Critical speed of rotation - why does gravity matter?
September 7, 2019 1:28 PM   Subscribe

The formula for critical speed of rotation includes the gravitational constant. Why? Including g seems to imply the unlikely idea that a rotating shaft won't go into resonant vibration in a zero-gravity environment, which is making me doubtful.
posted by clawsoon to Technology (10 answers total) 4 users marked this as a favorite
 
Well, in pretty much every physical equation, you tend to focus on only the largest forces in the system, since there are always a ton of smaller forces that produce such a tiny difference, that it doesn't really make sense to calculate them - for example the gravitational attraction of jupiter on an Earth-bound object or the difference in air pressure between the top and bottom of a tiny, much heavier than air object.

This equation, as you note, includes [i]g[/i] because it is assuming that gravitational deflection is going to be by far the biggest thing pulling the rotor off axis. In the case of something exceedingly small or light, or even just poorly machined and naturally off balance, this won't be the case and the equation given is not going to produce accurate predictions.

The stiffness of the shaft is kind of hidden in this equation - it's implied between the force of gravity and the deflection caused by the force of gravity. In a situation where gravity mattered less, the vibrational modes involving just the restoring force and the mass of the shaft and would be the relevant one to the situation.
posted by Zalzidrax at 1:50 PM on September 7, 2019 [1 favorite]


That's not the (universal) gravitational constant G, in that equation it is the acceleration due to gravity g.
posted by Rob Rockets at 1:51 PM on September 7, 2019 [3 favorites]


Poking aroudn a bit it looks like the Dunkerley's Method page has a bit more detailed treatment of elastic vibrations.
posted by Zalzidrax at 2:02 PM on September 7, 2019


Thanks for the Dunkerley's Method pointer, which I saw but didn't bother to follow. (My mistake.) Dunkerley's Method seems like it would at least work in space, since it doesn't involve g. It also... uh... looks like the equations I'm used to, at least once you convert from rpm to rad/s, for whatever looks are worth in engineering.
posted by clawsoon at 2:11 PM on September 7, 2019


(...or, rather, once you convert from rpm to revolutions per second.)
posted by clawsoon at 2:21 PM on September 7, 2019


That's not the (universal) gravitational constant G, in that equation it is the acceleration due to gravity g.

Yeah, that's what made it confusing, since g only applies near the surface of the earth and rotational vibration seems like it would be a universal problem.
posted by clawsoon at 2:22 PM on September 7, 2019


g is only a constant by convenience for us earth dwellers for most uses. It is actually variable, both on the planet (for extremely precise uses) and of course anywhere else in the universe.
posted by rockindata at 3:16 PM on September 7, 2019 [2 favorites]


I think the (unsourced) formula for Dunkerley's method in Wikipedia might include g rolled into that mystery 94.251 constant. It's surprisingly close to a value of nearer to the equator.

I'd say that g matters because all shafts are unbalanced, and gravity-based sag is an exciter for whirling in shafts. I've been around large rotating machinery that critical speed became a surprising issue. Many of the methods of estimation are no substitute for building and testing the thing and hoping it doesn't throw a wobbly.
posted by scruss at 3:28 PM on September 7, 2019


might include g rolled into that mystery 94.251 constant.

I've found better sources (e.g. this PDF) which suggest that 94.251 has something to do with 60*pi/2, with the 60 there to convert from minutes to seconds.
posted by clawsoon at 4:13 PM on September 7, 2019


Dunkerley's original paper (PDF) is freely available.
posted by clawsoon at 4:44 PM on September 7, 2019


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