Showing that sqrt(3) is the factor relating the phase to neutral voltage
to the phase to phase voltage is much easier than your method, and doesn't
even require knowing Oscar, or what a sine and a cosine are, all it takes
is a little simple geometry. Consider the following phasor diagram.

Eoa
|
|
|
|
|
|
o
/|\
/ | \
/ | \
/ | \
/------------|------------\
Eoc Eod Eob

Where
Eoa = Eob = Eoc

and we wish to know Ebc

We need to know only two geometric facts

1.) That the sum of the squares of the sides of a right triangle
equals the square of the hypotenuse.

2.) That in a right triangle that contains a 60 degree angle, the
side adjacent to the 60 degree angle is half the length of the
hypotenuse.

I forget how we know the first fact above, but the second is easy to
figure out thinking about equilateral triangles and knowing that the sum
of the interior angles of a triangle must total 180 degrees.

Now observing that if we extend the Eoa phasor downward it will form a new
phasor, let's call it Eod, which bisects Ebc and forms a right angle with
it.

Using the facts above we can say the following

(Eod ^ 2) + (Ebd ^ 2) = (Eob ^ 2)

Eod = Eob / 2

and

Ebd = Ebc / 2

Substituting for Eod we get

((Eob / 2) ^ 2) + (Ebd ^ 2) = (Eob ^ 2)

Substituting for Ebd we get

((Eob / 2) ^ 2) + ((Ebc / 2) ^ 2) = (Eob ^ 2)

and rearranging things

(Ebc ^ 2) / 4 = (Eob ^ 2) - (Eob ^ 2) / 4

combining terms we get

(Ebc ^ 2) = 3 * (Eob ^ 2)

And finally taking the square root of each side we have

Ebc = sqrt(3) * Eob

There you have it, the square root of three is the answer, and it didn't
take Oscar, or any sines and cosines to get there.


Regards,

John Byrns


In article , Chuck Harris
wrote:

This isn't right either :-( It's been too long since I have
done any phasor arithmetic. About 30 years.

So, I refreshed my memory by going back to some old texts, and
they say its easy!

Eoa + Eob = Eab so Eab = sqrt(3)Eoa

Why that's so obvious! No explaination is necessary! Thanks
alot Fitzgerald and Higginbotham!

Ok, how did they get there?
One way is to draw the 120 degree phasors with the Eoa on the
Y-axis, and Eob 30 degrees below the X-axis. Then calculate
the "X" component of each vector, and the "Y" component of each
vector. Add up the "X"s, and Add the "Y"s. Then the magnitude
of the resulting phasor is sqrt(Ex x Ex + Ey x Ey)

Eoa
|
|
|
o-----Ex
|\
| \ -30 degrees
| \
| \
| \
Eob

So, remembering my trig functions by using my favorite
method:

"Oscar Has A Hat On Always", sin, cos, tan

sin w = Opposite/Hypotenuse
cos w = Adjacent/Hypotenuse
tan w = Opposite/Adjacent

So, Eoa is easy, because it has only a y-component

Ey = Eoa + Eob sin(30) = Eoa + 1/2(Eob)
Ex = 0 + Eob cos(30) = 0 + 0.87

this is 3 phase power, and by definition the magnitude of Eoa = Eob

Ey = Eoa + Eoa sin(30) = Eoa + 1/2(Eoa)
Ex = 0 + Eoa cos(30) = 0 + 0.8

Eab = sqrt[Ex x Ex + Ey x Ey] = sqrt[(0.87**2)Eoa**2 + (1.5**2)Eoa**2]

Eab = Eoa sqrt[0.87**2 + 1.5**2] = Eoa sqrt[0.75 + 2.25]

so, as Professors Fitzgerald and Higgenbotham said

Eab = Eoa sqrt[3]

so, for 120V 3 phase "Y", the "line voltage" is

Eab = 120V sqrt[3] = 207.85V

There, the second time I have used phasors in 30 years. Hopefully
correctly this time.

