Author Topic: Please check my flywheel calculations  (Read 3320 times)

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mmagliaro

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Please check my flywheel calculations
« on: April 12, 2016, 12:17:51 PM »
+1
I may be able to put a double-shaft motor into my 0-6-0 project, allowing for a flywhee on the motor, spinning at 4x the speed of the output gearhead shaft.  I always thought that doing this would be a boon to smooth running because the flywheel would spin 4x faster than the worm, and thus provide lots of coasting momentum.  But after going through all the exhaustive calculations, now I'm not so sure (and I don't know if I want to spend 80 bucks on a double-shaft Faulhaber to find out!)

So if you can stand it, those of you versed in enough math and physics to follow this... can you tell me if the below chain of calculation is reasonable?  THANK YOU!

At the very end of all this, I figured out that with the flywheel I can fit in there, and with the load on the motor that I can measure at 5 mph, the drivers will coast for only 45 mSec on the flywheel when power is interrupted.  At 10 mph, they will coast for 180 mSec.
This doesn't sound like enough to be worth it.  What do you folks think?

The killer is that there isn't much room for the flywheel, so I can only use a thin one and it can only be about 9mm in diameter.

- Rotational speed of 51" drivers at 5 mph.
   Circumference of driver  =πd = 3.14159 x 51 = 160.22"
   5 mph / 60 *5280 *12 = 5280 in/min
   5280 / 160.22 = 32.95 rpm

- Angular velocity
   Gear ratio from motor to drivers is 80:1
   Motor is spinning at 32.95 * 80 = 2636 rpm at 5 mph
   Angular velocity ω of motor = 2636/60 * 2 *3.14159 = 276 rad/sec

- Moment of Inertia and Stored Energy in the Flywheel
   Flywheel is 9mm in diameter x 3mm thick, brass (density =8.553 g/cm3)
   Radius = 4.5mm = .45 cm
   Volume of cylinder = πhr^2 = 3.14159 x 3 x 4.5 x 4.5 = 190 cu mm = 0.19 cu cm
   Mass of flywheel = 8.553 g/cm3 x 0.19 = 1.62g
   Moment of inertia, I, for solid rotating disk = 1/2mR2 = 1/2 x 1.62 x .452 = 0.164g cm2 = 1.64e-08 kg m2

- Kinetic energy of rotating flywheel E = 1/2∙I ∙ω^2 = .0006246 joule
   Torque of 1 Nm (Newton-Meter) over 1 full rotation of the wheel uses an energy of 2π joules  (that's 2 PI)
   E = Torque x Angle  (2π = 1 Nm x 2π )
   We have .0006246 joule of stored kinetic energy
   .0006246 = T x 2π  , therefore T = 9.9408e-05 Nm.  That is the torque our flywheel can deliver so that it comes
                                       to a stop after only 1 rotation.
   T for 1 rotation before stopping:  9.9408e-05 Nm
   T for 10 rotations: 9.9408e-06 Nm
   T for 100 rotations: 9.9408e-07

- What is the torque of the motor?
  How long can our flywheel keep the motor turning?
  Look at Faulhaber data sheet:
   Torque constant = 1.7 mNm/A, and we know that at 5 mph, the motor is drawing about .030 A,
        so the torque = 1.7 x .030 = .051 mNm = .000051 Nm = 5.1e-05 Nm. 

        Therefore, with our torque / time value of 9.9408e-05 Nm for 1 rotation, we can sustain the motor's
        torque of 5.1e-05 for about 2 rotations of the motor before it stops.  (9.9408/5.1 = 1.94)
   With an 80:1 gear ratio, 2 rotations of the motor means the drivers will turn 1/40th of a turn coasting on the
        flywheel before they stop, about 9 degrees of rotation.   At 32.95 rpm (5 scale mph):

   1/40th of a rev / 32.95 rev/min = 7.5872e-04 min = .045 sec
   So the drivers will turn for only 45 mSec before they stop

   The stored energy in the flywheel increases as the square of the rotational speed.  So when the loco is
        running at 10 mph instead of 5, the stored energy in the flywheel (now .002498 joule)  is enough to turn the
        drivers a full 1/10 turn (180 mSec) before they stop (assuming the motor torque is about the same at that speed,
        and based on the current draw, it is). 


eja

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Re: Please check my flywheel calculations
« Reply #1 on: April 13, 2016, 12:47:21 AM »
+1
Anyone who actually remembered that much from their physics class deserves this ...

