Develop Reference

Calculating the angle between a velocity (particles motion) and a line

So I am creating a simulation of a bouncing ball, and the user can place lines the ball can collide with on the canvas by dragging from one point to another. There are essentially four lines that can be created:
So the object that stores a line is defined as such:
export interface pathSection {
xfrom: number;
yfrom: number;
xto: number;
yto: number;
length: number;
}
The first and third lines in the image for example dont give the same value from
Math.atan2(yto - yfrom, xto - from);
So given the (relative) complexity of the surfaces, I need to find the angle between a moving object and that surface at the point of collision:
The ball strikes the surface at an angle a, which is what I want!
However I am having trouble finding the angle between the two vectors. This is what I understood would work:
var dx = this.path[index_for_path_section].xfrom - this.path[index_for_path_section].xto;
var dy = this.path[index_for_path_section].yfrom - this.path[index_for_path_section].yto;
var posX = this.particle.pos.x;
var posY = this.particle.pos.y;
var posNextX = posX + this.particle.v.x;
var posNextY = posY + this.particle.v.y;
var angleOfRamp = Math.atan2(dy, dx);
var angleOfvelocity = Math.atan2(posNextY - posY, posNextX - posX);
var angleBetween = angleOfRamp - angleOfvelocity;
This is then used to calculate the speed of the object after the collision:
var spd = Math.sqrt(this.particle.v.x * this.particle.v.x + this.particle.v.y * this.particle.v.y);
var restitution = this.elasticity / 100;
this.particle.v.x = restitution * spd * Math.cos(angleBetween);
this.particle.v.y = restitution * spd * Math.sin(angleBetween);
However the angle calculated is around -4.5 Pi, about -90 degrees for the object directly down and the surface at what looks to be around 45-60 degrees…
The red arrow shows the path of the object moving through the surface - the white dots show where a collision has been detected between the surface and the object.
Any help on how to get the correct and usable angle between the two velocity and the line would be appreciated!
Note I have tried utilizing this solution, but have struggled to adapt it to my own work.

So it took me some time, and I am not 100% sure still of why it works because I think im finding the JavaScript angles system a bit tricky, but:
var dx = this.path[collided].xfrom - this.path[collided].xto;
var dy = this.path[collided].yfrom - this.path[collided].yto;
var spd = Math.sqrt(this.particle.v.x * this.particle.v.x + this.particle.v.y * this.particle.v.y);
var angleOfRamp = Math.atan2(dy, dx);
var angleOfvelocity = Math.atan2(this.particle.v.y, this.particle.v.x);
var angleBetween = angleOfRamp * 2 - angleOfvelocity; // not sure why :)
if (angleBetween < 0) { angleBetween += 2*Math.PI; } // not sure why :)
const restitution = this.elasticity / 100;
this.particle.v.x = restitution * spd * Math.cos(angleBetween);
this.particle.v.y = restitution * spd * Math.sin(angleBetween);
Thanks to all who looked :)

Improve discontinuity detection algorithm

I'm trying to create an algorithm that detects discontinuities (like vertical asymptotes) within functions between an interval for the purpose of plotting graphs without these discontinuous connecting lines. Also, I only want to evaluate within the interval so bracketing methods like bisection seems good for that.
EDIT
https://en.wikipedia.org/wiki/Classification_of_discontinuities
I realize now there are a few different kinds of discontinuities. I'm mostly interested in jump discontinuities for graphical purposes.
I'm using a bisection method as I've noticed that discontinuities occur where the slope tends to infinity or becomes vertical, so why not narrow in on those sections where the slope keeps increasing and getting steeper and steeper. The point where the slope is a vertical line, that's where the discontinuity exists.
Approach
Currently, my approach is as follows. If you subdivide the interval using a midpoint into 2 sections and compare which section has the steepest slope, then that section with the steepest slope becomes the new subinterval for the next evaluation.
