Author Topic: Flywheel calculations  (Read 5368 times)

Offline Jasonb

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Re: Flywheel calculations
« Reply #15 on: March 02, 2023, 04:48:32 PM »
I suppose it stops anyone who gets too close getting a whack from the crank :-[ Though may have been to reduce windage?

Always good on a model as they add a bit more flywheel effect for smooth running and if one or both sides are recessed to reduce material or add a counterbalance that adds a bit of eye candy to the model  8)


Offline internal_fire

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Re: Flywheel calculations
« Reply #16 on: March 02, 2023, 06:52:06 PM »
It’s also interesting for another reason.  As you have pointed out, the moment of inertia for spinning around the axis of the flywheel is the largest, thus is the one we want.  Note also that the other two principle axis nearly, but not quite, add up to the moment of the principle axis perpendicular to the plane of the flywheel.  The theory as outlined in the maths texts I am familiar with say this should be exact.

Both the model and the moment of inertia calculation are digitized at some level of coarseness. They are not purely analytic functions. Therefore the "theory" will not match the calculations in most cases.

Gene

Offline MJM460

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Re: Flywheel calculations
« Reply #17 on: March 03, 2023, 07:55:19 AM »
Hi Willy, good to have you on board.  I suspect by the look of the flywheel on the photo that you posted that the designer or builder was not thinking about the added moment of inertia at all, as it’s contribution would be very small compared with the the flywheel.  I suggest the motivation was more about the aesthetics and perhaps even ease of manufacture.  Being a full disk they were not even thinking about balance, which is the consideration driving most variations on the crank design.  Jason’s example pays some attention to balance as well as presenting a very elegant solution, but a bit harder to machine.

Hi Gene, the modern version of “slide rule error”.  Certainly it is possible that that could be contributing to part of what I am seeing, but I think that most of it is due to the program doing a three dimensional analysis.  The program is completely open to modelling three dimensional shapes so there would be no reason to make special provision for items that are approximately two dimensional, a simplification that is really useful for a pure manual calculation.

The three dimensional aspect of the flywheel means that the moment of inertia about the  Y-Y and Z-Z axes are both a little larger than they would be for a pure two dimensional object.  Thus the sum of those two ends up being a little greater than the moment about the X-X axis, instead being equal by the maths of a two dimensional form.  The X-X axis value is not affected by the issue, and is in fact the one we are normally most interested in.  I can come back to that further on in the thread if anyone is interested.

In the mean time Jason’s two examples show very clearly the effect of a relatively small change in diameter.  Thanks for putting in the Stewart design for comparison, Jason.

I have already mentioned kinetic energy, and momentum.  I suspect that everyone has heard of kinetic energy, and understands that kinetic energy is energy of an object in motion.   Kinetic energy is proportional to mass and the square of velocity.

I think that everyone also has an idea of momentum, that tendency of a moving object to keep moving unless you apply sufficient force to stop it, or even to make it move faster.

Our intuitive understanding of these concepts is generally related to linear motion.  We learn in Primary school about Newton’s first law, the one about a body continuing in uniform motion in a straight line unless acted on by an external force.  At least we did when I was in Primary school, though that was a long time ago.   The property that determines that tendency is called momentum.

Linear Momentum is dependent on the mass and the velocity, and can be quantified by multiplying the mass and velocity.

When it comes to rotary motion, we also have kinetic energy and angular momentum. 
Kinetic energy of rotation involves moment of inertia and angular velocity.   To quantify kinetic energy, we have the the same units for both linear and rotational motion, and the quantities can be added to get the total kinetic energy for the two types of motion.

You might at this stage correctly guess that angular momentum is quantified by multiplying the moment of inertia by the angular velocity.  The units for angular momentum are not the same as the units for linear momentum and you can’t add linear momentum and angular momentum, they are separate quantities that can exist at the same time.

We all have heard of the law of conservation of energy.  So far as we know, providing we allow for Einstein’s relativistic effects, it is a universal law.

