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.

Just puts me on a level with your CFI.

Bob

"buttman" > wrote in message
oups.com...
>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.
>

"buttman" > wrote in message
oups.com...
> 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.
>
Need some reading there. It's not a closed system, and energy does come
out the exhaust. Whether that energy is AVAILABLE is another question.

John Lowry, PhD (irreversible quantum statistical mechanics)
Flight Physics

> If the exhaust pipe wasn't
> there the heat would build up and reek havoc.

That depends how smelly the fuel is.

> No ENERGY is leaving the exhause pipe

False. This is evidenced by the fact that energy from the exhaust pipe
annoys the neighbors and can be used to drive a turbine. What is closer
to the truth is that the energy leaving through the exhaust pipe is too
much trouble to convert into thrust. (I will note however that the
exhaust pipe actually does add thrust, which often more than offsets the
drag caused by the pipe in the slipstream - I seem to remember 20-80
pounds of thrust from examples in ground school (mumble) years ago.)

Jose
--
I used to make money in the stock market,
now I make money in the basement.
for Email, make the obvious change in the address.

OK, forget the entropy. The point I was making is that energy is lost.
Energy that can't be used.

When you're doing a lazy 8 in a fixed pitch prop, when you slow down,
the engine produces less horsepower. Its like pulling the throttle back
in the maneuver. This problem doesn't exist with a constant speed prop
because engine RPM is maintained, therefore horsepower remains the
same. I think this is why everyone always says how its sooooo much
easier to do commercial maneuvers in the bonanza as opposed to our
skyhawks or cherokees.

I don't see how this is any more insignifigant than p-factor or any
other phenomenom.

buttman wrote:

> OK, forget the entropy. The point I was making is that energy is lost.
> Energy that can't be used.

If you're trying to analyze maneuvers by using energy, you're not going to
get anywhere fast. You are converting chemical energy into heat energy,
kinetic energy, grav. potential energy, etc. The air is sapping a varying
amount of energy from you as you climb, turn, change AOA, etc etc. As the
prop speeds up, or slows down, and/or the AOA of the prop changes,
effeciency ratios change. When you're climbing, the engine is getting
hotter and it can be argued that even that would affect the
efficiency/effectiveness of the engine. Thrust/power curves are constantly
changing. The list going on.


> When you're doing a lazy 8 in a fixed pitch prop, when you slow down,
> the engine produces less horsepower. Its like pulling the throttle back
> in the maneuver. This problem doesn't exist with a constant speed prop
> because engine RPM is maintained, therefore horsepower remains the
> same.

If the HP remains the same as you slow down, then the thrust must increase
linearly as you slow down. This doesn't happen, nor do we get an infinite
amount of thrust standing still before the takeoff roll.


> I think this is why everyone always says how its sooooo much
> easier to do commercial maneuvers in the bonanza as opposed to our
> skyhawks or cherokees.

It is?


> I don't see how this is any more insignifigant than p-factor or any
> other phenomenom.

This isn't a "How to get a Pilot Certificate" newsgroup, it's a "piloting"
newsgroup. Anything related to piloting including getting a certificate,
the theory of flight, crash analysis, and what if scenarios are perfectly
valid and I would strongly encourage such debate and discussion. If people
don't care about stuff that doesn't help them pass their checkride, they
don't have to read the theoretical posts.

Keep posting!

Hilton

Hilton wrote:

> If you're trying to analyze maneuvers by using energy, you're not
going to
> get anywhere fast. You are converting chemical energy into heat
energy,
> kinetic energy, grav. potential energy, etc. The air is sapping a
varying
> amount of energy from you as you climb, turn, change AOA, etc etc.
As the
> prop speeds up, or slows down, and/or the AOA of the prop changes,
> effeciency ratios change. When you're climbing, the engine is
getting
> hotter and it can be argued that even that would affect the
> efficiency/effectiveness of the engine. Thrust/power curves are
constantly
> changing. The list going on.

Thats my point. You're supposed to finish a lazy 8 at the same altitude
and airspeed as what you started with. The beauty of the lazy 8 is how
you start out in perfect equilibrium, disrupt that equilibrium by
changing all kinds of things, then returning back to the equilibrium
you started with. It tells you all sorts of things about the airplane.
Its probably my favorite commercial maneuver. Anyways, you need to add
power someway to replenish the energy lost due to induced drag, prop
drag, "changes in thrust/power curves" as you put it, in order to bring
yourself back to that equilibrium. And it's not just the engine either.
You lose energy from banking the wings too.

