Author Topic: Flywheel calculations  (Read 3020 times)

Offline Jasonb

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Re: Flywheel calculations
« Reply #30 on: March 06, 2023, 07:21:34 AM »
If you want to post some of the flywheel dimensions I'll run them through Alibre, will be interesting to compare results.

Online MJM460

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Re: Flywheel calculations
« Reply #31 on: March 07, 2023, 09:55:29 AM »
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

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Offline Jasonb

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Re: Flywheel calculations
« Reply #32 on: March 07, 2023, 10:24:58 AM »
Hole diameter?

Online MJM460

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Re: Flywheel calculations
« Reply #33 on: March 07, 2023, 09:57:40 PM »
Hi Jason, all the shafts are 6 mm diameter, so 6 mm holes.

I havenít 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 havenít done it yet.  I will cover the maths for such holes along with the spokes, possibly later today.

MJM460

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Offline Jasonb

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Re: Flywheel calculations
« Reply #34 on: March 08, 2023, 07:37:00 AM »
Thanks, it was more for the weight calc vs scales that I was thinking of.

Online MJM460

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Re: Flywheel calculations
« Reply #35 on: March 08, 2023, 07:58:48 AM »
Hi Jason,

You are absolutely right.  My weight calculation doesnít 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

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Online MJM460

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Re: Flywheel calculations
« Reply #36 on: March 08, 2023, 11:27:52 AM »
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


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

Offline Jasonb

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Re: Flywheel calculations
« Reply #37 on: March 08, 2023, 11:59:16 AM »
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.

Offline vtsteam

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Re: Flywheel calculations
« Reply #38 on: March 08, 2023, 01:13:40 PM »
Thanks again MJM460.  :popcorn: :ThumbsUp: :cheers:

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.
Steve

Online MJM460

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Re: Flywheel calculations
« Reply #39 on: March 09, 2023, 12:26:08 PM »
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

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Offline Zephyrin

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Re: Flywheel calculations
« Reply #40 on: March 10, 2023, 10:42:07 PM »
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 ?

Online MJM460

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Re: Flywheel calculations
« Reply #41 on: March 14, 2023, 08:52:38 AM »
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 itís 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 itís 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 donít hesitate to ask if you have more questions.

Thanks to everyone looking in,

MJM460



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

Offline Zephyrin

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Re: Flywheel calculations
« Reply #42 on: March 14, 2023, 05:37:47 PM »
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 !

Online MJM460

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Re: Flywheel calculations
« Reply #43 on: March 15, 2023, 10:58:03 AM »
Youríe 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 donít 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

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

Online MJM460

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Re: Flywheel calculations
« Reply #44 on: March 16, 2023, 11:17:31 AM »
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 itís 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, itís 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, letís 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.




« Last Edit: March 16, 2023, 11:23:59 AM by MJM460 »
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