Less well known are the law of conservation of momentum and the law of conservation of angular momentum which are another two separate universal laws.  To change the linear momentum of an object it requires an force.  Similarly to change the angular momentum of a rotating object, it requires a torque

In the context of this thread, the flywheel stores kinetic energy in its rotation.  The torque  generated by engine accelerates the flywheel while the torque needed to turn the engine through the parts of the revolution where the torque generated is negative slows the flywheel.  Clearly the increases and decreases in rotational speed due to those torques involve a change in angular momentum, consistent with Newtons law.

You can see that the moment of inertia comes into play in all the explanations of how flywheels work.

Tomorrow we can look at the formulae for the the moment of inertia for typical shapes that make up flywheels.

Thanks for looking in,

MJM460
 

The more I learn, the more I find that I still have to learn!

Offline Don1966

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Re: Flywheel calculations
« Reply #18 on: March 03, 2023, 05:55:42 PM »
Love this thanks for the explanations it’s a very interesting subject..

Regards Don

Offline vtsteam

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Re: Flywheel calculations
« Reply #19 on: March 03, 2023, 10:37:44 PM »
Your explanations are very clear. :cheers:
Steve

Offline MJM460

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Re: Flywheel calculations
« Reply #20 on: March 04, 2023, 11:11:47 AM »
Thanks Steve and Don.  The physics associated with flywheels and momentum gives rise to some really interesting results.  We all know a little about gyroscopes, and the same physics provides the mathematical model for understanding their behaviour.

We can probably pause the theory at this stage and come back to it later.  But first let’s do the calculation for the moment of inertia for a flywheel.
I have already talked about the definition of Moment of inertia, and the definition of the moment of inertia of a small particle. And how the moment of inertia for a more complex object is the sum of the moment of inertia of all the particles making up the object, a problem solved in calculus by integration.  We will skip the maths and look at the results for some elements which make up a flywheel.

Then calculation of Moment of Inertia for a complex object involves dividing the object into simple shapes for which a formula is available, then adding the calculated Inertia for each shape.

For a flywheel, a simple disc is a good starting point.

For a disc of radius r, and mass m the formula for I is -
   I = 1/2 x m x r^2

We should talk about units of measurement.  The formula applies for any units you want to use, so long as you are only comparing different flywheels.  The ISO metric system units would be kilogram for mass and metre for radius.  Thus the units for second moment of inertia would be kg.m^2.  The “.” Between kg and m implies multiplication, and the units would be spoken as “kilogram metre squared”.   This results in a quite reasonable numbers for a full size engine, but results in very small numbers for most of our models.  Thus I tend to use gram millimetres squared as so far I have made only quite small flywheels.  However I think it will be less confusing to use kg/mm^2 so that the figures are comparable with the ones Jason is using in Alibre.

Similarly if you prefer the fps system, the units would be lb.ft^2, though again for small flywheels, you might use ounce.inch^2.

Some simple beginners models are designed with a simple cylindrical flywheel, but more commonly, there is a rim, spokes and a hub.  No problem, and really only one more formula to use.

The attached photo shows four flywheels, three I have made from solid and a cast one I have not yet machined.  (Waiting for me to get on with another engine.). They will do as examples for easy calculation.

While the basic formula is based on the addition of the contribution of each little part of the object, you can just as easily subtract the contribution of the parts that are not there.  So for the rim of a flywheel, calculate the Inertia of a disc equal to the o.d. of the rim, then calculate the inertia of a disc the diameter of the inside of the rim and subtract it.  The difference is the moment of inertia of the rim.  Then add the moment of inertia of the spokes.  For my simple flywheels, with a plane web instead of spokes, remove a disk only of the thickness actually removed.

Similarly, we can approximate the shape of the hub to a small diameter cylinder, calculate that and add it to the other components.  You might be surprised at how little the hub contributes compared with the rim.

I hope it is obvious that in each of these formulae, the mass intended is the mass of that component, not the total mass of the flywheel.  Hence we also need to calculate the mass of the specific component being added or removed.

Next time, the calculation for the spokes.  Then we will look at the numbers for these small flywheels, and how much each part contributes to the whole.

Thanks to all who are looking in,

MJM460
The more I learn, the more I find that I still have to learn!

