Author Topic: P51 Question/Controversy  (Read 2619 times)

Offline BnZs

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Re: P51 Question/Controversy
« Reply #15 on: February 05, 2014, 02:45:35 PM »
What about the unloading factor though? And 400+mph Pony is essentially unloaded (absolute minimum drag) for a relatively long period of time when zooming vertically. Consider this maneuver-Gentle pull up to vertical, zoom (at which point the aircraft is unloaded) until the speed starts to bleed down, pull down to horizontal, roll level, and *then* unload your dive back to your original alt. Seems very energy efficient to me. I think in-game you can actually level out of such a maneuver going slightly faster than the speed at which you entered it, but I can't test it for sure right now...

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Offline Skyyr

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Re: P51 Question/Controversy
« Reply #16 on: February 05, 2014, 03:08:55 PM »
What about the unloading factor though? And 400+mph Pony is essentially unloaded (absolute minimum drag) for a relatively long period of time when zooming vertically. Consider this maneuver-Gentle pull up to vertical, zoom (at which point the aircraft is unloaded) until the speed starts to bleed down, pull down to horizontal, roll level, and *then* unload your dive back to your original alt. Seems very energy efficient to me. I think in-game you can actually level out of such a maneuver going slightly faster than the speed at which you entered it, but I can't test it for sure right now...

In my experience, "gently" pulling back means loading no more than 2-3G's. With only pulling 2-3G's, you'll be hard pressed to find much time to unload for any length of time that would be significantly beneficial. Even at 3G's, you're increasing the load of the aircraft threefold, which means you have to maintain that much more AoA to maneuver. To get vertical with enough speed to functionally benefit from unloading, you have to pull hard - very hard - very quickly.

Remember thermodynamics. You can't get more energy out than what you put in. Any increase in energy that you'll see is from the engine's output, which will largely depend on how long you "extend" after completing your aileron roll at the top of the Immelman and on the aircraft's acceleration. This isn't part of the maneuver, but rather an advantage of using the maneuver. The Immelman drains more energy than a shallow bank does, but it also allows it to be regained quicker, hence the efficiency of the maneuver. This is the portion that varies based on each aircraft and isn't part of the aerodynamics of the maneuver, but of the raw power output of the plane itself.

This brings us back to the statement that sparked this, that the Mustang poorly accelerates compared to other aircraft, which is why Immelman's are not necessarily effective when fighting a majority of late-war aircraft while flying it. It bleeds more energy in an Immelman, that in turn must regain, than other aircraft performing the same maneuver.

Keep in mind, an Immelman is the most efficient combat maneuver for reversing 180* - no argument there. The previous post (and point) was bringing out that even Immelmans burn a decent amount of energy.
« Last Edit: February 05, 2014, 06:04:58 PM by Skyyr »
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Offline bozon

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Re: P51 Question/Controversy
« Reply #17 on: February 06, 2014, 06:27:18 AM »
In my experience, "gently" pulling back means loading no more than 2-3G's. With only pulling 2-3G's, you'll be hard pressed to find much time to unload for any length of time that would be significantly beneficial. Even at 3G's, you're increasing the load of the aircraft threefold, which means you have to maintain that much more AoA to maneuver. To get vertical with enough speed to functionally benefit from unloading, you have to pull hard - very hard - very quickly.

Remember thermodynamics. You can't get more energy out than what you put in. Any increase in energy that you'll see is from the engine's output, which will largely depend on how long you "extend" after completing your aileron roll at the top of the Immelman and on the aircraft's acceleration. This isn't part of the maneuver, but rather an advantage of using the maneuver. The Immelman drains more energy than a shallow bank does, but it also allows it to be regained quicker, hence the efficiency of the maneuver. This is the portion that varies based on each aircraft and isn't part of the aerodynamics of the maneuver, but of the raw power output of the plane itself.
Skyyr,

The original situation was a zoom up and dive down, not a stick-to-the-belly immelman turn. It is true that in the initial phase, to get the nose up you have to increase AoA and pay with a little induced drag. I say "little" because normally you start from high speed and induced drag goes down with speed:

