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Math: Mercator Projection Convert 360 video pixel coordinate to sphere surface coordinate

I was trying to map the 360 video pixel coordinate to sphere surface coordinate but I couldn't get right result... It just mapped to the wrong position I already know the points of the XY data for 360 video pixels.
how map 2d grid points (x,y) onto sphere as 3d points (x,y,z)
I checked this link and I copied method from this but what I'm getting is not mapped to the desired position.
How can I get radius from the pixels?
I am not sure if I'm passing right radius for imageRadius but I thought it will be circumference/PI to get radius and the video ratio is 4096x2048. I also tried to pass the number 1 because UV is 0-1 but it was not right...
Is Method wrong?
Maybe the method is wrong. I passed random numbers into the imageRadius but couldn't get the right position... If I make X to negative number the seems like little bit closer to result....?
Current Result
https://youtu.be/t0I7Hlb-tbk
It mapped to up right position with the method that I found online...
Project File
https://drive.google.com/a/swordfish-sf.com/file/d/0B45RYzVs0t0_VVdaaHdmNHRWTk0/view?usp=sharing
If somebody can check the Unity project file that will be great...
Current Code
public class mapScript : MonoBehaviour {
public int input = 4098;
float imageRadius = 4098f / Mathf.PI; //2098? 3072? 4098?
float radius;
public GameObject testSphere;
void Start () {
radius = this.transform.localScale.x;
}
void Update () {
imageRadius = input / Mathf.PI;
int currentFrame = (int)this.GetComponent<VideoPlayer>().frame;
testSphere.transform.position = MercatorProjection(mapVals[currentFrame,0],mapVals[currentFrame,1]);
}
Vector3 MercatorProjection(float xVal, float yVal)
{
float lon = (xVal / imageRadius);
float lat = (2 * Mathf.Atan(Mathf.Exp(yVal / imageRadius)) - Mathf.PI / 2);
float calcX = radius * Mathf.Cos(lat) * Mathf.Cos(lon);
float calcY = radius * Mathf.Cos(lat) * Mathf.Sin(lon);
float calcZ = radius * Mathf.Sin(lat);
Vector3 result = new Vector3(calcX,calcY,calcZ);
Debug.Log(result);
return result;
}
float[,] mapVals = new float[,] {
{1969.21f, 928.625f},
{1969.6f, 928.533f},
{1968.92f, 928.825f},
{1968.68f, 929f},
{1968.47f, 929.067f},
{1968.41f, 929.025f},
{1968.48f, 928.992f},
....
};
}
Thank you.

As a side note, the radius is arbitrary. The pixel coordinates only map to the directional coordinates (polar [θ] and azimuthal [ϕ] angles).
We can do this by mapping each pixel to equal θ and ϕ intervals. The diagram below illustrates a low-resolution setup:
Let us adopt the convention that, for an image of with W, ϕ = 0 corresponds to:
Even W: half way between X = floor((W - 1) / 2) and X = ceil((W - 1) / 2)
Odd W: in the middle of the pixel column at X = floor((W - 1) / 2)
The pixel row at Y maps to the equilatitudinal line at θ = (Y + 0.5) / H * π.
To map all pixels in their entirety, let X start at -0.5 instead of 0, and end at W - 0.5; likewise for Y. Since integer coordinates map to the centers of the pixel regions shown above, this allows the whole area of any particular pixel to be addressed. You may need this later on if you plan on doing multi-sampling filtering for e.g. anti-aliasing.
Code:
Vector3 Mercator(float x, float y, int w, int h)
{
// outside of valid pixel region
if (x < -0.5f || x >= w - 0.5f || y < -0.5f || y >= h - 0.5f)
return new Vector3();
float theta = (y + 0.5f) / h * Math.PI;
float phi = ((x + 0.5f) / w - 0.5f) * 2.0 * Math.PI;
float c_t = Math.Cos(theta);
return new Vector3(c_t * Math.Cos(phi), c_t * Math.Sin(phi), Math.Sin(theta));
}
... and multiply the resulting direction vector by any "radius" you like, since it has (basically) nothing to do with the mapping anyway.

