { x: width * 2 / 41, y: height / 2 },
1981: { x: width * 3 / 41, y: height / 2 },
...
}
and then access it as follows:
function nodeYearPos(d) {
return yearCenters[d.year].x;
}
I have the following code, but its only setting the year...
var yearCenters = Array.from(new Array(2020-1919+1), (x, i) => i + 1919);
for (year = 1919; year <= 2020; year++) {
coords = getCentres(year); // this returns an object in the form {x : x, y : y}
yearCenters[year] = coords;
}
you can do as gorak commented but with the getCenters function
var yearCenters = Object.fromEntries(Array.from(new Array(2020-1919+1), (x, i) => [i + 1919, getCenters(i + 1919)]));
or you can also try
var yearCenters = {};
for (year = 1919; year <= 2020; year++) {
coords = getCenters(year);
yearCenters[year] = coords;
}
When you try to fetch by year in yearCenters array (e.g. yearCenters[year]) this won't work since the year is not the index in the array.
I would suggest you first convert the array into a JS object so that indexing on it works with years.
See below snippet -
// Create obejct from array
var yearCenters = Object.fromEntries(Array.from(new Array(2020-1919+1), (x, i) => [i + 1919, null]))
// This loop remains same
for (year = 1919; year <= 2020; year++) {
coords = getCentres(year); // this returns an object in the form {x : x, y : y}
yearCenters[year] = coords;
}
// Mock function
function getCentres(year) {
return {
x: Math.random() * 100,
y: Math.random() * 100
}
}
console.log(yearCenters)
I have a slider that has the following raw snap points:
[-100, -200, -300, -400, -500, -600]
And I would like to convert the sliding value to match the following snap points:
[0, 5, 10, 25, 50, 100]
A raw value in [-100, -200) should be mapped to a value in [0, 5)
A raw value in [-200, -300) should be mapped to a value in [5, 10)
A raw value in [-300, -400) should be mapped to a value in [10, 25)
And so on ..
How can I achieve that?
Edit: added my attempt (different raw values though)
// sliderValue is an integer obtained from the slider
const base = -70
const offset = -80
const limits = [
base + offset * 0, // -70
base + offset * 1, // -150
base + offset * 2, // -230
base + offset * 3, // -310
base + offset * 4, // -390
base + offset * 5, // -470
]
const points = [0, 5, 10, 25, 50, 100]
// I can't even begin to make sense of this
// don't know I came up with it, but it works ¯\_(ツ)_/¯
if (sliderValue <= limits[4]) {
percentage = scaleValue(sliderValue, limits[4], limits[5], 50, 100)
} else if (sliderValue <= limits[3]) {
percentage = scaleValue(sliderValue, limits[3], limits[4], 25, 50)
} else if (sliderValue <= limits[2]) {
percentage = scaleValue(sliderValue, limits[2], limits[3], 10, 25)
} else if (sliderValue <= limits[1]) {
percentage = scaleValue(sliderValue, limits[1], limits[2], 5, 10)
} else if (sliderValue <= limits[0]) {
percentage = scaleValue(sliderValue, limits[0], limits[1], 0, 5)
}
console.log(percentage)
// ..
function scaleValue(num, in_min, in_max, out_min, out_max) {
return ((num - in_min) * (out_max - out_min)) / (in_max - in_min) + out_min
}
You could take a function with a look up for the section. Then build the new value, based on the four values as a linear function.
function getValue(x) {
var a = [-100, -200, -300, -400, -500, -600],
b = [0, 5, 10, 25, 50, 100],
i = a.findIndex((v, i, a) => v >= x && x >= a[i + 1]);
return [x, (x -a[i])* (b[i + 1] - b[i]) / (a[i + 1] - a[i]) +b[i]].join(' ');
}
console.log([-100, -150, -200, -250, -300, -350, -400, -450, -500, -550, -600].map(getValue));
.as-console-wrapper { max-height: 100% !important; top: 0; }
Simple linear equation: add 100, divide by 20, then negate.