Termination
This repeats until it converges by either slope becoming undefined (reaching infinity) or the left side or the right side of the interval equaling the middle (I think this is because the floating-point decimal runs out of precision and cannot divide any further)
(1.5707963267948966 + 1.5707963267948968) * .5 = 1.5707963267948966
Example
function - floor(x)
(blue = start leftX and rightX, purple = midpoint, green = 2nd iteration midpoints points, red = slope lines per iteration)
As you can see from the image, each bisection narrows into the discontinuity and the slope keeps getting steeper until it becomes a vertical line at the discontinuity point at x=1.
To my surprise this approach seems to work for step functions like floor(x) and tan(x), but it's not that great for 1/x as it takes too many iterations (I'm thinking of creating a hybrid method where I use either illinois or ridders method on the inverse of 1/x as it those tend to find the root in just one iteration).
Javascript Code
/* Math function to test on */
function fn(x) {
//return (Math.pow(Math.tan(x), 3));
return 1/x;
//return Math.floor(x);
//return x*((x-1-0.001)/(x-1));
}
function slope(x1, y1, x2, y2) {
return (y2 - y1) / (x2 - x1);
}
function findDiscontinuity(leftX, rightX, fn) {
while (true) {
let leftY = fn(leftX);
let rightY = fn(rightX);
let middleX = (leftX + rightX) / 2;
let middleY = fn(middleX);
let leftSlope = Math.abs(slope(leftX, leftY, middleX, middleY));
let rightSlope = Math.abs(slope(middleX, middleY, rightX, rightY));
if (!isFinite(leftSlope) || !isFinite(rightSlope)) return middleX;
if (middleX === leftX || middleX === rightX) return middleX;
if (leftSlope > rightSlope) {
rightX = middleX;
rightY = middleY;
} else {
leftX = middleX;
leftY = middleY;
}
}
}
Problem 1 - Improving detection
For the function x*((x-1-0.001)/(x-1)), the current algorithm has a hard time detecting the discontinuity at x=1 unless I make the interval really small. As an alternative, I could also add most subdivisions but I think the real problem is using slopes as they trick the algorithm into choosing the incorrect subinterval (as demonstrated in the image below), so this approach is not robust enough. Maybe there are some statistical methods that can help determine a more probable interval to select. Maybe something like least squares for measuring the differences and maybe applying weights or biases!
But I don't want the calculations to get too heavy and 5 points of evaluation are the max I would go with per iteration.
EDIT
After looking at problem 1 again, where it selects the wrong (left-hand side) subinterval. I noticed that the only difference between the subintervals was the green midpoint distance from their slope line. So taking inspiration from linear regression, I get the squared distance from the slope line to the midpoints [a, fa] and [b, fb] corresponding to their (left/right) subintervals. And which subinterval has the greatest change/deviation is the one chosen for further subdivision, that is, the greater of the two residuals.
This further improvement resolves problem 1. Although, it now takes around 593 iterations to find the discontinuity for 1/x. So I've created a hybrid function that uses ridders method to find the roots quicker for some functions and then fallback to this new approach. I have given up on slopes as they don't provide enough accurate information.
Problem 2 - Jump Threshold
I'm not sure how to incorporate a jump threshold and what to use for that calculation, don't think slopes would help.
Also, if the line thickness for the graph is 2px and 2 lines of a step function were on top of each other then you wouldn't be able to see the gap of 2px between those lines. So the minimum jump gap would be calculated as such
jumpThreshold = height / (ymax-ymin) = cartesian distance per pixel
minJumpGap = jumpThreshold * 2
But I don't know where to go from here! And once again, maybe there are statistical methods that can help to determine the change in function so that the algorithm can terminate quickly if there's no indication of a discontinuity.
Overall, any help or advice in improving what I got already would be much appreciated!