« Last Edit: April 13, 2016, 12:50:13 AM by eja »

nkalanaga

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Re: Please check my flywheel calculations
« Reply #2 on: April 13, 2016, 01:52:02 AM »
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A faster flywheel will store more energy, but there are two other factors here:

1)  The flywheel is spun up by the gearing.  When the energy is returned to the motor, the speed is reduced by the same gearing, so it probably won't help much.

2)  You'll lose at least a little energy in the gears due to extra friction, which will reduce the efficiency of the flywheel.

If you have room for a double-shaft motor, I'd put the gearhead on the drive train side, and use a faster motor, with the flywheel directly attached to the motor shaft.  Since that would probably require redesigning the entire system, you might be better off ditching the gearhead entirely and using a longer flywheel, for more rotating mass.
N Kalanaga
Be well

mmagliaro

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Re: Please check my flywheel calculations
« Reply #3 on: April 13, 2016, 02:00:18 AM »
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A faster flywheel will store more energy, but there are two other factors here:

1)  The flywheel is spun up by the gearing.  When the energy is returned to the motor, the speed is reduced by the same gearing, so it probably won't help much.

2)  You'll lose at least a little energy in the gears due to extra friction, which will reduce the efficiency of the flywheel.

If you have room for a double-shaft motor, I'd put the gearhead on the drive train side, and use a faster motor, with the flywheel directly attached to the motor shaft.  Since that would probably require redesigning the entire system, you might be better off ditching the gearhead entirely and using a longer flywheel, for more rotating mass.

What you describe at the end of your post is exactly what I am doing.  I have a gearhead motor, and I want to use a double-shaft motor so that the flywheel can go on the backshaft and be spinning at 4x the speed of the output of the gearhead.   That's what I was trying to describe in my opening statement.

As it happens, I tested this with a spare motor lying around on the workbench: a 10mm maxon with a 4:1 gearhead.
I put a flywheel similar in thickness to what I'd have here (only about 1/8" thick), and my physical observations match what my mathematical calculations showed.  Even with the motor revved up pretty good, so the that the flywheel was really spinning fast and the output shaft was only 1/4 of that speed (because of the gearhead), when the power is cut, the motor/flywheel still stops quite rapidly and the output shaft of the gearhead appears to stop dead (which is what 45 mSec would look like to the eye).

I then tried it with a big honking flywheel like you'd find in a Kato diesel.  I get more momentum - enough to see the output shaft keep moving a little after the power is cut, but still pretty feeble considering how fast the motor was spinning and how big the flywheel is. 

The bottom line is that there just isn't enough room in this little engine to put on a big enough flywheel to really add much momentum.  I still wonder if it would add overall smoothness to the motion because it is constantly, instantaneously, smoothing out the rotational speed of the drivetrain.  I'm really interested in that more than in it coasting when power is completely cut off.

GaryHinshaw

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Re: Please check my flywheel calculations
« Reply #4 on: April 13, 2016, 02:27:01 AM »
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[Added - just saw the results of your experiment.  That certainly trumps my words below, but since they were already written, I thought I'd post anyway...]

