# Finding error on gravitational acceleration

1. Oct 6, 2012

### peripatein

How may the error on gravitational acceleration (g) be determined, given a set of measurements of time (t) and distance (d)? It is stated that the distances are measured precisely and time with an accuracy of 0.01 sec.
I have applied Least Squares on t = √(2/g) * √d and found (σ of √g) albeit was unsuccessful at deducing (σ of g). I have also tried using Error Propagation, hence (σ of t)^2 = [(∂t/∂g) * (σ of g)]^2 → (σ of t)^2 = (d/2g^3) * (σ of g)^2, yet what value would d have?

2. Oct 6, 2012

### Simon Bridge

How did you get the different measurements?

3. Oct 7, 2012

### Simon Bridge

Back to you ...

you have y=√g and σy and you want to find σg ... is that it?
so what is stopping you from putting g=y2 and finding the uncertainty on that the usual way?

4. Oct 7, 2012

### peripatein

Thank you! :-)

5. Oct 7, 2012

### Simon Bridge

No worries.

In general, if $y=f(x)$ and you have $x\pm \sigma_x$ and you need $\sigma_y$ then: $\sigma_y = \sigma_x f^\prime(x)$

Proof: expand f(x) in a power series $$y=f(x)=\sum_{n=0}^\infty a_n x^n$$ You know how to find the uncertainty on each of the terms: $$\frac{\sigma(x^n)}{x^n}=n\frac{\sigma_x}{x}$$ ...so $$\sigma_y = \sum a_n n\sigma_x x^{n-1} = \sigma_x\sum na_nx^{n-1} =\sigma_x\frac{d}{dx}\sum a_nx^n= \sigma_x f^\prime(x)$$

(Just for the people who read this later.)

6. Oct 9, 2012

### peripatein

Hence, does the uncertainty in the volume of a cylinder, the height of which is H+-ΔH and radius R+-ΔR, then equal SQRT[(πR^2*σH)^2 + (2πHR*σR)^2]?
May partial derivatives be applied here, the errors on both R and H expressed as Δ's notwithstanding?

7. Oct 9, 2012

### Simon Bridge

How would you normally estimate the error?
Compare and see.

8. Oct 10, 2012

### peripatein

I derived the error on the volume as SQRT[(πR^2*ΔH)^2 + (2πHR*ΔR)^2] using partial derivatives, namely SQRT[(dV/dH*ΔH)^2 + (dV/dR*ΔR)^2]. I am merely inquiring whether that would indeed be the right answer. May you please confirm?

9. Oct 10, 2012

### Simon Bridge

... and I have told you how to find out for yourself.

You wanted to know ...
if $z=f(x,y)$, can we find a general form for $\sigma_z$ in terms of $\sigma_x$ and $\sigma_y$ ... the trouble is that x and y may have any relationship and the derivation I gave for functions of one variable relies on dependent measurements.

eg. if $z=xy$

dz/dx=y, dz/dy=x so you'd expect $\sigma_z=y\sigma_x + x\sigma_y$

Well ... if x and y are statistically dependent measurements, then that would be the case.
However, if they were independent, then $$\sigma_z = \sqrt{y^2\sigma_x^2 + x^2\sigma_y^2}$$

Since H and R are (usually) independent measurements, the standard error on the volume of a cylinder would go: $$\sigma_V= \sqrt{ A^2\sigma_H^2+ H^2\sigma_A^2}$$... where A is the cross-sectional area: $A=\pi R^2$, $\sigma_A=2R\pi\sigma_R$ (using the rule) ... so: $$\sigma_V= \sqrt{ (\pi R^2)^2\sigma_H^2+ 2H^2R\pi\sigma_R}$$

Remembering that these are all approximations ;)
Following on from this, you should be able to work out if there is a general formula for the case $z=f(x,y)$ where x and y are statistically dependent. i.e. what if z=x+y?

10. Oct 12, 2012

### peripatein

Thak you! However, in your final evaluation of sigma_v, I believe you forgot to square 2*pi*R*sigma_R, did you not?

