Relative Centrifugal Force and RPM

A simple Google search revealed this:

https://engineeringunits.com/rcf-calculator-relative-centrifugal-force/https://engineeringunits.com/rcf-calculator-relative-centrifugal-force/

There are lots of calculators here that may save us all the trouble of starting from scratch.

And yes, units are important.

(yawn!)

a=rω², where ω is radians per sec

Convert RPM to radians per sec, 1 RPM = 2*pi/60 radians per second = .1047 rad/sec

Convert meters to inches, 1 meter = 39.37 in

a = r*(39.37 in/m)* (RPM²)/(.1047)²

Mmm.... a little dangerous!

The answer is technically correct BUT in practice g (acceleration due to gravity) is usually expressed as 32.2 ft/sec/sec in the imperial system, so better I think to concert meters to ft if moving a calculation from metric to the US or imperial system/s

And as an overall comment to the original post - it is absolutely important and easier and safer to convert everything to the standard units of the system being used at the start.

- so for mass its "kg" OR" lb", for Force its "lbal" OR "newton", for Time its always "sec", for angular speed its "rad/sec" and of course "g" in the relevant units is 32.2 for imperial and 9.81 for metric.

Yes, some people use "slug" for mass and "lb" for force in the imperial system, but no matter as long as the ratio of the two is g or 32.2

...it is absolutely important and easier and safer to convert everything to the standard units of the system being used at the start.

Agreed, but that's what the OP asked for. Mixing units is bad practice, you might just crash on Mars or something!

(Maybe this is homework.)

I believe your equation a = r*(39.37 in/m)* (RPM²)/(.1047)² should be

a = r*39.37 in/m *( rad/sec)² *(0.1047)² instead.

Ratch

Thanks. You are correct.

His input variables are RPM (in rev/min) and r (in meters).

Calculate a in inches/sec²

ω = RPM (2*pi *rad/min * 1 min/60 sec) = RPM * 0.1047 rad/sec

a = r*ω²

so a = r*(39.37 in/m)*(RPM*0.1047 rad/sec)² where r is in meters, RPM is revolutions/min and a is in/sec²

(0.1047)² should be in numerator.

g is gravitational force. It is the result of speed (rpm) and distance of the radius. A small radius requires a higher speed to achieve a specific g force. This can also be accomplished using a larger radius and fewer rpm (speed)

I have refreshed my knowledge from my days selling centrifuges where RCF (relative centrifugal force) was an important consideration. The relative centrifugal force (RCF) is given as a multiple of the acceleration of gravity (g). It is a unit-free value and serves to compare the separation and sedimentation performance.

The above was taken from the Hettich website

Ratch

The force on a point mass of m kg turning with radius R metres in a circular path about a point at angular velocity ω radians/second is...

mRω² Newtons [1 newton accelerates 1 kg mass at 1 metre per second per second]

Circumference of a circle is pi x diameter or 2*pi*R in terms of radius, angle per revolution = 2 x pi in radians. If you measure in RPM, you have to divide RPM by 60 to get rev/second. So the formula reduces to ...

mR*(RPM/60)² *(2*pi)² = mR*4*3.14159² * (RPM)²/3600 = mR*(RPM)²*0.0109662

However, if you want answer in kg weight per kg mass you have to divide by 9.81 to convert the Newtons (g= 9.81 m/s²), so you have 0.0109662/9.81 = 0.0011178 or 0.001118 without haggling over value of g, which varies with location & whether your brain comes up with 32.2 or 9.81 as a standard. If you want kg weight per kg mass & call it G, the m is removed & there is the OP's formula

G = 0.001118*R*(RPM)*(RPM) with R in metres.

If you want inches radius, you have to put inches/39.37 in place of R metres, which requires replacing 0.001118 by 0.001118/39.37 = 0.0000283

G = 0.0000283*r*(RPM)*(RPM) where r is radius in inches

It seems to me, a lot of replies have taken post #2 as correct, which appears to multiply by 39.37, not divide by 39.37.

Trying to get CR4 to deliver fact not error..

67model

Replying to Comment by

67model: (Use Copy & Paste to quote the original text.)

The force on a point mass of m kg turning with radius R metres in a circular path about a point at angular velocity ω radians/second is...

mRω² Newtons [1 newton accelerates 1 kg mass at 1 metre per second per second]

Circumference of a circle is pi x diameter or 2*pi*R in terms of radius, angle per revolution = 2 x pi in radians. If you measure in RPM, you have to divide RPM by 60 to get rev/second. So the formula reduces to ...

mR*(RPM/60)² *(2*pi)² = mR*4*3.14159² * (RPM)²/3600 = mR*(RPM)²*0.0109662

So far you are correct.

