# 'ampere feet' and (ohm cmil/ft)



## Zaped (Jul 6, 2008)

*Ampere foot* - definition of *Ampere foot* by the Free Online Dictionary *...*

A unit, employed in calculating fall of pressure in distributing mains, equivalent to a current of one ampere flowing through one foot of conductor.

An 'ampere foot' is unit of what ?

A cubic foot is a unit of volume.
An 'ampere foot' is a unit of what ?
- -- -

I looked in my Ferm's Fast Finder book. In a voltage drop table, one column is labeled 'Ampere Feet'. Well, my eyes went crossed, and I'm still trying to figure out what the heck it all means.

Anyhow, in voltage drop context, the 'k' constant (resistivity) of a conductor, is 11.83 (ohm cmil/ft) for copper at 50 degees C. How do you put that into words.

A 50 mph car will in one hour go fifty miles.

But what does 11.83 (ohm cmil/ft) tell me ?

What will a 11.83 (ohm cmil/ft) conductor do ?

I can relate to mph, but I can't relate to (ohm cmil/ft).

Please feel free to contribute any guess, speculation, analogy, explanation, prayer for Code insight, etc.

I'm just trying to scratch the surface a little. Every journey begins with one step somebody said.

--Zaped


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## Ultrafault (Dec 16, 2012)

For the record im not the most knowledgeable person here but i dont want you to go anwserless

Ok ill take a stab at the second question.
It really is 11.83ohms/cm/ft or if we transpose properly 11.83 ohms*ft/cm. In English 11.83 ohms for a copper (19 for aluminum) per foot for a one cmil strand. In that equation if feet is 1 and cm is 1 then ohms equal 11.83. Increase your wire size and the ohm value drops increase your length and the ohm value raises. This is called the approximate k method it gives you an approximate value. 

If you wanted to calculate the exact resistance you would look up the resistance in chapter 9 in ohms per 1000 ft divide by 1000 and use that value for k. The reason they are different is becuase of the skin effect combined with larger conductors bieng stranded can vary restiance.


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## Peewee0413 (Oct 18, 2012)

You deserve a cookie for that 1


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## uconduit (Jun 6, 2012)

if im not mistaken, amps-in is exactly equal to amps-out. if you put .0000000001 amps into a conductor 300 feet long (or 300 miles) you get .0000000001 amps out. voltage will suffer, but not current

"ampere feet" do not concern me.


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## backstay (Feb 3, 2011)

uconduit said:


> if im not mistaken, amps-in is exactly equal to amps-out. if you put .0000000001 amps into a conductor 300 feet long (or 300 miles) you get .0000000001 amps out. voltage will suffer, but not current
> 
> "ampere feet" do not concern me.


Kirchhoffs Law.


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