Often I am not sure when to make use of significant figures in a calculation.

For example at the moment. I have to calculate a certain equilibrium point which is given by:

$$m^* = \frac{\phi_{\mathrm o}^{*2} A}{g K^2 A_{\mathrm v}^{*2}}$$

And the values are: $\phi_{\mathrm o}^* = 0.001$, $A_{\mathrm v}^* = 0.001$, $\rho = 980$, $A = 1$, $K = 0.01$, $g = 9.81$. Now $m^* = 1.019367991845056e+03$

When applying the rules of significant figures the correct answer should be $m^* = 1e+03$, I believe. Since the amount of significant figures of the answer is equal to the number with the least amount of significant figures, which are $\phi_{\mathrm o}^*$, $A_{\mathrm{v}}^*$, $A$ and $K$, they all have 1 significant figure.

I am really questioning myself whether if I should use significant figures or not.

I hope some people can give some idea's/views when to use significant figures or not to use them.

Because $m^*$ is a equilibrium point I would say I would rather use $m^* \approx 1019.37$ rather then $1000$. But still one could argue also that I might use more digits, 1019.368.

  • $\begingroup$ Are the values of $\phi_{\mathrm o}^*$, $A_{\mathrm{v}}^*$, $A$ and $K$ measurements or are they exact values? $\endgroup$
    – Code-Guru
    Dec 14 '12 at 2:40
  • $\begingroup$ WG- apologies: I did not read your post carefully enough! $\endgroup$
    – amWhy
    Dec 14 '12 at 2:43

Are the values of $\phi_{\mathrm o}^*$, $A_{\mathrm{v}}^*$, $A$ and $K$ measurements or are they exact values? If the later, then I would claim that you shouldn't use them in determining the number of significant figures. This is because you can consider exact values as having infinite significant figures (i.e. you can add as many zeros as you wish after the decimal place) and therefore will never be chosen as having the minimum number of significant figures.

  • $\begingroup$ Yes they are stated as exact numbers, also $\rho$. Thus I should use the amount of significant digits of $g = 9.81$ which is a approximate of the earths gravitational acceleration. But still (maybe I should make a new question for this) often you have a result say $x * y = z \approx 1.232$. Now how do you not confuse the reader in thinking that a computation you do later with $z$ is not done with the approximate but with the actual value, $x * y$. You can ofcourse write down $x * y$ instead of $z$ but for lengthy equations this could get very messy. $\endgroup$
    – WG-
    Dec 14 '12 at 2:56
  • $\begingroup$ @WG- The physic textbook which I used in college did not seem to follow follow the rules of significant figures in the answers in the back of the book. In at least one of my classes, the professor required that we give all answers to 3 significant figures. I found that I got the "correct" answers by using intermediate values with as many digits as my scientific calculator held and only rounding the answer to the correct number of sig figs. $\endgroup$
    – Code-Guru
    Dec 14 '12 at 3:32
  • $\begingroup$ @WG- Also in your example, if $x$ and $y$ are exact values, then $z$ can be assumed to be exact as well. $\endgroup$
    – Code-Guru
    Dec 14 '12 at 3:35
  • $\begingroup$ Another small question, because we are working with 3 significant digits. Can one state that $m^* = 1.02e+3$ or that $m^* \approx 1.02e+3$? $\endgroup$
    – WG-
    Dec 14 '12 at 7:59
  • 1
    $\begingroup$ @WG- Technically, the later is correct. However, in a physics class $=$ is almost always understood to mean $\approx$. $\endgroup$
    – Code-Guru
    Dec 15 '12 at 19:04

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