Re: NEC Code Graphic
Originally posted by ranger12: Charlie B could you walk us through this calculation with a detailed explanation?
That?s a tall order, especially since I have no skill at creating a figure and posting it on an Internet site. So I?ll try just telling you how to draw the figure yourself. Take your compass, and draw a circle that has a radius of 2.77 inches. OK, make it two and three quarters inches, and we?ll call it good enough. From the center of the circle, draw a horizontal line to the edge of the circle, put an arrow tip where it hits the circle, and label that point ?A? (use a capital letter). Follow the circle around clockwise until you go one third of the way (120 degrees clockwise), mark the point with ?B,? and draw an arrow from the center to ?B.? Continue following the circle around clockwise until you have gone a total of two thirds of the way (240 degrees clockwise), mark the point with ?C,? and draw an arrow from the center to ?C.? Now erase the circle, leaving only the arrows.
Next, start from the center, and draw another arrow 30 degrees downwards (clockwise) from arrow ?A.? Make it?s length 1.20 inches (call it one and one quarter inches). Mark its arrow tip with a small case ?a.? Draw similar arrows, each 1.20 inches long, one that is 30 degrees clockwise from arrow ?B? (call this new one arrow ?b?) and the other 30 degrees clockwise from arrow ?C? (call this new one arrow ?c?). What you see is the voltage relationship between a 480/277 volt primary (the capital letter arrows) and a 120/208 volt secondary (the lower case letter arrows). Note that the angle between arrow ?A? and arrow ?b? is the 90 degrees that paul32 had mentioned in an earlier post.
You can measure each of the three ?cases? (see my earlier post) as follows. Imagine copying the arrow ?a? (include from its center dot to its arrow head), moving it without changing the direction in which it is pointing, and pasting the copy so that the center of arrow ?a? is at the arrow point of arrow ?A.? Now measure from the center of the original circle to the tip of the new location of arrow ?a.? You should get the 3.856 inches (just under 4 inches) that I have been discussing. You can also move a copy of arrow ?a? to the tip of arrow ?B? and to the tip of arrow ?C,? and you should be able to measure the 3.02 and the 1.83 inches that I mention in the earlier post.
As to the trigonometry, it?s more complicated than most electricians will ever need to use. But the ?simplest? way to do the calculation is to use the ?Law of Cosines.? Go back to the sketch of the first time you copied and pasted arrow ?a? to the tip of arrow ?A.? You have a triangle formed by the center of the original circle, point ?A,? and the new location of point ?a.? Label the 277 volt side as ?Side X,? then label the 120 volt side as ?Side Y,? and finally label the unknown voltage side as ?Side Z.? The angle between Side X and Side Y is 150 degrees (measure it, if you are unsure). The Law of Cosines states,
Z*2 = X*2 + Y*2 minus (2) times (X) times (Y) times (the cosine of the angle between X and Y).
If this looks a bit like the Pythagorean Theorem, it should. In a right triangle, the angle of interest is 90 degrees, and the cosine of 90 degree is zero. So that last term drops out, and leaves you with Z*2 = X*2 + Y*2.
In our case, Z*2 = 277*2 + 120*2 minus (2) times (277) times (120) times (cosine 150).
Solving gives you Z = 385.6.
QED