Drawing Parabola using DrawLine C#

Good evening, I know on the web there are similar questions and a few tutorials about it, but I'd like you to check my code and correct it. I mean, I'd like to know what's wrong with my project.
I have to draw a parabola graph given its equation on my main panel.
I also must include two buttons, zoom in and zoom out, which are used to reduce and enlarge the "view" panel's view (and so the parabola).
I was recommended to use a scale var.
This is my code:
note: x0, y0 are panel_main x center, y center.
I have x, y that are used to determine x,y from the equation.
xpc, ypc are converted for the window scale (so are pixels).
xmin, xmax are the extreme values that, with a certain scale, stay on the panel
I hope you can give me a hint, thanks a lot!
public void DisegnaParabola()
{
Graphics gs = panel_main.CreateGraphics();
pen.Color = Color.Black;
scale = (x0*2) / zoom; //Pixels equivalent to 1x or 1y
n_punti = (x0*2) / scale; //Number of x math points that are visible in window
xmin = -(n_punti / 2);
xmax = n_punti / 2;
precision = 1 / scale; //Increment of x to have 1px
if (asse_parabola.SelectedIndex == 0) //if Y axis
{
for (double i = xmin + precision; i < xmax; i += precision)
{
rifx = i - precision; //Old points
rifxpc = rifx * scale;
rify = (a * Math.Pow(rifx, 2)) + b * rifx + c;
rifypc = y0 - (rify * scale);
x = i; //New points
y = (a * Math.Pow(x, 2)) + b * x + c;
ypc = y0 - (y * scale);
gs.DrawLine(pen, (float)rifxpc, (float)rifypc, (float)xpc, (float)ypc);
}
}
else
{
scale = (y0*2) / zoom; //Pixels for 1y
n_punti = (y0*2) / scale; //Numbers of y in the window
ymin = -(n_punti / 2);
ymax = n_punti / 2;
for(double i=ymin+precision; i<ymax; i+=precision)
{
rify = y - precision;
rifypc = (y0*2) - rify * scale;
rifx = (a * Math.Pow(rify, 2)) + b * rify + c;
rifxpc = x0 + (rifx * scale);
y = i;
x = (a * Math.Pow(y, 2)) + b * y + c;
xpc = x0 + (x * scale);
gs.DrawLine(pen, (float)rifypc, (float)rifxpc, (float)ypc, (float)xpc);
}
}
lbl_canc.Visible = true;
}

Your question actually consists of several tasks and as usual the key is to take and break those apart..
One issue is getting the data, I will leave the details to you but show how to sparate it from the rest.
The next issue is to scale the data. I'll show you how to avoid this one altogether and scale the drawing tool instead.
And the third one is to draw them to a display surface. As you'll see this is really simple once the other issues are taken care of.
Let's start with the most important step: Collecting the data. You try to create and scale and draw them all in the same piece of code. This has many disadvantages..
Let's first collect the data in a suitable structure:
List<PointF> points = new List<PointF>();
List<T> is the collection of choice most of the time; certainly much nicer than arrays! In some method you should fill that list with your data, calculated from some formula.
Here is an example:
List<PointF> getPoints(float start, float end, int count, float ymax)
{
List<PointF> points = new List<PointF>();
float deltaX = (end - start) / count;
for (int i = 0; i < count; i++)
{
float x = i * deltaX;
// insert your own formula(s) here!
float y = ymax + (float)Math.Sin(x * somefactor) * ymax;
points.Add(new PointF(x, y));
}
return points;
}
Now for the second important part: How to scale the data? This can be done either when creating them; but again, separating the two taks makes them both a lot simpler.
So here is a function that, instead of scaling the data scales the Graphics object we will use to plot them:
void ScaleGraphics(Graphics g, List<PointF> data)
{
float xmax = data.Select(x => x.X).Max();
float ymax = data.Select(x => x.Y).Max();
float xmin = data.Select(x => x.X).Min();
float ymin = data.Select(x => x.Y).Min();
float width = Math.Abs(xmax - xmin);
float height = Math.Abs(ymax - ymin);
var vr = g.VisibleClipBounds;
g.ScaleTransform(vr.Width / width, vr.Height / height);
}
This method makes sure that all the data in our list will fit into the drawing surface. If you want to restrict them to a different size you can pass it in and change the code accordingly..
Finally we need to do the actual drawing. We do that where we should, that is in the Paint event of our drawing surface control..:
private void panel1_Paint(object sender, PaintEventArgs e)
{
if (points.Count < 2) return; // no lines to draw, yet
ScaleGraphics(e.Graphics, points);
e.Graphics.SmoothingMode = SmoothingMode.AntiAlias;
using ( Pen pen = new Pen(Color.Blue )
{ Width = 1.5f , LineJoin = LineJoin.Round, MiterLimit = 1f} )
e.Graphics.DrawLines(pen, points.ToArray());
}

Clockwise right hand turn angle in 2D?

