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Mostly going to be ham stores. Best bet hro, ham radio outlet. It's a franchise so chances a location close by. Many of 70cm 440mHz properties are close to GMRS. There is always us here to try and steer the train wreck. Sent from my SM-G975U using Tapatalk

Sadly, any overlaps or gaps that would result from overlaid circles would be fictitious at best. Coverages on maps today are shown as circles out of simplicity and convenience. It may or may not represent maximum tested range in one or two arbitrary directions. Coverages are actually jagged, random, irregular shaped blobs in the real world, they are only circular in outer space. I experimented with one just this past week in the mobile. Someone had informed me coverage was about 13 miles. Unreliable coverage began at about 3 miles and I lost complete comms at about 5 mile in my direction of travel. You can see that the circles would quickly become misleading if 13 miles were the official distance listed on the map. But I do get it, seeing the circles is nice visual aid. Michael WRHS965 KE8PLM

Every bit helps, but not significantly. Let me help by giving you a very, very crude illustration. A 5 watt radio provides 37dBm output. A typically radio has a receive sensitivity of -120dBm (the lowest level the radio needs to produce usable audio). That is 157dBm of difference. Now, imagine all 157dBm is lost in only 1 mile due to all the obstructions in the path, for an average of 1 dB loss per 33 ft. So now lets say you increase your radio power from 5 to 50 watts. That is an increase of 10dBm (37 to 47dBm). Ok, so now that you have increased your power 10 fold. How much further will you get if you assume the same linear average path poss of 1 dBm per 33 ft. You got it, 330’. So in this example, you increased your power by 10 fold yet your effective distance increased only from 5280 to 5610’. Now, if you were not battling the losses from all the obstructions in the path and went into outer space that same 5 watts would get you 225 miles, and 50 watts would get you 700 miles. There, signal level will drop based purely on inverse square law. The point I am trying to illustrate here is that presence of attenuation of signal caused by obstacles in the signal path plays a significant role in how far your signal will and will not travel. It takes a lot of extra power to “burn” through the obstacles. Much better to raise the antenna to remove the obstacles from the path in the first place. I hope this helps a bit. Michael WRHS965 KE8PLM

I have fixed the issues with the license importing script. I never heard back from the FCC, so I had to put in a hack to pre-process the files looking for a certain case. WRMM237 and WRMN710 are both in the system now.

Your own test seems to answer the question pretty well. Trees don't absorb 450MHz that badly unless they are just loaded with snow or something. Maybe not even then. Terrain is the dominant factor and higher power will just extend the fringe a small amount. Of course there are different factors in different situations - like cars or houses blocking your signal. You could also have beneficial reflections of your signal that help you reach over the horizon. In these cases the high power could make a bigger difference. I suspect in the city the higher power may be a bigger factor. But it's very tricky to put a number on. Vince

Good day DR. Power does play a factor, but it plays a minor factor in practice. Higher power can “burn” through the woods, but only if you put out enough power to “burn” through them. An increase from 5-50 watts is no where near enough to do that. While in outer space you could measure the increase in distance this extra 10dB would get you, on earth you have variable obstacles, terrain, curvature of the earth and variable RF noise conditions to contend with. All of these obstacles quickly chip away at what little power you have to give. This phenomena is truly why want you antenna a high as practically possible when long-range local communications is desired. The higher it is, the less obstacles the signal has to go through or around, thus the stronger the resulting signal will be at a given distance. Case in point. I can achieve achieve .6 miles reliable communication HT to HT in my heavily wooded area (level terrain) and unreliable communication out to 1.4 miles. Yet, using the same HT with same power I can open repeater 50 miles away when I stand in my front yard. What’s the difference? 1) The repeater antenna is perhaps 500-1000 foot higher in elevation than me. 2) There are no hills between me and the repeater to block my signal from reaching it. 3) There are no trees of consequence for the first 1-1/2 miles from my house in the exact direction of the repeater and 4) The performance of the repeater receiver is first rate. Hope this helps answer your question a bit. Michael WRHS965 KE8PLM