About constant speed props and commercial maneuvers

I asked this question to my CFI, but he just give me a blank stare, so
I'm hoping someone here could answer it.

As I understand it, all commercial maneuvers revolve around the idea of
converting between potential energy and kinetic energy, and the control
characteristics of the plane assiciated with these conversions.

Think about a lazy eight without the turn, to keep it simple. You're
just keeping a constant power setting and climbing then descending.
Before you start the climb you are cruising at 100 knots at 3000ft.
High kinetic energy, (relatively) low potential energy. At the top of
the climb you are now at 60 knots (low kinetic energy) and 3600 ft
(high potential energy). Now if you were to let go of the controls, the
plane will naturally nose down (if you trimmed it right) and level back
off at 3000 ft, and at the starting airspeed of 100 kts too.

The reason for this is the laws of thermodynamics. Energy converted
back and forth always equals the same in the end, with a small loss due
to entropy.

Now with that all said, imagine how a constant speed prop will perform
diffrently than a constant pitch prop. I don't know much about constant
speed props much since I've never flown one. When your airspeed
decreases in a constant pitch prop, engine RPM decreases, therefore
horepower decreases, right? But in a constant speed prop, the prop
governer will decrease the blade AOA, keeping the engine RPM the same,
but will horsepower remain the same? Would this result in less total
energy lost across the airspeed changes, therefore making it easier to
do commercial maneuvers?

You have obviously over analyzed this much more than necessary. You are
correct about nearly all of it. To answer you questions:

Yes as the RPM Decreases so does the Horsepower.

A constant Speed prop = constant RPM which = Constant Horsepower
(almost) The engine does not know that the airspeed has changed it is
only delivering power to the prop hub.
While the Horsepower delivered to the Propeller will be
constant the effecincy(sp) of the propeller at different speeds and
blade angles will vary some so the actual thrust produced will vary
some.

Yes I believe the constant speed prop results less energy lost across
the airspeed changes, but I doubt it is enough to be noticable.

I think the weight of the airplane has more to do making the maneuvers
easier to do. This is because you have more kinetic and potential
energy to work with through out the maneuver.


Brian

"buttman" wrote in message
oups.com...
I asked this question to my CFI, but he just give me a blank stare, so
I'm hoping someone here could answer it.

As I understand it, all commercial maneuvers revolve around the idea of
converting between potential energy and kinetic energy, and the control
characteristics of the plane assiciated with these conversions.

Did they eliminate the steep power turns, the 8's on pylons and slow flight
from the PTS?
[This addresses your "all commercial maneuvers" comment.]


Think about a lazy eight without the turn, to keep it simple. You're
just keeping a constant power setting and climbing then descending.
Before you start the climb you are cruising at 100 knots at 3000ft.
High kinetic energy, (relatively) low potential energy. At the top of
the climb you are now at 60 knots (low kinetic energy) and 3600 ft
(high potential energy). Now if you were to let go of the controls, the
plane will naturally nose down (if you trimmed it right) and level back
off at 3000 ft, and at the starting airspeed of 100 kts too.

The reason for this is the laws of thermodynamics. Energy converted
back and forth always equals the same in the end, with a small loss due
to entropy.

Perhaps the reason your CFI gave you a blank stare is because you use words
such as "entropy" without knowing what the word means.


Now with that all said, imagine how a constant speed prop will perform
diffrently than a constant pitch prop. I don't know much about constant
speed props much since I've never flown one. When your airspeed
decreases in a constant pitch prop, engine RPM decreases, therefore
horepower decreases, right? But in a constant speed prop, the prop
governer will decrease the blade AOA, keeping the engine RPM the same,
but will horsepower remain the same? Would this result in less total
energy lost across the airspeed changes, therefore making it easier to
do commercial maneuvers?


Perhaps when you start working on the complex aircraft and flying with a
constant speed prop you will get a feel for this all.

What you have written about "energy lost across airspeed changes" us
unconsistent with your comments about the kinetic vs. positional energy
issue.

I assume that you are not a troll but have been unable to find your name in
the FAA database so I can't tell whether you are a private pilot or a
student pilot.

"buttman" wrote in message
oups.com...
I asked this question to my CFI, but he just give me a blank stare, so
I'm hoping someone here could answer it.

As I understand it, all commercial maneuvers revolve around the idea
of
converting between potential energy and kinetic energy, and the
control
characteristics of the plane assiciated with these conversions.

Think about a lazy eight without the turn, to keep it simple. You're
just keeping a constant power setting and climbing then descending.
Before you start the climb you are cruising at 100 knots at 3000ft.
High kinetic energy, (relatively) low potential energy. At the top of
the climb you are now at 60 knots (low kinetic energy) and 3600 ft
(high potential energy). Now if you were to let go of the controls,
the
plane will naturally nose down (if you trimmed it right) and level
back
off at 3000 ft, and at the starting airspeed of 100 kts too.

The reason for this is the laws of thermodynamics. Energy converted
back and forth always equals the same in the end, with a small loss
due
to entropy.

Now with that all said, imagine how a constant speed prop will perform
diffrently than a constant pitch prop. I don't know much about
constant
speed props much since I've never flown one. When your airspeed
decreases in a constant pitch prop, engine RPM decreases, therefore
horepower decreases, right? But in a constant speed prop, the prop
governer will decrease the blade AOA, keeping the engine RPM the same,
but will horsepower remain the same? Would this result in less total
energy lost across the airspeed changes, therefore making it easier to
do commercial maneuvers?


Your propeller analyses were correct, but you've confused two different
systems: (1) work done by the engine; and (2) energy changes (between
gravitational potential and kinetic) of the airframe. In the climbing
phase the constant-speed prop will give you altitude quicker (since no
dropoff in power), but that time difference won't make any other
perceptible difference in the maneuver. Also I think your 'entropy' is
really only atmospheric friction. This is far from a "conservative"
system.