If I have two points p1 and p2 where p1 is the pivot point and p2 is the original direction the user was headed and they have a number of possible directions to go p3...pn in random sequence. How do I get the angles between the choices and the segment formed by p1,p2 as clockwise(right hand) positive values between 0 and 360 so that I can sort them from least to greatest?
Also the points p1...pn will be in any quadrant, I can’t assume they will always be in the positive x,y direction. The grid is a standard Cartesian grid not screen coordinates so Y gets smaller as you go down not larger.
So in this example (sorry for the poor drawing but Paint was all I had on my laptop) I need to get the angles:
(p2-p1-p3)
( p2-p1-p4)
( p2-p1-p5)
( p2-p1-p6)
In this order(smallest right hand turn to largest right hand turn):
[( p2-p1-p4), ( p2-p1-p6), ( p2-p1-p5), (p2-p1-p3)]
The points in my case are a class called Vertex:
public class Vertex
{
public double X = 0;
public double Y = 0;
public Vertex() { }
public Vertex(double x, double y)
{
X = x;
Y = y;
}
}
And the code for getting the angles and sorting looks like this right now but has a problem:
private static IEnumerable<Vertex> SortByAngle(Vertex original, Vertex pivot, List<Vertex> choices)
{
choices.Sort((v1, v2) => GetTurnAngle(original, pivot, v1).CompareTo(GetTurnAngle(original, pivot, v2)));
return choices;
}
private static double GetTurnAngle(Vertex original, Vertex pivot, Vertex choice)
{
var a = original.X - pivot.X;
var b = original.Y - pivot.Y;
var c = choice.X - pivot.X;
var d = choice.Y - pivot.Y;
var rads = Math.Acos(((a * c) + (b * d)) / ((Math.Sqrt(a * a + b * b)) * (Math.Sqrt(c * c + d * d))));
return (180 / Math.PI * rads);
}
The problem is the above is if I check it for:
original 66,-66
pivot 280,-191
choice 200,-180
I get an angle of 22.460643124 instead of 337.539356876 which means it went counter-clockwise from the original direction to get that angle. I need it to always go clockwise to get the angle.
What am I doing wrong and how do I fix it?
Update: OK so according to what you guys are saying I can probably use some cross product like math to determine CW vs CCW so the new method would look like this:
private static double GetTurnAngle(Vertex original, Vertex pivot, Vertex choice)
{
var a = original.X - pivot.X;
var b = original.Y - pivot.Y;
var c = choice.X - pivot.X;
var d = choice.Y - pivot.Y;
var angle = Math.Acos(((a * c) + (b * d)) / ((Math.Sqrt(a * a + b * b)) * (Math.Sqrt(c * c + d * d))));
angle = (180 / Math.PI * angle);
var z = (choice.X - pivot.X) * (original.Y - pivot.Y) - (choice.Y - pivot.Y) * (original.X - pivot.X);
if (z < 0)
{
return 360 - angle;
}
return angle;
}
Update 2:
Using the accepted solution it now looks like so:
private static double GetTurnAngle(Vertex original, Vertex pivot, Vertex choice)
{
var angle1 = Math.Atan2(original.Y - pivot.Y, original.X - pivot.X);
var angle2 = Math.Atan2(choice.Y - pivot.Y, choice.X - pivot.X);
var angleDiff = (180 / Math.PI * (angle2 - angle1));
if (angleDiff > 0)//It went CCW so adjust
{
return 360 - angleDiff;
}
return -angleDiff;//I need the results to be always positive so flip sign
}
So far as I can tell that works great so far. Thank you guys for the help!

Take a look at atan2 function. It takes delta y and delta x, so can distinguish all angles.
angle1 = atan2(p1.y-p0.y, p1.x-p0.x);
angle2 = atan2(p2.y-p0.y, p2.x-p0.x);
angle = angle2 - angle1;
If angle is negative, then CW, if positive CCW (or other way around depending on your axis orientation). Note |angle| may be > 180, in which case you may want to do 360-|angle| and reverse the CW CCW conclusion if you're after the shortest route.

