Let's further discuss flame front velocity in relation to charge density.
Doward said:
See, that air/fuel mixture will burn at a certain, steady rate. As you increase the density of the air/fuel mixture, the burn takes longer to complete, and the slower the flame front. This is why you get so much torque - the air/fuel mixture burns longer Think for a moment on that.
OK, according to this, you would need to ignite a more dense charge earlier, to maintain MTBT. (because the burn will take longer, you need max cyl pressure to occur at the right geometric piston location) The problem I have with that is that in fact you need to retard the timing as density increases. That is why when you lean out the mixture by scaling the afm, and the TCCS thinks that the density is
less than it really is, it advances the timing and results in detonation and ultimately rod extraction...
So, what's correct?
I'll do some digging and see if I can't be beneficial, rather than a baseless naysayer.
Also,
Doward said:
IIRC, optimum mechanical force from the piston to the crankshaft occurs 20 degrees AFTER TDC (which means you want the highest cylinder pressure at this point)
That said, you want to ignite the mixture at the point that gives you the highest pressure around 20 degrees ATDC. Keep in mind, when we say we are 'advancing' timing, we are actually firing the plugs further before TDC!
So does advancing the ignition help power? If before you advanced the timing, you were achieving the highest pressure after 20º ATDC, then yes - you should see some small torque gaines.
This optimum mechanical force occurs at an angle determined by the rod/stroke ratio. (which is funny, because I remember reading in SCC mag that rod/stroke ratio was one of the least important things to consider when building an engine) Peak torque transfer from the piston to the crank occurs when the crank and connecting rod are at 90 degree angles. Cylinder pressure is force per area, when you multiply by the surface area of the piston, you get a vertical force directed straight down the bore, transfered through the wrist pin, into the rod. Where this force meets the crank, it becomes a force times a moment arm, which we all know as torque. When the rod is perpendicular to the stroke of the crankshaft, the torque
Force X Lever Arm (stroke of crank/2) X sin (angle between force and lever arm) is maximized, since sin(90) = 1.
This image depicts this well
As you can see, the crank angle that results in a 90 degree rod-stroke angle is a triangle problem with the two legs being the stroke of the crank and the length of the rod. That being said, peak torque may occur at 20 degrees ATDC for some engines, but certainly not for all. in fact it's closer to 70 degrees in
most engines. I could do the math for a supra, but that would be trivial since all we know how to do is alter the factory settings, and we actually don't know when the peak cylinder pressures are occurring in the cylinder. Also, since sin(0) = 0, you can see that if peak pressure occurs at TDC, no torque at all is made, and instead, the rod bearings support the full force applied from the combustion pressure and that is what kills bearings. (That and the fact that average cyl pressures are 1,200 PSI or so and detonation pressures are closer to 12,000 PSI, so it's not hard to see why 10X the force on a bearing might cause some unforeseen damage...)
Anyway, lets linger on this for a bit and see what we can figure out...
Here's a great read about rod/stroke ratio that helps clarify anything I've missed
rod ratio
-Jake
