SigFigs in Volume and Uncertainty?

  • Thread starter Thread starter maxhersch
  • Start date Start date
  • Tags Tags
    Uncertainty Volume
Click For Summary
SUMMARY

The discussion focuses on calculating the change in volume and its uncertainty for a piston with a diameter of 7.500 ± 0.005 cm and a distance of 3.250 ± 0.001 cm. The calculated change in volume is 143.6 cm³, measured to four significant figures, while the uncertainty is determined to be ± 0.09 cm³. Participants emphasize that when explicit uncertainty values are provided, they should take precedence over the significant figures convention. Additionally, they discuss the appropriate methods for combining uncertainties, highlighting the root-sum-square approach as a more accurate alternative to simple addition.

PREREQUISITES
  • Understanding of significant figures in measurements
  • Familiarity with basic geometry, specifically the area of a circle
  • Knowledge of uncertainty propagation techniques
  • Ability to apply the root-sum-square method for combining uncertainties
NEXT STEPS
  • Study the principles of uncertainty propagation in physics
  • Learn how to apply the root-sum-square method for independent uncertainties
  • Explore the significance of significant figures in scientific measurements
  • Review the calculation of area and volume for cylindrical shapes
USEFUL FOR

Students in physics or engineering, educators teaching measurement and uncertainty, and professionals involved in precision measurements and data analysis.

maxhersch
Messages
20
Reaction score
0

Homework Statement


A car engine moves a piston with a circular cross section of 7.500 ± 0.005cm diameter a distance of
3.250 ± 0.001cm to compress the gas in the cylinder.
(a) By what amount is the gas decreased in volume in cubic centimeters?
(b) Find the uncertainty in this volume.

Homework Equations


Area = πr2
ΔVolume = Area × ΔDistance
%unc = (ΔA/A) × 100%

The Attempt at a Solution


The radius of the cross section of the piston is 3.75cm so the area comes out to 44.18 cm2
The change in volume then comes out to 143.6cm3
-This value is measured to 4 significant figures because both the diameter and the distance were given to 4 significant figures.​
The percent uncertainty in the diameter comes out to 0.0267%
The percent uncertainty in the distance comes out to 0.0308%
The percent uncertainty in the change in volume should then be these values added together, giving 0.0575%
-With the correct number of significant figures this should be 0.06% because the original uncertainties (0.005 and 0.001) both had only 1 significant figure.​
Then, to get the uncertainty for the change in volume you use %unc = (ΔA/A) × 100% and solve for ΔA
-For this I got 0.08618 which comes to ± 0.09cm3

My question is: If the change in volume (143.6cm3) has the correct number of significant figures but is measured only to the nearest tenth of a centimeter, then how can the uncertainty (±0.09cm3) be in hundredths of a centimeter?
 
Physics news on Phys.org
The "sigfig" rule is only a rule of thumb and should not be used if you have the option.
If you have actual uncertainty values, as is the case here, those are what you use.
 
Simon Bridge said:
The "sigfig" rule is only a rule of thumb and should not be used if you have the option.
If you have actual uncertainty values, as is the case here, those are what you use.
I agree, but would express it a little differently. The sig fig system is a convention for implying the accuracy. When it is given explicitly, the convention does not apply.
 
haruspex said:
I agree, but would express it a little differently. The sig fig system is a convention for implying the accuracy. When it is given explicitly, the convention does not apply.

Thanks, so just to be clear that would mean that I should express the change in volume and the uncertainty in the change in volume to the nearest thousandth of a centimeter because they were measured to that degree of accuracy in the question, right?
 
maxhersch said:
Thanks, so just to be clear that would mean that I should express the change in volume and the uncertainty in the change in volume to the nearest thousandth of a centimeter because they were measured to that degree of accuracy in the question, right?
Before we worry about such niceties, let's get the answer basically right. I can see one definite problem with your answer and another possible one.
You added an uncertainty percentage in one distance (diameter) to an uncertainty percentage in another distance (length) to get an uncertainty percentage in volume. Does anything strike you as rather doubtful in that?
Secondly, there are two approaches to adding up independent uncertainties. The approach you have used, simply adding them, finds the worst case result; the other finds a more likely range by using a root-sum-square rule. Which have you been taught to use?

Edit: one more problem maybe... how did you calculate the 0.0267%?
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
12
Views
28K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K