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Projection of a distance in rectangular coordinates

  1. May 22, 2013 #1
    robotarm.jpg

    My problem is that I believe I have a wrong concept somewhere, and I can't find what I'm doing wrong exactly. For this problem let's suppose what I want to do is find the rectangular coordinates of BC.

    I had two "possible solutions" I tried to achieve this, . First the correct one:
    (I apologize by drawing quality)
    Drawing.JPG

    By determining angles I can find that the angle between the vertical axis and BC is, therefore:
    [itex](300\sin(50),300\cos(50))[/itex]

    Now the wrong one:

    By looking at the original figure alone, first I project BC into AB, then project BC into either X axis or Y axis.
    Procedure:
    1. Projecting BC into AB -> [itex](300\sin(20),300\sin(20))[/itex]
    2. Projecting my new AB' into X and Y axis -> [itex]((300\sin(20))\sin(60),(300\sin(20))\cos(60))[/itex]

    So, why is my second procedure wrong?
     
  2. jcsd
  3. May 22, 2013 #2

    haruspex

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    You mean, C in relation to B, yes?
    I'm not at all sure what you mean by that. Is that supposed to be the coordinates, relative to B, of the projection of C onto the line AB (extrapolated)? If so, it's wrong.
     
  4. May 22, 2013 #3

    Simon Bridge

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    Just checking:
    You mean to find the vector from B to C in cartesian coords oriented to the ground/vertical?
    i.e. so the +j direction is OA?

    ##\vec{v}=\overrightarrow{BC} \\ \Rightarrow [\vec{v}]_{OA}= v_x\hat{\imath}+v_y\hat{\jmath} = (v_x,v_y)^t = 300(\sin 50, \cos 50)^t##

    (here I'm using ##[\cdots ]_{proj}## to label the projection)

    step 1: the angle is wrong.
    step 2: each projection of BC to AB needs to be projected onto i/j - so there should be a sum in there.
    Redo using unit vector notation and it should be clear.

    i.e.
    if ##\vec{u}=\overrightarrow{AB}## then the angle between ##\vec{v}## and ##\vec{u}## is 50+60=110deg.
    define ##\hat{\jmath}_u = \vec{u}/u## - i.e. it's a unit vector pointing in the same direction as ##\vec{u}##, and let ##\hat{i}_u## to be perpendicular to the right, to make cartesian coords aligned with the y-axis along ##\vec{u}##

    then .the projection onto AB goes:.. ##[\vec{v}]_{AB} = \hat{\imath}_u\sin 70 + \hat{\jmath}_u \cos 70##

    for step 2, you need to express ##\hat{\imath}_u## and ##\hat{\jmath}_u## in terms of ##\hat{\imath}## and ##\hat{\jmath}##
    so that ## [\vec{v}]_{OA} = [\hat{\imath}_u]_{OA}\sin 70 + [\hat{\jmath}_u]_{OA} \cos 70##


    i.e. ##[\hat{\jmath}_u]_{OA} = \hat{\imath}\sin 60 -\hat{\jmath}\cos 60##

    ... see what I mean? Each component of the projection onto AB gets another two components.
    (caveat: I'm doing this from mobile so I probably made a dumb mistake - the approach should be fine.)
     
  5. May 23, 2013 #4
    Yes, exactly. the +j direction would be OA and +i the , to the right.

    Ok so, I think I understood what you meant. Basically I was ignoring the j coordinate in my original idea. I didn't quite understood everything just yet, though. Please take a look in what I did in order for me to explain better what I can't quite understand:
    Solution.JPG
    Just explaining the picture, I made two drawings to ensure I understood the situation (Initially I was ignoring ##BC_j##) and not "polute" the first.

    My doubt is, I don't understand why ##BC_y = -(BC_i\cos(60)) + BC_j\cos(30)##, for instance. I discovered the "minus" because of empirical evidence (i.e. making sure it equals the answer I know is correct) and because of your tip. But... since it's a distance, shouldn't both of them be "positive"?

    I also don't understand ##BC_x##. In my second draw, for instance, wouldn't I be add the "same" part twice?

    Thanks in advance!
     
  6. May 25, 2013 #5

    Simon Bridge

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    You made the AC direction the +x direction in the BC coordinates where I made it the +y direction. No matter - you can make any definition you like as long as you are clear about it.

    Note: The projection of ##\vec{u}## in the direction of ##\vec{v}## is $$\frac{\vec{u}\cdot\vec{v}}{|\vec{v}|}=|\vec{u}|\cos\theta$$ ... which is not what you wrote.

    But, despite all the confusion over dot products, you got the correct relation by exploiting the geometry rather than using general formulas.

    In the next step you appear to have noticed that each AB-frame component resolves into two OA-frame components.

    That minus sign you got comes right off the graph - it is due to the direction of the projection.
    Without the minus sign, the resulting vector points the wrong way.

    In general:

    If ##\vec{v}=v_x \hat{\imath}##, ##v_x## is the amount of ##\vec{v}## in the direction of ##\hat{\imath}## ... if ##\vec{v}## points in the opposite direction to ##\hat{\imath}## then ##v_x## must be negative.

    Hope that helps :)
     
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