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OK, I'm not an engineer, no background in building engines (more in tearing them
up), and my math isn't all that strong. Several years ago and "old timer" gave
me the following chart for computing % of power.
Basically, you add Manifold Pressure and the first two number of the RPM giving
a number which you then use on the left side of the chart to give you % of
power. Here is the chart:
MP+RPM(1OO) % POWER
36 35
39 45
42 55
45 65
48 75
51 85
54 95
57 105
This jives with what Casey said for 24 squared=75% power and also works with the
idea of losing 1 inch of MP and adding 100 rpm, ie, 23" + 25[00].
I'm just presenting this, not defending it, but I think it does a reasonable job
on % of power. It doesn't address the question of horsepower which started the
thread, and doesn't address humidity, temperature, etc. but it's quick and
simple.
Matt McManus
LNC2 360
Quoting Sky2high@aol.com:
In a message dated 9/23/2007 9:15:29 A.M. Central Daylight Time,
glcasey@adelphia.net writes:
Here is rough way to calculate the power, probably within 10% and maybe 5%:
Starting with HP=Torque(ft-lb)xrpm/5252, replace torque with MAP:
HP=MAPxrpm/K where K is a constant ( 30" and 2700 rpm produces 180 hp giving
a K of
.00222, for example). But then none of that is really accurate. Torque
changes more than proportional to MAP because there is some friction HP to
be
accounted for. You could subtract something from 30 - a number between 5
and 10
would be appropriate. Then HP changes slower than a simple rpm multiplier
would suggest, partly because engine friction increases faster than rpm.
Then
torque will change with ambient temperature - a change of 50F will change
torque by about 10%. That isn't even too accurate because the air will be
heated
on its way to the cylinder - a number of maybe 5% per 50 degrees would be
closer (1% per 10 degrees). And all that is true only if you are running
best
power mixture. When LOP the best measure of power is fuel flow - I think
Paul has a number for this, which I forgot. What I do is memorize a number
for
75% power. Turns out it is about 24"-2400 rpm (picked a "square" number
because it is easier to remember). For every inch the manifold pressure
changes
the rpm has to change 100 to compensate, so 75% is at 23"-2500rpm and
25"-2300rpm). But because of the errors mentioned above 21"-2700rpm will
NOT
produce as much power as 26"-2200rpm. Then add or subtract 1% for every 10
degrees
the temp changes from 60F (NOT from standard conditions at that altitude).
Or one could create a chart with the above information to which one could
refer. Or, as in my case, with a NA engine (Naturally Aspirated, not
"Normally" as some would say. After all, the turbo enthusiasts would insist
that
boosted engines are more "normal".) I need all the power I can get most of
the
time, so it really doesn't matter - I can't do anything about it anyway.
Gary,
Try this calculator and see what the dew point does to % power.
_http://wahiduddin.net/calc/calc_hp_dp.htm_
(http://wahiduddin.net/calc/calc_hp_dp.htm)
Grayhawk
PS: I have an inexpensive instrument that I stick into the cabin air inlet
in flight to measure the dew point or % humidity. Temperature isn't enough.
PPS: The sq rt of .75 is .866. 86.6% of 30" is 26" and 86.6% of 2700 is
2300. But since I fly at 2500 rpm, using your adjustment, then a MAP of 24"
would approximate 75% power. Since I also like to fly at WOT and anytime I
see
a MAP that low, I must be at or below 75% power. Normally that would be at
or above 6500 MSL. Since I also use ram air induction at around 175 KIAS at
that altitude, I have to go up another 1000 feet and fly in the opposite
direction (towards O'Hare) in order to conduct the GAMI lean test. Bummer.
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