Interesting question.  I am with you up to the torque calculation: all of you numbers up to that point check out.   But I don't think the torque constant of your motor is the pertinent quantity for the rest of the calculation.  If I understand things correctly (and I am not a motor expert), you need to account for the back emf of the motor, Ve = ωKe, where Ke is the back-emf constant of the motor and ω is its angular velocity.  If the applied voltage to the motor is V0, then we have

V0 = IR + Ve = IR + ωKe

where I is the current draw and R is the resistance of the armature.  If the applied voltage drops to 0, then we would have a dissipation current given by

I'R = -ωKe

and the power dissipated by the motor would be P=I'2R, which steadily drops as the motor slows down.  I think it's a tricky calculation to determine how quickly ω decreases, but you can get a lower limit on the coast time by assuming that ω and I' are constant until the stored energy has dissipated, then they suddenly drop to zero. 

Without knowing R or Ke, I can't determine what I' would be, but you might have one or both numbers available from the manufacturer.   Once you know I', you can estimate the coast time, t, using t = E/P, where E = (1/2)Iω2 is the energy stored in the flywheel.
« Last Edit: April 13, 2016, 02:52:10 AM by GaryHinshaw »

peteski

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Re: Please check my flywheel calculations
« Reply #5 on: April 13, 2016, 03:17:55 AM »
0
Flywheels are pretty much useless when they are rotating slowly.

As far as motor goes, I assume you want the flywheel to be able to coast the locomotive when it encounters power drops (loses power). I also assume that you don't use DCC.  It is a coreless motor. This means that at the time you want the flywheel to do its job the motro is not connected to anything (its terminals are open circuit. If that is the case the motor's rotor will just freewheel (since no current will be flowing through the circuit). Unless the Back EMF is high enough to have the headlight LED conduct. Which is unlikely at slow speeds. So the motor will only present its bearing friction as a factor which resist the flywheel's momentum.
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Chris333

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Re: Please check my flywheel calculations
« Reply #6 on: April 13, 2016, 03:43:19 AM »
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Without math  :scared:  I just think the only way a flywheel would help is if you filled up the whole cab with it. I've seen Z scale locos re-powered with coreless motors and if they have a flywheel it is as wide/tall as the whole loco so they hide it in the cab. Like this:
http://www.die-minilok.de/V20_Bilder/Faulhaber2.jpg
http://www.die-minilok.de/

Plus this is a switcher. If it was a model of Amtrak's newest high speed loco, I say add the flywheel.

 :P

narrowminded

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Re: Please check my flywheel calculations
« Reply #7 on: April 13, 2016, 04:08:45 AM »
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It's too late to think that hard right now but without taking a pencil to an old envelope, as you have figured, in these scales there is very little to be had at low RPM with any size flywheel you might reasonably fit in there.  Not the kinds of numbers that we would wish for that could coast for as much as... seconds! :o ;)  But for your purposes of potentially smoothing the unit out just about any flywheel effect you can add would likely help and sure can't hurt (unless it's not spinning true) as the things that can cause noticeable noise and glitches can be occurring in those same microseconds you found and while small, the percentage increase of flywheel weight for the motor it's on will be measurable.  I would (and am doing it myself) put any flywheel in there that you can fit readily and without real expense.  If it's trouble or expensive to do it, let it go and then go back at it later if a problem surfaces but as you've found, there's just not a lot there and even IF a problem is discovered that you think a flywheel might help... there's STILL not a lot there so not TOO much can be expected. 

BTW, I like what you're doing and like how you go at it. 8)
Mark G.

mmagliaro

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Re: Please check my flywheel calculations
« Reply #8 on: April 13, 2016, 12:48:17 PM »
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Thank you, one and all!

Yes, I think I'm going to bail on this idea.  The double-shaft Faulhabers are not easy to find and they are very expensive.
It would drop-in where my single-shaft motor is now.  The only difference is that I would remove all the tungsten weight I now have in the nose of the engine, and put a flywheel on the motor shaft there.  So I would be losing weight, not gaining, because I would be replacing tungsten with brass.

I wasn't looking for whole seconds of coast.  I was hoping for something like 1/4 second, though. I don't care about the flywheel coasting the engine over dead spots in the track.  To me, that has always been a flawed line of thinking about flywheels.
The reason is that while it helps after the engine gets going, a big part of engine-stall issues is the engine refusing to start when it's been sitting on the track for a minute (or a week).  And if it doesn't have contact to get started, the flywheel is useless.

