Author Topic: A fundemental view on Weight and E state and its role in ACM  (Read 6784 times)

Offline dtango

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Re: A fundemental view on Weight and E state and its role in ACM
« Reply #60 on: June 09, 2009, 08:32:43 PM »
Thanks Badboy, Tango...I'd like to repeat my question though.

You have two airplanes that are similar in all of the "ratio" stuff (wing-loading, power-loading, thrust-drag). They'd have almost identical turn, climb, acceleration, and top speed right? But one is just plain bigger and heavier than the other. Would the heavier one possibly have some sort of advantage in zoom climb?
Let's just for grins use Badboy's example.  

His first fighter had 2000 lbs of thrust, 1000 lbs of drag, 7000 lbs of weight.
T/W = .286
T/D = 2
W/D = 7

a = 32(2000-1000-7000x0.9)/7000  = - 24 ft/s^2

Now let's take a fighter with 4000 lbs of thrust, 2000 lbs of drag, 14,000 lbs of weight.

T/W = .286
T/D = 2
W/D = 7

So ratio wise it's the same as the 1st plane but absolute weight-wise it's heavier.

a = 32(4000-2000-14,000x0.9)/14,000 = -24 ft/s^2

So no advantage (or disadvantage) due to the absolute differences in weight between the aircraft.

We must be careful here though.  These are point calculations to be illustrative.  Badboy picked a point flight regime that helps to explain particular concepts.  I'm doing the same with the hypothetical "2nd airplane" here.  In reality it gets even more complicated because thrust and drag vary with airspeed (and lift of course) which means that the ratios are dissimilar across the flight envelope and it would take more to fully evaluate the differences.  2ndly across the envelope there is actually a portion where greater mass helps but it doesn't make enough of a difference before we enter the region where greater mass is a detriment.  That's why I struggle to fully explain the concepts without the use of some broader modeling.

To me the easiest way to compare the impact of weight is to isolate weight by fixing everything else which we can do by comparing essentially the same aircraft/airframe with only weight being the difference.

See, you hear about big, aerodynamically clean aircraft having an advantage in zoom all the time, even against smaller aircraft that may even have superior power-loading, the best example probably being the mock dogfight with a Spit described in Thunderbolt!. Is this hogwash? Now it has been demonstrated that adding dead weight to a Jug wouldn't make a better Jug, but stacking a Jug up against a Spit, it ain't just a Spit with 8,000 lbs of weight added...is the ratio of mass to total drag a significant positive factor in vertical performance at all?
Yes, that is an example some folks sight but I don't know if that is just misquoted context or misconstrued understanding.  I don't have Thunderbolt! myself and so have no idea what the actual quote is.  Of course we've just shown that if folks are attributing the zoom ability to primarily a function of weight/mass differential then it's definitely a flawed conclusion.  It just doesn't come down to that or mass/drag ratio because you have to include thrust in the equation as well.  Ignoring thrust would well, make this ballistics and not aeronautics ;).

Tango, XO
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Offline dtango

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Re: A fundemental view on Weight and E state and its role in ACM
« Reply #61 on: June 09, 2009, 09:19:48 PM »

 Hey tango, sorry to cut this sentance out and pick at it! I have a another question. Seeing as the title states 'Weight and E state and it's role in ACMs'; Are my tests not a more accurate depiction of 'weight and E state' than an equal speed climb test?

 Here is my understanding of the theory.

1) The two P47s in BnZs' test started at equal altitude and equal speed. Thus the heavier loadout has less E as the only part that counts is climb rate and power/weight ratio.

2) If both P47s are at 15,000ft at equal speed, the heavier loadout has more E.


So applied to aces high, with regard to the boom and zoom tactic (almost always involving a diving attack) the heavier loadout on two identical aircraft performs better? If the two aircraft start at equal altitude and equal speed then the control is still present in the experiment. Apples to apples, so to speak.


edit: Also, if you watch the speeds in the test I posted, they are very similar at the zenith of the dive. The difference seems to be in momentum carried into the zoom?
 On the powered tests I even made the lighter loadout a slightly less G intensive pull out and it still topped out slightly before the heavy.


I'm not arguing, just trying to get my very basic understanding fine tuned by you guys.


Sorry batfink, I didn't look at your films (was at work :) ) so assumed the parameters of your flight tests.  I assumed you were at different airspeeds basis what you said to BnZ.  My bad. 

Not sure what to say about the results of your tests except perhaps variations in technique that make a difference.

Here's a particular point to note however which I think is a source for part of the confusion.

2) If both P47s are at 15,000ft at equal speed, the heavier loadout has more E.

Yes this is absolutely correct.  The heavier P-47 in this case has the greater TOTAL energy.  However if we stop there we can end up with wrong conclusions.  As former fighter pilot Dr. Ed Tomme says..

