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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.

Calculating the Area of a Closed Polygon on a Plane

I'm attempting to calculate the area of a polygon that lies on a plane (a collection co-planar points forming a non-intersecting closed shape), and I know a method that can calculate the area of an irregular (or any) polygon in two dimensions - but not three. My solution is to rotate the plane so that it's normal is 0 in the z direction (so I can treat it like it's 2D) and then run the 2D area function.
The problem is I have NO idea how to actually determine the rotation axes and amounts to flatten a plane on it's Z-axis. I do my rotation through the easiest method I could find for 3 dimensional rotation: Rotation Matrices. So, given that I'm trying to use rotation matrices to do my rotation, how do I figure out the angles to rotate my plane by to be oriented in the same direction as another vector? I don't actually know much calculus or Euclidean geometry, so whichever solution requires me to teach myself the least of both is the ideal solution. Is there a better way?
Here's my attempt below, which doesn't even come close to getting the plane flat on the Z axis. This is an instance method of my "Surface" class, which is a derivative of my "Plane" class, and has an array of co-planar points (IntersectPoints) forming a closed polygon.
public virtual double GetArea()
{
Vector zUnit = new Vector(0, 0, 1); //vector perprendicualr to z
Vector nUnit = _normal.AsUnitVector();
Surface tempSurface = null;
double result = 0;
if (nUnit != zUnit && zUnit.Dot(nUnit) != 0) //0 = perprendicular to z
{
tempSurface = (Surface)Clone();
double xAxisAngle = Vector.GetAxisAngle(nUnit, zUnit, Physics.Formulae.Axes.X);
double yAxisAngle = Vector.GetAxisAngle(nUnit, zUnit, Physics.Formulae.Axes.Y);
double rotationAngle = Vector.GetAxisAngle(nUnit, zUnit, Physics.Formulae.Axes.Z);
tempSurface.Rotate(xAxisAngle, yAxisAngle, rotationAngle); //rotating plane so that it is flat on the Z axis
}
else
{
tempSurface = this;
}
for (int x = 0; x < tempSurface.IntersectPoints.Count; x++) //doing a cross sum of each point
{
Point curPoint = tempSurface.IntersectPoints[x];
Point nextPoint;
if (x == tempSurface.IntersectPoints.Count - 1)
{
nextPoint = tempSurface.IntersectPoints[0];
}
else
{
nextPoint = tempSurface.IntersectPoints[x + 1];
}
double cross1 = curPoint.X * nextPoint.Y;
double cross2 = curPoint.Y * nextPoint.X;
result += (cross1 - cross2); //add the cross sum of each set of points to the result
}
return Math.Abs(result / 2); //divide cross sum by 2 and take its absolute value to get the area.
}
And here are my core rotation and get axis angle methods:
private Vector Rotate(double degrees, int axis)
{
if (degrees <= 0) return this;
if (axis < 0 || axis > 2) return this;
degrees = degrees * (Math.PI / 180); //convert to radians
double sin = Math.Sin(degrees);
double cos = Math.Cos(degrees);
double[][] matrix = new double[3][];
//normalizing really small numbers to actually be zero
if (Math.Abs(sin) < 0.00000001)
{
sin = 0;
}
if (Math.Abs(cos) < 0.0000001)
{
cos = 0;
}
//getting our rotation matrix
switch (axis)
{
case 0: //x axis
matrix = new double[][]
{
new double[] {1, 0, 0},
new double[] {0, cos, sin * -1},
new double[] {0, sin, cos}
};
break;
case 1: //y axis
matrix = new double[][]
{
new double[] {cos, 0, sin},
new double[] {0, 1, 0},
new double[] {sin * -1, 0, cos}
};
break;
case 2: //z axis
matrix = new double[][]
{
new double[] {cos, sin * -1, 0},
new double[] {sin, cos, 0},
new double[] {0, 0, 1}
};
break;
default:
return this;
}
return Physics.Formulae.Matrix.MatrixByVector(this, matrix);
}
public static double GetAxisAngle(Point a, Point b, Axes axis, bool inDegrees = true)
{ //pretty sure this doesnt actually work
double distance = GetDistance(a, b);
double difference;
switch (axis)
{
case Axes.X:
difference = b.X - a.X;
break;
case Axes.Y:
difference = b.Y - a.Y;
break;
case Axes.Z :
difference = b.Z - a.Z;
break;
default:
difference = 0;
break;
}
double result = Math.Acos(difference / distance);
if (inDegrees == true)
{
return result * 57.2957; //57.2957 degrees = 1 radian
}
else
{
return result;
}
}