UPDATE: Due to early-morning eye bleariness, I misread the question. (Sorry!) The general method for mapping linear relations to each other is to figure out the offset of the two sets and the scale factor.
I can't find a smooth relationship between the example points you gave, so I'm not sure how to find a single equation that would neatly and continuously map the points to each other. It looks like your solution (you said it works) might be the best: figure out which range each value maps to, and scale correspondingly.
You can just map the values:
var mapping = {
"-100": 0,
"-200": 5,
"-300": 10,
"-400": 25,
"-500": 50,
"-600": 100
}
function map_values(array){
return [mapping[array[0]], mapping[array[1]]];
}
var input = [-200,-300];
console.log(map_values(input));
I have a large cube composed of smaller cubes. The large cube consists of 10 cubes wide, by 10 cubes in length, by 10 cubes in height. For a total of 1000 cubes.
One cube will be randomly chosen to be blue
Three cubes will be randomly chosen to be green
I want to be able to determine which is the closest green cube to the blue cube.
One other thing that is important is that each side of the cube is connected to the opposite side (i.e. row 10 is considered next to row 1). This is the wraparound effect.
So, for example, if the blue cube is at coordinates 9:8:8 and the green cubes are each at 1:2:2, 5:5:3, and 6:3:4. Then the green cube at 1:2:2 should be considered the closest cube. If my calculations are correct, it should have a distance of 10 whereas the other two would each have a distance of 12.
Without the cube wraparound (side 1 connected with side 10) I have been able to come up with the following in JavaScript:
let lowest = 1000;
let lowest_index = -1;
for (i = 0; i < green_cube.length; i++){
let x_offset = Math.abs(blue_cube.x - green_cube[i].x);
let y_offset = Math.abs(blue_cube.y - green_cube[i].y);
let z_offset = Math.abs(blue_cube.z - green_cube[i].z);
let distance = x_offset + y_offset + z_offset;
if (distance < lowest){
lowest = distance;
lowest_index = i;
}
}
What is the proper way to code this when taking wraparound into effect?
Update
To clarify, the distance needs to be distance by number of cubes traveled to get from point A to point B. Distance must be traveled only along the X, Y, and Z axis, therefore, diagonal distance will not work. I believe this is referred to as taxicab distance in 3D space.
I believe it's often termed wraparound.
To take wraparound into account your distance measure, e.g. for the x dimension, should be:
let x_offset = Math.min((10 + blue.x - green[i].x) % 10, (10 + green[i].x - blue.x) % 10)
x_offset will always be positive.
Here is a stupid trick to keep your thinking straight.
Let v be the vector (5, 5, 5) - blue_cube. Add v to every cube's position, adding/subtracting 10 if it goes off an edge. Now the blue cube is at (5, 5, 5) and the shortest path to the other cubes no longer goes off the edge.
In your example, v = (5, 5, 5) - (9, 8, 8) = (-4, -3, -3). The first green cube moves to (1, 2, 2) + (-4, -3, -3) = (-3, -1, -1) = (7, 9, 9) and its distance is 10. The second green cube moves to (5, 5, 3) + (-4, -3, -3) = (1, 2, 0) and its distance is 12. The third green cube moves to (6, 3, 4) + (-4, -3, -3) = (2, 0, 1) and its distance is again 12. So the first is indeed closest.
In this code I am using distance calculation formula for 2 points in 3d (reference).