You find the Dn=direction from p1 to pn (x=pn.x-p1.x and y=pn.y-p1.y) by the formula:
Dn=f(x,y)=180-90*(1+sign(y))* (1-sign(x^2))-45*(2+sign(y))*sign(x)
-180/pi()*sign(x*y)*atan((abs(y)-abs(x))/(abs(y)+abs(x)))
So the angles are Angle(p2-p1-pn)=Dn-D2.

Rotating point by angle

I know that the theory of rotating a point by an angle is on the internet a million times, but I don't get my code to work properly.
I have a line with 2 points, when you click on 1 of the 2 points, you will rotate the point relative to the other point. In my testcase I have a upper left point and a bottom right point, so a diagonal line.
I want to make sure the line snaps to a 90 degrees rotation so It will always be a straight line (either vertically or horizontally). So I first get the current angle, then get the angle that It should be, and calculate the difference.
Point startPoint = obj2.Location;
Point currentEndPoint = new Point(obj2.Location.X + obj2.Size.Width, obj2.Location.Y + obj2.Size.Height);
Point newEndPoint = e.Location;
double angle = MathHelper.GetAngleOfVerticalLine(startPoint, newEndPoint);
double angleToBe = MathHelper.GetClosestNumber(angle, new double[] { 0, 90, 180, 270, 360 });
double angleToDo = 0.0; // -5
if (angle < angleToBe)
{
angleToDo = Math.Abs(angle - angleToBe);
}
else
{
angleToDo = angleToBe - angle;
}
angleToDo %= 360;
Point newSize = MathHelper.RotatePoint(newEndPoint, startPoint, angleToDo);
obj.Size = (Size)newSize;
public static Point RotatePoint(Point pointToRotate, Point centerPoint, double angleInDegrees)
{
double angleInRadians = angleInDegrees * (Math.PI / 180);
double cosTheta = Math.Cos(angleInRadians);
double sinTheta = Math.Sin(angleInRadians);
return new Point
{
X =
(int)
(cosTheta * (pointToRotate.X - centerPoint.X) -
sinTheta * (pointToRotate.Y - centerPoint.Y) + centerPoint.X),
Y =
(int)
(sinTheta * (pointToRotate.X - centerPoint.X) +
cosTheta * (pointToRotate.Y - centerPoint.Y) + centerPoint.Y)
};
}
But the results that I get, are not straight lines but they are kind of random. The angle, angleToBe and angleToDo work properly. The RotatePoint method should be the problem then, but I'm not 100% sure about that.

Can't you use the Matrix.Rotate class to do the heavy lifting? Source: http://msdn.microsoft.com/en-us/library/s0s56wcf.aspx (Of course the math is half the fun).

finding height on a heightmap stretched over a sphere C#

I'm looking for a bit of math help. I have a game were a 2D heightmap is generated and then stretched over a sphere using a length/direction formula. Now I need to know how to calculate the height between 2 points on my heightmap.
What I know:
The array that holds the heightmap
The angle in radians to my object
how many points there are on the heightmap
My problem look somewhat like so:
image
more images
The red and blue lines are the 2 heightmap points, and the light blue is where I'd like to calculate the height at.
Here's my current code to do it, but it doesn't work to well.
public double getheight(double angle)
{
//find out angle between 2 heightmap point
double offset = MathHelper.TwoPi / (heightmap.Length - 1);
//total brainfart attempt
double lowerAngle = offset * angle;
double upperAngle = offset * angle + offset;
//find heights
double height1 = heightmap[(int)lowerAngle];
double height2 = heightmap[(int)upperAngle];
//find offset angle
double u = angle - lowerAngle / (upperAngle - lowerAngle);
//return the height
return height1 + (height1 - height2) * u;
}
from my vegetation code, this seems to work okay, but is to rough to use for units and such, as they jump up/down as they move, due to it using only 1 heightmap point.
double[] hMap = planet.getHeightMap();
double i = hMap.Length / (Math.PI * 2);
this.height = hMap[(int)(angle * i)];

