Flywheel calculations

[MJM460]

At last, the promised thread on flywheels.  A little delayed by all the normal distractions of the summer holiday period, plus severe falls by two of the older relatives which led to additional travels.  Both on the long road to recovery now.

A flywheel is a large component that dominates the appearance of many of the engines we build.  Thus for a scale model of a full size prototype, the proportions of the model should be to scale, or the engine will look wrong.  But when material availability or machine limitations dictate a deviation from the plan dimensions, it is relatively simple to calculate alternative dimensions which will have the same critical properties as the specified dimensions, so the engine will run as expected.

This thread was prompted by a question in another thread about whether the other  dimensions of a flywheel could be changed to compensate for a necessary change in diameter as the available lathe was too small for the specified flywheel.  In my observation, it is not an uncommon question.

Of course the very question implies two assumptions, first that the flywheel size is somewhat critical, and second that the plan dimensions have been selected to achieve that right dimension. 

In practice, there is a very wide range of flywheel sizes that will be quite satisfactory, even though they might look very different.  In most cases, a dimensional change can be made to suit our circumstances without any effect on the engine operation. 

Any reciprocating engine has a torque characteristic that fluctuates.  For a single cylinder single acting oscillating engine, such as those many of us build as a first engine, the torque is negative for half the revolution, and a flywheel of some sort is necessary for the engine to run. 

For a double acting engine, the torque is zero twice per revolution, but never really negative, a much smaller flywheel will be adequate.  For a multi cylinder engine, where the torque is clearly positive throughout the complete revolution, a flywheel is strictly not required, though one is often included to reduce the speed fluctuation throughout each revolution.  In all cases, a wide range of flywheel sizes will give satisfactory performance.

I would suggest that “too small” would mean the flywheel is unable to store sufficient energy to carry past any zero or negative torque angles that occur in a particular engine.  Too large would mean the engine is unable to accelerate the engine through the positive torque angle, sufficiently for its momentum to carry through the minimums.  In between these two extremes, a larger flywheel reduces the variation in speed through each revolution when compared with a smaller one, without there being a particular “right size”.

Separate from the issue of the “right size”, the inertia of the flywheel to the plan dimensions can be calculated, and alternative dimensions can be selected which will have the same flywheel effect as the plan specified dimensions, perhaps best described as an equivalent flywheel.  But there are some basic concepts to grasp before we start the calculations.

Enough words for today.  Tomorrow, two fundamentally different types of motion and relevant flywheel properties.

I hope this is of interest, and that I can provide some help to the understanding of flywheels.

MJM460

[RReid]

Thank you for doing this. I think it will be quite interesting, and am looking forward to following along.

[Kim]

I'll certainly be following along!  Thanks for taking the time to do this MJM! 

Kim

[Jo]

 This is going to be really interesting 

Jo

[john mills]

I  will be following
John

[deltatango]

And me!
There was a suggestion that the Wyvern flywheels were too light for the purpose, it would be good to know what is really needed there.

David

[Jasonb]

I'll be following along too.

Though knowing what is needed when an engine does not perform or when starting from scratch is a bit different to making sure an alternative will give the same or better performance to an known working original.

I'll keep it to the James Coombes thread but this thread prompted me to look at the difference between a Stuart flywheel and the one I have just describes, quite interesting and it's only a click or two of the mouse in CAD to see the figures for the two.

[AVTUR]

I will also follow with interest. I dug out my college notes from 50+ years ago and did the sums for a couple of designs, both single cylinder, one IC and the other steam, a few years ago.

AVTUR

[MJM460]

Thank you for the positive response to the first post on this topic.  It’s a privilege to have you all following along.  Thanks also to all those who have simply looked in.

I am sure some of you could easily write the thread, so please feel free to look over my shoulder and correct or clarify as necessary.  It’s to all our advantage to have the  information as complete as possible and reliable for those who come along later.

Jason, does your programme actually calculate the moment of inertia for your flywheels?  Or have I misunderstood your meaning?

So, on to the  fundamental concepts.  It is necessary to understood what properties of a flywheel determine its effective size.

The “size” property of a flywheel is termed its inertia.  Strictly, it’s the second moment of mass.  But it’s usually abbreviated to Moment of Inertia, or just Inertia, which corresponds to our intuitive ideas associated with movement and momentum.  Moment of inertia is a property associated with rotational motion as opposed to linear motion. 

For linear motion, momentum and kinetic energy are determined by the mass and velocity.  The mass distribution or shape of the moving object is relatively unimportant. 

For rotation, mass is important as you expect, but even more important is the distribution of the mass, in particular, the distance of the mass from the axis of rotation.   The relevant velocity is the angular velocity or rotational speed.

Rotation and translation (or linear motion) both are similar but separate and independent motions in physics, and the differences lead to some cool concepts.  But we will get to that later.  First the requested calculation of inertia of a flywheel.

The “second moment” part of the definition means that the distance is squared, and hence has more than the proportionate effect on the result, thus it is not only the mass, but how that mass is distributed around the axis of rotation is important for the flywheel.

So for any small object, at distance r from its axis of rotation, the Inertia, I, is equal to the mass, m, times the radius, r, squared.  Small in this context, means dimensions small enough that all parts of the object can be considered to be at the same radius from the axis of rotation.

Written as an equation,
   I = m x r^2

For a larger object, the Inertia can be calculated by dividing the larger object into very small parts, multiplying the mass of each part by its radius squared, and adding all the results to get the inertia of the whole.  You can easily see from this equation that some mass located in the hub of a flywheel makes a much smaller contribution than the same mass located in the rim.  Similarly, a flywheel of larger diameter, will have a larger inertia, even when the rim dimensions are trimmed to have the same overall mass.

This calculation could be a tedious process, but the mathematics study called calculus makes it considerably easier.  Don’t worry, I don’t propose going through the detailed mathematical procedure.  Every student who has ever done more than a couple of years of calculus has already done all the examples we are interested in, and the results are tabulated in many maths and engineering text books for us to use.

Time for a break while that sinks in.   Tomorrow, we should look at how the flywheel moment of Inertia relates to kinetic energy and momentum before getting to the detailed calculation for typical flywheels.

Thanks to everyone for looking in,

MJM460

[Jasonb]

This is what Alibre gives me, as you can see it is possible to spin the flywheel on any of the three X,Y & Z axis but the best value as one would hope is on XX which is the one through the middle of the central bore. As the results are mm²kg that would seem to be radius in mm squared x mass in kg. This is for the 200mm dia flywheel used on the James Coombes I'm currently describing and is almost 4 times greater that what a standard Stuart 7" flywheel give. No wonder it runs so smooth at very low revs

[derekwarner]

Following on  MJM ....at the end, will see if I can reverse apply the engineering in the calculation 'result' on the flywheel of my Saito Y2DR engine

Derek

[Don1966]

Very interesting I wil follow also.

[MJM460]

Hi Don and Derek, good to have you on board.

Jason, that’s really cool, that Alibre does the calculation for you.  I assume that fusion 360 and other programs also do it.  It’s up there with the pipe stressing programs that calculate the natural frequencies as a byproduct that has its own uses.

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. 

Now the theory is based on the assumption that the flywheel is in a single plane so is a two dimensional object.  The calculations for a true three dimensional object are much more complex.  It is possible, even likely, that the program does a true three dimensional calculation.  A simpler explanation may be that the two dimensional axis are slightly displaced from the centre of the flywheel on the x axis.