A robust way to do this is to do a sum of the cross-products of the vertices of each edge. If your vertices are co-planar, this will produce a normal to the plane, whose length is 2 times the area of the closed polygon.
Note that this method is very similar to the 2D method linked in your question, which actually calculates a 2D equivalent of the 3D cross-product, summed for all edges, then divides by 2.
Vector normal = points[count-1].cross(points[0]);
for(int i=1; i<count; ++i) {
normal += points[i-1].cross(points[i]);
}
double area = normal.length() * 0.5;
Advantages of this method:
If your vertices are only approximately planar, it still gives the right answer
It doesn't depend on the angle of the plane.
In fact you don't need to deal with the angle at all.
If you want to know the plane orientation, you've got the normal already.
One possible difficulty: if your polygon is very small, and a long way away from the origin, you can get floating point precision problems. If that case is likely to arise, you should first translate all of your vertices so that one is at the origin, like so:
Vector normal(0,0,0);
Vector origin = points[count-1];
for(int i=1; i<count-1; ++i) {
normal += (points[i-1]-origin).cross(points[i]-origin);
}
double area = normal.length() * 0.5;

You need not to rotate the plane (or all points). Just calculate an area of polygon projection to Z-plane (if it is not perpendicular to polygon plane), for example, with you GetArea function, and divide result by cosinus of Poly-plane - Z-plane angle - it is equal to scalar product of zUnit and nUnit (I suggest that nUnit is normal vector to polygon plane)
TrueArea = GetArea() / zUnit.Dot(nUnit)

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;
}

c# rotation problem

I have 3 particles and one of them is the center particle. I want to rotate other two particle ( stored in particles list ) relative to the center particle with the formula q' = Θq + p where q' is the new position of the rotated particle, Θ is the orientation angle and p is the position of center particle. The initial position of other two particles is stored in initialParticlePosition list. THe problem is I think the angle I calculate is wrong because of the range. I thing I should take the range as [-pi, pi) or something like this. In some parts it calculates correct but sometimes it is wrong. Can someone help me with this code or give me another method of rotating.
{
angle = Math.Acos(Vector2.Dot(heading,new Vector2(0,-1) ));
for (int i = 0; i < 2; i++)
{
tempX = (double)initialParticlePositions[i].X * Math.Cos(angle) - (double)initialParticlePositions[i].Y * Math.Sin(angle) + centerParticle.position.x;
tempY = (double)initialParticlePositions[i].X * Math.Sin(angle) + (double)initialParticlePositions[i].Y * Math.Cos(angle) + centerParticle.position.y;
particles[i].position.x = tempX;
particles[i].position.y = tempY;
}
}

Some methods that might help (angles always in degrees, not rad):
public static double GetAngle(Vector v)
{
return Math.Atan2(v.X, -v.Y) * 180.0 / Math.PI;
}
public static Vector SetAngle(Vector v, double angle)
{
var angleInRads = angle * (Math.PI / 180.0);
var distance = v.Length;
v.X = (Math.Sin(angleInRads) * distance);
v.Y = -(Math.Cos(angleInRads) * distance);
return v;
}
static public Point RotatePointAroundCenter(Point point, Point center, double rotationChange)
{
Vector centerToPoint = point - center;
double angle = GetAngle(centerToPoint);
Vector centerToNewPoint = SetAngle(centerToPoint, angle + rotationChange);
return center + centerToNewPoint;
}
(You should start marking answers that help as answer, click the checkmark outline below the votes on the left, e.g. you could accept this answer)
Edit: Optimized the methods a bit.

The particle positions that are orbiting can be set with a single line of code each:
Assume p1, p2, & p3 are Vector2s and p2 & p3 are orbiting p1.
p2 = Vector2.Transform(p2 - p1, Matrix.CreateRotationZ(rotationChangeP2)) + p1;
p3 = Vector2.Transform(p3 - p1, Matrix.CreateRotationZ(rotationChangeP3)) + p1;
The Matrix.Create...() method will call the two trig functions for you.
edit. the Matrix & Vector2 structures & methods are XNA specific but included here because that's what the OP tagged his Q with.

angle = Math.Acos(Vector2.Dot(heading,new Vector2(0,-1)));
As you suspect, your combination of dot product and Acos will only give you angles in a 180
degree range.
Instead, use Atan2 on your unit vector to get a full range of angles from -pi to pi.
angle = (float)Math.Atan2((double)heading.Y, (double)heading.X);
You may need to negate the Y term if your Y axis is positive in the down direction.

