how can i draw circle in 3D space with 3 point ?

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shahin hashemi
shahin hashemi 2018 年 2 月 17 日
コメント済み: Ahmad Adee 2018 年 7 月 14 日
Dear all
I need to draw a circle in space with 3 points that I have.
For example in the picture below I need to through the circle among the purple orange and the yellow dots.
and the arrays are :
% for yellow dots :
x=[0.0614989006404664,0.113882215391292,0.150982198682794,0.168720920205744];
y=[0.00968012746750092,0.0348743496847940,0.0721490692945351,0.117011183107267]
z=[0.00433547467220984,-0.0182991382514783,-0.0513287453937180,-0.0902019922333220]
% for purple dots :
x=[0.0538953885999417,0.100380017854628,0.133603344329503,0.149651163313763]
y=[0.0181936243771012,0.0397661500353233,0.0724936983973233,0.112238051664588]
z=[-0.0140617255229237,-0.0345013279495237,-0.0642363653195066,-0.0992745656450756]
% for orange dots :
x=[0.0601617429784619,0.112244770523508,0.149580182011997,0.167771471007159];
y=[-0.00252268954107996,0.0216775496992773,0.0583666486260463,0.102963217155185]
z=[-0.0134985732822516,-0.0353846070611784,-0.0679669312855640,-0.106648973683681]
  1 件のコメント
shahin hashemi
shahin hashemi 2018 年 2 月 17 日
編集済み: shahin hashemi 2018 年 2 月 17 日
and i should mention that i consider that it is possible to draw the circle through these dots cause the radius is constant and it is 12.5e-3

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採用された回答

Roger Stafford
Roger Stafford 2018 年 2 月 18 日
Let your three points be P1=[x1;y1;z1], P2=[x2;y2;z2], P3=[x3;y3;z3], and call the as yet unknown center P0 = [x0;y0;z0]. I will give you a partial solution which leaves some work for you to do.
v1 = cross(P2-P1,P3-P1);
v1 = v1/norm(v1); %Unit vector orthogonal to plane of 3 points
The center, P0, must satisfy these three equations:
dot(P0-P1,v1) = 0
dot(P0-(p2+P1)/2,P2-P1) = 0
dot(P0-(p3+P1)/2,P3-P1) = 0
Your task is to express these equations in a standard linear form A*P0=B where A is 3-by-3 and B is 3-by-1, so that matlab can solve for P0 as P0 = A\B. That gives you the center.
Next do this
v2 = P1-P0;
R = norm(v2);
v2 = v2/R;
v3 = cross(v2,v1);
v3 = v3/norm(v3);
Now you have mutually orthogonal unit vectors v2 and v3 both parallel to the plane of the desired circle. Finally do this:
t = linspace(0,2*pi);
P = repmat(P0,1,numel(t)) + R*(v2*cos(t) + v3*sin(t));
plot3(P(1,:),P(2,:),P(3,:),'y-') % This should give you your circle
If you want only an arc of a circle, change the linspace limits appropriately.
  2 件のコメント
shahin hashemi
shahin hashemi 2018 年 2 月 18 日
it works for me tanx a lot
Ahmad Adee
Ahmad Adee 2018 年 7 月 14 日
How do you linearize these equations? Can you share the code? I have the equation but can separate the x0, y0 and z0 terms.

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