# rounding when doing calculation in physics

I came across a calculation for projectile motion in my physics book as follow:

$$r = \sqrt{x^{2} + y^{2}} = \sqrt{(4.5\,m)^{2} + (-1.2\,m)^{2}} = 4.7\,m$$

If I round the results in (1), (2) and (3), according to the significant figures:

\begin{align} (4.5)^{2} = 20.25 &\Rightarrow 20 \quad(1)\\ (-1.2)^{2} = 1.44 &\Rightarrow 1.4 \quad(2)\\ \sqrt{(20) + (1.4)} = \sqrt{(21.4)} \approx 4.6260134 &\Rightarrow 4.6 \quad(3)\\ \end{align}

If I only round in (3), I got:

$$\sqrt{(20.25) + (1.44)} = \sqrt{(21.69)} \approx 4.65725240888 \Rightarrow 4.7,$$

which is consistent with the answer of the book.

So why not just round at (3)?

I originally think that if the 'not significant figures' got multiplied together, then they might be carried to the 'significant figures'.

Why you make this guess? Isn't it a common sense?

Because I trust people here, then I found the book, which is a college-textbook 13-th edition, has another example:

$$v = \sqrt{{v_x}^{2} + {v_y}^{2}} = \sqrt{(22.2\,\frac{m}{s})^{2} + (10.0\,\frac{m}{s})^{2}} = 24.4\,\frac{m}{s}$$

If I only round at (III): \begin{align} (22.2)^{2} &= 492.84 \quad (I)\\ (10.0)^{2} &= 100.00 \quad (II)\\ \sqrt{(492.84) + (100.00)} = \sqrt{(592.84)} &\approx 24.3483059 \Rightarrow 24.3 \quad (III),\\ \end{align}

which is not consistent with the book.

Is the book wrong? If it is wrong, skip the following question.

If not, what's the actually steps that works for both?

Appreciated.

• "I think that if I round numbers in each step the error would become smaller" You expect the error to be smaller when you change the value to something else than what it was? Why would you think so?? Jan 20, 2017 at 15:44
• What does "step" and "last one" mean when you say "I only round at the last step" and "if I round only at the last one". Do you means "digit"? Jan 20, 2017 at 15:46
• This is more of a math.se question
– user140434
Jan 20, 2017 at 15:48
• @N1ng what do you mean with "last step"? As the text is now I unfortunately can't understand it. Therefore the downwards, I'm afraid Jan 20, 2017 at 15:52
• @Doc I don't think it really is? Is it? Jan 20, 2017 at 16:13

Rounding multiple times will make your answer less accurate. This is known as Round off error and should be fairly intuitive.

When you cut off a number you are changing the values in the operations. The more you cut numbers off the more different the values will be. As long as your original sources for the values are reliable, you will only be less accurate by rounding more times.

Significant figures (you mentioned in your edit you were using them) should only be applied after the math is all done. It is the result of knowing you can only be as accurate as the least accurate value you've been given. Doing this more doesn't make it more accurate. In fact signinfigant figures doesn't make the answer more accurate either, it just makes it a better reflection of the information it was based on.

• So I can make a conclusion that the book's wrong? I do follow your advice and I get $24.3$ instead of $24.4$. Jan 20, 2017 at 16:45
• This also depends on how you were taught to do significant figures. I was taught "you write down all of the digits that you're certain about, plus one more digit which is a guess." In other words when you're measuring how long something is with a ruler, you might measure it as $21.27\text{ cm}$ reflecting the fact that it got most of the way through that $2\text{ mm}$ mark but not quite all the way to the $3$: but I am also saying "it might have been 21.25 or 21.29, I don't know." If that is your standard then your "24.3 and 24.4" are both within each others' error margins. Jan 20, 2017 at 16:59