Develop Reference

Detect mouse is near circle edge

I have a function which gets the mouse position in world space, then checks to see if the mouse is over or near to the circle's line.
The added complication how ever is the circle is transformed at an angle so it's more of an ellipse. I can't see to get the code to detect that the mouse is near the border of circle and am unsure where I am going wrong.
This is my code:
function check(evt){
var x = (evt.offsetX - element.width/2) + camera.x; // world space
var y = (evt.offsetY - element.height/2) + camera.y; // world space
var threshold = 20/scale; //margin to edge of circle
for(var i = 0; i < obj.length;i++){
// var mainAngle is related to the transform
var x1 = Math.pow((x - obj[i].originX), 2) / Math.pow((obj[i].radius + threshold) * 1,2);
var y1 = Math.pow((y - obj[i].originY),2) / Math.pow((obj[i].radius + threshold) * mainAngle,2);
var x0 = Math.pow((x - obj[i].originX),2) / Math.pow((obj[i].radius - threshold) * 1, 2);
var y0 = Math.pow((y - obj[i].originY),2) / Math.pow((obj[i].radius - threshold) * mainAngle, 2);
if(x1 + y1 <= 1 && x0 + y0 >= 1){
output.innerHTML += '<br/>Over';
return false;
}
}
output.innerHTML += '<br/>out';
}
To understand it better, I have a fiddle here: http://jsfiddle.net/nczbmbxm/ you can move the mouse over the circle, it should say "Over" when you are within the threshold of being near the circle's perimeter. Currently it does not seem to work. And I can't work out what the maths needs to be check for this.

There is a typo on line 34 with orignX
var x1 = Math.pow((x - obj[i].orignX), 2) / Math.pow((obj[i].radius + threshold) * 1,2);
should be
var x1 = Math.pow((x - obj[i].originX), 2) / Math.pow((obj[i].radius + threshold) * 1,2);
now you're good to go!
EDIT: In regards to the scaling of the image and further rotation of the circle, I would set up variables for rotation about the x-axis and y-axis, such as
var xAngle;
var yAngle;
then as an ellipse can be written in the form
x^2 / a^2 + y^2 / b^2 = 1
such as in Euclidean Geometry,
then the semi-major and semi-minor axes would be determined by the rotation angles. If radius is the circles actual radius. then
var semiMajor = radius * cos( xAngle );
var semiMinor = radius;
or
var semiMajor = radius;
var semiMinor = radius * cos( yAngle );
you would still need to do some more transformations if you wanted an x and y angle.
so if (xMouseC, yMouseC) are the mouse coordinates relative to the circles centre, all you must do is check if that point satisfies the equation of the ellipse to within a certain tolerance, i.e. plug in
a = semiMajor;
b = semiMinor;
x = xMouseC;
y = yMouseC;
and see if it is sufficiently close to 1.
Hope that helps!

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.

How can I use .mousemove to fade a background image according to distance from center page?

I have this function set up
if (window.innerWidth && window.innerHeight) {
var winW = window.innerWidth;
}
var xM = winW/180;
var axis = 0;
$(window).bind('mousemove',function(e){
var xCoord = Math.floor(e.pageX/xM);
axis = 0.6 * Math.sin(xCoord);
var pageCoords = "( " + e.pageX + ", " + e.pageY + ", " + xCoord + " )";
$("span#showme").text(pageCoords);
});
setInterval(function() {
$("#welcome-background").fadeTo(0, 0.4 + axis);
}, 100);
(for additional reference and working visual- http://jsfiddle.net/ySjqh/2/ )
The code works in theory to divide the page evenly into segments from 0-180, then calculates which segment the mouse appears in. Then uses the Math.sin() function to derive how much opacity to apply, based on a padded starting point of 0.4 opacity (jQuery style), and should use the mouse position to determine how much of the remaining 0.6 to apply based on its distance from center, where mouse at center-page should yield full opacity.
What I don't get is why the script behaves this way, rolling through an entire sine wave when I've limited the input to the Math.sin(x) function to 1 < x < 180. If you replace xCoord with axis in the place where I build the jQuery text for #showme, you'll see that it throws negative numbers- which shouldn't be happening! ... so I don't get what the problem/behavior results from!!! Frustrating!!!

