Ohms Law E=IR (simplified here)
where P = I^2 x R:
I: say 2 amps
R: your estimation of CF rovings @ 25ohms
therefore:
P = 100W
Under typical circumstances:
I: 2.0A (same as above)
R: 36.0" of 24awg (7/32awg strand) @ 23.3ohm/1000 feet @ 20C, therefore
0.0690 ohms for 36.0"
therefore:
E = 0.276W
For a given resistance value the CF tow will have to dissipate ~100
watts of energy to conduct at 2A. A 36" lead of 24awg stranded will
dissipate 0.276 watts. The 24awg stranded lead outperforms the CF by
more than 2 magnitudes...
Lincoln Ross wrote:
see belowL
Well, what about the power wire? That's DC, isn't it? I don't know how
many tows equal a spar, though. It would have to be a lot.
Simon Van Leeuwen
RADIUS SYSTEMS
PnP SYSTEMS - The E-Harness of Choice
Cogito Ergo Zooom
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