Ok did some more analysis, Aad am quite comfy now with the fundamental ability of a 1 pulse per cyclinder scheme to place a spark pulse quite accurately in any reasonable cranksahft slew rate scenario while racing. It seems no slew rate compensation is even required.
Again, I think the reason driving higher resolution CRS schemes (36-1, etc) was to detect misfire via a crank position dynamics based approach, not to correct a glaring spark inaccuracy inherent in a 1 pulse per cylinder CRS scheme.
Here's a depiction of the CRS and EST on the stock Buick signaling:
The question is- when the crankshaft is accelerating, as in during a pass down the race track, how much of an affect does that have on the ability of a 1 pulse per cyclinder scheme to place the spark accurately?
We need a representative crankshaft acceleration rate. I chose an rpm plot from a 2000+ hp car on a 7.0 pass. The 1st gear acceleration rate should serve as a good example IMO. I estimate a peak acceleration rate "α" (alpha) of about 1667 rpm/s. Or 27.78 rev/s^2.
Here's the plot used:
So the ECU calculates the current angular speed "ω" (omega) from the crankshaft CRS_Period between the two 70 deg BTDC edges shown in the diagram above. Let's assume we're at 6000 rpm here, so we get a period of 10 ms and an ω = 100 rev/s.
The current 70 deg BTDC edge is the reference position pulse; when this comes in the ECU looks up the desired spark angle. Let's assume we want to spark at 30 deg BTDC. So we are currently at 70 BTDC and we want to spark 40 deg later at 30 BTDC. The ECU calculates the timer value it needs to load to make this happen.
In the case we're at 100 rev/s currently, so 1 deg is 10 ms/360 deg = 27.78 us/deg. To wait 40 deg then we need a counter value = 40 * 27.78 us = 1.111 ms.
If our cranksaft was rotating steady state at 100 rev/s with no acceleration, we would make a spark 40.0 deg later then at 30.0 deg BTDC, ignoring any small counter bit error for now (< 1/2 deg).
The question of the hour becomes- what is the
actual spark angle that we arrive at in 1.111 ms elapsed time, knowing that the crankshaft is also accerating at a constant rate of 27.8 rev/s? How far off will we be?
The kinematic equation for position angle θ (theta) resulting from rotating speed + rotating acceleration is:
θ_actual = (ω * t) + (1/2 * α * t^2)
We have:
ω = 100 rev/s
α = 27.8 rev/s^2
t = 1.111 ms
We get:
θ_actual = 40.0 deg + 0.006 deg = 40.006 deg.
Sounds like we're expecting a horror show here, but we're only
0.006 deg late because of the crankshaft angular acceleration. Even in 1st gear with 2000+ hp. No wonder the stock ECU has no slew rate compensation- it doesnt need it.
Don't think there's any need to worry about the 1 pulse per cylinder scheme being inadequate vs a higher resolution wheel- it seems it's fundamentally more than adequate.
We'll try to also do some real bench testing soon with a classic FAST box with the 1 pulse per cylinder scheme.
TurboTR
