Related
I found some useful coordinate conversion code at https://gist.github.com/govert/1b373696c9a27ff4c72a
However, there is a bit specifically in the EcefToEnu function that I'm not clear on
// Converts the Earth-Centered Earth-Fixed (ECEF) coordinates (x, y, z) to
// East-North-Up coordinates in a Local Tangent Plane that is centered at the
// (WGS-84) Geodetic point (lat0, lon0, h0).
public static void EcefToEnu(double x, double y, double z,
double lat0, double lon0, double h0,
out double xEast, out double yNorth, out double zUp)
{
// Convert to radians in notation consistent with the paper:
var lambda = DegreesToRadians(lat0);
var phi = DegreesToRadians(lon0);
var s = Sin(lambda);
var N = a / Sqrt(1 - e_sq * s * s);
var sin_lambda = Sin(lambda);
var cos_lambda = Cos(lambda);
var cos_phi = Cos(phi);
var sin_phi = Sin(phi);
double x0 = (h0 + N) * cos_lambda * cos_phi;
double y0 = (h0 + N) * cos_lambda * sin_phi;
double z0 = (h0 + (1 - e_sq) * N) * sin_lambda;
double xd, yd, zd;
xd = x - x0;
yd = y - y0;
zd = z - z0;
// This is the matrix multiplication
xEast = -sin_phi * xd + cos_phi * yd;
yNorth = -cos_phi * sin_lambda * xd - sin_lambda * sin_phi * yd + cos_lambda * zd;
zUp = cos_lambda * cos_phi * xd + cos_lambda * sin_phi * yd + sin_lambda * zd;
}
I get the inputs, the first 4 conversion lines, the 4 sin and cos lines and I get the matrix multiplication - there are numerous examples with that in the algorithm. But what I'm not clear on is the part
double x0 = (h0 + N) * cos_lambda * cos_phi;
double y0 = (h0 + N) * cos_lambda * sin_phi;
double z0 = (h0 + (1 - e_sq) * N) * sin_lambda;
double xd, yd, zd;
xd = x - x0;
yd = y - y0;
zd = z - z0;
I don't recognize this section from any of the algorithms I've seen. It appears to be some sort of offset, but aside from that, I'm unclear where the formulas came from or what exactly this code is doing. Can someone please enlighten me as to what this bit of code is doing? I just want to understand what I'm looking at.
They are the conversion from geodetic coordinates (lat,long,height) aka (phi,lambda,h0) to ecef cartesians (x0,y0,z0) and then the computation of the ecef vector from (x0,y0,n0) to (x,y,z).
For the first part, note that if the ellipsoid were a sphere (e==0) then the first part would be the conversion from spherical polars to cartesians
I have a set of 2D Cartesian points [b], which proceed clockwise from the start and form a closed shape. Each one of these has its own companion 2D Cartesian points q0 and q1 which define the Beziér curve around the point (along with the preceding and succeeding points). Together, all these points define a closed 2D composite Beziér curve.
I have a separate point p which is an arbitrary 2D Cartesian point on the same plane. Is there a simple algorithm for finding the (x, y) coordinates of a new 2D Cartesian point q which is the closest point on the path to p?
As illustrated here, I have the points b[0] to b[3] and their handles b[n].q0 and b[n].q1, and I have the arbitrary point p. I'm trying to calculate the point q, not as a floating-point position along the curve, but as a pair of (x, y) coordinates.
I tried searching this, but some seemed to only be for a very small curve, others were way over my head with abstract mathematics and scientific research papers.
Any help that leads me toward an algorithmic solution is greatly appreciated, especially if it can be translated into a C-like language rather than the pure math in the above SO answers.
By adapting the algorithm posted by Tatarize, I came up with this solution in Swift, which should be translatable to other languages:
struct BezierPoint {
let q0: CGPoint
let point: CGPoint
let q1: CGPoint
}
struct SimpleBezierCurve {
let left: BezierPoint
let right: BezierPoint
}
class BezierPath {
var pathPoints = [BezierPoint]()
func findClosestPoint(to targetPoint: CGPoint) -> CGPoint {
let segments = allSegments()
guard segments.count > 0 else { return targetPoint }
var closestPoint = (distance: CGFloat.infinity, point: CGPoint(x: CGFloat.infinity, y: CGFloat.infinity))
segments.forEach{ curve in
let thisPoint = BezierPath.findClosestPoint(to: targetPoint, along: curve)
let distance = findDistance(from: targetPoint, to: thisPoint)
if (distance < closestPoint.distance) {
closestPoint = (distance: distance, point: thisPoint)
}
}
return closestPoint.point
}
func allSegments() -> [SimpleBezierCurve] {
guard pathPoints.count > 0 else { return [] }
var segments = [SimpleBezierCurve]()
var prevPoint = pathPoints[0]
for i in 1 ..< pathPoints.count {
let thisPoint = pathPoints[i]
segments.append(SimpleBezierCurve(left: prevPoint, right: thisPoint))
prevPoint = thisPoint
}
segments.append(SimpleBezierCurve(left: prevPoint, right: pathPoints[0]))
return segments
}
static func findClosestPoint(to point: CGPoint, along curve: SimpleBezierCurve) -> CGPoint {
return findClosestPointToCubicBezier(to: point, slices: 10, iterations: 10, along: curve)
}
// Adapted from https://stackoverflow.com/a/34520607/3939277
static func findClosestPointToCubicBezier(to target: CGPoint, slices: Int, iterations: Int, along curve: SimpleBezierCurve) -> CGPoint {
return findClosestPointToCubicBezier(iterations: iterations, to: target, start: 0, end: 1, slices: slices, along: curve)
}
// Adapted from https://stackoverflow.com/a/34520607/3939277
private static func findClosestPointToCubicBezier(iterations iterations: Int, to: CGPoint, start: CGFloat, end: CGFloat, slices: Int, along curve: SimpleBezierCurve) -> CGPoint {
if iterations <= 0 {
let position = (start + end) / 2
let point = self.point(for: position, along: curve)
return point
}
let tick = (end - start) / slices
var best = CGFloat(0)
var bestDistance = CGFloat.infinity
var currentDistance: CGFloat
var t = start
while (t <= end) {
//B(t) = (1-t)**3 p0 + 3(1 - t)**2 t P1 + 3(1-t)t**2 P2 + t**3 P3
let point = self.point(for: t, along: curve)
var dx = point.x - to.x;
var dy = point.y - to.y;
dx *= dx;
dy *= dy;
currentDistance = dx + dy;
if (currentDistance < bestDistance) {
bestDistance = currentDistance;
best = t;
}
t += tick;
}
return findClosestPointToCubicBezier(iterations: iterations - 1, to: to, start: max(best - tick, 0.0), end: min(best + tick, 1.0), slices: slices, along: curve);
}
static func point(for t: CGFloat, along curve: SimpleBezierCurve) -> CGPoint {
// This had to be broken up to avoid the "Expression too complex" error
let x0 = curve.left.point.x
let x1 = curve.left.q1.x
let x2 = curve.right.q0.x
let x3 = curve.right.point.x
let y0 = curve.left.point.y
let y1 = curve.left.q1.y
let y2 = curve.right.q0.y
let y3 = curve.right.point.y
let x = (1 - t) * (1 - t) * (1 - t) * x0 + 3 * (1 - t) * (1 - t) * t * x1 + 3 * (1 - t) * t * t * x2 + t * t * t * x3
let y = (1 - t) * (1 - t) * (1 - t) * y0 + 3 * (1 - t) * (1 - t) * t * y1 + 3 * (1 - t) * t * t * y2 + t * t * t * y3
return CGPoint(x: x, y: y)
}
}
// Possibly in another file
func findDistance(from a: CGPoint, to b: CGPoint) -> CGFloat {
return sqrt(pow(a.x - b.x, 2) + pow(a.y - b.y, 2));
}
GitHub Gist
I have a line that is based on two (x,y) coordinates I know. This line has a starting and an end point. Now I want to add an arrowhead at the end point of the line.