In practical terms, three significant figures are more than adequate, and two probably enough.

The other thing to notice is that the moment about Y-Y and Z -Z are not quite equal.  This is expected because the flywheel pictured has six spokes, so the axis that lines up with two spokes is slightly less than the other due to the difference in spoke location with respect to the axis.  They should be equal with four or eight spokes.  The small differences also illustrate that the spokes themselves make a quite small difference to the total. 

Never the less, the figures given are more than adequate to compare different flywheels as you have done.  And you can see how the bit larger diameter flywheel is much more effective without being hugely heavier.  It would be interesting to see the actual figures for the Stewart flywheel.

I had intended to discuss kinetic energy, momentum and angular momentum but this post is getting long enough, so I will continue tomorrow.

Thanks to everyone looking in.

MJM460

[Jasonb]

I had a look at F360 but don't get that option with the free version, best I can do is ctr of mass

The slightly displaced figures could simply be a bit of rounding up or down somewhere though I generally use the symmetry and mirror functions so should be right.

I did make a bit of a boob with the comparrisson between the JC and the Stuart as I had not specified the material for the Stuart so it is actually about half the valve not 1/4

This is what I get for the Stuart, and looks like I drew that correctly as no slight offsets, drawn with ZZ as the central axis of rotation. Out of interest if I supress the spokes the figure drops slightly to 4740mm²kg so could almost be left out of the maths if a quick and dirty answer was needed

[steam guy willy]

Hi John , interesting article and info ..... Is there a reason for solid cranks as used in the crankshaft gear for this engine  ?? lots of extra metal being used ?

Willy

[Jasonb]

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 

[internal_fire]

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

[MJM460]

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
 

[Don1966]

Love this thanks for the explanations it’s a very interesting subject..

Regards Don

[vtsteam]

Your explanations are very clear.

[MJM460]

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

[TerryWerm]

Very interesting subject, I am following along as well! 

[MJM460]

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

[vtsteam]

Thanks again. 

[Kim]

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

[Jasonb]

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

[Kim]

Yes, that makes sense, Jason.  I just had to make it a really long spiral in my mind before it became clear. 

Kim

[jcge]

Great explanation of the serpentine spoke inertia calc !!

[vtsteam]

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.

[MJM460]

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

[Jasonb]

If you want to post some of the flywheel dimensions I'll run them through Alibre, will be interesting to compare results.

[MJM460]

Hi Jason, that is a very generous offer, thank you, which I will gratefully accept.  It is an ideal way to check my calculations.

The attached photo has a table of the relevant dimensions.  Remember that I am assembling these flywheels from a series of simple cylinders, so I think I have included all the necessary dimensions.

I have also included the mass by calculation and by kitchen scale, and my calculated moment of inertia.  If I have made an error, it will be there for all to see.

You can see I did not hit the mass for the two smaller flywheels as I did the first one.  It turns out the rims were a different thickness on the two sides.  I have used figure between the two  rather than adding complexity with additional calculations.  The final result is well within the accuracy required.

I think we will all be interested to see how the figures compare with those calculated by Alibre. 

Thank you,

MJM460

[Jasonb]

Hole diameter?

[MJM460]

Hi Jason, all the shafts are 6 mm diameter, so 6 mm holes.

I haven’t worried about that, as it is filled by the shaft so does not reduce the moment of inertia.

Of course if we were trying to calculate the total moment of inertia for the engine we would have to include the rest if the shaft, crank, etc.  but that goes beyond what I am trying to do at this stage.

Also no lightening or ornamental holes in the webs of those in the webs of those simple flywheels yet as you can see in the pictures.  Many threads on the forum have built my enthusiasm to try, but I haven’t done it yet.  I will cover the maths for such holes along with the spokes, possibly later today.

MJM460

[Jasonb]

Thanks, it was more for the weight calc vs scales that I was thinking of.

[MJM460]

Hi Jason,

You are absolutely right.  My weight calculation doesn’t allow for the hole, while the hole was not filled when I was weighing.

I will be delighted if the calculated weights are within the weight of 19 mm of 6 mm rod.  Or a bit more for the 65 mm one.  And always good to know where the potential sources of error are hiding.

Always good to have more eyes checking on things.

MJM460

[MJM460]

While talking about spokes I had intended to look at some other typical flywheel components, just to show that you can do a relatively easy calculation for most of the common arrangements.

Often people bore lightening holes in the web.  These not only reduce the weight, but I think are quite decorative, and might be a little easier than machining spokes on a manual machine.

Allowing for a round hole removed from the web is a similar process to the spokes.  The moment of inertia contributed by the material to be bored out consists of two parts.  First the moment of inertia of the disk about its centreline,  then using the parallel axis theorem to allow for it  rotating about an axis different from its centreline.

So for a disk of radius rand mass m, with centre d from the flywheel axis, the contribution to the inertia  of the flywheel is

I = 1/2 x m x r^2 + m x d^2

Obviously in this case it is subtracted from the total after multiplying by the number of holes.

I think with those combinations, you can make a reasonable calculation for any flywheel.

I have not considered some of the small decorative features that can really make a model stand out.  As details become physically smaller, the contribution to the inertia also becomes smaller, and for the accuracy we need in practice, they quickly become negligible.

It is interesting to look at the contribution by each component of a flywheel. 

If we look at the 75 mm flywheel I have already mentioned earlier.  I calculated a moment of inertia of 365.7 kg.mm^2.

Certainly there is no need to quote even one decimal place, and just the first three significant figures are more than enough.

I then calculated the moment of inertia of the rim alone.  The result was 338 kg.mm^2, so in this case, 92 % of the inertia is due to the rim.  If we calculated only the rim, we would have an error of less than 10%.

The hub contributes only 0.5 kg.mm^2.

Without the recessed web, the disk I started with has a moment of inertia of 414 kg.mm^2, about 13% more than the finished flywheel.

It rapidly becomes obvious that the majority of the moment of inertia is in the rim.  If you need to increase the moment of inertia by a useful amount, the most effective way is to increase the moment of inertia of the rim, preferably by increasing the outside diameter, which will increase the mass as well as the r^2.

If you increase the inside diameter of the rim as well, it is possible to increase the moment of inertia without increasing the mass.

However, reducing the diameter of a flywheel reduces the moment of inertia, which brings us back to the question that initiated this thread, a good topic for next time.

Thanks for looking in,

MJM460

[Jasonb]

Well the results are in and come out very close to the calculated ones. I expect any slight differences in weight ignoring the central hole was down to slightly different steel specs between what was calculated and what the scales read, Alibre gives me at least 12 different steel specs all with different mass

Regarding the hole for the first flyweel it came out 4.2g lighter once the hole was added or should that be subtracted? But as said the shaft will have inertia as will anything else rotating on it such as crank, eccentrics, gears, pullies etc and depending on what you want out of the engine these will either smooth things out for slow running if they add up to more. On the other hand if you want something that is going to change speed rapidly then keep them and the flywheel as small as possible.

[vtsteam]

Thanks again MJM460. 

I know this is probably premature, or even off topic enough to deserve another thread some day, but I would also be interested in an explanation of the method of estimating the stresses in a flywheel, since that's another important aspect of their design.

[MJM460]

Thanks Jason.  That is really excellent validation of the calculations which are clearly the same, and your help is great fully appreciated.

Short post tonight.  Had to help out my son.  Long story, but a four hour drive later we have things on track.  Definitely bed time here now.

