# Uncertainty Problem

1. Aug 30, 2004

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%

2. Aug 31, 2004

I did the calculation and I got 8%.

3. Aug 31, 2004

hmm well its supposedly like 4.45%, i have a feeling this has something to do with partial derivatives?

4. Aug 31, 2004

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. Sep 1, 2004

5280 feet in mile

6. Sep 1, 2004

I Still cant get it, I get something liek 3.8%.

7. Sep 1, 2004

### Chronos

See if this helps
$$A = \frac{1.226r}{t^2}$$
where
A = acceleration in g's
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. Sep 1, 2004

yeah but im not given any information about times

9. Sep 2, 2004

### Chronos

try substitution. T = d/v. The equation then becomes
$$A = \frac{1.226rv^2}{d^2}$$ 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
10. Sep 7, 2004