Factory installed inverter
Using a VDC of 14 which is most likely what the alternator is putting out on a running engine, with a copper stranded wire with a length of 20ft at a temp 100*F, the amperage would be 33.03. At 12VDC, same wiring length and temp the amperage would be 38.54. With that in mind using MIL-STD-975 the wire rating should be a #8. That would allow for 44.6 amps which would be below the highest requirement which would be 12VDC. I would say a #8 stranded copper wire with a 40amp fuse would fit this application safely.
Nice catch, and I agree he's pushing it, however after calculating myself, I come up with the following:
Using a VDC of 14 which is most likely what the alternator is putting out on a running engine, with a copper stranded wire with a length of 20ft at a temp 100*F, the amperage would be 33.03. At 12VDC, same wiring length and temp the amperage would be 38.54. With that in mind using MIL-STD-975 the wire rating should be a #8. That would allow for 44.6 amps which would be below the highest requirement which would be 12VDC. I would say a #8 stranded copper wire with a 40amp fuse would fit this application safely.
Using a VDC of 14 which is most likely what the alternator is putting out on a running engine, with a copper stranded wire with a length of 20ft at a temp 100*F, the amperage would be 33.03. At 12VDC, same wiring length and temp the amperage would be 38.54. With that in mind using MIL-STD-975 the wire rating should be a #8. That would allow for 44.6 amps which would be below the highest requirement which would be 12VDC. I would say a #8 stranded copper wire with a 40amp fuse would fit this application safely.
Not accounting for temp and length 800/12 is 57 amps. With the cable length and voltage drop and heat it will only go up. Min would be six gauge wire and a70 amp fuse. Also peak (although short) is 1600 watts.
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Sorry "Lorrel" I beg to differ!
One can not assume 14 vdc. !2 volts should be the minimum. So your calculations should start there.
As I have stated many times before. Calculating length of conductor is pure BS. Just simply calculate the full load and adjust the wire size to those specs.
I have not looked up what you are refering to here. I will later. But my referance comes from "The National Electrical Code" ( NEC ). In other words "The Electricians Bible".
X2 X2

Using a VDC of 14 which is most likely what the alternator is putting out on a running engine
with a copper stranded wire with a length of 20ft at a temp 100*F, the amperage would be 33.03. At 12VDC, same wiring length and temp the amperage would be 38.54.
With that in mind using MIL-STD-975
Originally Posted by ronjenx
Nicely done, but there are a few things to consider.
#10 wire to an 800 watt inverter is too small. Should be #6, maybe even #4.
A 30 amp fuse is too small for a 800 watt inverter. 70 amps would be more appropriate. (Assuming the proper wire size is used.)
Nicely done, but there are a few things to consider.
#10 wire to an 800 watt inverter is too small. Should be #6, maybe even #4.
A 30 amp fuse is too small for a 800 watt inverter. 70 amps would be more appropriate. (Assuming the proper wire size is used.)
Last edited by JK-Ford; Nov 1, 2010 at 11:43 AM.
Are you saying you agree its only going to draw 38.54 amps? I agree deff should be calculated at 12 volts with a 1 percent tol. Im just trying to learn why you feel it would only draw 38.54 amps at an 800 watt load.
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66.66 amps = 70 fuse / relay
70 amps = #4 THHN conductor
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