Hi Steve, Definitely on thread.  I will think about how to approach that after a few more basic things.  I wonder if Stewart (Propforward) is looking in, and whether his finite element program has a function for a spinning flywheel?

MJM460

[Zephyrin]

Very nice thread, thanks to share these lessons !
I do have a question about equilibration of an engine, as I cannot add a counterwheight on the crankshaft, I wonder if I can drill holes in the flywheel, to remove a similar wheight (I have seen such holes in many engines), or smaller ones if the position of the holes is larger than the position of the crankshaft couterwheigt, do I calculate this equivalent wheight using the square law as shown in post #36 ?

[MJM460]

Well, that weekend ended up bigger than Ben Hur!  Successful weekend all around.  Even visited my two sisters in law a further hour down the track, and finally home late last night.

Hi Zephyrin, sorry to be late in acknowledging your question.  In all the goings on, I was trying to at least keep up with the forum progress, and accidentally logged myself out.  And of course the key was at home.

The simple answer to your question is that you can certainly balance your engine with weights added to or removed from the flywheel, but it’s a slightly different calculation.  All the calculations so far are about calculating the moment of inertia for a flywheel.  Depending on the mass distribution, the flywheel might or might not be in balance.  For a typical round flywheel, it normally is balanced, as the rotating force due to the mass at any point is exactly balanced by a similar mass on the opposite side.  Even numbers of spokes mean each have an equal one the opposite side, while with odd numbers of spokes the numbers usually also result in a balanced flywheel.

On your engine, the crank pin rotates around the centre of the flywheel axis, it is not balanced by a similar mass on the opposite side, unless it is a multi cylinder engine with cranks at 180 degrees.  A round crank disk is itself in balance, but that does not allow for the crank pin.  Similarly other crank arrangements which are not totally symmetrical are not balanced.  Hence we often see cutouts in a crank disk or balance weights added to the crank shaft or crank arms.

If practical considerations mean balance weights cannot be added to the crank shaft, they can be added to the flywheel as you have suggested. 

Of course the rotating force due to the off balance of the crank pin is in line with the conrod, while the flywheel is not.  If you engine has twin flywheels, the solution is to add half the weight to each flywheel, so the resultant force is in line with the piston.  Otherwise, the mass which balances the rotating force introduces a rotating couple, which causes a rotary vibration in the plane of the engine.  At least this does not cause the engine to bounce away across the bench, and I suspect is a less unpleasant vibration.  And again, the reciprocating mass of the piston and little end of the conrod cannot be balanced by a rotating mass.

Now I have quite deliberately not mentioned the mass associated with the conrod.  The crank pin made moves in a circular path, and also rotates, so the effect on the moment of inertia is the same as any other of centre part of a flywheel.  However, the conrod is fixed at the little end on the piston which results in interesting dynamics and a few puzzles.

For balancing purposes, the big end and about a third of the rod section are lumped at the crank pin, and the balance weights calculated accordingly.  The exact masses allocated to the little end and big end depends on the geometry of the rod so depends on how accurately we want to calculate. But from the point of view of inertia, the conrod is more accurately assumed to be not rotating.  So the moment of inertia is calculated simply as mass times radius of its centre squared.  In fact the big end rotates a little, a few degrees of oscillatory motion, depending on the ratio of rod length to the stroke of the engine.

I will leave more detailed balance considerations to another time, but the question here was about calculating the mass to add or remove from the flywheel for balancing purposes.

The calculation is the simple see-saw balancing equation, where the moments of the two masses must be equal, so-

m1 x r1 = m2 x r2

Where m1 and r1 are the mass and it’s radius from the shaft centre of the crank pin and big end, (i.e crank throw), and m2 and r2 are the mass and radius of the weight on the flywheel.  Then by algebra,
m2 = m1 x r1/ r2

Obviously for a large flywheel compared with the crank throw, much smaller mass on the flywheel rim will be adequate.   

You can add the mass 180 degrees from the crank pin, or remove mass in line with the crank pin.

If you are interested in the magnitude of the rotating force, the formula is -

F = m x r x w^2

Where F is the force in Newtons, m is the mass in kg, r the radius of the mass from the shaft axis and w the angular velocity in radian/sec.  (Should be lower case omega for rotational velocity.)

I hope that answers the question, but don’t hesitate to ask if you have more questions.

Thanks to everyone looking in,

MJM460

[Zephyrin]

Thanks for such an exhaustive answer for my little problem, and yes, of course, i forgot to mention that i put two flywheels on it
I have to calculate the size of the holes to drill, with some tests between !

[MJM460]

Your’e welcome Zephyrin.  Two flywheels is a great arrangement from the balancing point of view, as by putting half the weight on each wheel, the effect can be kept in line with the centre of the crank mass, and thus avoid the couple which results in its own vibration.  If the flywheels are different distances from the centre of the crank pin, repeat the balancing calculation in that plane by putting more mass on the closer one and less mass on the one further away.

I also like the idea of removing weight instead of adding.  By drilling in steps as you have suggested, you can do the trials as you go so have good feedback along the way.  Adding and removal of mass have the same effect but 180 degrees apart.

The basic issue is that engine vibrations come from two elements, the piston and the crank pin (including the big end of the conrod.)

The circular path of the crank results in a rotating force.  This force can be fully balanced by a weight that is displaced 180 degrees to create an equal and opposite rotating force.

The piston motion is linear and cannot be balanced by a single rotating force.  One solution is two equal forces rotating in opposite directions.  Then the resultant force can be made linear to balance the piston.  The obvious complication of contra rotation plus the need to arrange the system so the resultant force is in the same plane as the piston make it clear why this solution is not often adopted.

If a little more weight is added than needed to exactly balance the big end, then a little more helps balance the piston in its line of travel, but introduces an extra force at right angles, so causes the vibration to increase in the perpendicular direction.  This is where your testing as you go might show real benefits.  I don’t know if your phone can report vibration levels with the appropriate Ap, but that might be a sensitive device to help you find the optimum.

I will be most interested to hear how you go.

MJM460

[MJM460]

Well, I hope that little detour was interesting.  It is relevant in the thread at this point for two concepts that help make the detailed picture.

First, you will remember that the moment of inertia is more accurately termed the second moment of mass.  Which opens the obvious questions about a first moment of mass, and what is its significance.

The first moment of mass for a point size object is simply the mass times the radius from the axis of rotation.  (The second moment is the mass times the radius squared.). It is the equation used for balancing equations.  For balance the centre of mass must be on the axis so the moments of all the elements of mass add up to zero.  Like moment of inertia, it can be calculated separately for directions at right angles, so for Zephryns engine, the equation is applied on the axis of the piston motion for balancing.  It is separately applied on the axis of the two flywheels, most importantly if the two flywheels are at different distances from the line of motion of the piston, to divide the mass required for balance between the two flywheels.

The second thing to note is the motion of the big end.  Remember that rotation or spin is quite separate from linear motion, and both can occur independently at the same time. 

The crank pin traced a circular path which is a combination of liner motions is two directions, around the crankshaft axis, and also rotates with the crank shaft.  However, the conrod, which also traces a circular path, does not rotate with the crankshaft.  It does rotate a few degrees each direction, as the little end is constrained to move with the piston.

The significance of this in in the contribution to the moment of inertia of the rotating system, and it’s contribution to the kinetic energy stored in the system.  The contribution of the big end to the moment of inertia is determined by its mass and distance from the axis only (I = m x r ^2) with no contribution from the rotation around itself.

