Develop Reference

Draw a perpendicular line through middle of two long/lat points

Hello i'm currently trying to draw a line through two long/lat lines to create a triangle. So far i have manged to draw a line through but the line is not perpendicular and looks skewed. Here is my code:
startPosition = [-0.17640, 51.426700];
endPosition = [0.17640, 51.796700];
triangleSizeY = (endPosition [1] - startPosition[1]) / 6;
/*subtract
end from start to work out direction and also use this divided by 6 to
get distance*/
triangleSize *= -1;
triangleSizeX = (endPosition [0] - startPosition[0]) / 6;
/*subtract
end from start to work out direction and also use this divided by 6 to
get distance*/
triangleSize *= -1;
var cx = (startPosition[0] + endPosition[0]) / 2;
var cy = (startPosition[1] + endPosition[1]) / 2;
var dx = (endPosition[0] - startPosition[0]) / 2;
var dy = (endPosition[1] - startPosition[1]) / 2;
positions[0] = [midPoint[0] + triangleSizeX, midPoint[1] +
triangleSizeY];
positions[1] = [cx - dy, cy + dx];
positions[2] = [cx + dy, cy - dx];
This is what it looks like:

First, lat/lon are angular so you can not do linear type distances. The steps you need to take to solve this:
Compute the distance between the 2 lat/lon pairs you want a perpendicular line from.
Take half the distance computed from the above step to get you the midpoint range.
Calculate the bearing between the 2 lat/lon pairs. (see reference below on computing a bearing from 2 lat/lon's)
With the half distance and bearing, you can compute the lat/lon of the midpoint. This is called computing a lat/lon based on a range and bearing. (See the reference below.)
Now you can go perpendicular from the midpoint by adding/subtracting 90 degrees from the bearing in step 3. Decide on a range you want to compute the new lat/lon from a range/bearing like in step 4.
This site (https://www.movable-type.co.uk/scripts/latlong.html) has the calculations you need to do this. Also, since the distance is relatively small, you can use the Equirectangular approximation over Haversine for distance calculation.

Converting between -180-180 to 0-360

I have a function that returns an angle from -180 to 180 and I need it to be 0 to 360. What is the formula to convert the angle?
Here's what I want:
0/360
270 90
180
Here's what I have:
-90
-180/180 0
90
Any help is greatly appreciated, working on an html/javascript game.

There are a number of ways this can be done while preserving the equivalent angle, with certain assumptions about the values.
If you are sure the values are actually in the range [-180,180), then you can use something like the following:
var x = Math.random()*360-180; // Generate random angle in range [-180, 180)
var y = x < 0 ? x+360 : x; // Normalize to be in the range [0, 360)
If you're doing this a lot, the branch operation may result in poor behavior (though maybe not in Javascript ... you'd need to profile it). So, it's sometimes better to do this without a branching operation. You can do this using the modulus (%) operator.
var x = Math.random()*360-180; // Generate random angle in range [-180, 180)
var y = (x + 360) % 360; // Normalize to be in the range [0, 360)
If you cannot guarantee the original value is in the range [-180, 180) you can essentially divide out the number of turns. This would look like
var x = getAngle(); // Some angle, potentially outside [-180, 180)
var y = x - 360 * Math.floor(x/360); // Normalizes to be in range [0, 360)
Some care will need to be taken for large absolute values of x due to the way floating point numbers are represented (which as I understand it is what Javascript uses, even when integer values would work well)
Edit: I just noticed you are also changing where you consider the 0 angle to be (above rather than to the right). In that case, you need to also add 90 degrees to rotate the 0 into the correct position. In this case, the first code segment becomes:
var x = Math.random()*360-180; // Generate random angle in range [-180, 180)
var y = x < -90 ? x+450 : x+90; // Normalize to be in the range [0, 360)
The next one becomes as Amadan indicated.
var x = Math.random()*360-180; // Generate random angle in range [-180, 180)
var y = (x + 450) % 360; // Normalize to be in the range [0, 360)
and the last one becomes
var x = getAngle(); // Some angle, potentially outside [-180, 180)
var y = (x+90) - 360 * Math.floor((x+90)/360); // Normalizes to be in range [0, 360)

