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Coordinate transformation

  1. Aug 17, 2015 #1

    diredragon

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    1. The problem statement, all variables and given/known data
    Transform the coordinates from the red c-system to the blue system. (Picture)

    2. Relevant equations
    Using(X Y) for the red cartesian system and (x y) for the blue system

    3. The attempt at a solution
    The solution to this problem gives
    x=Xcos▼ + Ysin▼
    y=-Xsin▼+Ycos▼
    Im not sure i understand in what way was this transformed. How can x coordinate in blue system contain Y coordinate from red system?
     

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  2. jcsd
  3. Aug 17, 2015 #2

    SteamKing

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    It doesn't look like the coordinates were translated between the two systems.

    What other manner of transformation is suggested by the diagram?
     
  4. Aug 17, 2015 #3

    diredragon

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    I think i get it now, the change in x is given by x=(xcosp)i-(xsinp)j and y=(ysinp)i-(ycosp)i
    r=xi + yj, getting the above result, this should be it.
     
  5. Aug 17, 2015 #4

    RUber

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    In you last post, you didn't differentiate between X and x which tells us which coordinate system you are referring to.
    Looking at the diagram, it looks like
    ##(1,0)_{(X,Y)} = (\cos \theta, \sin \theta)_{(x,y)} ##
    and
    ## (0,1)_{(X,Y)}=(- \cos \theta, \sin \theta)_{(x,y)}##
    Indicating that the angle from the x - axis is the same for each of the red basis vectors.
    Or is it that the red axes are orthogonal? In which case
    ##(1,0)_{(X,Y) }= (\cos \theta, \sin \theta)_{(x,y)} ##
    and
    ## (0,1)_{(X,Y)}=(\sin \theta, - \cos \theta)_{(x,y)}##
     
  6. Aug 17, 2015 #5

    diredragon

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    I think that the above equations correctly describe the transformation even if the red coordinate system is rotating, taking your example of (0,1) and putting the angle equal to end points of 0,pi/2,p we get the correct trabsformation. It agrees with the book answer, it goes with the proposed experiment so i think its right. Do you agree?
     
  7. Aug 17, 2015 #6

    RUber

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    My question was if the red coordinate system was based on orthogonal vectors.
    I did not think that the angle was a variable in the red system, is this some sort of polar coordinate system? I see the blue system as having the standard ##\hat i, \hat j## orthonormal basis vectors.
    If the red coordinate system has orthonormal basis vectors ##\hat e_1, \hat e_2## such that any point in the plane may be expressed as ##X\hat e_1 +Y \hat e_2## you are looking for a way to transform coordinates given as ##(X,Y)_{(red)}## into ##(x,y)_{(blue)}##
    The easiest way to do a coordinate transformation is to see what happens to the basis vectors themselves.
    Assuming that ##\hat e_1## is the vector pointing at about 45 degrees off the x axis, then ##(1,0)_{(red)}## would translate to ##(\cos \theta, \sin \theta)_{(blue)}##.
    Assuming that ##\hat e_2 ## is the vector pointing at about 135 degrees off the x axis and orthogonal to ##\hat e_1##, then ##(0,1)_{(red)}## would translate to ##(-\sin \theta, \cos \theta)_{(blue)}##.
    You can build the transformation matrix by stacking the transformations of the basis vectors like this:
    ##[T] = \begin{pmatrix} [Te_1]^T & [Te_2]^T \end{pmatrix} ##
    Which would look like:
    ##[T]_{red \to blue} = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}##
    Then your transformation would be a matrix multiplication of
    ## \begin{pmatrix} x\\y \end{pmatrix}_{(blue)} = [T] \begin{pmatrix} X\\Y \end{pmatrix}_{(red)}.##
     
  8. Aug 17, 2015 #7

    diredragon

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    Yes, they are orthogonal and i get what you mean, im not too familiar with matrices hut i get the idea. But what do you think about the equations i posted? They are correct right
     
  9. Aug 17, 2015 #8

    RUber

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    Is this the result you think is correct? What is r? Where are X and Y? Why is y only composed of i components and x composed of i and j?
    The matrix transformation I just posted (in #6) multiplies out to the result you posted originally as the solution. Try it out.
     
