Sender: <marv@lancaironline.net> (Marvin Kaye)
To: lml
Date: Fri, 22 Nov 2002 12:49:49 -0500
Message-ID: <redirect-1884483@logan.com>
Reply-To: <kycshann@kyol.net>
From: "Shannon Knoepflein" <kycshann@kyol.net>
Subject: RE: [LML] Back Flying
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You power is going to depend heavily on which mixture setting you are
using, ROP or LOP--which you didn't specify.  If you are running LOP,
you can calculate your power by just your fuel (too much air, so fuel is
limiting factor on power).  On a NA engine, 8.5:1 compression, a pretty
accurate number to use is 15.  Multiply your fuel burn by 15hp/gph and
you will get your HP.  Divide this by max continuous HP and you will
have your percent power.  This ONLY works LOP, as fuel is the only thing
effecting power.

On the ROP side, you have to use mass air flow, as the air is the
limiting factor in HP (too much fuel, so air becomes limiting factor).
Fuel flow doesn't matter on ROP side, as you have too much.  You must
use RPM and MAP to calculate the mass air flow.  You engine
documentation will usually come with these numbers.

For example, I run my 300hp IO-540 at about 25 LOP at 5000 feet, keeping
it out of the danger zone.  (I'd have to run it at about 125 ROP to
accomplish the same thing on the other side).  This results in about
15.3-15.6gph of fuel burn, depending on the temperature.  Multiplying
this by 15, I'm running at 230-234hp, or 77-78% power.  My CHT's are all
below 330.  For a note of reference, this is at 2550 rpm's, 25.5" map,
but that doesn't matter at all since I'm LOP.  If I go over to the ROP
side, 125ROP, over 18gph, all I do is pick up about 1-2 knots, and crud
up my engine with ROP crud.

(this all assumes an engine with a BSFC of about 0.39 #'s-fuel/hr/hp.
5.85#'s/gal divided by 0.39 #'s-fuel/hr/hp = 15 hp/gph.  For a 7.5:1
engine, it is 13.75 hp/gph due to the less efficient engine running
about 0.42 BSFC)

FUEL BURN DOES NOT MATTER ON ROP. :)

---
Shannon Knoepflein   <--->   kycshann@kyol.net

you wrote>>>
I cruised over at 1,500 ft, 7.6 gal/hr, 55% power and 175 ktas.

My engine uses 10 gal/hr at 75% power. Every .6 gal/hr below that 
figure is a 5% reduction in power. This comes from the engine manual. 
Therefore, to calculate my power, I subtract the fuel burn from 10 
and divide by .6. I then multiply that number by 5% and subtract it 
from 75%. In the above example that would be:

    1)   ( 10 - 7.6 ) = 2.4
    2)   2.4 / .6 = 4
    3)   4 * 5% = 20%
    4)   75% - 20% = 55%

I am open to suggestions as to any other way to calculate the power
used?
-- 
Lorn H. 'Feathers' Olsen, MAA, DynaComm, Corp.
248-478-4301, mailto:lorn@dynacomm.ws
LNC2, O-320-D1F, N31161, Y47, SE Michigan

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