I know that the arrow is an equilateral triangle, and therefore each angle has 60 degrees. Additionally, I know the length of one side, which will be 20. I also no one edge of the triangle (that is the end point of the line).
How can I calculate the other two points of the triangle? I know I should use some trigonometry but how?
P.s. The endpoint of the line should be the arrowhead's tip.
You don't need trig., just some vector arithmetic...
Say the line goes from A to B, with the front vertex of the arrowhead at B. The length of the arrowhead is h = 10(√3) and its half-width is w = 10. We'll denote the unit vector from A to B as U = (B - A)/|B - A| (i.e., the difference divided by the length of the difference), and the unit vector perpendicular to this as V = [-Uy, Ux].
From these quantities, you can calculate the two rear vertices of the arrowhead as B - hU ± wV.
In C++:
struct vec { float x, y; /* … */ };
void arrowhead(vec A, vec B, vec& v1, vec& v2) {
float h = 10*sqrtf(3), w = 10;
vec U = (B - A)/(B - A).length();
vec V = vec(-U.y, U.x);
v1 = B - h*U + w*V;
v2 = B - h*U - w*V;
}
If you want to specify different angles, then you will need some trig. to calculate different values of h and w. Assuming you want an arrowhead of length h and tip-angle θ, then w = h tan(θ/2). In practice, however, it's simplest to specify h and w directly.
Here's a sample LINQPad program that shows how to do that:
void Main()
{
const int imageWidth = 512;
Bitmap b = new Bitmap(imageWidth , imageWidth , PixelFormat.Format24bppRgb);
Random r = new Random();
for (int index = 0; index < 10; index++)
{
Point fromPoint = new Point(0, 0);
Point toPoint = new Point(0, 0);
// Ensure we actually have a line
while (fromPoint == toPoint)
{
fromPoint = new Point(r.Next(imageWidth ), r.Next(imageWidth ));
toPoint = new Point(r.Next(imageWidth ), r.Next(imageWidth ));
}
// dx,dy = arrow line vector
var dx = toPoint.X - fromPoint.X;
var dy = toPoint.Y - fromPoint.Y;
// normalize
var length = Math.Sqrt(dx * dx + dy * dy);
var unitDx = dx / length;
var unitDy = dy / length;
// increase this to get a larger arrow head
const int arrowHeadBoxSize = 10;
var arrowPoint1 = new Point(
Convert.ToInt32(toPoint.X - unitDx * arrowHeadBoxSize - unitDy * arrowHeadBoxSize),
Convert.ToInt32(toPoint.Y - unitDy * arrowHeadBoxSize + unitDx * arrowHeadBoxSize));
var arrowPoint2 = new Point(
Convert.ToInt32(toPoint.X - unitDx * arrowHeadBoxSize + unitDy * arrowHeadBoxSize),
Convert.ToInt32(toPoint.Y - unitDy * arrowHeadBoxSize - unitDx * arrowHeadBoxSize));
using (Graphics g = Graphics.FromImage(b))
{
if (index == 0)
g.Clear(Color.White);
g.DrawLine(Pens.Black, fromPoint, toPoint);
g.DrawLine(Pens.Black, toPoint, arrowPoint1);
g.DrawLine(Pens.Black, toPoint, arrowPoint2);
}
}
using (var stream = new MemoryStream())
{
b.Save(stream, ImageFormat.Png);
Util.Image(stream.ToArray()).Dump();
}
}
Basically, you:
Calculate the vector of the arrow line
Normalize the vector, ie. making its length 1
Calculate the ends of the arrow heads by going:
First back from the head a certain distance
Then perpendicular out from the line a certain distance
Note that if you want the arrow head lines to have a different angle than 45 degrees, you'll have to use a different method.
The program above will draw 10 random arrows each time, here's an example:
Let's your line is (x0,y0)-(x1,y1)
Backward direction vector (dx, dy) = (x0-x1, y0-y1)
It's norm Norm = Sqrt(dx*dx+dy*dy)
Normalize it: (udx, udy) = (dx/Norm, dy/Norm)
Rotate by angles Pi/6 and -Pi/6
ax = udx * Sqrt(3)/2 - udy * 1/2
ay = udx * 1/2 + udy * Sqrt(3)/2
bx = udx * Sqrt(3)/2 + udy * 1/2
by = - udx * 1/2 + udy * Sqrt(3)/2
Your points: (x1 + 20 * ax, y1 + 20 * ay) and (x1 + 20 * bx, y1 + 20 * by)
I want to contribute my answer in C# based on Marcelo Cantos' answer since the algorithm works really well. I wrote a program to calculate the centroid of a laser beam projected on the CCD array. After the centroid is found, the direction angle line is drawn and I need the arrow head pointing at that direction. Since the angle is calculated, the arrow head would have to follow the angle in any of the direction.
This code gives you the flexibility of changing the arrow head size as shown in the pictures.
First you need the vector struct with all the necessary operators overloading.
private struct vec
{
public float x;
public float y;
public vec(float x, float y)
{
this.x = x;
this.y = y;
}
public static vec operator -(vec v1, vec v2)
{
return new vec(v1.x - v2.x, v1.y - v2.y);
}
public static vec operator +(vec v1, vec v2)
{
return new vec(v1.x + v2.x, v1.y + v2.y);
}
public static vec operator /(vec v1, float number)
{
return new vec(v1.x / number, v1.y / number);
}
public static vec operator *(vec v1, float number)
{
return new vec(v1.x * number, v1.y * number);
}
public static vec operator *(float number, vec v1)
{
return new vec(v1.x * number, v1.y * number);
}
public float length()
{
double distance;
distance = (this.x * this.x) + (this.y * this.y);
return (float)Math.Sqrt(distance);
}
}
Then you can use the same code given by Marcelo Cantos, but I made the length and half_width of the arrow head variables so that you can define that when calling the function.
private void arrowhead(float length, float half_width,
vec A, vec B, ref vec v1, ref vec v2)
{
float h = length * (float)Math.Sqrt(3);
float w = half_width;
vec U = (B - A) / (B - A).length();
vec V = new vec(-U.y, U.x);
v1 = B - h * U + w * V;
v2 = B - h * U - w * V;
}
Now you can call the function like this:
vec leftArrowHead = new vec();
vec rightArrowHead = new vec();
arrowhead(20, 10, new vec(circle_center_x, circle_center_y),
new vec(x_centroid_pixel, y_centroid_pixel),
ref leftArrowHead, ref rightArrowHead);
In my code, the circle center is the first vector location (arrow butt), and the centroid_pixel is the second vector location (arrow head).
