Calculate Rotation matrix from 3 points

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Jimmy Neutron
Jimmy Neutron 2021-11-8
回答： William Rose 2021-11-8
I have 3 base coordinate points
p0 = [533,-422 -1];
px = [219 -57 -993];
py = [5 -70 -23];
I would like to calculate the transformation matrix above these points. How would I go about doing so? because I am quite lost...

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回答（2 个）

Matt J
Matt J 2021-11-8
编辑：Matt J 2021-11-8
You can use this File Exchange submission,
https://www.mathworks.com/matlabcentral/fileexchange/26186-absolute-orientation-horn-s-method
but it is not enough to have 3 points. You need 3 points both before and after the transformation (actually only 2 if you are only estimating rotation and not translation).
William Rose
William Rose 2021-11-8
Ran in:
@Jimmy Neutron,
A rotation in three dimensions has three degrees of freedom, so you need to know three "before" points and three "after" points.
I asssume you have given us the "after" points:
p0a = [533;-422;-1]; pxa = [219;-57;-993]; pya = [5;-70;-23];
I assume from the names of the ponts that the "before" points are
p0b=[0;0;0]; pxb=[1;0;0]; pyb=[0;1;0];
But those cannot be the real before points, because the "after" points are separated by more than 1 unit. Also, as @Matt J pointed out, the origin has been translated. You should check the "after" points to see that they form two equal-length legs of a right triangle. You may do this as follows:
fprintf('Length(pxa-p0a)=%.2f\n',norm(pxa-p0a));
Length(pxa-p0a)=1102.67
fprintf('Length(pya-p0a)=%.2f\n',norm(pya-p0a));
Length(pya-p0a)=634.96
fprintf('angle(pxa-p0a-pya)=%.2f radians\n',...
acos((pya-p0a)'*(pxa-p0a)/(norm(pya-p0a)*norm(pxa-p0a))));
angle(pxa-p0a-pya)=1.10 radians
We see that the sides are of unequal length and are not at right angles. Therefore you must reevaluate the original problem statement.

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