-Chuck




Chuck Harris wrote:
Doesn't work on anybody's calculator. I make the same mistake
over and over. By a fantastic coincidence sin(120 radians)
is very close to sqrt(3) And my calculator defaults to radians.

I do so little polyphase work, I forget to think in phasor
diagrams.


Va
|
|
|
o
/ \
/ \
/ \
Vb Vc


To figure out the voltage between two of 3 phases in a 120V "Y"
circuit, you have to add the phasors. If you call the phasors
Voa, Vob and Voc, (voltage from neutral to legs a,b,c) and you
want to calculate Voa + Vob it is:

Vab = (Voa + Vob) sin(120 degrees) - (120v + 120v) sin(120 degrees)

Vab = 2(120v)sin(120) = 240v sin(120 degrees) = 207.85v

Since sin(120 degrees) = sqrt(3)/2 you can also express this as:

Vab = sqrt(3) x 120v

-Chuck



Surf my web pages at, http://users.rcn.com/jbyrns/

On Sat, 12 Jul 2003 21:00:50 -0500 (John Byrns) wrote:


Showing that sqrt(3) is the factor relating the phase to neutral voltage
to the phase to phase voltage is much easier than your method, and doesn't
even require knowing Oscar, or what a sine and a cosine are, all it takes
is a little simple geometry. Consider the following phasor diagram.

Eoa
|
|
|
|
|
|
o
/|\
/ | \
/ | \
/ | \
/------------|------------\
Eoc Eod Eob

Where
Eoa = Eob = Eoc

and we wish to know Ebc

We need to know only two geometric facts

1.) That the sum of the squares of the sides of a right triangle
equals the square of the hypotenuse.

2.) That in a right triangle that contains a 60 degree angle, the
side adjacent to the 60 degree angle is half the length of the
hypotenuse.

I forget how we know the first fact above, but the second is easy to
figure out thinking about equilateral triangles and knowing that the sum
of the interior angles of a triangle must total 180 degrees.

Now observing that if we extend the Eoa phasor downward it will form a new
phasor, let's call it Eod, which bisects Ebc and forms a right angle with
it.

Using the facts above we can say the following

(Eod ^ 2) + (Ebd ^ 2) = (Eob ^ 2)

Eod = Eob / 2

I think to get this you have to know that sin(30deg) = 1/2, so there's
at least a bit of trig in here. I suspect there's a way around this,
however.

and

Ebd = Ebc / 2

Substituting for Eod we get

((Eob / 2) ^ 2) + (Ebd ^ 2) = (Eob ^ 2)

Substituting for Ebd we get

((Eob / 2) ^ 2) + ((Ebc / 2) ^ 2) = (Eob ^ 2)

and rearranging things

(Ebc ^ 2) / 4 = (Eob ^ 2) - (Eob ^ 2) / 4

combining terms we get

(Ebc ^ 2) = 3 * (Eob ^ 2)

And finally taking the square root of each side we have

Ebc = sqrt(3) * Eob

There you have it, the square root of three is the answer, and it didn't
take Oscar, or any sines and cosines to get there.

-
-----------------------------------------------
Jim Adney
Madison, WI 53711 USA
-----------------------------------------------

Hi John,

That makes sense. I'm not sure it is easier, but it does make
sense. The partial sums of the sin and cos are rolled up into
the identities you used. It's been so long since I've needed
to do any geometry, that I wasn't thinking in that direction.

When I did it with the sums of the x and the sums of the y...
method, the sqrt(3) came in two parts, one from the sum x, and
one from the sum y.

Thanks for the insite.

-Chuck

John Byrns wrote:
Showing that sqrt(3) is the factor relating the phase to neutral voltage
to the phase to phase voltage is much easier than your method, and doesn't
even require knowing Oscar, or what a sine and a cosine are, all it takes
is a little simple geometry. Consider the following phasor diagram.

Eoa
|
|
|
|
|
|
o
/|\
/ | \
/ | \
/ | \
/------------|------------\
Eoc Eod Eob

Where
Eoa = Eob = Eoc