I draw the arrow head by storing the vector values in the points for graphics.DrawPolygon() function in the System.Drawings. Code is shown below:
Point[] ppts = new Point[3];
ppts[0] = new Point((int)leftArrowHead.x, (int)leftArrowHead.y);
ppts[1] = new Point(x_cm_pixel,y_cm_pixel);
ppts[2] = new Point((int)rightArrowHead.x, (int)rightArrowHead.y);
g2.DrawPolygon(p, ppts);
You can find angle of line.
Vector ox = Vector(1,0);
Vector line_direction = Vector(line_begin.x - line_end.x, line_begin.y - line_end.y);
line_direction.normalize();
float angle = acos(ox.x * line_direction.x + line_direction.y * ox.y);
Then use this function to all 3 points using found angle.
Point rotate(Point point, float angle)
{
Point rotated_point;
rotated_point.x = point.x * cos(angle) - point.y * sin(angle);
rotated_point.y = point.x * sin(angle) + point.y * cos(angle);
return rotated_point;
}
Assuming that upper point of arrow's head is line's end it will perfectly rotated and fit to line.
Didn't test it =(
For anyone that is interested, #TomP was wondering about a js version, so here is a javascript version that I made. It is based off of #Patratacus and #Marcelo Cantos answers. Javascript doesn't support operator overloading, so it isn't as clean looking as C++ or other languages. Feel free to offer improvements.
I am using Class.js to create classes.
Vector = Class.extend({
NAME: "Vector",
init: function(x, y)
{
this.x = x;
this.y = y;
},
subtract: function(v1)
{
return new Vector(this.x - v1.x, this.y - v1.y);
},
add: function(v1)
{
return new Vector(this.x + v1.x, this.y + v1.y);
},
divide: function(number)
{
return new Vector(this.x / number, this.y / number);
},
multiply: function(number)
{
return new Vector(this.x * number, this.y * number);
},
length: function()
{
var distance;
distance = (this.x * this.x) + (this.y * this.y);
return Math.sqrt(distance);
}
});
And then a function to do the logic:
var getArrowhead = function(A, B)
{
var h = 10 * Math.sqrt(3);
var w = 5;
var v1 = B.subtract(A);
var length = v1.length();
var U = v1.divide(length);
var V = new Vector(-U.y, U.x);
var r1 = B.subtract(U.multiply(h)).add(V.multiply(w));
var r2 = B.subtract(U.multiply(h)).subtract(V.multiply(w));
return [r1,r2];
}
And call the function like this:
var A = new Vector(start.x,start.y);
var B = new Vector(end.x,end.y);
var vec = getArrowhead(A,B);
console.log(vec[0]);
console.log(vec[1]);
I know the OP didn't ask for any specific language, but I came across this looking for a JS implementation, so I thought I would post the result.
I have the following code to calculate points between four control points to generate a catmull-rom curve:
CGPoint interpolatedPosition(CGPoint p0, CGPoint p1, CGPoint p2, CGPoint p3, float t)
{
float t3 = t * t * t;
float t2 = t * t;
float f1 = -0.5 * t3 + t2 - 0.5 * t;
float f2 = 1.5 * t3 - 2.5 * t2 + 1.0;
float f3 = -1.5 * t3 + 2.0 * t2 + 0.5 * t;
float f4 = 0.5 * t3 - 0.5 * t2;
float x = p0.x * f1 + p1.x * f2 + p2.x * f3 + p3.x * f4;
float y = p0.y * f1 + p1.y * f2 + p2.y * f3 + p3.y * f4;
return CGPointMake(x, y);
}
This works fine, but I want to create something I think is called centripetal parameterization. This means that the curve will have no cusps and no self-intersections. If I move one control point really close to another one, the curve should become "smaller". I have Googled my eyes off trying to find a way to do this. Anyone know how to do this?
I needed to implement this for work as well. The fundamental concept you need to start with is that the main difference between the regular Catmull-Rom implementation and the modified versions is how they treat time.
In the unparameterized version from your original Catmull-Rom implementation, t starts at 0 and ends with 1 and calculates the curve from P1 to P2. In the parameterized time implementation, t starts with 0 at P0, and keeps increasing across all four points. So in the uniform case, it would be 1 at P1 and 2 at P2, and you would pass in values ranging from 1 to 2 for your interpolation.
The chordal case shows |Pi+1 - P| as the time span change. This just means that you can use the straight line distance between the points of each segment to calculate the actual length to use. The centripetal case just uses a slightly different method for calculating the optimal length of time to use for each segment.
So now we just need to know how to come up with equations that will let us plug in our new time values. The typical Catmull-Rom equation only has one t in it, the time you are trying to calculate a value for. I found the best article for describing how those parameters are calculated here: http://www.cemyuksel.com/research/catmullrom_param/catmullrom.pdf. They were focusing on a mathematical evaluation of the curves, but in it lies the crucial formula from Barry and Goldman.(1)
In the diagram above, the arrows mean "multiplied by" the ratio given in the arrow.
This then gives us what we need to actually perform a calculation to get the desired result. X and Y are calculated independently, although I used the "Distance" factor for modifying time based on the 2D distance, and not the 1D distance.
Test results:
(1) P. J. Barry and R. N. Goldman. A recursive evaluation algorithm for a class of catmull-rom splines. SIGGRAPH Computer Graphics, 22(4):199{204, 1988.
The source code for my final implementation in Java looks as follows:
/**
* This method will calculate the Catmull-Rom interpolation curve, returning
* it as a list of Coord coordinate objects. This method in particular
* adds the first and last control points which are not visible, but required
* for calculating the spline.
*
* #param coordinates The list of original straight line points to calculate
* an interpolation from.
* #param pointsPerSegment The integer number of equally spaced points to
* return along each curve. The actual distance between each
* point will depend on the spacing between the control points.
* #return The list of interpolated coordinates.
* #param curveType Chordal (stiff), Uniform(floppy), or Centripetal(medium)
* #throws gov.ca.water.shapelite.analysis.CatmullRomException if
* pointsPerSegment is less than 2.
