Tag Info

Hot answers tagged

44

The second and the speed of light are precisely defined, and the metre is then specified as a function of $c$ and the second. So when you experimentally measure the speed of light you are effectively measuring the length of the metre i.e. the experimental error is the error in the measurement of the metre not the error in the speed of light or the second. ...


15

To repeat Wikipedia: The speed of light in vacuum, commonly denoted c, is a universal physical constant important in many areas of physics. Its value is exactly 299,792,458 metres per second, a figure that is exact because the length of the metre is defined from this constant and the international standard for time. In other words, it's exact ...


14

Simple error analysis assumes that the error of a function $\Delta f(x)$ by a given error $\Delta x$ of the input argument is approximately $$ \Delta f(x) \approx \frac{\text{d}f(x)}{\text{d}x}\cdot\Delta x $$ The mathematical reasoning behind this is the Taylor series and the character of $\frac{\text{d}f(x)}{\text{d}x}$ describing how the function $f(x)$ ...


12

While appropriate in many important contexts, LeFitz's answer can fail in one important situation, and can lead you astray, for example, when plotting graphs in logarithmic scale. More specifically, LeFit'zs answer is only valid for situations where the error $\Delta x$ of the argument $x$ you're feeding to the logarithm is much smaller than $x$ itself: $$ ...


9

The idea is just that if the uncertainties are small enough you can approximate the function by its Taylor series $$ f(x_i + \delta_i) \approx f(x_i) + \sum_j \frac{\partial f(x_i)}{\partial x_j} \delta_j + \sum_{j,k} \frac{1}{2} \frac{\partial^2 f(x_i)}{\partial x_j \partial x_k} \delta_j \delta_k + \cdots. $$ If you neglect the second order terms the ...


9

This problem is generally called propagation of error / uncertainty. You can google it and find a lot of info (I'd also recommend Taylor's "Introduction to Error Analysis"). Here's the gist of it, though. If you have independent measured quantities $x, y, z, \ldots$ with errors $ \sigma_x, \sigma_y, \sigma_z, \ldots$, then the error on a function ...


8

Jerry Schirmer's right about why solving for $r$ first is the right procedure. One way to illustrate this is to notice that with the other procedure the uncertainty could go negative, which can't be right. But the main thing I wanted to point out is that, if the measurements of $V$ and $h$ are independent, and if the "errors" mean standard deviations as ...


8

You're confusing independent and dependent variables. When you propogate from uncertainties in the $x_{i}$ to some $f(x_{1},x_{2}...)$, the formula $\delta f(x_{1}...)=\sum \left|\frac{\partial f}{\partial x_{i}}\right|\delta x_{i}$ assumes that each of the $x_{i}$ is an independently measured variable and that $f$ is a dependent variable to be calculated ...


8

When quoting results, there are a few good rules to follow: Avoid rounding errors in intermediate calculations. Write your error to 1 significant figure if your data set is smaller than $10^2$, 2 if it's smaller than $10^4$ etc. Write your estimate and its error with the same number of decimal places. Rules 1. and 3. are simple to understand. Rule 2. ...


7

CODATA is a group that compares and combines all the most accurate experimental measurements of fundamental constants to give recommendations for best-guess values that should be used. They periodically update their values as new experiments are done. You are seeing that some wikipedia pages use old (not-updated) CODATA recommendations, while others have ...


7

The formula you've specified $$ \Delta k = \sqrt{(\Delta k_1)^2 + (\Delta k_2)^2} $$ is the formula to obtain error of quantity $k$, as being dependent on $k_1$ and $k_2$ according to the following expression $$ k = k_1 + k_2.$$ Generally, to obtain experimental error of a dependent quantity (and the expression stated in your question), you start with ...


7

Standard deviation adds uncertainties to the measured value: $23.3\pm 0.4\,{\rm m}$. One can quickly look at the error (which has units of ${\rm m}$ in my case) and think, The value could be as low as $22.9\,{\rm m}$ or as high as $23.7\,{\rm m}$ without much thinking. Modifying this to being a percentage of the value would be confusing. Plus it would be ...


7

No because none of them know the actual answer. The averaging process you describe only works if each estimate is of the exact answer plus noise. Otherwise it is known as the "Emperor's nose" problem. Nobody can see the Chinese emperor's face so they ask a million peasants how long his nose is, they average the results, and since they have such a large 'N' ...


6

You should always find an answer that is a formula, and then only apply significant figures once you get to the one final step of substituting your numbers back into the problem in place of variables. Avoid multiple intermediate steps of substituting numbers at all costs. Not only will this save your pencil a lot of work, but it will also cause your ...


