Others have already hit on what effect higher octane ratings had. Peter
Stickney will probably have one of his great replies coming along soon
too. But here is a quick rundown on what 104/150 octane should provide
for a Merlin 266.
The US had an empirical formula for calculating MAP limits at different
PN. It is a little conservative, but gives a good ballpark figure.
((old MAP -7) * new PN/old PN) +7 = new MAP
So, 66.6 in Hg on 100/130 octane would give:
((66.6 - 7) * 150/130) +7
59.6 * 1.154 + 7 = 75.76 in Hg
The RAF actually used +25 psi, about 80.9 inches.
We know the Merlin 266 was rated at 66.6 in Hg, 1705 hp @ 5750 ft in low
blower. That is enough information to approximate how much power the
engine provides at any altitude.
We also know static pressure at 5750 ft is approx 24.20 in Hg. So,
dividing 66.6 by 24.20 gives us approx 2.75 for the pressure ratio that
the Merlin 266 provides in low blower.
Multiplying static pressure by the pressure ratio gives the manifold
pressure available at any altitude. 80.9 in Hg would be attainable up
to about 500 ft unrammed, and approx 82.4 in Hg at SL.
Since we know it produces 1705 hp @ 66.6 in Hg we can figure how much
it makes at 80.9 in Hg. 1705 * 80/66.6 gives about 2071 hp. Then you
have to take the difference in temp into account. Sqrt of absolute temp
at 5750 ft / absolute temp at 500 ft times 2071 hp.
(sqrt (276.86 / 287.36)) * 2071 = 2032 hp @ 500 ft.
(I'm using the 1976 standard atmosphere for all calculations, older
atmosphere models might provide slightly different figures)
This should be accurate +- about 1%. You can do the same thing for just
about any engine, provided you have an accurate base altitude, power and
MAP rating to start with. I cheated and created an Excel spreadsheet
that does all the work for me.
You need to make sure and use static ratings, a lot of RAF ratings are
with 350 or 400 mph RAM which will screw things up. RAM will cause a
higher rated altitude from the ram pressure, but lower power due to
compression heating.
Greg Shaw
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