EDIT
As the above images explains, the more divergent the midpoints are the greater the need for more subdivisions for further inspection for that subinterval. While, if the points mostly follow a straight line trend where the midpoints barely deviate then should exit early. So now it makes sense to use the jumpThreshold in this context.
Maybe there's further analysis that could be done like measuring the curvature of the points in the interval to see whether to terminate early and further optimize this method. Zig zag points or sudden dips would be the most promising. And maybe after a certain number of intervals, keep widening the jumpThreshold as for a discontinuity you expect the residual distance to rapidly increase towards infinity!
Updated code
let ymax = 5, ymin = -5; /* just for example */
let height = 500; /* 500px screen height */
let jumpThreshold = Math.pow(.5 * (ymax - ymin) / height, 2); /* fraction(half) of a pixel! */
/* Math function to test on */
function fn(x) {
//return (Math.pow(Math.tan(x), 3));
return 1 / x;
//return Math.floor(x);
//return x * ((x - 1 - 0.001) / (x - 1));
//return x*x;
}
function findDiscontinuity(leftX, rightX, jumpThreshold, fn) {
/* try 5 interations of ridders method */
/* usually this approach can find the exact reciprocal root of a discountinuity
* in 1 iteration for functions like 1/x compared to the bisection method below */
let iterations = 5;
let root = inverseRidderMethod(leftX, rightX, iterations, fn);
let limit = fn(root);
if (Math.abs(limit) > 1e+16) {
if (root >= leftX && root <= rightX) return root;
return NaN;
}
root = discontinuityBisection(leftX, rightX, jumpThreshold, fn);
return root;
}
function discontinuityBisection(leftX, rightX, jumpThreshold, fn) {
while (true) {
let leftY = fn(leftX);
let rightY = fn(rightX);
let middleX = (leftX + rightX) * .5;
let middleY = fn(middleX);
let a = (leftX + middleX) * .5;
let fa = fn(a);
let b = (middleX + rightX) * .5;
let fb = fn(b);
let leftResidual = Math.pow(fa - (leftY + middleY) * .5, 2);
let rightResidual = Math.pow(fb - (middleY + rightY) * .5, 2);
/* if both subinterval midpoints (fa,fb) barely deviate from their slope lines
* i.e. they're under the jumpThreshold, then return NaN,
* indicating no discountinuity with the current threshold,
* both subintervals are mostly straight */
if (leftResidual < jumpThreshold && rightResidual < jumpThreshold) return NaN;
if (!isFinite(fa) || a === leftX || a === middleX) return a;
if (!isFinite(fb) || b === middleX || b === rightX) return b;
if (leftResidual > rightResidual) {
/* left hand-side subinterval */
rightX = middleX;
middleX = a;
} else {
/* right hand-side subinterval */
leftX = middleX;
middleX = b;
}
}
}
function inverseRidderMethod(min, max, iterations, fn) {
/* Modified version of RiddersSolver from Apache Commons Math
* http://commons.apache.org/
* https://www.apache.org/licenses/LICENSE-2.0.txt
*/
let x1 = min;
let y1 = 1 / fn(x1);
let x2 = max;
let y2 = 1 / fn(x2);
// check for zeros before verifying bracketing
if (y1 == 0) {
return min;
}
if (y2 == 0) {
return max;
}
let functionValueAccuracy = 1e-55;
let relativeAccuracy = 1e-16;
let oldx = Number.POSITIVE_INFINITY;
let i = 0;
while (i < iterations) {
// calculate the new root approximation
