Gordy,
I agree. It would be nice to have consensus.
Unfortunately, when I did my calculations it was one of those “can’t sleep at night so let’s do calculus” sessions and I didn’t save the paper I was working on. I also got lazy and used one of those online integral calculators. I don’t really trust machines.
The angular momentum for a point particle is classically represented as a pseudovector r × p, the cross product of the particle's position vector r (relative to some origin) and its momentum vector; the latter is p = mv. Using the center of rotation that equates to the center of mass will give you an approximation for the rotation vector. It was how I came up with the first estimate of 2. In that calculation I also simplified the velocity field around the loop to decrease linearly.
Of course the velocity field around the loop doesn’t decrease linearly. It can be derived using the cycloid equation or to use the transform that you showed earlier (add a constant to the x velocity). Since it’s linear it will be the same in any Cartesian coordinate system as long as the orientation of the axes doesn’t change, i.e. it’s okay to change the origin.
The rotation vector is another issue since the center of the loop in the loop centric frame and at the bottom of the loop in the fixed frame.it’s magnitude does not remain constant in that view.
"There can be only one." - The Highlander.
PS. I have a flying tank. Your argument is irrelevant.
PSS. How to generate a climbing loop through control of the casting stroke is left as a (considerable) exercise to the reader.