Looking for VD Calc from american electricians handbook

[aksparky]

Hi,

I took a class awhile back where we were taught a voltage drop calc from the american electricians handbook.
It was for several loads at different lengths and different loads.
We took each load and multiplied it by the footage to get "Foot Amps", then there was an average that you used to plug the numbers into the regular vd calc of 2xKxDxI divided by vd allowed to get the cm of a conductor that you would use for the whole run.
If anyone is familiar with this and has it handy I'd really appreciate the posting of it.
Thanks in advance

 

[cadpoint]

I think your looking for the load center of a circuit?

IE


Mulitipy each load by its distance from the supply end of the circuit.
Add these products for all the loads fed from the circuit and divide this sum by the sum of the individuals loads.
The result thus obtained is the distance fomr the supply end of the circuit ot the load center. It is this length that should be employed in computing the voltage drop of the circuit. Only the one measure out not twice.

3.4 AEH 14th Ed

IE

A averge of distance of all devices on the circuit (measure down the center) draw a two line circuit, or lay the circuit on its side, an "A" - on it's side, distandce x load...
keep adding a device by drawing backward "C" to the hat of the "A".

 

[cadpoint]

3.40

The math will looks like

distance x amp= S1
...d...... x ..a....=S2
...d...... x ..a....= S..
........(+)AmpT/(+)ST

= Distance to load center

 

[aksparky]

Thanks alot for the reply,

I never had the opportunity to see the calculation as presented in the textbook, just my sloppy note taking, which evidentally confused me.
That was what I was looking for, and the way the instructor explained this was its a good way to use 1 size conductor (average size) for a long run of multiple loads.

I usually will just use one of the online voltage calculators to figure wire size, but when there is multiple loads at different distances this way seemed like the way to go instead of doing it the long way.