let x3 = 0.5 * (x1 + x2);
let y3 = 1 / fn(x3);
if (!isFinite(y3)) return NaN;
if (Math.abs(y3) <= functionValueAccuracy) {
return x3;
}
let delta = 1 - (y1 * y2) / (y3 * y3); // delta > 1 due to bracketing
let correction = (signum(y2) * signum(y3)) * (x3 - x1) / Math.sqrt(delta);
let x = x3 - correction; // correction != 0
if (!isFinite(x)) return NaN;
let y = 1 / fn(x);
// check for convergence
let tolerance = Math.max(relativeAccuracy * Math.abs(x), 1e-16);
if (Math.abs(x - oldx) <= tolerance) {
return x;
}
if (Math.abs(y) <= functionValueAccuracy) {
return x;
}
// prepare the new interval for the next iteration
// Ridders' method guarantees x1 < x < x2
if (correction > 0.0) { // x1 < x < x3
if (signum(y1) + signum(y) == 0.0) {
x2 = x;
y2 = y;
} else {
x1 = x;
x2 = x3;
y1 = y;
y2 = y3;
}
} else { // x3 < x < x2
if (signum(y2) + signum(y) == 0.0) {
x1 = x;
y1 = y;
} else {
x1 = x3;
x2 = x;
y1 = y3;
y2 = y;
}
}
oldx = x;
}
}
function signum(a) {
return (a < 0.0) ? -1.0 : ((a > 0.0) ? 1.0 : a);
}
/* TEST */
console.log(findDiscontinuity(.5, .6, jumpThreshold, fn));
Python Code
I don't mind if the solution is provided in Javascript or Python
import math
def fn(x):
try:
# return (math.pow(math.tan(x), 3))
# return 1 / x
# return math.floor(x)
return x * ((x - 1 - 0.001) / (x - 1))
except ZeroDivisionError:
return float('Inf')
def slope(x1, y1, x2, y2):
try:
return (y2 - y1) / (x2 - x1)
except ZeroDivisionError:
return float('Inf')
def find_discontinuity(leftX, rightX, fn):
while True:
leftY = fn(leftX)
rightY = fn(rightX)
middleX = (leftX + rightX) / 2
middleY = fn(middleX)
leftSlope = abs(slope(leftX, leftY, middleX, middleY))
rightSlope = abs(slope(middleX, middleY, rightX, rightY))
if not math.isfinite(leftSlope) or not math.isfinite(rightSlope):
return middleX
if middleX == leftX or middleX == rightX:
return middleX
if leftSlope > rightSlope:
rightX = middleX
rightY = middleY
else:
leftX = middleX
leftY = middleY

Why does my Perlin Noise Generator make this squiggly pattern?

I've started learning about perlin noise generation, and I wanted to try to make my own generator in JavaScript. To get me started, I've been following along with this youtube tutorial. to try and copy their basic implementation. I've also been reading this article.
I've provided a jsFiddle of my implementation, which shows what I'm doing and what the output is. Instead of the smoothly-flowing, bubbling noise I see in the youtube tutorial, I get tightly-constricted black squiggles. Here's a picture:
Here's my generator code:
var p = [151,160,137,91,90,15,
131,13,201,95,96,53,194,233,7,225,140,36,103,30,69,142,8,99,37,240,21,10,23,
190, 6,148,247,120,234,75,0,26,197,62,94,252,219,203,117,35,11,32,57,177,33,
88,237,149,56,87,174,20,125,136,171,168, 68,175,74,165,71,134,139,48,27,166,
77,146,158,231,83,111,229,122,60,211,133,230,220,105,92,41,55,46,245,40,244,
102,143,54, 65,25,63,161, 1,216,80,73,209,76,132,187,208, 89,18,169,200,196,
135,130,116,188,159,86,164,100,109,198,173,186, 3,64,52,217,226,250,124,123,
5,202,38,147,118,126,255,82,85,212,207,206,59,227,47,16,58,17,182,189,28,42,
223,183,170,213,119,248,152, 2,44,154,163, 70,221,153,101,155,167, 43,172,9,
129,22,39,253, 19,98,108,110,79,113,224,232,178,185, 112,104,218,246,97,228,
