Voltage drop calculation for long feeder run

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Re: Voltage drop calculation for long feeder run

0.866 is approximately one half of the square root of three. The person who first put the formula into this form chose to combine two numbers into a single factor. It made the formula easier to work with, but made it a bit harder to see where the formula came from.
 
Re: Voltage drop calculation for long feeder run

sorry for the double posting, just getting use to the site.

Why half of the sq. root of 3
 
Re: Voltage drop calculation for long feeder run

The ?one half? part is there to get rid of the factor of ?2? that shows up at the beginning of the formula for a single-phase calculation. That factor of ?2? is used (in the single-phase formula) to account for voltage drop from the source to the load and back again to the source. In other words, you need to include the two-way distance, and the length ?L? is only a one-way distance, so you multiply by 2.

In the three-phase calculation, once you get rid of the initial factor of ?2,? all that remains is the one-way distance. But in any three-phase calculation, the factor of ?the square root of 3? is going to show up somewhere, before you are done. In this case, instead of multiplying the one-way distance by 2, you multiply it by 1.732 (the square root of 3).

Where does the factor of the square root of 3 come from? That is a bit tough to describe in words alone, and I can?t draw pictures. But I can give a hint. You may recall from sophomore Geometry class that a triangle with angles of 30, 60, and 90 degrees has a special property. That property is that the side opposite the 30 degree angle is shorter than the side opposite the 60 degree angle, and the ratio of their lengths is that very same ?one half the square root of 3.?
 
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