const calculateDistance3d = ({x: x1, y: y1, z: z1}, {x: x2, y: y2, z: z2}) => {
return Math.sqrt(Math.pow(x2 - x1, 2) + Math.pow(y2 - y1, 2) + Math.pow(z2 - z1, 2));
}
const calculateLoopedDistance = (cubeA, cubeB) => {
return calculateDistance3d(cubeA, {
x: cubeA.x + 10 - Math.abs(cubeB.x - cubeA.x),
y: cubeA.y + 10 - Math.abs(cubeB.y - cubeA.y),
z: cubeA.z + 10 - Math.abs(cubeB.z - cubeA.z)
});
};
const getClosest = (green_cube, blue_cube) => {
let minDistance = 1000;
let closestIndex = 0;
blue_cube.forEach((cube, index) => {
const distance = calculateDistance3d(green_cube, cube);
const loopedDistance = calculateLoopedDistance(green_cube, cube);
if (distance < minDistance || loopedDistance < minDistance) {
minDistance = Math.min(distance, loopedDistance);
closestIndex = index;
}
});
return closestIndex;
}
console.log(getClosest({x: 9, y: 8, z: 8}, [
{x: 1, y: 2, z: 2},
{x: 5, y: 5, z: 3},
{x: 6, y: 3, z: 4}
]));
console.log(getClosest({x: 9, y: 8, z: 8}, [
{x: 5, y: 5, z: 3},
{x: 1, y: 2, z: 2},
{x: 6, y: 3, z: 4}
]));
At the end of this script there are 2 logs with cube's data. You can test different data there.
I updated / fixed calculateLoopedDistance() function, which was incorrect.
Virtually replicate the green cubes as if they appeared at x, x-10 and x+10 and keep the minimum delta. This is done on the three axis independently.
I've come across another solution that also works:
let cube_width = 10;
let mid_point = cube_width / 2;
let x_offset = Math.abs(point1 - point2);
if (x_offset > mid_point){
x_offset = cube_width - x_offset;
}
I'm having a hard time figuring out whether this one or SirRaffleBuffle's solution is more efficient for time.
I have two vectors, the Y-aligned is fixed whereby the X-aligned is allowed to rotate. These vectors are connected together through two fixed-length segments. Given the angle between the two vectors (82.74) and the length of all segments, how can I get the angle of the two jointed segments (24.62 and 22.61)?
What is given: the magnitude of the vectors, and the angle between the X-axis and OG:
var magOG = 3,
magOE = 4,
magGH = 3,
magEH = 2,
angleGamma = 90;
This is my starting point: angleGamma = 90 - then, I will have following vectors:
var vOG = new vec2(-3,0),
vOE = new vec2(0,-4);
From here on, I am trying to get angleAlphaand angleBeta for values of angleGamma less than 90 degrees.
MAGNITUDE OF THE CONSTRAINED SEGMENTS:
Segments HG and HE must meet following conditions:
/
| OG*OG+ OE*OE = (HG + HE)*(HG + HE)
>
| OG - HG = OE - HE
\
which will lead to following two solutions (as pointed out in the accepted answer - bilateration):
Solution 1:
========================================================
HG = 0.5*(-Math.sqrt(OG*OG + OE*OE) + OG - OE)
HE = 0.5*(-Math.sqrt(OG*OG + OE*OE) - OG + OE)
Solution 2:
========================================================
HG = 0.5*(Math.sqrt(OG*OG + OE*OE) + OG - OE)
HE = 0.5*(Math.sqrt(OG*OG + OE*OE) - OG + OE)
SCRATCHPAD:
Here is a playground with the complete solution. The visualization library used here is the great JSXGraph. Thanks to the Center for Mobile Learning with Digital Technology of the Bayreuth University.
Credits for the circle intersection function: 01AutoMonkey in the accepted answer to this question: A JavaScript function that returns the x,y points of intersection between two circles?
