This article is a workup of replacement of my 2mm HDC G5RV and feedline with high strength 1.6 aluminium MIG wire to evaluate practical issues with use of an aluminium conductor.
The G5RV configuration is an inverted V, and although half a G5RV is 15m, the supports result in a 20m length of wire to the support. The configuration has a central support and simple spans for each leg of the G5RV to their respective supports.
Step 1: Calculate mass, strength, RF resistance of the wire
Density of aluminium is 2700kg/m^3, so the mass per unit length of 1.6mm wire is 0.00543kg/m.
5356 wire is specified as 39ksi which is 270MPa. 1.6mm 5356 wire then has a gross breaking strain of 543N.
Based on conductivity of 36e6S/m and skin effect, RF resistance at 3.5MHz is 0.126Ω/m, 27% higher than that of equivalent size plain copper.
The existing 9.5m open wire line (2mm copper spaced 50mm) has been calculated to have a loss under mismatch at 3.6MHz (load Z 35-j350Ω) of 0.26dB, the 1.6mm aluminium spaced 52mm would have a loss of 0.39dB, well less than the loss of commercial CCS windowed ladder line.
Step 2: Calculate catenary
Using Antenna wire catenary calculator, the catenary is calculated for the desired span with the wire parameters.
1.6mm 5356 Al MIG wire – inclined 20m span
|Wind velocity||40.0 m/s (144.0 km/h)|
|Wire diameter||0.001600 m|
|Mass / unit length||0.005430 kg/m|
|Gross Breaking Strength||543.0 N|
|Radial ice thickness||0.000000 m|
|End 1||(0.00,11.00) m|
|End 2||(18.70,4.00) m|
|Working Load Limit||155.1 N|
|Wind pressure||960.0 Pa|
|Wire length||20.016 m|
|Min vertical sag||0.642 m (3.2% of span)|
|Min normal sag||0.602 m (3.0% of span)|
|Lowest point||(18.700,4.000) m|
|Horizontal force at End 1 & End 2||138.0 N|
|Lateral force at End 1||70.9 N|
|Lateral force at End 2||-33.0 N|
|Slope at End 1||-0.514 (-27.20 °)|
|Intercept of End 1 slope with vertical below End 2||2.611 m (18.700,1.389) m|
|Max tension under wire weight force alone at End 1 (no wind, no ice)||4.4 N (2.8 % of WLL)|
|Min wave return time under wire weight force alone (no wind, no ice)||1.45 s (x3=4.34 s)|
|Catenary||y=7.27612011e+1 * cosh((x-3.59190136e+1)/7.27612011e+1) -7.08085382e+1|
Step 3: Insulator details
Above is a mockup of a dipole leg terminated on an insulator, and the feed line connected using the 3mm stainless steel tap connector. In use, the tap connector will be filled with marine grease or aluminium jointing compound to exclude oxygen and water.
Above is the 2mm stainless connector which will probably be used. The 1.6mm wire has to be bent to follow the path through this smaller connector body.
Step 4: Transmission line
Above, the feed line will be the pair or 1.6mm aluminium wires spaced 50mm, here using small pieces of 2mm clear acrylic and zip ties.
Performance is calculated at 3.6MHz with a load similar to that of the existing G5RV inverted V dipole.
1.6 x 50 aluminium feed line
|Twist rate||0 t/m|
|Length||44.24 °, 0.123 λ, 10.0000 m, 3.511e+4 ps|
|Line Loss (matched)||2.31e-2 dB|
|Line Loss||0.415 dB|
Loss under standing waves is about 0.4dB, less than 0.15dB higher than the existing 2mm copper feed line.
Step 4: Implementation
To be continued in a follow up article…