Flywheel calculations

[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