Of course, it’s a bit academic as the contribution of these parts is usually small due to the small radius from the axis.  In the case of a marine engine, where the flywheel is generally relatively small diameter compared with the typical stationary engines, and may be mainly or entirely the propellor it may be more significant.

To get back to the thread, in Post #37, I posted the result of the calculations for my 75 mm flywheel.  Jason was kind enough to put the same dimensions into Alibre, and so confirmed the validity of the calculations.  I also gave some breakdown of the components.  The rim contributed over 90% of the flywheel inertia.  Out of the total of 365 kg.mm^2. The hub contributed about 0.6, while the remainder is the remaining web after I had removed a little from each side to leave a 6 mm thick web which contributed about 24 kg.mm^2.

If I had been more brave, and continued to remove parts of the web to leave spokes, the result is even more clear.  Using the formula given earlier for the spokes, 17 mm long between the rim and the hub, say 6 x 8 mm in section, the contribution of each spoke would be 1.8 kg.mm^2.

For those of us doing the calculations manually, those proportions point to a useful simplification, that is to simply calculate the moment of inertia of the rim, especially for a spoked design where the rim would be approximately 98% of the total on this small flywheel.

Tomorrow, let’s look at the two smaller flywheels in my earlier picture, then a look at the MEM Corless conundrum that prompted the thread.

MJM460

PS JasonP, can you please send me the dimensions for the thickness and cross section of the rim of the 6.75 inch flywheel you eventually made.

[mklotz]

Everyone knows that the mass at the periphery of a flywheel makes more of a contribution to the moment of inertia than mass nearer the rotation axis.

I did a bit of math to quantize that statement.  In the fond hope that it may be of some help in this discussion...

Suppose you have a circular disk of material of density rho. Its radius is "R" and its thickness is "T". Then its mass, "M", is given by:

M = rho * pi * R² * T

and its moment of inertia, "I" is given by:

I = (1/2) * M * R²

Now suppose we cut this disk into two pieces by removing a circular disk of radius "r" from the middle, leaving an annulus of inner radius "r" and outer radius "R".

We have then:

md = mass of disk = rho * pi * r² * T

Id = moment of disk = (1/2) * md * r²

and for the annulus

ma = rho * pi * [(R² - r²)] * T

Ia = ma * (r² + R²)/2

As a sanity check, when r = 0 (i.e., no inner disk) , ma should equal M and Ia should equal I. A quick check will show that to be true. When r = R, (no annulus), md should equal M and Id should equal I. Again, they do. A further check, left as an exercise for the student, shows that:

Ia + Id = I

for any value of r, as it should.

We're interested in the ratio of Ia to I since it's this ratio that quantifies how much more important the outer portion of the flywheel is to the inertia.

Ia/I = {[R² - r²] * [r² + R²]} / R^4 = [R^4 - r^4] / R^4 = 1 - (r/R)^4

In the table below the number in the first row is r/R, the size of the disk removed in terms of the size of the original solid disk. The number in the second row is the moment of the resulting annulus expressed as a percentage of the moment of the original disk.

We can see that if we cut away a disk with a radius half the size of the original radius, the moment of the resulting annulus is still 93.75% of the original moment. Cutting away half the center of the disk loses us only 6.25% of the moment. That radius squared effect is very powerful.

Code: [Select]


   r/R      Ia/I (%)

  0.000  100.000
  0.100  99.990
  0.200  99.840
  0.300  99.190
  0.400  97.440
  0.500  93.750
  0.600  87.040
  0.700  75.990
  0.800  59.040
  0.900  34.390
  1.000   0.000



[MJM460]

Hi Marv, thanks for joining in, all help is appreciated.  I hope you are right about everyone knowing the mass at the periphery is more important than a similar mass near the axis, particularly if they have been following this thread.

Thanks for including the formula for the moment of inertia of an annular ring.  It can be easily derived from the repeated application of the disk formula as you have shown, but also appears in the lists of standard solutions in text books.  I left it out for simplicity, but I am glad you included it, as it is very useful and easy to use, particularly as the rim alone makes a very close approximation to the total for the flywheel.

I also like the last formula illustrated by the table at the end of your post.  It certainly shows the significance of the rim mass, though I note that for flywheels like the MEM Corless, r/R is greater than 90%, so the inertia is much less than the whole disk.  However, that formula shows how a small difference in r makes a big difference to the inertia.

That is about all the calculation we need to cover just about any flywheel we might want to make, and also enough to evaluate the effect of any changes we might contemplate to get around limitations of material availability or machinery available.

At this stage it’s worth looking at a few examples to see the effect of different sizes.  I attached a photo earlier, and repeated below.  They are quite small compared with many people use, but work well on my small engines.  Way bigger than necessary if anything.  Of the three solid ones, the smallest is 45 mm diameter and weighs 157 gm.  The moment of inertia is 41 kg.mm^2.  The 12 mm bore, single acting oscillator in my avatar runs very nicely with it.

The next one up is 65 mm diameter and weighs 310 gm, about double the 45 mm one, yet the inertia is 170 kg.mm^2, four times as much.  The third one is 75 mm diameter.  It weighs 440 gm, and the inertia is 365 kg.mm^2, nearly nine times as much, yet less than three times the mass, and 1.7 times the diameter of the smallest one.

And two more to show just how important to the inertia the diameter compared with the mass.  The second photo shows again the cast flywheel with spokes from the first photo, and a second, a water jet or laser cut (I’m not sure which), only 2.8 mm thick, and destined for a small Stirling engine.

The cast flywheel will clean up to 59 mm dia, 13 wide and will weigh 134 gm.  It’s moment of inertia is  76.9 kg.mm^2.

Compare that with the curved spoke model.  Diameter 107 mm, 2.8 mm wide and weighing 48 gm.  It’s moment of inertia is 92 kg.mm^2.  More inertia for just over a third of the mass.  Each spoke contributes about 4 kg.mm^2.

I hope that really confirms that for a flywheel, the important thing is not the mass, but how it is distributed.  Mass in and near the hub makes an insignificant contribution.

Tomorrow, let’s look at the MEM Corliss flywheel.

Thanks for looking in.

MJM460

[internal_fire]

It certainly shows the significance of the rim mass, though I note that for flywheels like the MEM Corless, r/R is greater than 90%, so the inertia is much less than the whole disk.  However, that formula shows how a small difference in r makes a big difference to the inertia.

I think that is somewhat misleading and leads to an incorrect conclusion.

When dealing with a rim-only flywheel one measures the mass of the actual remaining portion of the disk, not the 90% that is missing.

In this case the calculation becomes very easy. Merely use the actual mass and the outer diameter for calculating the inertia.

The error in doing so will be quite small, not 90%.

Gene

[MJM460]

Hi Gene, it looks like I did not express that very clearly.  Of course, you are quite right in that it s the moment of inertia of the remaining rim, not the missing part that we are interested in.

I was particularly referring to the little chart in Mary’s post, and near the bottom, where r is 90%, you can see that the Inertia is changing quite rapidly with small changes in the dimension of the inside of the rim.

In calculating the moment of inertia, the dimension for the inside of the rim is critical for a fixed outside diameter.

In calculating the inertia of a flywheel, the annulus formula Marv has quoted is a simple way to calculate the moment of inertia of the rim, which is a good approximation to the inertia of the whole flywheel.  This is because the contribution to the flywheel moment of inertia of the spokes  and hub is very small compared with the rim.