*/
public static List<Coord> interpolate(List<Coord> coordinates, int pointsPerSegment, CatmullRomType curveType)
throws CatmullRomException {
List<Coord> vertices = new ArrayList<>();
for (Coord c : coordinates) {
vertices.add(c.copy());
}
if (pointsPerSegment < 2) {
throw new CatmullRomException("The pointsPerSegment parameter must be greater than 2, since 2 points is just the linear segment.");
}
// Cannot interpolate curves given only two points. Two points
// is best represented as a simple line segment.
if (vertices.size() < 3) {
return vertices;
}
// Test whether the shape is open or closed by checking to see if
// the first point intersects with the last point. M and Z are ignored.
boolean isClosed = vertices.get(0).intersects2D(vertices.get(vertices.size() - 1));
if (isClosed) {
// Use the second and second from last points as control points.
// get the second point.
Coord p2 = vertices.get(1).copy();
// get the point before the last point
Coord pn1 = vertices.get(vertices.size() - 2).copy();
// insert the second from the last point as the first point in the list
// because when the shape is closed it keeps wrapping around to
// the second point.
vertices.add(0, pn1);
// add the second point to the end.
vertices.add(p2);
} else {
// The shape is open, so use control points that simply extend
// the first and last segments
// Get the change in x and y between the first and second coordinates.
double dx = vertices.get(1).X - vertices.get(0).X;
double dy = vertices.get(1).Y - vertices.get(0).Y;
// Then using the change, extrapolate backwards to find a control point.
double x1 = vertices.get(0).X - dx;
double y1 = vertices.get(0).Y - dy;
// Actaully create the start point from the extrapolated values.
Coord start = new Coord(x1, y1, vertices.get(0).Z);
// Repeat for the end control point.
int n = vertices.size() - 1;
dx = vertices.get(n).X - vertices.get(n - 1).X;
dy = vertices.get(n).Y - vertices.get(n - 1).Y;
double xn = vertices.get(n).X + dx;
double yn = vertices.get(n).Y + dy;
Coord end = new Coord(xn, yn, vertices.get(n).Z);
// insert the start control point at the start of the vertices list.
vertices.add(0, start);
// append the end control ponit to the end of the vertices list.
vertices.add(end);
}
// Dimension a result list of coordinates.
List<Coord> result = new ArrayList<>();
// When looping, remember that each cycle requires 4 points, starting
// with i and ending with i+3. So we don't loop through all the points.
for (int i = 0; i < vertices.size() - 3; i++) {
// Actually calculate the Catmull-Rom curve for one segment.
List<Coord> points = interpolate(vertices, i, pointsPerSegment, curveType);
// Since the middle points are added twice, once for each bordering
// segment, we only add the 0 index result point for the first
// segment. Otherwise we will have duplicate points.
if (result.size() > 0) {
points.remove(0);
}
// Add the coordinates for the segment to the result list.
result.addAll(points);
}
return result;
}
/**
* Given a list of control points, this will create a list of pointsPerSegment
* points spaced uniformly along the resulting Catmull-Rom curve.
*
* #param points The list of control points, leading and ending with a
* coordinate that is only used for controling the spline and is not visualized.
* #param index The index of control point p0, where p0, p1, p2, and p3 are
* used in order to create a curve between p1 and p2.
* #param pointsPerSegment The total number of uniformly spaced interpolated
* points to calculate for each segment. The larger this number, the
* smoother the resulting curve.
* #param curveType Clarifies whether the curve should use uniform, chordal
* or centripetal curve types. Uniform can produce loops, chordal can
* produce large distortions from the original lines, and centripetal is an
* optimal balance without spaces.
* #return the list of coordinates that define the CatmullRom curve
* between the points defined by index+1 and index+2.
*/
public static List<Coord> interpolate(List<Coord> points, int index, int pointsPerSegment, CatmullRomType curveType) {
List<Coord> result = new ArrayList<>();
double[] x = new double[4];
double[] y = new double[4];
double[] time = new double[4];
for (int i = 0; i < 4; i++) {
x[i] = points.get(index + i).X;
y[i] = points.get(index + i).Y;
time[i] = i;
}
double tstart = 1;
double tend = 2;
if (!curveType.equals(CatmullRomType.Uniform)) {
double total = 0;
for (int i = 1; i < 4; i++) {
double dx = x[i] - x[i - 1];
double dy = y[i] - y[i - 1];
if (curveType.equals(CatmullRomType.Centripetal)) {
total += Math.pow(dx * dx + dy * dy, .25);
} else {
total += Math.pow(dx * dx + dy * dy, .5);
}
time[i] = total;
}
tstart = time[1];
tend = time[2];
}
double z1 = 0.0;
double z2 = 0.0;
if (!Double.isNaN(points.get(index + 1).Z)) {
z1 = points.get(index + 1).Z;
}
if (!Double.isNaN(points.get(index + 2).Z)) {
z2 = points.get(index + 2).Z;
}
double dz = z2 - z1;
int segments = pointsPerSegment - 1;
result.add(points.get(index + 1));
for (int i = 1; i < segments; i++) {
double xi = interpolate(x, time, tstart + (i * (tend - tstart)) / segments);
double yi = interpolate(y, time, tstart + (i * (tend - tstart)) / segments);
double zi = z1 + (dz * i) / segments;
result.add(new Coord(xi, yi, zi));
}
result.add(points.get(index + 2));
return result;
}
/**
* Unlike the other implementation here, which uses the default "uniform"
* treatment of t, this computation is used to calculate the same values but
* introduces the ability to "parameterize" the t values used in the
* calculation. This is based on Figure 3 from
* http://www.cemyuksel.com/research/catmullrom_param/catmullrom.pdf
*
* #param p An array of double values of length 4, where interpolation
* occurs from p1 to p2.
* #param time An array of time measures of length 4, corresponding to each
* p value.
* #param t the actual interpolation ratio from 0 to 1 representing the
* position between p1 and p2 to interpolate the value.
* #return
*/
public static double interpolate(double[] p, double[] time, double t) {
double L01 = p[0] * (time[1] - t) / (time[1] - time[0]) + p[1] * (t - time[0]) / (time[1] - time[0]);
double L12 = p[1] * (time[2] - t) / (time[2] - time[1]) + p[2] * (t - time[1]) / (time[2] - time[1]);
double L23 = p[2] * (time[3] - t) / (time[3] - time[2]) + p[3] * (t - time[2]) / (time[3] - time[2]);
double L012 = L01 * (time[2] - t) / (time[2] - time[0]) + L12 * (t - time[0]) / (time[2] - time[0]);
double L123 = L12 * (time[3] - t) / (time[3] - time[1]) + L23 * (t - time[1]) / (time[3] - time[1]);
double C12 = L012 * (time[2] - t) / (time[2] - time[1]) + L123 * (t - time[1]) / (time[2] - time[1]);
return C12;
}
There is a much easier and more efficient way to implement this which only requires you to compute your tangents using a different formula, without the need to implement the recursive evaluation algorithm of Barry and Goldman.
