I'm writing a script that converts from a customer's preferred unit of $\frac{RPM}{s}$ and a device's input/output of $\frac{s}{100Hz}$ My Logic is as follows:

Converting $\frac{RPM}{s}$ to $\frac{s}{100Hz}$:

$$\frac{RPM}{s}=\frac{Revolutions}{(s)(min)}$$ $$\left(\frac{Revolutions}{(s)(min)}\right)\left(\frac{1 \;min}{60 \;s}\right)=\frac{Revolutions}{s^2}=\frac{Hz}{s}$$ $$\left(\frac{Hz}{s}\right)^{-1}=\frac{s}{Hz}$$ $$\left(1\frac{s}{Hz}\right)\left(\frac{100\;[1Hz]}{1\;[100Hz]}\right)=100\frac{s}{100Hz}$$


$$\left(\frac{RPM}{s}\right)^{-1}\left(\frac{60 \;s}{1 \;min}\right)\left(\frac{100\;[1Hz]}{1\;[100Hz]}\right)=\frac{s}{100Hz}$$

Therefore, I understand that my inverse function would also be: $$\left(\frac{\;s}{100Hz}\right)^{-1}\left(\frac{100\;[1Hz]}{1\;[100Hz]}\right)\left(\frac{60 \;s}{1 \;min}\right)=\frac{RPM}{s}$$

To me this FEELS wrong as, per these equations

$1\frac{RPM}{s} = 6000\frac{s}{100Hz}$


$1\frac{s}{100Hz} = 6000\frac{RPM}{s}$

Meaning I just need one equation to covert betweeen the two units. Am I missing something or do I really need to only do half the work?

  • $\begingroup$ can you say more about this device with the unusual unit? $\endgroup$
    – agentp
    Feb 26, 2018 at 21:49

1 Answer 1


$\frac{s}{100Hz}$ has basic unit dimensions of $[time]^2$, and as such is not a unit of angular acceleration. You cannot convert directly between these two quantities.

First, we can convert from $rpm/s$ into $Hz/s$. Using the relationship that $$60rpm=2\pi\ rad/s=1Hz$$ we can determine that $1rpm=\frac{1}{60}Hz$, and thus,

$$1\frac{rpm}{s}=\frac{1}{6000} \ \frac{100Hz}{s}$$

Therefore, to convert to from your customer's units to the input, you would use:

$$\left[input\ to\ device\ \frac{s}{100Hz}\right]=\frac{1}{6000\left[customer\ number\ in\ \frac{rpm}{s}\right]}$$

And to convert to your customers units from the output, you would use:

$$\left[customer \ number\ out\ \frac{rpm}{s}\right]=\frac{1}{\frac{1}{6000}\left[output\ from\ device\ \frac{s}{100Hz}\right]}$$

EDIT: see comments below for details - Removed rogue $\pi$'s.

  • $\begingroup$ Shouldn't $\frac{πHz}{30s}=\frac{π[100Hz]}{3000s}$? $\endgroup$
    – GentlemanS
    Feb 26, 2018 at 17:09
  • $\begingroup$ Also, isn't $\frac{1 Revolution}{minute} = \frac{1}{60}$ Hertz? Out of curiosity: Why is $π$ involved for the derivative of this equivalency? $\endgroup$
    – GentlemanS
    Feb 26, 2018 at 17:16
  • $\begingroup$ You're quite right, I've slipped a zero there! - Edited accordingly to correct. Re: your second comment - Acceleration is in units of $[distance][time]^2$, whether it is angular or not. In this case, the distance traveled per revolution (in radians) is $2\pi^c$. $Hz$, by definition are simply $\frac{1}{[time]}$, and as such the distance must be taken into account. $\endgroup$ Feb 26, 2018 at 18:00
  • $\begingroup$ I also don't get how pi shows up. There are no distances here. We are converting one angular rate to another. ( if you want a distance/time^2 measure you'd need a radius in there too no? ) $\endgroup$
    – agentp
    Feb 26, 2018 at 21:58
  • $\begingroup$ You must agree that $Hz/s$ has units of $[time]^-2$, and that $rpm/s$ has units of $[distance][time]^-2$. You might also agree that the radius of the shaft etc. would have no bearing on it's angular acceleration. It would affect the linear acceleration of a point on the surface of the shaft (what you're thinking of, @agentp), but none on the angular. Angular acceleration has SI units of $rad/s^2$, and the 'pulses' of the $Hz$ are being counted once per revolution. The distance travelled in one revolution, is $2\pi\ rad$. Don't just take my word for it: Wolfram|Alpha $\endgroup$ Feb 26, 2018 at 22:55

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