How do you figure?

I don't know whether detailed propeller efficiency is calculated in il2 FM. Someone told me it's only a simple value. It's ridiculous that a propeller a/c simulation game dosn't provide accurate efficiency curve. Simply setting 85% for all of CSP ? No, that's totally unacceptable.

People will spend quite a lot time to collect different propeller data such as prop diameter, reduction ratio, airfoil section shape, angle, etc. Next step is to use Xfoil/Ansys(software) to calculate complete efficiency curve for every propeller. It's worthy because <<cliff of Dover>> could also benefit from this work. Don't forget 10% efficiency difference will cause 100-200 HP error.

like this:
Attachment 9813

BTW, efficiency drops as altitude increases. If a CSP get 85% at sea level, there is only 85%*80%=68% at 6000m altitude(800KM/H TAS).
Attachment 9814

Well, I know for a fact from playing that it takes more elevator trim to maintain level flight the higher the altitude. So, I think we can conclude air density effects on wing lift are modeled. I will take a guess Oleg somehow imbedded the prop efficiency losses in the air density values as well. In other words, he came up with a simplified way of doing a complex operation to save cpu cycles.

So now you want TD to rewrite the physics modelling because of something that "Someone told" you... ?

If you can provide verifiable evidence that IL-2 is wrong, do so. But bear in mind that even if you do, this is a ten-year-old game, and is hardly likely to undergo a substantial rewrite that would make little practical difference in terms of relative aircraft performance - at lest, from the evidence I've seen so far. IL-2 gets it wrong at high Mach numbers: but this isn't news. It seems to be fairly consistent in the 'wrongness' anyway, so why worry about it...

As for CloD benefiting from the work, do you have any information at all regarding how this entirely new simulation models such things?

I think this debate has shown a lot more than just a single number WRT real life performance.
It is, however, wrong to assume that the game's model is overly simple and wrong.

http://digital.library.unt.edu/ark:/...dc62616/m1/19/


3-blade Hamilton standard 6507A-2 on P47D, efficiency varies from 83%-63% when TAS is between 0.25-0.7 Mach.


Diagram 1, Cp/blade=0.9, should be at high altitude.

Diagram 2, Cp/blade=0.8, should be at medium altitude.

Diagram 3, Cp/blade=0.6, should be at low altitude.

"Someone" is a man involved in il2 mod developing.

As for relative aircraft performance, if your Tempest MKV(9lbs) get caught by a la7@3000m altitude with same speed, could you just dive in a 60 degree angle to the ground (<720Km/h)and simply get far away from la7 shooting range and then come back to 2500m with a much better "energy saving" zoom ? No, in my experience, you can't achieve that. How could those light--2.5 tons--tiny aircraft--smaller 3-blade ClarkY prop.---la7 dive with same accelaration as a 5-tons-huge aircraft-much bigger 4-blade prop. Tempest MKV?

When TempestMKV/La7 dives to 720km/h=200m/s=0.59Mach on the deck, what's the efficiency?
la7 Shvetson M-82FN 14 cyl. with 2-stage supercharger and direct fuel injection rated at 1,850 hp at 2,500 rpm. VISh-105V-4 3 bladed controllable-pitch metal prop of 10.17 ft (3,10 m) diameter

only 2400-2500rpm engine, merely 3.1meter prop, I don't know the reduction ratio but let's assume 1350rpm for propeller. We know 3-blade 1350rpm 4m Hamilton efficiency is 77% or so at 0.59Mach @ low altitude. Advance ratio for Hamilton=2.22, advance ratio for la7 VISh-105V-4=2.87. wow, 2.87? this ratio is for P47 @0.77Mach.

It's very reasonable for la7 to get only 50% efficiency in this situation. La7 loses extra 35% efficiency? lost 650HP? A piece of Sh*t for La's high speed diving. This soviet monster shows his weakness, haha.

I just suspect 10-year-old il2 FM how to treat prop efficiency. If "someone" tell me what formular il2 uses in FM, everything will be clear.

Not in the envelope of the aircraft. It is a CSP.

CSP could maintain constant efficiency roughly when speed of propeller tip is below 0.85. Please check 3-blades efficiency curve, when advance ratio is far greater than 2.2, namely when prop tip is approaching 1 Mach, the story changes. A CSP will lose efficiency inevitably at high speed diving(a/c noise louder and louder).

Btw, CSP will also lose efficiency when TAS is very very low.

You are correct but don't misapply it as it has little bearing on the game shapes in IL2.

The efficiency is nearly constant in any portion of the envelope that design can sustain flight....

That is the beauty of a CSP.

The very nature of power producers is such that the faster they go, the less thrust they produce. The reverse is also a characteristics of power producers. The lower the velocity, the more thrust they produce. That efficiency drop occurs because the propeller blades are stalled just like in very high speed flight. The reason is different but believe me, both realms, high and low speed, produce stalled blade portions. In the low speed realm, we are looking at speeds at taxi and the first part of take off but our thrust force is extremely high at low velocity. Therefore, in the scheme of things, it is a useless detail to include the reduction in efficiency in a dive. The performance is not sustainable in the first place and our reduction in thrust with velocity is already well approximated by:

In other words, the details are included when you make the standard assumption of .85 efficiency.
You could also incorrectly conclude that all subsonic propeller theory violates the very definition of lift because it does not include the fact lift force develops at right angles to the relative wind. This means that in all propellers, regardless of blade stalling will not produce thrust. Why? As the velocity increases the relative wind gradually shifts and eventually lift produced by our propeller no longer parallels the flight path but is deflected upward.
Fortunately we don't have to do that or at least we would not be adding any accuracy by deriving our own approximation of the effect. It is one more thing rolled up in our standard formulation.

Crumpp, this thread is about DIVE, in a high speed dive, you can easily reach very high TAS which, of course, you can not sustain. Finally, as you pull out of dive and begin to level, the aircraft will slow down to its max. level speed, isn't it? The late dive peroid is a temporary flight process in which you may have 25% efficiency advantage over your opponent bacause both propellers could not maintain constant efficiency(85%), namely your 4-blade prop get 80% and your enemy just have 55%, this is a huge tactic factor. Forthermore, from dive terminal speed to max. level speed, that is the slowdown peroid, you still have efficiency advantage. So how long is your advantage peroid? 40seconds? 1 minute? It depends on your altitude to dive from. You are bleeding his energy in this peroid even if your enemy is the best pilot in the world! You are stealing his energy for 40 seconds by the "hand of God". :)Now you can run away with ease, out of his shooting range. This is called "OUTDIVE".


As I said above, if you are stealing 500HP from your enemy for 30 seconds, he loses "hundreds meters altitude" energy! Imagine that in a il2 combat you decrease throttle to 65% for 30 seconds, that's suicide.

In a high speed flight, a little thrust will give you a lot power, don't forget:

Power=thrust*speed

It is output power not thrust determines your energy state.

come on now. 30 seconds...40 seconds??? in la7 or tempest or whatever, you cruise at 3000 meters then enter dive, it takes just few seconds to hit max level speeds and few seconds more to hit max dive speeds. not that long.

Who is stealing 500 hp???

How much excess excess thrust does a P47 have at 20,000 feet in level flight at Vmax?

NONE

How much excess thrust does any propeller airplane have at Vmax?

NONE

If our airplanes Vmax is 420mph EAS and we dive to Vne at what speed does our Excess Thrust produced by the Propeller = 0?

Vmax

How much Excess power does a CSP propeller aircraft have from cruise flight to Vmax to devote to a dive excess thrust?

All of the excess thrust produced to Vmax...at Vmax our excess propeller thrust = ZERO

What speed does a CSP equipped aircraft traveling at cruise flight reach zero excess propeller thrust at in a dive?

At Vmax....the same speed as in level flight

What is the design propeller efficiency for a CSP in an aircraft envelope from Vs to Vmax?

n = .85 from Stall to Vmax

Crumpp, that doesn't seem quite right based on the diagram you posted a few pages back. Peak efficiency 2.2 at Vmax. but, you can still move down and to the right of the curve towards less efficient (advance ratio increasing for a given blade pitch) as the tip speeds increase and approach mach??? As long as the efficiciency value is > zero, don't you have some thrust being produced by the propellar?

edit:
referring to post #143.

Sure you have some thrust being produced but it is not EXCESS thrust. It is excess thrust that accelerates the airplane.

In a dive from Vmax, the only excess thrust is the component of weight.


Propeller thrust only accelerates the aircraft within its envelope.

okay, I think I get it. you said it three posts back, I just didn't sink when I read that the first time.

btw, this has been a very interesting thread. thnks for your input guys. i learned some things.

Crumpp, :grin: I totally disagree with you. Imagine that you are now flying a la7 with 110% throttle and you get Vmax speed. The efficiency of propeller is 80%, yes, you propeller are producing thrust(not big) which is equal to La7's drag force. Namely, La7 is in equilibrium. You have no excess propeller thrust, right? Till now, we should have no bifurcation.

And then you suddenly turn off your engine and feather your propeller, It's obvoius that you are about to lose speed because your engine is dead, but you push the stick forwards and begin to dive. A portion of gravity will help you counteract the dargforce and make you maintain Vmax during your "dead enigine" dive. If you dive in a more steep angle, you could even faster than Vmax. in another word, you are spending altitude to maintain your Vmax or get even more speed.

During your "dead enigine" dive, if you suddenly turn on your 1850HP engine, will you dive faster? Will engine give you excess thrust?

In my opinion, definitely, you will get greater dive acceleration because your propeller efficiency is 80-60%.

In your opinion, prropeller efficiency suddenly become zero from 80% when aircraft speed is somehow higher than Vmax.



Propeller thrust could accelerate the aircraft beyond its envelope, as long as gravity force comes to help aircraft-----DIVE.

Come on, when my Tempest MKV dives to deck and level fly reaching 780km/h, how long will it take to slow down my Tempest to 600km/h? Only just few seconds?

