Re: A new computation of G from the Cavendish experiment

On Oct 15, 8:32 am, pioneer1 <1pione...@xxxxxxxxx> wrote:

How to translate these into initial conditions?

I tried to explain this already. Think about what you have in
mathematical terms. The minimum set of info we need is:
x1, (x2,t2), x3, and (x4,t4) and the fact that x1 and x3 correspond to
positive and negative extrema. The even numbered subscripts are for
intermediate data. The "rest point" doesn't count; that's just an
estimate Cavendish made, and we're trying to do better than Cavendish.
It's annoying that we don't have the times for the extrema, but that's
understandable. The arm is moving so slowly then, it's hard to
estimate that time accurately anyway. Fortunatly, the intermediate
values make up for it.

Now think about what you *want*. If you're solving the "exact" dif eq,
you want x(0), x'(0), G, torsion constant, and damping constant. Those
are *all* needed to solve it numerically. But, since it's solved
numerically, really all you can do it iterate these values until you
converge on a solution where all your knowns are satisfied. You will
find that to be very difficult to code.

However, if you use my linearized solution, you want to find A,
B, ,,,, E from the known parameters. The extremum statements can be
expressed as x'=0 (you need to know how to differentiate an analytic
equation). Other than that, it's "just" algebra and a bit of numerical
root finding in the end. It gets pretty sticky, but it can be done. I
can't make it any simpler than that unless I just went and did it
myself.

.