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Matthias Elvenkemper <elvenkemper@unidui.uni-duisburg.de> wrote:

The balancing procedure where you balance wheels independently (with a tapered master weight sticking in the crankpin hole) is in effect a simple way of dynamical balancing! Provided that all work is done with utmost accuracy this procedure will yield the same results like the dynamical balancing I think it is more practical and more accurate to use a balancing machine for the following reasons: the only special thing to make is the master weight, the master weight is easy to make (simple cylindrical shape), less weighing and calculating involved (less risk of errors), procedure of assembly and truing of wheels with master weight is exactly the same as usual with rods (less risk of errors).

The method of individual balancing of each flywheel involves making a special arbor to pass through the flywheel driveshaft hole (left wheel) and pinion shaft hole (right wheel). This arbor must have a tapered hub which must fit perfectly into the flywheel tapered hole. Also for calculating the master weight you have to determine the weight of the crankpin, rollers, retainers etc. because this is also rotating weight. This is not nessessary when using a balancing machine because the crankpin, rollers etc. are installed when the balancing is done. You have to make a special master weight which must be tapered to fit perfectly into the crankpin tapered hole.

Let's compare the procedures; Dynamical balancing on a machine:

After having calculated total reciprocating mass (pistons and recipr. mass of rods), assemble flywheels (with proper thickness thrustwashers) and true them.

Make the master weight. This master weight is cylindrical with same inside diameter as rods and same length as female rod width. The weight of this must be: rotating mass of rods plus % of total reciprocating mass according to the balance factor you want to have.

Disassemble flywheels.

Assemble again with the master weight (incl. thrustwashers) instead of the rods.

True the flywheels.

Bring flywheels to a shop that has a balancing machine and a good name. The shop will put the flywheel assembly on a machine and turn it fast. This simulates the flywheels rotating in a motor. Even a perfectly trued (and perfectly statically balanced) flywheel assembly can stagger due to different center of gravity or different angle between of center of gravity and center of crankpin hole. Dynamical balancing will detect this and by drilling the flywheels individually until no staggering occurs you can cure it.

Pick flywheels up at the shop, disassemble masterweight and assemble flywheels with rods, true them. End of story.
 
You think this is complicated? Well read on for the procedure involving separate balancing of flywheels:

After having calculated total reciprocating mass (pistons and recipr. mass of rods), assemble flywheels (with proper thickness thrustwashers) and true them.

Disassemble flywheels and weigh crankpin, crankpin nuts, crankpin thrust washers, all crankpin rollers, retainers, lockwashers, lockwasher screws.

Determine master weight: Calculate the sum of rotating mass of rods plus weight of crankpin, rollers, retainers, thrustwashers,etc., plus % of reciprocating weight according to the balance factor you want to have. Divide by two and this is the correct weight of your master weight.

Make master weight with tapered hole.

Make special tapered arbor that passes through driveshaft (pinion shaft) hole.

Assemble one flywheel with arbor and master weight put on parallel ways or knife edges and balance it.

Do the same with the other flywheel.

Assemble flywheels with rods and true them. End of this story.

How boys, can you ever be absolutely sure that your arbor sits as perfect in the driveshaft (or pinionshaft hole) as the drivepin and pinion pin  tightened with the nut and the recommended torque? How can you be sure that the masterweight sits as perfect in the crankpin hole as the crankpin tightened with nuts? Plus, the making of arbor and master weight is something for a very skilled machinist. For the less talented this is additional cost. (ok, you save what the shop charges for dynamical balancing but I think you still end up cheaper, and with a result as perfect it can be).

Still, for the hard-core do-it-yourself guy with advanced machining skills the individual balancing (quasi-dynamical) method might be worth considering.
 

Response
 
From: Guy <guyiii@home.com>
Another great post, Mat
> Guy (and Moen):
> The balancing procedure where you balance wheels
> independently (with a tapered master weight sticking
> in the crankpin hole) is in effect a simple way of
> dynamical balancing! Provided that all work is
> done with utmost accuracy this procedure will
> yield the same results like the dynamical balancing
> on a machine.