function deg2rad(deg) {
return deg * Math.PI / 180;
}
function rad2deg(rad) {
return rad * 180 / Math.PI;
}
function lessThanEpsilon(x) {
return (Math.abs(x) < 0.00000000001);
}
function angleBetween(point1, point2) {
var x1 = point1.X(), y1 = point1.Y(), x2 = point2.X(), y2 = point2.Y();
var dy = y2 - y1, dx = x2 - x1;
var t = -Math.atan2(dx, dy); /* range (PI, -PI] */
return rad2deg(t); /* range (180, -180] */
}
function circleIntersection(circle1, circle2) {
var r1 = circle1.radius, cx1 = circle1.center.X(), cy1 = circle1.center.Y();
var r2 = circle2.radius, cx2 = circle2.center.X(), cy2 = circle2.center.Y();
var a, dx, dy, d, h, h2, rx, ry, x2, y2;
/* dx and dy are the vertical and horizontal distances between the circle centers. */
dx = cx2 - cx1;
dy = cy2 - cy1;
/* angle between circle centers */
var theta = Math.atan2(dy,dx);
/* vertical and horizontal components of the line connecting the circle centers */
var xs1 = r1*Math.cos(theta), ys1 = r1*Math.sin(theta), xs2 = r2*Math.cos(theta), ys2 = r2*Math.sin(theta);
/* intersection points of the line connecting the circle centers */
var sxA = cx1 + xs1, syA = cy1 + ys1, sxL = cx2 - xs2, syL = cy2 - ys2;
/* Determine the straight-line distance between the centers. */
d = Math.sqrt((dy*dy) + (dx*dx));
/* Check for solvability. */
if (d > (r1 + r2)) {
/* no solution. circles do not intersect. */
return [[sxA,syA], [sxL,syL]];
}
thetaA = -Math.PI - Math.atan2(cx1,cy1); /* Swap X-Y and re-orient to -Y */
xA = +r1*Math.sin(thetaA);
yA = -r1*Math.cos(thetaA);
ixA = cx1 - xA;
iyA = cy1 - yA;
thetaL = Math.atan(cx2/cy2);
xL = -r2*Math.sin(thetaL);
yL = -r2*Math.cos(thetaL);
ixL = cx2 - xL;
iyL = cy2 - yL;
if(d === 0 && r1 === r2) {
/* infinite solutions. circles are overlapping */
return [[ixA,iyA], [ixL,iyL]];
}
if (d < Math.abs(r1 - r2)) {
/* no solution. one circle is contained in the other */
return [[ixA,iyA], [ixL,iyL]];
}
/* 'point 2' is the point where the line through the circle intersection points crosses the line between the circle centers. */
/* Determine the distance from point 0 to point 2. */
a = ((r1*r1) - (r2*r2) + (d*d)) / (2.0 * d);
/* Determine the coordinates of point 2. */
x2 = cx1 + (dx * a/d);
y2 = cy1 + (dy * a/d);
/* Determine the distance from point 2 to either of the intersection points. */
h2 = r1*r1 - a*a;
h = lessThanEpsilon(h2) ? 0 : Math.sqrt(h2);
/* Now determine the offsets of the intersection points from point 2. */
rx = -dy * (h/d);
ry = +dx * (h/d);
/* Determine the absolute intersection points. */
var xi = x2 + rx, yi = y2 + ry;
var xi_prime = x2 - rx, yi_prime = y2 - ry;
return [[xi, yi], [xi_prime, yi_prime]];
}
function plot() {
var cases = [
{a: 1.1, l: 1.9, f: 0.3073},
{a: 1.0, l: 1.7, f: 0.3229}
];
var testCase = 1;
var magA = cases[testCase].a, magL = cases[testCase].l;
var maxS = Math.sqrt(magA*magA+magL*magL), magS1 = maxS * cases[testCase].f, magS2 = maxS - magS1;
var origin = [0,0], board = JXG.JSXGraph.initBoard('jxgbox', {boundingbox: [-5.0, 5.0, 5.0, -5.0], axis: true});
var drawAs = {dashed: {dash: 3, strokeWidth: 0.5, strokeColor: '#888888'} };
board.suspendUpdate();
var leftArm = board.create('slider', [[-4.5, 3], [-1.5, 3], [0, -64, -180]]);
var leftLeg = board.create('slider', [[-4.5, 2], [-1.5, 2], [0, -12, -30]]);
var rightArm = board.create('slider', [[0.5, 3], [3.5, 3], [0, 64, 180]]);
var rightLeg = board.create('slider', [[0.5, 2], [3.5, 2], [0, 12, 30]]);
var lh = board.create('point', [
function() { return +magA * Math.sin(deg2rad(leftArm.Value())); },
function() { return -magA * Math.cos(deg2rad(leftArm.Value())); }