"L" the lift is proportional to "G" for a given mass, so you can treat it as G. Yes, the drag goes up strongly with G, but it also goes down with the square of the speed. At the typical 300 mph speeds of the pony, parasitic drag is still a significant part of the drag as long as you do not pull more than 3-4 G (so G^2 factor does not go wild). This will take you about 6 seconds to point the plane from level to 70-90 degrees up. However, the moment you started to lose speed, you also reduce the parasitic drag which is a very significant part of your drag at those speeds. The net drag *almost* always goes down and shifts the energy balance to a net positive. You spend long seconds at speeds in the vicinity of 100-200 mph (up and down) where the parasitic drag is low, prop pulls well and induced drag is nearly zero because you are zooming. This is where you build the energy. By the time you stall at the top of the zoom, you have more energy than you started with - unless you were totally hamfisted with the stick.

Easy exercise:
go offline, fly at 0 feet above  the water at 300 mph and pull into a vertical zoom. You can watch the G meter in the cockpit. Do your hammerhead/whatever at the top, come straight down and pull at wave-top level. What's you speed now?


A flat turn is generally inefficient energy-wise because you are using part of your lift to maintain level flight (and pay the induced drag), and keep the high-speed high parasitic drag component throughout the maneuver. This generally leaves very little room for net energy gains.
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Offline Skyyr

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Re: P51 Question/Controversy
« Reply #18 on: February 06, 2014, 12:42:25 PM »
Skyyr,

The original situation was a zoom up and dive down, not a stick-to-the-belly immelman turn. It is true that in the initial phase, to get the nose up you have to increase AoA and pay with a little induced drag. I say "little" because normally you start from high speed and induced drag goes down with speed:
(Image removed from quote.)
"L" the lift is proportional to "G" for a given mass, so you can treat it as G. Yes, the drag goes up strongly with G, but it also goes down with the square of the speed. At the typical 300 mph speeds of the pony, parasitic drag is still a significant part of the drag as long as you do not pull more than 3-4 G (so G^2 factor does not go wild). This will take you about 6 seconds to point the plane from level to 70-90 degrees up. However, the moment you started to lose speed, you also reduce the parasitic drag which is a very significant part of your drag at those speeds. The net drag *almost* always goes down and shifts the energy balance to a net positive. You spend long seconds at speeds in the vicinity of 100-200 mph (up and down) where the parasitic drag is low, prop pulls well and induced drag is nearly zero because you are zooming. This is where you build the energy. By the time you stall at the top of the zoom, you have more energy than you started with - unless you were totally hamfisted with the stick.

Easy exercise:
go offline, fly at 0 feet above  the water at 300 mph and pull into a vertical zoom. You can watch the G meter in the cockpit. Do your hammerhead/whatever at the top, come straight down and pull at wave-top level. What's you speed now?


A flat turn is generally inefficient energy-wise because you are using part of your lift to maintain level flight (and pay the induced drag), and keep the high-speed high parasitic drag component throughout the maneuver. This generally leaves very little room for net energy gains.


Regarding drag, yes, parasite drag is a source of drag, but it's a not a functional problem in maneuvering because it's most significant at cruise speeds, where the airspeed is of the least concern to the pilot (if you're fast enough for parasite drag to be significant, you're fast enough that your energy state is not a problem). Slowing down to the point that it's minimized is unreasonable in the scope of energy conservation, as it requires the pilot to maintain a speed that makes vertical maneuvering impractical (corner speed). In other words, the only time it matters on paper is when it doesn't matter in flight, because your airspeed would be increasing significantly. The average WWII-sized aircraft aircraft has a parasite drag profile smaller than the size of dinner plate. This can't be changed, so therefore it's not something we actually consider in a combat environment (or a flight environment, for that matter), outside of Va and Vne speeds (and flap/gear extension speeds, but I digress).

Induced drag, however, varies directly with the load put on the aircraft, which is why it matters in the scope of energy conservation. Pilot input does affect it through every phase of flight, whether slow or fast, low or high.

Regarding the Immelman, you're confusing the acceleration given by the aircraft's engine with energy conservation of the maneuver. Try what's described above in an aerobatic glider and look at your airspeed after performing a climb, hammerhead, and extension back to the deck - you will show a loss in overall speed. I know, I've done it more times than I can count.

Further, even straight vertical, you will have induced drag unless you completely unload the aircraft to a 0G configuration, which is impossible to fully do in a WWII aircraft and still maintain straight vertical orientation. Even when vertical, you will have a 1G load on the wings. It's not that much of a load nor is the induced drag significant, but it's not the pure energy gain you're implying it to be.