"Total energy includes mass, so it is not an accurate indicator of maneuverability [or zoom performance in our case].  (The C-5A has the most total energy!)  It is energy per pound of aircraft weight, or Specific Energy (ES) which indicates an aircraft's capability to maneuver."

More relevant to our discussion is Specific Excess Power, PS that is of interest to us which represents the change of energy with respect to time.  In other words though a pilot can decide to re-arrange an airplane’s total energy, the rate at which the energy exchanged is limited and determined by the PS relationship.

See this thread for more:
http://bbs.hitechcreations.com/smf/index.php/topic,209163.msg2488778.html#msg2488778

And it's this energy rate of change that we're really trying to understand which tells us the rate of energy bleed or gain.  And it's this very relationship that also tells us that given the same aircraft, that apples to apples the lighter aircraft has the greater average PS vs. the heavier which means the lighter aircraft will zoom higher.

Tango, XO
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« Last Edit: June 09, 2009, 10:11:02 PM by dtango »
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Offline mechanic

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Re: A fundemental view on Weight and E state and its role in ACM
« Reply #62 on: June 09, 2009, 09:25:11 PM »
Thank you for that explanation sir, you make it easy to understand. The difference in technique sounds very likely. I had not even considered that the heavier/lighter aircraft would get to 4G differently due to the mass.
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Offline BnZs

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Re: A fundemental view on Weight and E state and its role in ACM
« Reply #63 on: June 09, 2009, 09:26:39 PM »
 Ignoring thrust would well, make this ballistics and not aeronautics ;).

Tango, XO
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So, to make it short, any advantage that an aircraft might have when considered as a projectile is not likely to be a significant factor in vertical performance?

Second, what IS the explanation for being able to get nearly identical zoom climbs out of airplanes with widely varying power-loading, I.E, SpitXVI and F4U-1 for example?
« Last Edit: June 09, 2009, 09:28:20 PM by BnZs »
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Offline mechanic

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Re: A fundemental view on Weight and E state and its role in ACM
« Reply #64 on: June 09, 2009, 09:55:06 PM »
I watched the P47 test again and noticed something else interesting.


The second -50% 6guns non-powered test topped out at almost exactly the same altitude as the third -100% 8guns non-powered test. both were almost exactly 12.7k

- lighter aircraft stalled out and turned over at 30mph

- double the fuel load and another pair of guns, the heavy stalled and almost lost control at 55mph but a few feet higher.


That would lead us to believe that in virtual combat the lighter aircraft is still the better choice as you would need to be agile at the top to convert to an advantage, not wallowing over 20mph faster at the same altitude. This is what the experts here are making clear with math and great explanations.



The first -50% 8guns non powered test stomped the other two right up to 12.9k and dropped over neatly at 32mph.

So in moderation the extra weight may be of value. Is it fair to assume if you intend to BnZ your prey in a p47N you would be not helping yourself much taking the six gun loadout?

again though, my test was very basic and lacking multiple results for some kind of average.



« Last Edit: June 09, 2009, 09:58:44 PM by mechanic »
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Offline dtango

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Re: A fundemental view on Weight and E state and its role in ACM
« Reply #65 on: June 09, 2009, 10:09:48 PM »
So, to make it short, any advantage that an aircraft might have when considered as a projectile is not likely to be a significant factor in vertical performance?
Not exactly but when you account for thrust in the equation it makes a significant contribution which overshadows the ballistics only way of looking at it.  That's why batfink's engine idle tests show the heavier airplane zooming higher.  That's what you would expect.  Throw thrust into the equation and things change.  For the range of thrust we're talking about for WW2 aircraft, it has a big impact.

Second, what IS the explanation for being able to get nearly identical zoom climbs out of airplanes with widely varying power-loading, I.E, SpitXVI and F4U-1 for example?
Hehe, too many factors here to make any general statements that would be correct.  In other words there isn't an easy explanation I can think of that explains it for these variety of cases.

Tango, XO
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Offline Agent360

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Re: A fundemental view on Weight and E state and its role in ACM
« Reply #66 on: June 09, 2009, 11:41:30 PM »
Just read the whole thread.

FREAKEN AWSOME!!!  :aok


Offline BnZs

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Re: A fundemental view on Weight and E state and its role in ACM
« Reply #67 on: June 10, 2009, 12:02:15 AM »


So in moderation the extra weight may be of value. Is it fair to assume if you intend to BnZ your prey in a p47N you would be not helping yourself much taking the six gun loadout?



Heh, well you ought to take the guns for reasons having nothing to do with performance...their weight doesn't help near as much as the extra killing capacity helps. I suppose you could reach some point where extra .50s weren't worth the weight...maybe around 25 or so...