I hope that is clearer, but for clarity, the annulus formula is -

I = m / 2 x (R^2 + r^2)

Where m is the mass of the annulus, R and r are the outside and inside radii of the annular ring.

I have assumed everyone can calculate the mass, but Marv has included it in his post if it is needed.

I started this thread in response to JasonP’s problem with the MEM Corliss flywheel.  So I have looked out the drawings and in passing looked at some of the early builds.  While the design intent was to select a size most would be able to handle, and I think they mostly succeeded, however, there are also quite a few builders with mini lathes, facing the same issue as JasonP.

So I think the next step is to calculate the moment of inertia of the original design, and then look at the alternative solution Jason P selected, and perhaps some alternatives.

I had hoped to present this in this post, but time ran away from me this afternoon so I will try again for tomorrow.

MJM460

[internal_fire]

My point is that referencing the moment of inertia loss when employing a large r/R ratio (from Marv's post above) is not the typical method one would use.

An r/R ratio of 0.9 means that 81% of the mass has been removed, leaving only 19% of the mass from the original disk. If all of that mass is added back to the remaining rim then the moment of inertial will be more than 180% of the original full-disk moment.

The MEM Corliss design shows exactly this strategy, with a flywheel having a diameter of 7.25 inches and an r/R ratio of nearly 95%. The "missing" mass has been at least partially added back in the form of 1.5 inch breadth.

Gene

[mklotz]

It seems like the first step is to determine how to calculate the flywheel moment needed by the engine and the error band of that computation. With that information in hand, it's easy to use the annulus moment equation...

ma = rho * pi * [(R² - r²)] * T

Ia = ma * (r² + R²) / 2 = (rho *pi/2) * (R^4 - r^4) * T

which, by adjusting R, r and T, can be made to match Ia to the calculated requirement.

The hub and web will add a bit to Ia, but, as we've seen, it's only a minor, usually negligible amount.

My gut feel tells me that determining the required moment with any accuracy will be practically impossible for most small engines.  The minor contribution of the web and hub will be completely overshadowed by the error band of the requirement computation.

[steamer]

Wot Marv said

The error is small and totally negligible.
Further Rocket Scientists..with a capital S.. are pretty good at this calculation thing by the way.....just sayin

Dave

[internal_fire]

It seems like the first step is to determine how to calculate the flywheel moment needed by the engine and the error band of that computation.

That was what I expected when I first saw the topic appear.

Calculating the moment of inertia is a standard textbook problem. Figuring out what is needed is a whole different beast. 

Gene

[Jasonb]

That was what I expected when I first saw the topic appear.

Read the first two paragraphs of the opening post again.

As the post was prompted by another thread about altering the flywheel it seems to have started in the right place to me, working out what is shown on a drawing and then working out what you need to do if you alter one size  to keep the same moment of the known working original.

[Jasonb]

PS JasonP, can you please send me the dimensions for the thickness and cross section of the rim of the 6.75 inch flywheel you eventually made.

I don't know if we will hear back from the other Jason, might be worth sending him a PM as there was a bit of ambiguity about his 1.375" rim width which he said was wider than original rather than less, could have been a typo as 1 3/8" would have been a likely cleaned up size from the rough 1.5" that he brought home..

I've done a rim profile based on the 6.75" x 1.375" that gives the same moment but won't steal your fire MJM but happy to post after and see if we come up with something similar

[MJM460]

Thanks Jason, my results are shown below together with a copy of the drawing I used.  It will be interesting to see how our answers compare.  It is also very valuable check for any numerical errors.  Did you include the narrow “inner ring” where the spokes sit?

Thanks Gene, Marv and Dave, good to have you all keeping me on my toes.

I do hope I have not conveyed the impression that I am going to present a calculation for the required moment of inertia for a flywheel for a given engine.  I did point out way back at the beginning that by calculating the moment of inertia of a given flywheel design, we can evaluate the likely effect of any changes in design that we might contemplate.

This thread was started in the attempt to help out a model builder who wanted to make his flywheel smaller than the design originally called for.  It would be useful to have a basis to evaluate those changes.  In the end, there is a very wide range of flywheel sizes that will work, with little more effect than a greater speed variation within each revolution.  I have suggested that model flywheels are probably designed on the basis of what looks right on the basis of other similar engines or full size prototypes.

I agree totally that calculating the required inertia is a much more difficult problem.  Also agree that the maths I have presented so far is all standard text book solutions.  But we are not all rocket scientists, and we have not all even done the basic maths.   I won’t make any promises about calculating a required size as there are too many unknowns.  I was thinking of looking at the physics of rotating objects which are quite interesting and lead to cool effects, and after that see where things go.  I hope the topic is of interest to those who are not familiar with these calculations.  Of course, if anyone has detailed information on the torque characteristic of both their engine and it’s load …….

I used the annulus formula already quoted, to calculate the moment of inertia of the rim of the MEM Corliss engine flywheel.  I have attached a copy to help make clear the terminology.  I think there may also be a metric version, but this is the one I used.  I have ignored the hub and spokes as being insignificant in comparison with the rim as described earlier, but I did include the “narrow inner rim”, which was too big to ignore, about 17% of the main rim.

To summarise the results, the 7 1/4 inch original flywheel design is 1 1/2 inches wide and the rim I.D is 6.875 diameter.  My spreadsheet scratchpad converts to mm and divides diameters by 2 to get the required radii.  The mass is 928 g and the moment of inertia is 7370 kg.mm^2.  Note the mass of this flywheel is just over twice that of my 75 mm (3 inch) flywheel, while the moment of inertia is twenty times, such is the importance of locating the mass at the largest possible radius from the axis.

JasonP proposed making his flywheel 6.5 inches dia and 1.375 inches wide.  He did not mention his proposed inside diameter, so I have assumed that he used the same thickness and detail for the inner rim detail where the spokes sit.

For this flywheel the mass is 770 g and the moment of inertia 4865 kg.mm^2.  So 80% of the mass but only 66% of the inertia of the original design.

Now I can’t say this will not work, in fact I feel the engine will run quite satisfactorily with this flywheel, though it may not idle quite as slowly as with the original design.
However, the proposed width was less than the original, keeping the width the same proportion of the diameter, which would help to maintain the appearance, but it removes mass from that critical outside diameter, and that shows in the reduction of Inertia.

So what alternative dimensions could he have selected with that same outside diameter.

I did a couple of trials and came up with increasing the width of the rim to 1.75 inches and thickened the rim  to make 6.0 in. I.D.  As a result of these changes the mass increased to 1141 g and the moment of inertia increased to 7177 kg.mm^2.  Ninety seven percent of the inertia of the original design, for a relatively small increase in weight.

Everyone will have a different idea on the effect on the appearance of the engine with these different flywheels.  I believe they will all work.  To me, they all look ok, (but I’m no artist).  I expect the lowest moment of inertia will limit low speed operation, but I can’t put a number on that.  But for me, it is worth at least having a number that is valid for comparison of the alternative designs.  And for flywheels, the relevant number is the moment of inertia.

I hope that answers the original question about whether the flywheel dimensions can altered, and the associated question of whether we can adjust other dimensions to compensate, and so have the same moment of inertia as the original.

I plan to go on looking at the physics of rotating objects and how the flywheel does its job.  Nothing new for the rocket scientists, but I’m not one of those either, I intend to try and keep it simple.  Hopefully my ramblings will interest others who just like to understand more about their models.