251,34,242,193,238,210,144,12,191,179,162,241, 81,51,145,235,249,14,239,107,
49,192,214, 31,181,199,106,157,184, 84,204,176,115,121,50,45,127, 4,150,254,
138,236,205,93,222,114,67,29,24,72,243,141,128,195,78,66,215,61,156,180,
151,160,137,91,90,15,
131,13,201,95,96,53,194,233,7,225,140,36,103,30,69,142,8,99,37,240,21,10,23,
190, 6,148,247,120,234,75,0,26,197,62,94,252,219,203,117,35,11,32,57,177,33,
88,237,149,56,87,174,20,125,136,171,168, 68,175,74,165,71,134,139,48,27,166,
77,146,158,231,83,111,229,122,60,211,133,230,220,105,92,41,55,46,245,40,244,
102,143,54, 65,25,63,161, 1,216,80,73,209,76,132,187,208, 89,18,169,200,196,
135,130,116,188,159,86,164,100,109,198,173,186, 3,64,52,217,226,250,124,123,
5,202,38,147,118,126,255,82,85,212,207,206,59,227,47,16,58,17,182,189,28,42,
223,183,170,213,119,248,152, 2,44,154,163, 70,221,153,101,155,167, 43,172,9,
129,22,39,253, 19,98,108,110,79,113,224,232,178,185, 112,104,218,246,97,228,
251,34,242,193,238,210,144,12,191,179,162,241, 81,51,145,235,249,14,239,107,
49,192,214, 31,181,199,106,157,184, 84,204,176,115,121,50,45,127, 4,150,254,
138,236,205,93,222,114,67,29,24,72,243,141,128,195,78,66,215,61,156,180]
function generatePerlinNoise(x, y){
// Determine gradient cell corners
var xi = Math.floor(x) & 255;
var yi = Math.floor(y) & 255;
var g1 = p[p[xi] + yi];
var g2 = p[p[xi + 1] + yi];
var g3 = p[p[xi] + yi + 1];
var g4 = p[p[xi + 1] + yi + 1];
// Get point within gradient cell
var xf = x - Math.floor(x);
var yf = y - Math.floor(y);
// Get dot products at each corner of the gradient cell
var d1 = generateGradient(g1, xf, yf);
var d2 = generateGradient(g2, xf - 1, yf);
var d3 = generateGradient(g3, xf, yf - 1);
var d4 = generateGradient(g4, xf - 1, yf - 1);
var xFade = fade(xf);
var yFade = fade(yf);
var x1Interpolated = linearInterpolate(xFade, d1, d2);
var x2Interpolated = linearInterpolate(xFade, d3, d4);
var noiseValue = linearInterpolate(yFade, x1Interpolated, x2Interpolated);
return noiseValue;
}
function generateGradient(hash, x, y){
switch(hash & 3){
case 0: return x + y;
case 1: return -x + y;
case 2: return x - y;
case 3: return -x - y;
default: return 0;
}
}
// This is the fade function described by Ken Perlin
function fade(t){
return t * t * t * (t * (t * 6 - 15) + 10);
}
function linearInterpolate(amount, left, right){
return ((1-amount) * left + (amount * right))
}
I'm utilizing the generator function by dividing the pixel x and y values by my canvas height, and scaling by a frequency variable:
var freq = 12;
var noise = generatePerlinNoise((x/canvas.offsetHeight)*freq, (y/canvas.offsetHeight)*freq);
noise = Math.abs(noise);
I'm currently just using the noise value to generate a greyscale color value:
var blue = Math.floor(noise * 0xFF); // Scale 255 by our noise value,
// and take it's integer portion
var green = Math.floor(noise * 0xFF);
var red = Math.floor(noise * 0xFF);
data[i++] = red;
data[i++] = green;
data[i++] = blue;
data[i++] = 255;
The point of this for me is to learn more about noise generation and javascript. I've tried to think through the problem and made some observations:
There are no visible artifacts, so it seems like my fade function is working fine.
There don't seem to be any repeating patterns, so that's a good sign.
