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I made a program that creates parallel lines from the mouse coordinates with a certain distance away which can be modified by the distance variable at the beginning of the code. The problem is that it does not work as it should when drawing.
Line 1 and Line 2 are the lines that are parallel to the line formed from the mouse coordinates, the "pointSlope" variable.
distance = 30
function setup() {
createCanvas(600, 600);
}
function draw() {
lineCreate([pmouseX,pmouseY], [mouseX,mouseY])
}
function lineCreate(point1, point2) {
fill(0)
stroke(0)
x0 = point1[0];
x1 = point2[0];
y0 = point1[1];
y1 = point2[1];
if (abs(x1 - x0) > abs(y1 - y0)) {
if (x0 > x1) {
let t = x0; x0 = x1; x1 = t;
t = y0; y0 = y1; y1 = t;
}
for (let x = x0; x <= x1; x++) {
let slope = (y1-y0) * (x-x0) / (x1-x0)
let y = y0 + (y1-y0) * (x-x0) / (x1-x0);
line1 = y + distance*Math.sqrt(1+pow(slope,2))
line2 = y - distance*Math.sqrt(1+pow(slope,2))
circle(x, line1, 2, 2);
circle(x, line2, 2, 2);
}
} else {
if (y0 > y1) {
let t = x0; x0 = x1; x1 = t;
t = y0; y0 = y1; y1 = t;
}
for (let y = y0; y <= y1; y++) {
let x = x0 + (x1-x0) * (y-y0) / (y1-y0);
circle(x, y, 2, 2);
}
}
}
I think that using vectors may help you in this situation, try this code as example:
let dist = 40;
function setup() {
createCanvas(600, 600);
v = createVector();
noStroke();
}
function draw() {
lineCreate();
}
function lineCreate() {
v.x = mouseX-pmouseX; v.y = mouseY-pmouseY;
h = v.heading();
LX = dist*cos(h+PI/2); LY = dist*sin(h+PI/2);
RX = dist*cos(h-PI/2); RY = dist*sin(h-PI/2);
for (let i=0; i<v.mag(); i++) {
fill(0).circle(pmouseX+i*cos(h),pmouseY+i*sin(h),2);
fill(160).circle(pmouseX+LX+i*cos(h),pmouseY+LY+i*sin(h),2);
fill(160).circle(pmouseX+RX+i*cos(h),pmouseY+RY+i*sin(h),2);
}
}
Here we are drawing two parallel lines in road-like path, but as you can see there are gaps in turns caused by using pmouse. I think, that it will not be possible to get rid of these artifacts, if you not move away from using the pmouse to more complex ways of calculating the trajectory.
I've been trying to rotate a square for a project, I've done research and think I have the right formula to calculate the rotated points. I calculate the points as if they're individual around the center of the square. How to fix it?
//Declaring variables
float x0, y0, xo, yo,x1,y1,x2,y2,x3,y3, theta, newx, newy, s, c;
void setup() {
size (800,800);
//To debug
//frameRate(1);
fill(0);
//Initializing variables
xo = 400;
yo = 400;
x0 = 350;
y0 = 450;
x1 = 350;
y1 = 350;
x2 = 450;
y2 = 350;
x3 = 450;
y3 = 450;
theta = radians(5);
s = sin(theta);
c = cos(theta);
}
void draw() {
//Reseting the background
background(255);
//Drawing the square
quad(x0,y0,x1,y1,x2,y2,x3,y3);
//Doing the rotations
x0 = rotateX(x0,y0);
y0 = rotateY(x0,y0);
x1 = rotateX(x1,y1);
y1 = rotateY(x1,y1);
x2 = rotateX(x2,y2);
y2 = rotateY(x2,y2);
x3 = rotateX(x3,y3);
y3 = rotateY(x3,y3);
}
//Rotate x coordinate method
float rotateX(float x, float y) {
x -= xo;
newx = x * c - y * s;
x = newx + xo;
return x;
}
//Rotate y coordinate method
float rotateY(float x, float y) {
y -= yo;
newy = x * s - y * c;
y = newy + yo;
return y;
}
There are two things:
1) You have a sign error in rotateY(). The y term should have a positive sign:
newy = x * s + y * c;
2) When you do this:
x0 = rotateX(x0,y0);
y0 = rotateY(x0,y0);
... then the first call modifies x0, which the second call then uses. But the second call needs the original coordinates to rotate correctly:
float x0Rotated = rotateX(x0, y0);
y0 = rotateY(x0, y0);
x0 = x0Rotated;
The same thing for the other points.
I have lena image in Matlab. First I need to find the centroid C, and then divide the image into equal number of parts. I can calculate the centroid of the image but from that how can I divide the image into equal number of parts as shown below. Please anyone help me.
