Re: A new computation of G from the Cavendish experiment

I've found your mistakes and corrected them below. Incidentally, I
myself have personnally measured G to within 1% in 1980 in Junior Lab
when I was a physics major at Case Western Reserve University. I made
improvements to the Cavendish balance and analysis techniques that
resulted in a more accurate measurement than any previous student had
ever obtained on that instrument, according to the profession. So, I
can personnally assure you that G is within 1% of its moderrn quoted
value without any need for conformity with the hide-bound reactionary
physics establishment :-)

I will scan and upload my final report to my reprints page at
http://home.comcast.net/~rudenbiz/physics/pubs/index.htm
Give me a week or so.

On Aug 29, 11:37 pm, pioneer1 <1pione...@xxxxxxxxx> wrote:

theta as theta = gyration arm / r = 0.0547 radians.

The above is an example of exactly what I mean when I say you don't
define things clearly. You define r in your Fig. 1 as the distance on
the scale of the arm with the mass in the M' location to the center of
the scale (which you assume to correspond to theta=0). But what is
"gyration arm"? The arm is a physical object, not a measurement.
Furthermore, r should be in the numerator. All your variables should
be defined in terms of elementary measurements of distance and mass
for people to understand you.

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. 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 = 24.04 divisions and S' =
18.01 divisions, respectively. Since the apparatus is symmetric, that
means the gravity-free equilibrium position is half-way between
(21.025 div). r, then should be defined as r=(S-S')/2= 3.015 div =
0.383508 cm. Theta, then is r divided by the radius of the arm. For
this, you have an inconsistency in the definition of L. In the linked
calculation of moment of inertia it's defined as the length of the arm
L=186.18 cm, but on the main page, its numerical value is half that
(ie, the radius). I'll take the former definition, so theta = 2r/
L=0.00412 cm. This itself only provides a minor to you G calculation,
reducing it to 1.341E-7.

A larger error apparently results from you confusion about L and/or
what the torque is. The torque due to Gravity expression in you main
page is correct, but only if you assume that L is the *length*. Recall
that the wire is torqued by *two* masses on each end of the arm,
resulting in a torque of (L/2)*2GMm/s^2. Since you, however, use the
radius definition, you need to divide your G estimate by two to give
G=6.708E-8. This is about 1% from it's modern value, just as Cavendish
obtained.


.

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