I go generate a complete range of values in the greyscale - not just black and white.
The general issue seems to be how the gradient at each pixel affects its neighbors: Mine seem to clump together in snake-like ropes of fixed widths. It seems like the gradient vector options supplied and the permutation table used to randomly-ish select them would govern this, but mine are an exact copy from the tutorial: The same 4 vectors each pointing into a quadrant at 45 degrees, and the standard permutation table.
This leaves me stumped as to figuring out what the cause is. My general suspicions boil down to:
I've messed up the algorithm somewhere in a subtle way that I keep overlooking. (Most likely)
There's a subtle difference in the way JavaScript does something that i'm overlooking. (Maybe)
I'm generating noise correctly, but incorrectly applying the result to the RGB values used in my canvas image data. (Least likely)
I'd really like to get to the bottom of this. Thanks in advance for your help! :)
Also: I DO think this pattern is cool, and this is a learning exercise, so if anyone can share insight into why I'm getting this pattern specifically, that'd be great!

Canvas circle collision, how to work out where circles should move to once collided?

I am having a go at building a game in html canvas. It's a Air Hockey game and I have got pretty far though it. There are three circles in the game, the disc which is hit and the two controllers(used to hit the disc/circle).
I've got the disc rebounding off the walls and have a function to detect when the disc has collided with a controller. The bit I am struggling with is when the two circle's collide, the controller should stay still and the disc should move away in the correct direction. I've read a bunch of article's but still can't get it right.
Here's a Codepen link my progress so far. You can see that the puck rebounds off the controller but not in the correct direction. You'll also see if the puck comes from behind the controller it goes through it.
http://codepen.io/allanpope/pen/a01ddb29cbdecef58197c2e829993284?editors=001
I think what I am after is elastic collision but not sure on how to work it out. I've found this article but have been unable to get it working.
http://gamedevelopment.tutsplus.com/tutorials/when-worlds-collide-simulating-circle-circle-collisions--gamedev-769
Heres is my collision detection function. Self refer's to the disc and the controller[i] is the controller the disc hits.
this.discCollision = function() {
for (var i = 0; i < controllers.length; i++) {
// Minus the x pos of one disc from the x pos of the other disc
var distanceX = self.x - controllers[i].x,
// Minus the y pos of one disc from the y pos of the other disc
distanceY = self.y - controllers[i].y,
// Multiply each of the distances by itself
// Squareroot that number, which gives you the distance between the two disc's
distance = Math.sqrt(distanceX * distanceX + distanceY * distanceY),
// Added the two disc radius together
addedRadius = self.radius + controllers[i].radius;
// Check to see if the distance between the two circles is smaller than the added radius
// If it is then we know the circles are overlapping
if (distance <= addedRadius) {
var newVelocityX = (self.velocityX * (self.mass - controllers[i].mass) + (2 * controllers[i].mass * controllers[i].velocityX)) / (self.mass + controllers[i].mass);
var newVelocityY = (self.velocityY * (self.mass - controllers[i].mass) + (2 * controllers[i].mass * controllers[i].velocityX)) / (self.mass + controllers[i].mass);
self.velocityX = newVelocityX;
self.velocityY = newVelocityY;
self.x = self.x + newVelocityX;
self.y = self.y + newVelocityY;
}
}
}
Updated
Deconstructed a circle collision demo & tried to implement their collision formula. This is it below, works for hitting the puck/disc forward & down but wont hit the back backwards or up for some reason.