Are you kidding me?

it depends on what dive angle you took after you turned off your engine at Vmax level and entered your dive. if the dive angle was steep enough, you will continue to accelerate due to gravity. if it is too shallow a dive angle, you will start to slow down. in the latter case, if you are below Vmax in the dive due to too shallow a dive angle and turn on your engine, you have potential excess thrust to tap into and get more acceleration from your prop. However, if you are above Vmax because your dive angle was steep, turning on your engine won't give more acceleration because you can not create excess thrust from your propeller at that point. Your TAS is too high. You are moving down and to the right on your efficiency/advance ratio curve for the given blade angle. Any excess thrust is coming from your dive angle/gravity. When your TAS is high enough that you get no thrust, you should lower your rpms to reduce your drag profile, provided you still have room to your Vne speed.

Umm, who is chasing who here? Tempest has much more dive room than LA7. Push it to ~ 850 kph in your dive extension and zoom climb out from there. Don't waste the zoom energy to 600 kph unless your exiting the fight. Also, Tempest performs better at lower alt.

Face it BlackBerry, Crumpp won.:-P

You are wrong, LOL. I'll show you with La7's data. When La7 get Vmax=610km/h at low altitude with 1850 HP engine output.

power=thrust*speed

thrust=power/speed=(engine output)*(prop-efficiency)/speed=1850*0.735KW*85%/(610000/3600)=8KN=816kgf

The thrust force is equal to 816kg weight, that is 33% of la7's weight(2.5 tons).

The air drag force is also 816kg force, so la7's speed is steady=610km/h=170m/s.

If La7 dive in a angle of "A" in order to get a portion of gravity for help.

sin(A)=0.33

A=19.3 degree.

So you can turn off your engine/feather your prop and dive in a 20 degree angle, I'll bet that you can sustain 610km/h WITHOUT ENGINE and WITH THE HELP OF GRAVITY.

If you dive in 45 degree, I promise your "dead engine" La7 will be faster and faster untill lost your wing(>730km/h). When you reach 650km/h, you turn on your engine, the thrust is not as big as 816kg, but still around 500kgf, that is, you add "half ton" thrust to your diving La7.

Believe it or not.

In your opinion, la7's propeller will provide zero thrust @650km/h because this is out of "envelope"/Vmax.

Try it in il2 4.11m with your buddy's la7, you shut down engine when speed is above 610km/h and he is still using 110% power, and check if he could pulll away from you or not.

And try to shut down your enigne when you want to escape from the battle field by high speed diving(above Vmax) and when the enemy is chasing you with his 110% WEP. If you dare do that, you'll be caught by him even his aircraft is slower than you@level flight.

Cruising at 3000 meters and dive! :)In order to prevent the poor la7 from losing his wings, my Tempest MKV will carefully control the speed below 740km/h, and I'll show you how long I can stay in the high speed flying status.

[/QUOTE] In your opinion, la7's propeller will provide zero thrust @650km/h because this is out of "envelope"/Vmax.[QUOTE]


no, that is not what I'm saying. it may provide thrust. but it is not "excess" thrust. that is the key here. excess thrust, excess thrust, excess thrust. it is not excess thrust because Crumpp posted a diagram on csp propellar that shows you can not have peak efficiency beyond Vmax. The only way to get beyond vmax and create excess thrust is to dive at the necessary angle. go back and look how he defined excess thrust. it's the difference between the two force vectors. in level flight, the force vector from gravity has no forward direction. at vmax and level flight, there is no more opportunity to create excess thrust from the prop. you have to dive to create excess thrust and acceleration.

MadBlaster, When your la7 diving at 20 degree with constant speed of 610km/h, the air drag force is completely counteracted by g*sin(20), and you turn on engine, so that you add extra thrust from engine, although the engine thrust is smaller than airdrag force, this is the excess thrust.

Both gravity---g*sin(20) and engine thrust are allied, the sum of these two, are counteracting against air drag force, since the sum of them is greater than air drag( at 610km/h), your la7 speed increases, when you reach 700km/h, air drag is quite more than g*sin(20), so a portion of engine thrust(eg 40%) will be used in completely 100% counteracting air drag. The excess thrust is from the left portion of engine thrust(60%). The more efficiency, the more excess thrust you get, understand?


If you dive at 60 degree angle reaching 700km/h, of course, your engine thrust could not 100% conteract the air drag by itself, so you need gravity--g*sin(60) to help you, if propeller efficiency is high, you need less gravity to help you, and more gravity will be used as excess thrust. if you turn off engine(efficiency=0), you need the most gravity to help, thus minimum excess thrust.

Therefore, excess thrust comes from the sum of gravity and engine thrust. Gravity and engine thrust help each other, this is teamwork, if one of them performs better, the other will have more ability to counteract air drag, Understand? The more propeller efficiency, the more sum, the faster dive.

I've got your logic, Crumpp. If you still believe that engine could NOT provide excess thrust in a high speed dive, simply shut down your engine. There is a saying:

You got it!

One small tweak though....

It is not that you cannot have peak efficiency beyond Vmax. You could design such a propeller but there is little point as all performance the airplane cannot sustain is instantaneous. Of course there would be some serious design trade off's to gain such performance as well. A supersonic propeller design would not work very well on a WWII Piston fighter.

http://www.aerospaceweb.org/question...s/q0031b.shtml

The CSP is designed to maintain peak efficiency through the designs sustainable envelope.

I drew these pictures. Hope you understand my opinion about excess thrust.

When speed above Vmax, engine thrust is always smaller than air drag force.

Attachment 9825

Attachment 9824

Speed in IAS.

This is my test in Il2 4.11m, Tempest cruising 250km/h @3000 m altitude, and dive to 700km/h, then I tried to maintain 700km/h by lowing my altitude in a shallow dive, after getting on deck at 700km/h, Tempest was slowed down to Vmax-600 km/h.


In this whole process, there is 70 seconds during which tempest speed is between 650km/h and 700km/h(IAS). And it took 40 seconds to slow down Tempest from 700km/h to 600km/h on the deck. Don't forget this is in low altitude where the air is thick--high density, if you dive at high altitude where air density is much less, you could hold a longer time in high speed.


If a La7 follows me from the very beginning with same energy, I can easily drag him down to his low efficciency zone----650-700 km/h IAS for 70 seconds. During this period, I could steal several hundreds of HP from him for more than 1 minutes, it's a HUGE energy loss for La7.

Attachment 9826

See here, 3-blade NACA16 propeller on P47D, propeller tip speed is 1.0-1.1 Mach while efficiency is still quite high.

http://digital.library.unt.edu/ark:/...adc62616/m1/7/

Attachment 9828

here, effect of compressibility on propeller efficiency.

http://digital.library.unt.edu/ark:/...dc62616/m1/25/

Right...

Correct.

You getting lost in the trees Blackberry and cannot see the forest.

Maybe if you go back to my very first post, it will help you to gain a better understanding. You understand what is going on with the actual forces but do seem to be able to recognize it in the math.

http://forum.1cpublishing.eu/showpos...3&postcount=98

At Vmax, that initial excess force is composed entirely of a component of weight.

We don't have to break anything down. The detail is already there in our calculations. To determine aceleration, we need the amount of excess force along our vector of motion. It is that excess force that causes the aceleration.

If we start our dive at a velocity below Vmax, then our initial thrust force is the difference between that specific velocity propeller thrust and zero at Vmax. Then we add the additional component of weight that shifts to thrust.

The derivative between that and equilibrium is your average excess force along that vector...

Your argument that your game would benefit from a more "detailed" propeller model such as Blade Element Theory in dive performance is not valid.

You are confusing the mathmatical process of summing the forces with what is actually going on with those forces in a moment in time.

We have already considered that moment in time when we determined our derivative.

yes blackberry. this example is rigged though. if i spawn from 3000 meters and V max with the engine off, I can also create acceleration or de-cceleration simply by changing the angle of the dive and the result is a new terminal velocity/equilibrium. it's the change in the sum of the two vector forces, not simply "the sum of" the two vector forces. the thrust vector is still zero in this case.

But we have been assuming up to now the engine is on, level flight and v max.

well said.

You understand Blackberry that is the excess force that moves the aircraft it is new equilibrium point.

The summing of the forces determines the vector of motion but the rate of change in motion along that vector is a function of the excess force.

For example, when the engine is off our thrust = 0 but our drag force remains. The summing of the forces results in a negative vector and our aircraft slows down as it seeks a new equilibrium point.

If the pilot pushes the stick forward to control the angle that he can shift a component of weight in order to counter act that drag force to maintain velocity or even speed up.

Right.

However, what is equilibrium point? In my opinion, the equilirium point is the point where all forces are well balanced.

You dive in 45 degree with 110% WEP, your speed is always increasing, BEFORE you reach the so called " equilibrium point", you lost your wings.

Try il2 4.11m, use Tempest, P51D, P47, fw190d,bf109K, La7, to dive in 45 degree with full throttle, you can NEVER find a equilirium point where the speed stops increase.Could you? Could you? Why? The equilirium point is sth 1300km/h! Those planes will explode before reaching 1300km/h, believe it or not. They are not supersonic a/c. :)

If you dive in a very shallow angle, for example, dive at 10 degree, the equilirium point is sth.700-800km/h which piston planes could withstand. Yes, you can reach this equilirium point.


As I said again and again, dive steeply. If you dive at 45-60% degree, you are always accelarating, you are always increasing your speed, therefore, before you pull to level flight, there is no equilirium point at all in your dive.

Since there is no equilirium point, the Newton 2nd rule tells us:

Force=acceleration*mass

Thus:

dive acceleration=(Excess Thrust in steep dive)/aircraft weight=(engine thrust + gravity along dive- air drag)/(aircraft weight.)

Imagine that there are two la7, "A" is equipped with high efficiency propeller, the other "B" with low efficiency prop, anything else being equal.

