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Turning on snowboard or surfboard

  1. Oct 30, 2009 #1
    In a turn, a rider might be leaning over at around 45 degrees at times. What is the g-force at that point going down into his feet ? I'm thinking that it is essentially independant of the speed and turn radius. Also, I'm thinking the rider is close enough to standing straight - no bent knees or hips, arms near to by his or her sides. What about getting really low - say 30 degrees ? Is it a simple sin cos or tan calculation ?
     
  2. jcsd
  3. Oct 30, 2009 #2
    the force due to gravity is his mass times the local gravitational field strength- neither depend on any angles. In snowboarding the rider himself doesn't lean, you still keep your weight more or less over the board (maybe a little in a fast tight turn), mostly you angle the board so that the "bloated" ends bend upwards slightly, and then you trace a curved path.
     
  4. Oct 30, 2009 #3

    russ_watters

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    Staff: Mentor

    That's not correct. You can calculate the g-force (acceleration) required for a balanced turn for a snow board, surf board - even an airplane, knowing only the bank angle.

    For a snowboard and surfboard, in order to not fall over, the resultant effective gravitational acceleration vector of your center of gravity must pass through your board. Ie, if you lean to far to the right, you fall over to the right. So to balance the forces, you draw a diagram showing the resultant vector pointing from your CoG to the board, the gravitational force vector straight down, and a third vector horizontal into the turn. You have all three angles and the magnitude of one side, so you can solve for the magnitude of the other sides.

    For a 45 degree bank angle, you have a 45/45/90 right triangle, so the horizontal and vertical components are both 1g and the resultant force (the one you "feel") is 1.4g.
     
  5. Oct 30, 2009 #4
    I think about this when I'm riding my boards all the time!

    How many g's you are pulling is dependent on your velocity and radius of curvature.

    Take the constant velocity circular motion equation for normal accelleration:

    a=(velocity)^2/(radius of your turn)

    to find how many g's you pull divide 'a' by 'g', gravity constant.
     
  6. Nov 2, 2009 #5
    Thanks , Russ , I definitely like yours the most.
     
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