 ```To All from PA0SE Andy Talbot wrote: ``````I find that rather impossible to believe - 300m of thick cable being a ``````dummy ``````load at 137kHz ! Go back to the fundamental equations and calculate properly rather than ``````rely ``````on tables and software used for the wrong purpose . This may be a better way to estimate the performance.... Inductive reactance of a shorted length of line Xl = Zo.TAN(2 . PI . L ``````/ ``````vf / Wavelength) With Wavelength = 2188m, velocity factor = 0.67, Zo = 50 this gives 171 ohms (= 200uH at 137k) you must have slipped a digit somewhere to get 2mH. This is the reactance looking into the coax, shorted at the far end and neglecting any losses. To get 2mH Xl = 1722 ohms, L = 0.245 wavelength (in air) = 373m of ``````coax. ``````Not very much of an increase on 300m and shows how critical the length is and how fast Xl will change with frequency. (In fact,since a lot of the numbers above have been rounded and we are ``````very ``````close to a shorted quarter wave, a back calculation using the rounded ``````values ``````to check gave Xl = 1600 rather than the 1720 ohms used in the forward calculation - that's how twitchy this technique will be) For an estimate of losses : Skin depth of copper at 137kHz is approximately 0.18mm From D = ``````503 ``````SQRT(Resistivity / Freq / uo) For Cu Resistivity = 1.7E-8 Ohms / m, and uo (magnetic permeability) ``````= 1 ``````Diameter of centre conductor = 2.5mm (near enough anyway) so cross sectional area of conducting path is 0.18mm * 2.5mm = 0.45E-6 m^2 RF Resist = Resisivity * Length / Area = 1.7E-8 * 370m / 0.45E-6m^2 = 14 ohms. For a quick estimate assume the braid losses are a lot less than the ``````centre ``````conductor as they have a much larger surface area, so can be ignored (although that may not necessaily be the case) and we can also ignore dielectric losses (a reasonable assumption at these freqs) so Q = Xl / R = 1722 / 14 = 123 Which is about what I got on my 5mH conventional coil of 1.5mm wire, ``````300mm ``````diameter and 400mm long. In other words, a very expensive, very large and heavy 'coil' - making it from coax `````` Andy says "Go back to the fundamental equations and calculate properly". A good advice but Andy did not follow it himself. The equation for Xl he used is fine for a lossless transmission line but for a lossy line the full equation for the input impedance of a transmission line line of length l, characteristic impedance Zo and terminated by an impedance Zl should be used: Zin = Zo x {Zlcosh(gamma) * l + Zosinh(gamma * l)} / { Zlsinh(gamma * l) + Zocosh(gamma * l)} I wrote "gamma" in full because my computer refuses to produce the Greek character for it. I happily leave it to others to apply this rather formidable equation to Alan Melia's cable. I feel not competent to do so but also no need as N6BV's computer program does that for me as it is based upon that formidable equation. To check the validity of the computer result I performed a simple test. When the 984 ft long RG-213 cable is shortcircuited at its end the computer finds for the impedance (Zin)sc at the input of the cable at 137 kHz: (Zin)sc = 49.42 + j 172.52 as mentioned before. When the cable ends in an open circuit the input impedance (Zin)oc is according to the computer: (Zin)oc = 2.60 - j 13.72 Now compute the product of these impedances: (Zin)sc * (Zin)oc = (49.42 + j 172.52) * (2.60 - j 13.72) = 2418 - j 207 According to another basic equation this should be equal to the square of the characteristic impedance of the cable: Zo^2 = (50.0 - j 2.3)^2 = 2505 - j 230; pretty near the expected result. I leave it to experts to judge whether this test is a sufficient proof on the validity of the computer results. For me it is enough to stick to my opinion that Alan's cable would be fine as a dummy load, provided a capacitor with a reactance of -j 172.52 Ohms (6727 pF) is put in series to tune out the reactance. 73, Dick, PA0SE ```