At any piont on the 45 degree dive, A always has more engine thrust than B, thus A always has bigger excess thrust than B, finally, A always has better dive acceleration than B.

What are you talking about here?

Do you need a method to estimate the velocity of the new equilibrium point given the increase in forces?

Sure there is blackberry, you just have not reached it yet.

Your equilibrium point velocity is found by converting the Vmax TAS to EAS. We are going to change altitude and we don't have to constantly mess with density effects.

Now we can use the relationship our parasitic drag component to find our new velocity. You already know the Cdo of the design.

So using the relationship of parasitic drag force and velocity:

Dp2 = Dp1(V2/V1)^2

Re-arrange to solve for velocity:

At 100 KEAS our airplane produces 5000 lbs of parasitic drag. At what velocity will it produce 10000lbs of parasitic drag?

100 KEAS * SQRT(10000 / 5000) = 141 KEAS

If we suddenly gained 5000lbs of thrust (5000lbs plus 5000 lbs) our new equilbrium velocity would be 141 KEAS.

When P51 at Vmax=700km/h, what's the air drag force? 1500lbs?

When P51 dive in 45 angle with full throttle, what's the next equilibrium piont speed?

1200km/h!

When Bf109K dive in 45 angle with full throttle, what's the next equilibrium piont speed?

1100km/h!

When P47D dive in 45 angle with full throttle, what's the next equilibrium piont speed?

1300km/h!

When TempestMKV dive in 45 angle with full throttle, what's the next equilibrium piont speed?

1200km/h!


When Dora dive in 45 angle with full throttle, what's the next equilibrium piont speed?

1100km/h!

Since those planes never reach so called the next equilibrium piont, the equilibrium is totally useless in analysis of 45 degree dive acceleration.Since you have to pull your aircraft out of 45 degree angle far before reaching the equilirium point, you are always accelerating in dive, you always have excess thrust during dive, higher efficiency propeller always provides higher thrust.

The sum of gravity and engine thrust is always bigger than air drag force.


I post the picture again, do you understand me?

Attachment 9831

blackberry, the sim would be very boring if all the forces were balanced in equilibrium. there would be nothing to do!:-P I can only think of one time that occurs, when your on the ground and the engine is off and the wind is zero. equilibrium point is not something that has to manifest to exist. obviously, there are constraints that may make it impossible to achieve (e.g., max dive speed). It is the process of moving from set of conditions to another. Dynamic verses Static. This is the concept that is being pointed out.

Yes，in 45 degree dive, aircrafts are always unbalanced , aircrafts are always increasing speed, always accelerating.

This is a dynamics process, in this process, if my la7 throttle is 100% while your la7 throttle is 60%, I can always outdive you when moving to the equilibrium point which impossible for real flight.

Why? My la7 100% engine produce more thrust than your 60% engine in 45 degree dive, so I have more excess thrust during the whole dive, finally, I can get better dive acceleration than you. understand?

Even if you turn off fuselage damage in difficulty setting, both of us could achieved "equilibrium" speed:

I, with throttle 100%, equilibrium speed is 1000km/h
You, with throttle 60%, equilibrium speed is 950km/h

I can still pull away from you because my engine give me more thrust. Look this picture, when both of us reach equilibrium point, my air drag force is higher than you, this is to say, my speed is bigger than yours.
Attachment 9832

Now , do you still believe that engine could not produce excess thrust during high speed>Vmax dive?

Sorry, I need more information because the example keeps changing. What is the speed of the two planes before they go into the dive? Is the one at 100% throttle already at Vmax? Or are they both at Vmax because the the one at 60% dove down from higher altitude previously? Honestly, this example doesn't seem good one. The information is sketchy. Are you trying to trick me?:)

Imagine there are two tempestMKV named 'X' and 'Y' respectively.

X is equipped with a good CSP, while Y is equipped with a bad CSP. Everything else being equal. both a/c weight equal. drag coefficient equal, engine is equal ......

Within the envelop, within Vmax, both good propeller and bad propeller share same efficiency, they performs identically. good prop=bad prop=85% efficiency

But when dive into 0.6-0.7Mach which is out of a/c envelop, good propeller=80% efficiency, bad propeller=60% efficiency.

At first, both X and Y use 100% throttle in level flight SIDE BY SIDE, because they share same efficiency within evelop, both speeds are Vmax.

Then, both begin to dive at 45 angle, and after a while, both speed are entering 0.6-0.7 Mach, suddenly, "Y" CSP lose efficiency to 60%, while "X" efficiency is still 80%. That is to say, "X" engine could provide more thrust.

Question: Will "X" begin to pull away from "Y" from now?

doesn't make sense. the props can not be identical. the curves would have to be different slopes. "suddenly"...is this realistic if they are identical? no.

But assuming it is realistic, i say tenatively 'yes', Y falls behind from this 'sudden' loss of thrust on the bad prop and gravity has not completely taken over yet, so I guess there is an excess reverse thrust. but I have to think about it some more.

You are beginning to understand me.

"X" vs "Y" ,Everything ELSE being euqal, propellers are different: good prop vs bad prop.

During TAS 0.2-0.6 Mach, good prop=bad prop=85%

During TAS 0.6 -0.8 Mach, good prop> bad prop; ie 80%>60%.

"Y" has no more gravity for help because both dive in same angle=45 degree, and both weight=5 tons.

"Y" has no reinforcement.

the example works better for me if you assume the 'sudden' drop in efficiency for the bad prop happens at the margin of the csp peak envelope at/near Vmax/level. And to make it even more emphasis, the efficiency drops from .85 peak to 0 no thrust when 1 more kph is added above Vmax/level in 1 second, as soon as you enter the 45 dive. also, drag force just kicked you back into the peak envelope. okay, yes the bad prop now has excess negative thrust. because in that 1 second you went from max thrust to no thrust from the prop and drag force pushed you back and not enough time has elapsed for excess thrust from 45 weight vector to overcome the momentary loss from the prop. I'm probably missing something.


Now, we know there are no 'sudden' drops in these curves. they have slopes and they are a functions of TAS, RPM, reduction ratio, blade diameters...etc. this example, rigged.

Good prop and bad prop share same diameter, same rpm, same advance ratio@same TAS. Their DIFFERENCE IS THE SHAPE OF AIRFOIL SECTION.

TAS Mach / good CSP / bad CSP
0.6 85% 85%
0.61 84% 83%
0.62 83% 80%
0.63 82% 77%
0.64 82% 73%
0.65 81% 70%
0.66 81% 67%
0.67 81% 64%
0.68 80% 60%
0.69 80% 56%
0.70 80% 52%

//////////////////////

Your TempestMKV equipped with bad propeller, my Tempest with good propeller.

If I drag you into a high speed dive, around 0.7 Mach TAS for 40 seconds, what is your energy loses?

To simulate this sharply efficiency drops for bad propeller, you can simply use il2 4.11m, quick mission, spawn @ 5000 m with Tempest mkv, you intently decrease your throttle to 60% during 0.6-0.7 Mach 45 degree dive and 45 degree zoom.

You will feel the energy loss.

putting the weight vector/dive angle to the side for a bit, I think what is missing here in the discussion is the drag vector. Now, beyond peak/vmax. what is happening is the drag vector is slowly overwhelming the thrust of the propeller as it gets more and more inefficient until efficiency is at zero and advance ratio reaches some high number at some given high TAS. So, it is negative excess thrust between those two vectors because drag force is becoming greater than the prop thrust vector (still ignoring the weight vector for now). So, the question is how is the interplay between these two vectors modeled in IL-2. Maybe it is already accounted for in the modeling of the drag force? Like an effectiveness factor or something that is attached to the drag coefficient as the thrust begins to degrade beyond Vmax? I really have no idea, only speculate. :)


edit:
of course, the only way to get beyond vmax is to dive, and its the sum of all forces...etc. so this interplay between opposing thrust and drag vectors gets masked over by weight/dive vector. am i on the right track?

Airplanes certainly do reach their equilibrium point in a dive.

However the most common restriction to dive performance is dynamic pressure limits <flutter> and mach limits.

Completely irrelevant though as the equilibrium point estabilishes the rate of aceleration in the dive.

You can debate it all day long but it does not change the fact it is how performance is predicted.

The excess force is what drives the aircraft to equilibrium. It is all accounted for in the formulation.

The interplay is covered in the derivative.

You start out with the maximum force which is the moment in transitioning that a component of weight has shifted to thrust and the propeller thrust is still at level flight velocity. That is the maximum excess force you will have available.

Let's look at the rectilinear motion equations and solve both a zoom climb problem and a dive problem using the same airplane at the same entry speeds.

We will end our zoom at Vy or best rate of climb speed and end our dive at the equilibrium point.

We are going to just simplify the drag and thrust values to illustrate the mechanics of solving the problem.

Characteristics of our Airplane:

Weight 9000lbs
Thrust in lbs = 1000lbs
Drag in lbs = 500

Zoom climb from 300mph to Vy at a 45 degree angle:
The airplane will move to equilibrium

Entry speed = 300mph = 441fps
Zoom Angle 45 degrees
Vy = 150mph = 220.5fps

Zoom height:

Sum the forces on the flight path -

9000lbs * sin 45 = 6364lbs
1000lbs – 500lbs - 6364lbs = 5864lbs

a = F/m

m = 9000lbs/32.2 = 279.5 lb-s^2/ft
a= 5864lb/279.5lb-s^2/ft
a = 20.98 ft/s^2

s = (V1^2 – V2^2 ) / 2a

s = (441^2 – 220.5^2)/(2 * 20.98ft/s^2) = 3476.18 ft

3476.18 ft * sin 45 = 2458 ft

Now let's dive under the same conditions:

Characteristics of our Airplane:

Weight 9000lbs
Thrust in lbs = 1000lbs
Drag in lbs = 500

Zoom climb from 300mph to Vy at a 45 degree angle:

Entry speed 300mph = 441fps
Angle of Dive 45 degrees

We need a ballpark of our equilibrium speed. A quick method is to use the relationship of Parasitic drag. It is the major drag component at high speed. More detailed analysis will give better results but this is accurate within 10%. 10% is acceptable for climb/dive performance.