Wellll - I consider one-flywheel-at-a-time as static balancing, period.  IMHO balancing both wheels together is not worthy of my time....think about this - if the driveside wheel is more massive, how do you know which one to drill to achieve balance with two-at-a-time??  ...I think dynamic also incorporates an additional check in that it is done at a particular RPM....if the weights calculated and placed on the wheels are "incorrect" it WILL NOT balance at the desired RPM....

Balancing involves more than the unbalanced wts, it addresses the unbalanced FORCES as in F= ma, so higher RPM involves much greater reciprocating acceleration / deceleration, yes?

As to which is "easier" between dynamic & static - I respectfully disagree (based partly upon my desire to do it myself..).....as to the procedure for static, I'll address that below:

> The method of individual balancing of each
> flywheel involves making a special arbor to
> pass through the flywheel driveshaft hole
> (left wheel) and pinion shaft hole (right wheel).
> This arbor must have a tapered hub which must
> fit perfectly into the flywheel tapered hole.

Commercial shafts & bobweights are available for HD, or can be cooked up at home (with a lathe)...if one has a lathe and can't do this they have no business messing about with the flywheels anyway....

> Well read on for
> the procedure involving separate balancing
> of flywheels:
> After having calculated total reciprocating mass
> (pistons and recipr. mass of rods), assemble
> flywheels (with proper thickness thrustwashers)
> and true them.

Au contaire - no need to assemble wheels at all for one-at-a-time balancing:

1. One removes the nice shiny new parts from boxes, degreases them, polishes rods if desired, weighs pistons & equalize piston wts;

2. Weigh rods (rotat & recip), pistons, rings, wristpins, circlips, crankpin, rollers & cages, nuts & retainers, etc for determination of 100% rotating plus ___% reciprocating;

3.  Add 100% rotating + ___% recip; divide sum by 2 (balance wt");

4.  Using commercial static balance kit, make up "balance wt" and install on wheel, put wheel on commercial shaft, balance on knife edge; drill side of wheel rim trial & error (or add "Mallory metal") until it balances;

5. Put same balance wt on other wheel, repeat;

6. Clean, assemble & true wheels.  End of Story (except for happy
riding)

> How boys, can you ever be absolutely
> sure that your arbor sits as perfect in the
> driveshaft (or pinionshaft hole) as the
> drivepin and pinion pin tightened with the
> nut and the recommended torque?

I do damn good work, if I do say so myself...

> Still, for the hard-core do-it-yourself guy
> with advanced machining skills the individual
> balancing (quasi-dynamical) method might
> be worth considering.

OK
 
 
From: "Moen" <moen@get2net.dk>
Guy,Could you expand on the rpm aspect? I'm not sure I get that. Sure F = ma, but why would the m's change with rpm?
 
 
From: Guy <guyiii@home.com>
Actually, this wasn't a very well developed thought (dang, Moen -you call me on all my baloney), but with this balance factor discussion we've kind of focused on the wts of components (the
"m" of F=ma) and when the wheels are statically balanced and at rest, a=0 so F=0 ....but each time the rod/piston cycle they start from dead stop, accelerate to warp speed, decelerate to zero, etc...as RPMs increase piston speed will likewise increase and accel/decel rate "increase" ...so at 1000 RPM piston acceleration is X, at 2000 RPM piston acceleration 2X, etc

Likewise, at 1000 RPM reciprocating component acceleration is X, at 2000 RPM recip component acceleration 2X, etc

So, it merely seems to me that within the stew that is "balance factor" that we discussed is a compromise on the the variable forces that the rods exert on the crankpin....

a "derivative" of F = ma is impulse & momentum, that is: F dt = change in mv....each cylinder firing gives an impulse to the flywheel

so during the flywheel rotation, the flywheels rotation is not uniform or smooth, but accellerates under firing, decelerates from load and friction

(side note - if firing impulse changes flywheel momentum and affects "balance", then compression ratio and camming will as well)

Another side note: Cotten (I think) stated that lower balance factors go with faster accelerating flywheels, and I agree because I think we're basically making the wheels a little lighter, and a higher balance factor makes the wheels effectively a little heavier  (back to F=ma with light / heavy flywheels tradeoff)
 

From: "Cotten" <Liberty@npoint.net>
Guyii wrote:
> Another side note: Cotten (I think) stated that lower balance factors
> go with faster accelerating flywheels, and I agree because I think
> we're basically making the wheels a little lighter, and a higher
> balance factor makes the wheels effectively a little heavier  (back to
> F=ma with light / heavy flywheels tradeoff)
> Did this babble make sense?

