I must of been half asleep that day, the question was badly written, I'll try again.
The fuel rack controls the rotation of the plungers in their sleeves and in turn when the spill port in the plunger is exposed, controlling injection volume.
What I was trying to figure out, and have now realised, is the relationship between rack position and injection volume. Initially I was reading the graph (I have the same manual) as RPM vs injection volume, and in that case it makes little sense. Now I have my brain switched on, its reading RPM vs rack position, and from there we can calculate injection volume.
So after looking at a plunger, the ratio of rack position / plunger rotation angle and fuel delivery volume would appear linear, so:
If 9.5mm rack = 8cc / 1000st (approx)
and 11.0mm rack = 78cc / 1000st
so 1.5mm rack = 70cc change in volume.
so 0.02mm rack = 1cc change in volume.
so 9.34mm rack = 0cc / 1000st (approx)
Going by the graph that means that if you increase the idle speed by around 200RPM the governor will cut the fuel completely. Also explains why these engines are so difficult to stall, go below 400RPM (engine) and the fuel rack goes to almost full fuel.
Is this correct?, seems very sensitive to me.
Back to what this has to do with LPG:
In relation to A/F ratio, how does 162:1 air/diesel compare to 0.4% lpg? I'll assume 200:1 air/LPG but thats a guess only.
If the rise in RPM by injecting LPG at 0.4% causes the IP to reduce the diesel by half, that goes to 324:1 air/diesel and still 200:1 LPG, more gas than diesel.
Still don't have a clue what this will cause the engine to do, any ideas?
I'm interested in this stuff but I should really stick with electrics, normally only need ohms law :cool:
Lyle.

