sci.physics.research

On Aug 30, 10:55 am, Edward Ruden wrote:

> I've found your mistakes and corrected them below.

This is great! Many thanks for your help with this. I agree with
everything you wrote and I corrected the computations in the wiki and
I gave you credit for the corrections (I hope this is ok).

> Now, as for theta, you can't assume that theta=0 when the scale reads
> 20 divisions of 1/20'th of an inch. You'd have to rotate the weights
> to the neutral positiion half-way between the two extremes and measure
> the scale to determine that.

Ok. But Cavendish neglected this step. He computed the rest points
from 3 successive measurements and he included them in his chart.

>You said, however, only that measurements
> where taken with M' and M at the near positions, at which point the
> equilibrium position on the scale read S = divisions and S' =
> divisions, respectively.

I made a couple of mistakes with numbers here. Now I corrected the
figure 1 in the wiki. Correct numbers are:

S = and S' =

>...r, then should be defined as r=(S-S')/2= div =
> cm.

Ok. With the corrected numbers I got:

r = (S-S')/2= div = " = cm

>Theta, then is r divided by the radius of the arm.

Right, my mistake. Corrected:

theta = / =

> ... the wire is torqued by *two* masses on each end of the arm,
> resulting in a torque of (L/2)*2GMm/s^2.

Ok. I corrected this, and I get

*10^-8

This is of the recommended value. So, very close. But I think
that 5 digit accuracy is not justified here. Cavendish gave the
density of the earth to only two digits, so maybe the result of this
computation should be

*10^-8.

Thanks again for your help with this. I'll check your computation when
it is up on your site. I am glad that someone else is interested in
the original Cavendish experiment. I have other questions about the
experiment that I hope to post soon.