Thanks for looking in

MJM460

[Jasonb]

I did include the central web in my alternative designs but not the spokes or hub. JasonP mentioned cutting teh spokes from a solid disc so likely to be square or rectangular section which will make up for their slightly shorter length.

Fist image shows the original MEM flywheel in solid red and then an outline of a basic replacement using the 6.76" x 1.375" dimensions. If the rim is thickened to 0.25" and the web made 0.281" rather than 0.25" then the moment is 7131mm²kg which is near enough the originals 7121mm²kg

If that makes the rim look a bit thick then my second image shows that adding a draft angle of 0.062" from an 0.218" outside edge to the web and internal fillet to the 6.75 x 1.375 rim makes for a better looking rim and also a slight increase in moment to 7137mm²kg

[vtsteam]

I'm greatly appreciative of your discussion here MJM460, whether the same is available in textbooks or not. In fact practically everything we do in model engine construction has been covered many times in standard and model engineering textbooks, periodicals (some over a century old), and a great variety of other sources, yet we still continue to do them and each of us continues to add to our own personal knowledge.

Not everyone has the same level of understanding or experience. The patient and respectful explanatory method for this technical subject that you've made here may not be new or interesting to everyone. But it is for many, and, hey we're all free to read, or not, any particular thread subject. I'm glad people are still posting a variety of interesting material here about both construction and theory and covering all levels of experience.  Please do keep on writing what you are thinking about, and thanks for your patience and generosity in doing so. 

[Kim]

Here here!  What Steve said! 
Enjoying it and learning from this discussion regardless.  And I have had this basic physics, but it still helps me to think it through and put it all together relating to our hobby.

Thanks, MJM,
Kim

[MJM460]

Thanks Steve and Kim, your support is greatly appreciated.

Hi Jason, thanks for putting those designs in Alibre.  I really like your solution with the sloping inner ring.  This ensures the apparent rim thickness is in keeping with the original design and is a very elegant solution with the same inertia as the original.  Clearly an alternative practical solution for anyone building the engine on a mini lathe.

We seem to have diverged slightly in the inertia figures compared with the previous examples.  The only thing I can think of is which of the available steel densities you are using.  I am using 7840 kg/m^3 (or 7.84 x 10^-6 kg/mm^3).  Or perhaps one of us have rounded a dimension somewhere. But in any case we are clearly calculating the same thing well within any required level of accuracy, and the difference is only 4%.  It’s great to have confirmation, thank you.

I have sent the suggested pm to JasonP.

So far, I have presented the formula necessary to calculate the inertia of a given flywheel by combining the common elements by either addition or subtraction.  I hope that I have also shown that the hub and spokes contribute minimally to the total so we can get an adequate answer by considering the rim alone.

Then I have, with Jason’s help, calculated the inertia for a range of small flywheels as used on models and hope that this gives an idea of the relative importance of the sizes.

The purpose of the flywheel is to store energy as kinetic energy during the positive part of the torque characteristic, and return it to keep the engine turning during the negative torque part of the cycle.  It does this by speeding up during the positive torque angles and slowing down during the negative angles.  If the engine is driving a load, and after all, that is what it is for, the positive torque has to overcome the load, and accelerate the load as well as the flywheel, and during negative torque angles, the flywheel has to supply the energy to drive the load as well as keep the engine turning.  And of course the inertia of the load also contributes to flywheel inertia.

In the question of whether there is a “right size” for a flywheel, I really don’t think there is a precise answer.  Rather there is a range over which the engine performance will be satisfactory.  In my experience for example the 45 mm flywheel on my avatar seems to work fine, and the unloaded engine runs at about 2000 rpm with the little methylated spirit burner.  This can be compared with the 65 mm and 75 mm ones which are both on double acting engines of the same 12 mm bore though slightly larger stroke. 

The double acting engines have minimal portion of the revolution with negative torque compared with the little oscillator which has negative stroke for the entire exhaust half of the revolution.  So it would seem for the double acting ones, a smaller flywheel would be adequate, but in making ones from the available materials (off cuts from the local Handy Steel store), I have actually made them much larger.  This should enable them to maintain a more even speed.

At the other end of the scale, many will remember the years of lockdown, all the shows were shutdown so we held a virtual show on the forum.  As my contribution, I made a little video of my small Mamod engine running as it might if it was running in a show, but without the restrictions on firing to run on real steam.  I was quite surprised when at one point the flywheel started moving along the shaft, and fell right off, right while I was filming.  Quite a surprise, but the engine kept running, and noticeably faster.  Definitely not rehearsed.  At that point most of the inertia was contributed by the crank disk which has the typical balance cutouts.  Unfortunately the video is a bit longer than I like, but I hope appropriate for the show.  That is a very wide range of flywheels for those small engines.

It has already been pointed out that calculating the right size for a flywheel is quite difficult for a model, and perhaps the main reason is the difficulty of determining the torque characteristic for the engine, that is a graph or perhaps a table summary of the variation of torque throughout each revolution, or in the case of a four stroke engine, over two revolutions.

Now it is not uncommon for an engineering problem to have to be solved with insufficient information for precision.  One approach is to estimates some limits for the variables and explore the limits of those variables.  That should give me something to exercise the brain cells.  I wonder how far we can get.  I am thinking that we can start with an acceptable range of speed variation, and calculate the torque required to accelerate from the minimum to the maximum.  It will help us to see how the theory leads to equations of motion that help us understand how it all goes together.

I’m no politician, I don’t yet know the answer, and I’m not a rocket scientists so it will be pretty rough, but I’m game to give it a try for a short time anyway.  Fools rush in and all that.  And any helpful suggestions gratefully accepted.

Thanks for looking in and for all the contributions so far.

MJM460

[vtsteam]

I'll be very interested to see where your explorations go MJM460. What's the worst that can happen in just thinking aloud about something like this? 

So, in line with your example, I'm guessing you'll be considering moment of inertia values for internal counterweights, if any.

Interestingly, if small engines of some particular type turn out to need relatively small moments of inertia to run acceptably, and we optimize flywheel size, various masses close to the center of rotation become much more important.

[Jasonb]

looks like I drew the original central web at 3/16" wide not 1/4" once changed we have the same figures.

Yes it's interesting how little moment is needed particularly if the engine is going to spend it's time running at speed. When I did some of those small engines last year I did not think they would run well at low speeds with their tiny flywheels but they did.

[RReid]

As one of those who learned much of this stuff many years ago but haven't used it since, I appreciate having this well presented review available without digging out the old texts. Thank you, MJM!

And as an Alibre user and a lazy bones when it comes to hand calcs, I appreciate Jason's contributions as well. Thank you, Jason!

[MJM460]

Hi Steve, have to agree.  The worst is usually a spreadsheet of figures that I can’t find my way through a year later.  It’s always worth a try, as even finding out I can’t get any useful results is still a result, right?

But internal counter weights?  With the formulas discs and for spokes and the parallel axis theorem, most things can be calculated, so it is easy to check if the numbers are significant.  Of course, once the flywheel fell off my little Mamod engine, only the disk and an insignificant big end remained, so those little weights might be more significant with a very small flywheel.

Hi Jason, thank you for checking that.  It is so easily done, but nice to find a reason for discrepancies, it’s the value of checking calculations, more important than the small difference in the numbers.  Checking of the computer output is a more difficult issue, but the agreement of those numbers suggests at least we are both calculating the same thing.  Probably the first time anyone has checked the Alibre moment of inertia calculations.