441(SQRT 1000/500) = 624fps

At 624 fps, the acceleration about the CG will be zero.

Dive:

Sum the forces on the flight path -

9000lbs * sin 45 = 6364lbs
1000lbs – 500lbs + 6364lbs = 6864lbs

a = F/m

m = 9000lbs/32.2 = 279.5 lb-s^2/ft
a= 6864lb/279.5lb-s^2/ft
a = 24.6 ft/s^2

s = (V1^2 – V2^2 ) / 2a

s = (441^2 – 624^2)/(2 * 24.6ft/s^2) = -3961ft (negative is a vector direction)

3961ft * sin 45 = 2801 ft of altitude lost!!

There are very few airplanes with a "bad propeller". It does happen but is caught very quickly as nothing will line up between predictions and the realized.

Including dive performance....which is how they caught it.

That is why Hawker initiated the study.

Okay that is helpful. What Blackberry is wondering, we know the software does these calculations continuously every few milliseconds. In a high speed dive beyond Vmax where TAS is increasing, does the software recognize the degradation of this value? -> Thrust in lbs = 1000lbs. Or maybe it handles it by applying effectiveness factor to this value ->Drag in lbs = 500? Or maybe it ignores it and keeps both opposing forces as static (i.e., "maximum excess force you will have available")?

I can't answer any specific questions about IL2 FM. If it uses standard methods to predict aircraft performance, then it does account for it.

I would be willing to bet it does use standard methods.

Now you know the specific numbers for thrust and drag change. For example here is the P47D22 at Take Off weight from Vs to Vmax:

Thrust available in Pounds
6353.75
6353.75
6353.75
6051.190476
5776.136364
5294.791667
4887.5
4538.392857
4235.833333
3971.09375
3737.5
3630.714286
3529.861111
3434.459459
3344.078947
3258.333333
3176.875
3099.390244
3025.595238
2955.232558
2888.068182
2823.888889
2762.5
2703.723404
2647.395833
2593.367347
2541.5
2491.666667
2443.75
2425.09542

Thrust required in Pounds
1537.476184
1537.476184
1537.476184
1458.574233
1396.29734
1311.942922
1269.557568
1259.393629
1274.840011
1311.284626
1365.430601
1398.36277
1434.870602
1474.747484
1517.813624
1563.911983
1612.9049
1664.671298
1719.104337
1776.109457
1835.602718
1897.509403
1961.762824
2028.303316
2097.07736
2168.036844
2241.138416
2316.342935
2393.614985
2425.09542

yes, from your earlier post on parasitic drag force verse velocity and EAS method.

I suppose you got those numbers from customizing udpgraph tool? I think you must manually configure it somehow, or do calculation off the output file? I don't see those parameters in my version.

Crumpp，when airplane zoom 45 degree from 300mph to 150mph, It takes a quite long time: 20 seconds?

In the whole zoom process, Is the acceleration a constant or a variable ? All your formular is of "constant accelerated motion". That's not correct.

In fact, when airplane begin to zoom, the "a"=20.98ft/s^2, however, after a few seconds, as speed drops, the engine thrust increase to 1100lbs, and the air drag will be 450lbs, so the "a" is no long 20.98, "a" will be smaller than 20.98. The acceleration changes during the whole zoom process, so it's a "variable accelerated motion".

I am surprised, your math model is too simple to get correct result.

Again, constant acceleration formulars are wrong in calculating airplane's dive.

Do you mean when the airplane reach 624fps(424mph), the acceleration is zero?


OMG, why don't you open il2 4.11m and choose any piston aircraft, spawn at 7500m altitude and dive at 45 degree with full throttle?

I promise your aircraft will lose wings sooner or later. You could NEVER reach the equilibium point where the speed stops increase.

Crumpp, plz run il2 4.11m and try some tests, it's very easy for you to find the truth.

This is the expression for the average acceleration:

a = F/m

m = 9000lbs/32.2 = 279.5 lb-s^2/ft
a= 5864lb/279.5lb-s^2/ft
a = 20.98 ft/s^2

It is all in the formula, blackberry.

Think about it. Do you know what the acceleration would be if just gravity is acting alone on the airplane?

32.2ft/s^2.....

Do you think our acceleration is going to greater or smaller than the acceleration of gravity at the beginning of the dive when we have the most excess thrust?

GREATER THAN

Why is our average acceleration lower than that of gravity???

20.98 ft/s^2 < 32.2ft/s^2......

Because there is an aerodynamic braking force acting on the aircraft that lowers the average acceleration.

Yes the formulation requires constant acceleration but that does not mean it does not accurately describe the motion of the aircraft.

For this airplane under these conditions...YES.

It is called terminal velocity.

g=32.2ft/s^2=9.8m/s^2

At the beginning of 45 degree dive, dive acceleration is greater than g due to bigger engine thrust and smaller air braking force, and at the end of dive, acceleration is smaller than g due to smaller engine thrust and bigger air braking force.

But we need to calculate acceleration very accurate. In a 15 seconds long 45 degree dive, if my average acceleration is slightly than yours, for example, is 3ft/s^2 or 0.9m/s^2 more than yours. I'll be 50km/h faster than you in the end. That's a hugh advantage.

Do you remember the 60 degree dive between P47D and fe190G in 1943 summer? At the end of dive, P47D had a much greater speed than fw190G, sth. like 50km/h difference is huge enough.

I have a question for you:
At 1st second, you begin to 45 degree dive,
At 2nd second, your speed increase a little, acceleration is greater than g.
at 3rd second, your speed increase a little more, and acceleration is still greater than g.

.....

at 20th second, you reach the equilibrium point, and acceleration is ZERO. Your speed remains same.
at 21st second, your speed is same as 20th second.
at 22nd second, hour speed is same as 20th second.
......

How about the 19th second? what's your acceleration? It should be smaller than g, but I need more accurate data.

Is the acceleration at 19th second nearly zero or around 0.5*g?

Look at what the aircraft are placarded for in terms of dive speeds.

The aerodynamic forces will tell you the FW-190A can outdive the P47. Just look at the sea level speeds and power required. At sea level, EAS = TAS and EAS is the speed the airplane always feels.

The P47 generates ~272 THP more than the FW-190A8 to travel ~20mph slower. It takes a lot of power to push that big heavy P47 through the air.

However, the relationship of thrust and drag is not the primary limiting factor in a dive for these airplanes. Dynamic pressure limits and mach limits tend to set the speed limits in WWII fighters.

In the case of the FW-190 vs P 47, the FW-190 is limited to ~466mph at low altitude while the P47 is limited to 500 mph at low altitude.

Those placard limits are not set arbitrarily nor is there any wiggle room or safety factor. A Focke Wulf pilot exceeding 466 mph is taking a huge risk he will not survive the dive. There are plenty of incidents of FW-190 pilots diving straight into the dirt barrier because mach effects made the elevator control ineffective. There are also incidents of the pilot turning the aircraft to confetti by aggressive use of the trim to recover.

In a dive to Vne, the P47 will always gain ~34mph over the Focke Wulf FW-190A.

Come on, Achilles and the Tortoise story is very simple.

Sum of time=1+0.5+0.25+0.125+........ is limited, will never bigger than 2.
of course, Achilles can't over take tortoise WITHIN 2 seconds because tortoise is some distance ahead of Achilles in the beginning.

In ancient time, people didn't know the property of "infinite series", so they were puzzled. They drew the conclusion that Achilles would never overtake tortoise because there are infinite numbers of time series in the sum, We modern people will laugh at them because the sum of infinite numbers is limited to "2".

My story, is totally different, from 1st to 20th seconds, a 45 degree diving a/c acceleration changes from g(usually higher) to 0. During the dive process, a/c moves to equilibium point. The acceleration is always decreasing. If you treat it as a constant acceleration motion, you'll make a mistake.

I am a little disappointed that we havn't known each other's opinion after many posts.


Crumpp, P47 Vne=500mph, Fw190's Vne=466mph, the difference is 34mph, so if both of them dive steeply and pull to level flight at Vne speed, P47 is always 34mph faster than fw190. That' very simple, all of us know it. However, that is "terminal speed difference", not dive acceleration difference.

Although X plane's Vne is more than Y plane's, it is possible that X is outdived by Y WITHIN Y's Vne speed. The test in Italy 1943 between P47D an Fw190G is very useful to demonstrate this.

That is my explanation:

g=9.8 m/s^2=32.2 ft/s^2

When both Fw190G and P47 dive in 65 degree angle, their acceleration along path(65 degree) is calculated in this formular:

Acceleration=g*sin(65)+g*(thrust/weight)-air braking force/weight

Fw190A has better thrust/weight, P47 has less braking force/weight, but at low speed, fw190's advantage is more profound, so P47 was outdived by fw190g INITIALLY, although P47's Vne (500mph)higher than fw190(466mph). This perfectly demonstrates that terminal speed and dive acceleration are totally two different concepts.

In later dive period, P47 passed fw190, both arrived 3000ft but P47 has much greater speed.Both speeds of them are well BELOW 466mph. Furthermore, P47 got higher speed BEFORE reaching 3000ft, isn't it? Don't forget fw190 was ahead of P47 after initial dive.

For example, when fw190 reached 8000ft altitude, P47 was at 8500 ft, so P47 was outdived by fw190 initially.On what altitude did P47 get same speed as fw190? Perpahps at 5000ft, at that time, fw190 was still ahead of P47, so fw190 probably at 4500ft. BEFORE P47 got same speed as fw190, P47's acceleration was better than fw190, otherwise, how could P47 get same speed? Don't forget P47 was slower than fw190 after initial dive. So P47 time line is below:

Altitude
@8500ft, was slower and after fw190
@6500ft, began to get higher acceleration than fw190, but still slower than fw190
@5000ft, get same speed as fw190, but still after fw190
@4000ft, faster than fw190, but still after fw190
@3000ft, passed fw190 with higher speed and higher acceleration. At this time, both P47 and Fw190 speed are BELOW 466mph.

