Re: [Toon-members] SE3: relation of input-translation to output-translation

[Tom Drummond]

Rafael

If you exponentiate a mix of a translation and a rotation, the easiest way to understand it is using the following formula for exp:

exp(x) = limit as n goes to infinity of  (1+x/n) ^ n

or imagine taking a tiny step of 1/100 of your translation and 1/100 of your rotation and then repeat this 100 times (or whatever big number you prefer).  In other words the translation and rotation get mixed up in each other.

Imagine exponentiating the vector (1,0,0,0,0,1) = 1 unit of translation in the x direction and 1 unit of rotation around the z axis.  What you do is travel along a curved path that starts in the x direction and is 1 unit long and curves through 1 radian in the x - y plane.  This is equivalent to a rotation of 1 radian about an axis parallel to z through the point x=0, y=1.

Hope that helps your understanding

Tom

On Tue, Jun 23, 2009 at 4:10 PM, Rafael Spring <address@hidden> wrote:

Hello gentlemen,

I am using TooN for a while now and I really appreciate it. Scope, usability and speed are really great and I couldn't imagine working without it.
However, today, I've had some weird behavior and I'm unsure if TooN is supoosed to behave like that (Btw. I am using TooN beta3) or if my understanding of SE3s is insufficient. The question boils down to what happens to an input translation after exponentiating.
Consider the following sourcecode:

----
Vector<3> rotation = makeVector(0.4, 0.6, 0.15);        //arbitrary numbers..
Vector<3> translation = makeVector(-1, 0, 0.5); //arbitrary numbers..
Vector<6> mu;
mu.slice(0,3) = translation;
mu.slice(3,3) = rotation;
       
SE3<> se = SE3<>::exp(mu);
Vector<3> t = se.get_translation();

----
Now, I was expecting:

t == (-1, 0, 0.5)

following the assumption that a euclidean transformation by an SE3 equals to an orthogonal transformation (exp'ing the rotation part of mu) + a translation.
The true value, however, is:

t == (-0.789750 , -0.19886, 0.734775)

So I reversed my assumption: It could be a translation by t followed by an orthogonal transform (this would correspond to the translation-part of mu being the inverse camera translation as seen from camera 1 and get_translation() being the inverse translation as seen from camera 2). This assumption, however, was neither confirmed:

----
// make a rotation-only SE3
Vector<6> mu_ = mu;
mu_.slice(0,3) = makeVector(0,0,0);
SE3<> se_ = SE3<>::exp(mu_);

// make a translation-only SE3  
Vector<6> mu__ = mu;
mu__.slice(3,3) = makeVector(0,0,0);
SE3<> se__ = SE3<>::exp(mu__);
       
// 1st translate then rotate
Vector3 t_ = (se_ * se__).get_translation();

//  t_ == (-0.5293 , -0.412372 , 0.89431)

----

Now to the part beyond my understanding: The second assumption holds (roughly) if I halve the rotation amount in mu:

----
Vector<6> mu_;
mu_.slice(0,3) = makeVector(0,0,0);
mu_.slice(3,3) = rotation / 2.0;
SE3<> se_ = SE3<>::exp(mu_); //se_ is half the rotation
       
Vector<6> mu__ = mu;
mu__.slice(3,3) = makeVector(0,0,0);
SE3<> se__ = SE3<>::exp(mu__); //se__ is only translation
       
Vector3 t__ = (se_ * se__).get_translation();   // 1st translate then rotate

// t__ == (-0.8024 , -0.19517 , 0.75371)
// t__ is roughly  == t

----
Is this within the general logic of SE3 members (am I missing something?) or might there be something wrong with the SE3 class?

Thanks!
Rafael


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--
Dr T Drummond                               http://mi.eng.cam.ac.uk/~twd20
Machine Intelligence Laboratory
Department of Engineering
University of Cambridge