this.discCollision = function() {
for (var i = 0; i < controllers.length; i++) {
// Minus the x pos of one disc from the x pos of the other disc
var distanceX = self.x - controllers[i].x,
// Minus the y pos of one disc from the y pos of the other disc
distanceY = self.y - controllers[i].y,
// Multiply each of the distances by itself
// Squareroot that number, which gives you the distance between the two disc's
distance = Math.sqrt(distanceX * distanceX + distanceY * distanceY),
// Added the two disc radius together
addedRadius = self.radius + controllers[i].radius;
// Check to see if the distance between the two circles is smaller than the added radius
// If it is then we know the circles are overlapping
if (distance < addedRadius) {
var normalX = distanceX / distance,
normalY = distanceY / distance,
midpointX = (controllers[i].x + self.x) / 2,
midpointY = (controllers[i].y + self.y) / 2,
delta = ((controllers[i].velocityX - self.velocityX) * normalX) + ((controllers[i].velocityY - self.velocityY) * normalY),
deltaX = delta*normalX,
deltaY = delta*normalY;
// Rebound puck
self.x = midpointX + normalX * self.radius;
self.y = midpointY + normalY * self.radius;
self.velocityX += deltaX;
self.velocityY += deltaY;
// Accelerate once hit
self.accelerationX = 3;
self.accelerationY = 3;
}
}
}

I'm not great at these types of math problems, but it looks like you need to rotate your vectors around sine and cosine angles. I will point you at a working example and the source code that drives it. I did not derive this example.
I solved just the circle collision detection part of this problem recently, but one solution I came across includes code for establishing new vector directions. Ira Greenburg hosts his original source at processing.org. Ira further cites Keith Peter's Solution in Foundation Actionscript Animation: Making Things Move!
I copied Ira's code into Processing's Javascript mode then pushed it to Github Pages so you can see it before you try it.

The main issue with my code was the user controller was attached to the mouse. When a collision would happen, the function would run constantly because the circles where still touching due to the mouse position. I changed my code so the controller is controlled by the users keyboard.
I also asked for help on reddit & got some help with my collision code. Some good resources where linked.
(http://www.reddit.com/r/javascript/comments/3cjivi/having_a_go_at_building_an_air_hockey_game_stuck/)

Calculate roll angle for Google Maps Street View

Preamble: there's an issue logged with the Google Maps API, requesting the ability to correct the roll angle of street view tiles to compensate for hills. I've come up with a client-side workaround involving some css sorcery on the tile container. Here's my rotate function:
rotate: function() {
var tilesLoaded = setInterval(function() {
var tiles = $('map-canvas').getElementsByTagName('img');
for (var i=0; i<tiles.length; i++) {
if (tiles[i].src.indexOf(maps.panorama.getPano()) > -1) {
if (typeof maps.panorama.getPhotographerPov != 'undefined') {
var pov = maps.panorama.getPhotographerPov(),
pitch = pov.pitch,
cameraHeading = pov.heading;
/**************************
// I need help with my logic here.
**************************/
var yaw = pov.heading - 90;
if (yaw < 0) yaw += 360;
var scale = ((Math.abs(maps.heading - yaw) / 90) - 1) * -1;
pitch = pov.pitch * scale;
tiles[i].parentNode.parentNode.style.transform = 'rotate(' + pitch + 'deg)';
clearInterval(tilesLoaded);
return;
}
}
}
}, 20);
}
A full (and more thoroughly commented) proof-of-concept is at this JSFiddle. Oddly, the horizon is just about perfectly level if I do no calculation at all on the example in the JSFiddle, but that result isn't consistent for every Lat/Lng. That's just a coincidence.
So, I need to calculate the roll at the client's heading, given the client heading, photographer's heading, and photographer's pitch. Assume the photographer is either facing uphill or downhill, and pov.pitch is superlative (at the min or max limit). How can I calculate the desired pitch facing the side at a certain degree?
Edit: I found an equation that seems to work pretty well. I updated the code and the fiddle. While it seems to be pretty close to the answer, my algorithm is linear. I believe the correct equation should be logarithmic, resulting in subtler adjustments closer to the camera heading and opposite, while to the camera's left and right adjustments are larger.

I found the answer I was looking for. The calculation involves spherical trigonometry, which I didn't even know existed before researching this issue. If anyone notices any problems, please comment. Or if you have a better solution than the one I found, feel free to add your answer and I'll probably accept it if it's more reliable or significantly more efficient than my own.