Offline TerryWerm

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Re: Flywheel calculations
« Reply #21 on: March 04, 2023, 08:46:21 PM »
Very interesting subject, I am following along as well! 
----------------------------
Terry
Making chips when I can!

Offline MJM460

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Re: Flywheel calculations
« Reply #22 on: March 05, 2023, 10:13:21 AM »
Welcome aboard, Terry.

Yesterdays post was definitely a bit long, but I hope it was clear enough.  It seemed difficult to split.

You can see how the single formula for a circular disk is able to cope with most elements of a flywheel.  An additional formula for spokes will just about complete what is required.

All those early texts on calculus include a formula for a rod spinning about an axis perpendicular to its length.  That definitely seems useful for spokes. However, the formula is for an object spinning about that principle axis, while a spoke in a flywheel is spinning/rotating about the centre of the flywheel which is at the centre of the hub. 

Fortunately those same text books also have a theorem for calculating the moment of inertia about an axis that is parallel to the principle axis.  In words, as it applies to a flywheel spoke, the moment of inertia of the spoke about the centre of the flywheel is equal to the moment of inertia about its own centre line plus the moment of inertia of a point mass (equal to the mass of the spoke and located at the centre of the rod) about the centre of the flywheel.

As a formula,

I = 1/12 x m x L^2 + m x d^2

Where m is the mass of the spoke, L is the length of the spoke and d is the distance of the centre line of the spoke from the centre line of the flywheel.

I hope that all makes sense.  Obviously d is half the length of the spoke plus half the diameter of the hub.

Once we have the moment of inertia of one spoke, we can multiply by the number of spokes and add the total to the other components of the flywheel.

Curved spokes are very little different.  Imagine a rod sliced into a stack of biscuits. Which are then pushed into the curved shape.  The developed length and hence the mass of the spoke is a little larger than the straight one, but the centre line is the same distance from the centre of the flywheel and the moment of inertia about its centre line is the same as for the straight rod.  The net effect is the same as a straight spoke of larger cross section, so mass of the developed length but distributed over the same length as the radial spoke.

That’s just about all we need to make a sufficiently accurate estimate of the moment of inertia of any flywheel that interests us.  Next time I will do the calculations for those little flywheels in my collection.  It will help put everything in proportion.

Thanks to everyone for looking in,

MJM460

The more I learn, the more I find that I still have to learn!

Offline vtsteam

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Re: Flywheel calculations
« Reply #23 on: March 05, 2023, 03:12:37 PM »
Thanks again.  :popcorn: :cheers:
Steve

Offline Kim

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Re: Flywheel calculations
« Reply #24 on: March 05, 2023, 04:13:58 PM »
Curved spokes are very little different.  Imagine a rod sliced into a stack of biscuits. Which are then pushed into the curved shape.  The developed length and hence the mass of the spoke is a little larger than the straight one, but the centre line is the same distance from the centre of the flywheel and the moment of inertia about its centre line is the same as for the straight rod.  The net effect is the same as a straight spoke of larger cross section, so mass of the developed length but distributed over the same length as the radial spoke.

I don't see how slicing a cylinder into biscuits and distributing them radially around the flywheel would change the mass at all...  Seems like the mass of the curved spoke would be the same as that of a similar diameter straight spoke, and therefore, the moment of inertia would be the same.  What am I missing?  Wy would a curved spoke be larger?

Hmm....

After re-reading my own question, then re-reading your explanation, I think I see now.  If the spoke really was made up of only the biscuits cut from the initial spoke, its mass would be identical to the straight spoke. But once you fill in the stairsteps between the biscuits, it gets more mass.  I did the thought experiment of pushing this to the extreme, where each biscuit offset enough that you made a complete spiral around the flywheel.  You couldn't maintain the same diameter as that single spoke without increasing the material used, and thus, the mass.  And, as you said, MJM, the length of the spoke is CLEARLY much longer than that of the straight spoke.

This thought experiment really helped me follow.