@1000ft(if don't pull to level), ahead of fw190, faster than fw190, higher acceleration than fw190 and the distance between them are enlargening.

@-5000ft(deep valley! Air density =SL), P47 still has a little acceleration, while fw190 reaches equilibium point at 1200km/h (if firm enough). Distance between them is big.

@-7000ft (deeper valley, air density=SL), p47 reaches equilirium point with 1400km/h(if firm enough)and fw190 is still 1200km/h, distance between them is bigger.

@-10000ft(deepest valley,air density=SL), p47 is still 1400km/h, fw190 is still 1200km/h, distance between them is even bigger and increasing for ever.


if @1000ft level flight(pull his stick to avoid loss), ahead of fw190, faster than fw190, and LOWER SLOW DOWN RATE than fw190, distance between them are enlargening.


The most interesting is the level flight out of envelop, above Vmax. When both P47 and Fw190 fly at Vmax at sea level, P47's 7ton weight couldn't himself at all, and P47 has a bigger wing area while same 2000HP engine with fw190, so P47's Vmax is less than fw190.

But if both of them level fly above Vmax, the 7 tons weight helps P47 a lot because the heavier, the more ability to retain high speed above Vmax.

Acceleration=g*(thrust/weight)-air braking force/weight

Acceleration is always negative when you are slow down from 800km/h to 600km/h on the deck, and it usually takes a whole minute/60 seconds to slow down to 600km/h(Vmax).

Thus if P47's big 4-blade propeller efficiency is 20% higher than fw190's small 3-blade at the speed range from 680 to 800km/h IAS(out of envelop), fw190 will suffer a lot of ENERGY BLEEDING.

We all know bf109k4 has a excellent climb rate over La7, it usually takes k4 a whole minute to establish 500 m altitude advantage if both have same energy in the beginning. P47/P51/Tempest could also establish 500 m higher altitude if they use high speed boom and zoom tactics-----their own method! AS EFFECTIVE AS BF109K4.

It is obvious for all il2 players if 109K4' s climb rate is degraded to la7 level. People will shout:"Sh*t, where is 109K4's climb advantage? how can I do a energy fighting?" Unfortunately, if il2 neglects the propeller efficiency difference/sharp drops between various aircrafts at high speed(680-800km/h), P51P47Tempest will lose half of their energy fight tactics effect------energy-saving high speed boom and zoom, and not many players will notice it, many players don't know that P51P47Tempest are tigers WITHOUT teeth in il2. The teeth had been taken by someone who, for some reason, didn't pay enough attention to high Mach number/compressibility effect which leads to efficiency decrease of a CSP.

Blackberry, it's complex calculus algorithm. But in computer, this is handled/simulated by optimizing the algorithm every few milliseconds. He already said, he doesn't have access to the flight model. What exactly to you expect him to do? His formula describe at a moment in time when forces are max. He would have to expand the formula from Tzero to Tn when terminal velocity is reached.

Really, I shouldn't talk about math or physics. I had to copy off my buddy's papers to get through high school. Just a heads up.;)

Ah, what the hell.

T=time. If you take your formula below and integrate it over T(intial) until T(terminal), where T(terminal) is time when terminal velocity is reached and T(initial) is when forces are at maximum, I think that gives you your answer. But I make no promises.

Acceleration=g*sin(65)+g*(thrust/weight)-air braking force/weight

I see you were still writing.

You have to audit the numbers coming out of the sim using devicelink tool. That is the only way you could prove it. I tend to think that would be rather difficult. There's probably some stuff you need visibility on that devicelink numbers don't tell you. Maybe some randomness. For example, rudder yaw. But you have the input parameters show in this thread a few posts back, so you could do reasonableness approximation to get comfortable with it. You could build your own algorithm in C++, put in those input parameters, and run it with output to a file. then check that output to devicelink numbers from the sim. if the numbers are close, then it is accounted for. if not, then your statement could be true. otherwise, it is speculation. Crumpp says he thinks it is accounted for in standard formulation and generated some output that does not disprove his conjecture.

Accurate calculation should be performed by Computer. As il2 players, we just need to know on what speed will we dive with better acceleratin than opponents.

In the example of P47 vs Fw190G, the P47 did three steps to overtake Fw190.

1) to get better dive acceleration

And after some time,

2) to get a higher speed

And after some time,

3) to over take fw190G

Therefore, P47 acquired better dive acceleration firstly. So we can estimate roughly:

A)From 400km/h to 500km/h IAS, fw190A has better 65 degree dive acceleration.

B)from 500km/h to 750km/h and above, P47 dives faster.

It's important for P47 to keep high speed in battle field. If a fw190a rides on you at 3000m altitude, with same altitude and IAS, he is ready for shooting at you because there is 350 meters distance between you. THE SPEED OF YOU IS VERY IMPORTANT. If both of you are 500km/h, your P47d could outdive fw190a immediately in a 65degree dive, when you dive to 900m altitude, you could pull around 150m away from him, thus you are out of his shooting range(500m) when both of you begin to level fly near the sea level.

Furthermore, you may 60km/h(17m/s) faster than fw190a in the end of dive(750--800km/h), during the slow down level flight of 30 seconds, fw190a speed decreases more quickly than you, roughly you are round 1000 meters ahead of fw190a when both of you become slower than 650km/h.

I've never seen that significant dive advantage in il2, so I suspect the propeller efficiency calculating (700km/h-800km/)in il2 FM.

I just got a flashback of chasing p47 in a steep dive in 109 g-10. I got some light hits with machine gun and he was getting away, then he pulled hard g and the wings flew off. No EAD, but it was pretty funny. All he had to do was fly straight.

You know the 4.11m simulates the G effect on airplane structure, if you pull your stick too hard at high speed, you may lose your wings or get damaged.

I think the 4.11m is lack of pilots strength simulation. Before 4.11m, we had a "super" airplane structure and "super man" pilot with great strength.

"super" plus "super" is "ordinary" result: aircraft is quite firm at high speed maneuver.

However, in 4.11m, we have a "ordinary" plane structure and "super" pilot.

"ordinary" plus "super" is "oddness": aircrafts are too fragile. At high speed, the stick should be very heavy, so heavy that a pilot could not pull out 8-11g even with his full strength. So we need "ordinary" plane structure and "ordinary" pilots, the result should be "ordinary": aircrafts are not so fragile at high speed.

Interesting. Do you think this could be achieved through software alone?

Aviar

I don't think I was clear.

The FW-190 has a higher terminal velocity than the P47. It developes almost as much power but has lower drag production.

Niether aircraft though is limited by the relationship of thrust and drag. Dynamic pressure limits and mach limits set the boundary.

This is not necessarily due to any aerodynamic limitations, Blackberry. There were human being operating these airplanes for the test in question.

Some operational realities of these two aircraft:

1. The FW-190 pilot closes the cooling gills, which are closed anyway for level flight, ensures the propeller is in automatic, which it is for normal operation unless the KG is operating in emergency mode, pushes the nose forward and dives. His engine is automatic and adjust's as required for actual conditions under air load. The supercharger system responds instantly to power changes. In short, he pushes the stick forward and goes....

2. The P47 pilot does not have a automatic system. He has to trim the aircraft tail heavy, reduce his manifold pressure, reduce rpm to 2600, close cowl flaps, and then dive. He has a throttle he must baby because of the turbocharger. A turbocharger does not respond instantly to power changes and the P47 is much more restricted in its use. Reducing it too fast and he can lose pressure in the system which will take him ~30 seconds of level flight to get back. He cannot dive until he reduces it. In short, he cannot just push the stick forward and go......

Ternimal velocity (speed of equilibium piont)depends on many factors.

A: fw190a8 with full oil tank and ammo, 4 tons weight
B: fw190a8 almost out of oil and ammo, 3.5 tons weight.

Max level flight at sea level: A=B
Dive in 45 degree: A>B

The heavier of aircraft, the more benefit in dive. in a 0 degree "dive", that's level flght, fw190a8 has more terminal speed than p47d, eg, 580km/h vs 560km/h. In a 45 degree dives, obviously P47 has more.


.

Perhaps you are right about the operational difference between P47 and Fw190 in summer 1943. Perhaps the fw190 pilot was waiting for p47 until p47's pilot finished his preparation before dive. Don't forget that was a "test", not a real "combat". Btw, there is evidence that p47 propeller has more efficiency at 2600rpm than 2800rpm in a high speed dive. Does P47 need to reduce manifold before dive?

Anyway, for some reasons, P47 was outdived by fw190G initially. P47 was slower and behind of fw190 after a short time of 65 degree dive. In order to over take fw190 in later dive stage, P47 need to get higher acceleration at first, that is the proof of P47 has better dive acceleration WITHIN fw190's allowable dive limits----466mph.

True but the power required is cubed...

Sorry Blackberry but the even the P47 does not generate enough power to overcome the lower power required advantage.

Do the math and you will see.

Yes, if he does not the airload will increase his manifold pressure and the waste gate will open to depressurize the system.

You understand the turbocharger requireds very gentle throttle movements. If you reduce power too fast, you can depressurize the system and you basically have a very poorly normally aspirated engine for the next 30 seconds until the turbine spools back up to repressurize the system.

Question Spitfire vs 109: dive performance and 4 vs 3 blades propeller
 

Since this interesting topic is about propeller types and dive efficiency, i just wanted to ask if, for instance, the fact that the Spit switched from three to four blades prop did allow it to improve its dive performance and to reach better dive speeds than the 109 which kept its 3 blades prop through the entire war period, as we know. Since both designs are very similar in the concept i'm sure the use of a different propeller would show in terms of DIVE performance?