John Lowry
Flight Physics

I was under the impression that whenever energy is transformed from one
form to another, some is lost due to entropy. When fuel (chemical
energy) is transformed into mechanical energy, some energy is lost
because heat escapes the system through the exhaust pipe and oil lines.
I just looked up entropy and it said "For a closed thermodynamic
system, a quantitative measure of the amount of thermal energy not
available to do work", in other words, the energy that has been "lost".
So the heat through the exhause pipe could be called entropy. Is this
not correct?

Now if you have X joules of kinetic energy and Y joules of potential
energy at the begining of the maneuver, you mathematically should end
with the same KE and PE values all throughout the maneuver. Given
you're power settings are so that you're KE and PE are constant,
meaning the energy the engine is outputting is equal to all the drag
forces of the airframe, i.e. straight and level.

Now, in nature that perfect energy conversion is not possible. Some
energy is going to be lost. This lost energy is entropy. As your
airspeed decreases, your engine is outputting less energy, so at the
top of our no-turning lazy 8, we actually have less TOTAL energy than
when we started. Thats because the prop is experiencing more induced
drag and as a result that energy lost is entropy. Does that make sense?
My question is will a constant speed prop, since it does not lose
horsepower at slower airspeeds, not lose as much eneregy throughout the
maneuver making it easier?

By the way I'm a private pilot with an instrument rating, and am about
15 hours away from getting my commercial. My real name isn't Buttman
(although I WISH IT WAS), its just the name I use on the internet, so
it shouldn't be in the FAA database.

Consider, if you will, the many little old ladies and little old men, and
mathophobics of both genders, who have successfully achieved the commercial
ticket without worrying about such things. Way too much analysis. If I was
preparing you for your CFI and you hit me with that you and I would have a
long talk about what is important and what is not important.

Bob Gardner

"buttman" wrote in message
oups.com...
I was under the impression that whenever energy is transformed from one
form to another, some is lost due to entropy. When fuel (chemical
energy) is transformed into mechanical energy, some energy is lost
because heat escapes the system through the exhaust pipe and oil lines.
I just looked up entropy and it said "For a closed thermodynamic
system, a quantitative measure of the amount of thermal energy not
available to do work", in other words, the energy that has been "lost".
So the heat through the exhause pipe could be called entropy. Is this
not correct?

Now if you have X joules of kinetic energy and Y joules of potential
energy at the begining of the maneuver, you mathematically should end
with the same KE and PE values all throughout the maneuver. Given
you're power settings are so that you're KE and PE are constant,
meaning the energy the engine is outputting is equal to all the drag
forces of the airframe, i.e. straight and level.

Now, in nature that perfect energy conversion is not possible. Some
energy is going to be lost. This lost energy is entropy. As your
airspeed decreases, your engine is outputting less energy, so at the
top of our no-turning lazy 8, we actually have less TOTAL energy than
when we started. Thats because the prop is experiencing more induced
drag and as a result that energy lost is entropy. Does that make sense?
My question is will a constant speed prop, since it does not lose
horsepower at slower airspeeds, not lose as much eneregy throughout the
maneuver making it easier?

By the way I'm a private pilot with an instrument rating, and am about
15 hours away from getting my commercial. My real name isn't Buttman
(although I WISH IT WAS), its just the name I use on the internet, so
it shouldn't be in the FAA database.

"buttman" wrote in message
oups.com...
I was under the impression that whenever energy is transformed from one
form to another, some is lost due to entropy. When fuel (chemical
energy) is transformed into mechanical energy, some energy is lost
because heat escapes the system through the exhaust pipe and oil lines.
I just looked up entropy and it said "For a closed thermodynamic
system, a quantitative measure of the amount of thermal energy not
available to do work", in other words, the energy that has been "lost".
So the heat through the exhause pipe could be called entropy. Is this
not correct?

No. By its very nature any sytem with a exhaust pipe isn't a closed
thermodynamic system.

By the way I'm a private pilot with an instrument rating, and am about
15 hours away from getting my commercial. My real name isn't Buttman
(although I WISH IT WAS), its just the name I use on the internet, so
it shouldn't be in the FAA database.


You really wish your name was Buttman? Hell, go get it changed.

Whew! I was hoping somebody would say that.

"Bob Gardner" wrote in message
...
Consider, if you will, the many little old ladies and little old men, and
mathophobics of both genders, who have successfully achieved the
commercial
ticket without worrying about such things. Way too much analysis. If I was
preparing you for your CFI and you hit me with that you and I would have a
long talk about what is important and what is not important.

Bob Gardner

I never said it was important. I was thinking about this and that, and
came up with the idea out of curiousity. I remember a while back there
was a huge thread on whether the frickin stall horn would work when
flying inverted, so I thought it'd at least make good discussion.

Anyways, why is it not important? Is it because all that you'll ever
NEED to know about the commercial maneuvers is how to do them correctly
and not what's happening and why is happening? If so, then I disagree.
Or is it because then diffrence between fixed pitch and fixed speed
performance is neglegible? If thats the case then I see your point, but
still I think the thinking behind it is at least something to gain
from. I've learned more in the past few hours I've spent thinking about
this topic then I ever would've spent just memorizing the steps in
doing a chandelle.

It is only exhausted because it is entropy. If the exhaust pipe wasn't
there the heat would build up and reek havoc. The closed system is the
inside of the engine. Energy comes in as fuel, energy comes out as
shaft rotation. No ENERGY is leaving the exhause pipe, only the entropy
associated with the chemical reaction in the cylinders.

Anyways, if I could change my name it would be to Cornelius Charles
Buttman III, but I can't do that because my kids (when I have them)
will get made fun of.