Pretend you're in a plane which has a low horsepower engine and a high
fixed blade AOA. This gives you a (relativly) fast cruise speed, but
hardly any torque left over to handle climbs. If you were to do a lazy
8, it would be a lot harder. As you pull up the airspeed slows down
causing the prop to impose more stress on the engine. This slows down
engine RPM and horsepower drops sharply down the performance curve.
You're putting the plane in a condition where it is horribly
inefficient. The results of all this is a very little altitude
increase, along with a huge decrease in airspeed at the 90 degree point
of the maneuver. Now for the second half, you can't return to that
equilibrium you started with because you lost too much. you're going to
have to live with being outside PTS by either diving lower than your
starting altitude, or leveling off too slow.

Now if you have a plane that was specifically designed to handle these
changes, then you're not going to lose as much power so its going to be
easier. You'll still lose power in the climb even with a constant speed
prop for various reasons, but its not going to be nearly as much. You
still "lose energy" by banking and increased drag.

So I guess my point is that it's impossible to do a lazy 8, unless
either your plane is 100% efficient (which is physically impossible),
or you somehow add power. This is spliting hairs, and you may only lose
3 or 4 knots of "energy", but theoretically its true.

>
> It is?
>

I've been lead to believe so. That might be due to the fact that even
though the control surfaces are roughly the same surface area as
compared to a Skyhawk, the bonanza is faster, allowing the plane to be
more responsive. I start my training in the bonanza in about another
week, so I'll see for myself soon.

>
> Keep posting!
>

OK I will!

"buttman" > wrote in message
ups.com...
> Thats my point. You're supposed to finish a lazy 8 at the same altitude
> and airspeed as what you started with. The beauty of the lazy 8 is how
> you start out in perfect equilibrium, disrupt that equilibrium by
> changing all kinds of things, then returning back to the equilibrium
> you started with.

I agree with those who suggest you are over-complicating the issue.

Yes, the lazy eight should finish up at about the same altitude and airspeed
as you had when you started the maneuver. But IMHO, the biggest factor in
aiding you to that goal, other than flying the maneuver correctly, is
selecting an appropriate power setting.

Throughout the maneuver, your power setting is "wrong" for the flight
attitude and configuration (clean). You are either slowing down or speeding
up. Ideally, you'll wind up as much slower than your equilibrium point as
you wind up faster than it, and in the end it all comes out even. To
accomplish this, you either need to compensate by spending more or less time
in the decelerating or accelerating portion of the maneuver (as
appropriate), or you need to select a power setting that puts those end
point roughly the same distance from the middle equilibrium point.

I am, of course, oversimplifying the whole "middle, low end, high end" part
of this discussion. The end points may or may not be exactly the same
distance from the ideal middle point, from an absolute airspeed point of
view. But the basic idea is true, regardless: there's an appropriate
"center point" around which you fly the maneuver, and your power setting
determines that center point (assuming the rest of the maneuver is flown the
same...you can rush or slow down portions of the maneuver to compensate for
a "wrong" center point, of course).

I believe that you are right, that a constant speed prop provides a more
constant contribution of power throughout the maneuver. But, for one thing,
the maneuver is as much about drag (force) as it is about power (force over
distance over time). Since your airspeed is constantly changing, you're not
really producing a constant balance between engine output and drag anyway.

For another thing, I believe that in the context of the maneuver, the
difference in power output between a fixed pitch prop and a constant speed
prop isn't significant. Not compared to the other issues surrounding the
maneuver (flying it smoothly, and choosing an appropriate power setting for
the maneuver).

I think of it this way: as far as preserving your equilibrium, it's not
really that important how much energy the engine is providing at any given
point during the maneuver. The only thing that's important there is how
much extra energy the engine provides, TOTAL, throughout the maneuver.
Regardless of the type of prop installed, you control this directly through
the choice of the power setting used. If you finish the maneuver fast, your
power setting was too high; slow, your power setting was too low. It works
the same regardless of the type of prop.

> [...]
> So I guess my point is that it's impossible to do a lazy 8, unless
> either your plane is 100% efficient (which is physically impossible),
> or you somehow add power. This is spliting hairs, and you may only lose
> 3 or 4 knots of "energy", but theoretically its true.

