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On this page: Basic
Theory.
Page2: Balancing.
Page 3: Chassis Stiffness. Page
4: Balance Factors. Page 5: Crankshaft
Windup.
From: Cotten <liberty@npoint.net>
When I read the military manual method,
which was apparently the same as your 741/640B manual, I proceeded to back
calculate factors whenever a Chief came in for rebuild. Usually they figured
about 65%, which our posts last year confirmed as appropriate. The arbitrary
addition of weight in the Q&A book is no more confusing than some of
the primitive Milwaukee speed balancing recipes, where a prescribed number
of holes of a given size were to be drilled, independent of piston mass,
and actual knife-edge trials weren't even mentioned.
Bless the forgiving nature of the V-twin
design! Harmonics and vibrations aside, balancing seems to affect the performance
of a motor as much as anything. I have found that low factored Milwaukee
machines jump like dirtbikes with a short powerband, while high factored
ones pull rpm slower, but useable over a much larger range of rpm. Balancing
is apparently a tuning procedure for simple single crankpin motors, whereas
opposed or multiple throw motors require accurate factoring to prevent
them from pulling themselves apart.
Anyone else see it this way?
From: Duff <MICHIGANDER@Worldnet.att.net>
Na, you (Cotten) are not the last to find
this out! I'll give you a hint, you may take a set of Indian rods, weigh
them by supporting the rod by the wrist pin. Let the big end sit on a gram
scale and get it's weight, do the same for both rods. The weight is then
used for the recip total
weight like piston complete etc. This
get ya around 64% which is about what Indian's end's up with.
From: Cotten <liberty@npoint.net>
That's been my experience also. There
is a lot of methods of supporting the rod for weighing: platforms,thread,
even roller bearings on a lab apparatus. I have even whipped up a spirit
level with nylon pucks to slip into the rod ends to set up for perfect
level when weighed. It's a bit of overkill, when I can often resort to
what I call the "Clymer" method: After washing all lube out of the crank,
I found that I can set the entire flywheel assembly on a v-block on my
scale table, and then rotate the crankpin position to where the rod is
level when laying on the scale pan. Care must be taken to center the rods,
and the other rod must not interfere. And the rod end must sit square and
center on the pan. I obtain rodtop weights that are quite close to readings
on the disassembled rod. (Face it, a coupla grams inaccuracy isn't a lot
of final factor)
Of course, I usually only resort to this
when someone brings me a trued assembly, and then usually for a check of
balance factor.
From: Matthias Elvenkemper <elvenkemper@unidui.uni-duisburg.de>
Listers, Cotten offers a good example
with the comparison of low and high factored Milwaukie bikes. Multiple
throw motors (e.g. four cylinder inline) have little problems with vibrations
caused by first order momentum because these are balanced by the offset
crank. What counts in multiple throw motors is picky choosing rods and
pistons for equal weight because differences cause vibrations caused by
second (and higher) order momentum that can be very nasty.
A single crankpin motor produces vibrations
caused by reciprocating (up and down) forces and rotating (fore and aft)
forces. The up and down forces are produced by the pistons (with pin, retainers,
rings) and the reciprocating mass (upper part) of the rods. The longer
the rods (weight equal) the higher the reciprocating mass.
The balance factor determines the ratio
of "up and down" versus "fore and aft" vibrations.
In practice (apart from the r.p.m. range
where the "worst" vibrations occur) the "desirable" ratio of up and down
vs. fore and aft vibrations depends to a large extend on the frame design
and how the motor is fixed to the frame.
In all single crankpin motors ALL of the
rotating masses are balanced by the flywheel counterweight. Plus part of
the reciprocating masses are balanced.
The "balance factor" is the percentage of the reciprocating masses being balanced by flywheel counterweight.
A motor with low balance factor thus vibrates
up and down a lot and vibrates fore and aft a little. The extreme case
would be a vertically mounted single cylinder motor with ZERO balance factor
(only rotating masses balanced). This motor vibrates only up and down.
