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Converting between revolutions per minute (RPM) and revolutions per second (RPS) is not straightforward for me in Mathematica.

I know that $1\; \text{RPM} = \frac{2\; \times\; \Pi\; \text{Radian}}{\text{Second}\; \times\; 60}$. To get RPS then, it is just a matter to remove the factor of $60$ in the denominator.

Taking a numerical example, I can convert say $500$ RPM to Hertz as follows:

$[\text{Convert}[\frac{500\; \times\; 2\; \times\; \Pi\; \text{Radian}}{\text{Second}\; \times\; 60}, \text{Hertz}],9]$

which gives me $52.3597776\; \text{Hertz}$

But now if I want to convert $500$ RPM to RPS, the following does not work as in the Convert command, the second argument has also numerical values in it:

$[\text{Convert}[\frac{500\; \times\; 2\; \times\; \Pi\; \text{Radian}}{\text{Second}\; \times\; 60}, \frac{2\; \times\; \Pi\; \text{Radian}}{\text{Second}}],9]$

which gives me as output: $\frac{52.3598776\; \text{Radian}}{\text{Second}}$ instead of the desired answer of about $8.33\:$ . The factor of $2\; \Pi$ is the troublemaker.

So how to do an appropriate conversion from RPM to RPS?

update:

I am using version 9.

m_goldberg
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Chrissy
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  • Which version of Mathematica are you using? – Mr.Wizard Dec 18 '13 at 09:45
  • If You are using Mathematica 9 You can try UnitConvert[] function. I'm using ver. 7.0 so I can't check if it works in Your case. – Wojciech Dec 18 '13 at 09:48
  • @Wojciech yes, N@UnitConvert[Quantity[500,"Revolutions"/"Minutes"], "Revolutions"/"Seconds"] does give the result 8.333... – cormullion Dec 18 '13 at 11:28
  • That one was pretty straightforward, since it is in the documentation center :) – Wojciech Dec 18 '13 at 11:49
  • As does UnitConvert[Quantity[500., "RPM"], "RPS"] – bobthechemist Dec 18 '13 at 13:55
  • It works because the units are recognized, but that doesn't answer the general question of defining a custom unit – Rojo Dec 18 '13 at 14:07
  • The title is about custom defined units, but the rest of the question is rather about specific problem, especially with a question 'how to do an appropriate conversion from RPM to RPS?' at the end. I can't see how defining new units could help in converting rpm to rps. – Wojciech Dec 18 '13 at 14:51
  • @Wojciech It can. If, instead of all this mess with units, one simply writes down 1r/min /. min -> 60s /. r -> RPS*s, where r stays for "rotation", min for "minutes" and s for "seconds" one immediately gets the answer. – Alexei Boulbitch Dec 18 '13 at 15:12
  • @AlexeiBoulbitch Then the easiest way would be to create a package with custom units, where this might help http://mathematica.stackexchange.com/questions/15338/simpler-input-for-the-new-unit-support , or download AutomaticUnits package from here: http://library.wolfram.com/infocenter/MathSource/7655/ , I think rpm should be there. – Wojciech Dec 18 '13 at 15:44
  • @Wojciech I disagree. It is much shorter to write "s" then "Second". If one makes serious calculations he ends up with tens of different variables with dimensions. To write them all along with with their dimensions may, of course, be a pleasure, though a masochistic one. In a package one necessarily has to generalize on one hand, and to think about possible use of already reserved names on the other hand. The cost is a great awkwardness, like the one with complete names of units I pointed out above. – Alexei Boulbitch Dec 19 '13 at 08:51
  • @Wojciech Continuation: In contrast, in any real-science calculation one only needs a part of the full set of units, and, consequently, only a part of the full set of rules for their transformations. Indeed, one does not need any package to fast write down and fast apply a simple rule, as I did above. – Alexei Boulbitch Dec 19 '13 at 08:55

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