Eqivalent rigid body rotation from translation and rotation

Hi,

In 2D, I can see that any combinations of translation plus rotation about the origin can be transformed into a pure rotation but perhaps about a different point other than the origin.

I wonder in 3D, is this possible? And, how to find out the equivalent pure rotation then? thanks!

Answer 1

[Matt J]

If the roto-translation is

y=R*x+t

where R is a 3x3 rotation matrix and t is a 3x1 translation vector, then to find the center of rotation, one must solve the fixed point equation

xc=R*xc+t

One solution is

xc=pinv(eye(3)-R)*t

Then the original roto-translation can equivalently be written

 (y-xc)=R*(x-xc)

or in other words a pure rotations with respect to the origin, xc. However, the redefined origin xc is not unique. It can be any point along the axis

 xc + null(eye(3)-R)*scalar

Answer 2

[Ryan G]

I'm not sure how you convert a translation into a rotation, but there are a number of rotational transformations available in MATLAB.

Answer 3

[Jan]

The relative position of two 3D bodies can be described by a rotation around and a translation along the helical axis, see e.g. http://en.wikipedia.org/wiki/Screw_axis and http://www.kwon3d.com/theory/jkinem/helical.html.

A translation is a rotation around a point with an infinite distance. While in 2D these can be the two points on the line orthogonal to the translation (if one can count points in an infinite distance at all), in 3D these are all point in the orthogonal plane.

Answer 4

[francesco]

how can I describe in 2d the combination of a circle rotation (certain spindle) and a fixed translation over x axis feed?

n=1 deg=1 for i=1:5 for deg=0:360 sample(n:1)=n sample(n:2)=deg n=n+1 end end comet(sample(:,1)+400*cosd(sample(:,2)),40*sind(sample(:,2)))

this is wrong there is no real translation works as a scatter plot :(