If you take the Barry-Goldman parametrization (referenced in Ted's answer) C(t) for the knots (t0,t1,t2,t3) and the control points (P0,P1,P2,P3), its closed form is pretty complicated, but in the end it's still a cubic polynomial in t when you constrain it to the interval (t1,t2). So all we need to describe it fully are the values and tangents at the two end points t1 and t2. If we work out these values (I did this in Mathematica), we find
C(t1) = P1
C(t2) = P2
C'(t1) = (P1 - P0) / (t1 - t0) - (P2 - P0) / (t2 - t0) + (P2 - P1) / (t2 - t1)
C'(t2) = (P2 - P1) / (t2 - t1) - (P3 - P1) / (t3 - t1) + (P3 - P2) / (t3 - t2)
We can simply plug this into the standard formula for computing a cubic spline with given values and tangents at the end points and we have our nonuniform Catmull-Rom spline. One caveat is that the above tangents are computed for the interval (t1,t2), so if you want to evaluate the curve in the standard interval (0,1), simply rescale the tangents by multiplying them with the factor (t2-t1).
I put a working C++ example on Ideone: http://ideone.com/NoEbVM
I'll also paste the code below.
#include <iostream>
#include <cmath>
using namespace std;
struct CubicPoly
{
float c0, c1, c2, c3;
float eval(float t)
{
float t2 = t*t;
float t3 = t2 * t;
return c0 + c1*t + c2*t2 + c3*t3;
}
};
/*
* Compute coefficients for a cubic polynomial
* p(s) = c0 + c1*s + c2*s^2 + c3*s^3
* such that
* p(0) = x0, p(1) = x1
* and
* p'(0) = t0, p'(1) = t1.
*/
void InitCubicPoly(float x0, float x1, float t0, float t1, CubicPoly &p)
{
p.c0 = x0;
p.c1 = t0;
p.c2 = -3*x0 + 3*x1 - 2*t0 - t1;
p.c3 = 2*x0 - 2*x1 + t0 + t1;
}
// standard Catmull-Rom spline: interpolate between x1 and x2 with previous/following points x0/x3
// (we don't need this here, but it's for illustration)
void InitCatmullRom(float x0, float x1, float x2, float x3, CubicPoly &p)
{
// Catmull-Rom with tension 0.5
InitCubicPoly(x1, x2, 0.5f*(x2-x0), 0.5f*(x3-x1), p);
}
// compute coefficients for a nonuniform Catmull-Rom spline
void InitNonuniformCatmullRom(float x0, float x1, float x2, float x3, float dt0, float dt1, float dt2, CubicPoly &p)
{
// compute tangents when parameterized in [t1,t2]
float t1 = (x1 - x0) / dt0 - (x2 - x0) / (dt0 + dt1) + (x2 - x1) / dt1;
float t2 = (x2 - x1) / dt1 - (x3 - x1) / (dt1 + dt2) + (x3 - x2) / dt2;
// rescale tangents for parametrization in [0,1]
t1 *= dt1;
t2 *= dt1;
InitCubicPoly(x1, x2, t1, t2, p);
}
struct Vec2D
{
Vec2D(float _x, float _y) : x(_x), y(_y) {}
float x, y;
};
float VecDistSquared(const Vec2D& p, const Vec2D& q)
{
float dx = q.x - p.x;
float dy = q.y - p.y;
return dx*dx + dy*dy;
}
void InitCentripetalCR(const Vec2D& p0, const Vec2D& p1, const Vec2D& p2, const Vec2D& p3,
CubicPoly &px, CubicPoly &py)
{
float dt0 = powf(VecDistSquared(p0, p1), 0.25f);
float dt1 = powf(VecDistSquared(p1, p2), 0.25f);
float dt2 = powf(VecDistSquared(p2, p3), 0.25f);
// safety check for repeated points
if (dt1 < 1e-4f) dt1 = 1.0f;
if (dt0 < 1e-4f) dt0 = dt1;
if (dt2 < 1e-4f) dt2 = dt1;
InitNonuniformCatmullRom(p0.x, p1.x, p2.x, p3.x, dt0, dt1, dt2, px);
InitNonuniformCatmullRom(p0.y, p1.y, p2.y, p3.y, dt0, dt1, dt2, py);
}
int main()
{
Vec2D p0(0,0), p1(1,1), p2(1.1,1), p3(2,0);
CubicPoly px, py;
InitCentripetalCR(p0, p1, p2, p3, px, py);
for (int i = 0; i <= 10; ++i)
cout << px.eval(0.1f*i) << " " << py.eval(0.1f*i) << endl;
}
Here is an iOS version of Ted's code. I excluded the 'z' parts.
.h
typedef enum {
CatmullRomTypeUniform,
CatmullRomTypeChordal,
CatmullRomTypeCentripetal
} CatmullRomType ;
.m
-(NSMutableArray *)interpolate:(NSArray *)coordinates withPointsPerSegment:(NSInteger)pointsPerSegment andType:(CatmullRomType)curveType {
NSMutableArray *vertices = [[NSMutableArray alloc] initWithArray:coordinates copyItems:YES];
if (pointsPerSegment < 3)
return vertices;
//start point
CGPoint pt1 = [vertices[0] CGPointValue];
CGPoint pt2 = [vertices[1] CGPointValue];
double dx = pt2.x - pt1.x;
double dy = pt2.y - pt1.y;
double x1 = pt1.x - dx;
double y1 = pt1.y - dy;
CGPoint start = CGPointMake(x1*.5, y1);
//end point
pt2 = [vertices[vertices.count-1] CGPointValue];
pt1 = [vertices[vertices.count-2] CGPointValue];
dx = pt2.x - pt1.x;
dy = pt2.y - pt1.y;
x1 = pt2.x + dx;
y1 = pt2.y + dy;
CGPoint end = CGPointMake(x1, y1);
[vertices insertObject:[NSValue valueWithCGPoint:start] atIndex:0];
[vertices addObject:[NSValue valueWithCGPoint:end]];
NSMutableArray *result = [[NSMutableArray alloc] init];
for (int i = 0; i < vertices.count - 3; i++) {
NSMutableArray *points = [self interpolate:vertices forIndex:i withPointsPerSegment:pointsPerSegment andType:curveType];
if ([points count] > 0)
[points removeObjectAtIndex:0];
[result addObjectsFromArray:points];
}
return result;
}
-(double)interpolate:(double*)p time:(double*)time t:(double) t {
double L01 = p[0] * (time[1] - t) / (time[1] - time[0]) + p[1] * (t - time[0]) / (time[1] - time[0]);
double L12 = p[1] * (time[2] - t) / (time[2] - time[1]) + p[2] * (t - time[1]) / (time[2] - time[1]);
double L23 = p[2] * (time[3] - t) / (time[3] - time[2]) + p[3] * (t - time[2]) / (time[3] - time[2]);
double L012 = L01 * (time[2] - t) / (time[2] - time[0]) + L12 * (t - time[0]) / (time[2] - time[0]);
double L123 = L12 * (time[3] - t) / (time[3] - time[1]) + L23 * (t - time[1]) / (time[3] - time[1]);
double C12 = L012 * (time[2] - t) / (time[2] - time[1]) + L123 * (t - time[1]) / (time[2] - time[1]);
return C12;
}
