Date: Wed, 21 Jan 1998 07:01:45 -0800
From: Robert Harris 
Subject: Math needed.
To: 'Dave Williams' 

Fiction Check.  Think I have fingered out how rod offset works.  Need math to
see if its worth it. Bear with me.

Piston Top Dead Center - when the piston don't go no higher.
Rod Verticality - when the rod is perfectly parallel to the cylinder sides and
"vertical" with respect to a piston top "horizontal"
Crank Verticality - when the crank is perfectly parallel to the cylinder sides
and "vertical" with respect to a piston top "horizontal"

With no offset, Piston Top Dead Center occurs when both the rod and crank
simultaneously achieve verticality.

With pin offset, the rod goes vertical post/prior(depending on direction of
offset) to the crank going vertical. Bisecting the angle between the
verticality's gives Piston Top Dead Center.

Since the piston reach's top dead center before the crank, it therefore reach's
bottom dead center after the crank. This is because with the pin offset, the
rod can achieve verticality only on the side the pin is offset to.  We bisected
the angle to get Piston TDC, and did the same to get Piston BDC. If we sum the
two angle's, or simple use the crankshaft degree's between Rod Verticality and
Crank Verticality for piston TDC, we get " Offset Delta " in degrees.

With no offset, the classic Otto cycle is divided into four parts with 180
degree duration.  (Course cam timing smudges that up pretty badly, but the
crank still thinks 180). With pin offset, we subtract the " Offset Delta" from
the two cycles that occur in the direction of the offset and add "Offset Delta"
to the remaining two cycles.

Assuming the engine turns clockwise facing it, offsetting the pin to the left
results in shortening the Compression and Exhaust cycle by "Offset Delta" and
lengthening the Intake and Power strokes by the same amount of "Offset Delta".
This extra rotational degrees are added at to both Top (offset delta/2) and
Bottom (offset delta/2) dead center times.

First lets deal with Heywoods slap crap.  Since the rod angularity is decreased
during compression and exhaust, we get less skirt friction during this time.
However, this reduction is countered by increases during power and intake
because of greater rod angularity - thus more "slap".  No matter which way the
pin is offset, the "slap" is increased in two strokes and decreased in the
other two - so obviously since production is offset to the right, the most slap
must occur during power and this offset reduces that slapping sound and quote
friction.  The friction argument sounds like feces examination to cover an
esthetic decision by long dead engineers to make the damn thing quieter.

My thought is since rod angularity is increased during the power stroke, this
increased angularity is seen by the engine as an increase in effective stroke
during the power cycle, resulting in more torque (longer lever arm effect). The
shortened period of time during compression is also a good thing, as it lessons
the time portion of the detonation equation (time, pressure, temp). The
shortened exhaust cycle is made up by opening valve tad earlier if needed.  The
lengthened intake cycle allows more time for charge to fill cylinder.

An unfortunate side effect is a modest decrease in displacement which occurs no
matter which side its offset to.  This is because the effective stroke is sum
of the crank throw, minus the effect of the offset delta. The piston neither
rises as high nor depresses as low in the cylinder as if there was no offset.

But wait - for only 19.95 we include the ginzo knives with your purchase.
Maximum pressure during combustion is developed near top dead center -
typically around 12 degrees after TDC, with combustion extinguishing by about
20 degrees ATDC.  Increasing the time spent near TDC increases the BMEP, dus
more power.  Offsetting the pin to the left increases the time at TDC by
slowing down the piston movement after piston TDC (more rotational degrees
remember). So we pick up power with increased rod angularity when we want it,
increased BMEP and we give up some "friction" loss.

How much pin offset?  Need math to finger that out and create some interest
making graphs etc.  Suspect that since stroking increase's rod angularity
significantly without slap crap disaster, that the pin can be offset
significantly before getting into danger.

Increasing rod length results in decreasing rod angularity decreasing torque
lever arm, however it increased dwell time near TDC, more than compensating.
The combination of the two seems to result in something sweet.

So am I full of shit or does this begin to make sense??


Date: Thu, 22 Jan 1998 00:20:31 -0800
From: Robert Harris 
Subject: RE: Math needed.
To: 'Dave Williams' 

Most people need numbers to visualize, I generally need them only to measure
and quantify. Bet you thought I was kidding when I said become one with
machine.  No rush on numbers -

-----Original Message-----
From:   Dave Williams [SMTP:dave.williams@chaos.lrk.ar.us]
Sent:   Wednesday, January 21, 1998 9:58 AM
To:     bob@bobthecomputerguy.com
Subject:        Math needed.


-> So am I full of shit or does this begin to make sense??

I'd say you've managed to get the whole thing figured out from first
principles.  A-number-1!

The trick, as you found, is that a cranked link doesn't move in a
symmetrical path wrt rotation.  A move/degree curve is egg-shaped, with
the pointy end up.  When you offset the pin you incline the egg, which
affects the cyclic events, etc.

I really am going to find and copy that stuff, but I'm against the wall
right now.