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#1
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#2
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Seems like we want to say 8 watts/(4*pi*r^2) =
..0027w/m^2, and that yields a radius of about 15.4 meters, which would actually be a good approximation of far field for 2M wavelength. Tom- You caught me! I don't have a calculator handy, but it seems to me that the correct formula yields an even lower power when spread over three times the area. 73, Fred, K4DII |
#3
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Tom-
You caught me! I don't have a calculator handy, but it seems to me that the correct formula yields an even lower power when spread over three times the area. Caught again! I took another look at my earlier calculations and found that not only did I have the wrong formula for the area of a sphere, I had my equation upside down! I ended up with something like watts per square meter equals square meters per watt. Yes, the power is spread over a greater area, but that did not compensate for the really stupid mistake. So, if one volt per meter equals 2.7 milliWatts per square Meter, that field strength would be generated by a five watt handheld radio at a distance of 15.4 Meters, which is about 50 Feet. This is sounding serious unless the defibrillator has a greater tolerance. I certainly hope it does, since we are often irradiated by passing mobile radios running 50 watts or so, at closer distances than 50 feet. Lets see - 50 watts is ten dB greater, so the distance would be about 3 times as great as 5 watts, or about 150 feet for one volt per meter. Or, maybe I should just keep my mouth shut to keep from screwing up again! 73, Fred, K4DII |
#5
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One of the tech support people at Guidant actually sent us some
information regardint the testing limits on my wife's device. In the area of interest to hams, it thas been tested at a field strength of 140 V/m RMS over a frequency of 500 KHz to 6 GHz.. Thanks for the discussion and information, Ed Bailen - N5KZW On 04 Jun 2004 00:17:10 GMT, (Fred McKenzie) wrote: Tom- You caught me! I don't have a calculator handy, but it seems to me that the correct formula yields an even lower power when spread over three times the area. Caught again! I took another look at my earlier calculations and found that not only did I have the wrong formula for the area of a sphere, I had my equation upside down! I ended up with something like watts per square meter equals square meters per watt. Yes, the power is spread over a greater area, but that did not compensate for the really stupid mistake. So, if one volt per meter equals 2.7 milliWatts per square Meter, that field strength would be generated by a five watt handheld radio at a distance of 15.4 Meters, which is about 50 Feet. This is sounding serious unless the defibrillator has a greater tolerance. I certainly hope it does, since we are often irradiated by passing mobile radios running 50 watts or so, at closer distances than 50 feet. Lets see - 50 watts is ten dB greater, so the distance would be about 3 times as great as 5 watts, or about 150 feet for one volt per meter. Or, maybe I should just keep my mouth shut to keep from screwing up again! 73, Fred, K4DII |
#6
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I'm not going to check you gentlemen's math -- it sounds like you've
gone through that pretty thoroughly. But I'll point out that you're calculating the field strength from an isotropic source in free space. Let me remind the readers that this is the absolute lowest possible gain an efficient antenna can have. A dipole in free space produces a field strength about 2 dB greater than this in some direction; an isotropic source over ground produces a field strength 3 dB greater. So this rough estimate of field strength is always lower than the field that a real antenna can produce in some direction. _RF Exposure and You_ (Hare, ARRL) has some excellent information on how to approximate field strength of real antennas. And of course modeling programs do the same thing. But under no circumstances would I depend on either one, or even better-than-average amateur measurements, to make any determination that might impact human life or safety. Roy Lewallen, W7EL Fred McKenzie wrote: Tom- You caught me! I don't have a calculator handy, but it seems to me that the correct formula yields an even lower power when spread over three times the area. Caught again! I took another look at my earlier calculations and found that not only did I have the wrong formula for the area of a sphere, I had my equation upside down! I ended up with something like watts per square meter equals square meters per watt. Yes, the power is spread over a greater area, but that did not compensate for the really stupid mistake. So, if one volt per meter equals 2.7 milliWatts per square Meter, that field strength would be generated by a five watt handheld radio at a distance of 15.4 Meters, which is about 50 Feet. This is sounding serious unless the defibrillator has a greater tolerance. I certainly hope it does, since we are often irradiated by passing mobile radios running 50 watts or so, at closer distances than 50 feet. Lets see - 50 watts is ten dB greater, so the distance would be about 3 times as great as 5 watts, or about 150 feet for one volt per meter. Or, maybe I should just keep my mouth shut to keep from screwing up again! 73, Fred, K4DII |
#7
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I'm not going to check you gentlemen's math -- it sounds like you've
gone through that pretty thoroughly. But I'll point out that you're calculating the field strength from an isotropic source in free space. Let me remind the readers that this is the absolute lowest possible gain an efficient antenna can have. A dipole in free space produces a field strength about 2 dB greater than this in some direction; an isotropic source over ground produces a field strength 3 dB greater. So this rough estimate of field strength is always lower than the field that a real antenna can produce in some direction. _RF Exposure and You_ (Hare, ARRL) has some excellent information on how to approximate field strength of real antennas. And of course modeling programs do the same thing. But under no circumstances would I depend on either one, or even better-than-average amateur measurements, to make any determination that might impact human life or safety. Roy Lewallen, W7EL Fred McKenzie wrote: Tom- You caught me! I don't have a calculator handy, but it seems to me that the correct formula yields an even lower power when spread over three times the area. Caught again! I took another look at my earlier calculations and found that not only did I have the wrong formula for the area of a sphere, I had my equation upside down! I ended up with something like watts per square meter equals square meters per watt. Yes, the power is spread over a greater area, but that did not compensate for the really stupid mistake. So, if one volt per meter equals 2.7 milliWatts per square Meter, that field strength would be generated by a five watt handheld radio at a distance of 15.4 Meters, which is about 50 Feet. This is sounding serious unless the defibrillator has a greater tolerance. I certainly hope it does, since we are often irradiated by passing mobile radios running 50 watts or so, at closer distances than 50 feet. Lets see - 50 watts is ten dB greater, so the distance would be about 3 times as great as 5 watts, or about 150 feet for one volt per meter. Or, maybe I should just keep my mouth shut to keep from screwing up again! 73, Fred, K4DII |
#8
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Tom-
You caught me! I don't have a calculator handy, but it seems to me that the correct formula yields an even lower power when spread over three times the area. Caught again! I took another look at my earlier calculations and found that not only did I have the wrong formula for the area of a sphere, I had my equation upside down! I ended up with something like watts per square meter equals square meters per watt. Yes, the power is spread over a greater area, but that did not compensate for the really stupid mistake. So, if one volt per meter equals 2.7 milliWatts per square Meter, that field strength would be generated by a five watt handheld radio at a distance of 15.4 Meters, which is about 50 Feet. This is sounding serious unless the defibrillator has a greater tolerance. I certainly hope it does, since we are often irradiated by passing mobile radios running 50 watts or so, at closer distances than 50 feet. Lets see - 50 watts is ten dB greater, so the distance would be about 3 times as great as 5 watts, or about 150 feet for one volt per meter. Or, maybe I should just keep my mouth shut to keep from screwing up again! 73, Fred, K4DII |
#9
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Seems like we want to say 8 watts/(4*pi*r^2) =
..0027w/m^2, and that yields a radius of about 15.4 meters, which would actually be a good approximation of far field for 2M wavelength. Tom- You caught me! I don't have a calculator handy, but it seems to me that the correct formula yields an even lower power when spread over three times the area. 73, Fred, K4DII |
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