-(NSMutableArray*)interpolate:(NSArray *)points forIndex:(NSInteger)index withPointsPerSegment:(NSInteger)pointsPerSegment andType:(CatmullRomType)curveType {
NSMutableArray *result = [[NSMutableArray alloc] init];
double x[4];
double y[4];
double time[4];
for (int i=0; i < 4; i++) {
x[i] = [points[index+i] CGPointValue].x;
y[i] = [points[index+i] CGPointValue].y;
time[i] = i;
}
double tstart = 1;
double tend = 2;
if (curveType != CatmullRomTypeUniform) {
double total = 0;
for (int i=1; i < 4; i++) {
double dx = x[i] - x[i-1];
double dy = y[i] - y[i-1];
if (curveType == CatmullRomTypeCentripetal) {
total += pow(dx * dx + dy * dy, 0.25);
}
else {
total += pow(dx * dx + dy * dy, 0.5); //sqrt
}
time[i] = total;
}
tstart = time[1];
tend = time[2];
}
int segments = pointsPerSegment - 1;
[result addObject:points[index+1]];
for (int i =1; i < segments; i++) {
double xi = [self interpolate:x time:time t:tstart + (i * (tend - tstart)) / segments];
double yi = [self interpolate:y time:time t:tstart + (i * (tend - tstart)) / segments];
NSLog(#"(%f,%f)",xi,yi);
[result addObject:[NSValue valueWithCGPoint:CGPointMake(xi, yi)]];
}
[result addObject:points[index+2]];
return result;
}
Also, here is a method for turning an array of points into a Bezier path for drawing, using the above
-(UIBezierPath*)bezierPathFromPoints:(NSArray *)points withGranulaity:(NSInteger)granularity
{
UIBezierPath __block *path = [[UIBezierPath alloc] init];
NSMutableArray *curve = [self interpolate:points withPointsPerSegment:granularity andType:CatmullRomTypeCentripetal];
CGPoint __block p0 = [curve[0] CGPointValue];
[path moveToPoint:p0];
//use this loop to draw lines between all points
for (int idx=1; idx < [curve count]; idx+=1) {
CGPoint c1 = [curve[idx] CGPointValue];
[path addLineToPoint:c1];
};
//or use this loop to use actual control points (less smooth but probably faster)
// for (int idx=0; idx < [curve count]-3; idx+=3) {
// CGPoint c1 = [curve[idx+1] CGPointValue];
// CGPoint c2 = [curve[idx+2] CGPointValue];
// CGPoint p1 = [curve[idx+3] CGPointValue];
//
// [path addCurveToPoint:p1 controlPoint1:c1 controlPoint2:c2];
// };
return path;
}
Thanks for the reply of Ted and cfh.
Sorry to my poor English and I am not very sure if my understanding is right.
It confused me before that what is the relation between τ in http://graphics.cs.cmu.edu/nsp/course/15-462/Fall04/assts/catmullRom.pdf with α in
"Parameterization of Catmull-Rom Curves"[Yuksel et al. 2009].
It finally seems that τ has little relation to α.
In Ted's reply we can find that as long as t(i+1)-t(i)=1, we call it "uniform" Catmull-Rom curve.
So, all curves in http://graphics.cs.cmu.edu/nsp/course/15-462/Fall04/assts/catmullRom.pdf are "uniform" curves, while parameterized Catmull-Rom can only produce one "uniform" curve when α=0.τ just affects how sharply the curve bends at the (interpolated) control points and can produce a series of "uniform" curves.
Here, if we take the equation from cfh into consideration, i.e.:
C(t1) = P1
C(t2) = P2
C'(t1) = (P1 - P0) / (t1 - t0) - (P2 - P0) / (t2 - t0) + (P2 - P1) / (t2 - t1)
C'(t2) = (P2 - P1) / (t2 - t1) - (P3 - P1) / (t3 - t1) + (P3 - P2) / (t3 - t2)
When α=0, we know that t(i+1)-t(i)=1. Substituting t(i+1)-t(i)=1 to C'(t1) and C'(t2), we can get:
C'(t1)=1/2(P2-P0)
C'(t2)=1/2(P3-P1).
That is to say, the only one "uniform" curve generated by parameterized Catmull-Rom curve meets the special case τ=1/2 http://graphics.cs.cmu.edu/nsp/course/15-462/Fall04/assts/catmullRom.pdf. More different "uniform" curves can be generated by the changes of τ, while α pays more attention to something else.
I coded something in Python (adapted form Catmull-Rom Wikipedia page) that compares uniform, centripedal, and chordial CR Splines (though you can set alpha to whatever you'd like) using random data (you can use your own data and the fucntions work fine). Note that for the endpoints I just stuck in a quick 'hack' that maintains the slope from the first and last 2 points, although the distance between this point and the first/lost known point is arbitrary (I set it to 1% of the domain... for no reason at all. So keep that in mind before applying to something important):
# coding: utf-8
# In[1]:
import numpy
import matplotlib.pyplot as plt
get_ipython().magic(u'pylab inline')
# In[2]:
def CatmullRomSpline(P0, P1, P2, P3, a, nPoints=100):
"""
P0, P1, P2, and P3 should be (x,y) point pairs that define the Catmull-Rom spline.
nPoints is the number of points to include in this curve segment.
"""
# Convert the points to numpy so that we can do array multiplication
P0, P1, P2, P3 = map(numpy.array, [P0, P1, P2, P3])
# Calculate t0 to t4
alpha = a
def tj(ti, Pi, Pj):
xi, yi = Pi
xj, yj = Pj
return ( ( (xj-xi)**2 + (yj-yi)**2 )**0.5 )**alpha + ti
t0 = 0
t1 = tj(t0, P0, P1)
t2 = tj(t1, P1, P2)
t3 = tj(t2, P2, P3)
# Only calculate points between P1 and P2
t = numpy.linspace(t1,t2,nPoints)
# Reshape so that we can multiply by the points P0 to P3
# and get a point for each value of t.
t = t.reshape(len(t),1)
A1 = (t1-t)/(t1-t0)*P0 + (t-t0)/(t1-t0)*P1
A2 = (t2-t)/(t2-t1)*P1 + (t-t1)/(t2-t1)*P2
A3 = (t3-t)/(t3-t2)*P2 + (t-t2)/(t3-t2)*P3
B1 = (t2-t)/(t2-t0)*A1 + (t-t0)/(t2-t0)*A2
B2 = (t3-t)/(t3-t1)*A2 + (t-t1)/(t3-t1)*A3
C = (t2-t)/(t2-t1)*B1 + (t-t1)/(t2-t1)*B2
return C
def CatmullRomChain(P,alpha):
"""
Calculate Catmull Rom for a chain of points and return the combined curve.
