single vs. 3 phase

[mivey]

complex proof

complex proof

Since carl finds this not so easy, I will put forth the considerable effort and countless hours of calculations to offer the following complicated proof:
using the two known identities sin(C+D)=sinC*cosD+cosC*sinD and sin(C-D)=sinC*cosD-cosC*sinD

we can subtract them to get sin(C+D)-sin(C-D)= sinC*cosD+cosC*sinD-sinC*cosD+cosC*sinD = 2*cosC*sinD

now let A=C+D and B=C-D

then A+B=C+C+D-D=2C or C=1/2*(A+B)
and A-B=C+D-C--D=2D or D=1/2*(A-B)

then we have the "given": sinA-sinB=2*cos(1/2*(A+B))*sin(1/2*(A-B))

I'm exhausted.:smile:

 

[coulter]

Yes, we will never know because I choose not to waste my time on such a pointless exercise. ...

I agree - that is a fine reason to not do something that is, in your words, "easy".:roll:

carl

 

[rattus]

I agree - that is a fine reason to not do something that is, in your words, "easy".:roll:

carl

P.S. And also because Carl is just trying to bait me instead of contributing in a positive manner.

 

[gpedens]

single vs. 3 phase

I wonder if Grantcool got his question answered? The answer is if it is 2 phases or 200000000 phases and they are added together to supply a single load, the result is a single phase load. The common mistake is that in the average home circuit 120/240 a lot of folks seem to think you have 2 phases 120 volts that are 180 apart and get 240 volts. If that were true the result would be 120 + (-120)= 0 volts. The 120 legs are each just 1/2 of the single 240 winding of the transformer. In other words how much plowing would you get done with a 2 mule plow if one mule was pulling east at 500 pounds and the other west at 500 pounds? Could I prove to you that the two 120 legs are 180 degrees apart? Sure, just like I can show you that two D cell batteries are 180 apart. Its a matter of how you choose your reference point. In the 120/240 transformer with the center as reference at any instance in time one "hot" leg is the positive with respect to the center and the other "hot" leg except at 0. When at 0 both "hot" legs and the center or neutral will be 0. Look at how the "phases" add. Does 1+1=2, or does 1+1=1.414, or 1+1= 1.732? In a two phase system there are 2 voltages 90 degrees apart and 120+120= 169.7 volts 45 degrees from each of the 120 legs. If it is 2 phases of a three phase system 120+120=207.8 volts 60 degrees from each of the 120. In Grantcools question 1+1=2 therefore both "phases" or half windings are in phase with each other at 0 degrees apart. In other words both mules are moving east with a total pull of 1000 pounds. Just think of the voltage vectors as mules plowing or youngins playing tug of war. This is the simple answer I have found that works for most folks. The actual math is much more involved and if you rode them mules and they moved like two sine waves moving east, ever bone in your body would be broken from the G forces. Zero to 60 in 4 milliseconds and back to 0 in 4 more milliseconds is rough. Most folks are only interested in the RMS result or the DC voltage equivalent. I loved math in high school and college engineering. Like it so much I had to take some of them 2 or 3 times to make a B.

 

[rattus]

Mules?

Mules?

I am one of those who claims the voltages on L1 and L2 of a split-phase system are 180 degrees apart. It is like two mules hitched to opposite ends of a beam operating an ancient hay baler or syrup mill. Although they are pulling in opposite directions, they both contribute equally to the job at hand.

Fact is, you do not add, you subtract to find the difference in voltage, i.e.,

V12 = V1n - V2n = 240V

This is true for any voltage measurement.

 

[mivey]

1/10 Borax

1/10 Borax

If that were true the result would be 120 + (-120)= 0 volts...Its a matter of how you choose your reference point

When choosing your reference point, just make sure you are hitching the right end of the mule to the beam. With a neutral reference, the math result is 120 - (-120)= 240.

Instead of tug of war, think of your youngins pushing a merry-go-round from opposite sides. They are pushing in opposite directions but have twice the power.

 

[coulter]

P.S. And also because Carl is just trying to bait me instead of contributing in a positive manner.

Interesting choice of words - certainly not positive:-?

Mostly I was calling BS on a statement I didn't think you could back up. Without grabbing a magic formula out of .... ahhhh, thin air:roll: the math is pretty complicated.

I think I made my point.

carl

 

[winnie]

I presume that we are talking about the theorem that the sum of any two sine functions, of arbitrary amplitude and phase, but the _same_ frequency is another sine function of the same frequency. Or in other words, proving that:

given any A, B, C, and D, you can find E and F so that
A * sin(omega * T + B) + C * sin(omega * T + D) = E * sin(omega * T + F)

This proof is trivial if you are willing to accept Euler's formula (e^ix = cos(x) + i * sin(x)) as a given

1) I rewrite one of my sine functions as the imaginary part of the corresponding complex exponential, using Euler's formula.
A * sin(omega * T + B) = Im( A * e^{i(omega * T + B) } )

2) I use the laws of exponents, in particular the fact that X^(A+B) = (X^A) * (X^B) to split up frequency and phase:
Im( A * e^{i(omega * T + B) } ) = Im( e^{i*omega*T} * A * e^{iB})

3) I do the same thing for my other sine functions, and rewrite the original equation that I'm trying to prove is true:
Im( e^{i*omega*T} * A * e^{iB}) + Im(e^{i*omega*T} * C* e^{iD}) = Im( e^{i*omega*T} * E* e^{iF})

4) Now I rearrange a bit:
Im( e^{i *omega*T} * (A * e^{iB} + C * e^{iD}) = Im(e^{i*omega*T} * E * e^{iF})

And that's it. The left hand side is clearly the same periodic function as the right hand side [e ^{i * omega * T} ] simply multiplied by some _constants_.
Solve A * e^{iB} + C * e^{iD} to get E and F, using the rules of complex addition (vector addition).