Well, I'm not aware of any airplane in which it's impossible to do a lazy
eight without changing the power during the maneuver. So I'd say that
"point" of yours is obviously incorrect. In your example of a
low-horsepower, high-prop-pitch airplane, all you should need to do is use a
slightly higher power setting for your entry into the maneuver. Of course,
that presumes such an airplane and frankly, people don't usually go around
putting high-pitch fixed-pitch props on low horsepower airplanes.

>> [talk about ease of performing the maneuver between
>> fixed-pitch and CS props]
>>
>> It is?
>
> I've been lead to believe so. That might be due to the fact that even
> though the control surfaces are roughly the same surface area as
> compared to a Skyhawk, the bonanza is faster, allowing the plane to be
> more responsive. I start my training in the bonanza in about another
> week, so I'll see for myself soon.

My experience has been that of the three airplanes I flew a lazy eight in --
a C172, a C177RG, and my Lake Renegade -- the easiest airplane for the
maneuver was the Cardinal and the hardest was the Lake, both of which have
constant speed props. IMHO, control feel on the Cessnas is better, and I
especially like the stabilator on the Cardinal (not everyone feels this way
:) ). Even in the Lake, once I got the power setting figured out, the
maneuver went pretty smoothly, as it should. (I needed a lower power
setting...flying the maneuver near Va, as I was doing with the Cessnas, was
too fast an entry and I wound up too fast at the end).

Frankly, even in the C172 the maneuver is fairly easy as long as you fly it
by the numbers and don't try to rush it. Like you, I find the lazy eights
to be one of the more enjoyable of the commercial maneuvers, but it does
require a relaxed, smooth hand on the controls. Provide that, and I don't
see why it shouldn't go well in just about any airplane, constant speed prop
or not.

Pete

buttman wrote:

> Thats my point. You're supposed to finish a lazy 8 at the same altitude
> and airspeed as what you started with. The beauty of the lazy 8 is how
> you start out in perfect equilibrium, disrupt that equilibrium by
> changing all kinds of things, then returning back to the equilibrium
> you started with. It tells you all sorts of things about the airplane.

Define equilibrium. Technically speaking, at the moment you arrive at your
'starting point', your pitch is changing, your bank angle is changing, your
altitude is changing, your VSI is changing, your energy state is changing,
your airspeed is changing, your governer is changing...

I think it's way too deep to look at the energy aspects of a lazy-8. I
think Peter's advise is the best; i.e. nail your power settings which is
what you've also alluded to. Personally, I think the key to flying the
lazy-8 is to ensure that your pitching and banking moments are out of sync
with one another. I think too many CFIs teach them to be in sync which is
wrong. I 'fired' my CSEL CFI for that reason and another reason related to
my 'comfort level' while flying with him.

Hilton

"Hilton" > wrote in message
ink.net...
> [...] Personally, I think the key to flying the
> lazy-8 is to ensure that your pitching and banking moments are out of sync
> with one another. I think too many CFIs teach them to be in sync which is
> wrong.

I wouldn't be surprised if there's a different way to teach the maneuver for
every CFI that's out there. :)

However, the CFI that had the most success (out of two) teaching me the lazy
eight was the one who showed me that you pretty much only have to use
aileron right at the beginning of the maneuver. A little bit of roll input,
and then just a nice slow increase in back pressure (with ailerons neutral),
causes just the right amount of increase in bank angle throughout the first
half of the maneuver.

I'm not sure what "out of sync with each other" means (or "in sync" for that
matter), and maybe it just means the same as what I have found to work well.
Can you elaborate on your terminology?

Pete

Peter wrote:

> I'm not sure what "out of sync with each other" means (or "in sync" for
that
> matter), and maybe it just means the same as what I have found to work
well.
> Can you elaborate on your terminology?

Sure. At the start of the maneuver, no pitch, no bank, etc... Start
banking and pitching. Now, the maximum pitch occurs at the 45 degree point,
at which point the pitch is reduced to level at the 90 degree point in the
maneuver, and then the minimum pitch (max down) occurs at the 135 degree
point. So, the minimum and maximum pitch up and down occur at the 45, 135,
etc degree positions. However, the bank is zero at the start, and maximum
at the 90 degree position, and then zero again at the 'bottom'. So the
minimum and maximum bank left and right occur at the 0, 90, 180, etc
positions.

IMHO, if a student is simultaneously pitching and banking 'in sync', they
are doing it wrong. Recap: At the start, you start pitching and banking.
When you get to the 45 degree position, you keep increasing the bank, but
start decreasing the pitch etc. I believe the Jeppesen book shows this
well.

Hilton