How to find the orientation of the line of the intersection between two planes?
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Is there any method/indiacator that i can use to know the orientation of the the intersection line between two planes( using Dual Plucker Matrix )?
I used the follwoing code get the line:
P1 =[177668442.453315 ,-102576923.076923, 0];
P2 =[ -102576923.076923 ,177668442.453315 ,-102576923.076923];
P3= [0, -102576923.076923, 88834221.2266576];
P11= [152763459.308716 , -102576923.076923, 0];
P22=[ -102576923.076923, 183536536.231793 , -102576923.076923];
P33= [0, -102576923.076923, 91768268.1158967];
A=null([[P1;P2;P3],ones(3,1)]); %plane 1
B=null([[P11;P22;P33],ones(3,1)]); %plane 2
L=A*B.' - B*A.' %line of intersection
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Matt J
2023년 2월 5일
편집: Matt J
2023년 2월 5일
P1 =[177668442.453315 ,-102576923.076923, 0];
P2 =[ -102576923.076923 ,177668442.453315 ,-102576923.076923];
P3= [0, -102576923.076923, 88834221.2266576];
P11= [152763459.308716 , -102576923.076923, 0];
P22=[ -102576923.076923, 183536536.231793 , -102576923.076923];
P33= [0, -102576923.076923, 91768268.1158967];
A=null([[P1;P2;P3],ones(3,1)]); %plane 1
B=null([[P11;P22;P33],ones(3,1)]); %plane 2
L=A*B.' - B*A.' %line of intersection;
N=null(L);
a=N(:,1)+N(:,2);
b=N(:,1)+2*N(:,2);
direction=normalize( b(1:3)/b(4)-a(1:3)/a(4) ,'n')
or,
direction=normalize( cross(A(1:3),B(1:3)) ,'n')
추가 답변 (1개)
Torsten
2023년 2월 5일
편집: Torsten
2023년 2월 5일
If you look at the next lines in Matt's code, he creates 100 points on the line.
Thus in the modified code below, Pstart could be taken as a point on the line and d as a direction vector for the line emanating from Pstart.
Did you mean something like this ?
P1 =[177668442.453315 ,-102576923.076923, 0];
P2 =[ -102576923.076923 ,177668442.453315 ,-102576923.076923];
P3= [0, -102576923.076923, 88834221.2266576];
P11= [152763459.308716 , -102576923.076923, 0];
P22=[ -102576923.076923, 183536536.231793 , -102576923.076923];
P33= [0, -102576923.076923, 91768268.1158967];
A=null([[P1;P2;P3],ones(3,1)]); %plane 1
B=null([[P11;P22;P33],ones(3,1)]); %plane 2
L=A*B.' - B*A.'; %line of intersection
N=null(L);
t=linspace(-.1,.1);
xyz=N(:,1) + N(:,2)*t;
xyz = xyz(1:3,:)./xyz(4,:);
P1 = xyz(:,1);
P2 = xyz(:,2);
Pstart = P1
d = (P2-P1)/norm(P2-P1)
댓글 수: 13
Torsten
2023년 2월 6일
편집: Torsten
2023년 2월 6일
The usual representation of the plane of intersection is given by all solutions x of a linear system of the form
A*x = b
Here, A is a 2x4 matrix and b is a 2x1 vector.
The rows of A are the normal vectors of 2 hyperplanes in the 4d space.
The vector b somehow represents the distance of these hyperplances to the origin.
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