Those serious about turbocharging should read this!

[asavage]

"It is better to be thought a fool, than to open your mouth and remove all doubt".

I think you've omitted a key part of that aphorism.

"Tis better to remain silent and be thought a fool, than to speak and remove all doubt," is the way I recall it. Perhaps Google would reveal yet something else.

[Knucklehead]

Only 15 lbs air. No extra fuel. Torque will increase, but the rpm at which it will max may or may not change. Pyro will be cooler.

[davehoos]

1. A VE type injection pump.
2. During engine operation, the RPM limits at maximum torque. (dyno roller)
4. no "boost compensator"

pressurized air charge ... from natural atmosphere to 15 psi.

What should be the reaction? incease in tourque[possibly] due if insuficent air to atomise and burn fuel quickly in the cyl.
note-with a VE pump Ive assumed that is at full throttle and the overspeed governor is not reducing the fuel quantity.

RPM rises? RPM remains? RPM drops?
quote-Torque will increase, but the rpm at which it will max may or may not change." I agree also ignition timing has an effect and assume that the NA air intake VE was good .

Pyrometer expectations?
if unburnt fuel was present it will rise.if not it should level or drop.[/quote]

[philip]

Only 15 lbs air. No extra fuel. Torque will increase, but the rpm at which it will max may or may not change. -snip

In this setting, does the 15 lbs intake air significally raise cylinder compression, resulting torque increase?

And/or does the 15 lbs intake air pressure (no fuel change) also reduce unburnt gases in the combustion chamber that increases torque?

[Knucklehead]

I do not have the equipment nor the mockup to make such measurements obviously, and not being a properly educated mechanical engineer, neither can I make proper calculations, so I can only theorize that doubling the initial intake density would significantly increase compression pressure to more than double according to my understanding of Boyle's law. But I would also theorize that the force required to compress the charge would more than negate the force rebounded on the power stroke. So, that would leave all other increases to be from improvements in combustion.

Something more concrete anybody? Since I have effectively said nothing, maybe I shouldn't have said it.