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Inertia Calculations
calculations for the moment of inertia for a solid cylinder.
By Dave down under on 25 October, 2005 - 8:13 pm

I'm doing motor sizing calculations. I have two formulas for calculating the moment of inertia for a solid cylinder.

J = 1/2 * m * r^2

OR

J = pi/32 * p * l * r^4

Where

J = inertia
m = mass
p = density
l = length of cylinder

Can anyone tell me what the missing part of this equation is please? Is it for metric or imperial density?

Your first equation is correct. The second one is almost correct, but divide by 2 instead of 32. The two equations are equivalent. Either system of units can be used (metric or imperial), just be consistent.

The second formula should be diameter NOT radius.

Use diameter and it works.

By Dave Down Under on 6 November, 2005 - 8:01 pm

Thanks Phil

You're the man!!

1 out of 1 members thought this post was helpful...

Responding to Dave Downunder's Oct 25, 8:12pm query... conversion between metric & English units is often a source of error because of the Flywheel Effect is determination.

In the following formulae, please note my deviation from "SI" terminology by labeling mass as "m" while meters as "mt". I found that it eliminated much confusion among my students.

Metric Units
J=(G/g)x(D^2/4), kgf-mt^2
English Units
J=(W/g)xR^2, lbf-ft^2

Definitions:
o J Moment of Inertia "kg mt^2" or lb ft^2.
o m Mass "kg" or "lb".
o k Radius of rotation "mt" or "ft".
o D 2xR "mt" or "ft".
o G Metric weight "kgf".
o W English weight "lbf".
o g Gravitational acceleration "9.8 mt/s^2" or "32.2 ft/s^2"

Also note that a straight conversion of kg m^ to lb ft^2" will not yield the correct answer because the metric system uses the "diameter" dimension, but the English formula uses radius! However, the correct conversion is,

WR^2 = 5.933 x GD^2, and
GD^2 = 0.166 x WR2

Addressing the 'mass' determination of a solid-disk or cylinder having a radius, R, and length, L!. Solving the density equation, rho = mass / vol, for mass,

mass = m = rho x vol = rho x pi x R^2 x L / g

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[tal-2@webtv.net] (Cepsicon@aol.com)

My supplier converted GD^2 = 1.788 kgf.m^2
to WR^2 = 1527.49 lbm.in^2.

I don't understand how this was done. Also instead of "lbf" he has "lbm".

Dip

1 out of 1 members thought this post was helpful...

Dip... The value your supplier came up with is correct if he used inches instead of feet for the Radius of Gyration.

But the real error is that he ignored the true definition relating Metric Units to English Units. The definition was given in my 03-Nov-05 response to the Thread's originator!

Regards, Phil Corso

1 out of 1 members thought this post was helpful...

Hi Phil,

your post of 5th Nov 2005 has nearly sorted me out, thank you. On it's own I would probably not be worried but I am trying to relate to a set of figures from a book by K.I.T.Richardson called 'The Gyroscope Applied' and convert them to metric.

He uses an example to which I have added my conversions:
Disc radius 2 ft (0.61 mt)
Disc weight 50 lb (22.68 kg or not - see below)
Couple applied at right angles to plane of spin 200 lb.ft (27.65 kg.mt)

Inertia = 0.5 x 50 x 2^2 = 100 lb.ft^2 (0.5 x 22.68 x 0.61^2 = 4.22 kg.mt^2 or not - see below)
From your post I assume the disc 'weight' should in fact be mass and the disc weight is actually 50 x 32.2 = 1610 lbf so the metric conversion would be to 728.5 kgf making the metric mass 728.5 / 9.81 = 74.26 kg. This being the case the inertia would be 0.5 x 74.26 x 0.61^2 = 13.82 kg.mt^2

Mr Richardson then goes on:
K (torque moment) = 200 x 32 = 6400 pdls.ft (27.65 x 9.81 = 271 N.mt)
Nb: here he uses 32 instead of 32.2 which is very confusing considering the speed he chose!

Precession rate = K /Inertia/ang vel = 6400 / 100 / 32 = 2 rad./sec
Precession rate = K /Inertia/ang vel = 271 / 4.22 / 32 = 2 rad./sec
Not:
Precession rate = K /Inertia/ang vel = 271 / 13.82 / 32 = 0.61 rad./sec

This seems to have worked out using the weight instead of the mass, so where is the misunderstanding?

