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I assume that the D.E. is one wire in the model you have now. I
suppose it has an odd number of segments, so that the source can go just in the middle. I'm assuming the hairpin parallel tube segments are center-to-center 10cm apart. I will assume the D.E. is currently 250cm long, all at some x value, and y extending from -1.25m to +1.25m, as one "wire" of some diameter. To be able to connect the hairpin at the right places, we need wire ends at -.05m and +.05m, so we can change the D.E. wire to be one segment between y=-.05m and +. 05m, and then add 2 wires the same diameter as the D.E., one from y=-1.25m to -.05m, and one from y=+.05m to +1.25m, and each with about half as many segments as the original D.E., all at the original x value. (Or exchange x and y if I have guessed wrong.) So if the original D.E. had 11 segments, put 5 segments on each of the two new wires. Now add three new wires to represent the hairpin U. Assuming the hairpin is oriented perpendicular to the plane of the antenna, it will be all at the same x value as the D.E. I'll assume the antenna is at z=0. So the wires will have ends at x,-.05,0;x,-.05,.35 and x,-. 05,.35;x,+.05,.35 and x,+.05,.35;x,+.05,0 -- and all with diameter 8mm or .008m. Thanks Tom, I did the EZNEC model exactly as you suggested, but I now have another doubt (I must re-read the EZNEC help one of these days...) I would say that I shall still feed the driven element at its center, and not separately feed the two junctions between the hairpin and the driven element Is that right? 73 Tony I0JX |
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