Uh, no, ....your total flymass is independent of factor...Cotten
 

From: "Moen" <moen@get2net.dk>
Guy,NOW you've made my head hurt...
> with this balance factor
> discussion we've kind of focused on the wts of components (the
> "m" of F=ma) and when the wheels are statically balanced and at
> rest, a=0 so F=0 ....but each time the rod/piston cycle they start
> from dead stop, accelerate to warp speed, decelerate to zero,
> etc...as RPMs increase piston speed will likewise increase and
> accel/decel rate "increase"...so at 1000 RPM piston acceleration is
> X, at 2000 RPM piston acceleration 2X, etc

OK. Linear acc.

> Likewise, at 1000 RPM reciprocating component acceleration is X,
> at 2000 RPM recip component acceleration 2X, etc

OK, but angular this time.

Our domesticated ME student will need to make a kinematic equation for how x (linear) & x (angular) goes together. Sadly my mind is blank on this...

> So, it merely seems to me that within the stew that is "balance
> factor" that we discussed is a compromise on the the variable
> forces that the rods exert on the crankpin....

Yes, quite possibly!

> a "derivative" of F = ma is impulse & momentum, that is: F dt =
> change in mv....each cylinder firing gives an impulse to the flywheel
> so during the flywheel rotation, the flywheels rotation is not uniform
> or smooth, but accellerates under firing, decelerates from load and
> friction

Yes. But that's a whole other thing! I don't think this is very important from a vibration perspective, but for the whole picture our student needs another kinematic equation, and to
superimpose it on top of the first (I think).

> (side note - if firing impulse changes flywheel momentum and
> affects "balance", then compression ratio and camming will as
> well)

Where did "balance" come from all of a sudden?
 
> Another side note: Cotten (I think) stated that lower balance factors
> go with faster accelerating flywheels, and I agree because I think
> we're basically making the wheels a little lighter, and a higher
> balance factor makes the wheels effectively a little heavier  (back to
> F=ma with light / heavy flywheels tradeoff)

As Cotten just said, flymass is not tied to the balancing factor (extreme case: no flywheel, just a "bob weight". The factor could still be real high, and the wheels real light!)

> Did this babble make sense?  What we need is a practical ME who
> deals in machine design day-to-day...

Well, yes. But where did the "rpm variable balance factor" enter the picture?
 

From: TGSawyer48@aol.com
Moen, Are you considering the distance between main bearing and rod bearings or the overall crankshaft length? Both are important but not in exactly the same way. Are you considering the journal diameter as a percentage of the throw and separately the crankpin diameter as a percentage of the throw. Both of these are important too. As you must have noticed by now there are many more factors involved than you are contemplating.
 
 
From: "Moen" <moen@get2net.dk>
Tom,Omigawd, no. I wasn't considering any of those! Moen
 

From: Matthias Elvenkemper <elvenkemper@unidui.uni-duisburg.de>
Hi Guy,
> > if you use HD 45" flywheels with 741 barrels you will
> > not be able to get a displacement of 50 cubic inches.
>
> Right, I was assuming 45 barrels....

OK, these will indeed give you 50 cu.inches with HD 45 wheels.Max Bubeck told me
that his experience is that the 50inchers
are the fastest Scouts because they are still able to stand high
r.p.m. Going higher with stroke only increases low end torque
but piston speeds get critical.

> > A friend will lend (borrow) me a pair of Chief flywheels,
> > rods and an original piston. Will have balance factor
> > checked with the same procedures.
> Great, can you also post the basic data on those??