Anyway, if the tile canvas is a sphere, 0 pitch (horizon) is a plane, and camera pitch is another plane intersecting at the photographer, the two planes project a spherical lune onto the canvas. This lune can be used to calculate a spherical triangle where:
polar angle = Math.abs(camera pitch)
base = camera heading - client heading
one angle = 90° (for flat horizon)
With two angles and a side available, other properties of a spherical triangle can be calculated using the spherical law of sines. The entire triangle isn't needed -- only the side opposite the polar angle. Because this is math beyond my skills, I had to borrow the logic from this spherical triangle calculator. Special thanks to emfril!
The jsfiddle has been updated. My production roll getter has been updated as follows:
function $(what) { return document.getElementById(what); }
var maps = {
get roll() {
function acos(what) {
return (Math.abs(Math.abs(what) - 1) < 0.0000000001)
? Math.round(Math.acos(what)) : Math.acos(what);
}
function sin(what) { return Math.sin(what); }
function cos(what) { return Math.cos(what); }
function abs(what) { return Math.abs(what); }
function deg2rad(what) { return what * Math.PI / 180; }
function rad2deg(what) { return what * 180 / Math.PI; }
var roll=0;
if (typeof maps.panorama.getPhotographerPov() != 'undefined') {
var pov = maps.panorama.getPhotographerPov(),
clientHeading = maps.panorama.getPov().heading;
while (clientHeading < 0) clientHeading += 360;
while (clientHeading > 360) clientHeading -= 360;
// Spherical trigonometry method
a1 = deg2rad(abs(pov.pitch));
a2 = deg2rad(90);
yaw = deg2rad((pov.heading < 0 ? pov.heading + 360 : pov.heading) - clientHeading);
b1 = acos((cos(a1) * cos(a2)) + (sin(a1) * sin(a2) * cos(yaw)));
if (sin(a1) * sin(a2) * sin(b1) !== 0) {
roll = acos((cos(a1) - (cos(a2) * cos(b1))) / (sin(a2) * sin(b1)));
direction = pov.heading - clientHeading;
if (direction < 0) direction += 360;
if (pov.pitch < 0)
roll = (direction < 180) ? rad2deg(roll) * -1 : rad2deg(roll);
else
roll = (direction > 180) ? rad2deg(roll) * -1 : rad2deg(roll);
} else {
// Fall back to algebraic estimate to avoid divide-by-zero
var yaw = pov.heading - 90;
if (yaw < 0) yaw += 360;
var scale = ((abs(clientHeading - yaw) / 90) - 1) * -1;
roll = pov.pitch * scale;
if (abs(roll) > abs(pov.pitch)) {
var diff = (abs(roll) - abs(pov.pitch)) * 2;
roll = (roll < 0) ? roll + diff : roll - diff;
}
}
}
return roll;
}, // end maps.roll getter
// ... rest of maps object...
} // end maps{}
After rotating the panorama tile container, the container also needs to be expanded to hide the blank corners. I was originally using the 2D law of sines for this, but I found a more efficient shortcut. Thanks Mr. Tan!
function deg2rad(what) { return what * Math.PI / 180; }
function cos(what) { return Math.cos(deg2rad(what)); }
function sin(what) { return Math.sin(deg2rad(what)); }
var W = $('map-canvas').clientWidth,
H = $('map-canvas').clientHeight,
Rot = Math.abs(maps.originPitch);
// pixels per side
maps.growX = Math.round(((W * cos(Rot) + H * cos(90 - Rot)) - W) / 2);
maps.growY = Math.round(((W * sin(Rot) + H * sin(90 - Rot)) - H) / 2);
There will be no more edits to this answer, as I don't wish to have it converted to a community wiki answer yet. As updates occur to me, they will be applied to the fiddle.