"""
sz = len(P)
# The curve C will contain an array of (x,y) points.
C = []
for i in range(sz-3):
c = CatmullRomSpline(P[i], P[i+1], P[i+2], P[i+3],alpha)
C.extend(c)
return C
# In[8]:
# Define a set of points for curve to go through
Points = numpy.random.rand(12,2)
x1=Points[0][0]
x2=Points[1][0]
y1=Points[0][1]
y2=Points[1][1]
x3=Points[-2][0]
x4=Points[-1][0]
y3=Points[-2][1]
y4=Points[-1][1]
dom=max(Points[:,0])-min(Points[:,0])
rng=max(Points[:,1])-min(Points[:,1])
prex=x1+sign(x1-x2)*dom*0.01
prey=(y1-y2)/(x1-x2)*dom*0.01+y1
endx=x4+sign(x4-x3)*dom*0.01
endy=(y4-y3)/(x4-x3)*dom*0.01+y4
print len(Points)
Points=list(Points)
Points.insert(0,array([prex,prey]))
Points.append(array([endx,endy]))
print len(Points)
# In[9]:
#Define alpha
a=0.
# Calculate the Catmull-Rom splines through the points
c = CatmullRomChain(Points,a)
# Convert the Catmull-Rom curve points into x and y arrays and plot
x,y = zip(*c)
plt.plot(x,y,c='green',zorder=10)
# Plot the control points
px, py = zip(*Points)
plt.plot(px,py,'or')
a=0.5
c = CatmullRomChain(Points,a)
x,y = zip(*c)
plt.plot(x,y,c='blue')
a=1.
c = CatmullRomChain(Points,a)
x,y = zip(*c)
plt.plot(x,y,c='red')
plt.grid(b=True)
plt.show()
# In[10]:
Points
# In[ ]:
original code: https://en.wikipedia.org/wiki/Centripetal_Catmull%E2%80%93Rom_spline
How do I calculate the intersection points of two circles. I would expect there to be either two, one or no intersection points in all cases.
I have the x and y coordinates of the centre-point, and the radius for each circle.
An answer in python would be preferred, but any working algorithm would be acceptable.
Intersection of two circles
Written by Paul Bourke
The following note describes how to find the intersection point(s)
between two circles on a plane, the following notation is used. The
aim is to find the two points P3 = (x3,
y3) if they exist.
First calculate the distance d between the center
of the circles. d = ||P1 - P0||.
If d > r0 + r1 then there are no solutions,
the circles are separate. If d < |r0 -
r1| then there are no solutions because one circle is
contained within the other. If d = 0 and r0 =
r1 then the circles are coincident and there are an
infinite number of solutions.
Considering the two triangles P0P2P3
and P1P2P3 we can write
a2 + h2 = r02 and
b2 + h2 = r12
Using d = a + b we can solve for a, a =
(r02 - r12 +
d2 ) / (2 d)
It can be readily shown that this reduces to
r0 when the two circles touch at one point, ie: d =
r0 + r1
Solve for h by substituting a into the first
equation, h2 = r02 - a2
So P2 = P0 + a ( P1 -
P0 ) / d And finally, P3 =
(x3,y3) in terms of P0 =
(x0,y0), P1 =
(x1,y1) and P2 =
(x2,y2), is x3 =
x2 +- h ( y1 - y0 ) / d
y3 = y2 -+ h ( x1 - x0 ) /
d
Source: http://paulbourke.net/geometry/circlesphere/
Here is my C++ implementation based on Paul Bourke's article. It only works if there are two intersections, otherwise it probably returns NaN NAN NAN NAN.
class Point{
public:
float x, y;
Point(float px, float py) {
x = px;
y = py;
}
Point sub(Point p2) {
return Point(x - p2.x, y - p2.y);
}
Point add(Point p2) {
return Point(x + p2.x, y + p2.y);
}
float distance(Point p2) {
return sqrt((x - p2.x)*(x - p2.x) + (y - p2.y)*(y - p2.y));
}
Point normal() {
float length = sqrt(x*x + y*y);
return Point(x/length, y/length);
}
Point scale(float s) {
return Point(x*s, y*s);
}
};
class Circle {
public:
float x, y, r, left;
Circle(float cx, float cy, float cr) {
x = cx;
y = cy;
r = cr;
left = x - r;
}
pair<Point, Point> intersections(Circle c) {
Point P0(x, y);
Point P1(c.x, c.y);
float d, a, h;
d = P0.distance(P1);
a = (r*r - c.r*c.r + d*d)/(2*d);
h = sqrt(r*r - a*a);
Point P2 = P1.sub(P0).scale(a/d).add(P0);
float x3, y3, x4, y4;
x3 = P2.x + h*(P1.y - P0.y)/d;
y3 = P2.y - h*(P1.x - P0.x)/d;
x4 = P2.x - h*(P1.y - P0.y)/d;
y4 = P2.y + h*(P1.x - P0.x)/d;
return pair<Point, Point>(Point(x3, y3), Point(x4, y4));
}
};
Why not just use 7 lines of your favorite procedural language (or programmable calculator!) as below.
Assuming you are given P0 coords (x0,y0), P1 coords (x1,y1), r0 and r1 and you want to find P3 coords (x3,y3):
d=sqr((x1-x0)^2 + (y1-y0)^2)
a=(r0^2-r1^2+d^2)/(2*d)
h=sqr(r0^2-a^2)
x2=x0+a*(x1-x0)/d
y2=y0+a*(y1-y0)/d
x3=x2+h*(y1-y0)/d // also x3=x2-h*(y1-y0)/d
y3=y2-h*(x1-x0)/d // also y3=y2+h*(x1-x0)/d
Here's an implementation in Javascript using vectors. The code is well documented, you should be able to follow it. Here's the original source
See live demo here:
// Let EPS (epsilon) be a small value
var EPS = 0.0000001;
// Let a point be a pair: (x, y)
function Point(x, y) {
this.x = x;
this.y = y;
}
// Define a circle centered at (x,y) with radius r
function Circle(x,y,r) {
this.x = x;
this.y = y;
this.r = r;
}
// Due to double rounding precision the value passed into the Math.acos
// function may be outside its domain of [-1, +1] which would return
// the value NaN which we do not want.
