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cone plot (request for help)

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Or Peer
Or Peer el 4 de Mayo de 2019
Comentada: Or Peer el 4 de Mayo de 2019
I have 4 normalized vectors that represent the edges of error ellipse, how do I plot a cone between them that represent the error area up until a certain point(x*10,y*10,z*10) for example.
in other words, a semi cone with no head ( there is an "head" if the error in the head is tiny enought to count as the same point for every vector).
for example I have a data of vectors 2x3 where first line is the direction vector and line 2 is point of origin
left:
0.000777551633326096 -0.324338092709069 -0.904293501718821
-0.0155631822372405 6.49182977305030 18.1
right:
-0.0740720973626052 -0.324366480207892 -0.904433163158820
1.48237041373031 6.49139541860418 18.1
up:
-0.0366252121589512 -0.373079389279788 -0.904694864673741
0.732751302082479 7.46410442862344 18.1
down:
-0.0366693335703280 -0.275625183637174 -0.904031684016283
0.734172208073828 5.51840816205618 18.1
right now I have a code of drawing ellipse at a given height which then I loop to give me the ellipse error over length
function ellipse(left,right,up,down,height)
left(1,1:3)=left(1,1:3)*(height/left(1,3))+left(2,1:3);
right(1,1:3)=right(1,1:3)*(height/right(1,3))+right(2,1:3);
up(1,1:3)=up(1,1:3).*(height/up(1,3))+up(2,1:3);
down(1,1:3)=down(1,1:3).*(height/down(1, 3))+down(2,1:3);
x=mean([min([left(1),right(1),up(1),down(1)]),max([left(1),right(1),up(1),down(1)])]);
y=mean([min([left(1,2),right(1,2),up(1,2),down(1,2)]),max([left(1,2),right(1,2),up(1,2),down(1,2)])]);
t=0:0.1:2*pi;
Xt=0.5*norm(left-right)*cos(t)+x;
Yt=0.5*norm(up-down)*sin(t)+y;
Z=t-t+height;
plot3(Xt,Yt,Z,'color','green');
hold on;
  7 comentarios
darova
darova el 4 de Mayo de 2019
If you have all profiles for surface, just rebuild your function header
function [Xt, Yt, Z] = ellipse(left,right,up,down,height)
Write all profiles in matrix
i = 1;
for height=1:-50:-1000
[x,y,z] = ellipse(left,right,up,down,height)
X(:,i) = x;
Y(:,i) = y;
Z(:,i) = z;
i = i + 1;
end
And create surface
surf(X,Y,Z)
Or Peer
Or Peer el 4 de Mayo de 2019
thank you, the code worked.

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