Measurement uncertainty question

[Sheneron]

Homework Statement

Lets say m1 = 80 +- 0.1 g and m2 = 60 +- 0.2g and p = 1.0 g/cm^3 +- 0.1 g/cm^3

I want to solve for V and V = (m1-m2)/p

What would the uncertainty on this be?
When you subtract the two masses, it becomes 20 +- 0.3g.
You divide that by 1g/cm^3 +- 0.1 g/cm^3.

0.3 is 1.5% of 20.
0.1 is 10% of 1.
So it would be 11.5% uncertainty in the final. 20cm^3 +- 2.3 cm^3

Is that right? Would the uncertainty at the end be in cm^3 as well?

 

[Sheneron]

one more time...

bump

 

[LowlyPion]

Homework Statement

Lets say m1 = 80 +- 0.1 g and m2 = 60 +- 0.2g and p = 1.0 g/cm^3 +- 0.1 g/cm^3

I want to solve for V and V = (m1-m2)/p

What would the uncertainty on this be?
When you subtract the two masses, it becomes 20 +- 0.3g.
You divide that by 1g/cm^3 +- 0.1 g/cm^3.

0.3 is 1.5% of 20.
0.1 is 10% of 1.
So it would be 11.5% uncertainty in the final. 20cm^3 +- 2.3 cm^3

Is that right? Would the uncertainty at the end be in cm^3 as well?

The units of uncertainty of your result would be as you suggest ±cm³

As for your calculation I believe that for addition or subtraction the rule is the square root of the sum of the squares of the absolute uncertainties of the quantities.
Applied to your case it would be the root of (.1² +.2²) or (.05)^1/2

The division multiplication rule is similar but it employs the Relative uncertainty, where Relative uncertainty is absolute uncertainty / quantity.

For the numerator this should be (.05)^1/2/20
For the denominator that would yield (.1/1)

The resulting uncertainty then of your sample would be (.05/400 + .01)^1/2 expressed as the nominal value of your calculation (e.g 20 ± 20*(the calculated relative uncertainty) in cm³.

 

[Sheneron]

For the first part though that would only give an uncertainty of 0.22, which can't be right.

In our example, 80 +- 0.1 and 60 +- 0.2 could be as low as 59.8+79.9 = 139.7 or could be as high as 80.1 + 60.2 = 140.3

The uncertainty would be to low by the way you suggested. The real value would be somewhere between 139.7 and 140.3 or equal to those. You can't make those bounds any smaller.

 

[Sheneron]

By the way thanks for the response

 

[LowlyPion]

For the first part though that would only give an uncertainty of 0.22, which can't be right.

In our example, 80 +- 0.1 and 60 +- 0.2 could be as low as 59.8+79.9 = 139.7 or could be as high as 80.1 + 60.2 = 140.3

The uncertainty would be to low by the way you suggested. The real value would be somewhere between 139.7 and 140.3 or equal to those. You can't make those bounds any smaller.

What I provided you are the rules for calculating error propagation given the uncertainty that you provided.

Here is a quick note on how to approach it:
http://www.physics.unc.edu/~spm/Uncertainty%20Tricks.pdf

Other texts may treat uncertainty differently, possibly more conservatively by not employing the RSS, but there is one thread of agreement in treating addition and subtraction propagation using the absolute uncertainties and multiplication and division operations with relative uncertainties. This was what I attempted to illustrate with your problem.

Without knowing what methodology may be being taught in your text or coursework, I chose what I consider to be the more rigorous and exact.

If you are not using the RSS, but merely the absolute and relative according to the operation, (+,-,*,/), then your way of calculating the sum of the absolutes for the numerator and converting to relative and then adding the relative of the denominator yields a more conservative answer 11.5% - which as you noted is a larger bound to the one I calculated using the root sum of the squares.
So I think in the case of your first example it could be expressed as 20.0 ± 2.3 cm³

By way of illustration in the second case where you would sum to 140.0 ± .3
Your relative error there is considerably smaller and that result would be (using the division rule):
.21% + 10% = 10.21% or 140 ± 14.3 cm³

Hope that helps.

Cheers.

 

[Sheneron]

Thanks for the link, I will have to look into it more when I get a chance. I suppose I should first get a better idea of absolute and relative uncertainty.