6

The ultimate answer is the JCGM 100:2008 guide followed by most of the metrology institutes around the world. The specific chapter on combining uncertainties is Chapter 5. Specifically, for a two-variable function $f(t_1, t_2)$ of two random variables, Eq. (16) of Section 5.2.2. gives $$\Delta f^2= \left (\frac{\partial f}{\partial t_1} \right )^2 \Delta ...


6

It's not, at least not in the statistical sense. What you are doing is finding the (linearly approximated) change in $y$ obtained by changing the inputs by their standard deviations. This is okay as a rough approximation, but you can do better with almost no extra work. If you want the actual variance and standard deviation of $y$, the formula is different. ...


6

When you divide numbers with uncertainties, the relative uncertainties of the two numbers add in quadrature (pdf). If one of the relative uncertainties is much lower than the other, than you can ignore it. Given the wording of the problem that you quote, it appears that you can treat the radius of the rod as having a negligible uncertainty. So your reasoning ...


6

It's down the fact that different properties are used to calculate the ages and if you look at the margin of error in the calculations they're not incompatible. The lower limit for the age of HD140283 is: 14.46 - 0.8 = 13.68 billion years which is within the range for the age of the universe. Once better measurements of HD140283 are made it's age ...


6

The typical gravitational acceleration on the surface of the Earth, $g \approx 9.8\: \mathrm{m/s^2}$, has uncertainty. That's one of the reasons why the $\approx$ symbol is used. The Earth's gravitational field varies a lot due to oceans, the thickness of the crust, mountains, non-uniform density in the crust and mantel, etc. A pair of satellites was ...


6

I believe you are thinking of the Central Limit Theorem. The mean and variance of the averages of many measurements are better estimates of the precision of your measuring rule, but don't tell you anything about the accuracy of your measuring rule. Your measuring rule may be biased. The Central Limit Theorem is a part of mathematics. IMO you should also ...


5

Yes, the only sensible formula for the total error is the sum in quadrature, $$ \Delta X_{\rm total} = \sqrt { \Delta X_{\rm syst}^2 + \Delta X_{\rm stat}^2 } $$ The key assumption behind the validity of the formula is that the two sources of error are independent i.e. uncorrelated. $$ \langle \Delta X_{\rm syst} \Delta X_{\rm stat} \rangle = 0$$ Because of ...


5

I think you're exercising an incorrect picture of statistics here - mixing the inputs and outputs. You are recording the result of a measurement, and the spread of these measurement values (we'll say they're normally distributed) is theoretically a consequence of all of the variation from all different sources. That is, every time you do it, the length of ...


5

As you can read in this Wikipedia article, it was decided recently to base all SI units on seven constants of nature. To be able to do so, these constants have to be set to absolute values. Therefore it was decided, that these constants are fixed without error margin at their commonly accepted values to derive all other SI units from those now fundamental ...


5

The error would be in the order of 10^-14. This is mathematically similar to the sense of errors you have on your hand watch, caused by mechanical inaccuracy - probably in the range of 1 second per week, or 1 second per year if its a Rolex :) One should note however, that such a very small inaccuracy in time measurement in atomic clocks is perhaps less than ...


5

No, of course not. Yes, some people will overestimate and others will underestimate. Averaging would cancel out the bias to some extent, but there's no reason to expect it to cancel out the bias perfectly. We all have similar eyes and brains. We are all deceived by the same optical illusions, in the same way. We all have a shared cultural understanding of ...


4

That depends entirely on what you consider to be "expected range of values." When you see a value like $3.43\pm 0.04$ (I will omit units for brevity), in many cases, it actually represents a normal probability distribution with a mean of $3.43$ and a standard deviation of $0.04$. If the $3.43\pm 0.04$ is the result of an experiment, for example, then the ...


4

Here's the general derivation of the commonly used, and often (but not always) valid, uncertainty propagation formula for independent small Gaussian errors. $\newcommand{\bbv}[1]{\mathbf{#1}}$ Consider a quantity $y$, calculated from measured quantities $\bbv{x}$ $$ y + \Delta{y} = f(\bbv{x}+\Delta\bbv{x}) = ...


4

You're measuring the quantity $X$ and you got results $+1,0,-1$ and perhaps $+1,-1$ again. Assuming that your systematic error is zero, these numbers are randomly generated around the right value you want to know. That's why you want to estimate the right value as the average of the results you obtained. That's $$ \overline{X}= \frac{(-1)+0+(+1)}{3} = ...


4

I think the real question is actually posed most directly in your comment (so you might want to consider editing some of this into the original question): I'm more concerned about understanding whats going on and making sure that I know how to do it. I've heard that you shouldn't worry about significant figure rules until you have your final answer. How ...



Only top voted, non community-wiki answers of a minimum length are eligible