-Jon

 

[crossman]

Now I see Rattus' point that the solution is so simple it is hardly worth doing. I mean, the whole proof is rather obvious.

 

[mivey]

...This proof is trivial if you are willing to accept Euler's formula (e^ix = cos(x) + i * sin(x)) as a given...

But that's the rub. Carl appears unwilling to accept common identities and formulas because they are "magic". It appears he is claiming you can't make your proof without deriving basic formulas.

I'm just not sure how far back in time he wants us to go before he is satisfied that the math we are using is not some new-fangled "magic" math. Like rattus said, why re-invent the wheel?

Carl, come towards the light...don't be afraid...drop that sliderule and use your new HP...but you will need some of those new things they call batteries. :grin:

 

[coulter]

Now I see Rattus' point that the solution is so simple it is hardly worth doing. I mean, the whole proof is rather obvious.

Absolutely - Well, I was trying to point out that the derivation of the magic formula is not exactly easy

Jon's is one of the only two I know. The other is somewhat simpler - geometric, mostly by inspection.

mivey -
I'm going to have to follow your recommendation and come to the light. I just broke off the hand crank on my kurta

carl

 

[mivey]

...I just broke off the hand crank on my kurta...

What's a kurta? I did a search and found some pajama-looking thing.

Are you talking about the Kurta tablets? It is getting hard to find parts for them anymore. We used to have a nice 36x48 for mapping work.

 

[rattus]

What's a kurta? I did a search and found some pajama-looking thing.

Are you talking about the Kurta tablets? It is getting hard to find parts for them anymore. We used to have a nice 36x48 for mapping work.

The correct spelling is "Curta".

 

[mivey]

curta

curta

That's some interesting history reading.

 

[coulter]

The correct spelling is "Curta".

rattus is right - fat fingered that one.

carl

 

[rattus]

My three points:

My three points:

Interesting choice of words - certainly not positive:-?

Mostly I was calling BS on a statement I didn't think you could back up. Without grabbing a magic formula out of .... ahhhh, thin air:roll: the math is pretty complicated.

I think I made my point.

carl

1. If one knows trig, the derivation of the sum and difference formulas is easy as evidenced by a couple of responses.

2. If one doesn't know trig, it is hard.

3. No one cares about this pointless exercise anyway, so just what is your point?

 

[coulter]

... No one cares about this pointless exercise anyway, ...

Okay

You have a good day:smile:

carl

 

[engy]

Just to keep this thread going...

V12 = V1n - V2n = 240V QUOTE]

It's that simple.
I like to use the "240' hill" analogy.

If you stand at the top or bottom of the hill, all would agree there is one hill 240' in height.

If you stand at the middle of the hill, the hill has not changed, it is still one 240' hill, however somone could argue there are actually two hills - one 120' that goes uphill, and one that goes 120' downhill.

With 120/208 the hill might get a bit funky, but I digress...

 

[rattus]

What we have learned--well some of us anyway:

What we have learned--well some of us anyway:

After several threads, hundreds of posts, a couple of closures, and some excellent help from mivey and winnie, the following TRVTHs have been demonstrated:

1. Sinusoidal voltages at opposite ends of a center-tapped secondary exhibit a 180-degree phase difference. The fact that this difference is not the result of a time delay is immaterial. Furthermore, it matters not that these voltages are induced in a common winding. If it looks like a phase difference, then it must be a phase difference. What you see is what it is.

2. It is acceptable, even conventional, to use the CT as the reference point in a 120/240V split-phase service. Note that we split a single phase; we do not create a second phase.

3. A static phasor is a constant complex number which provides the RMS value (magnitude) of a sinusoidal voltage or current and its phase angle. Phasors may be represented in several different forms, e.g., polar, trigonometric, rectangular, or exponential. Some authors extend this definition to impedances as well.

4. The term ?Negative magnitude? is an oxymoron and should never be used. The real and imaginary components of phasors may take on negative values but the magnitude is always positive.

5. A rotating phasor is similar to a static phasor with the major exceptions that it is a function of time and its magnitude is the peak value rather than the RMS value. It provides instantaneous values?real and imaginary.

6. Static phasor arrows can be drawn in either of two directions. For example, in the phasor diagram of the 4-wire delta, there are five arrows, therefore there are 32 correct ways to draw the diagram. That being said, only two of these would normally be used.

7. In summing voltages around a loop, addition is performed if the direction of summation agrees with the direction of the arrow. Subtraction is performed if the arrow points against the direction of summation. Since this is an algebraic summation, we merely change signs and add to perform subtraction. This process is equivalent to reversing the direction of the arrow and shifting the phase angle by 180 degrees.

8. The order of summation of phasors is arbitrary as long as the relative direction of summation is maintained. However, for convenience and clarity, we normally sum them in the order they appear in the diagram.

 

[crossman]

Sounds good to me Rattus. I must say that these "theory" threads are certainly enlightening and worthwhile. And very entertaining. Sure, there may be a certain amount of "hostility" but it is all worthwhile for me.

I hope we can continue having these passionate discussions.