], {size: 3, name: 'lh'});
var LA = board.create('line', [origin, lh], {straightFirst: false, straightLast: false, lastArrow: true});
var cLS1 = board.create('circle', [function() { return [lh.X(), lh.Y()]; }, function() { return magS1; }], drawAs.dashed);
var lf = board.create('point', [
function() { return +magL * Math.sin(deg2rad(leftLeg.Value())); },
function() { return -magL * Math.cos(deg2rad(leftLeg.Value())); }
], {size: 3, name: 'lf'});
var LL = board.create('line', [origin, lf], {straightFirst: false, straightLast: false, lastArrow: true});
var cLS2 = board.create('circle', [function() { return [lf.X(), lf.Y()]; }, function() { return magS2; }], drawAs.dashed);
var lx1 = board.create('point', [
function() { return circleIntersection(cLS1, cLS2)[0][0]; },
function() { return circleIntersection(cLS1, cLS2)[0][1]; }
], {size: 3, face:'x', name: 'lx1'});
var lx2 = board.create('point', [
function() { return circleIntersection(cLS1, cLS2)[1][0]; },
function() { return circleIntersection(cLS1, cLS2)[1][1]; }
], {size: 3, face:'x', name: 'lx2'});
/* Angle between lh, lx1 shall be between 0 and -180 */
var angleLAJ = board.create('text', [-3.7, 0.5, function(){ return angleBetween(lh, lx1).toFixed(2); }]);
/* Angle between lf, lx1 shall be between 0 and 180 */
var angleLLJ = board.create('text', [-2.7, 0.5, function(){ return angleBetween(lf, lx1).toFixed(2); }]);
var rh = board.create('point', [
function() { return +magA * Math.sin(deg2rad(rightArm.Value())); },
function() { return -magA * Math.cos(deg2rad(rightArm.Value())); }
], {size: 3, name: 'rh'});
var RA = board.create('line', [origin, rh], {straightFirst: false, straightLast: false, lastArrow: true});
var cRS1 = board.create('circle', [function() { return [rh.X(), rh.Y()]; }, function() { return magS1; }], drawAs.dashed);
var rf = board.create('point', [
function() { return +magL * Math.sin(deg2rad(rightLeg.Value())); },
function() { return -magL * Math.cos(deg2rad(rightLeg.Value())); }
], {size: 3, name: 'rf'});
var RL = board.create('line', [origin, rf], {straightFirst: false, straightLast: false, lastArrow: true});
var cRS2 = board.create('circle', [function() { return [rf.X(), rf.Y()]; }, function() { return magS2; }], drawAs.dashed);
var rx1 = board.create('point', [
function() { return circleIntersection(cRS1, cRS2)[1][0]; },
function() { return circleIntersection(cRS1, cRS2)[1][1]; }
], {size: 3, face:'x', name: 'rx1'});
var rx2 = board.create('point', [
function() { return circleIntersection(cRS1, cRS2)[0][0]; },
function() { return circleIntersection(cRS1, cRS2)[0][1]; }
], {size: 3, face:'x', name: 'rx2'});
var angleRAJ = board.create('text', [+1.3, 0.5, function(){ return angleBetween(rh, rx1).toFixed(2); }]);
var angleRLJ = board.create('text', [+2.3, 0.5, function(){ return angleBetween(rf, rx1).toFixed(2); }]);
board.unsuspendUpdate();
}
plot();
<!DOCTYPE html>
<html>
<head>
<link rel="stylesheet" type="text/css" href="//cdnjs.cloudflare.com/ajax/libs/jsxgraph/0.99.7/jsxgraph.css" />
<link rel="stylesheet" href="style.css">
<script type="text/javascript" charset="UTF-8" src="//cdnjs.cloudflare.com/ajax/libs/jsxgraph/0.99.7/jsxgraphcore.js"></script>
</head>
<body>
<div id="jxgbox" class="jxgbox" style="width:580px; height:580px;"></div>
</body>
</html>
According to your sketch, the coordinates of E and G are:
E = (0, -magOE)
G = magOG * ( -sin(gamma), -cos(gamma) )
Then, calculating the position of H is a trilateration problem. Actually, it is just bilateration because you are missing a third distance. Hence, you will get two possible positions for H.