The same motor 100% balanced would only vibrate fore and aft. The basics
apply to V-Twins as well, with other things equal the wider the cylinder
angle the more the motor vibrates fore and aft. Thus: An identical flywheel
assembly put in different cylinder angle motors would have a different
balance factor.
Kinematics (the angle of the rod on the
downstroke) play the role here.
Practical example: Late Indian Sport Scouts
have a balance factor of 82%. If you put Sport Scout flywheels rods and
pistons in 45 degree cases the balance factor rises to a little over 84%.
This small difference would be only noticable in racing motors if at all.
However: Theory fits Cotten's Milwaukie
bikes comparison perfectly. The low factored bike accelerates fast (other
things equal due to the weight of flywheels being lower with less counterweight)
but has a limited r.p.m. range because reciprocating masses cause bad up
and down vibrations when reaching higher r.p.m. The same bike high factored
accelerates slower (other things equal due to higher weight of flywheels
with heavier counterweight) but has a broader usable r.p.m. range because
bad vibrations occur at much higher r.p.m.
Also (I think) this explains why Indian
had a lower balance factor on the Chief compared to the Sport Scout. The
essence of the Chief is "you can't beat cubes" low end torque which makes
it a good puller at low r.p.m. The long stroke of the Chief would prohibit
high r.p.m anyway because of dangerous piston speeds. So it would not make
sense to choose a high balance factor.
The essence of the 45 101 and Sport Scout is trading cubes and low end torque for power at high r.p.m. This is no threat for motor life because piston speed in a 45" Scout is approx. 27% lower than in a 74" Chief. So it was logical to choose a high balance factor to support the (piston speed wise) ability to stand high r.p.m.
From: Guy <guyiii@home.com>
Mat - thank you for your efforts, and
you do a great job in English, so please don't take this as criticism,
just a helping hand..."momentum" = mass x velocity, you may have meant
"moment" (force x distance), but I think first order "couples" is a more
proper terminology (actually I think V-twins have an engine vibration phenomena
described as a "rocking couple")...
I looked up my info on balance factors,
a dynamic shop used 52% to 56% (but I think this is apples & oranges)...S&S
uses 60% for static...of course their wheels are lighter than stock....so
it seems reasonable that "low" factors are used with lighter wheels for
quicker revs, and vice-versa, heavy wheels and higher factor....I still
have an unsettled feeling that some of these "factor" equations are using
different input, so that 85% may be70% by a different definition.
From: "Dave Clements" <clements@la-tierra.com>
Guy: Rocking couple is only found in v
twins that have the rods side by side not in engines with male and female
rods.
From: Matthias Elvenkemper <elvenkemper@unidui.uni-duisburg.de>
It is clear that the balancing method
shown in the Indian manuals is better than nothing. And it might be overengineering
to do more on the low r.p.m. Chief motor if you are sure that both flywheel
sides are a matching pair. If left and right flywheel have different center
of gravity you can not find out with the static method. Even if you do
the job perfect the flywheels will stagger when running and cause more
vibrations than nessessary.
From: Cotten <liberty@npoint.net>
I found a copy of a page out of a loose-bound
Milwaukee manual that outlines a static balance method that requires the
centerpoint of 'balance' for each rod be measured so as to determine the
weight of the top of the rod by proportional distance. The rod is balanced
upon a knife edge and marked at its center of gravity. The distance between
this and the wristpin bore center is measured, as well as the total center-to-center
(wristpin to crankpin) distance. The reciprocating weight of the rod is
then determined by multiplying the total weight of the rod by the marked
distance, and then dividing by the total center-to-center distance.
Am I the last to learn of this? It does
not seem to reproduce my 'hanging' method of determining the weight of
the rod. (Even more confusing is that it calls for a factor of 50%.) Any
clues?