I would like to see the relative dive performance of the aircraft between 109, Spit MkV and later MkIX. I think it could help answering the question we are asking ourselves? If someone knows about this it would be very nice!

PS My prejudice so far is that the 109 always dove better than the Spit (three or four blades propeller) Please correct me if i'm wrong :)

Good post. There is dive acceleration between spit IX/XIV and bf109G6as(m).
http://www.wwiiaircraftperformance.org/sl-wade.html

http://www.wwiiaircraftperformance.org/wade-dive.jpg

Notice that the boost pressure of bf109G is 1.75ata, so that's bf109G equipped with MW50.

The source says the dive acceleration (looks like 45 degree)of spitfire IX and BF109G6as is roughly same, but it also points outs that spitfire IX was outdived by bf109g6as INITIALLY.

It's very interesting to find initial dive advantage of German aircraft for many times . :) History is telling us some clue that we should pay enough attention to.

My comprehension of this report is:

When both spitfireIX and BF109G6as fly side by side with same speed/altitude, and begin to dive from low-medium speed, bf109g6as will outdive spitfireIX. The reasons are obvious,:

1) bf109 has better engine thrust/weight rario because same engine output/prop efficiency at low-medium speed but 109 is lighter.

2)bf109 has smaller wing area, so less air drag than spit.

It was widely accepted that bf109 could outdive spitfire INITIALLY among RAF pilots during WWII.

However, as speed building up, spitfire IX begin to catch up with bf109! Roughly near max allowable, spitfire IX could become "side by side" again with bf109. therefore, both dive accelerations are roughly same. And spitfire XIV will slightly ahead of bf109g6as in the end of high speed dive, although still be outdived INITIALLY.

There is also indirect comparation:
http://www.hawkertempest.se/TacticalTrials.htm

When comparing tempestmkv and spitfire XIV, it says:
Comparing with bf109G
So we can also include that bf109G outdive spitfire XIV initially, but spitfireXIV could slightly outdive 109G at high speed.

Why 4-blade propeller spitfires dives quite fast at high speed? More weight than bf109? Perhaps.More propeller efficiency at high speed? Perhaps.

http://www.spitfireperformance.com/spit9v109g.html

F/Lt. Irving "Hap" Kennedy did a stint with No. 185 flying Spitfire IXs from Malta in June 1943 and wrote:

The real compensation was that I was strapped into a brand new Spitfre IX. The Malta squadrons were being re-equipped with "Nines" after a couple of years, including the blitz year of 1942, during which the Spit V was the defense of the Island. Recently the Luftwaffe had moved their latest Messerschmitt Me 109G into Sicily along with some Focke-Wulf 190 squadrons. These latter were superb aircraft and the old Spit V just couldn't keep up to them. The Spit IX, a heavier brute in the engine but the same airframe with the lovely loose ailerons, an additional 250 h.p., a four-bladed prop, and a supercharger that came in with a tremendous kick at 21,000 feet, once more gave us the advantage of a superior performance. We were full of enthusiasm.

... At 1440 Hours, a red flare went up from the dispersal hut, arching over the strip, and my mechanic jumped to his battery. I pulled on my helmet, fastened the oxygen mask, put on my gloves, turned the oxygen valve on, and primed twice. The engine broke into a roar. The mechanic pulled out the battery cable and gave me a "thumbs up" and I was tearing down the strip with full throttle and 3000 R.P.M. Airbourne, gear up, throttle back a little to let the lads catch up, at 4500 f.p.m. climb.

... I had the throttle open and I rolled over and headed on a course to cut the angle toward the 109s, which had separated a little. I wound on nose-heavy trim so essential to keep the aircraft in a high-speed dive. The Spit responded eagerly as I dove more steeply than the 109s, with Red Two and Three no doubt following, although I could not see them. The controls got very heavy as the airspeed needle moved far right at 480 mph. (Corrected for altitude, true airspeed approached 600 mph.) I could see that I was gaining on the nearest Me 109. That was something new. We were already half-way to Sicily; that was no problem. We knew from years of experience, dating back to the boys who had been in the Battle of Britain, that the 109 with its slim thirty-two foot wing was initially faster in a dive than we were. But we accepted that compromise happily in exchange for our broad superior-lift wing with its better climb and turn. One couldn't have it both ways. In any case, I was closing on this Me 109, which I recognised as a G. Perhaps he wasn't using full throttle.

We were down to 5,000 feet and our dive had become quite shallow. I could see the Sicilian coast a few miles ahead. Now I was within range at 300 yards, and I let him have a good squirt. The first strikes were on the port radiator from which white smoke poured, indicating a glycol coolant leak. I knew I had him before the engine broke out in heavy black smoke. (Bf 109 G-4 "Black 14" of 2(H)/14, flown by Leutnant Friedrich Zander, shot down 10 June 1943)

Squadron Leader I.F. Kennedy DFC & Bar, Black Crosses of my Wingtip, (General Store Publishing House, Ontario, 1995), pp.58-61.



//////////
You fly ahead, and your wingman 500m behind of you with same speed/altitude. If you dive first, your wing man could NOT overtake you in a dive because he is same dive acceleration with you. Isn't it? So if your bf109G6as is caught by a spitfire IX with same energy and 400m behind of you, you could dive to escape from spit, in the beginning you could pull away from spit but in the end, the distance between you is roughly same as beginning--400m.

With same energy, if you want to catch/get closer to your opponent who dives ahead of you, your aircraft should has the ability of outdiving him. If you are in a P51, P47 or Tempest, congratulations! You could get closer to his bf109 or fw190 in a high speed dive.

There is information about the relationship between stick position and strength needed at high speed. Perhaps some aircrafts data is absent. :(

To simulate pilot strength? It depends on the decision of mod developing team.

Ok, let's do the math. It seems that you've neglected one fact that P47 achives almost same Vmax with only 50% enginethrust/weight ratio at level flight.

Fw190A8, 3.5 tons weight, 2000HP, prop efficiceny 80%, Vmax at sea level=580km/h=161m/s

P47D,7 tons, 2000HP, prop efficiceny 80%, Vmax at sea level=550km/h=153m/s

Vmax is a equilirium point. Formular is below, on the left side of formular is the thrust part, including engine thrust and gravity vector, on the right side,is air braking force which is proportional square of speed.

g*sin(0)+enginethrust/mass == dragcoefficent*(580^2)

Enginethrust=engineoutput*efficiecny/speed

For fw190a8 at Vmax, level flight =dive angle with 0

Total thrust=enginethrust=2000*735*80%/161=7300N=21.3% of weight

that is to say, with total thrust/ratio of 21.3%, fw190 could achieve 161m/s.

When dive at 45 degree and assume double speed=322m/s. Add 70.7% gravity thrust/weight ratio, minus 10.6% engineThrust/weight ratio, finally we get 81.35% thrust/weight ratio, thus 81.35/21.3=3.82 times thrusts/weight as before, and 1.95 times of airspeed is needed to counteract.

1.95 is roughly equal to 2, so the new equilirium speed is around 580*2=1160km/h.

Let's see P47, when Vmax level flight.

Total thrust=enginethrust=2000*735*80%/153=7686N=11.2% of weight

When dive at 45 degree ,assume the speed is 322m/s=2.1 times Speed before.

Add 70.7% gravity thrust/weight ratio, minus 5.87% engineThrust/weight ratio, finally we get 76% thrust/weight ratio, thus 76/11.2=6.79 times thrusts/weight as before, and 2.6 times of airspeed is needed to counteract

2.6>2.1

so 322m/s air is not enough to counteract the increased thrust/weight ratio, so 322m/s is below P47 new equilirium speed.

To sum up, when dive at 45 degree, P47 get more total thrust/weight ratio increase (6.79 times vs 3.82 times) than fw190a8, thus higher equilibrium speed.

Do the math with the actual numbers, blackberry.

All units in SI. Assume there is no air compressibility during dive, and 80% prop efficiency all along dive.

Fw190A8

m----mass,3500kg
A----dive angle
p----engine output,HP
r----propeller efficeincy=0.8=80%
Vmax ----max level speed at SL, 580km/h=161m/s
d----drag coefficient
g----gravity, 9.8 m/s^2
t---- engine thrust, N


when level flys at Vmax, fw190a8 in equilirium, zero acceleratiom, all forces are balanced.

t=P*r/V=2000*735*0.8/161=7300N

m*g* sin(0)+t=d*Vmax^2

d=t/Vmax^2=7300/161^2=0.2816

when dive in 45 degree, let the new equilirium speed as V:

new engine thrust should be 7300*(161/V)

m*g* sin(45)+t=d*V^2

3500*9.8*0.707+7300*(161/V)=0.2816*V^2

thus

v^3-86115V-4173651=0

Finally, we get V=315.2m/s=1134.7km/h.
And BMW801 still produce thrust as 7300*(161/315.2)=3729N=380kgf>10% aircraft weight


P47D

m----mass,6000kg
A----dive angle
p----engine output,HP
r----propeller efficeincy=0.8=80%
Vmax ----max level speed at SL 550km/h=153m/s
d----drag coefficient
g----gravity, 9.8 m/s^2
t---- engine thrust, N



when level flys at Vmax, P47D in equilirium, zero acceleratiom, all forces are balanced.

t=P*r/V=2000*735*0.8/153=7686N

m*g* sin(0)+t=d*Vmax^2

d=t/Vmax^2=7686/153^2=0.3283>0.2816

Obviuosly ,P47d has more air drag coefficenct due to bigger wing area.


when dive in 45 degree, let the new equilirium speed as V:

new engine thrust should be 7686*(153/V)

m*g* sin(45)+t=d*V^2

6000*9.8*0.707+7686*(153/V)=0.3283*V^2

v^3-126628V-3581962=0

Finally, we get V=369.2m/s=1329.1km/h>1134.7km/h.
And R2800 still produce thrust as 7686*(153/369.2)=3187N=325kgf>5% aircraft weight

Obviuosly, 6-ton p47d gets more benifit than 3.5-ton fw190A8 in a 45 degree dive.

