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jread:
A vector has a magnitude and a direction.
Consider a two dimensional plane with X and Y rectangular coordinates. Draw a vector from X=0, Y=0 to X=1, Y=0. This is a vector of unit length pointing to the right along the X axis.
Rotate this vector counter clockwise around the 0,0 point. The variation of the Y value of the vector as a function of angle as you rotate the vector is the sine of the angle. This you can see from trigonometry.
Create a second vector that is 120 degrees counterclockwise from the original vector and that starts at the 0,0 point. As you rotate this second vector around the 0,0 point it also produces a sine variation in the Y axis, but its Y values are shifted in angle relative to those of the first vector.
Put the first vector at 0 degrees and the second vector at 120 degrees and determine the distance between the non-zero ends of the vectors. Note that if the two vectors are maintained at the 120 degree difference and rotated together, then the magnitude of the vector sum between them remains constant as the two vectors are rotated about 0,0. Call this sum of the first two vectors the third vector.
If you project the end points of the third vector on the Y axis you will also find that the difference on the Y axis of the two end points is a sine wave but at an angle shifted by 60 degrees from the other two vectors. This difference angle would change if the magnitudes of the vectors 1 and 2 were not the same.
The first vector can be your voltage from N (neutral -- center point of your Y transformer -- 0,0 point on the X-Y plane) to line A. This is phase AN. Let your second vector be phase BN. Assume that the magnitude of both voltages AN and BN are 120 V. The vector length from A to B will be your voltage for phase AB.
How do you calculate the magnitude of phase AB. From trigonometry. Put vector 1 at angle 0, thus vector 2 is at 120 deg. The X axis projection of vector 2 is 120/2 = 60 V from trig. Thus, there is a total distance on the X axis of 180. The Y axis component of the 120 deg vector is 120*sq-rt of 3/2 (120 V * sin 60 deg). We now have a right triangle of 180 for the long side and 120*0.866 for the short side.
Vector 3 (the voltage from line A to line B) is the hypotenuse of the right triangle. This can be solved for by trigonometry or the Pythagorean theorem. Using the theorem Vab = Van * sq-root ( (3/2) squared + ( (sq-rt 3) / 2) squared ) = Van * sq-rt ( (9/4) + (3/4) ) = Van * sq-rt ( 12/4 ) = Van * 1.732.
I used (sq-rt 3) / 2 for the cos of 30 deg instead of 0.866025404 because I know this is the exact value and it works nice in the Pythagorean theorem.
I have tried to provide an intuitive link between a sine wave and vectors, and then the solution for your specific question. If this is not clear to you, then indicate where you need clarification.
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