A question about Knife Switches

[N6YW]

Greetings to the group...
I was just given a very nice Knife Switch from WWII, apparently one that was used with many types
of transmitters of the era. Being that the modern day 450 ohm window line wasn't around back then,
the spacing of the conductors are much narrower than the posts of the switch.
My question is, does this affect the impedance or upset things in any way? Should I just abandon the idea and build my own line of the correct spacing?
My reasoning is simple, doing things right as far as practicable. The switch would be located inside of
my shack, quite close to the output of the Johnson KW MB. To me, the use of the 450 line while being
practical, just looks silly.
What would you ?
Thanks!

[W7TFO]

In my experience, that switch reresents too short a piece of line to worry about in OWL service.

73DG

[KF1Z]

For HF it isn't going to make any difference.

[KA2DZT]

If you think the switch won't work, you can send it to me.

At HF frequencies that switch will work OK FB.

Fred 

[K1JJ]

No problem at all.

When matching open wire with a balanced tuner, the open wire can vary in spacing.  Even if the spacing started at 8" at the antenna and narrowed down to 1" at the matcher output terminals, it will not affect your loss or antenna pattern.  This assumes you have the ant matcher adjusted for a decent match at the transmitter.

Open wire loss, for a given wire diameter and insulator material, is not dependent on reasonable spacing.  Even if the feedline to antenna match is worse as a result of the spacing, the actual loss of good quality OWL is insignificant.


T

[Steve - K4HX]

Sometimes feedline spacing is varied (tapered) to do matching balun functions.

[N6YW]

One of the very reasons why this site rules...
You guys always come through with the answers. Thanks!
Due to the small footprint of my property, a short folded dipole (Cobra Ultralite Jr.) will be used
with the KW MB. I have no choice until we sell and move to the country or a large city lot outside
of Los Angeles. While we have very little lightning in LA, I am happy to have this just in case.
Shunt the dead end to ground via a large piece of copper ribbon.

[W3RSW]

Perfect.  In my shack, smaller version of your dpdt switch common center goes to the ant. Tuner and thence to the rigs. One pole side to the antenna line the other pole side to outside ground which is also tied to E ground about five feet away.

[AB2EZ]

When you cut a transmission line to insert a lossless device (like a knife switch or a connector or a piece of transmission line with a different characteristic impedance) in series... you are inserting two equal and opposite mismatches: one at the input to the inserted device, and one at the output of the inserted device.

The first mismatch causes a reflection, and the second mismatch causes a reflection that is equal in size, but 180 degrees out of phase with the first reflection.

As long as the propagation distance between these two mismatches is small compared to a wavelength at the frequency of operation (i.e. a small 1-way propagation phase shift), the two reflections will almost cancel, and the inserted device will be almost invisible from the perspective of disturbing the transmission line that has been cut.

How small is small enough?

For most ham radio intents and purposes, 1/40th of a wavelength (i.e. 360 degrees/40 = 9 degrees of 1-way phase shift) is short enough*.

On 40 meters, a propagation distance between the two mismatches of 1 meter, or less, is okay.

*With 9 degrees of 1-way phase shift (i.e. 18 degrees of total phase shift from the location of the 1st reflection to the location of the 2nd reflection, and back), the second reflection will be (180+18) degrees out of phase with the first reflection. 18 degrees corresponds to 0.314 radians. The total reflected power... compared to that of either of the two reflections, considered by itself... will be reduced by a factor of |1-exp(j0.314|**2 = [1- cos(.314 radians)]**2 + [sin(.314 radians)]**2 = .0024 + .0955 = .0979. I.e. the total power reflected by the combination of the input and the output mismatch from the inserted device will be about 9.8% of the power that is reflected by either of the mismatches considered individually... because the two reflections mostly cancel each other.

Stu