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I recently bought a new scroll saw and was commenting to someone about how it was a relatively slow saw... low ... RPMs (thinking like a circular saw). Then it occurred to me that not being a circle, the blade movement wouldn't really be measured in RPMs.

So I was trying to think what it would be measured in, and I realized it is moving up and down, oscillating, essentially like a wave... so thought maybe it would be measured in Hertz?

I looked up measures of frequency, and this article says that Hertz apply to physical waves such as sound (as well as EM waves). Technically this is not actually a sine wave, but it is essentially an oscillating movement.

So I was just wondering, would Hertz be the correct unit of measure for the movement of the blade on a scroll saw?

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Indeed the Hertz unit is the correct unit to use. Hertz is a measure of oscillatory phenomena in

$$\frac{\mathrm{cycles}}{\mathrm{second}}$$

It doesn't matter that your scroll saw doesn't trace out a perfect sine wave, it's still oscillatory and Hertz is the right unit to measure that.

The Wikipedia article spells this out is better detail and uses a heart beat as an example of non-sinusoidal oscillation that can be measured in Hertz.

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  • $\begingroup$ Technically $\mathrm{Hz}$ is $s^{-1}$ (inverse seconds) however cycles per second is the easiest way to understand the unit. $\endgroup$ Commented May 29, 2013 at 15:42
  • $\begingroup$ And the convention in physics is that $\mathrm{s}^{-1}$ is usually understood to mean radians per second rather than cycles per second. $\endgroup$
    – user10851
    Commented May 29, 2013 at 16:12
  • $\begingroup$ Hmm, technically, if Hz is simply cycles/second, then Hz could - technically - be used to describe the motion of a circular saw too, couldn't it? With the "cycle" simply being defined as one complete revolution of the saw blade? $\endgroup$
    – eidylon
    Commented May 30, 2013 at 21:17
  • $\begingroup$ @eidylon yes however $\mathrm{Hz}$ applied to rotation is usually in radians per second (typically just $\omega$). $\endgroup$ Commented May 30, 2013 at 21:24

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