I'm not arguing that you won't accelerate in a dive with a running engine; this thread is diverging away from the original discussion, to which my original point was that any time you maneuver, energy is lost, including when maneuvering to the vertical. The scenario of diving back down or extending to gain back from the engine is not part of the maneuver, as the Immelman ends the second that you roll wings level after coming out of the top of the half-loop. If you killed your engine and dove back to your starting altitude with a wings-level orientation after performing the maneuver, you'd find that you would have lost airspeed. Just because you are in a better position to regain energy doesn't mean that you actually gained energy - this is one of the fundamental concepts in more advanced energy-fighting tactics.
« Last Edit: February 06, 2014, 12:49:18 PM by Skyyr »
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Offline bozon

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Re: P51 Question/Controversy
« Reply #19 on: February 06, 2014, 06:26:49 PM »
I am sorry Skyyr, in another section of the forum I would have ignored this, but the help and training forum is read by new players.

Regarding drag, yes, parasite drag is a source of drag, but it's a not a functional problem in maneuvering because it's most significant at cruise speeds, where the airspeed is of the least concern to the pilot (if you're fast enough for parasite drag to be significant, you're fast enough that your energy state is not a problem).
Parasitic and induced drag are equal at 1G at your best climbing speed, which for the P-51 is what 160 mph? anything above that parasitic will rule. At 300+ you have to pull quite a few Gs for it not to rule.

Quote
The average WWII-sized aircraft aircraft has a parasite drag profile smaller than the size of dinner plate. This can't be changed, so therefore it's not something we actually consider in a combat environment
What is fixed is the parasitic drag coefficient. The drag varies with speed - a lot.

Quote
Regarding the Immelman, you're confusing the acceleration given by the aircraft's engine with energy conservation of the maneuver.
Acceleration is one manifestation of net energy gain. It can either go into increasing speed (aka acceleration) or into increasing alt (aka climb).

Quote
Try what's described above in an aerobatic glider and look at your airspeed after performing a climb, hammerhead, and extension back to the deck - you will show a loss in overall speed. I know, I've done it more times than I can count.
That is your problem right here - WWII fighters had an engine. The engine is part of the energy equation as a source. Gliders have only a sink if the air itself is still.

Quote
Further, even straight vertical, you will have induced drag unless you completely unload the aircraft to a 0G configuration, which is impossible to fully do in a WWII aircraft and still maintain straight vertical orientation. Even when vertical, you will have a 1G load on the wings.
That is simply wrong when normal terms are used.
** If you look at pieces of the plane you may find a surface that produce some lift, which is countered by the lift of some other piece of the plane - thus you get net 0 lift but still have the drag. That is NOT considered induced drag by anyone that I ever talked to about aerodynamics.

Quote
I'm not arguing that you won't accelerate in a dive with a running engine; this thread is diverging away from the original discussion, to which my original point was that any time you maneuver, energy is lost, including when maneuvering to the vertical.
Wrong. I bold-faced the problematic statement for you - You have an engine.


In AH you fly WWII planes with engines, not gliders. This changes things.
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Offline Skyyr

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Re: P51 Question/Controversy
« Reply #20 on: February 06, 2014, 09:41:16 PM »
Parasitic and induced drag are equal at 1G at your best climbing speed, which for the P-51 is what 160 mph? anything above that parasitic will rule. At 300+ you have to pull quite a few Gs for it not to rule.

You seem to be caught up on the fact that parasite drag increases with speed that you're neglecting to address that there's nothing you can do to minimize it without reducing your airspeed to a level where vertical maneuvering cannot be functionally achieved. If you slowed to 160mph, could you do anything except variations of horizontal turns? No, at least not in a Mustang anyway. No one is going to fly around at a minimum drag speed because the loss in overall energy is too great - at least, they shouldn't.

