Uncertainty question

  • #1
This isn't a homework question, but instead a question about an example in a book I'm reading, in prep for next semester. As such using the posting template is a bit of a miss. Hope that can be forgiven.

1. Homework Statement


I'm reading "An Introduction to Error Analysis" by John R Taylor during my spring vacation as brush up for my next semester. I encountered an example that doesn't make sense to me though. It goes through teaching 3 rules as follows:

upload_2018-8-17_14-8-6.png
(3.8)
upload_2018-8-17_14-8-16.png
(3.9)
upload_2018-8-17_14-8-35.png
(3.10)

Now the example it gives is as follows (leaving out units for ease):

##t = 1.6\pm0.1##

##h = 46.2\pm0.3##

Now it calculates ##g = \frac{2*h}{t^2}## and more importantly its uncertainty as follows:

$$\frac{\delta g}{g}=\frac{\delta h}{h}+2*\frac{\delta t}{t} = 0.007+2*0.063 = 0.133$$

as justifed by 3.8 and 3.10. However here is my problem; What about 3.9?

The Attempt at a Solution



As far as I can see, the example completely forgets about the factor 2. With 3.9 in mind, ##x = \frac{h}{t^2}## shouldn't it be:

$$\frac{\delta g}{g}=2*\left(\frac{\delta h}{h}+2*\frac{\delta t}{t}\right)$$

What am I missing?
 

Attachments

Answers and Replies

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,652
Honestly, you should not be using 3.8 because it essentially makes the assumption that all the errors are correlated (which may be conservative, but usually not the case). If the errors are uncorrelated, it would be more appropriate to add the relative errors in quadrature.

However, given 3.8-10: No, there should not be a factor of 2. What is given in 3.9 is ##\delta q##, not ##\delta q/q##. The factor of 2 from ##\delta q## cancels the factor of 2 from ##q##.

Edit: To expand on that, 3.9 states ##\delta q = |B| \delta x## and by definition ##|q| = |B| |x|## and therefore
$$
\frac{\delta q}{|q|} = \frac{|B|\delta x}{|B| |x|} = \frac{\delta x}{|x|}.
$$

Also, do not write ##\delta * x##, it is ##\delta x## which is the error in ##x##. The ##\delta## and the ##x## are part of the same symbol representing the error in ##x##.
 
  • Like
Likes NicolaiTheDane
  • #3
Honestly, you should not be using 3.8 because it essentially makes the assumption that all the errors are correlated (which may be conservative, but usually not the case). If the errors are uncorrelated, it would be more appropriate to add the relative errors in quadrature.

However, given 3.8-10: No, there should not be a factor of 2. What is given in 3.9 is ##\delta q##, not ##\delta q/q##. The factor of 2 from ##\delta q## cancels the factor of 2 from ##q##.

Edit: To expand on that, 3.9 states ##\delta q = |B| \delta x## and by definition ##|q| = |B| |x|## and therefore
$$
\frac{\delta q}{|q|} = \frac{|B|\delta x}{|B| |x|} = \frac{\delta x}{|x|}.
$$

Also, do not write ##\delta * x##, it is ##\delta x## which is the error in ##x##. The ##\delta## and the ##x## are part of the same symbol representing the error in ##x##.
Of course! Embarrassing oversight on my part. Thanks a bunch!
 

Related Threads on Uncertainty question

  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
4
Views
6K
  • Last Post
Replies
7
Views
581
  • Last Post
Replies
5
Views
14K
  • Last Post
Replies
12
Views
762
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
6
Views
2K
Top