11. Oct 12, 2012

### Simon Bridge

I like to leave little mistakes in there to see if the person I'm talking to is paying attention and that's my excuse and I'm sticking to it. ;)

12. Oct 12, 2012

### HallsofIvy

Well, put!:tongue:

13. Oct 18, 2012

### peripatein

Hi Simon,
Do the standard deviations (in the formula for error propagation) denote the overall uncertainties with respect to the their parameters (thus including the addition in quadrature of the resolution), or rather merely the statistical error (=sigma/sqrt(N))? E.g. let us take (σ_H)^2 which appeared in the above example. Is its value equal to
sqrt{[sigma/sqrt(N)]^2 + [resolution of device used]^2}?

14. Oct 19, 2012

### Simon Bridge

Hmmm...
Would the little sigmas include uncertainties introduced by the resolution of the instrument doing the measuring?

That is a fair question: the short answer is "yes".

The wee sigmas are supposed to represent statistical uncertainties in measurement - whatever the source.
The overall uncertainty is usually the only one you can get anyway, and, even when it is possible to separate the different sources out, there is seldom any advantage in doing so.

If repeated measurement gives identical results then the resolution of the instrument dominates and some estimation must be made for that. If there is variation in repeated measurements, then the statistics will already account for the uncertainties in the resolution... and from any other sources.

Under what circumstances would you need to include both terms?

15. Oct 19, 2012

### peripatein

(1) Supposing a set of measurements of table's length is produced by using a ruler of 1mm resolution. I could obviously calculate the mean of the table's length and thence find the variance and the standard deviation. That standard deviation is NOT equal to the total uncertainty in that dimension. I could find the statistical error by dividing the standard deviation found by sqrt(N). In order to calculate the total uncertainty in centimeters let's say, I would have to add, in quadrature, 0.1^2, and then take the sqrt of that, yielding the overall uncertainty.
Now, my question was simply whether the wee sigmas in the error propagation formula represent the total uncertainty or merely the statistical error.

(2) When I tried calculating the mean and variance in the table's length using this time the formulae for continuous, uniform distribution I got different results. Why is that? They are expected to be (relatively) similar, are they not?

16. Oct 19, 2012

### Simon Bridge

If you are still unsure ... feel free to work the derivation yourself.

17. Oct 19, 2012

### peripatein

When I tried calculating the mean and variance in the table's length using, this time, the formulae for continuous, uniform distribution I got different results. Why is that? The results are expected to be (relatively) similar, are they not?

18. Oct 19, 2012

### Simon Bridge

Why would you expect different methods of estimating the uncertainty to yield similar results and what has that got to do with the formula you asked about?

If the method of estimating uncertainties is such that the uncertainty on the power of a measurement is the power times the uncertainty on the measurement, and the function can be represented by a power series, then the derivation holds and it is valid to use the formula. Otherwise it is not.

What's the problem?

19. Oct 20, 2012

### peripatein

I am not sure I follow your argument. I have tried finding the table's length using two methods. Both should be applicable, as using a ruler implies uniform distribution, does it not? The first method involved calculating the mean length from the measurements, finding the variance, and substituting it in the formula for the total uncertainty, namely sqrt((variance/N) + (ruler's resolution)^2).
The second method involved using the formulae for uniform distribution to find the mean and deviation.
The results were different. I am 100% certain the first method is correct. But why would the uniform distribution formulae not be applicable in this case?

20. Oct 20, 2012

### peripatein

Another question I have concerns the table's surface area.
If L=122.14±0.14[cm] and W=24.30±0.57[cm], I got that S=2968.00±69.70[〖cm〗^2], using ∆S=√((∂S/∂L)^2 〖∆L〗^2+(∂S/∂W)^2 〖∆W〗^2 ).
Would you kindly confirm this result? Is it plausible that ∆S would be nearly 70 cm^2??

21. Oct 20, 2012

### Simon Bridge

I'm sorry - could it be that you have another question about uncertainties that is off-topic for this thread?

Note that all measurement statistics are discrete - we use continuous stats as an approximation in special cases. If you have a new question, I'd suggest starting a new topic and posting the link to avoid cluttering this one.

22. Oct 21, 2012

### peripatein

Thank you, Simon. I shall try that.