You're quoting formulas without realizing that they have little functional application in ACM or piloting in general. Parasite drag is ONLY a significant concern to aircraft designers. Why? Because once an aircraft is designed, it's stuck with whatsoever parasite drag coefficient that it has. The pilot doesn't worry about it outside of being aware of it, using it to calculate Vx, Vy, and similar performance speeds. Do you know how large the dedicated section of parasite drag is that we teach to student pilots over the course of a four year curriculum? ONE chapter. It's that relatively insignificant. Know how large the section on induced drag is? It spans the course of multiple maneuvers, flight lessons, lesson materials, and aerodynamic topics - it is a HUGE concern to the pilot because it's a large source of drag and because it not only varies with airspeed, but with maneuvering.

What is fixed is the parasitic drag coefficient. The drag varies with speed - a lot.

See above.

Acceleration is one manifestation of net energy gain. It can either go into increasing speed (aka acceleration) or into increasing alt (aka climb).

Acceleration is not an energy gain. Acceleration is a measurement of the ability to regain energy, but it is not regained energy. It is technically defined as the rate at which the velocity of an object changes over time.

That is your problem right here - WWII fighters had an engine. The engine is part of the energy equation as a source. Gliders have only a sink if the air itself is still.

The engine is an energy source over time. Drop a plane from a dead start into the air and its engine will give it 0 airspeed at 0 seconds into the flight. It needs time to build up thrust. It's potential energy, but it's not an absolute measurement of energy, except from what it takes to climb vertically to achieve that altitude.

Also, the air "moving" makes no difference whatsoever to a glider or any other plane. I can show you examples of 200mph winds that produce zero lift to an aircraft. The ONLY time a wing will produce lift from air passing over it is when it's making a less-than-critical AoA against the chord line of the wing. I'm not sure where you get that "air" keeps a glider or any aircraft from sinking.


That is simply wrong when normal terms are used.
** If you look at pieces of the plane you may find a surface that produce some lift, which is countered by the lift of some other piece of the plane - thus you get net 0 lift but still have the drag. That is NOT considered induced drag by anyone that I ever talked to about aerodynamics.

Any time lift is produced, drag is produced. They're interlinked. Now sure, if an entire airfoil/wing has a zero lift component, then there's no drag. But just because the aircraft has a net zero lift component does not mean the airfoils have no lift. The angle of incidence in most aircraft ensures that there is almost always some lift being generated.

Wrong. I bold-faced the problematic statement for you - You have an engine.  In AH you fly WWII planes with engines, not gliders. This changes things.

But it doesn't change the aerodynamics of the maneuver, hence why it was brought up. An Immelman consists one a half loop and an aileron roll, NOTHING else. You're trying to claim that the energy you could gain from diving (which is using additional power from the engine) is the same as the energy you have when you exit - this is incorrect. The potential energy gained from an Immelman is measured as the power it would take to zoom climb to that altitude in the same aircraft - that it a direct measurement of the energy you have conserved. That, combined with your current airspeed, is your instantaneous energy state.
« Last Edit: February 06, 2014, 10:17:26 PM by Skyyr »
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Offline katanaso

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Re: P51 Question/Controversy
« Reply #21 on: February 06, 2014, 09:54:56 PM »
Not to derail your guys' conversation, but I had time to edit some films and post them for Muzzy as well as others regarding turn fighting a 51 and using all of the flaps:

http://bbs.hitechcreations.com/smf/index.php/topic,358957.0.html

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Offline Tank-Ace

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Re: P51 Question/Controversy
« Reply #22 on: February 06, 2014, 11:14:23 PM »
One key is going to be to use the vertical and "play the egg" so to speak.  Beyond simply dropping flaps at the top of a loop or oblique turn it's sometimes advantageous to actually drop throttle on the way up depending an speed, E state and where your opponent is.  In other words use the climb to drop speed and tighten your turn radius while using flaps to retain lift and bring you around.

Most MA opponents are going to keep the throttle firewalled, particularly on the way up.  Use that to your advantage.

Saw this just felt it warranted a comment.


Be EXTREMELY careful about cutting throttle in the vertical. Especially against a real vertical fighter like the G-14, K-4, or Spit 14. You may wind up handing them the E and alt advantage if you cut too soon, or just chop throttle.
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Offline bozon

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Re: P51 Question/Controversy
« Reply #23 on: February 07, 2014, 09:05:35 AM »
You seem to be caught up on the fact that parasite drag increases with speed that you're neglecting to address that there's nothing you can do to minimize it without reducing your airspeed to a level where vertical maneuvering cannot be functionally achieved.
Skyyr, you keep repeating that and you keep thinking like a glider pilot. You can do something about parasitic drag, it is called converting speed into alt. Then, when the total drag is low, your engine has enough spare power (after paying for the drag that is left) to build your energy. This is why you gain energy in a zoom.

