how to create an arc path from 3 points(x, y, z) in plane?

14 views (last 30 days)
I want to create an arc trajectory cross over n=3 points P(n)=(x, y, z), I decided to draw a circle over 3 points in plane. so I have center, radius, theta (angle in x, y plane) and phi(angle around z axis), and I know the position of 3 points (x, y, z), How can I calculate theta in parametric equation of the circle for each points? then extract an arc between p1 , p2 and p3 from this circle? I implemented this program in MATLAB.. Thanks a lot.

Accepted Answer

Torsten
Torsten on 17 May 2022
Determine the center of the circle:
Determine the parametric equation of the circle:
Determine the range of theta to represent only the arc connecting the three points.
Up to you.
  6 Comments
hadis ensafi
hadis ensafi on 22 May 2022
Hi !! thanks for your answer! my circle equation is: (center+sin(thetai)*rad.*v1+cos(thetai)*rad.*v2;) in 3D plane for each point on the circle!
I want to calculate theta for P1 and P2 and P3 to extract only sector from P1 to P3 that passes through P2, generally!
According to you, should I put my point (P!,P2,P3) coordinate instead of pi/2 thar you wrote? like this:
funX = @(thetai) center(:,1)+sin(thetai)*rad.*v1(:,1)+cos(thetai)*rad.*v2(:,1);
theta = fsolve(funX,p1);
I did it, but did not work.
please please guide me ! thanks a lot!
Torsten
Torsten on 22 May 2022
Edited: Torsten on 22 May 2022
theta1init = ...;
theta2init = ...;
theta3init = ...;
funX = @(thetai) center(:,1)+sin(thetai)*rad.*v1(:,1)+cos(thetai)*rad.*v2(:,1)-P1;
theta1 = fsolve(funX,theta1init);
theta1 = mod(theta1,2*pi);
funX = @(thetai) center(:,1)+sin(thetai)*rad.*v1(:,1)+cos(thetai)*rad.*v2(:,1)-P2;
theta2 = fsolve(funX,theta2init);
theta2 = mod(theta2,2*pi)
funX = @(thetai) center(:,1)+sin(thetai)*rad.*v1(:,1)+cos(thetai)*rad.*v2(:,1)-P3;
theta3 = fsolve(funX,theta3init);
theta3 = mod(theta3,2*pi);
thetamin = min(theta1,theta2,theta3);
thetamax = max(theta1,theta2,theta3);
If this does not work, please report the error message or explain why the result is unexpected.
Here is some code in 2d:
P1 = [0.5*sqrt(2) 0.5*sqrt(2)];
theta1init = 0.7;
fun = @(theta) cos(theta)*[1 0] + sin(theta)*[0 1] - P1;
theta1 = fsolve(fun,theta1init);
theta1 = mod(theta1,2*pi)
fun(theta1)

Sign in to comment.

More Answers (0)

Categories

Find more on General Physics in Help Center and File Exchange

Products


Release

R2019b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!