 ```To All, As a answer to a question by Mike, G3XDV, I have sent the following e-mail to the reflector: ``````Mike, G3XDV wrote: ``````1. No top load Efficiency: 10.8% ``````2. One horizontal top load wire of 5 m `````` Efficiency: 12.5% ``````4. One horizontal top loading wire of 10 m ``````Efficiency: 11.9% ``````6. One horizontal top loading wire of 20 m Efficiency: 10.0% `````` Dick, There's something wrong here isn't there?. You have a reducing efficiency for increasing top load length. Surely that can't be right. `````` I replied: ``````Yes, I agree it looks strange. But it results from the way efficiency is calculated: radiation resistance divided by the sum of radiation and loss resistance. When the loss resistance increases more than the radiation resistance efficiency goes down. In the real world the actual efficiency will certaily get better as the small increase of wire loss is almost swamped by the total resistance of the system: earth loss + coil loss + wire loss. What really matters is the increase of Rs. It is better to ignore the efficiency figures; I used them only to ``````calculate ``````Rs from the real part in Z = R + jX `````` After sending the e-mail I realised I should have explained why I mentioned efficiency at all if it was better ignored. The reason is that the computer program does not produce radiation resistance directly. What comes out of the machine is Z = R + jX for an antenna over perfect ground, plus the efficiency. The only loss involved is then due to the resistance of the antenna wires, so R = wire resistance + radiation resistance. ``````From this follows: Efficiency = {Radiation resistance/(wire resistance + ``````radiation resistance)} x 100%. So I needed the efficiency to calculate the radiation resistance. Peter, DF3LP, in his e-mail suggested to set the wire resistance to zero so directly obtaining Rs. But my program AO does not allow that; one has to insert the material the wires are made of. 73, Dick, PA0SE ```