Uncertainty Problem

  • Thread starter purduegrad
  • Start date
  • #1
11
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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:
 

Answers and Replies

  • #2
698
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I did the calculation and I got 8%.
 
  • #3
11
0
hmm well its supposedly like 4.45%, i have a feeling this has something to do with partial derivatives?
 
  • #4
698
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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?
 
  • #5
11
0
5280 feet in mile
 
  • #6
698
0
I Still cant get it, I get something liek 3.8%.
 
  • #7
Chronos
Science Advisor
Gold Member
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See if this helps
[tex]A = \frac{1.226r}{t^2}[/tex]
where
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
 
  • #8
11
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yeah but im not given any information about times
 
  • #9
Chronos
Science Advisor
Gold Member
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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:
  • #10
11
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I'm still not following how to get that 4.45 answer, can you please elaborate for me?
 

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