  10. Aug 17, 2015 #9

    diredragon

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    The post has a mistake it ahould read X and Y equal to the stated and y's second component is j not i. Your way is right as well.
     
  11. Aug 17, 2015 #10

    RUber

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    I think I might have messed up my signs--Sorry.
    Let's look at this without matrices.

    In the red coordinates, (X,Y) is a point that is ##X \hat e_1 + Y \hat e_2##
    In the blue coordinates, ##\hat e_1 = \cos \theta \hat i + \sin \theta \hat j ## and ##\hat e_2 = \cos (\theta+\pi/2) \hat i + \sin ( \theta+\pi/2) \hat j ##
    ## \cos (\theta+\pi/2) = - \sin \theta ## and ##\sin ( \theta+\pi/2) = \cos \theta ##
    So, ##\hat e_2 = -\sin \theta \hat i + \cos \theta \hat j ##
    Plug these in to the first equation in terms of (X,Y) and you should have your points described by ##\hat i ## and ##\hat j ##.
    Your x will be the sum of the i components and your y will be the sum of the j components.
     
  12. Aug 17, 2015 #11

    diredragon

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    Yup, thats right, so the transformation from red to blue is
    x=Xcos∅+Ysin∅
    y=Ycos∅-Xsin∅
     
  13. Aug 17, 2015 #12

    RUber

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    For some reason I am getting different signs.
    I get
    ## x = X \cos \theta - Y \sin \theta \\ y = X \sin \theta + Y \cos \theta##
     
  14. Aug 18, 2015 #13

    diredragon

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    It is true, however not the answer the book gives. I get the same result as this though, but its confusing as everybody seems to be using the -Xsin() involving formula
     
  15. Aug 18, 2015 #14

    RUber

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    Maybe there is some direction or order assumption that is being made. Do you have a coordinate diagram that is labeled?
     
  16. Aug 18, 2015 #15

    diredragon

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    In fact i have a scatch done last year by my friend which solves the problem. I added it below. Can you see where our mistreat was?
     
    Last edited: Aug 18, 2015
  17. Aug 18, 2015 #16

    diredragon

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    Here it is.
     
  18. Aug 18, 2015 #17

    RUber

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    Still not seeing it. Is the image format compatible with the site?
     
  19. Aug 18, 2015 #18

    Ray Vickson

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    There are two slightly different versions of such transformations.

    Say you have on original coordinate system and a point ##p = (x,y)## in it. Then you can do two things:
    (1) fix the point in place but rotate the coordinate system counterclockwise through the angle ##\theta## (which is actually clockwise if ##\theta < 0##). In this case the same physical point ##p## has new components ##(x',y')## in the new coordinate system.
    (2) Fix the coordinate system but physically rotate the point ##p## clockwise through angle ##\theta##, to get a new point ##p' = (x',y')##.

    One of these scenarios will have
    [tex] x' = x \cos \theta - y \sin \theta\\
    y' = x \sin \theta + y \cos \theta [/tex]
    while the other one will have
    [tex] x' = x \cos \theta + y \sin \theta\\
    y' = - x \sin \theta + y \cos \theta [/tex]
    I will leave it to you to figure out which is which.
     
  20. Aug 18, 2015 #19

    diredragon

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    Based on this picture it seems like the equations presented on the blackboard depict a point rotation rather than whole syatem, is it not?
     

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  21. Aug 18, 2015 #20

    RUber

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    I see now. I was simply translating (X , Y) in the red coodinates to an (x, y) in the blue coordinates. Where (x , y) and (X, Y) refer to the same location.
    I was not rotating them.
     
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