[Mgs] Engine maths...and spare time

Sat Mar 28 07:36:37 MDT 2020

"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 
  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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