IMO, P47D has two advantage over fw190a8 in a steep dive.

1) more weight, more benifit. This is proved by calculations above. I believe the il2 FM could handle this kind of calculation very well.
2) higher propeller efficiency at high speed(>Vmax) dive.I suspect the il2 FM could NOT handle this kind of calculation very well.

I suggest that we try to reproduce the P47D vs Fw190G in a 65 degree dive from 10000ft altitude in il2 4.11m, at initial dive stage, let fw190G go ahead, pull away rapidly. This can be achieved by decreasing P47 throttle. And then we can check whether the p47D(with full throttle) could overtake fw190G in the end of dive, ie, at 3000ft altitude with much greater speed.

If not, I prefer that il2 FM has undermodelled 4-blade aircrafts/overmodelled 3-blade dive acceleration for many years. That's totally ridiculous for allied aircrafts such as p47, p51, tempest, spitfire,f4u, etc.

You have their power output as the same. They are not the same.

The P47 achieves a Vmax of 339mph on 2300shp and the FW-190A8 achieves 359mph on 1980 shp.

when level flys at Vmax, fw190a8 in equilirium, zero acceleratiom, all forces are balanced.

t=P*r/V=1980*735*0.8/161=7231N

m*g* sin(0)+t=d*Vmax^2

d=t/Vmax^2=7231/161^2=0.2790

when dive in 45 degree, let the new equilirium speed as V:

new engine thrust should be 7231*(161/V)

m*g* sin(45)+t=d*V^2

3500*9.8*0.707+7231*(161/V)=0.2790*V^2

thus

v^3-86918V-4172728=0

Finally, we get V=316.4m/s=1139km/h.


P47D

339mph=151.6m/s

when level flys at Vmax, P47D in equilirium, zero acceleratiom, all forces are balanced.

t=P*r/V=2300*735*0.8/151.6=8921N

m*g* sin(0)+t=d*Vmax^2

d=t/Vmax^2=8921/151.6^2=0.3882>0.2816

Obviuosly ,P47d has more air drag coefficenct due to bigger wing area.


when dive in 45 degree, let the new equilirium speed as V:

new engine thrust should be 8921*(151.6/V)

m*g* sin(45)+t=d*V^2

6000*9.8*0.707+8921*(151.6/V)=0.3882*V^2

v^3-107089V-3483833=0

Finally, we get V=342.4m/s=1232.6km/h>1139km/h.

http://en.wikipedia.org/wiki/Pratt_%...00_Double_Wasp

R-2800-59 - 2,300 hp (1,700 kW)

http://www.wwiiaircraftperformance.o...ifferences.jpg

http://www.wwiiaircraftperformance.o...omp-p47dmn.jpg

2300HP ---345mph@SL


Specifications (P-47D Thunderbolt)

General characteristics
Crew: 1
Length: 36 ft 1 in (11.00 m)
Wingspan: 40 ft 9 in (12.42 m)
Height: 14 ft 8 in (4.47 m)
Wing area: 300 ft² (27.87 m²)
Empty weight: 10,000 lb (4,536 kg)
Loaded weight: 17,500 lb (7,938 kg)

Specifications (Fw 190 A-8 )

Data from Fw 190 A 8
General characteristics
Crew: 1
Length: 9.00 m (29 ft 5 in)
Wingspan: 10.51 m (34 ft 5 in)
Height: 3.95 m (12 ft 12 in)
Wing area: 18.30 m² (196.99 ft²)
Empty weight: 3,200 kg (7,060 lb)
Loaded weight: 4,417 kg (9,735 lb)

Regarding the Fw-190A8 details, this may be of some use to you, since your figures are slightly off.

http://i7.photobucket.com/albums/y29...1Dec270119.jpg
http://i7.photobucket.com/albums/y29...2Dec270120.jpg

http://i7.photobucket.com/albums/y29...1Jun061404.jpg

Fruitbat, fw190A8 has extra exhaust thrust when max level fly at sea level. So fw190a8 could achieve 580 km/h with the help of (1950HP*0.8+exhaust thrust).


Attachment 9893

580km/h is quite a high speed, so fw190 A8 is boosted by more than 2000HP.

P47 loses its exhaust thrust due to exhaust turbocharger.

The Fw 190 A-8 entered production in February 1944, powered either by the standard BMW 801 D-2 or the 801Q (also known as 801TU). The 801Q/TU, with the "T" signifying a Triebwerksanlage unitized powerplant installation, was a standard 801D with improved, thicker armour on the front annular cowling, which also incorporated the oil tank, upgraded from 6 mm (.24 in) on earlier models to 10 mm (.39 in). Changes introduced in the Fw 190 A-8 also included the C3-injection Erh?hte Notleistung emergency boost system to the fighter variant of the Fw 190 A (a similar system with less power had been fitted to some earlier Jabo variants of the 190 A), raising power to 1,980 PS (1,953 hp, 1,456 kW) for a short time. The Erh?hte Notleistung system operated by spraying additional fuel into the fuel/air mix, cooling it and allowing higher boost pressures to be run, but at the cost of much higher fuel consumption. From the A-8 on, Fw 190s could be fitted with a new paddle-bladed wooden propeller, easily identified by its wide blades with curved tips

http://www.wwiiaircraftperformance.o...en-1-10-44.jpg

So fw190a8 drag coefficient is no less than 0.2790.

Or we could calculate the loaded weight fw190A8, that's 4500kg.


when dive in 45 degree, let the new equilirium speed as V:

new engine thrust should be 7231*(161/V)

m*g* sin(45)+t=d*V^2

4500*9.8*0.707+7231*(161/V)=0.2790*V^2

thus

v^3-111752V-4172728=0

Finally, we get V=352m/s=1267km/h>1232.6km/h. Slightly bigger than a 6-ton P47D. If we use a 7-ton P47D, again, P47 has more equilibrium speed.

So, the aircraft weight plays a important role in a 45 degree dive.

Blackberry, the reason i posted the charts i did, is because the weights you are using appear to be off.

empty is 3050, max load is 4272 for a standard A8, it also lists the HP as 1700.

how you want to interpret that is up to you.

Oh, you might want to rethink your view on available HP at altitude, @5,700m it increases the engine power to 1,440HP at sea level it does what you say, but your not going to be diving much at sea level me thinks!

http://i7.photobucket.com/albums/y29...1Jun061537.jpg
http://i7.photobucket.com/albums/y29...2Jun061537.jpg

as for when the A8 was produced, i am well aware, this may interest you regarding production.

factory produced, serial numbers, numbers and dates.

http://i7.photobucket.com/albums/y29...3Jun061539.jpg

See my edited post. Fw190A8 is Notleistung emergency boost system which produce 1953HP, I use 1980HP in calculation. There is some exhaust boost(bigger than 27HP at high speed). 578km/h, I use 580km/h.

So fw190a8 drag coefficient is no less than 0.2790

Ok, let's be more accurate with 4272kg

when dive in 45 degree, let the new equilirium speed as V:

new engine thrust should be 7231*(161/V)

m*g* sin(45)+t=d*V^2

4272*9.8*0.707+7231*(161/V)=0.2790*V^2

thus

v^3-106090V-4172728=0

Finally, we get V=344m/s=1238km/h. roughly euqal to 1232.6km of 6-ton P47D.

1)There should be 100HP+ exhaust boost for fw190A8@580km/h@SL, so darg efficienct should be more than 0.2790.

2) max load is 4272kg for a standard A8, max load is 8000kg for a standard P47D, even with 7-ton weight, P47D could also has higher equilirium speed.

I just compare P47D and fw190A8 at sea level.

If we compare the data at 7000m altitude, P47D has more advantage because fw190a8 lose a lot of engine output there while P47D is still 2000-2300HP.

Furthermore, the air density is only 1/2-1/3 of sea level, and aircrafts should maintain a longer time at high speed above Vmax. It's more easy to dive to a high Mach number where the 4-blade propeller has efficiency advantage.

BTW, german broad chord 3-blade propeller loses 8% efficiency at Vmax compared with old 3-blade. Therefore, allied 4-blade prop should has more efficiency than german new wide blades. not 80% vs 80%, that's sth. 85% vs 75% at Vmax.

To sum up, at 7000m, a high speed diving P47D is more dreadful than at sea level.

Yes it does as does the lower drag of the FW-190.

If you use Take Off Weights, the equilibrium speed of the FW-190 is higher than the P47.

It does not matter though as neither aircraft can achieve equilibrium velocity as they are dynamic pressure and mach limited.

The P47 always has higher limits.

Yes his weights are off and it makes a big difference. Thanks for catching that.

No but don't think of it as sea level. It is EAS when we use sea level performance. Adjust it for density effects and it becomes actual performance.

As such it is good approximation of Indicated Airspeeds and delivers a very good prediction of the performance trends you should see.

At SL @1.42ata@2700U/min with allowance for dynamic pressure gains at equivilent airspeeds.

The 2300 hp for the R-2800 is at War Emergency Power and the 1980 hp for the BMW801D2 is at Erhohte Notleistung 1.58ata@2700U/min.

You will not be comparing fighters....

You guys need a weight and balance chart for the P47 series?

I'd be very interested in seeing one thanks:)

for the FW to clarify, is it 1,980 PS or 1,980hp?

this has turned into a very interesting thread imo.:cool:

That is Horsepower.

http://img805.imageshack.us/img805/4...andbalance.png

BTW at the velocity we are talking about, the conversion form PS to HP is really irrelevant.

Thanks for posting the weight and balance chart:cool:

Fw190 A8 578km/h@SL 1953HP ,and what 's the weight you want to use? weight and balance chart for fw190A8?

p47D(R2800-59) 345mph=555km/@SL ,2300HP, and what 's the weight you want to use?