Sure I will.

> Au contaire - no need to assemble wheels at all for one-at-a-time
> balancing:

Disagree, see below.

> 1. One removes the nice shiny new parts from boxes, degreases
> them, polishes rods if desired, weighs pistons & equalize piston
> wts;
>
> 2. Weigh rods (rotat & recip), pistons, rings, wristpins, circlips,
> crankpin, rollers & cages, nuts & retainers, etc for determination of
> 100% rotating plus ___% reciprocating;

You forgot the crankpin thrustwashers and thrustwasher pins.These are also rotating weight. Different thickness means different weight (not much but we want to minimize all sources
of errors, don't we) that's why I said first assemble flywheels until you know which thrustwashers give you proper rod sideplay.
 
 
From: Fred Dufrene <Nominoee@compuserve.com>
>>I'm not sure I can see why a stroked engine in a stock chassis would need another balance factor, but maybe I'm missing something?<<

I've surely missed something me too but :

1) at the same rpm, the speed of piston is higher with a stroked motor. right ?
2) for a given object of a given weight its weight increase with speed. right ?

so with a higher speed an object will be heavier. so the balance factor will need to be different to "adjust" this difference of weight.
 
 
From: Duff <MICHIGANDER@Worldnet.att.net>
Fred, You have some of it right, but there will be a increase in weight because the rod combo is different from origional and the piston weight can change because you use some other piston other than origional ones.
And of course the flywheel weight will be different if you use a different set of wheels other than origional to the motor in question. The piston travel will increase, however the rpm of the original motor may not increase just because you stroked the motor. It is possible your rpm would not have to be as high if the cubic inches net more power, thus more hp from a given motor or more torque.
Indian was rather insiteful when they only stroked the 74 motor to 80 cuin, they kept the same rods the pistons were almost the same weight ( Ind only had to turn off some extra length at the bottom of the piston) the flywheels weigh the same. They only went from the standard 74 head
to the Bonn type head where the piston came up into the head. Pretty crafty! So the balance factor did not have to change much at all, infact the 80 motor will run rather nicely with 64% same as the 74 motor does.
Because there is a stroking factor doesn't mean there is a automatic weight increase nor does higher rpm mean that the wheels and assembly become heavier because of it.
 
 
From: Fred Dufrene <Nominoee@compuserve.com>
Duff,when I said stroker I forgot to say I included the "standard" 80" In fact I thought only about the 80 but this morning I hadn't very clear ideas :-) about the rpm. I didn't say to stroke an engine gives you more rpm (it will even be the contrary), I said stroking an engine will increase piston speed and so it will increase its moving weight.
Even if Indian stroked the 74 to only 80 with same piston's weight it would give a different balancing... I don't know in what percentage it "plays"
 
 
From: Matthias Elvenkemper <elvenkemper@unidui.uni-duisburg.de>
Duff wrote:
>  Indian was rather insiteful when they only stroked the 74 motor to 80
> cuin, they kept the same rods the pistons were almost the same weight (
> Ind only had to turn off some extra length at the bottom of the piston)
>  the flywheels weigh the same. They only went from the standard 74 head
> to the Bonn type head where the piston came up into the head. Pretty
> crafty! So the ballence factor did not have to change much at all,
> infact the 80 motor will run rather nicely with 64% same as the 74 motor
> does.
Even with identical pistons and rods the balance factor changes when you change stroke. Of course this can be corrected by according change flywheel counterweight. I have no data on 80" balance factor but rule of thumb would make me guess it should be factored lower than
the 74". You very well may be right that the 80" factor isn't much different though.
 
Duff,if you increase the stroke without changing anything else, you will lower the factor. This is so logical that I think I do not have to describe it. However - and I said that in my former message - it is of course possible to reach the same balance factor with the longer stroke flywheels, you just need more counterweight.
 
 
From: Duff <MICHIGANDER@Worldnet.att.net>
Hi Mat
The factor is the same 64% I got that percentage from Paul Gambaccini old Ind dealer and close friend now gone.