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Re: Real data from real experiments.
At 05:25 AM 11/11/96 +0000, you wrote:
>From dbell-at-baygate.bayarea-dot-netSun Nov 10 21:33:27 1996
>Date: Sat, 9 Nov 1996 22:10:53 -0800 (PST)
>From: Dave Bell <dbell-at-baygate.bayarea-dot-net>
>To: Tesla List <tesla-at-poodle.pupman-dot-com>
>Subject: Re: Real data from real experiments.
>
>Hi, Brent!
>
> Exactly... Actually, I mentioned the hydraulic analogy
>to Jack a while back, with no better comprehension resulting.
>I suggested comparing the two caps to two water tanks, one
>filled, the other, identical tank empty. Open a valve that
>connects the two at the bottom. After oscillating a bit,
>the water levels end up at 1/2 the original height, and
>a total of 1/2 the original POTENTIAL energy is available.
>Certainly, nothing was "radiated", and we can't easily
>believe the water was heated that much by friction as it
>passed through the valve and pipes.
>
>> The best analogy I can think of relating to the 'missing 1/2 of the
>> charge' puzzle is that of water stored behind a dam:
>
---------------------------------------
>Dave
One of the problems with "the water levels end up at 1/2 the original
height" is that this is not always correct. The levels can be different than
1/2 original height when the tanks are of different shapes or are not equal.
How would you calculate the difference in heights compared to the original
height for unequal tanks?
Also,"a total of 1/2 the original Potential energy is available" can be
misleading because it is not clear what you mean. Do you mean that the other
half of the energy is in losses? Note that when the tanks are unequal the
energy in the tanks are not 1/2 the original energy. How would you calculate
the difference in energy when the tanks are unequal?
The water level in two unequal tanks would be level under certain conditions
but not 1/2 the original height except under certain conditions of tank
shapes. Another interesting problem. The same is true of unequal
capacitors. The voltage across the two capacitors would be the same but not
1/2 the original voltage as the Coulomb equation might be interpreted. Both
the energy and Coulomb equations must be properly used to arrive at a
correct answer. Tesla coils must also meet the laws of conservation of
energy and conservation of charge.
Jack C.