Anyway though, even if ballast did give some small edge in a pure zoom, as you know actually fighting with the airplane, even in a so-called "bnz" or energy fighting style isn't going to consist solely of a series of max-speed dives and unloaded zooms, not by a long shot. One will have to put loads on the airplane often during maneuvers (increased weight=increased wingloading), climb, accelerate, reduce throttle in dives to keep your speed reasonable, make use of turn performance, etc.
« Last Edit: June 10, 2009, 12:05:36 AM by BnZs »
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Offline bozon

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Re: A fundemental view on Weight and E state and its role in ACM
« Reply #68 on: June 10, 2009, 05:33:36 AM »
So, to make it short, any advantage that an aircraft might have when considered as a projectile is not likely to be a significant factor in vertical performance?

Second, what IS the explanation for being able to get nearly identical zoom climbs out of airplanes with widely varying power-loading, I.E, SpitXVI and F4U-1 for example?
Starting with the second, the differences in power-loading are just not that big to be significant as all WWII planes have low absolute power/mass ratio. In addition, power/mass is a deceiving number. It is not the same as thrust/mass=acceleration. Your engine may be producing 2000HP, but that does not mean you will be able to put it into kinetic energy. Consider that P=F*v=m*a*v. It means that if power P is constant and you could get all this power into kinetic energy, at v~0 you get infinite acceleration, or thrust. Basically, at very low speeds your 2000HP go to kinetic energy of the AIR that goes through the prop - your HP are literally blowing in the wind. How does this translate into actual THRUST depend on the prop diameter and aerodynamics. If there is some advantage to being light and having high power/mass it is at the low speeds, but it is not as significant as one might expect.

On the other side of the scale, at high speed, the combined weight and high parasitic drag overwhelm the thrust. At these speeds, the excess power left after overcoming the drag is very low and the plane behaves more like a projectile. This gives a small advantage of low drag/mass (likely the heavy) plane, but the deceleration in this part of the zoom is highest so it passes quickly.

So now lets think in terms of absolute energy and altitude the planes will reach. When they stall at the top, all the energy is in potential energy - this total energy at the end of the zoom is what we are after. For specific energy this is proportional to the hight of the zoom.
why is it not much larger for the higher power loaded plane?
Remember that the energy gain from the engine is power*time - so two things happen here: first, the fraction of raw engine power that your airframe gains vs. the part that goes to create wind, gets smaller at low speeds. The other is that the time is short. Because the thrust/weight<<1 for all WWII planes, the time it takes for the speed to drop to zero is not very different and in any case - short, a few seconds. In this short time, even with a little better engine-power/mass you don't get to build a lot more energy/mass, or altitude, using the small fraction of engine power that goes to propelling the plane and not creating wind.


Now the first, following the discussion above. Higher mass may be advantageous in a certain range of parameters. For high enough initial speeds and low enough thrust/weight planes, it may actually work - but not likely in real life. The matter of fact is that planes are not set vertically and released and same velocities. A heavy, high wing-loaded plane will loose a lot more energy just to go from level into vertical in a maneuver that will be sharp enough not to resemble a climb through most of it.

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Offline BnZs

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Re: A fundemental view on Weight and E state and its role in ACM
« Reply #69 on: June 10, 2009, 06:08:24 AM »
So basically any WWII prop plane zooming almost exactly the same from a given speed is to be expected?


Starting with the second, the differences in power-loading are just not that big to be significant as all WWII planes have low absolute power/mass ratio. In addition, power/mass is a deceiving number. It is not the same as thrust/mass=acceleration. Your engine may be producing 2000HP, but that does not mean you will be able to put it into kinetic energy. Consider that P=F*v=m*a*v. It means that if power P is constant and you could get all this power into kinetic energy, at v~0 you get infinite acceleration, or thrust. Basically, at very low speeds your 2000HP go to kinetic energy of the AIR that goes through the prop - your HP are literally blowing in the wind. How does this translate into actual THRUST depend on the prop diameter and aerodynamics. If there is some advantage to being light and having high power/mass it is at the low speeds, but it is not as significant as one might expect.

On the other side of the scale, at high speed, the combined weight and high parasitic drag overwhelm the thrust. At these speeds, the excess power left after overcoming the drag is very low and the plane behaves more like a projectile. This gives a small advantage of low drag/mass (likely the heavy) plane, but the deceleration in this part of the zoom is highest so it passes quickly.

