W5DXP wrote:
Ian White, G3SEK wrote:
As you yourself say, the Bird senses current and voltage samples; and
these are then added (forward) or subtracted (reverse). There is no
physical mechanism inside the Bird by which E x I multiplication takes
place, so power is not directly sensed or measured.

I'm sorry, but that is incorrect. Refresh your memory on how a slide
rule multiplies. Two linear values of length are physically added but
a product is read from the results. The secret to multiplying using
linear scales is to calibrate that linear scales logarithmically.
It's easy to understand for a slide rule, is it not? The same applies
to the Bird wattmeter scale.

You are mistaken. You may want it to be so, but it ain't.

The Bird "power" scale is almost a pure square law, indicating power as
being proportional to V-squared, where V is the voltage detected by the
diode in the slug. No log function is involved.

This isn't just theory. I and another person have independently verified
the Bird 43's scale law by experiment, measuring the meter current at
various power levels and then determining the mathematical law of the
meter scale. We even took the current measurements to the level of
precision where we could see that a few of the Bird's individual scale
marks weren't drawn in quite the right place relative to all the rest.
We confirmed that the calibration fits the square law almost perfectly
near the top end of the scale, but deviates by a few percent near the
bottom end where diode's threshold voltage becomes significant.


The Bird takes two linear voltages and adds them.

Agreed. One voltage is a sample of the voltage on the centre conductor
of the Bird's internal transmission line, and the other is derived by
transformer action from the current on the line. When sensing in the
forward direction, these voltages are made to add in phase; then the sum
is detected by the diode.

When the slug is turned around, the current sample reverses its phase
but the phase of the voltage sample stays the same, so now the two
subtract. With a matched load, the two voltages are arranged to be equal
and opposite so you get a null in the reverse direction. That's what
gives the Bird its directional properties.

But the power scale on
the Bird meter face is scaled logarithmically just like it is on a slide
rule.

Sorry, that is factually incorrect, so everything you concluded from it
is incorrect too.



--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
Editor, 'The VHF/UHF DX Book'
http://www.ifwtech.co.uk/g3sek

Ian White, G3SEK wrote:
Wrong. The Bird has no second meter coil, and no other physical
mechanism to multiply the E sample by the I sample.

Remember when nomograms were popular? Do you think you could
design a nomogram with variables of voltage and current to yield
power? Where's the physical multiplication mechanism?
--
73, Cecil http://www.qsl.net/w5dxp



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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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Ian, G3SEK wrote:
"The detected voltage (now DC) is squared by the meter scale to
indicate power."

That`s right. It only takes about 1.4X the deflection to represent 2X
the power.

Cecil was saying the same with his slide rule illustration. To square a
number, you multiply the logarithm of the number by 2 and take the
antilogarithm of the product. The slide rule is designed to do this for
you with its "engine divided scales".

Each sample taken in the Bird wattmeter, line volts and line amps, is
exactly half the total sample, so one of them is really redundant to the
power indication, were it not needed to cancel sensing of the unwanted
direction of power transmission. The 1:1 ratio between samples is
ensured by the lock Zo places on the ratio between volts and amps in the
transmission line. A dynamometer has no such lock so both both volts and
amps must be sampled in the absence of constant volts or constant amps.

Best regards, Richard Harrison, KB5WZI