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Uncertainty Problem

  1. Aug 30, 2004 #1
    Here's the problem:

    A magazine publishes lateral acceleration capability from cars it tests. Measurements are made using a 150' diameter skidpad, the vehicle path deviates from the circle +/- 2 ft and the vehicle speed is read from a fifth wheel sensor measuring the system to +/- .5 mph. Then it says to estimate the experimental uncertainty if the reported lateral acceleration is .7g.

    All I can think of is A=v^2/r...and some how i would have to cut the 150 diameter into 75 as a radius. I got 75 +/- .0133. then i found v = to 22.69, and you have to square that...I'm basically stumped...the answer is like 4.XX% :frown:
  2. jcsd
  3. Aug 31, 2004 #2
    I did the calculation and I got 8%.
  4. Aug 31, 2004 #3
    hmm well its supposedly like 4.45%, i have a feeling this has something to do with partial derivatives?
  5. Aug 31, 2004 #4
    no, you dont need partial derivatives to solve it. Ill get back to you with the answer, by the way, how many feet in a mile?
  6. Sep 1, 2004 #5
    5280 feet in mile
  7. Sep 1, 2004 #6
    I Still cant get it, I get something liek 3.8%.
  8. Sep 1, 2004 #7


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    See if this helps
    [tex]A = \frac{1.226r}{t^2}[/tex]
    A = acceleration in g's
    r = radius of track
    t = time [in seconds] to complete 1 lap
    This is how g's are calculated on a skidpad
    re: http://www.carcraft.com/howto/53698/index7.html
  9. Sep 1, 2004 #8
    yeah but im not given any information about times
  10. Sep 2, 2004 #9


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    try substitution. T = d/v. The equation then becomes
    [tex]A = \frac{1.226rv^2}{d^2}[/tex] Does that help?
    Given the uncertainty of measurements in the quantities r and v, the answer is there. I agree with the 4.xxx% result, see what you get.
    Last edited: Sep 2, 2004
  11. Sep 7, 2004 #10
    I'm still not following how to get that 4.45 answer, can you please elaborate for me?
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