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Normal vector of a 2d-surface

  1. Nov 21, 2014 #1
    Can you guys show me how to calculate the normal vector n, given that i know the coordinates of the center c (c1,c2),
    the side of the square (let's call it s), and the length of the plank (l) below the square?

    i am taking the assumption that n is (0,1), which seems to be incorrect, because the result i get from using this value does not tally with what i see in the results that i measured. vector_n.jpg
     
  2. jcsd
  3. Nov 21, 2014 #2

    Mark44

    Staff: Mentor

    The normal vector to what? Is n the normal to the plank? The plank appears to be horizontal, but no coordinate system is shown, so I can't say whether the plank is horizontal.

    What does the square have to do with anything? If you want an answer, you need to provide more information.
     
  4. Nov 21, 2014 #3
    Sorry for my lacks of info. Yes, n is normal to the plank, and the plank is horizontal, the square is just an object where i get the coordinate of its center of mass c, a circle or a polygon will do as well.
     
    Last edited: Nov 21, 2014
  5. Nov 21, 2014 #4

    Mark44

    Staff: Mentor

    Why do you care about the square or its center of mass? A normal to the plank is the vector <0, 1>, which could be written as 0i + 1j.
    What results are you measuring?
     
  6. Nov 21, 2014 #5
    It's actually used to measure post-impact velocity (http://en.wikipedia.org/wiki/Collision_response), when the object bounces off, i am trying (0,1) for the normal vector as well, but the result does not really tally with what i see when i measure it, will it matter if the plank is not on the ground but higher up?
     
  7. Nov 21, 2014 #6

    Mark44

    Staff: Mentor

    Since the plank is three-dimensional, a normal would be <0, 0, 1>, a vector that points straight up. If the plank is horizontal, then this would also be a normal to the plank, whether the plank is on the ground or elevated above the ground.

    If you are dropping an object on the plank and it's not bouncing straight up, it could be that the object is rotating when it hits, which could change the direction of the recoil. It's hard to say, because you still haven't given many details about what you are doing.
     
  8. Nov 22, 2014 #7
    It am having an experiment where a flat square piece is dropped onto a small plank (almost flat as well, probably 5mm thick) and the camera on top of the set up can give information on the displacement in x,y of the square, angle of rotation, speed in x and y as well as angular velocity. I am trying to calculate the post impact velocities of the square based the values measured right before the impact.
     
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