CapetaniosG
Active member
- Joined
- Apr 12, 2005
- Messages
- 168
- Hatteras Model
- 53' MOTOR YACHT (1969 - 1988)
To properly size electrical cables, there is a formula we can use. Here it is.
Acmil= (KxIxL)/V Where:
Acmil is the cross-sectional area of the wire in circular mils
K=10.75 representing the resistance of copper in mil-foot
I= is the maximum Amps in the circuit we're designing
L= the length of the conductor (wire) in the circuit in feet.
V= is the maximum allowable voltage drop at max. load
Here is an example: We want to size the electrical wire for our new windlass drawing 100 amps and we want to limit the voltage drop from the battery
(24 Volts) to the windlass to 5% or 1.2 volts (0.05x24=1.2volts). We measured the routing of the wiring from the battery to the windlass to be 20 feet or 40 feet of cable run. The required cross-sectional area of the wire is:
Acmil= (10.75x100x40)/ 1.2= 35,883.333 circular mils. We will choose AWG #4
wire which has Acmil= 37,360.
You can apply the formula to any circuit as long as you know the max. amps, cable run and the max voltage drop you want to have in the circuit. The rest of it is simple Arithmetic.
For reference:
AWG14 = 3,702 Circular mils
AWG12 = 5,833 Circular mils
AWG10 = 9,943 Circular mils
AWG 8 = 14,810 Circular mils
AWG 6 = 25,910 Circular mils
AWG 4 = 37,360 Circular mils
AWG 2 = 62,450 Circular mils
Have I confused some of you with the new math? If I did, my sincere apologies. I didn't mean to. Pull out your Algebra book and refresh and stop spending so much time boating. Guys I just consumed a dozen clams and half a dozen beers. The clams were local, the beer from Holland.
CapetaniosG
Hatteras 53MY
Acmil= (KxIxL)/V Where:
Acmil is the cross-sectional area of the wire in circular mils
K=10.75 representing the resistance of copper in mil-foot
I= is the maximum Amps in the circuit we're designing
L= the length of the conductor (wire) in the circuit in feet.
V= is the maximum allowable voltage drop at max. load
Here is an example: We want to size the electrical wire for our new windlass drawing 100 amps and we want to limit the voltage drop from the battery
(24 Volts) to the windlass to 5% or 1.2 volts (0.05x24=1.2volts). We measured the routing of the wiring from the battery to the windlass to be 20 feet or 40 feet of cable run. The required cross-sectional area of the wire is:
Acmil= (10.75x100x40)/ 1.2= 35,883.333 circular mils. We will choose AWG #4
wire which has Acmil= 37,360.
You can apply the formula to any circuit as long as you know the max. amps, cable run and the max voltage drop you want to have in the circuit. The rest of it is simple Arithmetic.
For reference:
AWG14 = 3,702 Circular mils
AWG12 = 5,833 Circular mils
AWG10 = 9,943 Circular mils
AWG 8 = 14,810 Circular mils
AWG 6 = 25,910 Circular mils
AWG 4 = 37,360 Circular mils
AWG 2 = 62,450 Circular mils
Have I confused some of you with the new math? If I did, my sincere apologies. I didn't mean to. Pull out your Algebra book and refresh and stop spending so much time boating. Guys I just consumed a dozen clams and half a dozen beers. The clams were local, the beer from Holland.
CapetaniosG
Hatteras 53MY