Hi Ron, I think Jason has done us all a favour by showing us that the calculations are there in the program.  I assume you only have to find the right button in one of the tool bars to display the result without further effort.  I’m with you, knowing it is there, I wouldn’t bother calculating manually, even with the help of a spreadsheet.  Though I hope that going through the theory has prompted the memory or helped people understand what is behind the number.

When I mention manual calculation, instead of back of an envelope, or even a note book, I actually open a spreadsheet, and put the inputs and formulas into that.  The computer does the arithmetic for me, and I name the range so I can find it again.  And I can easily explore the sensitivity to some of the variables without starting to make mistakes in the calculation.  This way I put many calculations into one sheet instead of ending up with a multitude of files with names I can’t associate with the subject.  Surprising how often I come back to the same calculation.  But basically it is still manual input of the required formulas.  And the file is called Scratchpad!

I am making good progress with some flywheel calculations but I need to look out some old files with real run figures for my engines.  Then we can see what we can learn with realistic inputs. 


I hope to have more tomorrow,

MJM460

[MJM460]

I learned long ago that flywheels in full size practice are sized on the basis of limiting the instantaneous speed variation within one revolution.  So I did some calculations, just using the basic physics, to get average rotational speed, accelerations, torques and power based on an assumed speed variation.  The number most difficult to find is the actual speed variation for one of our model engines.  I don’t have an answer to that yet, which is another issue.  But it’s instructive to do the calculations anyway, and then explore the sensitivity of the results to the initial assumption.

I rummaged around my notes and found a number of log sheet records of test runs that I have done on my models.  Unfortunately, most of the ones that would be most useful now, relating to that little single acting oscillating engine, were carried out before I bought the digital tacho.  At the time I was I was most interested in the boiler performance, dictated in part because I had no equipment able to measure the engine output.  But I did find just one result from 2017 that had some engine speed readings, including a maximum reading of 1000 rpm.  And we have already calculated the flywheel inertia.  Should be all we need if we assume a speed variation, just to see what the calculations mean.

The printed result is attached below.  I have assumed a constant torque and hence acceleration.  Not true of course, but the total kinetic energy into the flywheel for a given speed change is the same whether the torque is constant or variable, so the result is adequate for our purposes.

The average speed is 1000 rpm as measured by the digital tacho, and converted to radians/sec.

With the assumed speed variation, the increase in speed from minimum to maximum occurs over half a revolution for this engine, and we can calculate the time (t) for half a revolution in seconds.

With a little algebra, the angular acceleration is calculated using w2 = w1 + alpha x t where w2 and w1 are the initial and final speeds and alpha is the angular acceleration in radians/sec^2.  (As usual, on my iPad I am using w instead of the text book lower case omega)

The the average torque is I x alpha, note that I have switched the units for I to Kg.m^2.  This avoids the need to keep track of multiple powers of 10 so that torque is in Newton metres, or N.m

Similarly Power for this half revolution is torque times angular velocity.

Note the parallels between those formulas for rotational motion compared with the equivalent ones for linear motion.  Torque and angular velocity and acceleration replace force and linear velocity and acceleration, otherwise the formula is the same.  Resulting directly on Newtons laws.

The result of those simple calculations is interesting!  The engine is running at constant average speed, so it must slow as much during the exhaust cycle as it speeds up during the steam inlet, hence the flywheel gives the same energy back during the exhaust half of the revolution.

The engine is running free with no connected load, so the same power is given back during the exhaust, which means that is the power required to turn the engine at that speed.  The engine has to supply that average power, so the peak power from the engine is about twice that.  About but not exactly, as it takes power to push the exhaust steam out of the cylinder, which is not exactly the same as the inlet half of the revolution.  However, if the power absorbed in the exhaust stroke was less that supplied during steam inlet, the engine would speed up.  As it runs at constant speed, at least until the burner starts running out of fuel, that maximum speed provides a reasonable estimate of the engine power, providing of course that our assumed speed variation is correct.  I hope that logic is clear, but don’t hesitate to ask for clarification, or if you disagree, please add to the discussion.

(Note, it is ok for my little meths powered plant and similar low power plants, but not a good idea to run a larger plant unloaded at full power.  You can see from the above calculation that the speed will increase until all the power is absorbed and this is potentially dangerously destructive.  Make sure you do have a suitable load coupled up before running full throttle.)

We can return to that later, but while we are looking at the calculation, there is more we can learn from that calculation.

Time for a break while that much gets digested, then a little more from this exercise tomorrow.

MJM460

[MJM460]

Continuing from yesterday, let’s explore what else we can learn from that calculation.

How important is that initial assumption of 5% speed variation?  A spreadsheet is great for exploring “What if?”.  I changed the input cell to 2.5%, which. with all else constant, reduced the  power  to 0.75 W.  (Some might argue that even this is optimistic for such an engine.) 

Then, back to the initial 5% speed variation, and reduced the flywheel inertia to 21 kg.mm^2, half the original value.  Again, the power reduced to 0.75 Watts.  The same engine and steam conditions, so the engine is presumably still capable of the original 1.5 Watts, so reducing the flywheel inertia leaves more capacity to drive an external load.  An extra 0.75 over the original 1.5 is a very significant difference.  A heavier than necessary flywheel gives improved speed regulation, but absorbs power that could otherwise drive the external load.

Finally, with the reduced flywheel inertia of 20.5 kg.mm^2, I increased the number in the speed cell until I matched the original power of 1.5 Watts.  The new speed was 1260, so the increase in rpm for half the flywheel inertia was about 1/4 of the original speed.

If we want to run an engine very slowly for display purposes, it is probably preferred to have a very even speed so it is less likely to stop at the slow point of the revolution.  You can see from these trials that a larger moment of inertia reduces the speed variation, but at the same time, increases the power required to accelerate the flywheel during the steam inlet.  This implies more pressure required. 

On the other hand, if you want to run at the lowest possible pressure, a good measure of a well built engine, a smaller flywheel inertia requires less power for the acceleration part of the revolution, but the resulting greater speed variation makes it more likely that the engine will stop at the slow point.

At higher speed for a working engine, the risk of stopping is less, and a lower moment of inertia requires less power for acceleration, and leaves more power available to drive the load.

As always, you have to find the balance between competing requirements to optimise the flywheel size.

These explorations are easier on a real computer, where a spread sheet normally has a goal seeking tool which finds the answer for you, however that function is not available on my iPad so I have had to do it manually by trial and error.

But some interesting conclusions that hold, even if the real speed variation is different from that initial guess.  It would be really useful if we could actually measure that variation, but that would be a project for another thread.

We can also use the information so far to get a little more information about that engine.  Perhaps I should do a test run or two first to check those figures.

Thanks for looking in,

MJM460

[MJM460]

As an aside, there is some more really cool maths/physics around flywheels.  We are all familiar with linear motion, and the associated forces and velocities.  We know that forces and velocities are vectors which simply means that to describe them fully, you have to include not only magnitude, but also direction.

Also that to add the effect of two vector quantities, we generally visualise this graphically by drawing a triangle, where two sides of the triangle represent two forces or velocities, and the third side is the resultant of the two.  This helps us solve problems like adding our boat speed to the wind speed to get the apparent wind, and similarly for aeroplanes and cars.  (As we generally know the apparent wind, we actually subtract the vectors to get the true wind).  We also know that we can divide a vector into two components at right angles to each other and that a vector has no effect at right angles to itself.

