Cec, you have YOUR explanation and I have MY explanation.

Which is the most simple?

There is a bridge.

When the variable arm, the load, is 50 ohms the bridge is balanced and
the meter indicates SWR = 1:1

When the variable arm is either 0 ohms or infinite ohms, the meter
indicates SWR = infinity :1

What can be more simple than that? How it works can be visualised.

But the meter is ambiguous. It cannot distinguish between loads of 0
ohms and infinite ohms. Additional information is required.

This serious ambiguity also applies to your weird contraption. ;o)
----
Regards, Reg.

Reg Edwards wrote:
Cec, you have YOUR explanation and I have MY explanation.

Mine is a lot simpler. The Heath HM-15 has two pickup elements.
If you install a Z0 resistor load at one end it "picks up" the forward
wave. If you install a Z0 resistor load at the other end it "picks up"
the reflected wave. The two pickup voltages are rectified and compared
through a calibration procedure.

The parts that came with the HM-15 kit in the 50s-60s included two
72 ohm resistors. RG-ll was very popular at the time. If one wanted
a 72 ohm SWR meter, one installed the 72 ohm resistors. If one wanted
a 50 ohm SWR meter, one installed the 50 ohm resistors. A switch could
be installed that switched between 50 ohms and 72 ohms calibration.

This serious ambiguity also applies to your weird contraption. ;o)

Actually, the Heathkit design concept is easier to understand than
is the bridge explanation or the toroid-pickup/phasor-addition
explanation. The first SWR meter I built in the 50s, used two
lengths of insulated wire shoved under the braid of the coax.
It worked but, at the time, I had no idea why it worked. Heath's
little slotted line pickup device was pretty slick. I sometimes
see them for sale at hamfests.
--
73, Cecil http://www.qsl.net/w5dxp

On Tue, 27 Sep 2005 15:48:35 +0000 (UTC), "Reg Edwards"
wrote:
But the meter is ambiguous. It cannot distinguish between loads of 0
ohms and infinite ohms. Additional information is required.

Hi Reggie,

Without recourse to that "additional information," explain how you
achieve the unambiguous by your method of probing lines (be they
parallel, coaxial, or waveguide).

In other words, your objection is a non sequitur, it is meaningless
because you need the same additional information and you cannot
demonstrate any measurable difference between the manifold methods of
coming to the same determination.

Of course, if you throw a spanner in the other guy's gear-box, you
might win the race.

73's
Richard Clark, KB7QHC

Rich, your abuse of the English language renders it impossible for me
or anybody else to make any sense of what you are waffling about.
----
Punchinello, G4FGQ

On Tue, 27 Sep 2005 19:13:16 +0000 (UTC), "Reg Edwards"
wrote:

Rich, your abuse of the English language renders it impossible for me
or anybody else to make any sense of what you are waffling about.

Aw Reggie,

Are you using a prescription grade wine glass when you were trying to
read it? Or can we blame it on the grape? No ifs ands or buts now
because with each new post the question becomes more remote and harder
for you to answer.

73's
Richard Clark, KB7QHC