EDIT: example at end based on additional question info
Sounds to me like a linear interpolation - if you look at it from a 2d point of view, you've got two points:
(x1, y1) = point one on heightmap
(x2, y2) = point two on heightmap
and one point somewhere between (x1,x2) at an unknown height:
pu = (xu, yu)
A generic formula for LERP is:
pu = p0 + (p1 - p0) * u
where:
p0 = first value
p1 = second value
u = % where your unknown point lies between (p0,p1)
Here, we'll say p0 == y2 and p1 == y1. Now we need to determine "how far" the unknown point is between x1 and x2 - if you know the angles to the two heightmap points, this is easy:
u = ang(xu) - ang(x1) / (ang(x2) - ang(x1))
Alternatively, you could project your angle out to Max(y1,y2) and get the "unknown x pos" that way, then calculate the above.
So, let's try a contrived example:
p1 = point one in map = (1,2) therefore ang(p1) ~ 57 degrees
p2 = point two in map = (2,4) therefore ang(p2) ~ 114 degrees
note that here, the "x axis" is along the surface of the sphere, and the "y-axis" is the distance away from the center.
pu = object location = py #angle 100 degrees ~ 1.74 radians
px = (1.74 rad - 1 rad ) / (2 rad - 1 rad) = 0.74 / 1.0 = 0.74 => 74%
py = y0 + (y1 - y0) * u
= 2 + (4 - 2) * 0.74
= 2.96
Hopefully I didn't drop or misplace a sign there somewhere... :)
Ok, your example code - I've tweaked it a bit, here's what I've come up with:
First, let's define some helpers of my own:
public static class MathHelper
{
public const double TwoPi = Math.PI * 2.0;
public static double DegToRad(double deg)
{
return (TwoPi / 360.0) * deg;
}
public static double RadToDeg(double rad)
{
return (360.0 / TwoPi) * rad;
}
// given an upper/lower bounds, "clamp" the value into that
// range, wrapping over to lower if higher than upper, and
// vice versa
public static int WrapClamp(int value, int lower, int upper)
{
return value > upper ? value - upper - 1
: value < lower ? upper - value - 1
: value;
}
}
Our Test setup:
void Main()
{
var random = new Random();
// "sea level"
var baseDiameter = 10;
// very chaotic heightmap
heightmap = Enumerable
.Range(0, 360)
.Select(_ => random.NextDouble() * baseDiameter)
.ToArray();
// let's walk by half degrees, since that's roughly how many points we have
for(double i=0;i<360;i+=0.5)
{
var angleInDegrees = i;
var angleInRads = MathHelper.DegToRad(i);
Console.WriteLine("Height at angle {0}°({1} rad):{2} (using getheight:{3})",
angleInDegrees,
angleInRads,
heightmap[(int)angleInDegrees],
getheight(angleInRads));
}
}
double[] heightmap;
And our "getheight" method:
// assume: input angle is in radians
public double getheight(double angle)
{
//find out angle between 2 heightmap point
double dTheta = MathHelper.TwoPi / (heightmap.Length);
// our "offset" will be how many dThetas we are
double offset = angle / dTheta;
// Figure out two reference points in heightmap
// THESE MAY BE THE SAME POINT, if angle ends up
// landing on a heightmap index!
int lowerAngle = (int)offset;
int upperAngle = (int)Math.Round(
offset,
0,
MidpointRounding.AwayFromZero);
// find closest heightmap points to angle, wrapping
// around if we go under 0 or over max
int closestPointIndex = MathHelper.WrapClamp(
lowerAngle,
0,
heightmap.Length-1);
int nextPointIndex = MathHelper.WrapClamp(
upperAngle,
0,
heightmap.Length-1);
//find heights
double height1 = heightmap[closestPointIndex];
double height2 = heightmap[nextPointIndex];
// percent is (distance from angle to closest angle) / (angle "step" per heightmap point)
double percent = (angle - (closestPointIndex * dTheta)) / dTheta;
// find lerp height = firstvalue + (diff between values) * percent
double lerp = Math.Abs(height1 + (height2 - height1) * percent);
// Show what we're doing
Console.WriteLine("Delta ang:{0:f3}, Offset={1:f3} => compare indices:[{2}, {3}]",
dTheta,
offset,
closestPointIndex,
nextPointIndex);
Console.WriteLine("Lerping {0:p} between heights {1:f4} and {2:f4} - lerped height:{3:f4}",
percent,
height1,
height2,
lerp);
return lerp;
}