game programming in C#: sprite collision

I have a C# snippet code about sprite collision in C# game programming, hope you guys will help me to clarify it.
I dont understand method IsCollided, especially the calculation of d1 and d2 to determine whether the sprites collide or not, the meaning and the use of Matrix Invert() and Multiphy in this case as well as the use of Color alpha component to determine the collision of 2 sprites.
Thank you very much.
public struct Vector
{
public double X;
public double Y;
public Vector(double x, double y)
{
X = x;
Y = y;
}
public static Vector operator -(Vector v, Vector v2)
{
return new Vector(v.X-v2.X, v.Y-v2.Y);
}
public double Length
{
get
{
return Math.Sqrt(X * X + Y * Y);
}
}
}
public class Sprite
{
public Vector Position;
protected Image _Image;
protected Bitmap _Bitmap;
protected string _ImageFileName = "";
public string ImageFileName
{
get { return _ImageFileName; }
set
{
_ImageFileName = value;
_Image = Image.FromFile(value);
_Bitmap = new Bitmap(value);
}
}
public Matrix Transform
{
get
{
Vector v = Position;
if (null != _Image)
v -= new Vector(_Image.Size) / 2;
Matrix m = new Matrix();
m.RotateAt(50.0F, new PointF(10.0F, 100.0F));
m.Translate((float)v.X, (float)v.Y);
return m;
}
}
public bool IsCollided(Sprite s2)
{
Vector v = this.Position - s2.Position;
double d1 = Math.Sqrt(_Image.Width * _Image.Width + _Image.Height * _Image.Height)/2;
double d2 = Math.Sqrt(s2._Image.Width * s2._Image.Width + s2._Image.Height * s2._Image.Height)/2;
if (v.Length > d1 + d2)
return false;
Bitmap b = new Bitmap(_Image.Width, _Image.Height);
Graphics g = Graphics.FromImage(b);
Matrix m = s2.Transform;
Matrix m2 = Transform;
m2.Invert();
Matrix m3 = m2;
m3.Multiply(m);
g.Transform = m3;
Vector2F v2 = new Vector2F(0,0);
g.DrawImage(s2._Image, v2);
for (int x = 0; x < b.Width; ++x)
for (int y = 0; y < b.Height; ++y)
{
Color c1 = _Bitmap.GetPixel(x, y);
Color c2 = b.GetPixel(x, y);
if (c1.A > 0.5 && c2.A > 0.5)
return true;
}
return false;
}
}

It's a little convoluted. The two parts:
Part One (simple & fast)
The first part (involving v, d1 + d2) is probably confusing because, instead of doing collision tests using boxes, the dimensions of the image are used to construct bounding circles instead. A simple collision test is then carried out using these bounding circles. This is the 'quick n dirty' test that eliminates sprites that are clearly not colliding.
"Two circles overlap if the sum of there(sic) radii is greater than the distance between their centers. Therefore by Pythagoras we have a collision if:
(cx1-cx2)2 + (cy1-cy2)2 < (r1+r2)2"
See the "Bounding Circles" section of this link and the second link at the foot of this answer.
commented code:
// Get the vector between the two sprite centres
Vector v = this.Position - s2.Position;
// get the radius of a circle that will fit the first sprite
double d1 = Math.Sqrt(_Image.Width * _Image.Width + _Image.Height * _Image.Height)/2;
// get the radius of a circle that will fit the second sprite
double d2 = Math.Sqrt(s2._Image.Width * s2._Image.Width + s2._Image.Height * s2._Image.Height)/2;
// if the distance between the sprites is larger than the radiuses(radii?) of the circles, they do not collide
if (v.Length > d1 + d2)
return false;
Note: You may want to considering using an axially aligned bounding box test instead of a circle here. if you have rectangles with disparate widths and lengths, it'll be a more effective/accurate first test.
Part Two (slower)
I haven't got time to check the code 100%, but the second part --having established that the two sprites are potentially colliding in step one-- is performing a more complicated collision test. It is performing a per-pixel collision test using the sprite's image source -- specifically its alpha channel. In games, the alpha channel is often used to store a representation of transparency. The larger the value, the more opaque the image (so 0 would be 100% transparent, 1.0f would be 100% opaque).
In the code shown, if the pixels in both sprites are overlapping and both have alpha values of > 0.5f, the pixels are sharing the same space and represent solid geometry and are thus colliding. By doing this test, it means that transparent (read: invisible) parts of the sprite will not be considered when testing for collisions, so you can have circular sprites that do not collide at the corners etc.
Here's a link that covers this in a bit more detail.