[Mgs] Engine maths...and spare time

Wed Apr 1 17:56:27 MDT 2020

It takes about 3ms to burn the fuel in a cylinder from when the spark
ignites it.  The engine designers want the peak pressure to occur at
between 5-10 ATDC.
At idle speed 3ms is about 15 degrees.
So if you want your peak pressure at 5 ATDC, your base timing is 10 BTDC,
if  you want  your peak pressure at 10 ATDC your timing is 5 BTDC.
If you look at tune up specs for non smog controlled engines, most have
base timing in the 5-10 BTDC range.
Now if you double the engine speed, it still take 3ms to burn, but to keep
the max pressure at 5-10 ATDC, you have to light the mixture sooner.
You do eventually get to a point where no more advance will help.  That
point varies with the engine design,  cam profile, combustion chamber
design, headers etc.
Rick

On Sat, Mar 28, 2020 at 5:55 AM Richard Lindsay via Mgs <mgs at autox.team.net>
wrote:

> 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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