basic question about units analysis

[dave...@gmail.com]

In this video: https://www.youtube.com/watch?v=BcJGgG4HmwE the author works through a very simple moment computation, with a length represented as a vector wedged with a force represented as a vector, yielding a torque represented as a bivector.  So I can see how a vector with units of meters wedged with a vector with units of Newtons yields a bivector with units of Newton-meters.  So far so good.

But... when things get more complicated, how do the units all work out, and what kind of units are attached to a multivector?  Do the different components of a multivector have different units all together in the general case?  I am thinking about how you might represent a wrench (3 DOF forces in Newtons, 3 DOF torques in Nm) as a multivector, being applied to a mass where the inertial tensor is another multivector of appropriate units.  Since f=ma (still, even in GA...) I would expect a multivector where the vector has units of m/s^2, and the bivectors have units of rad/s^2.  I am not sure how to get there.  For that matter, I am not sure how to represent the point of attachment for the wrench w.r.t. the CG of the mass?

-dave

[Manfred]

You may be assuming too much.  You cannot arbitrarily choose bits of a multivector to identify as parts of a wrench, and in particular, your assumed vector/bivector split seems unlikely to behave like a wrench in the algebra.  It would make sense to first find a text where wrenches are modelled in GA.  On units, it is not unusual to have a basis that allows us to express components of a vector in different units, e.g. metres and seconds for components of a spacetime offset vector in special relativity.

On Tuesday, 19 March 2019 18:28:58 UTC-4, dave...@gmail.com wrote:

[...]  I am thinking about how you might represent a wrench (3 DOF forces in Newtons, 3 DOF torques in Nm) as a multivector, [...]  I would expect a multivector where the vector has units of m/s^2, and the bivectors have units of rad/s^2.  [...]