On Wed, 05 Sep 2007 17:02:38 -0700, Jim Kelley
wrote:

Not "my claim," my report.

So be it. The absence of porous plates in operating radiometers tends
to cast doubt on your report that the plates must be porous.

Hi Jim,

You should distinguish between reporting, claims, and what you see.
The point about being porous is to substantiate the expectation of
explaining the full energy budget (and specifically for the Crookes
radiometer). I don't know how many times I have to emphasize this,
but NO METHOD achieves that balance.

The claim they must be porous arrives
through the math necessary to balance the kinetic forces.

But a balance of forces would result in the absence of an observable
effect. An imbalance in forces is required in order to produce
movement.

The balance is in the energy applied and the energy expended. You put
an HP into a car, and it will accelerate 550 foot-pounds/sec. You put
x photons into ANY radiometer, and the change in inertia WILL NOT
balance.

[This is why I expressed my question in Newtonian terms for the
benefit of the twins who are so devoted to the master (that they are
wholly lost in a simple 2 variable computation). The difference
between that computation and performance is extreme. What is more
compelling, is that it is quite a departure from what Quantum
Mechanics would predict. NO METHOD achieves that balance.]

Now, if we simply move to another radiometer (Nichols, Tear, Hull, and
Webb already recited) without that partial vacuum, the vanes still
move, and expressely by Radiation Pressure.

By a different mechanism and in the opposite direction, yes.

In essence, these instruments indicate,
not measure.

A description which applies beautifully to power meters as well, don't
you agree? ;-)

No. Power meters to even uncommonly high accuracy still conform to
Newtonian mechanics.

The coy context of the thread was measuring the mass of a Photon.
Absolutely no SI Units have been named or any quantitative values
offered (the rather standard omission from claims made here).
However, feel free to introduce your own side thread's goal or even offer a
guess (your own quatitative value for the mass).

I'd like to offer m = E/c^2 as a guess.

The photo-electron appears to even depart from that. More to follow.

73's
Richard Clark, KB7QHC