Fw190A8

m----mass,4272 kg,max load for a standard A8,100% fuel
A----dive angle
p----engine output,1953 HP@sea level
r----propeller efficeincy=0.8=80%
Vmax ----max level speed at SL, 578km/h=160.6m/s
d----drag coefficient
g----gravity, 9.8 m/s^2
t---- engine thrust, N


when level flys at Vmax, fw190a8 in equilirium, zero acceleratiom, all forces are balanced.

t=P*r/V=1953*735*0.8/160.5=7155N

m*g* sin(0)+t=d*Vmax^2

d=t/Vmax^2=7155/160.5^2=0.2778

P47D

m----mass,5675kg, 12500lb, normal combat load(55% fuel)
A----dive angle
p----engine output,2300 HP
r----propeller efficeincy=0.8=80%
Vmax ----max level speed at SL 345mph=555km/h=154.3m/s
d----drag coefficient
g----gravity, 9.8 m/s^2
t---- engine thrust, N

when level flys at Vmax, P47D in equilirium, zero acceleratiom, all forces are balanced.

t=P*r/V=2300*735*0.8/154.3=8765N

m*g* sin(0)+t=d*Vmax^2

d=t/Vmax^2=8765/154.3^2=0.3681



55% fuel P47D vs 100% fuel Fw190A8 in 45 dive degree

for fw190A8, let the new equilirium speed as V:

new engine thrust should be 7155*(160.5/V)

m*g* sin(45)+t=d*V^2

4272*9.8*0.707+7155*(160.5/V)=0.2778*V^2

thus

v^3-106548V-4133828=0

we get fw190A8 V=344.5m/s=1240km/h.
And BMW801 still produce thrust as 7155*(160.5/344.5V)=3333N

for P47D let the new equilirium speed as V:

new engine thrust should be 8765*(154.3/V)

m*g* sin(45)+t=d*V^2

5675*9.8*0.707+8765*(154.3/V)=0.3681*V^2

v^3-106818V-3674109=0

we get P47D V=342.8m/s=1234km/h=1240km/h of Fw190A8

And R2800-59 still produce thrust as 8765*(154.3/342.8 )=3945N



55% fuel P47D vs 100% fuel Fw190A8 in 65 dive degree

for fw190A8, let the new equilirium speed as V:

new engine thrust should be 7155*(160.5/V)

m*g* sin(65)+t=d*V^2

4272*9.8*0.906+7155*(160.5/V)=0.2778*V^2

thus

v^3-136538V-4133828=0

we get fw190A8 V=384m/s=1382km/h.


for P47D let the new equilirium speed as V:

new engine thrust should be 8765*(154.3/V)

m*g* sin(45)+t=d*V^2

5675*9.8*0.906+8765*(154.3/V)=0.3681*V^2

v^3-136884V-3674109=0

we get P47D V=383m/s=1379km/h=1382km/h of Fw190A8


Conclusion:

1) 55% P47D and 100% fuel Fw190A8 share same equilirium speed at 45-65 degree dive.
2) 5675kg P47D=normal combat load, 4272kg Fw190A8=max load A8. There is only 0.454(12500-10700)=817kg difference between empty and normal load P47, but there is (4272-3050)=1222kg difference for fw190A8. In fact, 5675kg P47D is only 200 US galon fuel, not 100% fuel. if internal fuel is full(375 US gal), 6152kg for P47D which has better dive acceleration.

This is wrong. You are confusing terminal velocity with aceleration.

At Vmax:

FW190

9418lbs * sin 45 = 6660lbs excess thrust

a = F/m

m = 9418lbs/32.2 = 292 lb-s^2/ft

a = 6660lbs/292lb-s^2/ft

a = 22.8 ft/s^2 for the Focke Wulf

P47D22:

13500lbs * sin 45 = 9546lbs

a = F/m

m = 13500lbs/32.2 = 419 lb-s^2/ft

a= 9546lb/419lb-s^2/ft

a = 22.78 ft/s^2 for the P47D22

That is the best case scenario for the P47D22.

If we dive from say, 260 KEAS, then we see 25.35 ft/s^2 from the FW190 and 24.4ft/s^2 from the P47.

The FW190 has a .95ft/s^2 advantage in aceleration rate.

anyone know what the Vne limits are for these planes?

Yes, they are listed in the Operating Notes....

At low altitude:

FW190 - 466 mph IAS

P47 - 500 mph IAS

Not everyone has the operating notes for these planes! I knew they would be in them though!!

thanks for the figures, appreciated. :cool:

In your opinion, when dive from Vmax at 45 degree(from SL to a deep valley),

excess thrust=weight vector=weight*sin(45)

for any aircraft with mass=m, there is

a=weight*sin(45)/mass=32.2*sin(45)=22.8ft/s^2

for a 10lb plane, a=22.8
for a 100lb plane,a=22.8
for a 1000lb plane, a=22.8

In my opinion, when dive from Vmax at 90 degree(from SL to a deep valley),

excess thrust=weight vector+ engine thrust - drag force

at Vmax:
For full loaded fw190A8:

a = 9.8+ 7155/4272-0.2778*(160.5 ^2)/4272= 9.8+1.67- 1.67=9.8m/s^2= 32.2ft/s^2

For full loaded P47D:

a = 9.8+ 8767/6152-0.3681*(154.3 ^2)/6152= 9.8+1.425- 1.425=9.8m/s^2= 32.2ft/s^2

when speed building up to 720km/h=200m/s.......

For full loaded fw190A8:

a = 9.8+ 7155*(578/720)/4272-0.2778*(200 ^2)/4272= 9.8+1.35- 2.6=8.55m/s^2


For full loaded P47D:

a = 9.8+ 8767*(555/720)/6152-0.3681*(200 ^2)/6152= 9.8+1.10- 2.39=8.51m/s^2


when speed building up to 800km/h=222m/s......


For full loaded fw190A8:

a = 9.8+ 7155*(578/800)/4272-0.2778*(222 ^2)/4272= 9.8+1.21- 3.2=7.81m/s^2


For full loaded P47D:

a = 9.8+ 8767*(555/800)/6152-0.3681*(222 ^2)/6152= 9.8+0.989- 2.95=7.84m/s^2


when speed building up to 850km/h=236m/s......


For full loaded fw190A8:

a = 9.8+ 7155*(578/850)/4272-0.2778*(236 ^2)/4272= 9.8+1.14- 3.62=7.32m/s^2


For full loaded P47D:

a = 9.8+ 8767*(555/850)/6152-0.3681*(236 ^2)/6152= 9.8+0.93- 3.33=7.4m/s^2


Conclusion, P47D has slightly better dive acceleration when reaching 750-850km/h if both propeller efficiency=80%. I think this is the il2 FM method, if you test both in il2 4.11m, You'll find slightly dive acceleration difference.

However, if fw190A8 propeller efficiency drops from 80% to 50% at 850km/h,

For full loaded fw190A8:

a = 9.8+ (50%/80%)7155*(578/850)/4272-0.2778*(236 ^2)/4272= 9.8+0.71- 3.62=6.89m/s^2 quite smaller than 7.4m/s^2 of P47D.

if fw190A8 propeller efficiency drops from 80% to 50% at 800km/h,

For full loaded fw190A8:

a = 9.8+ 50%/80%*7155*(578/800)/4272-0.2778*(222 ^2)/4272= 9.8+0.76- 3.2=7.36m/s^2 quite smaller than 7.84m/s^2 of P47D.

From fw190A8, wide chord propeller was used in fw190, the new wide propeller outperforms old narrow chord cousin at low -medium speed, but is inferior to old one at high speed(>= vmax).

If we assume that there is 0.5 m/s^2 difference between Fw190A8 and P47D along the 20 seconds dive, in the end of dive, P47D is 10m/s=36km/h faster. In fact, there is quite some air compressibility at 800kmh(500mph) where aircrafts need 200HP+ to overcome the increase of air darg coefficient which is NOT a constant any more. The more air darg coefficent, the more dive acceleration advantage for P47D due to bigger weight.


Conclusion, the il2 FM(without Mach number) is NOT suitable for simulating the high speed dive because of air compressibility which influences the propeller efficiency and air-wing drag coefficient.

Duh...:oops:

damn Ti89 and my old brain.

Go back and read the flight test results. This gives a good approximation of the performance you will see in the air.

And the P47's CSP is going to drop too, blackberry. I like the idea you present with modeling efficiency. Problem is you are making the same mistake the NACA did in adopting the 16 series. The calculations just did not bear out and give good agreement with the air.

Just a thought, a generic modification of efficiency based on some basics of propeller design might enhance things. Problem is we don't have enough information on the propellers to implement any kind of detail.

My estimate is below, please correct me if I'm wrong.

Efficiency: assume 3-blade naca16= 3-blade clark= 3-blade gotingen @0.7Mach

4-Blade naca of P47=88% at 0.4 Mach

3-Blade naca of P47=82% at 0.4 Mach
(See P47 data I'v posted)

3-Blade naca of P47d=63% at 0.7 Mach (out of envelop,so <80%) see data posted

3-Blade naca/gottingen of fw190A5= 50% at 0.7 Mach (3.3 m diameter, bigger advance ratio)

3-Blade naca/gottingen of fw190A8= 40% at 0.7 Mach(wide blade even lose 8% at Vmax)

4-Blade naca of P47=?????% at 0.7 Mach????? It should be bigger than 63% or not? It seems that US has never unclassified the 4-blade NACA16 high Mach number wind tunnel data. :( Do they want to cover something? But they did make the decision of choosing 4-blade In WWII.

If 70%, there is 30% efficiency advantage over fw190A8, this could explain a lot of 45 degree dive test in history.

In Museum, there are real fw190A8 and P47, p51, bf109, BTW, Crumpp you have real fw190A8 and P51. So someone could get the airfoil shape of propellers.

And then, someone could simulate propeller efficiency by Xfoil or Ansys(software).

I agree with you this is a hard work, 1C developping team needs a lot of money and time to do so.