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From: Gary Casey <casey.gary@yahoo.com>
Subject: Re: power cower
In-Reply-To: <list-3650240@logan.com>
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Grayhawk,=0AOops - you're right.  I was being sloppy with my math.  The rea=
l equation would have to be something like  %power =3D ((1- (RPMmax-RPM) * =
RPMmax * 0.8) * (MAP-10)/(MAPmax-10)) * 100.=0ANo wonder I can't do it in m=
y head {:-).  There is also the effect of back pressure (also true of turbo=
charged engines, but a little more obscure) - The BMEP of an engine at sea =
level might be something like 163psi (in the case of a 200hp, 360cu.in. eng=
ine turning 2700 rpm).  Increase the altitude by 10,000 ft, keeping the sam=
e MAP and it would increase by about 5psi, or 3%, ignoring friction.  Inclu=
de friction and the power would increase by maybe 4%.  Very roughly, at the=
 same MAP, power will go up by about 0.5% per 1,000 ft of altitude.  Fuel f=
low will stay the same(this time neglecting the effect of volumetric effici=
ency), so the engine is actually more efficient at high altitude.  This is =
the same effect as that of throttling losses - the difference between Manif=
old Absolute Pressure (MAP) and Exhaust Absolute Pressure (EAP?) is what th=
e engine has to pump against and that subtracts from useful work.=0A=0AMust=
 be a slow news day...=0AGary=0A=0A=0A=0A________________________________=
=0A=0AGary,=0A =0AOK.=0A =0ABut - in your second formulation, the max calcu=
lated power is=0Aonly 80% when I know that, with everything fire walled at =
sea level and dry cold=0Aair, it is 100% - unless I am flying at 200 KIAS, =
then the new 100% is=0Aprobably 105% times the old. :-) Thank heavens the S=
AE has a different standard=0Afor calculating HP.=0A =0APerhaps an Excel ap=
p on your I-phone would help you in the air.=0A =0AGrayhawk=0A =0AIn a mess=
age dated 5/20/2009 1:59:20 P.M. Central Daylight Time,=0Acasey.gary@yahoo.=
com writes:=0AGrayhawk,=0AToo late {:-) I already looked at it.  I compared=
 it to the rough formula I have always used - % power =3D RPM/RPMmax * MAP/=
MAPmax  - and it comes within 1% of it.  I stopped there and didn't try to =
"reverse engineer" the formula, except I have hunch it has to do with assum=
ing the max RPM is 2500 (the 2.5) and the max MAP is 35 inches (the 3.5).  =
Regardless, the formulas are not quite right.  As rpm increases, engine fri=
ction increases and volumetric efficiency decreases so the increase in powe=
r won't go up directly with RPM.  The opposite is true of MAP.  It takes a =
certain MAP just to power the engine (overcome friction) so power will incr=
ease more than the increase in MAP.  How much are these effects?  Without r=
eal data it would be hard to say, but in the RPM and MAP ranges that are us=
eful in aircraft engines the factor for rpm might be something like 0.8. Fo=
r MAP there is a subtractive term of about 10 inches.  The equation would t=
hen look like  %power
 =3D (RPM/RPMmax*0.8) * (MAP-10)/MAPmax-10).  Gets hard to do in your head =
(well, mine anyway), so I just remember that a 10% reduction in RPM will dr=
op the power less than 10%, but a 10% drop in MAP will drop the power by mo=
re than 10%.  Then there is the effect of altitude..=0AGary=0A=0A=0A      
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<html><head><style type=3D"text/css"><!-- DIV {margin:0px;} --></style></he=
ad><body><div style=3D"font-family:'times new roman', 'new york', times, se=
rif;font-size:12pt"><div>Grayhawk,</div><div>Oops - you're right. =A0I was =
being sloppy with my math. =A0The real equation would have to be something =
like =A0%power =3D ((1- (RPMmax-RPM) * RPMmax * 0.8) * (MAP-10)/(MAPmax-10)=
) * 100.</div><div>No wonder I can't do it in my head {:-). =A0There is als=
o the effect of back pressure (also true of turbocharged engines, but a lit=