So now lets think in terms of absolute energy and altitude the planes will reach. When they stall at the top, all the energy is in potential energy - this total energy at the end of the zoom is what we are after. For specific energy this is proportional to the hight of the zoom.
why is it not much larger for the higher power loaded plane?
Remember that the energy gain from the engine is power*time - so two things happen here: first, the fraction of raw engine power that your airframe gains vs. the part that goes to create wind, gets smaller at low speeds. The other is that the time is short. Because the thrust/weight<<1 for all WWII planes, the time it takes for the speed to drop to zero is not very different and in any case - short, a few seconds. In this short time, even with a little better engine-power/mass you don't get to build a lot more energy/mass, or altitude, using the small fraction of engine power that goes to propelling the plane and not creating wind.


Now the first, following the discussion above. Higher mass may be advantageous in a certain range of parameters. For high enough initial speeds and low enough thrust/weight planes, it may actually work - but not likely in real life. The matter of fact is that planes are not set vertically and released and same velocities. A heavy, high wing-loaded plane will loose a lot more energy just to go from level into vertical in a maneuver that will be sharp enough not to resemble a climb through most of it.


"Crikey, sir. I'm looking forward to today. Up diddly up, down diddly down, whoops, poop, twiddly dee - decent scrap with the fiendish Red Baron - bit of a jolly old crash landing behind enemy lines - capture, torture, escape, and then back home in time for tea and medals."

Offline bozon

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Re: A fundemental view on Weight and E state and its role in ACM
« Reply #70 on: June 10, 2009, 06:23:57 AM »
So basically any WWII prop plane zooming almost exactly the same from a given speed is to be expected?
There will be differences of course, but look at the plane statistics. The range of power/mass, drag/mass values is not very big and as long as all the planes are underpowered, the differences are reduced even more. It is not like one will zoom twice as high as the other and disappear into the great blue yonder. We are talking about scale of 100ft or slightly more differences.
Mosquito VI - twice the spitfire, four times the ENY.

Click!>> "So, you want to fly the wooden wonder" - <<click!
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Offline chewiex

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Re: A fundemental view on Weight and E state and its role in ACM
« Reply #71 on: June 10, 2009, 06:57:35 AM »
A Zoom Climb is basically any climb with a rate of climb is greater than your sustained climb rate (the one you can see on the charts in hangar). You are using your kinetic energy (speed) to go up, it's basically a conversion of kinetic energy into potential energy (altitude).

Of course, you will get slower and slower, and at one point this kinetic energy (speed) is exausted. Then your engine alone is responsible for taking you higher, you are now in a sustained climb.


It's like in a roller coaster: At the beginning, your car is pulled up the 1st hill by cable (=engine), until it reaches the peak. This part is a sustained climb
When the car goes over the top, dives down, and goes up again on the next hill without any cable, just by it's own momentum - that's zoom climb.

Thank you sir, that explains it perfectly.  :salute
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Offline Dawger

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Re: A fundemental view on Weight and E state and its role in ACM
« Reply #72 on: June 10, 2009, 08:47:19 AM »
Nothing wrong with a cordial discussion.  The basic concept that the OP was getting at is correct and the transformation of kinetic into potential and back again is fundamental to maneuvering performance.  Compared to multiple ongoing threads on ho's, ack hugging, and gang banging, etc., etc., I think this is an overall good thing.   :D

While the resulting discussion might have some value, if false information is not clearly repudiated it taints further discussion.

This is a glaringly obvious posting of bad information so it is very easy to spot.

In many other discussions it is very difficult for a beginner to differentiate between good and bad information.

I see horribly bad information posted on this forum daily that is praised as good advice.

I don't think it serves the game well for that to occur. It would be fine if this was not a venue for beginners seeking help but it is.

Offline BaldEagl

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Re: A fundemental view on Weight and E state and its role in ACM
« Reply #73 on: June 10, 2009, 09:10:39 AM »
It would be fine if this was not a venue for beginners seeking help but it is.

Fankly, I highly doubt that a beginner seeking help is here to learn the finer points of weight, thrust and drag as they relate to zoom climbs.  More likely; what's the best plane, how do I land or how do I shake a guy off my tail.

Even though the opening premise was incorrect, or at least based on incorrect mathematics, this has developed into a great thread.  If false information is posted in this forum, which, as you point out it often is, then it's the responsibility of those who have the correct answers to correct it.  It seems to me that's exactly what's been done here so, no harm done and, in fact, Humble invited correction in his OP.
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Offline humble

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Re: A fundemental view on Weight and E state and its role in ACM
« Reply #74 on: June 10, 2009, 09:13:27 AM »
While the resulting discussion might have some value, if false information is not clearly repudiated it taints further discussion.

This is a glaringly obvious posting of bad information so it is very easy to spot.

In many other discussions it is very difficult for a beginner to differentiate between good and bad information.

I see horribly bad information posted on this forum daily that is praised as good advice.

I don't think it serves the game well for that to occur. It would be fine if this was not a venue for beginners seeking help but it is.

Kiss my grits you pompous little .....

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