Re: A new computation of G from the Cavendish experiment

On Aug 28, 8:34 pm, Edward Ruden <rudenbz...@xxxxxxxxx> wrote:
On Aug 27, 7:47 pm, pioneer1 <1pione...@xxxxxxxxx> wrote:
..

http://www.densytics.com/wiki/index.php?title=Cavendish_experiment_and_G

You need to make your calcuations more clear and self-contained. It
requires too much detective work to even figure out what your
variables are defined as in terms of measurements. Like, how do you
define theta? Deflection from equilibrium without gravity from either
big ball would be appropriate for it's occurance in the equations, but
is that how you define it, or is it deflection between going from M to
M' attraction (which should be 2*theta)? Such a confusion could be the
key to the discrepancy. It's not clear what *you* think the variables
should represent, so we'll never know.

Ok. I would like to make the calculations self-contained. Let me know
what is not clear. For theta, you might want to check Figure 1 here:

http://www.densytics.com/wiki/index.php?title=Cavendish_experiment_and_G

In Cavendish's pendulum, as shown in the figure, the scale division 20
was where the pendulum arm was at rest. At 20 divisions theta equals
zero. In this particular experiment he moved the weights from M to M'.
At M the rest point of the arm was at 18.01 divisions. When he moved
the weights to M' the rest point of the arm moved (as calculated by
Cavendish) to 24.04. I used r = 24.04 - 20 as the angle of
displacement to compute restoring torque tau = k theta. I computed
theta as theta = gyration arm / r = 0.0547 radians.

I don't understand why I should be using 2 theta. Can you explain?

Thanks.

.