From: Guy <guyiii@home.com>
Good info, Cotten - I've not seen it....I
think the problem with determining the rod centroid this way is with actually
being able to balance it on a knife edge, marking it, and then measuring....before/during/after
balancing my Shovelhead stroker, I was paying attention to my car hot rod
mags....the hot rod aftermarket performance industry (per the photos -
no text) uses
the hanging method....I was able to use
the hanging method and weigh each end of the rod and the sum of those ends
equalled the total rod weight - mathematically, both methods should give
the same result....50% factor doesn't surprise me (the quoted 85% does!)
some balance shops use 52-56% of reciprocating wt in dynamic balance approach,
S & S quotes 60% of recip wt for their wheels....I think there is still
much confusion (in my mind anyway) as to whether these various static balance
recipes use consistent definitions and methodology.....so, I'm still looking
for some info to establish what static balance % to use on Indians when
using what I consider to be a "correct" definition (ie, clear separation
of rotating vs reciprocating wt....100% rotating wt plus ____% recip
wt, the sum divided by two and bob wt placed on each wheel).
Also, I'm unsure of the practical effect
on balancing when we have a heavy drive-side wheel & lighter pinion-side
wheel (stock HD) vs both wheels weighing the same (T&O, S&S, modified
HD)...
From: Matthias Elvenkemper <elvenkemper@unidui.uni-duisburg.de>
Cotten, the formula you mention gives
you the ROTATING mass of the rod not the reciprocating!
This method has been used for decades,
it is a good approximation for determining reciprocating and rotating mass
of rods though there are more accurate methods. Of course it does NOT call
for a balance factor of 50%.
From: Cotten <liberty@npoint.net>
Now I am really confused! Is not the top
of the rods (in effect) the reciprocating mass, and the crank end (in effect)
the rotating mass?
From: Matthias Elvenkemper <elvenkemper@unidui.uni-duisburg.de>
Cotten,you are right that top end of rod
is (in effect) reciprocating mass and crank end is in (effect) rotating
mass. I don't understand why you are confused. Let's put forth your formula
mathematically:
You described the formula you found in
the HD manual as follows:
"The rod is balanced upon a knife edge
and marked at its center of gravity. The distance between this and the
wristpin bore center is measured, as well as the total center-to-center
(wristpin to crankpin) distance. The reciprocating weight of the rod is
then determined by multiplying the total weight of the rod by the marked
distance, and then dividing by the total center-to-center distance."
I used the following symbols:
"REC.W" means reciprocating weight (of
rod)
"ROT.W" means rotating weight (of rod)
"TW" means total weight
"D-REC" means distance from center of
gravity to center of wristpin bore
"TL" means total length of rod (from center
to center of bores)
Thus your formula is:
ROT.W = TW x D-REC / TL
Assume rod A of my example: TW = 500 grams,
TL = 20 cm
D-REC = 14 cm
With your formula you get:
ROT.W = 500 grams x 14 / 20
ROT.W = 350 grams
For reciprocating weight you get the same
as in my "direct" formula for REC.W :
REC.W = TW - ROT.W
REC.W = 500 grams - 350 grams
REC.W = 150 grams
The "direct" formula for calculating the reciprocating weight of the rod which is widely used gives (of course the same results):
REC.W = TW x ( 1 - D-REC / TL )
REC.W = 500 grams x ( 1 - 14 / 20 )
REC.W = 500 grams x 0.3
REC.W = 150 grams
quod erat demonstrandum
So as I said the formula you mentioned gives rotating mass of rod. Subtract from total rod weight and you have reciprocating mass of rod. Get the same result from the "direct" formula for reciprocating mass I mentioned.
Hope this stops the confusion...Mat
From: Cotten <liberty@npoint.net>
Mat! My confusion hits me at this step:
The H-D method figured "The reciprocating
weight of the rod is then determined by multiplying the total weight of
the rod by the marked distance, and then dividing by the total center-to-center
distance."
But you symbolized it as:
Thus your formula is:
ROT.W = TW x D-REC / TL
It seems like I said:
REC.W = TW x D-REC / TL
From: Matthias Elvenkemper <elvenkemper@unidui.uni-duisburg.de>
The formula in your HD manual has a typo
(slip). This error is the source of confusion and I should have pointed
out the slip.