Simple thought experiment:
You are flying on the deck at your max level speed. Can you gain energy in a straight line? no - parasitic drag is eating all the power that the engine can give. Can you gain energy in a flat turn? no you are guaranteed to lose some. Can you gain energy by zooming? YES.

By the way, the original Immelman turn was not a half loop at all. It was a zoom climb followed by a hammerhead/wingover turn at the top, which is very energy effecient and usually nets a positive energy gain.
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Offline pembquist

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Re: P51 Question/Controversy
« Reply #24 on: February 07, 2014, 10:36:44 AM »
How about this: 2 identical aircraft flying say 1/2 way between stall speed top speed, same altitude. One zooms up till just below stall speed and then dives down to its starting altitude, one aircraft stays level. At the end of the maneuver which aircraft is faster, which aircraft is further ahead.

Now same thing only starting at top speed.

Next scenario: 2 identical aircraft as above, top speed. One does an immelman to reverse diving back to the starting altitude and then flying level, accelerating to the same speed as at the start. The other aircraft does a standard rate flat turn to reverse. to keep it simple ignoring the lateral displacement of the the aircraft, when the aircraft are abeam of each other which one will be going faster?

 
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Offline Skyyr

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Re: P51 Question/Controversy
« Reply #25 on: February 07, 2014, 01:00:23 PM »
Skyyr, you keep repeating that and you keep thinking like a glider pilot. You can do something about parasitic drag, it is called converting speed into alt. Then, when the total drag is low, your engine has enough spare power (after paying for the drag that is left) to build your energy. This is why you gain energy in a zoom.

Simple thought experiment:
You are flying on the deck at your max level speed. Can you gain energy in a straight line? no - parasitic drag is eating all the power that the engine can give. Can you gain energy in a flat turn? no you are guaranteed to lose some. Can you gain energy by zooming? YES.

By the way, the original Immelman turn was not a half loop at all. It was a zoom climb followed by a hammerhead/wingover turn at the top, which is very energy effecient and usually nets a positive energy gain.


I'm not saying you can't gain energy after performing a maneuver - I'm stating that the maneuver itself does not net you an energy gain; it simply positions you to gain more energy within a certain time frame - there IS a difference.

I don't (and haven't) disagreed with what you're saying - you're simply misunderstanding the other half of the argument. I also fly and instruct in airplanes (not just gliders), so I'm not talking from the perspective of a glider pilot. It's simple physics that ANY maneuvering will result in more energy loss than straight-and-level flight - that's the nature of entropy. The only wildcard is the engine, but the engine itself does not add energy; the engine produces thrust, which must accelerate the aircraft - that's where the additional energy comes from, it's not instantaneous. Therefore, any speed gain due to descending after an Immelman isn't from the maneuver, it's from acceleration over time due to the engine and gravity and that extra altitude simply helps us make that more efficient.

If we could graph what you're talking about, putting time on the X axis and an instantaneous energy state on the Y axis (where the energy state is a converted value representing the sum of both airspeed and altitude), you'd see that straight and level flight produces a straight line that doesn't deviate. Performing an Immelman would produce a line that dips below the line produced by straight and level flight, and then climbs above it during a dive. However, when averaged out, you'd find that the straight and level line would hold a higher average energy state, as well as a higher sum of energy when measured over time on the X axis. This is because the aircraft can only ever produce a fixed amount of energy, whether straight and level or otherwise. Any energy gained from "unloading" and reducing speed to reduce parasite drag will be offset moreso by induced drag.

The energy graph of the aircraft performing the Immelman would show higher peaks of energy, but they would be offset by the "valleys" where energy is lost in the transitional stages of maneuvering from horizontal to vertical. Anything else would violate the laws of thermodynamics. Now sure, we can operate aircraft at different altitudes, or different airspeeds, to make them more efficient, but the rule still holds true for two aircraft at the same altitude, airspeed, etc.