Thanks
Using poly2mask to create binary sectors and using the resulting sectors for indexing
Code:
im = imread('peppers.png');
r = 300;
out1 = ones(max(size(im,1),r*2)+2,max(size(im,2),r*2)+2,3).*255;
xoffset = floor((size(out1,2)-size(im,2))/2);
yoffset = floor((size(out1,1)-size(im,1))/2);
out1(yoffset:yoffset+size(im,1)-1,xoffset:xoffset+size(im,2)-1,:) = im(:,:,:);
im = out1;
cy = floor(size(im,1)/2);
cx = floor(size(im,2)/2);
figure;
imshow(uint8(im));
hold on
pos = [cx-r+1 cy-r+1 r*2 r*2];
rectangle('Position',pos,'Curvature',[1 1]);
x1 = [-r, 0, -r*cosd(45), -r*cosd(45); r, 0, r*cosd(45), r*cosd(45)]+cx+1;
y1 = [0, -r, -r*sind(45), r*sind(45); 0, r, r*sind(45), -r*sind(45)]+cy+1;
plot(x1,y1);
hold off
figure;
for i = 0:45:315
t = linspace(-i,-i-45,128);
x = [cx, cx+r*cosd(t), cx];
y = [cy, cy+r*sind(t), cy];
bw = poly2mask( x, y, size(im,1),size(im,2));
bw = repmat(bw,1,1,3);
out = ones(size(im,1),size(im,2),size(im,3)).*155;
out(bw) = im(bw);
subplot(2,4,(i/45)+1); imshow(uint8(out));
end;
Results:
Original Image
Partitions drawn over Original Image
Segments of the image
Update
for getting pixel values of the lines, by using Bresenham function from here
figure;
bw1 = zeros(size(im,1),size(im,2));
outmat = zeros(size(bw1));
[X,Y] = bresenham(cx+1-r,cy+1,cx+1+r,cy+1);
ind = sub2ind(size(outmat), Y, X);
outmat(ind) = 1;
[X,Y] = bresenham(cx+1,cy+1-r,cx+1,cy+1+r);
ind = sub2ind(size(outmat), Y, X);
outmat(ind) = 1;
[X,Y] = bresenham(cx+1-r*cosd(45),cy+1-r*sind(45),cx+1+r*cosd(45),cy+1+r*sind(45));
ind = sub2ind(size(outmat), Y, X);
outmat(ind) = 1;
[X,Y] = bresenham(cx+1-r*cosd(45),cy+1+r*sind(45),cx+1+r*cosd(45),cy+1-r*sind(45));
ind = sub2ind(size(outmat), Y, X);
outmat(ind) = 1;
se = strel('disk',5); %// change the '5' value to affect thickness of the line
outmat = imdilate(outmat,se);
outmat = repmat(boolean(outmat),1,1,3);
outmat1 = zeros(size(outmat));
outmat1(outmat) = im(outmat);
imshow(uint8(outmat1));
Pixel values under each lines
Check the following code. I just did it for a grayscale image. You can now change it to a color image as well. Check and pls confirm this is what you wanted.
clear all;
i = rgb2gray(imread('hestain.png'));
imshow(i);
cr = floor(size(i,1)/2);
cl = floor(size(i,2)/2);
r = min(cr, cl);
a = 90;
r1 = cr;
c1 = size(i,2);
v1=[c1 r1]-[cl cr];
i2 = zeros(size(i,1),size(i,2),ceil(360/a));
for ri = 1:size(i,1)
for ci = 1:size(i,2)
v2=[ci ri]-[cl cr];
a2 = mod(-atan2(v1(1)*v2(2)-v1(2)*v2(1), v1*v2'), 2*pi) * 180/pi;
d2 = pdist([ci ri; cl cr],'euclidean');
if d2<=r
if ceil(a2/a)==0
a2 =1;
end
i2(ri,ci,ceil(a2/a)) = i(ri,ci);
end
end
end
figure;
for i=1:360/a
subplot(2,180/a,i);
imshow(mat2gray(i2(:,:,i)));
end
Sample output:
I created a 3D stick man walking in matlab and I want to add an image as the floor that he is walking on. I'm not sure how to go about doing this. I looked at an example of someone adding an image as a background. That is kinda what I want but I want it to appear as the floor. I going to give the stick man a trajectory and make him walk across the floor. Can anyone point me in the right direction.
Ok now I got it thanks to Andrey.
