Function returns different outputs with same inputs
7 views (last 30 days)
Show older comments
Hello everyone,
Context (not relevant for Problem):
For a school project, im playing around with "rotation matrix" and "Quaternions", i already calculated the Quaternions with a different methode, and now i want to compare the results i get from rotationsmatrix2quaterionen(numeric_result) with the ones i calculated earlier. The function seems to work, rr is nearly zero (not exactly because of numeric imprecision)
Problem:
I don't understand why 'rr' and 'gh' give me different results. The calculation should be the same both times, only that one time at first i write the output in a seperate array.
I assume i made some stupid mistake, but i can't find it.
numeric_result = [
0.8138 -0.4063 0.4155 10.0000;
0.4698 0.8808 -0.0590 12.5000;
-0.3420 0.2432 0.9077 64.0000;
0 0 0 1.0000]
e0 = 0.9490
e1 = 0.0796
e2 = 0.1996
e3 = 0.2308
[e0_, e1_, e2_, e3_] = rotationsmatrix2quaterionen(numeric_result)
rr=[e0, e1, e2, e3]-[e0_, e1_, e2_, e3_]
% Output:
% rr = 1×4
% 1.0e-04 *
%
% 0.1370 -0.1127 0.4496 0.0109
gh=[e0, e1, e2, e3]-rotationsmatrix2quaterionen(numeric_result)
% Output:
% gh = 1×4
% 0.0000 -0.8694 -0.7494 -0.7182
function [e0, e1, e2, e3] = rotationsmatrix2quaterionen(RotMat)
% e = e0 + e1*i + e2*j + e3*k
R=RotMat;
e0 = 0.5*sqrt(1+R(1,1)+R(2,2)+R(3,3));
e1 = (R(3,2)-R(2,3)) / (4*e0);
e2 = (R(1,3)-R(3,1)) / (4*e0);
e3 = (R(2,1)-R(1,2)) / (4*e0);
end
0 Comments
Accepted Answer
Stephen23
on 5 Jan 2024
Edited: Stephen23
on 5 Jan 2024
The difference is very simple:
Here you return FOUR output arguments from your function call:
[e0_, e1_, e2_, e3_] = rotationsmatrix2quaterionen(numeric_result)
Here you only return ONE output argument from your function call:
gh=[e0, e1, e2, e3]-rotationsmatrix2quaterionen(numeric_result)
2 Comments
Stephen23
on 5 Jan 2024
Edited: Stephen23
on 5 Jan 2024
The first output argument has a value of approx. 0.9490, so your second attempt simply subtracts one scalar from a vector of four values. Lets try it right now (due to floating point accuracy the result will only be similar to what you showed):
[0.9490,0.0796,0.1996,0.2308] - 0.9490
You seem to be writing your MATLAB code as if it were Python.
More Answers (0)
See Also
Categories
Find more on Logical in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!