Here is code as suggested by #Amadan
function convert(deg) {
return ((deg + 450) % 360);
}
window.console.log(convert(-180));
window.console.log(convert(-90));
window.console.log(convert(0));
window.console.log(convert(90));
window.console.log(convert(180));

Canvas Graph plotting data incorrectly

I have made a simple graph in a canvas but am having difficulty with two issues.
The first issue is setting the vertical axis with an appropriate scale automatically with enough room for each data value in an array. Ideally i'd like the numbers to be more rounded to the nearest million or thousand etc depending on it's actual value ranges rather than a value like 33145 as the first scale line.
Currently one value is too high for the scale and is not being drawn on the canvas because it is out of bounds.
The second issue, is the points don't seem to be plotting in their correct location, which I am unsure where my mistake was.
I made a JSFiddle as for the most part it might be a bit confusing without seeing it in action:
http://jsfiddle.net/ezttywzr/
This is how i plot my data and draw my vertical axis:
Vertical Axis:
var x = 0,
y,
range = data.max() - data.min(),
valueStep = range / 10,
// get width of largest number
margin = 3 + ctx.measureText(data.min() + (valueStep*10)).width,
pixelStep = (graph.height-40) / 10,
verticalP = pixelStep,
output;
// draw left hand values
for(var i = 0; i < 11; i++){
output = data.min() + (valueStep*i);
y = graph.height-20 - (verticalP + i*pixelStep);
ctx.fillText(output,x,y+6);
ctx.beginPath();
ctx.moveTo(margin, y);
ctx.lineTo(x2,y);
ctx.stroke();
}
Data Plotting:
var y = graph.height,
x = margin,
pos,
valueStep = (graph.width-(margin*2)) / data.length,
pixelRange = graph.height-20,
pp = range / pixelRange;
for(var i = 0; i < data.length; i++){
x += valueStep;
pos = x - (valueStep/2);
ctx.beginPath();
ctx.moveTo(x, graph.height-20);
ctx.lineTo(x, graph.height);
ctx.stroke();
ctx.fillText('Week '+(i+1),pos-(ctx.measureText('Week '+(i+1)).width/2),y);
ctx.beginPath();
ctx.arc(pos,(graph.height-20)-(verticalP+(data[i]/pp)),2,0,2*Math.PI);
ctx.stroke();
ctx.fill();
}

Nice job so far.
I made a few changes: http://jsfiddle.net/ezttywzr/2/
To get the scale I used
STEP = data.max() / NUM_HORIZONTAL_LINES
Where NUM_HORIZONTAL_LINES is the number of horizontal lines you want above the x-axis. In this case I used 10.
This means the first line will be 1 * STEP, the second will be 2 * STEP, the third will be 3 * STEP and so on..
This scale is convenient because it guarantees that the max value fits on the graph. In fact, the max value is on the top line because of the way we defined the scale.
Once we have our scale it's easy to calculate the position of the points relative to the x-axis. It's simply:
(PIXELS_PER_STEP / STEP) * VALUE
To go a step further you can do some math to round the top point of the graph up and pick a scale with that has nice round numbers.

Incorrect angle, wrong side calculated

I need to calculate the angle between 3 points. For this, I do the following:
Grab the 3 points (previous, current and next, it's within a loop)
Calculate the distance between the points with Pythagoras
Calculate the angle using Math.acos
This seems to work fine for shapes without angels of over 180 degrees, however if a shape has such an corner it calculates the short-side. Here's an illustration to show what I mean (the red values are wrong):
This is the code that does the calculations:
// Pythagoras for calculating distance between two points (2D)
pointDistance = function (p1x, p1y, p2x, p2y) {
return Math.sqrt((p1x - p2x)*(p1x - p2x) + (p1y - p2y)*(p1y - p2y));
};
// Get the distance between the previous, current and next points
// vprev, vcur and vnext are objects that look like this:
// { x:float, y:float, z:float }
lcn = pointDistance(vcur.x, vcur.z, vnext.x, vnext.z);
lnp = pointDistance(vnext.x, vnext.z, vprev.x, vprev.z);
lpc = pointDistance(vprev.x, vprev.z, vcur.x, vcur.z);
// Calculate and print the angle
Math.acos((lcn*lcn + lpc*lpc - lnp*lnp)/(2*lcn*lpc))*180/Math.PI
Is there something wrong in the code, did I forget to do something, or should it be done a completely different way?

