Small vertical loop math revisited - 980
Like many other references, I've been harping on the 1/10th wavelength maximum for cutting your circumference of wire if you want the characteristics of a small vertical loop.
*** REMINDER - these are RX-ONLY loops directly connected ***
But what is the reference point you cut from mathematically? I have never actually seen it printed as a formula - just references to the .10 wavelength and then a project which seems to use less than the maximum desired.
Do you just double the classical half-wave dipole formula and use 936 / f mHz * 0.10 ??
Or do you use the one for cubical quads which uses 1005 ? Or my earlier reference to using 1000 ?
For years, I've been using these and getting good results, but the loops always seemed just a bit big after prolonged evaluation. As much as I hate to reduce my aperture area, I am now using loops just a tad smaller.
The best I could do was interpret the results from the ARRL Antenna Handbook where it describes a 20-foot circumference loop as being .037 wavelengths long at 1.81 mhz.
So, from that I am very happy with some new loops cut with:
980 / f mHz * 0.10
In fact, the ARRL tips off that while 0.10 is good, 0.085 is even better, since even though 0.10 wavelength loop is a current-node only type, the phase starts change above 0.085. I'll leave it at that, but now I think I have found something I can hang my loop-hat on:
980 / f Mhz * 0.085 = circumference in feet.
This is the formula I'm using and currently having good results with. Discernable difference? Can't tell yet since the great S/N ratio and no real lab equipment to measure rx-current makes it hard for me to say one way or the other.
Like many other references, I've been harping on the 1/10th wavelength maximum for cutting your circumference of wire if you want the characteristics of a small vertical loop.
*** REMINDER - these are RX-ONLY loops directly connected ***
But what is the reference point you cut from mathematically? I have never actually seen it printed as a formula - just references to the .10 wavelength and then a project which seems to use less than the maximum desired.
Do you just double the classical half-wave dipole formula and use 936 / f mHz * 0.10 ??
Or do you use the one for cubical quads which uses 1005 ? Or my earlier reference to using 1000 ?
For years, I've been using these and getting good results, but the loops always seemed just a bit big after prolonged evaluation. As much as I hate to reduce my aperture area, I am now using loops just a tad smaller.
The best I could do was interpret the results from the ARRL Antenna Handbook where it describes a 20-foot circumference loop as being .037 wavelengths long at 1.81 mhz.
So, from that I am very happy with some new loops cut with:
980 / f mHz * 0.10
In fact, the ARRL tips off that while 0.10 is good, 0.085 is even better, since even though 0.10 wavelength loop is a current-node only type, the phase starts change above 0.085. I'll leave it at that, but now I think I have found something I can hang my loop-hat on:
980 / f Mhz * 0.085 = circumference in feet.
This is the formula I'm using and currently having good results with. Discernable difference? Can't tell yet since the great S/N ratio and no real lab equipment to measure rx-current makes it hard for me to say one way or the other.
Last edited by a moderator: