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I have made function in Matlab, that shale calculate the rotation matrix for fi degrees around unit vector k.

function R = rot(k,fi)
    rad = fi*(pi/180);

    % This is just to make it easyer to read!
    x = k(1);
    y = k(2);
    z = k(3);

    % Create a 3x3 zero matrix
    R = zeros(3,3);
    % We use the formual for rotationg matrix about a unit vector k

    R(1,1) = cos(rad)+x^2*(1-cos(rad));
    R(1,2) = x*y*(1-cos(rad))-z*sin(rad);
    R(1,3) = x*z*(1-cos(rad))+y*sin(rad);

    R(2,1) = y*x*(1-cos(rad))+z*sin(rad);
    R(2,2) = cos(rad)+y^2*(1-cos(rad));
    R(2,3) = y*z*(1-cos(rad))-x*sin(rad);

    R(3,1) = z*x*(1-cos(rad))-y*sin(rad);
    R(3,2) = z*y*(1-cos(rad))+x*sin(rad);
    R(3,3) = cos(rad)+z^2*(1-cos(rad));
end

But it don't seem to work, i have a test rules

rot(k,0) == I
rot(k,fi) == rot(k,fi+360)
rot(-k,fi) == transpose(rot(k,fi))
rot(k,f1)*rot(k,f2) == rot(k,f1+f2)
rot(k,fi)*k == k

My function is always successfull on rule nr 1, 3 and sometime on 2. What am i doing wrong here? is my function not correct?

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How different are the results? Is it a case of rounding error? Also, $k$ must be a unit vector. –  Daryl Oct 24 '12 at 21:13
    
If it's the test rot(k,fi) == rot(k,fi+360) that's failing, this is caused almost certainly by rounding errors -- you're only using the second argument as an argument for the sine and cosine, which should be invariant. One would expect rounding errors, though -- are these comparisons exact? If so, I'd be surprised if the test didn't fail. –  joriki Oct 24 '12 at 21:24
    
The diffrence are small! example sometime are 0 compared with a -0 –  DoomStone Oct 24 '12 at 22:17

1 Answer 1

You are using equality checks on floating point values.

I suggest you do not check for equality but see whether the absolute difference is less than epsilon.

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