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 Originally Posted by K9STH
Wait a minute!
The inverse log of (0.855/10) is 0.821296552. Multiplying by 50 watts gets 41.06482764 watts. That is 0.000172358 watts more output! 
Glen, K9STH
That extra bit might overload my receiver.
A government which robs Peter to pay Paul can always depend on the support of Paul.
-- George Bernard Shaw
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Originally Posted by K4SAV
It probably depends on who's brand of RG213 you have, as well as the calculator used.
RG-213/U carries a Mil spec. Therefore, the brand does not matter or, it's not RG-213/U.
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Originally Posted by WA7PRC
RG-213/U carries a Mil spec. Therefore, the brand does not matter or, it's not RG-213/U.
The question was about RG213, not RG-213/U.
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I see what I did. I used two different calculators which had different specs for "RG-213". I didn't check my math, must've been the sun spots.
50 watts in you'll get around 41-42 watts out not taking into account connectors, relative humidity, phase angle of the sun and day of week.
Electricity takes ALL paths to ground, not just the least resistive. -Guido
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Use LMR-400 even less loss or hard line many times can be picked up for free the best.
73 de Fred N0AZZ
_____________________________________
The License is Only Your Starting Point in Radio!
MVDX/CC of SW MO., DX Hogs, OARS, NARC, NCDXF
ARRL member, ARRL and W5YI VE
DX the thrill of the chase
""D-STAR making use of the 2/ 440m repeaters for real world Digital Voice usage around town and around the world""
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Originally Posted by K4SAV
It probably depends on who's brand of RG213 you have, as well as the calculator used.
Jerry. K4SAV
TRUE RG-213/U should have the same loss, (within mil-spec tolerance) regardless of manufacturer. That's the whole point of mil-spec cables.
A designation of RG213 is meaningless.
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deleted - correcting errors ...
Last edited by K4SAV; 05-01-2012 at 06:48 PM.
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Originally Posted by WA9SVD
TRUE RG-213/U should have the same loss, (within mil-spec tolerance) regardless of manufacturer. That's the whole point of mil-spec cables.
The mil spec (M17/74-RG213) only specifies the maximum attenuation. If you are trying to calculate the exact attenuation you need the data from the manufacturer of the cable.
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Actually, even within various production runs of cable the attenuation may very well differ slightly. Therefore, the best way to determine the attenuation is to actually measure the loss in the cable.
Glen, K9STH
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Originally Posted by K9STH
Actually, even within various production runs of cable the attenuation may very well differ slightly. Therefore, the best way to determine the attenuation is to actually measure the loss in the cable.
Glen, K9STH
Of course. The manufacturer's published specs will give a "nominal" value for such features as attenuation, impedance, and even diameter, all of which can vary slightly from lot to lot. But most Amateurs don't have sufficiently accurate equipment to determine loss in a short, or even moderately long, piece of coax. 0.1 dB or so difference from published data would be difficult for most to measure.
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