clear all
cyl = UnitCylinder(2);
sph = UnitSphere(2);
% Head
L1 = 2;
Head = translate(scale(sph,L1/2, L1/2, L1/2),0,0,L1+4.5);
Head.facecolor = 'yellow';
%Shoulder
r2 = 0.3;
L2 = 3;
Shoulder = translate(rotateX(scale(cyl,r2/2,r2/2,L2/2),90),0,0,5);
Shoulder.facecolor = 'red';
%Left Upper Arm
w1_s = [-20:4:20 20:-4:-20];
r3 = 0.3;
L3 = 2;
Upper_Arm_left = translate(scale(cyl,r3/2,r3/2,L3/2),0,0,-L3/2);
Upper_Arm_left.facecolor = 'red';
%Right Upper Arm
Upper_Arm_right = translate(scale(cyl,r3/2,r3/2,L3/2),0,0,-L3/2);
Upper_Arm_right.facecolor = 'red';
%Left Forearm
w2_s = [-5:1:5 5:-1:-5];
L3_f = 2.5;
Fore_Arm_left = translate(scale(cyl,r3/2,r3/2,L3_f/2),0,0,-L3_f/2);
Fore_Arm_left.facecolor = 'red';
%Right Forearm
Fore_Arm_right = translate(scale(cyl,r3/2,r3/2,L3_f/2),0,0,-L3_f/2);
Fore_Arm_right.facecolor = 'red';
%Chest
r4 = 2;
L4 = 2;
Chest = translate(scale(cyl,r4/2,r4/2,L4/2),0, 0, 5-L4/2);
Chest.facecolor = 'yellow';
%Weist
r5 = 1;
L5 = 2;
Weist = translate(scale(cyl,r5/2,r5/2,L5/2),0, 0, 5-L4-L5/2);
Weist.facecolor = 'yellow';
%Hip
L6 = 1.5;
Hip = translate(rotateX(scale(cyl,r2/2,r2/2,L6/2),90),0,0,5-L4-L5-r2/2);
Hip.facecolor = 'green';
%Left Upper Leg
r7 = 0.4;
L7 = 2.5;
L71 = (L6/2+r7/2);
L72 = 5-L7/2-L4-L5;
Upper_Leg_left = translate(scale(cyl,r7/2,r7/2,L7/2),0,0,-L7/2);
Upper_Leg_left.facecolor = 'green';
%Right Upper Leg
Upper_Leg_right = translate(scale(cyl,r7/2,r7/2,L7/2),0,0,-L7/2);
Upper_Leg_right.facecolor = 'green';
%Left Lower Leg
L7_f = 3;
Lower_Leg_left = translate(scale(cyl,r7/2,r7/2,L7_f/2),0,0,-L7_f/2);
Lower_Leg_left.facecolor = 'green';
%Right Lower Leg
Lower_Leg_right = translate(scale(cyl,r7/2,r7/2,L7_f/2),0,0,-L7_f/2);
Lower_Leg_right.facecolor = 'green';
angle1 = 0;
angle2 = 0;
for i = 1:120
angle1 = w1_s(rem(i,length(w1_s))+1);
angle2 = w2_s(rem(i,length(w1_s))+1);
Arm_left = combine(translate(rotateY(Fore_Arm_left,angle2),0,0,-L3), Upper_Arm_left);
Arm_right = combine(translate(rotateY(Fore_Arm_right,-angle2),0,0,-L3), Upper_Arm_right);
Arm_left = translate(rotateY(Arm_left,angle1),0,-L2/2,(5-L3/2)+L3/2);
Arm_right = translate(rotateY(Arm_right,-angle1),0,L2/2,(5-L3/2)+L3/2);
Leg_left = combine(translate(rotateY(Lower_Leg_left,-angle2),0,0,-L7), Upper_Leg_left);
Leg_right = combine(translate(rotateY(Lower_Leg_right,angle2),0,0,-L7), Upper_Leg_right);
Leg_left = translate(rotateY(Leg_left,-angle1),0,-L71,L72+L7/2);
Leg_right = translate(rotateY(Leg_right,angle1),0,L71,L72+L7/2);
Upper_Body = combine(Head, Shoulder, Arm_left, Arm_right, Chest, Weist);
Lower_Body = combine(Hip, Leg_left, Leg_right);
walker = combine(Upper_Body, Lower_Body);
cla
img = imread('peppers.png');
[X,Y] = ndgrid([-10 10],[-10 10]);
zImage = [-5 -5; -5 -5];
surf(X,Y,zImage,'CData',img,'FaceColor','texturemap');
% view([1 1 1]);
hold on
view(3)
set(gca,'xlim',[-10 10],'ylim',[-10 10],'zlim',[-6 6]);
renderpatch(walker);
camlight
box on
drawnow
pause(0.04)
end
I want to get something like this floor but I don't know how to do it. This person used a .fig in their code.This is a piece of what they used.
F1 = open('background1.fig');
background_gca = gca;
F2 = figure(2);
.
.
.
clf(F2)
copyobj(background_gca,F2);
view([-40,25])
set(gca,'xlim',[-5 5],'ylim',[-5 5],'zlim',[0 10]);
renderpatch(walker);
camlight
box on
% axis off
drawnow;
You had a small syntax error:
First, you reversed z and y :
surf(xImage,zImage,yImage,...)
Also, you should think about surf as a surface. Thus, all z should be 0
img = imread('peppers.png');
[X,Y] = ndgrid([-5 5],[-5 5]);
zImage = [0 0; 0 0];
surf(X,Y,zImage,'CData',img,'FaceColor','texturemap');
view([1 1 1]);
make that background image as GIF image, it feels like man walking on floor
otherwise visit here, useful video:http://www.youtube.com/watch?v=Ztb6_kIkXb8
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)