HI there your math and calculations are perfect. Your running into the same problem most people do on calculators, which is orientation. What I would do is find out if the point lies to the left or right of the vector made by the first two points using this code, which I found from
Determine which side of a line a point lies
isLeft = function(ax,ay,bx,by,cx,cy){
return ((bx - ax)*(cy - ay) - (by - ay)*(cx - ax)) > 0;
}
Where ax and ay make up your first point bx by your second and cx cy your third.
if it is to the left just add 180 to your angle

I've got a working but not necessarily brief example of how this can work:
var point1x = 0, point1y = 0,
point2x = 10, point2y = 10,
point3x = 20, point3y = 10,
point4x = 10, point4y = 20;
var slope1 = Math.atan2(point2y-point1y,point2x-point1x)*180/Math.PI;
var slope2 = Math.atan2(point3y-point2y,point3x-point2x)*180/Math.PI;
var slope3 = Math.atan2(point4y-point3y,point4x-point3x)*180/Math.PI;
alert(slope1);
alert(slope2);
alert(slope3);
var Angle1 = slope1-slope2;
var Angle2 = slope2-slope3;
alert(180-Angle1);
alert(180-Angle2);
(see http://jsfiddle.net/ZUESt/1/)
To explain the multiple steps the slopeN variables are the slopes of the individual line segments. AngleN is the amount turned at each junction (ie point N+1). A positive angle is a right turn and a negative angle a left turn.
You can then subtract this angle from 180 to get the actual interior angle that you want.
It should be noted that this code can of course be compressed and that five lines are merely outputting variables to see what is going on. I'll let you worry about optimizing it for your own use with this being a proof of concept.

You need to check boundary conditions (apparently, if points are colinear) and apply the proper calculation to find the angle.
Also, a triangle can't have any (interior) angle greater than 180 degress. Sum of angle of triangle is 180 degrees.