So rather, I was just looking to get super-smooth, super-uniform speed and motion from the motor and the engine, overall, by having the flywheel in there to even out the motion.  And I am pretty sure that 1/4 second or so of flywheel time would do a great job.  If fact, probably a lot less than that.  But 45 mSec.   Boy, I don't know if that would even out anything. 

peteski

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Re: Please check my flywheel calculations
« Reply #9 on: April 13, 2016, 02:10:26 PM »
0

I wasn't looking for whole seconds of coast.  I was hoping for something like 1/4 second, though. I don't care about the flywheel coasting the engine over dead spots in the track.  To me, that has always been a flawed line of thinking about flywheels.
The reason is that while it helps after the engine gets going, a big part of engine-stall issues is the engine refusing to start when it's been sitting on the track for a minute (or a week).  And if it doesn't have contact to get started, the flywheel is useless.


But it is the coasting through dead spots that N scale flywheels are really good for (due to their small size).

I also don't see how a flywheel will help in getting the engine going. The flywheels are good fo rsmoothing out the motion (that is why pretty much all internal combustion engines use flywheels - to smooth out the rough impulses of torque coming from the explosion in the cylinders).  But the flywheel has inertia which makes it resist changes in speed.  So when the mode; is stopped, the flywheel's inertia  will resist the motor's torque to start moving.   How is that helping in getting the model going?
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mmagliaro

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Re: Please check my flywheel calculations
« Reply #10 on: April 13, 2016, 02:16:30 PM »
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But it is the coasting through dead spots that N scale flywheels are really good for (due to their small size).

I also don't see how a flywheel will help in getting the engine going. The flywheels are good fo rsmoothing out the motion (that is why pretty much all internal combustion engines use flywheels - to smooth out the rough impulses of torque coming from the explosion in the cylinders).  But the flywheel has inertia which makes it resist changes in speed.  So when the mode; is stopped, the flywheel's inertia  will resist the motor's torque to start moving.   How is that helping in getting the model going?
It doesn't.  I said *after* the engine gets going.  The flywheel is only useful for smoothing out the motion of the engine after it gets going.  But while that helps with coasting over dead spots in the track, I don't look at that as its primary usefulness, even though it does help there.  And that's because it doesn't solve the problem of bad pickup, in general.  If there is dirty track, dirty wheels, or an engine with spotty pickup, the flywheel is only going to help after the engine starts moving.  When it won't start because of dirty track, you are still dead in the water.

I am more interested in how a flywheel would improve the motion of the engine, not in helping it overcome dirty track.
So let's assume pickup and dirt are not a problem.  The flywheel still helps to even out the motion of the engine because little variations caused by friction, gear mesh, rod binds, etc, won't affect briefly speed up/slow down the engine as much ... at least, if the flywheel is big enough.

So my quandry is still this: would a flywheel that can store enough energy to keep the engine moving for only 45 mSec be useful for making the overall motion of the drive look smoother?  I kind of doubt it, but I really don't  know.

narrowminded

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Re: Please check my flywheel calculations
« Reply #11 on: April 13, 2016, 05:24:48 PM »
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That really depends on whether there's a problem or not.  If it's already a smooth device then there's nothing for a flywheel to fix.  Wouldn't hurt but also wouldn't help because there was nothing TO help.  There was no problem.  So if we look at the device, what might be helped?  A motor that has distinct cogging action would benefit with some flywheel action, the more distinct the problem the more it would help.  The driven parts and the mass of the loco rolling is also a flywheel of sorts and will tend to smooth out the motion of the whole assembly rolling down the tracks but because it happens further and further towards the end of the chain of events, a cogging motor might not show in rolling performance at the axle but might be noisy back early in the drive train showing up as noise as it accelerates and decelerates, taking the lash out of the gears at that point. A very minor instantaneous glitch could also see some benefit which isn't out of the realm of possibilities but that's really a pickup problem.  While that might be helped by a flywheel carrying it through a dead spot it could be argued that the flywheel is just a belt added to a pair of suspenders (good pickups).  But if none of those factors exist there's nothing to help and why I suggested that if it was easy and minimal cost, go for it.  It can't hurt and for all of the reasons that principles in play suggest are POSSIBLE.  Then, if you do nothing and testing suggests it needs help in those areas you can go back at it and with something a little more specific that you're looking for. And even then, if a flywheel is looked to as a fix for something you already know that it would be of limited benefit, only helping with pretty minor issues.  It can't fix a frog switch stall. ;)