Robin

1 out of 1 members thought this post was helpful...

Robin... can you scan and forward the pertinent pages of Mr. Richardson's book to me?

Regards, Phil Corso (Cepsicon [at] aol [dot] Com

Hi Phil, thanks for your reply. A scan is on it's way.

Robin

By Robin Child on 22 April, 2010 - 2:34 pm
1 out of 1 members thought this post was helpful...

Hi Phil,

With many thanks for all the work you put in on this here is the example corrected for 2 mistakes made by the author which canceled out, and with the correct unit conversions:

Disc radius 2 ft (0.61 mt)

Disc weight 50 lb (this is weight so should be 50 lbf so does convert to 22.68 kgf)

Angular velocity 32 rad./sec (not to be confused with g which is 32.17 ft/sec^2 - see later)

Couple applied at right angles to plane of spin 200 lb.ft (also should be lbf.ft so is 27.65 kgf.mt)

Inertia = 0.5 x 50 x 2^2 = 100 lb.ft^2 - here was mistake 1 as the 50lbf weight should be a mass of 50/g = 1.5542 lb(m) or slugs, so should be:

Inertia = 0.5 x 50/g x 2^2 = 3.1084 lb.ft^2 where g is 32.17 ft/sec^2 (which converts to 0.5 x 22.68/9.8067 x 0.61^2 = 0.4297 kg.mt^2)

K (torque moment) = 200 x 32.17 = 6400 pdls.ft - this is mistake 2 which cancels out mistake 1! 200 lbf.ft as it should be written is already in force(or weight) units and so does not need to be converted to poundals which is also a force unit.

To be consistent the torque moment must be left in kgf.mt units and not converted to N.mt
Hence:

Precession rate (Imp) = K /Inertia/ang vel = 200 / 3.1084 / 32 = 2.01 rad./sec

Precession rate (Metric) = K /Inertia/ang vel = 27.65 / 0.4297/ 32 = 2.01 rad./sec

So there we are, simple when you know how, and I hope you all do too!

Robin

Note: Phil put in some serious time on this question which resolved a lot of doubts I had. He is one of those guys who take pleasure from helping people, and that makes me feel just great.

By S Sharath on 20 May, 2015 - 11:13 am

Hi,

I tried to understand your comparison by running some numbers in both unit system. Can you help me understand where I am going woring
Assume :
G = 100 kgs (W=220.462 lbs)
D = 4 mt (R= 13.123/2= 5.6 ft)
g = 9.8 mt/sec^2 = 32.2 ft/sec^2

plugging in the values , J (in metric) = (100/9.8)*((4*4)/4)=40.81 kgf-mt^2

Therefore, J (in English Units) = (220.462/32.2)*5.6^2 = 217.71 lbf-ft^2

Your thread also says , GD^2 = 0.166 WR^2 and WR^2 = 5.933 GD^2
From above values , GD^2=(100*4^2)=1600 and WR^2 = 220.46*5.6^2

From the arrived Values GD^2 is not equal to 0.166 WR^2 and neither is WR^2 is not equal to 5.933 GD^2.

So I feel I am missing something and overseeing some conversion somewhere. help me understand this .

Thanks

1 out of 1 members thought this post was helpful...

S. Sharath,

I believe you have overlooked the difference in the definition of flywheel-effect:

By definition, the German term, (GD²/4) is equal to the English term, (Wk²)! Note the parentheses! That is to indicate the terms are Symbolic, and should not be taken literally, because, unfortunately, they omit the "g" terms.

Starting with a simple units-conversion, 1.0kg(f)-mt²) = (2.2lb(f))(3.28ft)² = 23.7 lb(f)-ft²! And, as pointed out in earlier posts to this thread, this factor was erroneously used as THE conversion-factor. Instead the actual conversion- factor should have been 23.7/4 or 5.9.

In conclusion, the terms (GD²/4) and (WR²) are just symbolic. Always use the equation with g terms included. Meaning, GD²/4g = 5.9*WR²/g', where g represents metric units, and g', English units.