First, let us define a new coordinate system, where E lies at the origin and G lies on the x-axis. The x-axis direction in our original coordinate system is then:
x = (G - E) / ||G - E||
The y-axis is:
y = ( x.y, -x.x )
The coordinates of E and G in this new coordinate system are:
E* = (0, 0)
G* = (0, ||G - E||)
Now, we can easily find the coordinates of H in this coordinate system, up to the ambiguity mentioned earlier. I will abbreviate ||G - E|| = d like in the notation used in the Wikipedia article:
H.x* = (magGH * magGH - magEH * magEH + d * d) / (2 * d)
H.y* = +- sqrt(magGH * magGH - H.x* * H.x*)
Hence, we have two solutions for H.y, one positive and one negative.
Finally, we just need to transform H back into our original coordinate system:
H = x * H.x* + y * H.y* - (0, magOE)
Given the coordinates of H, calculating the angles is pretty straightforward:
alpha = arccos((H.x - G.x) / ||H - G||)
beta = arccos((H.y - E.y) / ||H - E||)
Example
Taking the values from your example
magOG = 3
magOE = 4
magGH = 3
magEH = 2
angleGamma = 82.74°
we first get:
E = (0, -4)
G = 3 * ( -sin(82.74°), -cos(82.74°) )
= (-2.976, -0.379)
Our coordinate system:
x = (-0.635, 0.773)
y = ( 0.773, 0.635)
In this coordinate system:
E* = (0, 0)
G* = (0, 4.687)
Then, the coordinates of H in our auxiliary coordinate system are:
H* = (2.877, +- 0.851)
I will only focus on the positive value for H*.y because this is the point that you marked in your sketch.
Transform back to original coordinate system:
H = (-1.169, -1.237)
And finally calculate the angles:
alpha = 25.41°
beta = 22.94°
The slight differences to your values are probably caused by rounding errors (either in my calculations or in yours).
I need to find a library that allows me to get interpolated values from irregular 2d data. Imagine having something like this:
var data = [{{x: 0, y: 0, value: 0},
{x: 0.5, y: 1, value: 1},
{x: 1, y: 0, value: 2}}, .... Many more elements]
var value = interpolate(data, 0.24, 0.3); // 0.24 = x, 0.3 = y
What the interpolate method does is that it finds the element, in this case a triangle, that the coordinate is inside. Then it interpolates the value between the corners of the element it is contained in.
I do realize that there are lots of aspects in it to optimize performance like building up a tree that allows fast narrowing of elements by having preprocessed bounding boxes. All of this would be great as well, but I am just trying to get started.
There must be some library out there that I can use for it instead of writing my own.
Since search results for barycentric interpolation in javascript were inconclusive, here's some code that might help you get started.
This code takes as input a data set of 2D points, each with a "value", and a "new point" with an unknown value. It first finds the smallest triangle in the data set that contains the "new point", then performs barycentric interpolation using that triangle to find a value for the "new point".
This runs reasonably quickly with a data set of a few hundred points. There are many opportunities for testing, error checking, and optimization - for example, don't look at every possible triangle in the data set. N choose 3 grows with the cube of N, so optimizing to look at triangles made with only points "close to" the "new point" could show significant performance gains.