Just use:
xCoord = (xCoord * Math.PI) / 180; // Convert value to Radians
and it works..
Sample

http://jsfiddle.net/ySjqh/4/
axis = 0.6 * (1 - Math.abs(e.pageX - winW/2)/(winW/2));
Using the X distance from the center instead of sin.

Canvas, drawing a line segment

My trigonometry is more than weak, and therefore I do not know how to draw a line segment shorter than full lines start point and end point.
http://jsfiddle.net/psycketom/TUyJb/
What I have tried, is, subtract from start point a fraction of target point, but it results in a wrong line.
/*
* this is an excerpt from fiddle, that shows
* the actual calculation functions I have implemented
*/
var target = {
x : width / 2 + 60,
y : 20
};
var start = {
x : width / 2,
y : height
};
var current = {
x : 0,
y : 0
};
var growth = 0.5;
current.x = start.x - (target.x * growth);
current.y = start.y - (target.y * growth);
My bet is that I have to use sin / cos or something else from the trigonometry branch to get it right. But, since my trigonometry is not even rusty, but weak in general, I'm stuck.
How do I draw a proper line to target?

If I understand you correctly, then this should give you what you're looking for:
current.x = start.x + (target.x - start.x) * growth;
current.y = start.y + (target.y - start.y) * growth;

The equation is a linear interpolate, its the same as linear easing. You take the delta of the start and end (min and max), multiply it by a percent (the normal) of how far along delta you are and then you add it back to the start value. Incredibly essential algorithm :)

How to convert a point to polar coordinates in ECMAScript?

I need to turn a click location into a polar coordinate.
This is my current algorithm. Location is the location on the canvas of the click ({x:evt.clientX, y:evt.clientY}), center is the offset of the origin from 0,0. For example, if the circle is centered on 250, 250, center is {x:250, y:250}. Scale is the scale of the radius. For example, if the radius of a circle from the center would normally be 50 and the scale is .5, the radius becomes 25. (it's for zooming in/out)
this.getPolarLocation = function(location){
var unscaledFromCenter = {
x: location.x - center.x,
y: location.y - center.y
};
var angle = this.getAngleOnCircle(unscaledFromCenter);
var dist = Math.sqrt(unscaledFromCenter.x * unscaledFromCenter.x + unscaledFromCenter.y * unscaledFromCenter.y) * this.ds.scale;
return {
angle:angle,
dist:dist,
toString: function(){
return "Theta: ".concat(angle).concat("; dist: ").concat(dist);
}
};
}
this.getAngleOnCircle = function(location){
var x = location.x;
var y = location.y;
if(x == 0 && y > 0)
return Math.PI / 2;
if(x == 0 && y < 0)
return 3 * Math.PI / 2;
if(y == 0 && x > 0)
return 0;
if(y == 0 && x < 0)
return Math.PI;
var angle = Math.atan(y/x);
if(x > 0 && y > 0)
return angle;
if(x < 0)
return Math.PI + angle
return Math.PI * 2 + angle;
}
Screenshots of the issue. The left is what happens zoomed out (and is not supposed to happen). The right is zoomed in (scale >= 1), and is what is supposed to happen.
I'm under the impression that my center coordinates are being shifted slightly off. It seems to work fine for scale >= 1, but not for scale < 1
Source:
circos.html: http://pastie.org/private/cowsjz7mcihy8wtv4u4ag
circos.js: http://pastie.org/private/o9w3dwccmimalez9fropa
datasource.js: http://pastie.org/private/iko9bqq8eztbfh8xpvnoaw
Run in Firefox
So my question is: why doesn't this work?

For some reason, the program automagically works when I close firebug. It doesn't seem to work on Firefox 5, only the version I have (in the 3s somewhere). Either way, I'm scrapping the project for something more object oriented. There's no way the current algorithm could handle a genome. (which is exactly what I'm going to be mapping)
UPDATE:
I figured out the problem... I was measuring the distance from the top left of the page, not the top left of the canvas. Thus, when firebug was enabled, the screen was shifted, making the problems worse. The solution is the use canvas.offsetLeft and canvas.offsetTop to calculate the position on the canvas.