Keith Dysart wrote:
On Dec 4, 9:06 pm, "Tom Donaly" wrote:
50 ohm Shorted line 8.5655 meters long. Frequency = 7 Mhz. 2 volt 50 ohm
generator. Current at input = 12.361 milliamps. Current at short =
40 milliamps. Divide the current at the imput of the line by the current
at the short and take the arc sine (in radian mode)

arc cosine, perhaps?

of the result. This
is 1.2566. Now take Beta = .14671 and multiply it by the length 8.5655.
This also equals 1.2566, which is the angular length of the shorted
line. Will someone explain how this works to Cecil?

Cecil either knows this, in which case it is unnecessary to explain
it,
or he does not, in which case it will be impossible.

...Keith
Right. I lied.
73,
Tom Donaly, KA6RUH

Tom Donaly wrote:
Keith Dysart wrote:
On Dec 4, 9:06 pm, "Tom Donaly" wrote:
50 ohm Shorted line 8.5655 meters long. Frequency = 7 Mhz. 2 volt 50 ohm
generator. Current at input = 12.361 milliamps. Current at short =
40 milliamps. Divide the current at the imput of the line by the current
at the short and take the arc sine (in radian mode)

arc cosine, perhaps?

of the result. This
is 1.2566. Now take Beta = .14671 and multiply it by the length 8.5655.
This also equals 1.2566, which is the angular length of the shorted
line. Will someone explain how this works to Cecil?

Cecil either knows this, in which case it is unnecessary to explain
it,
or he does not, in which case it will be impossible.

...Keith
Right. I lied.
73,
Tom Donaly, KA6RUH
To amplify:
Arc Cosine is correct. And the comment on Cecil is right on the mark.
The same thing can be accomplished, above, using an open stub and
measuring the voltage at both ends. All this is just theoretical,
though, because line loss will skew the results. Besides, why try to
measure current or voltage when all you have to do is measure length
and frequency?
73,
Tom Donaly, KA6RUH

Tom Donaly wrote:
The same thing can be accomplished, above, using an open stub and
measuring the voltage at both ends. All this is just theoretical,
though, because line loss will skew the results. Besides, why try to
measure current or voltage when all you have to do is measure length
and frequency?

Tom, you are not going to understand what I am saying until
you perform the stub exercise I provided. Just do one at a
time. Assume ideal lossless conditions with VF=1.0.

---600 ohm line---+---10 deg 100 ohm line---open

How many degrees of 600 ohm line does it take to make the
above stub look like 1/4 wavelength, i.e. 90 degrees?

Until you perform the exercise, you are just creating
diversions and avoiding the technical truth.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Tom Donaly wrote:
The same thing can be accomplished, above, using an open stub and
measuring the voltage at both ends. All this is just theoretical,
though, because line loss will skew the results. Besides, why try to
measure current or voltage when all you have to do is measure length
and frequency?

Tom, you are not going to understand what I am saying until
you perform the stub exercise I provided. Just do one at a
time. Assume ideal lossless conditions with VF=1.0.

---600 ohm line---+---10 deg 100 ohm line---open

How many degrees of 600 ohm line does it take to make the
above stub look like 1/4 wavelength, i.e. 90 degrees?

Until you perform the exercise, you are just creating
diversions and avoiding the technical truth.

If you know how to do it, Cecil, don't be coy about it. Just state
your case and be done with it. Since you already stated that
the total electrical length is 90 degrees, you're just asking me
to prove your point. Do it yourself, and then I'll tell you
whether I agree with you or not.
73,
Tom Donaly, KA6RUH

Tom Donaly wrote:
If you know how to do it, Cecil, don't be coy about it. Just state
your case and be done with it. Since you already stated that
the total electrical length is 90 degrees, you're just asking me
to prove your point. Do it yourself, and then I'll tell you
whether I agree with you or not.

Does that mean you don't know how to do it?

I already did it on another thread, Tom. Adding 43 degrees
of Z0=600 ohm feedline to the 10 degrees of Z0=100 ohm
feedline will turn the stub into an electrical 1/4
wavelength (90 degree) open stub. And that's exactly
how base-loaded mobile antennas work.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Tom Donaly wrote:
If you know how to do it, Cecil, don't be coy about it. Just state
your case and be done with it. Since you already stated that
the total electrical length is 90 degrees, you're just asking me
to prove your point. Do it yourself, and then I'll tell you
whether I agree with you or not.

Does that mean you don't know how to do it?

I already did it on another thread, Tom. Adding 43 degrees
of Z0=600 ohm feedline to the 10 degrees of Z0=100 ohm
feedline will turn the stub into an electrical 1/4
wavelength (90 degree) open stub. And that's exactly
how base-loaded mobile antennas work.

It will, will it? I'm waiting for you to prove it. Do you
really expect it to be resonant at the right frequency?
This is getting tiresome, Cecil. I guess you think that
stating things over and over again will make them come
true. Oh, well, it's my fault. I shouldn't waste my time
trying to get blood out of a turnip.
73,
Tom Donaly, KA6RUH

Tom Donaly wrote:

...
It will, will it? I'm waiting for you to prove it. Do you
really expect it to be resonant at the right frequency?
This is getting tiresome, Cecil. I guess you think that
stating things over and over again will make them come
true. Oh, well, it's my fault. I shouldn't waste my time
trying to get blood out of a turnip.
73,
Tom Donaly, KA6RUH

You mean at the frequency where the 600 line length is 53 degrees and
the 100 line 10 degrees length ... well, I guess that already answers
your own question, doesn't it?--but then, you should have already knew
that ...

JS

Tom Donaly wrote:
Cecil Moore wrote:
I already did it on another thread, Tom. Adding 43 degrees
of Z0=600 ohm feedline to the 10 degrees of Z0=100 ohm
feedline will turn the stub into an electrical 1/4
wavelength (90 degree) open stub. And that's exactly
how base-loaded mobile antennas work.

It will, will it? I'm waiting for you to prove it. Do you
really expect it to be resonant at the right frequency?

I have proved it in a reply to Dan and verified it
with MicroSmith. I don't know what else you are
asking for. Yes, it will resonate at the design
frequency. Are you incapable of those simple
calculations? Note that everything is rounded
off to the nearest degree.
--
73, Cecil http://www.w5dxp.com