# How to convert density ratio of grams/ mm^3 into metric tonnes/ m^3

RayDonaldPratt
For the dimensions of a right cylinder, I am given three significant digits for the diameter (17.4 mm) and the height (50.3 mm). The formula for the volume of a right cylinder is V = Pi x r^2 x h, which would lead here to Pi x (17.4 mm / 2)^2 x 50.3 mm = 11,960.69354 mm^3 before rounding to 3 significant digits (i.e., 12,000 mm^3).

I am also told that this right cylinder has a mass of 49 grams, and I am told that the "density" is the mass (49 g) divided by the volume (11,960.69354 mm^3), which is .004096752402 g / mm^3 (before rounding).

My first question, or request for affirmation, is that the "49" grams reduces the number of significant digits in the final result to two digits, true?

I am then asked to express (convert) this density ratio into metric tonnes per cubic meter (Tn / m^3). And, I am told that a metric ton is 1,000 kilograms, or, as it were, 1,000 x 1,000 grams. That would be a Mg (Mega-gram). However, some online calculators that I have looked at for converting grams per cubic millimeter to metric tonnes per cubic meter would simply multiply 49 grams from above by 1,000, i.e., 49,000 tonnes per cubic meter. Right or wrong, the math does not make sense to me.

I assume that 1,000 cubic millimeters is 1 cubic meter. And, raising 49 grams proportionately would give me 49,000 grams, or, 49 kilograms. So, raising 49 kilograms to a metric tonne without proportionately changing from cubic meters would move the decimal place back three spaces to .049 metric tonnes per cubic meter.

With two significant digits, the answer after rounding would be .05 metric tonnes per m^3, but that answer has been marked wrong. I will try again with the answer of .049, but if that fails (it will be my fourth attempt), I will post this question to the Internet and ask how to correctly solve this problem. I am taking a Coursera MOOC about physics solely for my personal use. (I want to actually use math and physics to solve a more complex problem that I am personally interested in solving, and I am therefore more interested in understanding all the math and physics than in getting a passing grade with missed answers.)

Where and why does my thinking go wrong?

Welcome to PF.
My first question, or request for affirmation, is that the "49" grams reduces the number of significant digits in the final result to two digits, true?
Or is it 49.00 gram accurate to 4 digits ?
I assume that 1,000 cubic millimeters is 1 cubic meter.
1000 cubic mm are one cubic cm.
1000 cubic cm are one litre.
1000 litres are 1 cubic metre.

Lnewqban
Mentor
My first question, or request for affirmation, is that the "49" grams reduces the number of significant digits in the final result to two digits, true?
Correct.

As for the conversion factor, consider the units as mathematical entities, along with the fact that multiplying by 1 doesn't alter any equality. Therefore,
\begin{align*} 1 \frac{\mathrm{g}}{\mathrm{mm}^3} &= 1 \frac{\mathrm{g}}{\mathrm{mm}^3} \times \frac{1\, \mathrm{Tn}}{1000\, \mathrm{kg}} \times \frac{1\, \mathrm{kg}}{1000\, \mathrm{g}} \times \left( \frac{1000\, \textrm{mm}}{1\, \textrm{m}} \right)^3 \\ &= 1000 \frac{\mathrm{Tn}}{\mathrm{m}^3} \end{align*}
So the conversion factor you found online is indeed correct. The mistake you made is
I assume that 1,000 cubic millimeters is 1 cubic meter.
1,000 mm is 1 m, so you get 1,000,000,000 mm3 for 1 m3.

berkeman and Lnewqban
Homework Helper
Gold Member
2022 Award

Imagine a cube 1m x 1m x 1m.

In mm, its dimensions are 1000mm x 1000mm x 1000mm. So in mm³ its volume is 1000x1000x1000 mm³ = 10⁹ mm³.

You might like to ask yourself what an area of 1m² is in units of mm².

Lnewqban
Homework Helper
Gold Member
Steve4Physics
RayDonaldPratt
Welcome to PF.

Or is it 49.00 gram accurate to 4 digits ?

1000 cubic mm are one cubic cm.
1000 cubic cm are one litre.
1000 litres are 1 cubic metre.
Thank you, it was given just as 49, so it was just 2 significant digits.
And, I was able to visualize 1,000 cubic millimeters as only creating the length (in cubic millimeters) of one edge of a cubic meter, but not the full volume of a cubic meter. Thank you.

Lnewqban
RayDonaldPratt
Correct.

As for the conversion factor, consider the units as mathematical entities, along with the fact that multiplying by 1 doesn't alter any equality. Therefore,
\begin{align*} 1 \frac{\mathrm{g}}{\mathrm{mm}^3} &= 1 \frac{\mathrm{g}}{\mathrm{mm}^3} \times \frac{1\, \mathrm{Tn}}{1000\, \mathrm{kg}} \times \frac{1\, \mathrm{kg}}{1000\, \mathrm{g}} \times \left( \frac{1000\, \textrm{mm}}{1\, \textrm{m}} \right)^3 \\ &= 1000 \frac{\mathrm{Tn}}{\mathrm{m}^3} \end{align*}
So the conversion factor you found online is indeed correct. The mistake you made is

1,000 mm is 1 m, so you get 1,000,000,000 mm3 for 1 m3.
Thank you. I tried to do what you did above, but because my assumptions were wrong, my results were wrong. When I corrected my approach to yours above, I got 4.09 x 10^-3 x 10^9 Tn in the numerator, and I got 10^6 m^3 in the denominator. The net result after rounding to two significant digits was 4.1 Tn/m^3 and it was marked correct. Thank you!

Lnewqban and DrClaude