[R-390] From the horse's mouth

[rbethman]

The horse crosseth the line - a winner - but he finally died.

I apologize for my exhumation, but I felt a point or two had to be made.  In
AC the voltage peaks to the same value on either half of the cycle.  What I
felt was "really" missing was the concept of power and time.  There is no
time factor in a power equation.  That was the hardest thing to get across.
Once we got that established we could move on and get to the "real" meat!

Bob - N0DGN
(I'm not much for titles - I feel that they are useless...but)
Board Certified Nuclear Power Plant Operator and Maintenance Technician -
and Instructor - Since September 1974



----- Original Message -----
From: "Joe Foley" <redmenaced@yahoo.com>
To: <r-390@mailman.qth.net>
Sent: Sunday, January 13, 2002 2:54 PM
Subject: [R-390] From the horse's mouth


>
>
>  > I see the problem...we are on the same side but
>  > speaking a different
>  > language. I am talking about cumulative
>  dissipation
>  > divided by time,
>  > or avaerage power. As I said, the power delivered
>  > during that half
>  > wave would be at twice the rated current, which
>  the
>  > tubes MIGHT
>  > be able to handle for a little while at a 50% duty
>  > cycle. But the
>  > continous dissipation would also be too high as
>  > well, and the filaments
>  > would still be running too hot. We agree.
>  >
>  > Those that argue for the diode would probably
>  think
>  > that if you could
>  > somehow just eliminate nine of ten cycles, you
>  could
>  > use ten times the
>  > voltage... Maybe we could use twelve million volts
>  > for one microsecond
>  > per second, wouldn't that deliver twelve watts
>  > continous?
> +++++++++++
>  Yes,
>
>  The language took a beating in this discussion
>  because
>  of the diverse backgrounds and education levels.  A
>  very detailed description starting from the basics
>  would have put everyone on the same page and level.
>
>  The patience of those who know is appreciated.
>
>  So sayeth the dead horse!
>
>
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