However, if you want answer in kg weight per kg mass you have to divide by 9.81 to convert the Newtons (g= 9.81 m/s²), so you have 0.0109662/9.81 = 0.0011178 or 0.001118 without haggling over value of g, which varies with location & whether your brain comes up with 32.2 or 9.81 as a standard. If you want kg weight per kg mass & call it G, the m is removed & there is the OP's formula

G = 0.001118*R*(RPM)*(RPM) with R in metres

The OP did not give any mass, so why should we worry about the force. Anyway, a 1 kg of weight is the force of 1 kg at Earth gravity.

f you want inches radius, you have to put inches/39.37 in place of R metres, which requires replacing 0.001118 by 0.001118/39.37 = 0.0000283

G = 0.0000283*r*(RPM)*(RPM) where r is radius in inches

It seems to me, a lot of replies have taken post #2 as correct, which appears to multiply by 39.37, not divide by 39.37.

Trying to get CR4 to deliver fact not error..

Not so, inches = meters*39.37 inches/1 meter. Therefore, when meters are in in the numerator, you have to multiply by 39.37 to get inches.

To convert inches/sec^2 to "g", we do inches/sec^2 * 1 ft/12 inches * 1 g/32.2 ft/sec^2

Ratch

Ratch,

Why worry about the force? The OP wrote about "relative centrifugal force" and post #6 shows essentially same formula with "1118" from a centrifuge company as RCF = relative centrifugal force.

I was wrong, converting acceleration m/s² to in/s² requires multiplying by 39.37. (your post #8 is also wrong, replacing RPM by rad/second & leaving RPM to rad/s multiplier in place).

My mistakes were to think the OP wanted to put inches radius into his RCF formula, which had R in metres (when what he really wanted was a formula for acceleration in inch/sec/sec - but he did ask what his RCF formula came from), & not read post #2 properly (it calculates "a" - without defining it as acceleration - & ends with a formula without writing in its units of inch/sec/sec).

I guess the correct answer to the OP is that his RCF formula is already in "customary USA units" because it is just a ratio and does not have any units until you multiply by a standardised acceleration due to gravity, which would be 32.174 feet/second/second x 12 inches/foot = 386.088 for inch units. This could be expressed as a formula...

acceleration = 0.001118 x R x (RPM)² x (0.10472)² x 386.088 = 0.0047336 x R x (RPM)² inches/second/second with R in metres.

If R were in inches, then this would avoid the many zeros.....

acceleration = R x (RPM)² /8317 in/s²

My first boss told me "Only one man never made a mistake - and he was crucified for it!"

A Happy Christmas to all.

My understanding is that accelleration in the context of RCF is a bit of a red herring. The idea of a centrifuge is to get up to speed as soon as possible and maintain that speed constantly for a period of time. The RCF then becomes a function of the diameter of the head and the constant speed.

Why worry about the force? The OP wrote about "relative centrifugal force" and post #6 shows essentially same formula with "1118" from a centrifuge company as RCF = relative centrifugal force.

What is RCF? I never heard of such a thing and no definition was given. If no mass units were specified by the OP, then how can force be calculated? Only radius and RPM were given to find in/sec^2.

I was wrong, converting acceleration m/s² to in/s² requires multiplying by 39.37. (your post #8 is also wrong, replacing RPM by rad/second & leaving RPM to rad/s multiplier in place).

I did not replace RPM by radians/sec. I was correcting someone's mistake. Read post #8 again.

I did calculate in post #11 that 1 in/sec^2 = 1/(32,2*12) = 2.59E-3 g .

Ratch

Ratch,

This is copy Rixter post #9 conclusion.....

a = r*ω²

so a = r*(39.37 in/m)*(RPM*0.1047 rad/sec)²where r is in meters, RPM is revolutions/min and a is in/sec²

That is a = r*(39.37 in/m)*(RPM)²*(0.1047)^{2 [1]}

your Post #8......

I believe your equation a = r*(39.37 in/m)* (RPM²)/(.1047)² should be

a = r*39.37 in/m *( rad/sec)² *(0.1047)² instead. [2]

Ratch

__________________
Hopelessly Pedantic

Compare [2] with [1] - you replaced RPM by rad/sec - leaving(0.1047)² in place.

Being pedantic, that makes a formula which gives the wrong answer.

67 model

67 Model,

Yes, you are correct. I miscopied the units. Sorry for the mistake.

Ratch