[Mgs] Engine maths...and spare time

Richard Lindsay richardolindsay at gmail.com
Sat Mar 28 06:54:59 MDT 2020


Hello friends,

   When one is a geek, one thinks of geeky things. I am a geek and this
house-bound morning I woke up thinking about ignition timing. Here are the
details.

   We know that the charge (fuel plus air) in a cylinder doesn't burn
instantly, despite our perception to the contrary. Rather, it takes a
finite length of time from the occurance of the 'spark', the flame front to
cross the combustion chamber, and to raise the MEP (Mean Effective
Pressure) to a maximum - the point where it does the most work. But how
much time?
   Physics problems always start by listing the 'known' and the property to
'find'. So in this case,

KNOWN:
   Idle speed: 900rpm
   Idle timing advance: 4° BTDC
   Speed at maximum advance: 3500rpm
   Maximum timing advance: 32° BTDC

FIND:
   Time from spark to MEP

   The first thing one might know is that the goal at idle is not to
produce maximum power. In fact, at idle 100% of the available power is used
to overcome the friction and other forces that exist at idle speed. Stated
another way: Idle speed is the fastest the engine can achieve given the
available charge. That fact is evident (with carbureted engines) when one
notices that engine speed gradually increases, even for a fixed throttle
setting, as the engine warms and friction forces decrease. But back to the
problem.

   Because the goal at idle is smooth running and progression off of idle
(e.g. speeding up), not maximum power, the calculated wavefront speed may
not be correct at idle. But let's see.

   At idle speed, 900rpm in this MG TD example, the XPAG engine is turning
900rpm or 900rpm / 60mps = 15rps (revolutions per second).

   Distributor speed is 1/2 engine speed so at idle the distributor is
turning only 7.5 revolutions per second. But timing numbers are specified
in degrees of crank rotations so we will stick with 15rps.

   We don't know how fast the flame front travels across the combustion
chamber but we do know that maximum work occurs when the piston is half way
down the cylinder. And we also know that work isn't an instantaneous
parameter so it must begin before the half way point and last past that
point. Lots of unknowns and theory doesn't always work in practice. But if
we use the average piston position at half-way down the bore, where most
work is most effective, and the MEP (Mean Effective Pressure), since Mean
is average, calculations begin.

   A single revolution is 360° so half-way down the power stroke is 90°.
Add the idle spark timing of 4° BTDC (Before Top Dead Center) and we get
94° of crank rotation from spark to MEP at half-way down. That's 94/360 or
about 0.26 of an engine revolution. And the engine is turning 15
revolutions per second or 67ms (milliseconds) per revolution. So 67 x 0.26
= 17ms from spark to MEP at half-way down the power stroke, at idle.

   If we repeat the calculations for operating engine speed and at maximum
advance, we get 3500rpm / 60mps = 58rps (revolutions per second). Maximum
advance is 32° BTDC so 90° + 32° = 122°, spark to MEP or 122°/360° = 0.34
of a revolution

   58rps is 17ms/r so 17ms/r x 0.34r = 5.78ms from spark to MEP at half-way
down the power stroke. This is a more representative number than the 17ms
at idle. One might even divide the idle elapsed time minus the optimal time
across the strike's midpoint. Doing so would mean at idle, the pressure at
idle becomes most effective 5.6ms before half-way and for another 5.6ms
after midpoint. Interesting that the idle pressure application time is
about the same as the maximum pressure application time, or is that
circular logic?

   Yes everything above is ripe with assumptions and perhaps even
apocryphal and resplendent with errors, but it is only 7am after all.

   Anyone with extra house-bound time on their hands, please check my maths
and share your corrections, including the logic of the whole
experiment...or perhaps even why geeks think these ways!

Rick
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