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Shifting Coordinates question

  1. Sep 3, 2011 #1
    I'm stuck on this question.. I have achieved the answer by fluke but would love help understanding the process its driving me nuts...

    The velocity of a ball in an x-y coordinate system is (10, -5) where distance is measured in metres. A second coordinate system, p-q, uses units of feet (1 ft = 0.3048 m). The p-axis is oriented at alpha = 15 degrees relative to the x-axis. The origin of the p-q system is located at (10m, 2m) in the x-y system. What is the velocity vector of the ball in the p-q coordinate system?

    So the method I went about this is to bring the p-q axis back to the origin of the x - y co ordinate system by effectivley shifting the point (10,-5) by (10,2)
    giving me the point (0,-7)... then I did a rotation transformation of 15 degrees which gave me

    p = -ysin(15) and q=ycos(15) then multiplied both by 3.28ft/m to give me the ft/s
    but this gives me the point (5.94,-22.18) which is completely incorrect.

    whats the proper method of approaching this ??
  2. jcsd
  3. Sep 3, 2011 #2


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    hi rambo5330! :smile:
    hint: why would changing the coordinate origin make any difference to a velocity vector? :wink:
  4. Sep 3, 2011 #3
    Well that is indeed how I arrived at the answer the first time, is by making a transformation matrix and not shifting the co-ordinates origin.

    however this is whats really confusing me, I thought for sure (especially with a velocity vector) it would make a big difference where the coordinate system p-q is with respect to the x-y system.

    having the p-q axis shifted.. and then projecting the point from the x-y system onto this.. wouldnt the location of the origin affect the magnitude of the velocity vector and or direction??
  5. Sep 3, 2011 #4
    the more i think about it the more it bothers me...

    Moving the origin of a coordinate system without moving the point is exactly the same as moving the point,while keeping the origin fixed. which is exactly the same as creating a new vector with a different magnitude and direction?
  6. Sep 3, 2011 #5


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    velocity vector = distance vector minus distance vector over time …

    shifting the origin changes both distance vectors by the same amount, os it doesn't change the difference :wink:

    to put it another way: if you're in a car racing round a circuit, does your speed and direction depend on where the origin of coordinates is?

    (if the circuit is on a moving ship, then your speed and direction does depend on the speed and orientation of the ship, but still not on its actual position)
  7. Sep 3, 2011 #6
    Interesting, thank you very much. I will think about that one for awhile !
  8. Sep 5, 2011 #7
    I have to revisit this thread. I'm still having a difficult time conceptually/graphically understanding this. So we're dealing with a graph of velocity. Which would be distance vs time graph correct? or do you interpret this as a velocity versus time graph? im having a really hard time understanding how an origin shift does not affect this situation. because if a shift doesnt affect it then if i were to shift the origin of p-q to the origin of x-y I should get the same solution, which I dont. any insite would be greatfully appreciated...or perhaps is this to be read as the velocity is (10i -5j)m/s as in both x and y axis are in m/s
    Last edited: Sep 5, 2011
  9. Sep 5, 2011 #8


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    hirambo5330! :smile:
    there's no graph here :confused:

    the position of the ball is a curve in x,y space

    the velocity at any point is a vector tangent to that curve, with length proportional to the speed

    so if you shift the origin, then you shift the whole curve sideways, but you don't change the direction of the tangents​
  10. Sep 5, 2011 #9
    that makes a lot more sense maybe the graphic that goes along with the question was throwing me off... it shows a stadnard x,y and then the p-q axes rotated and shifted.

    we are taught to do this with a rotation matrix etc..

    the way you explained it makes perfect sense though if i think about the tangents
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