So what about rotational motion.  Well, rotations also involve vectors.  Torques and angular velocities are vectors, so to describe them we need to define not only the magnitude, but also the direction.  The direction of a torque is defined by a “rule of thumb”, if you curl the fingers of your right hand around to make a thumbs up sign, with fingers going the direction of rotation which results from the torque, the direction of the torque is defined by the direction pointed by your thumb.  Now I am sure this definition is a bit arbitrary.  In a left handed world, we might use the left hand and the maths would still work, but it would be very confusing if there were some using each.  So in this world, the mathematical definition used the right hand.  Angular velocity is also a vector.

We see a very interesting result of this vector behaviour in those little gyroscopes that we often see in a good toy shop, or in your local science museum shop.  Basically a flywheel suspended on two bearings and set spinning by winding a string around the shaft, then pulling the string as hard as you are able to set the wheel spinning.  The cage has non-rotating extensions to the shaft, often one with a point, and the other end with a little ball shape.

If you set the wheel spinning and stand it with the shaft vertical on the pointed end, and you set it perfectly vertical is sits spinning without falling over, but more likely, it will be not quite vertical, which results in a torque due to the weight not being quite in the same vertical line as the equal and opposite supporting force on the point.  If the flywheel was stopped it would just fall over.  But when it is spinning, instead of falling, the flywheel axis will progress with the top of the shaft tracing out a small circle.  And this continues for quite a long time.  Just like a spinning top.

We can use that rule of thumb to determine the direction of the torque due to the support point and the weight of the wheel, and sure enough, it is at right angles to the direction of the spin.  Being at right angles, it has no effect on the magnitude of the spin, but simply changes its direction.  So that torque causes a change in the direction of the spin in the  horizontal direction, indicated by that rule of thumb, causing that progression.

Those little gyroscopes are usually supplied with a little stand with a little hollow spherical depression on top.  You can stand the point in that depression and it will balance there as it progresses.  An extreme of this situation is if you put the spherical end of of the cage on top of the stand in that hollow, with the spinning shaft horizontal, and the other end unsupportd.  Even then, the gyroscope does not immediately fall as you might expect, but will just progress around its support. 

Of course, it eventually falls, but the manner of falling is also interesting, but we will get back to that.  That gyroscope action is behind the safety advice for portable circular saws, to line up the saw with the cut before switching on, and wait for it to stop before you move it away.  If you have a job with a large number of successive cuts you might be tempted to move the saw before it stops.  The unexpected direction of movement in response can result in nasty accidents.  If you must move the saw while still rotating, make sure you are holding the body of the saw firmly with both hands and move it slowly while you get to understand the motion.

On the other hand, if the sphere is installed in a ball and set floating so there is no torque resulting from the support, it will just keep spinning with the axis maintaining a constant direction, except for the motion imparted from the rotation of the earth.  If a motor is provided to maintain the rotation against friction, we have the makings of a gyroscopic compass.

All based on the same vector physics as our flywheels.

Thanks for looking in.

MJM460

[Kim]

Thanks, MJM for this fun little physics refresher!  I've really enjoyed following along with this thread!

Kim

[MJM460]

Thanks Kim, I’m glad you are enjoying it.  I hope there is also something there for those seeing it for the first time.

On re-reading my last post, I notice some missing words when talking about the gyroscope.  I intended to say put the ball end of the cage on top of the stand with the spinning shaft horizontal, and the other end unsupported.  I have corrected that though it was probably obvious to those familiar with a gyroscope.

The rotational equivalent of the linear F = m x a is similar in appearance,

Torque = moment of inertia x angular acceleration

It is a mathematical expression of  Newton’s second law and a consequence of the conservation of angular momentum.

Now for that gyroscope sitting horizontal with one end sitting on the support, we have already worked out the direction of movement, which makes a horizontal circle.  The torque is equal to the weight of the wheel times the distance to the support, so that formula gives us the acceleration.  However, in this case, the rotation is around the vertical axis (plus it is also spinning around its horizontal axis) so for the progression calculation, we need the moment if inertia about that vertical axis.  Remember the additional ones in the result of Jason’s calculation?  It is a little more than half the moment about the primary axis.

And the gyroscope does not continue to accelerate but quickly reaches a steady angular velocity of progression.  Basically this is due to the friction at the support point, which results in a torque that by the rule of thumb is at right angles to the axis of spin and resists the progression.  In turn, this friction torque changes the direction of the axis of spin downwards resulting in the graceful slow tilting down of the gyroscope until it falls off the tower.  Nothing like the motion you would expect if you tried to support it by one end only when it is not spinning.

The purpose of this little side track is to illustrate the differences and similarities between linear and rotational motion.  Besides, gyroscopes are fascinating things

Thanks for looking in

MJM460

[MJM460]

Back in post no 65 I used the flywheel inertia and an assumed speed variation to calculate the power into the system from the flywheel during the exhaust half of the revolution of my little single acting oscillating engine.  It obviously takes the same power during the steam inlet half of the revolution.

 With the engine running unloaded at constant speed, the energy from the flywheel is the total energy into that system during the exhaust cycle.  With the engine running at constant speed, all that energy is absorbed in friction and in pushing out the exhaust steam.  This is approximately the same as the energy needed to run the engine for the inlet part of the cycle.  The steam also has to accelerate the engine during that steam inlet.  Hence the total energy input from the steam is twice the energy returned by the flywheel during the exhaust cycle.

If we use the standard formulas relating power, torque and Brake Mean Effective Pressure, with a little algebra, with the torque already calculated, we can arrive at a figure for the BMEP.  Remember that BMEP is an average pressure for the power stroke, and is the mean differential between the two sides pressure of the piston. 

Just for interest, I did the calculation, and for my little single acting oscillating engine.

The calculated power is 0.56 watts.  Then I calculated a BMEP of 26 kPa.   This is based on an average speed of 720 rpm, a speed variation of +/-2.5%, flywheel inertia of 41 kg.mm^2.  The engine is 12 mm bore and 16 mm stroke.  The peak pressure in the cylinder will be somewhat higher than the Mean, but it’s hard to guess the pressure variation in an oscillating engine.  If it is approximately sinusoidal, the peak would be about 40% higher than the mean.

For reference, the boiler temperature for that test run was 110 C which means a saturation pressure of 143 kPa absolute.  My steam plant has a 3/16 tube to connect the engine to the boiler compared with 5/32 which is possibly more common, so there might be minimal pressure loss between the boiler and the engine inlet.  With atmospheric pressure on the other side of the piston, about 43 kPa differential.   This is only about 60% higher than the calculated BMEP.

Unfortunately, I have no way of measuring the actual speed variation, it is just a guess.  For a relatively high inertia flywheel on such a small engine, it is reasonable to expect the speed variation to be small.  All things considered that BMEP estimate is surprisingly close to the boiler pressure, but I can make no claim to the accuracy other than that the peak pressure cannot be higher than the boiler pressure.  Similarly that calculated power is no more accurate than the guessed speed variation, similarly it is an upper limit.

I think the main conclusion is that the actual speed variation, if we could measure it, would be something less than the +/- 2.5% I have assumed here.  However, if we could measure it ..…

 I think that is about as far as I can take those calculations.  We started with calculating the moment of inertia of a flywheel, and how it is more than just mass.  Did a little exploration of rotational motion and linear motion principles, side tracked into the vector maths associated with rotational motion and gyroscopes, and finally showed how flywheel inertia and instantaneous speed variation are related to torque and power of an engine running unloaded at constant speed. 

I do hope that is of interest.

MJM460