Rod/stroke ratio, piston speeds, etc. (was: Hey my Chevy has Ford Rods !!!!!!)

[The Dupuis]

Any point on a circle can be described by radius squared = x squared plus y
squared.  Exactly the same as Pythagorean's theorem, z squared = x squared
plus y squared.  In this case, the radius is half of the stroke of your
crankshaft.  The connecting rod also forms a right triangle when referenced
to the centreline of the bore and the rod journal, and can be described by
P's theorem.  Look at the crank first.

For your stroke of 1.74", in order for the crank to gain 0.010" of y value
(come up the bore), X must increase by [the root of 1.740" squared minus
1.730" squared], or 0.186".  Let's look at the rod.

Since the rod is a triangle too, moving the rod bearing end sideways will
lower the crown of the piston.  If you move the rod bearing 0.186" sideways,
you will drop the piston [the root of 5.700" squared - 0.186" squared], or
0.003".  The piston will drop the same amount as a function of x at either
end of the stroke, whether y is increasing or decreasing.  Therefore, at the
top of the stroke the piston drops 0.013" with 0.010" of crankshaft stroke,
and at the bottom of the stroke the piston moves up by only 0.007" with
0.010" of crankshaft stroke.  Both examples use the same amount of
crankshaft stroke, therefore the same amount of rotation, or in this case,
one tenth of a degree.

Answer your question?
Matt

> -----Original Message-----
> From: owner-diy_efi at diy-efi.org [mailto:owner-diy_efi at diy-efi.org]On
> Behalf Of elcamino73
> Sent: January 14, 2002 9:34 PM
> To: diy_efi at diy-efi.org
> Subject: Re: Rod/stroke ratio, piston speeds, etc. (was: Hey my Chevy
> has Ford Rods !!!!!!)
>
>
> I do not understand how dwell time can be different at BDC than TDC!
> Enlighten me please.
> And we all have too much time on our hands.
>
> ----- Original Message -----
> From: "The Dupuis" <dupuis10 at telusplanet.net>
> To: <diy_efi at diy-efi.org>
> Sent: Wednesday, January 09, 2002 9:04 PM
> Subject: Rod/stroke ratio, piston speeds, etc. (was: Hey my Chevy has Ford
> Rods !!!!!!)
>
>
> > Just for kicks, I've drawn two 350 chevs in AutoCAD, accurate
> to 6 decimal
> > places.  One has 5.7" rods and the other has 6.2" rods.  I've broken
> > crankshaft rotation into 5 degree chunks.  Also, I found the point that
> the
> > crank and the rod are at right angles to each other, indicating peak
> piston
> > speed.  Finally, I've plotted how many degrees it takes for both engines
> to
> > "dwell" within .010" of TDC and BDC.  I'll give it to you in
> inches first:
> >
> > At the bottom of the stroke, a piston on 5.7" rods travels
> 0.004870" and a
> > piston on 6.2" rods moves 0.005043".  At the top of the stroke
> the piston
> on
> > a 5.7" rod travels 0.009139" and the piston on a 6.2" rod travels
> 0.008967"
> > Maximum piston speed for the 5.7" rod is at 163.024501 degrees and with
> the
> > 6.2" rod it happens at 164.323482 degrees.  Maximum piston velocity for
> the
> > 5.7" rod is 0.158697" in the 5 degrees centered on max velocity, and the
> > same value with the 6.2" rods gives 0.157648".  At 90 degrees, both
> pistons
> > are moving the same speed - 0.151795" in 5 degrees.
> >
> > Dwell time, as defined by the above, is 10.75982 degrees for
> the short rod
> > motor at TDC and 10.862748 degrees for the long rod.  At BDC, the short
> rod
> > dwells for 14.737538 degrees and the long rod motor dwells for 14.48436
> > degrees.
> >
> > So, let's take a 6000 rpm motor.  Each revolution takes .01 seconds, and
> > 5/360 of that is .000139 seconds.  TDC to 5* ATDC average speed for 5.7"
> rod
> > = 65.8"/sec and 6.2" rod = 64.6"/sec.  Peak piston velocity +/- 2.5
> degrees
> > for 5.7" rod = 1142.6"/sec and 6.2" rod = 1135.0"/sec.  Piston velocity
> for
> > both engines at 90* = 1092.9"/sec.  Average piston velocity for
> 5* BBDC to
> > BDC for 5.7" motor = 35.0"/sec and 6.2" rod = 36.3"/sec.
> >
> > I'm not trying to be anal, only to indicate that piston velocity is also
> > related to crank angle and rod/stroke ratio.
> >
> > All this proves ONE thing for sure - I've got WAY too much time on my
> hands!
> > LOL!  Another thing is that in order to drop peak piston velocity from
> > 1142.6"/sec to 1135.0"/sec costs LOTS of money, in the form of pistons,
> > rods, machining, etc.  Does the extra .102928 degrees at TDC justify the
> > long rod motor?  An extra 0.000000286 seconds at TDC???  You
> decide.  I'm
> > too tired.
> >
> > Matt
> >
> > ----- End of forwarded message from owner-diy_efi at diy-efi.org -----
> >
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