Regards,
Phil Corso

Hello Phil,

Thanks for responding back. Based on values expressed in my previous post

GD²/4 = 100*4²/4= 400 kg-mt²
WR²= 220.462*5.65² = 6913 lbs-ft²
and GD²/4g=400*(1/9.8)= 40.81
5.9*WR²/g'=5.9*6913*(1/32.2) = 1,266

GD²/4g = 5.9*WR²/g'

but from above answers, they don't look to be equal

WR² = 5.933 x GD², and
GD² = 0.166 x WR²

if GD²/4 is in kgf-mt² and 1 kgf-mt²=23.7 lbs-ft², shouldn't it be 5.933*GD²=WR²..or something of that sort?

So confusing to understand

-S

1 out of 1 members thought this post was helpful...

S. Sharath

Given the Moment-of-Inertia, Jm (in metric-units), then the Moment-of-Inertia, Je (in English units) is 5.93 times greater.

Example:
Given Jm = 10,000 kg(f) - mt^2 units, then Je = 5.93xJm or 59,300 lb(f) - ft^2 units.

Phil

By anonymous on 1 June, 2015 - 10:22 am

Hi.

Does (f) stands for "force" ? Then it shouldn't be there.

Regards,

Jacek Dobrowolski

> Example:
> Given Jm = 10,000 kg(f) - mt^2 units, then Je = 5.93xJm or 59,300 lb(f) - ft^2 units.

1 out of 1 members thought this post was helpful...

Jacek, you are absolutely correct!

However, I have always advised using the 'f', which is based on the German DIN system, to help remember that the value is based on weight and not mass!

In a related subject I also advise the use of the term 'mt' for meters, ad not 'm'!

Regards,
Phil

Surely inertia (whether rotating or straight-line) has nothing to do with "weight" (which is actually a force) - it is solely a function of mass. I'm sure the guys in the International Space Station have to take that into account!

Bruce...

Surely you remember that "Weight" divided by "Gravity" equals "Mass"!

Phil

Of course - but "weight" = "force exerted on a mass due to gravity", and both weight and "g" are variables, whereas mass is constant. Any calcs including a constant g=9.81m/s^2 but with the "weight" determined with a local spring balance, for example, will be invalid for anywhere except at the earth's surface - and even in that case can be up to 1 % out.

If the effect involved is dependent on the local force due to gravity, it's OK to use gravity-based units, but if not they can lead to a lot of confusion and significant errors.

By Anonymous on 3 June, 2015 - 3:33 am

Off topic:
This brings to my mind my discussion with some ... engineer.

The topic was vacuum pumps. Their name plates were stating something like -0.2 bar and second one 0.6 bar. According to that guy, the first one was "creating" better vacuum. Or maybe more vacuum in vacuum?

And why did I mention it? Well, mathematical description of things requires models. They are build one on top of the other from very simple (like F=m*g) to more complicated (which is in fact usually the order the humanity discovered them). The trick is to know why they are so simplified and can still work while the reality is much more abstruse.

Regards,
Jacek Dobrowolski

Hi Friends,

Can we calculate thickness of tube if following parameter is provided-

Outer Diameter of Tube- 159 mm
Tube Length - 4231 mm
Continuous Torque @2.0 SF- 621 N-m
Peak Overload Torque - 1864 N-m
Tube Assembly - Composite Material

Thanks & Regards,
Prashant

Hi Phil, I hope you read this

I read a lot on the subject and I still have not reached a conclusion, maybe you can help me:

I use software for motor starting calculations in which I am asked to enter the following data:

"Enter the rated speed in revolutions per minute (RPM) and WR2 in either lb-ft2 for English unit system or in Kg-m2 for Metric unit system or H in MW-sec/MVA for the Motor, Coupling Gear, and Load. ETAP calculates WR2 or H when one of them is known and RPM has been entered based on the following equation:

`H = 5.48 * 10-9 * WR2 * RPM2 / MVA   (for WR2 = Moment of inertia in kg-m2)`

So... my doubt is whether I should insert GD^2 or GD^2/4 as WR^2

I am using metric units, and I only have a "box" to enter a value for WR^2

What do you think?

Best regards,
Juan