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Basic Trigonometry Help Please - Phasor Diagram

  1. Jun 22, 2013 #1
    1. The problem statement, all variables and given/known data
    Hello everyone,

    I am having difficulty with something that should be very basic knowledge for most engineers here. I cannot see how this trigonometric relationship is being made from the phasor diagram I attached...

    I remember the SohCahToa rule, but it's just not working here for me. Please if anyone can explain how these two equations were set up and how I can analyze the phasor diagram to see how.. it would be very much appreciated. Thank you in advance.
     

    Attached Files:

  2. jcsd
  3. Jun 22, 2013 #2

    Simon Bridge

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    ... basically, ##\vec{I}\!_0 = \vec{I}\!_q + \vec{I}\!_d##
    ... notice that Iq and Id are prependicular to each other?
     
  4. Jun 23, 2013 #3
    Yes I see that they're perpendicular. But I can't see how they are all related with sin and cosine..
     
  5. Jun 23, 2013 #4
    Ok so I know these general formulas for a vector:

    Ax=A cosθ
    Ay=A sinθ
    A=Ax+Ay (vectors)

    Before I assumed "Id" to be our Ax and "Iq" to be our Ay here, which is where I think I made my mistake right? Since we are taking the angle with respect to the "y-axis" it seems to make more sense to use the Iq component as our Ax right? Or just use the parallelogram theorem of equivalency.. If so then I can see the relation through SohCahToa.

    But I still don't see how -Io fits into the Id equation trigonometrically from the vector diagram..why is Io negative?

    If someone can confirm my doubts I would be greatly appreciative.. Thank you.
     
    Last edited: Jun 23, 2013
  6. Jun 23, 2013 #5

    Simon Bridge

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    You are getting there.
    Id is your Ax ... but, using your formulas, theta is to the "x axis".
    The angles to the x axis is 90 minus the angle to the y-axis.

    I notice that the sketch is not to scale.
    Anyway - you really need to stop thinking in terms of fitting problems to memorized formulas.
    You should be making the formulas fit the problem ... when you add Id and Iq, head-to-tail, you get a right-angled triangle with I0 as the hypotenuse and you know all the angles.
     
  7. Jun 23, 2013 #6
    Thanks, but I still don't see in the phasor diagram why Io would be negative...
    I can produce the equations perfectly, besides the fact that I get a positive Io, which I know is incorrect.
     
  8. Jun 23, 2013 #7

    Simon Bridge

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    I0 is not negative.
    The minus sign comes from the trig identity... cos(90-A)=...
     
  9. Jun 23, 2013 #8
    I'm confused because this is how it's shown in the book, as I wrote the equations. The book didn't take 90-theta and I'm trying to understand how these relations are written using the original value of theta without subtracting from 90..
     
    Last edited: Jun 23, 2013
  10. Jun 23, 2013 #9
    I see now. In the book they direct the d-axis in the opposing direction and the negative sign is carried over to the Io side of the equation!
     
  11. Jun 23, 2013 #10

    Simon Bridge

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    The book took a shortcut. The shortcut confused you - so, to understand it, you have to go the long way around.

    Considering the way you have been thinking about the problem, you will understand it better if you use the relations you wrote properly. This means taking θ=90-<angle> in order to get the right theta for your equations. Can you see why this is the right theta for your equations?

    Books won't always use the variable symbols to mean the same thing that you are used to.
    This is allowed - and it is why you should really be figuring out the equations that go with the problem instead of, what you are doing, trying to fit the problem into the equations.
     
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