Ian White, G3SEK wrote:
V(v) and V(i) are *added* - vectorially, as two RF voltages connected in
series - and then the sum voltage is detected. No logs are taken and
they are not multiplied together in any other way. The detected voltage
(now DC) is squared by the meter scale law to indicate power.

Yes, when you add two values on a slide rule, you are accomplishing
multiplication because the scales are logarithmic. When you add two
values of voltage, you are multiplying them if the meter scale is
properly calibrated. Log(V1^2) = Log(V1*V1) = Log(V1) + Log(V1)

You were caught in a totally incorrect statement about the meter scale
law, but you can't stop dodging. Just read again what you wrote:

I still don't understand why log(V1)+log(V1) doesn't equal log(V1*V2)
which doesn't equal log(V1^2).

Here's an example: Let's say we have a slide rule where the distance
between the '1' index and the '2' is one inch. We set the index on
'2' and read the value of the product off the '2' on the other scale.
All we have done is add one inch to one inch and everyone knows that
is two inches. But because of the logarithmic scale we read '4' as the
answer. Thus the answer to '2' times '2' is '4' even though all we did
was physically add one inch to one inch. How can one inch plus one inch
yield an answer of '4'? The same way one volt plus one volt can yield
an answer of 200 watts.

Let's say one volt inside the Bird equals 100 volts on the transmission
line and one volt inside the Bird equals 2 amps on the transmission line.
When the meter measures two volts, we calibrate the Bird at 200W. When the
meter measures four volts, we calibrate the Bird at 800W. That's similar
to the logarithmic calibration of the slide rule. We are adding one volt
to one volt but we are reading the product off the wattmeter scale.

are you saying that log(V^2) is not
equal to 2*log(V)? I say log(V^2)=2*log(V), i.e. the square law power
function is a linear function using logarithms. That's the basis of
the Bird wattmeter design. I would certainly like to see a proof that
log(V^2) is not equal to 2*log(V). (So would a lot of people wearing
white coats. :-))

So "The square law power function is a linear function using logarithms".

Of course. Log(V1^2)= Log(V1)+Log(V1) does it not? Is that not a linear
function? That's how slide rules work. You add logarithms to accomplish
multiplication. What am I missing?
--
73, Cecil http://www.qsl.net/w5dxp



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