Thanks for this, MJM,  very interesting!
Kim

Offline Jasonb

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Re: Flywheel calculations
« Reply #25 on: March 05, 2023, 04:46:23 PM »
Simple way to look at it is that a curved spoke will need to be longer (when straightened out) than a straight spoke to get from hub to rim so if the same cross section the curved one will weigh more

Offline Kim

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Re: Flywheel calculations
« Reply #26 on: March 05, 2023, 05:00:15 PM »
Yes, that makes sense, Jason.  I just had to make it a really long spiral in my mind before it became clear.  ;)

Kim

Offline jcge

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Re: Flywheel calculations
« Reply #27 on: March 05, 2023, 10:18:39 PM »
Great explanation of the serpentine spoke inertia calc !!

Offline vtsteam

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Re: Flywheel calculations
« Reply #28 on: March 05, 2023, 10:35:12 PM »
Looking at it from another perspective, even if the spoke was circular in cross section, once bent into a curve, a section through it in the radial (to the hub) direction will now be an ellipse, with more area than the original circular section.

Likewise any other spoke shape will be similarly elongated as a sliced section.

If a section's area is multiplied by the biscuit height, you will get the approximate volume of the biscuit, (though with some error). The smaller and more numerous the biscuits cut, the closer the approximation.

Steve

Offline MJM460

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Re: Flywheel calculations
« Reply #29 on: March 06, 2023, 04:58:00 AM »
I knew as I wrote it that I was getting into trouble on that one, and more words only made it worse.  But you have so eloquently sorted it out for me.  Kim your first inclination was correct, spreading out the biscuits would not change the moment of inertia at all, but filling in the steps adds more mass as you have concluded.  As Steve says the biscuits will be filled out to an elliptical form.  Of course, the actual straight spokes may not be circular, and the bent spokes may not be circular if they were sliced and restocked to be straight, but I hope you can now see the principle, and why I suggest that with a “modified” mass per unit length, they can be can be calculated as an equivalent straight spoke, with no complex formula required.  It is only necessary that each element of the mass is the same radial distance from the centreline of the flywheel as in the curved design for the calculated inertia to be correct.

As we move on, I hope to show that for practical purposes, any difference is not important, but I find it is helpful to understand what approximations, if any, have been made in a calculation in case I come across something unusual that makes those approximations less accurate.

So thanks to Kim, Steve and Jason for helping sorting that out.  Much appreciated.

And welcome aboard, jcge.  Good to have you along.

So let’s move on to put some numbers against those flywheels in my earlier photo.  The largest one is only 75 mm in diameter so will be interesting beside Jason’s larger ones.

I set up a little spread sheet with a row for the dimensions, then put the mass calculations and moment of inertia calculations in some rows in the space below, thus making a rectangular array.  After I had checked the calculations, I did a copy and paste to additional areas below to accomodate additional flywheels by simply changing the dimensions.  I used a density of 7840 kg/m^3 for steel but divided by 10E9 (or 10^9) to get kg/mm^3 for comparison with Jason’s figures.

I calculated 365.7 kg/mm^3.  Dramatically smaller than the size range in Jason’s examples, but more than adequate for my little models which are only 12 mm bore.  That one lives on the mill engine which has 19 mm stroke.  It’s the one I used in my digital Governor thread, and it ran quite nicely even at low speed.  Probably still much larger than needed.

As a rationality check for accuracy I put the flywheel on the digital kitchen scales and got 445 gm compared with the calculated 444 gm.  Much better than I expected, must be a few compensating errors in my measurements and assumed density, but close enough.  As the mass calculation contains all the physical dimensions of the flywheel used in the inertia calculation, including the material density, calculating the mass accurately is a big step towards having the inertia calculation correct.

I have no way of directly measuring the moment of inertia.  There are probably experiments that could be carried out for someone really keen, but I am confident of the formula used, so I settled for carefully checking the spreadsheet maths, an important step that is in my experience, sometimes overlooked.

Next time I will show some of the results and the interesting features of the results of those calculation, before looking at the difference spokes would have made, and moving on to the smaller one.

Thanks to everyone just looking in, I hope you are enjoying the ride.

MJM460
The more I learn, the more I find that I still have to learn!

 

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