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

15 次查看（过去 30 天）
hadis ensafi
hadis ensafi 2022-5-17
编辑： Torsten 2022-5-22
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.

请先登录，再回答此问题。

采纳的回答

Torsten
Torsten 2022-5-17
Determine the center of the circle:
https://math.stackexchange.com/questions/1076177/3d-coordinates-of-circle-center-given-three-point-on-the-circle
Determine the parametric equation of the circle:
https://math.stackexchange.com/questions/73237/parametric-equation-of-a-circle-in-3d-space
Determine the range of theta to represent only the arc connecting the three points.
Up to you.
  6 个评论
显示 4更早的评论
hadis ensafi
hadis ensafi 2022-5-22
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 2022-5-22
编辑：Torsten 2022-5-22
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)

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 General Physics 的更多信息

标签

产品


版本

R2019b

Community Treasure Hunt

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

Start Hunting!

Translated by