A little more fun:

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After all of your other posts, I thought this one would be over my head again.Not so, the answer to this one is easy...

The ankle bone is connected to the shin bone, the shin bone is connected to the .... :D

When I try to follow yours and the other bright guys posts, I cannot get past the "shin bone".
 
Paid my dues:

Paid my dues:

76nemo said:
Where do you guys come up with this stuff? Better yet, where do you find the time????
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First, it is a valid question, and second, I have paid my dues for the past 60 years and now amuse myself by amusing others--well sometimes.
 
Its so far over my head I not only dont know the answer, I dont even understand the question. I thought a phasor was something that Captain Kirk used against the Klingons on star trek.:smile:
 
rattus said:
First, it is a valid question, and second, I have paid my dues for the past 60 years and now amuse myself by amusing others--well sometimes.
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Please understand Sir, I NEVER said it wasn't valid. I am not that arrogant!
Actually, I skimmed by it so quick, I thought it wasn't genuine and you were prodding at it:grin:
 
I am:

I am:

76nemo said:
Please understand Sir, I NEVER said it wasn't valid. I am not that arrogant!
Actually, I skimmed by it so quick, I thought it wasn't genuine and you were prodding at it:grin:
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I am prodding. Prodding some of these guys with AS, BS, MS, Phd. or PE on their business cards.
 
Best semi educated guess ? ?

Best semi educated guess ? ?

It's a trick. Be it an optical illusion.
It's the A-C phase angle equaling 120?'s of 3 phase service.
(hint N to B or Vbn is the 120? line- off of Van- flop about NA line, it'd be
A-B phase angle equaling 120?)
No, sorry I'm not going to supply other algebra or calculations just a link.
Here
 
OK, I'll bite:
using @ to represent angle,
Vab = Van - Vbn
and
Van = |Van| at an angle of @a
Vbn = |Vbn| at an angle of @b
and
Van = |Van|*(cos@a+jsin@a)
Vbn = |Vbn|*(cos@b+jsin@b)
so
Vab = [|Van|*cos@a-|Vbn|*cos@b] +j|Van|*sin@a -j|Vbn|*sin@b

OR
Van = Re{Van} +j*Im{Van}
Vab = Re{Vab} +j*Im{Vab}
so
Vab = Re{Van}-Re{Vab} + j*Im{Van}-j*Im{Vab}

EDIT: I forgot to go back to magnitude and direction so:

Vab = |Vab| at an angle of @ab
so

|Vab| = sqrt([Re{Van}-Re{Vab}]^2 + [Im{Van}-*Im{Vab}]^2)
or
|Vab| = sqrt([|Van|*cos@a-|Vbn|*cos@b]^2 + [|Van|*sin@a -|Vbn|*sin@b]^2)

and

@ab = tan^-1([|Van|*sin@a -|Vbn|*sin@b]/[|Van|*cos@a-|Vbn|*cos@b])
or
@ab = tan^-1([Im{Van}-*Im{Vab}]/[Re{Van}-Re{Vab}])
 
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mivey said:
OK, I'll bite:
using @ to represent angle,
Vab = Van - Vbn
and
Van = |Van| at an angle of @a
Vbn = |Vbn| at an angle of @b
and
Van = |Van|*(cos@a+jsin@a)
Vbn = |Vbn|*(cos@b+jsin@b)
so
Vab = [|Van|*cos@a-|Vbn|*cos@b] +j|Van|*sin@a -j|Vbn|*sin@b

OR
Van = Re{Van} +j*Im{Van}
Vab = Re{Vab} +j*Im{Vab}
so
Vab = Re{Van}-Re{Vab} + j*Im{Van}-j*Im{Vab}

EDIT: I forgot to go back to magnitude and direction so:

Vab = |Vab| at an angle of @ab
so

|Vab| = sqrt([Re{Van}-Re{Vab}]^2 + [Im{Van}-*Im{Vab}]^2)
or
|Vab| = sqrt([|Van|*cos@a-|Vbn|*cos@b]^2 + [|Van|*sin@a -|Vbn|*sin@b]^2)

and

@ab = tan^-1([|Van|*sin@a -|Vbn|*sin@b]/[|Van|*cos@a-|Vbn|*cos@b])
or
@ab = tan^-1([Im{Van}-*Im{Vab}]/[Re{Van}-Re{Vab}])
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So now I know what the shin bone is connected to.
This is what I meant before, you guys are on a different level.