KinectJS: Algorithm required to determine new X,Y coords after image resize

BACKGROUND:
The app allows users to upload a photo of themselves and then place a pair of glasses over their face to see what it looks like. For the most part, it is working fine. After the user selects the location of the 2 pupils, I auto zoom the image based on the ratio between the distance of the pupils and then already known distance between the center points of the glasses. All is working fine there, but now I need to automatically place the glasses image over the eyes.
I am using KinectJS, but the problem is not with regards to that library or javascript.. it is more of an algorithm requirement
WHAT I HAVE TO WORK WITH:
Distance between pupils (eyes)
Distance between pupils (glasses)
Glasses width
Glasses height
Zoom ratio
SOME CODE:
//.. code before here just zooms the image, etc..
//problem is here (this is wrong, but I need to know what is the right way to calculate this)
var newLeftEyeX = self.leftEyePosition.x * ratio;
var newLeftEyeY = self.leftEyePosition.y * ratio;
//create a blue dot for testing (remove later)
var newEyePosition = new Kinetic.Circle({
radius: 3,
fill: "blue",
stroke: "blue",
strokeWidth: 0,
x: newLeftEyeX,
y: newLeftEyeY
});
self.pointsLayer.add(newEyePosition);
var glassesWidth = glassesImage.getWidth();
var glassesHeight = glassesImage.getHeight();
// this code below works perfect, as I can see the glasses center over the blue dot created above
newGlassesPosition.x = newLeftEyeX - (glassesWidth / 4);
newGlassesPosition.y = newLeftEyeY - (glassesHeight / 2);
NEEDED
A math genius to give me the algorithm to determine where the new left eye position should be AFTER the image has been resized
UPDATE
After researching this for the past 6 hours or so, I think I need to do some sort of "translate transform", but the examples I see only allow setting this by x and y amounts.. whereas I will only know the scale of the underlying image. Here's the example I found (which cannot help me):
http://tutorials.jenkov.com/html5-canvas/transformation.html
and here is something which looks interesting, but it is for Silverlight:
Get element position after transform
Is there perhaps some way to do the same in Html5 and/or KinectJS? Or perhaps I am going down the wrong road here... any ideas people?
UPDATE 2
I tried this:
// if zoomFactor > 1, then picture got bigger, so...
if (zoomFactor > 1) {
// if x = 10 (for example) and if zoomFactor = 2, that means new x should be 5
// current x / zoomFactor => 10 / 2 = 5
newLeftEyeX = self.leftEyePosition.x / zoomFactor;
// same for y
newLeftEyeY = self.leftEyePosition.y / zoomFactor;
}
else {
// else picture got smaller, so...
// if x = 10 (for example) and if zoomFactor = 0.5, that means new x should be 20
// current x * (1 / zoomFactor) => 10 * (1 / 0.5) = 10 * 2 = 20
newLeftEyeX = self.leftEyePosition.x * (1 / zoomFactor);
// same for y
newLeftEyeY = self.leftEyePosition.y * (1 / zoomFactor);
}
that didn't work, so then I tried an implementation of Rody Oldenhuis' suggestion (thanks Rody):
var xFromCenter = self.leftEyePosition.x - self.xCenter;
var yFromCenter = self.leftEyePosition.y - self.yCenter;
var angle = Math.atan2(yFromCenter, xFromCenter);
var length = Math.hypotenuse(xFromCenter, yFromCenter);
var xNew = zoomFactor * length * Math.cos(angle);
var yNew = zoomFactor * length * Math.sin(angle);
newLeftEyeX = xNew + self.xCenter;
newLeftEyeY = yNew + self.yCenter;
However, that is still not working as expected. So, I am not sure what the issue is currently. If anyone has worked with KinectJS before and has an idea of what the issue may be, please let me know.
UPDATE 3
I checked Rody's calculations on paper and they seem fine, so there is obviously something else here messing things up.. I got the coordinates of the left pupil at zoom factors 1 and 2. With those coordinates, maybe someone can figure out what the issue is:
Zoom Factor 1: x = 239, y = 209
Zoom Factor 2: x = 201, y = 133

OK, since it's an algorithmic question, I'm going to keep this generic and only write pseudo code.
I f I understand you correctly, What you want is the following:
Transform all coordinates such that the origin of your coordinate system is at the zoom center (usually, central pixel)
Compute the angle a line drawn from this new origin to a point of interest makes with the positive x-axis. Compute also the length of this line.
The new x and y coordinates after zooming are defined by elongating this line, such that the new line is the zoom factor times the length of the original line.
Transform the newly found x and y coordinates back to a coordinate system that makes sense to the computer (e.g., top left pixel = 0,0)
Repeat for all points of interest.
In pseudo-code (with formulas):
x_center = image_width/2
y_center = image_height/2
x_from_zoom_center = x_from_topleft - x_center
y_from_zoom_center = y_from_topleft - y_center
angle = atan2(y_from_zoom_center, x_from_zoom_center)
length = hypot(x_from_zoom_center, y_from_zoom_center)
x_new = zoom_factor * length * cos(angle)
y_new = zoom_factor * length * sin(angle)
x_new_topleft = x_new + x_center
y_new_topleft = y_new + y_center
Note that this assumes the number of pixels used for length and width stays the same after zooming. Note also that some rounding should take place (keep everything double precision, and only round to int after all calculations)
In the code above, atan2 is the four-quadrant arctangent, available in most programming languages, and hypot is simply sqrt(x*x + y*y), but then computed more carefully (e.g., to avoid overflow etc.), also available in most programing languages.
Is this indeed what you were after?