Flywheel Design?

[miner49r]

Hiya Folks,
On my first engine (Elmers Wobbler #25) I pressed a piece of copper water pipe over my aliminum flywheel. I looks great. I eventually want to make more elaborate flywheels like I see some of the craftsmen make here when I found Troutsqueezer's post showing his flywheels using a heavy steel ring over an aluminum hub. Still simple, yet a step up from a solid disc.

Now this got me thinking. Is having the greater mass on the circumference more or less efficient than a solid unit?

Regards,
Alan

 

[JorgensenSteam]

I think that sometimes builders us a heavier outer rim to give an engine more momentum as it is trying to get over dead centers, since in general the driving force drops to zero on the dead centers.

The flywheel stores energy during the power part of the stroke and delivers that power back to keep the reciprocating parts moving during the non-power parts of the stroke. The heavier the flywheel rim, the more energy it can store, and the smoother the engine will furnish power to a belt or gear train.

Pat J

 

[Chazz]

Here's a thought, (if you have clearance) you can "Kill two birds" as it were, both add outer mass and look cool (maybe?) since you already have an Aluminum shaft, ringed by copper, press a steel ring over the copper?

As to flywheel design, as any component in an engine design, there are engineering considerations to consider, I am sure you have seen flywheels which are 'anchor' shaped on one side (IE a BIG honkin' off-center-weight) such as an old train locomotive drive wheel.

If the flywheel is too heavy, the motor will not be able to turn it (the flywheel) over, if the flywheel is too light, as Pat J said, there will not be enough energy to complete the cycle, hence a stall. (That is if you want an engine to start and maintain without intervention )

Cherry Missmas
Chazz

P.S. Machinery's Handbook + other tomes will get you as deep into engine\flywheel design as you care to go, but most of the 'small' engines running on air can overcome "Flywheel-sag" with 'Fly-agra' (sorry couldn't resist) :big:

 

[Ken I]

miner49r, the formaula for inertia is mass x radius squared - so the further out you can place the mass - the greater the inertia for the same mass.

So typical flywheel designs have virtually all their inertia in the rim - the hub and "spokes" being suficient only to do the job - hence an overall saving in mass and material (for a large cast flywheel).

If you are turning a flywheel from solid - then all that material removal is going to save some weight with the loss of some inertia - so its mostly a cosmetic consideration under those circumstances.

Ken

 

[Blogwitch]

Just to give you a bit of an insight, I made an engine, and when it was run, I found that there wasn't enough mass on the flywheel to give a smooth running engine, so this is what I came up with.

http://www.homemodelenginemachinist.com/index.php?topic=2839.msg85955#msg85955


John

 

[Wrist Pin]

Fly-agra......groannnnn......

 

[mklotz]

This discussion of flywheel moments of inertia intrigued me so I went off and did a bit of calculating...

Imagine you have a solid metal disk. It has a mass m0, a radius r0 and a resulting moment of inertia I0 (=0.5*m0*r0^2). Now suppose we cut a circular disk of radius r1 out of the center of this disk leaving us with an annulus vaguely similar to a flywheel rim. Given the loss of mass to the original disk, how much has the moment of inertia decreased?

In the code box below the number in the first row is r1/r0, 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.

  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

 

[Troutsqueezer]

The fact that we are dealing with radius squared in the equation tells us a lot.

 

[miner49r]

Great information. I already had it in my head that the flywheel needs to be heavy enough to sustain energy, yet not so heavy that the engine won't be able to turn it.

That being said... If the flywheel needs to have a mass of "x", would it matter if the weight is distributed throughout the flywheel or concentrted toward the circumference?

Regards,
Alan

 

[Tin Falcon]

That being said... If the flywheel needs to have a mass of "x", would it matter if the weight is distributed throughout the flywheel or concentrted toward the circumference?

Regards,
Alan

Keeping the bulk of the mass on the rim keeps the moment of inertia high while lowering total mass. that is why most fly wheels have a thin web or spokes .
Tin

 

[mklotz]

That being said... If the flywheel needs to have a mass of "x", would it matter if the weight is distributed throughout the flywheel or concentrted toward the circumference?

Alan, you might want to read my post (reply #6) above to answer your question.

 

[miner49r]

mklotz,
Even at 50 I still need the teacher to whack me in the back of the head to get me to use my brain. I re-read the text several times and I think I get it now. If I have a flywheel with a radius of x and I remove the center at 0.50x I retain 93.75% of it's original enertial moment. For a loss of only 6.25%.
Furthermore, if 0.70x of the center radius is removed on the same flywheel the enertial moment is reduced to 75.99%.

Your Humble Student

 

[mklotz]

mklotz,
Even at 50 I still need the teacher to whack me in the back of the head to get me to use my brain. I re-read the text several times and I think I get it now. If I have a flywheel with a radius of x and I remove the center at 0.50x I retain 93.75% of it's original enertial moment. For a loss of only 6.25%.
Furthermore, if 0.70x of the center radius is removed on the same flywheel the enertial moment is reduced to 75.99%.

Yes, you've got it. The numbers vividly demonstrate that the central part of the disk contributes far less to the moment of inertia than the rim portion does. One more nuance - it's spelled "inertia", not "enertia".