You are right if you formalize what the
HD manual says as:
REC.W = TW x D-REC / TL
But this is wrong! TW x D-REC / TL gives
the rotating mass!
Each rod that I know of has higher rotating
mass than reciprocating mass which is logical because more mass is concentrated
near and around the massive (rotating) crankpin. A lot less mass is concentrated
near and around the skinny (reciprocating) wristpin.
Again, I am sure the source of the confusion
is the typo in the
HD manual.
Cotten wrote:
> Mat! I did overlook the obvious! D-REC
will always be the longer portion of the rod. (and thanx for saving my
self-esteem!)
From: Guy <guyiii@home.com>
Thanks, Mat....now we're getting somewhere....This
is exactly the method that I use as recommended by S&S (except
I use "hanging" method of rod weighing)...I then multiply the recip wt
by 60%, add it to 100% of the rotating wt, divide the sum by 2 and put
that wt on each flywheel via bob wts to balance them separately....
As an engineer I appreciate that this is
the most "scientific" approach to STATIC balancing, but we must keep in
mind that underlying assumptions - such as that the lower 50% of the rod
wt is rotating wt and not reciprocating wt are not strictly true but useful
simplifications so we can solve the problem.....and with those assumptions,
as well as other engine component characteristics and RPM range, we EMPIRICALLY
determine acceptable balance factors....
So, now that we know we're both using
the term "balance factor" the same way....my questions remain:
what is there about Indian Chiefs that requires an 85% balance factor when HD, T&O, S&S, and others use 50% to 55% to 60% on HD big twins (and sporties)??
What does S&S recommend for big Indians??
Do Scouts really need to be balanced differently from Chiefs?
If the referenced "50%" is not a balance
factor, what is it?
From: Matthias Elvenkemper <elvenkemper@unidui.uni-duisburg.de>
Guy, many misunderstandings are due to
a wrong understanding of the term "balance factor". The one and only worldwide
used definition is: The percentage of total reciprocating weight balanced
by the counterweight in the flywheels.
Total reciprocating weight is both pistons (plus rings, pins, retaining rings) plus the reciprocating weight of the rods.
This is the process:
It is essential that you know the balance
factor that the motor should have.
- weigh the rods
- measure length of rods (from center
of
crankpin bore to center
of wristpin bore)
- balance rod on knife edge
- mark the center of gravity
- measure distance between this and center
of wristpin bore
Now:
"REC.W" means reciprocating weight (of
rod)
"ROT.W" means rotating weight (of rod)
"TW" means total weight
"D-REC" means distance from center of
gravity to center of wristpin bore
"TL" means total length of rod (from center
to center of bores)
The formula for calculating the reciprocating weight of the rod is:
REC.W = TW x ( 1 - D-REC / TL )
Subtract reciprocating weight from total weight gives you rotating weight:
ROT.W = TW - REC.W
Calculate reciprocating weight of both
rods.
Total reciprocation weight is the sum
of REC.W of both rods
plus the weight of both pistons with wristpin,
rings, retaining rings.
The BALANCE FACTOR is the percentage of total reciprocating weight balanced by the flywheel counterweight.
It is clear that when you balance a rod on a knife edge 50% of the weight will be on the left and the other 50% will be on the right, don't get it wrong here: What counts is the weight distribution of the rod!
For example: Imagine two rods of identical
weight:
TW = 500 grams
and identical length:
TL = 20 cm (centimeters).
You put both rods on a knife edge and
measure D-REC 14 cm for rod A and D-REC 12 cm for rod B.
Calculate reciprocating weight of rod A and rod B with the formula:
rod A:
REC.W = 500g x ( 1 - 14 / 20 )
REC.W = 500g x ( 1 - 0.7 )
REC.W = 500g x 0.3
REC.W = 150g
rod B:
REC.W = 500g x ( 1 - 12 / 20 )
REC.W = 500g x ( 1 - 0.6 )
REC.W = 500g x 0.4
REC.W = 200g
The results are intuitively evident. Rod
A has a lower reciprocating weight because more of its weight is
distributed around the big end (thus higher rotating weight)