Making a 2-port coaxial power divider for 144MHz from 70ohm coaxial cable. 70ohm coax is hard to find so 75ohm coax can be used with a slight mis-match (SWR 1.125:1 due to the power divider ... if the antennas are exactly 50ohms!) Example using RG-11 75ohm coaxial cable. Required over-all length of finished item : 150/144.2 x 0.66 = 686mm (a 1/2 wl in coax). 1. Cut 2 lengths of RG-11, each about 750mm long. Fit N-Type female connectors as in the diagram 2. Measure from the tip of each connector 343 + 5 = 348mm and cut 3. Carefully remove 20mm of outer jacket from each cable keeping the braid in place 4. Tin the braid (quickly to prevent heat damage to the dilectric) 5. Measure out 8mm from the outer jacket end and carefully score / remove the braid ends 6. Remove 10mm of dilectric so there is 8mm tinned braid + 2mm dilectric + 10mm inner 7. Place the two cables end-to-end, overlap the 10mm lengths of inner conductor and solder 8. Check the tip-to-tip length is 686mm 8. Prepare the 50ohm station feedline end according to the diagram 9. Hook the loop of the feedline inner over the join in the phasing line inners and solder 10. Wrap the 150mm feedline braid ends around the tinned phasing line braids 2.5 times. Do not cut the excess braid ends as they will be used in step 12. 11. Quickly solder the wrapped braids without melting the inner dilectric 12. Fold the two feedline braid ends side-by-side across the top if the joint and solder 13. Optional : Solder a 20-23mm wide "U" shaped piece of brass shim over the braids 14. Inject the joint with hot-melt glue to seal 15. Place the "T" piece strengthener over the joint 16. Wrap the whole joint with several layers of self-amalgamating tape to weather proof 17. Add equal lengths of phasing lines to the antenna feedpoints and work DX ... Note : The "50ohm station feedline" can be a halfwave length of coax going to the preamp box, etc. Built with re-calculated lengths for 147MHz by ZL1RS 2005 - to repair the Far North Radio Club 2M repeater (replace a noisey power divider on the 2 phased vertical arrays). SWR OK, still working fine. Hi Jeff, 2 Port 6M Power Divider: ... why not use 70 coax cable/CATV cable? See attached plans/instructions for 2M, but you can easily re-calculate for a halfwave length at 6M with the velocity factor of the particular 70/75ohm coax you have available. ... otherwise, for a 2 port 1/4 WL power divider you have the required inner tube diameter and square tube dimentions to get the correct impedance transformation (same as used with your 2 port 2M and 70cm power dividers), all you need is to recalculate the inner tube length between the connector pins for the 6M frequency. This is 1/4 wavelength X the velocity factor of an air dilectric coax line. ... or simply scale the length between the connector pins. I would bet my life that the correct length between the pins of the 6M 2 port 1/4 wavelength power divider is 144.2/50.1 times the length used in the 2M 2 Port 1/4 wl powwer divider. The rest of the figures stay the same as they are just mechanical mountings for the connectors. 6 Port 2M Power Divider: If you have already built 4 port 2M dividers you know the lengths of inner and outer to use (I assume the 4 port 2M or 70cm power dividers have been 1/2 wavelength power dividers?? i.e. with 2 connectors at each end an one in the middle??). All you is to just change the ratio of inner tube OD to the Square tube ID (Inner Dimention!) to be correct to match the six antennas to the feedline port. Check the info and table at: http://fermi.la.asu.edu/w9cf/articles/square/index.html For a half wavelength 6 port divider you are taking 3 antennas in parallel up to 100 ohms on each half of the power divider, so from transmission line theory ... Z-line = SQRT(Z-in x Z-out) = SQRT(50/3 x 100) = 40.82 ohms From the W9CF table, 40.82 ohms needs a ratio between 1.825 (40.62980ohms) and 1.85 (41.44725ohms) so, interpolating the table, 1.83 would be good. Example : If your 1" square tube has a wall thickness of 1/16" the ID (Inner Dimention) is 7/8", so you need an inner conductor round tube with an OD of (7/8" divided by 1.83 =) 0.478". 1/2" OD would be close (but not perfect) ... the ratio would be (7/8" / 1/2" =) 1.75 thus, from the W9CF table the impedance of this ratio in a square outer/round inner power divider is 38.10766ohms. Apply that line impedance to the 1/4wl transmission line formula and the impedance of 3 x 50 ohms antennas at the center of the power divider becomes 87ohms. 87ohms in parallel with 87 ohms from the other half of the power divider = 43.5 ohms (or a 1.14:1 SWR at the feedline port). A thinner wall square tube or a thinner center conductor tube is needed to get the exact 1.83 ratio and exact impedance transformation. Hope that is of use. 73, Bob