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I conducted an experiment and measured the values of $R$ and $H$ to calculate $v_i$.

The equation used is: $$v_i=\sqrt{{gR^2}\over{2H}}$$.

My average values ($v_i$ is calculated only for the average values over multiple trials) of $R$ and $H$ were 1.90 m 1.02 m respectively.

I estimated that my measured values were within $\pm$ 0.01 m of the true value.

Theorizing that the largest difference in result would occur when one value was on the high end and the other was on the low end, I did the following to calculate the absolute error:

$$|Error|=\sqrt{{(9.8)(1.91)^2}\over{2(1.01)}}-\sqrt{{(9.8)(1.90)^2}\over{2(1.02)}}=0.04$$

So, I would say that a measurement error 1 cm in either quantity could produce an error in the calculated value of $v_i$ of 0.04 m/s, but something seems sketchy about this methodology. Is my conclusion/reasoning correct?

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You should divide by the smaller quantity to get the 'maximum' or end of the range. – Zain Patel Jan 30 at 23:49
    
I did, but I apparently typed out the equation incorrectly. Updating question now. – HCBPshenanigans Jan 30 at 23:50
    
Updated it. Equation is now correctly typed as it was calculated. – HCBPshenanigans Jan 30 at 23:51
2  
This is a very standard error propagation problem. – Rob Jeffries Jan 31 at 0:31
up vote 1 down vote accepted

Your method is perfectly correct for estimating the error, especially at a high school or early university level. As Farcher correctly points out, it's unlikely for both errors to be at their extreme concurrently, so your error is more like a maximum bound on the error than a typical representative error.

The further you go in physics, the more you'll want to use a more rigorous statistical method of error analysis.

It's also worth noting that this method only estimates error due to measurement, when there could be other sources of error.

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A maximum bound is what I was going for, so that is good to hear. Statistical analysis of data hasn't been covered so I doubt I'm expected to use it. The given instruction was to "Estimate how errors of measurements are transformed in the calculation of $v_i$; that is, estimate (quantitatively) the error in the measurements of R and H and calculate how these errors would affect your final answer." I would say given that info, I did the correct thing, right? – HCBPshenanigans Jan 31 at 0:27

Your method assumes that when one measurement is at one extreme the other is also at an extreme. This is not likely to happen very often and therefore you have overestimated your error.

If you have two quantities $A\pm a\%$ and $B\pm b\%$ a better estimate of the percentage error in $A \times B$ or $\frac A B$ is $\sqrt {a^2+b^2}$. The percentage error in $A^2$ would be $2a$ and in $\sqrt a$ it would be $\frac a 2$

So if the percentage error in $R$ was $r\%$ and in $H$ was $h\%$ then the percentage error in your result would be $\frac 1 2 \sqrt{(2r)^2+h^2}$

This is for one set of readings.
If you take $n$ sets of readings then you percentage error will go down by $\sqrt n$. Note this gives you a diminishing return as you take more and more readings.
Take 10 readings the error goes down by about a factor 3. Take 100 readings and the error goes down by a factor 10. So by taking 10 times more readings you have only reduced the error by a factor of about 3.

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This answer seems to relate more to statistical percentages, whereas my question was about this one specific measurement set. – HCBPshenanigans Jan 31 at 0:09
    
Your have got a series of measurements and have to find the best estimate of an unknown quantity. That is what statistics do for you. @Alexander has derived an equivalent equation and we both get that the estimate of the error in your value of $v_i$ is $\pm 0.03$. – Farcher Jan 31 at 0:28

Formula for value - $$v_i=\sqrt{{g}\over{2}}RH^{-\frac{1}{2}}$$

General formula for error -

$$ \delta v_i = \sqrt{(\frac{\partial v_i}{\partial R}\delta R)^2+(\frac{\partial v_i}{\partial H}\delta H)^2} $$

First Calculate the derivatives - $$ \frac{\partial v_i}{\partial R} = \sqrt{{g}\over{2}}H^{-\frac{1}{2}} $$ $$ \frac{\partial v_i}{\partial H} = -\frac{1}{2}\sqrt{{g}\over{2}}RH^{-\frac{3}{2}} $$

Now plug the derivatives into the Error formula and simplify a little (using the relation $\delta R=\delta H=0.01$) - $$ \delta v_i =\delta R \sqrt{(\frac{\partial v_i}{\partial R})^2+(\frac{\partial v_i}{\partial H})^2} = \delta R \sqrt{{g}\over{2}}\sqrt{\frac{1}{H}+\frac{R^2}{4H^3}} $$

$$ \delta v_i = \delta R \sqrt{{g}\over{2H}}\sqrt{1+\frac{R^2}{4H^2}} =0.01\sqrt{\frac{9.8}{2\cdot1.02}}\sqrt{1+(\frac{1.9}{2\cdot 1.02})^2}=0.03$$

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Care to offer an answer with your answer? – HCBPshenanigans Jan 31 at 0:12
    
It's a straightforward calculation – Alexander Jan 31 at 0:20
    
en.wikipedia.org/wiki/Propagation_of_uncertainty#Simplification. Yes, this appears to be a simple and correct answer. +10. – Gert Jan 31 at 0:22

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