[Mgs] Engine maths...and spare time

Sat Mar 28 08:23:30 MDT 2020

As I have said before, I don't drive much but I do enjoy understanding and
building engines. Anyone interested is said topic, especially LBC engines,
consider buying and ingesting this book.

https://www.amazon.com/sports-car-engine-tuning-modification/dp/0837600448/ref=sr_1_1?keywords=the+sports+car+engine+its&qid=1585403679&sr=8-1

Did I say 'consider'? I strongly recommend buying it. Complex, dynamic
engine processes explained in common language.

Rick

On Sat, Mar 28, 2020, 9:17 AM Richard Lindsay <richardolindsay at gmail.com>
wrote:

> Thank you Paul. As you know, I have always appreciated your depth of
> knowledge and experience. In fact, as you point out, there are SO many
> variables. Theory just can't keep up with practice. The factory's tuning
> recommendations for our beloved engines are almost certainly empirical.
> That is, decided upon by trial and error.
>
> And of course, tuning for maximum power is not the only maker's goal. Fuel
> economy, emissions, build cost, and a plethora of other constraints
> contribute.
>
> Again, thank you Paul,  *et al.*, for indulging my house-bound mental
> exercises.
>
> Rick
>
> On Sat, Mar 28, 2020, 8:54 AM PaulHunt73 <paulhunt73 at virginmedia.com>
> wrote:
>
>> "at idle 100% of the available power is used to overcome the friction
>> and other forces that exist at idle speed"
>>
>> Most of what the engine is doing at idle is as a vacuum pump, generating
>> about 16 in Hg. or so in the intake manifold, and is why when you introduce
>> an intake vacuum leak the idle speed goes up.  This may be included in your
>> 'other forces' above.
>>
>> Whist there may well be a most efficient point to start combustion and
>> the flame front, the prime consideration has to be avoiding
>> spontaneous combustion at any point, i.e. pinking or detonation.  As the
>> flame front travels pressure inside the engine is rising, but after TDC the
>> volume available is reducing, which tends to counteract the pressure
>> increase.  There is also the effect of leverage i.e. the angle the con rod
>> makes relative to the piston.
>>
>> I'm certainly not going to check your maths, a specific engine is what it
>> is, and the timing has to be set taking those specifics into account plus
>> other factors like fuel grade and type.
>>
>> PaulH.
>>
>>
>> ----- Original Message -----
>>
>> *From:* Richard Lindsay via Mgs <mgs at autox.team.net>
>> *To:* mgs at autox.team.net List
>> *Sent:* Saturday, March 28, 2020 12:54 PM
>> *Subject:* [Mgs] Engine maths...and spare time
>>
>> 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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