Difference between line vector and free vector

[mytch]

Hi,

I started to study Roy Featherstone's book "Rigid Body Dynamics Algorythms".

The book starts off by explaining Spatial Algebra, where translations and rotations are gathered in a 6-D vector using Plucker's coordinates.

At some point the book says;

"A line vector is a quantity that is characterized by a directed line and a magnitude.
A pure rotation of a rigid body is a line vector, and so is a linear force
acting on a rigid body. A free vector is a quantity that can be characterized by
a magnitude and a direction. Pure translations of a rigid body are free vectors,
and so are pure couples. A line vector can be specified by five numbers, and
a free vector by three. A line vector can also be specified by a free vector and
any one point on the line."

Can someone explain the difference between line vector and free vector in different words, especially the part where a line vector can be specified by five numbers.

my current understanding would be that a free vector is the common euclidean vector, but then in the formulation above the two seem to differ by the fact that one is characterized by a directed line and the other by a direction. What's the difference ?

Disclaimer; most of my algebra is self taught so if say some non-sense, that's why :)

Hopefully this is the right location for such a post.

Michael

 

[jedishrfu]

Welcome to PF!

My understanding is a free vector has direction and magnitude.

and it sounds like a line vector has additional components for theta and phi rotational angles (think spherical coordinate angles).

So a line vector could be used to describe a physical object in space including its rotation about the vector.

Does that make sense?

This may help too:

http://en.wikipedia.org/wiki/Rigid_body_dynamics

http://en.wikipedia.org/wiki/Rigid_body

 

[mytch]

Thank you, it makes sense for the most part :)

so if a line vector is defined by a vector plus theta and phi (relative to the basis axis?) the text would mean;

"A pure rotation of a rigid body is a line vector"; the vector would be 0 and theta and phi would be the rotation of the body relatively to the basis of the space.

"so is a linear force acting on a rigid body"; that one i don't get, why would a linear force need to be a line vector and not just a free vector.

"Pure translations of a rigid body are free vectors"; Ok

"so are pure couples"; OK but then how is it different from the linear force?

 

[mytch]

Here is the following part of the text that may help too, it's quite confusing to me.

"Let ^s be any spatial vector, motion or force, and let s and sO be the two 3D coordinate vectors that supply the Plucker coordinates of ^s."
[me; in Plucker coordinates s are 3 rotations angles around the basis axis, and sO a vector, though here it seems to be saying that they are 2 vectors]

"Here are some basic facts about line vectors and free vectors.
 If s = 0 then ^s is a free vector."
[me: yes only sO is left and that's is a vector]

"If s.sO = 0 then ^s is a line vector. The direction of the line is given by
s, and the line itself is the set of points P that satisfy OP X s = sO."
[me: in order to do a dot product here, it has to be 2 vectors, but then that's different from Plucker coordinates?]

"Any spatial vector can be expressed as the sum of a line vector and a free vector. If the line vector must pass through a given point, then the expression is unique.

Any spatial vector, other than a free vector, can be expressed uniquely as
the sum of a line vector and a parallel free vector. The expression (for a
motion vector) is


|     s    |     |  0   |
| sO - hs  |   + |  hs  | where h =  (s.sO) / (s.s)

[me: The | above are supposed to be big [] ]

This last result implies that any spatial vector, other than a free vector, can be described uniquely by a directed line, a linear magnitude and an angular magnitude. Free vectors can also be described in this manner, but the description is not unique, as only the direction of the line matters. "

Any help with this would be much appreciated.