function acossafe(x) {
if (x >= +1.0) return 0;
if (x <= -1.0) return Math.PI;
return Math.acos(x);
}
// Rotates a point about a fixed point at some angle 'a'
function rotatePoint(fp, pt, a) {
var x = pt.x - fp.x;
var y = pt.y - fp.y;
var xRot = x * Math.cos(a) + y * Math.sin(a);
var yRot = y * Math.cos(a) - x * Math.sin(a);
return new Point(fp.x+xRot,fp.y+yRot);
}
// Given two circles this method finds the intersection
// point(s) of the two circles (if any exists)
function circleCircleIntersectionPoints(c1, c2) {
var r, R, d, dx, dy, cx, cy, Cx, Cy;
if (c1.r < c2.r) {
r = c1.r; R = c2.r;
cx = c1.x; cy = c1.y;
Cx = c2.x; Cy = c2.y;
} else {
r = c2.r; R = c1.r;
Cx = c1.x; Cy = c1.y;
cx = c2.x; cy = c2.y;
}
// Compute the vector <dx, dy>
dx = cx - Cx;
dy = cy - Cy;
// Find the distance between two points.
d = Math.sqrt( dx*dx + dy*dy );
// There are an infinite number of solutions
// Seems appropriate to also return null
if (d < EPS && Math.abs(R-r) < EPS) return [];
// No intersection (circles centered at the
// same place with different size)
else if (d < EPS) return [];
var x = (dx / d) * R + Cx;
var y = (dy / d) * R + Cy;
var P = new Point(x, y);
// Single intersection (kissing circles)
if (Math.abs((R+r)-d) < EPS || Math.abs(R-(r+d)) < EPS) return [P];
// No intersection. Either the small circle contained within
// big circle or circles are simply disjoint.
if ( (d+r) < R || (R+r < d) ) return [];
var C = new Point(Cx, Cy);
var angle = acossafe((r*r-d*d-R*R)/(-2.0*d*R));
var pt1 = rotatePoint(C, P, +angle);
var pt2 = rotatePoint(C, P, -angle);
return [pt1, pt2];
}
Try this;
def ri(cr1,cr2,cp1,cp2):
int1=[]
int2=[]
ori=0
if cp1[0]<cp2[0] and cp1[1]!=cp2[1]:
p1=cp1
p2=cp2
r1=cr1
r2=cr2
if cp1[1]<cp2[1]:
ori+=1
elif cp1[1]>cp2[1]:
ori+=2
elif cp1[0]>cp2[0] and cp1[1]!=cp2[1]:
p1=cp2
p2=cp1
r1=cr2
r2=cr1
if p1[1]<p2[1]:
ori+=1
elif p1[1]>p2[1]:
ori+=2
elif cp1[0]==cp2[0]:
ori+=4
if cp1[1]>cp2[1]:
p1=cp1
p2=cp2
r1=cr1
r2=cr2
elif cp1[1]<cp2[1]:
p1=cp2
p2=cp1
r1=cr2
r2=cr1
elif cp1[1]==cp2[1]:
ori+=3
if cp1[0]>cp2[0]:
p1=cp2
p2=cp1
r1=cr2
r2=cr1
elif cp1[0]<cp2[0]:
p1=cp1
p2=cp2
r1=cr1
r2=cr2
if ori==1:#+
D=calc_dist(p1,p2)
tr=r1+r2
el=tr-D
a=r1-el
b=r2-el
A=a+(el/2)
B=b+(el/2)
thta=math.degrees(math.acos(A/r1))
rs=p2[1]-p1[1]
rn=p2[0]-p1[0]
gd=rs/rn
yint=p1[1]-((gd)*p1[0])
dty=calc_dist(p1,[0,yint])
aa=p1[1]-yint
bb=math.degrees(math.asin(aa/dty))
d=90-bb
e=180-d-thta
g=(dty/math.sin(math.radians(e)))*math.sin(math.radians(thta))
f=(g/math.sin(math.radians(thta)))*math.sin(math.radians(d))
oty=yint+g
h=f+r1
i=90-e
j=180-90-i
l=math.sin(math.radians(i))*h
k=math.cos(math.radians(i))*h
iy2=oty-l
ix2=k
int2.append(ix2)
int2.append(iy2)
m=90+bb
n=180-m-thta
p=(dty/math.sin(math.radians(n)))*math.sin(math.radians(m))
o=(p/math.sin(math.radians(m)))*math.sin(math.radians(thta))
q=p+r1
r=90-n
s=math.sin(math.radians(r))*q
t=math.cos(math.radians(r))*q
otty=yint-o
iy1=otty+s
ix1=t
int1.append(ix1)
int1.append(iy1)
elif ori==2:#-
D=calc_dist(p1,p2)
tr=r1+r2
el=tr-D
a=r1-el
b=r2-el
A=a+(el/2)
B=b+(el/2)
thta=math.degrees(math.acos(A/r1))
rs=p2[1]-p1[1]
rn=p2[0]-p1[0]
gd=rs/rn
yint=p1[1]-((gd)*p1[0])
dty=calc_dist(p1,[0,yint])
aa=yint-p1[1]
bb=math.degrees(math.asin(aa/dty))
c=180-90-bb
d=180-c-thta
e=180-90-d
f=math.tan(math.radians(e))*p1[0]
g=math.sqrt(p1[0]**2+f**2)
h=g+r1
i=180-90-e
j=math.sin(math.radians(e))*h
jj=math.cos(math.radians(i))*h
k=math.cos(math.radians(e))*h
kk=math.sin(math.radians(i))*h
l=90-bb
m=90-e
tt=l+m+thta
n=(dty/math.sin(math.radians(m)))*math.sin(math.radians(thta))
nn=(g/math.sin(math.radians(l)))*math.sin(math.radians(thta))
oty=yint-n
iy1=oty+j
ix1=k
int1.append(ix1)
int1.append(iy1)
o=bb+90
p=180-o-thta
q=90-p
r=180-90-q
s=(dty/math.sin(math.radians(p)))*math.sin(math.radians(o))
t=(s/math.sin(math.radians(o)))*math.sin(math.radians(thta))
u=s+r1
v=math.sin(math.radians(r))*u
vv=math.cos(math.radians(q))*u
w=math.cos(math.radians(r))*u
ww=math.sin(math.radians(q))*u
ix2=v
otty=yint+t
iy2=otty-w
int2.append(ix2)
int2.append(iy2)
elif ori==3:#y
D=calc_dist(p1,p2)
tr=r1+r2
el=tr-D
a=r1-el
b=r2-el
A=a+(el/2)
B=b+(el/2)
b=math.sqrt(r1**2-A**2)
int1.append(p1[0]+A)
int1.append(p1[1]+b)
int2.append(p1[0]+A)
int2.append(p1[1]-b)
elif ori==4:#x
D=calc_dist(p1,p2)
tr=r1+r2
el=tr-D
a=r1-el
b=r2-el
A=a+(el/2)
B=b+(el/2)
b=math.sqrt(r1**2-A**2)
int1.append(p1[0]+b)
int1.append(p1[1]-A)
int2.append(p1[0]-b)
int2.append(p1[1]-A)
return [int1,int2]
def calc_dist(p1,p2):
return math.sqrt((p2[0] - p1[0]) ** 2 +
(p2[1] - p1[1]) ** 2)