 

[miner49r]

if 0.70x of the center radius is removed on the same flywheel the inertial moment is reduced to 75.99%.

Now the question arises of what to do about the 24.01% loss in the above quote. If the radius needs to remain constant, the weight could be added as width to the flywheel. But I don't think it is that simple.

How do I put this in words? A wheel is a simple machine that consists of an untold number of levers with a common fulcrum. Less force is needed to do the same work the further we get from the fulcrum. Let's say that the 24.01% loss comes to 12g. Because the weight will be added to the greater radius of the of the lever, I surmise that a mass less than 12g. is needed to attain the original 100% inertial moment.

Just thinking. And before you ask... Yes, it hurts. ;D

 

[mklotz]

Now the question arises of what to do about the 24.01% loss in the above quote. If the radius needs to remain constant, the weight could be added as width to the flywheel. But I don't think it is that simple.

How do I put this in words? A wheel is a simple machine that consists of an untold number of levers with a common fulcrum. Less force is needed to do the same work the further we get from the fulcrum. Let's say that the 24.01% loss comes to 12g. Because the weight will be added to the greater radius of the of the lever, I surmise that a mass less than 12g. is needed to attain the original 100% inertial moment.

Just thinking. And before you ask... Yes, it hurts. ;D

The formula for the moment of inertia of an annulus about its axis of symmetry is:

I = m * (r1^2 + r2^2)/2

where:

m = mass
r1 = inner radius of annulus
r2 = outer radius of annulus

If we preserve r1 and r2 and increase the mass by making the ring 24% thicker (in the axial direction), the moment will indeed increase by 24%.

There are many other ways to adjust the moment, eg. by adding a ring to flywheel circumference or installing dense plug weights in the rim. The mathematics involved in calculating the new moment depend on the technique selected so I can't provide a general solution here.

However, in my example there was nothing sacred about the moment of the original circular slab. Therefore there is no implied need to restore the moment to that of the original slab.

In the world of model engines, it almost never occurs that the designer specifies a moment of inertia of the flywheel. Most of the time it's done by the TLAR (That Looks About Right) method. While one might increase the moment of a flywheel in order to make an engine run more slowly, it's seldom done mathematically.

 

[mklotz]

While watching TV last night I thought of a slightly better way to demonstrate the relative importance of the flywheel rim to the overall moment. I made a few calculations and the results are presented below...


Consider a simplistic flywheel that consists of nothing more than a solid slab of metal cut from a cylinder. It has mass 'M' and radius 'R'. Its moment of inertia about its axis of symmetry is I=MR^2/2.

Let's divide this slab into four regions by quartering the radius. Thus, region 1 extends from the center (r=0) to r=R/4. It's the solid disk that forms the center of the slab. Region 2 extends from r=R/4 to r=R/2. It's an annulus surrounding region 1. Region 3 is another annulus and extends from r=R/2 to r+3R/4. Region 4 is the outermost annulus and extends from r+3R/4 to r=R.

Now let's compute the mass, 'm', of each of these regions and the moment of inertia, 'i', for the region. We'll express these as fractions of the total mass, 'M', and total moment, 'I'.

region		m/M		i/I		i/I (%)

1		1/16		1/256		0.4
2		3/16		15/256		5.9
3		5/16		65/256		25.4
4		7/16		175/256		68.3

 

[miner49r]

Hiya Marv,
I can understand that most designers use the "TLAR" method because you are probably one of the few that knows the science behind it all. I hope you are not losing any sleep on account of me.

One quick question. What is meant by this figure (^) in your equations?

Alan

 

[bezalel2000]

Hi Alan

'^' means 'to the power of' in this case it is followed by 2 so R^2 = R2 = (R x R) or R squared

if its followed by 3 as in R^3 = R3 = (R x R x R) or R cubed


I=MR^2/2 is the same as I = M x R2 x 1/2

Bez

 

[SmoggyTurnip]

Here is another way to look at it. If we take a simple solid flywheel disk of mass m and radius r we have a moment of inertia as mentioned:

I = m * r^2 /2


Now lets cut a small disk of radius k*r out of the center of that flywheel where k is a number between 0 and 1.
For example if our small disk is half the diameter of the flywheel then k = .5


Take the mass of that small disk, melt it down and add it evenly to what is left of the outside ring by making the ring thicker but not changing its inside or outside radius.
We now have a ring with a new moment of inertia:

I2 = m*( (k*r)^2 + r^2 )/2
= m (k^2+1) r^2 / 2

I2 = (k^2+1)*I

Since k can be anywhere from 0 to 1 we can increase the moment of inertia to a maximum of 2*I or double the starting inertia without adding any mass (this would be an infinitely thin ring though).

 

[Ken I]

Gee - you guys like your numbers.

A connumdrum

A hole 100mm long (use 4" if you like) is drilled through the centre of a sphere - leaving a ring like a tube olive, [100mm long (wide) which is the length of the hole after drilling].

What is the volume of the ring ?

Can be solved by logic.

If you know the solution don't spoil it by revealing it too soon.

Ken