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RPM to G Unit
 

RPM to G Unit
by Ken Myers on 04/28/05 at 13:12:11

Is there a way of converting revolutions per minute to G units?


Re: RPM to G Unit
by Robert Fogt on 04/29/05 at 02:33:36

There probably is a way, though I have no idea how that would be.

g is a unit of acceleration, and RPM is a unit of angular velocity.

So there is no direct conversion, but I think its possible with additional unknowns. Hopefully another visitor can answer your question.


Re: RPM to G Unit
by Gabriel on 05/09/05 at 17:03:16

I've just made the calculation while writting this, so check it for possible errors:

Summary:
LF = 1/893.6 * N^2/r * (min^2/m)
where:
LF is the load factor (i.e. the ratio of the actual acceleration to the acceleration due to the gravity)
N is the angular speed in rev/min (RPM)
r is the radius in meters of the circle the particle is describing.
And the factor min^2/m will make the whole thing unitsless as it should (because it is a ratio of two accelerations).

Calculation:

The radial acceleration of a particle describing a circular arc is:

a=w^2*r
acceleration = angular speed squared / radius

a is acceleration in [distance/time^2]
w (omega) is angular speed in [1/time^2] units

Be careful in using consistent units.

w is 1/time, where 1 is "radians" (dimensionless angular unit), so it can be radians per seconds, per minutes, etc.
One turn = 2*pi radians, so if we call N the angular speed in [rev/time] units we have w=N*2*pi

g is the acceleration of a free falling object due to the gravity. The acceleratin measured as the number of g's that a body is subject to is ussually called load factor (ussually N, but let's let's call it LF because we already used N for the angular speed in rev/time). LF=a/g is an actual acceleration divided by a reference accelration (g), so LF is dimensionless. So we have a=LF*g.

Putting all this together we get:
LF*g=(2*pi*N)^2/r, since (2*pi)^2=39.48:
LF=(39.48*N^2/r)/g

Now all you have to do is use consistent units in N, r and g. For example, using the metric system we will take time in seconds (so N will be in [rev/s]) and the distance in meters, so the radius will be in [m] and g in [m/s^2], more precisely g=9.8 m/s^2.

Finally, if you want N in rev/min (i.e. RPM) you only have to take into account that 1 second=1/60 minutes, so g will be g=9.8 m/s^2=9.8 m/(1/60min)^2=9.8*3600 m/min^2=35280 m/min^2. So just be carful to put N in RPM and the radius in meters and you get

LF=(39.48*N^2/r)/35280 m/min^2

LF = 1/893.6 * N^2/r * (min^2/m)
where:
LF is the load factor (i.e. the ratio of the actual acceleration to the acceleration due to the gravity)
N is the angular speed in rev/min (RPM)
r is the radius in meters of the circle the point is describing.
And the factor min^2/m will make the whole thing unitsless as it should (because it is a ratio of two accelerations).


Re: RPM to G Unit
by Robert Fogt on 12/13/05 at 08:18:34

I was sent this URL here:
http://www.msu.edu/~venkata1/gforce.htm


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