tle more obscure) - The BMEP of an engine at sea level might be something l=
ike 163psi (in the case of a 200hp, 360cu.in. engine turning 2700 rpm). =A0=
Increase the altitude by 10,000 ft, keeping the same MAP and it would incre=
ase by about 5psi, or 3%, ignoring friction. =A0Include friction and the po=
wer would increase by maybe 4%. =A0Very roughly, at the same MAP, power wil=
l go up by about 0.5% per 1,000 ft of altitude. =A0Fuel flow will stay the
 same(this time neglecting the effect of volumetric efficiency), so the eng=
ine is actually more efficient at high altitude. =A0This is the same effect=
 as that of throttling losses - the difference between Manifold Absolute Pr=
essure (MAP) and Exhaust Absolute Pressure (EAP?) is what the engine has to=
 pump against and that subtracts from useful work.</div><div><br></div><div=
>Must be a slow news day...</div><div>Gary</div><div style=3D"font-family:t=
imes new roman, new york, times, serif;font-size:12pt"><br><div style=3D"fo=
nt-family:times new roman, new york, times, serif;font-size:12pt"><font siz=
e=3D"2" face=3D"Tahoma"><hr size=3D"1"></font><br>=0A=0A =0A =0A<font id=3D=
"role_document" color=3D"#000000" size=3D"2" face=3D"Arial">=0A<div>Gary,</=
div>=0A<div>=A0</div>=0A<div>OK.</div>=0A<div>=A0</div>=0A<div>But - in you=
r second formulation,=A0the max calculated=A0power is=0Aonly 80% when I kno=
w that, with everything fire walled at sea level and dry cold=0Aair, it is =
100% -=A0unless I am flying at 200 KIAS, then the new 100% is=0Aprobably 10=
5% times the old. :-) Thank heavens the SAE has a different standard=0Afor =
calculating HP.</div>=0A<div>=A0</div>=0A<div>Perhaps an Excel app on your =
I-phone would help you in the air.</div>=0A<div>=A0</div>=0A<div>Grayhawk</=
div>=0A<div>=A0</div>=0A<div>=0A<div>In a message dated 5/20/2009 1:59:20 P=
.M. Central Daylight Time,=0Acasey.gary@yahoo.com writes:</div>=0A<blockquo=
te style=3D"BORDER-LEFT:blue 2px solid;PADDING-LEFT:5px;MARGIN-LEFT:5px;"><=
font style=3D"BACKGROUND-COLOR:transparent;" color=3D"#000000" size=3D"2" f=
ace=3D"Arial">=0A  <div style=3D"FONT-FAMILY:'times new roman', 'new york',=
 times, serif;FONT-SIZE:12pt;">=0A  <div>Grayhawk,</div>=0A  <div>Too late =
{:-) I already looked at it. =A0I compared it to the rough=0A  formula I ha=
ve always used - % power =3D RPM/RPMmax * MAP/MAPmax =A0- and it=0A  comes =
within 1% of it. =A0I stopped there and didn't try to "reverse=0A  engineer=
" the formula, except I have hunch it has to do with assuming the max=0A  R=
PM is 2500 (the 2.5) and the max MAP is 35 inches (the 3.5).=0A  =A0Regardl=
ess, the formulas are not quite right. =A0As rpm increases,=0A  engine fric=
tion increases and volumetric efficiency decreases so the increase=0A  in p=
ower won't go up directly with RPM. =A0The opposite is true of MAP.=0A  =A0=
It takes a certain MAP just to power the engine (overcome friction) so=0A  =
power will increase more than the increase in MAP. =A0How much are these=0A=
  effects? =A0Without real data it would be hard to say, but in the RPM and=
=0A  MAP ranges that are useful in aircraft engines the factor for rpm migh=
t be=0A  something like 0.8. For MAP there is a subtractive term of about 1=
0 inches.=0A  =A0The equation would then look like =A0%power =3D (RPM/RPMma=
x*0.8) *=0A  (MAP-10)/MAPmax-10). =A0Gets hard to do in your head (well, mi=
ne anyway),=0A  so I just remember that a 10% reduction in RPM will drop th=
e power less than=0A  10%, but a 10% drop in MAP will drop the power by mor=
e than 10%. =A0Then=0A  there is the effect of altitude..</div>=0A  <div>Ga=
ry</div>=0A  <div style=3D"FONT-FAMILY:times new roman, new york, times, se=
rif;FONT-SIZE:12pt;"><br></div></div></font></blockquote></div></font>=0A</=
div></div><div style=3D"position:fixed"></div></div><br>=0A=0A=0A=0A      <=
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