60 years at this...now I understand.:cool:
 
Pierre C Belarge said:
So now I know what the shin bone is connected to.
This is what I meant before, you guys are on a different level.


60 years at this...now I understand.:cool:
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Fact is, 60 years ago, I heard about power factor. Didn't believe it then. Trying to understand it ever since.
 
mivey said:
OK, I'll bite:
using @ to represent angle,
Vab = Van - Vbn
and
Van = |Van| at an angle of @a
Vbn = |Vbn| at an angle of @b
and
Van = |Van|*(cos@a+jsin@a)
Vbn = |Vbn|*(cos@b+jsin@b)
so
Vab = [|Van|*cos@a-|Vbn|*cos@b] +j|Van|*sin@a -j|Vbn|*sin@b

OR
Van = Re{Van} +j*Im{Van}
Vab = Re{Vab} +j*Im{Vab}
so
Vab = Re{Van}-Re{Vab} + j*Im{Van}-j*Im{Vab}

EDIT: I forgot to go back to magnitude and direction so:

Vab = |Vab| at an angle of @ab
so

|Vab| = sqrt([Re{Van}-Re{Vab}]^2 + [Im{Van}-*Im{Vab}]^2)
or
|Vab| = sqrt([|Van|*cos@a-|Vbn|*cos@b]^2 + [|Van|*sin@a -|Vbn|*sin@b]^2)

and

@ab = tan^-1([|Van|*sin@a -|Vbn|*sin@b]/[|Van|*cos@a-|Vbn|*cos@b])
or
@ab = tan^-1([Im{Van}-*Im{Vab}]/[Re{Van}-Re{Vab}])
Click to expand...

So why did you quit? WHAT IS THE ANSWER? RATTUS lays awake at night thinking of this stuff.
 
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Yeah!

Yeah!

bob said:
So why did you quit? WHAT IS THE ANSWER?
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Come on Bob, let's hear your solution as well.

If we start with 120V @ 0 and 120V @ -120, what is the magnitude and phase of Vab??
 
bob said:
So why did you quit? WHAT IS THE ANSWER? RATTUS lays awake at night thinking of this stuff.
Click to expand...
That was as far as I could go without any actual values.

rattus said:
Come on Bob, let's hear your solution as well.

If we start with 120V @ 0 and 120V @ -120, what is the magnitude and phase of Vab??
Click to expand...
And now we have it. So Vab becomes:
207.85<30 or 180+j103.92
 
Close enough:

Close enough:

mivey said:
That was as far as I could go without any actual values.

And now we have it. So Vab becomes:
207.85<30 or 180+j103.92
Click to expand...

Now that wasn't hard was it?
 
Summary:

Summary:

Let me summarize what mivey has done. Correct me if I make a typo.

First, Kirchoff to the rescue. We sum the phasors in a CCW direction, but CW would work just as well.

Vbn + Vab - Van = 0

Vab = Van - Vbn

Now let,

Van = 120V @ 0
Vbn = 120V @ -120

Now convert to rectangular and collect terms,

Vab = 120V -(-60 -j104)V = 120V + 60V + j104V = 180V + j104V

And then back to polar (courtesy of hp)

Vab = 120V @ 30

Note that we had to use subtraction because one of the phasors was pointing against the direction of summation. Note also that algebraic subtraction is nothing more than changing the sign and adding.

Now isn't someone going to ask about sqrt(3)?
 
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