Consider this:

You have two aircraft at maximum cruise speed. Both are going the same direction. One continues going straight, while the other goes up, does an loop with a shallow dive back down to our starting altitude (which is effectively an Immelman), and then tries to catch the original aircraft. The aircraft that performed the loop will be higher for a short time, faster when it comes back down to altitude... but it will never catch the original aircraft and it will actually be further behind than it was before. There is no net energy gain, only a loss. This is the nature of induced drag.

And yes, I know the definition of an Immelman has changed over time. I'm referring to the more modern depiction of it.
« Last Edit: February 07, 2014, 01:31:54 PM by Skyyr »
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Offline Skyyr

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Re: P51 Question/Controversy
« Reply #26 on: February 07, 2014, 01:44:28 PM »
How about this: 2 identical aircraft flying say 1/2 way between stall speed top speed, same altitude. One zooms up till just below stall speed and then dives down to its starting altitude, one aircraft stays level. At the end of the maneuver which aircraft is faster, which aircraft is further ahead.

Now same thing only starting at top speed.

Next scenario: 2 identical aircraft as above, top speed. One does an immelman to reverse diving back to the starting altitude and then flying level, accelerating to the same speed as at the start. The other aircraft does a standard rate flat turn to reverse. to keep it simple ignoring the lateral displacement of the the aircraft, when the aircraft are abeam of each other which one will be going faster?

 

And this is where energy states come in.

In your first scenario, the aircraft that stays level will be further ahead and the climbing aircraft will come back down faster. However, the distance lost by the climbing aircraft can never be regained and the gain in airspeed that it had will bleed off due to parasite drag. The plane that stayed straight and level will have a higher energy state until the plane that climbed and then descended accelerates back to cruise speed.

The speed loss in the first example will be more pronounced for the slower aircraft, and less noticed in the faster flying aircraft.

In the second scenario, the climbing/diving aircraft might be faster, depending on the angle descended at and the aircraft used. However, you can't just look at speed for an energy state comparison. See my reply above.
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Offline pembquist

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Re: P51 Question/Controversy
« Reply #27 on: February 07, 2014, 02:19:42 PM »

Consider this:

You have two aircraft at maximum cruise speed. Both are going the same direction. One continues going straight, while the other goes up, does an loop with a shallow dive back down to our starting altitude (which is effectively an Immelman), and then tries to catch the original aircraft. The aircraft that performed the loop will be higher for a short time, faster when it comes back down to altitude... but it will never catch the original aircraft and it will actually be further behind than it was before. There is no net energy gain, only a loss. This is the nature of induced drag.

I think that what is confusing is the use of the word energy. If it is just the combination of speed and altitude at any instant than the plane that has looped and come down has more energy than the one that stayed level. The fact that the plane that stayed level is miles away and the plane that looped will be unable to catch up to it despite its higher "energy state"  doesn't matter unless somebody is thick headed and believes in perpetual motion. Consider the same two planes but flying head on, the looper will meet the leveler with more "energy" despite the fact that the leveler has traveled farther in the same time span. I read some post in which the author was bemused any time he heard somebody say they were climbing to get some E. I don't understand why it seems functionally accurate.

What I would like to understand better is how to decide whether tis better to climb or accelerate upon spotting a con.
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Offline pembquist

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Re: P51 Question/Controversy
« Reply #28 on: February 07, 2014, 02:39:52 PM »
My last post crisscrossed so I didn't read your reply to the first, if that makes sense. I think my second post is still reasonably lucid, what say you?
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Offline Vraciu

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Re: P51 Question/Controversy
« Reply #29 on: February 07, 2014, 03:25:55 PM »
Parasitic drag is a function of the design*. There isn't much you can do about it.   Total drag increases with speed (and if you get slow enough, it will increase with deceleration as well).  It can also increase with maneuver.   Lift is the culprit in induced drag. What changes is the ratio.

Assuming level flight to keep it simple....  At slower speeds induced is the problem, at higher speeds parasitic. You can't do much about the latter short of chopping parts off like antennae, cannon bulges--which we cannot do in the game--or rocket rails--which we can in the hangar before we roll.

Knowledge is power but in the case of parasite drag...I am not really sure how it can help you short of saving fuel or gliding dead stick to get home (L/D Max).



* Parasite drag has multiple sub-categories: friction, form, interference, etc.
« Last Edit: February 07, 2014, 04:25:22 PM by Vraciu »
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