What you're really trying to determine, while all the direct result of very exact and explainable principles, is almost impossible to predict because you don't nor ever will have the data for the near infinite number of subtleties involved.  Those subtleties are what's causing something to happen but you can't realistically predict them or measure them.  That's why some things in practicality are only really determined by test.  It helps to understand the basics and they can predict many problems or just the potential for problems as well as prevent them when they are glaringly obvious  (an out of true, out of balance flywheel) but to know that there IS a subtle problem or the severity of it is just beyond the real world scope of things.  And the understanding of the principles will also help to identify a problem that DOES surface in testing as well as a reasonable direction to head in for a fix.  But even then, it may not be assured other than more testing.  I think you have all of the information that you CAN have at this point and will only be sure once you get it running and see what, if any, physics surprises are lurking in that wonderful mass of clever bits.   8) 

I think that you've thought it through thoroughly and have taken it as far as you can at this point.  You have identified possible areas for attention IF some problems surface that may, but most probably won't.  In light of what you DO know, I can't think of anything you can or should do differently that would incur additional cost or create much additional work until you've had a good test run.  How's that for a bunch of BS?... that I really hope isn't. :)
Mark G.

mmagliaro

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Re: Please check my flywheel calculations
« Reply #12 on: April 13, 2016, 06:32:39 PM »
0
Well, thank you for that intense stream of consciousness, narrow.

Really, if I didn't have to lay out $80 to get the motor, I would "give it a go" and see what happens without hesitation.

Here's the state of things.  I have it running.  I have a test tender in place.  I have test rods (just simple brass strips)
drilled and installed on the rear and center drivers, so I have gotten them quartered really close and gotten the rod holes drilled really close.  How close?  Close enough that I can run it at 0.8 mph.   I will move on to drilling the 3rd hole, and fine-tuning quartering on the front driver set next, and then I've have a fully functional mechanism.

But...
(there's always a but)...

The rotational motion of the drivers isn't perfect.  Now, believe me, it's pretty darn good, and if I left it like this, I don't think I would be complaining.  It never jack-rabbit starts.  It doesn't jam.  By itself or shoving a cut of 10 cars, on a curve or on a straight, it doesn't behave any differently (Can you run it in a box?  Would you, could you, with a fox? LOL!).   

It just has occasional points where it slows down or speeds up ever-so-slightly.  By 5 or 7 mph, this side-effect completely goes away.  So I am being very fussy here.
This is something that, most likely, could be impossible to improve or "fix".  We have a hand-made mechanism here, and there are so many potential sources of tiny errors that could cause such a small effect that I would go mad trying to "fix" it.

And I do not intend to do that.

But if a flywheel could "sweep it under the rug", I'd be all for it.  :D That's what got me thinking this way.


GaryHinshaw

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Re: Please check my flywheel calculations
« Reply #13 on: April 13, 2016, 06:54:49 PM »
0
Have you considered DCC? ;) Decoders with back-emf feedback deal with the variable load issue electronically (but not full power dropouts).

peteski

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Re: Please check my flywheel calculations
« Reply #14 on: April 13, 2016, 06:56:23 PM »
0
Have you considered DCC? ;) Decoders with back-emf feedback deal with the variable load issue electronically (but not full power dropouts).

...and also provide variable momentum.


I suspect that "electronic flywheels" are against Max's DC religion.  :D :trollface:
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