// Calculate the area of a triangle
function triangle_area(vertexA, vertexB, vertexC) {
return Math.abs(((vertexA.x - vertexC.x) * (vertexB.y - vertexA.y) - (
vertexA.x - vertexB.x) * (vertexC.y - vertexA.y)) * 0.5)
}
// Given a number N, return a list of all possible triples from the list [1..N]
// credit: http://stackoverflow.com/a/5752056/1612562
function list_triples(N) {
var fn = function(n, src, got, all) {
if (n == 0) {
if (got.length > 0) {
all[all.length] = got;
}
return;
}
for (var j = 0; j < src.length; j++) {
fn(n - 1, src.slice(j + 1), got.concat([ src[j] ]), all);
}
return;
}
var triples = [];
// Generates the list [0, ..., N]
// credit: http://stackoverflow.com/a/20066663/1612562
var indices =
Array.apply(null, {length: N}).map(Number.call, Number);
fn(3, indices, [], triples);
return triples;
}
// Given three vertices of a triangle and a point, determine if
// the point falls in the triangle
// credit: https://koozdra.wordpress.com/2012/06/27/javascript-is-point-in-triangle/
// credit: http://www.blackpawn.com/texts/pointinpoly/default.html
function is_in_triangle(newPoint, vertexA, vertexB, vertexC) {
var v0 = [vertexC.x - vertexA.x, vertexC.y - vertexA.y];
var v1 = [vertexB.x - vertexA.x, vertexB.y - vertexA.y];
var v2 = [newPoint.x - vertexA.x, newPoint.y - vertexA.y];
var dot00 = (v0[0] * v0[0]) + (v0[1] * v0[1]);
var dot01 = (v0[0] * v1[0]) + (v0[1] * v1[1]);
var dot02 = (v0[0] * v2[0]) + (v0[1] * v2[1]);
var dot11 = (v1[0] * v1[0]) + (v1[1] * v1[1]);
var dot12 = (v1[0] * v2[0]) + (v1[1] * v2[1]);
var invDenom = 1 / (dot00 * dot11 - dot01 * dot01);
var u = (dot11 * dot02 - dot01 * dot12) * invDenom;
var v = (dot00 * dot12 - dot01 * dot02) * invDenom;
return ((u >= 0) && (v >= 0) && (u + v < 1));
}
// Perform barycentric interpolation on a point in a triangle
function barycentric_interpolate(newPoint, vertexA, vertexB, vertexC) {
var area = triangle_area(vertexA, vertexB, vertexC);
var sub_area_1 = triangle_area(newPoint, vertexB, vertexC);
var sub_area_2 = triangle_area(vertexA, newPoint, vertexC);
var sub_area_3 = triangle_area(vertexA, vertexB, newPoint);
return ((sub_area_1 * vertexA.v) + (sub_area_2 * vertexB.v) + (sub_area_3 *
vertexC.v)) / area;
}
// Find the smallest triangle in the data set containing the new
// point, and perform barycentric interpolation using that triangle
function interpolate(newPoint, data) {
var triangles = list_triples(data.length);
var smallest_triangle_area = Number.MAX_VALUE;
var smallest_triangle;
for (t in triangles) {
var vertexA = data[triangles[t][0]];
var vertexB = data[triangles[t][1]];
var vertexC = data[triangles[t][2]];
var in_triangle = is_in_triangle(newPoint, vertexA, vertexB, vertexC);
if (in_triangle) {
if (triangle_area(vertexA, vertexB, vertexC) < smallest_triangle_area) {
smallest_triangle = [vertexA, vertexB, vertexC];
}
}
}
return smallest_triangle
? barycentric_interpolate(newPoint, smallest_triangle[0], smallest_triangle[1], smallest_triangle[2])
: "Interpolation failed: newPoint isn't in a triangle";
}
var newPoint = {'x': 0.24, 'y': 0.3};
var data = [
{'x': 0, 'y': 0, 'v': 0},
{'x': 0.5, 'y': 1, 'v': 1},
{'x': 1, 'y': 0, 'v': 2},
{'x': 1.5, 'y': 2.5, 'v': 1.5},
{'x': 2, 'y': 1, 'v': 0.5}
];
console.log(interpolate(newPoint, data));
There are also other kinds of spatial interpolation, e.g. kriging, which does have at least one ready-made .js library.