Sustained Turn Performance Of Piston Engine Aircraft

[badger45]

Hi All,

since I noticed that there are not only many fans of WWII aircraft, but also many people actually interested in calculation of basic performance indicators, I would like to post some useful equations. I have found the exact approach of aerodynamics engineers quite bothersome to use and tried (successfully) to reduce the equations as much as possible, while retaining sufficient degree of accuracy.

The simplest approach to calculate time to perform sustained turn at 1000 meters height:

t = (W/b) * (Vmax/P)^0,5 * (1/8,80)

where W is weight in kg, b is span in meters, V is max sea level speed in km/h and P is power at the maximum sea level speed (measured in metric horspower, i.e. 735.5 W). The constant is empirically derived, but the rest of the equation follows from following reasoning.

Time to perform sustained turn is inversely proportional to turn rate, which is defined as g*[(n^2)-1)]^0.5/V (g is 9.81 m/s2, n is load factor in g's and speed in m/s in sustained turn).

The speed in sustained turn is
V = [(P*eff*735.5)/(2*ro*S*Cd0)]^(1/3) m/s
and the load in sustained turn is
n = 0.687 * {[(P*eff*735.5)^2*E*ro*S*K]/W}^(1/3) m/s2

P is metric horsepower, eff is propeller efficiency coefficient, ro is air density in kg/m3, S is wing area in m2, Cd0 is zero lift drag coefficient, E is aircraft maximum lift-to-drag ratio, K is induced drag coefficient and W is weight in kg.

K = pi*AR*oswald (oswald means Oswald span efficiency)
AR = b^2/S
E = [(K/Cd0)^0.5]/2

Assuming that propeller efficiency and Oswald is constant for all aircraft (not exactly true) and we are moving at sea level, we can rewrite the above equations without numerical constants (i.e. rewrite just variables, ~ denotes "is in proportion to") as:

V ~ P^1/3 * S^(-1/3) * Cd0^(-1/3)
n ~ P^2/3 * S^(1/3) * Cd0^(-1/6) * K^0.5 * W^(-1)

Cd0 can be estimated numerically, using formula Cd0 = Cd - Cl^2/K, but for the purpose of further simplification it can be assumed as equal (in max speed level flight) to total coefficient of drag. Error from this assumption is negligible (on the order of less than 1% accuracy in final time to turn computation). This is because of the 90 percent share of Cd0 in total drag (as long as you take max speed level flight conditions). Total coefficient of drag is proportional to P * Vmax^(-3) * S^(-1), where the Vmax denotes sea level max speed. Inserting this into equations above (and assuming, that the turn is done at maximum power) we get:

V ~ Vmax

This is not that much surprising. Speed for best load sustained turn (at SL) occurs at constant percentage of max speed for that altitude, usually 64 percent. Variations in actual aircraft seem to be under 1%, when the result was computed exactly with all coefficients. The second equation (after further substitution for K ~ (b^2)/S) yields:

n ~ P^0.5 * Vmax^0.5 * b * W^(-1)

Now because of the subtraction in [(n^2)-1]^0.5, there is not much to simplify here. But since the aircraft I am mostly interested in (WWII pistons, usually fighters) are all making the sustained turns at 2-4 g's, I can simply ignore the subtraction. The ordering of aircraft (i.e. who turns better) should not be affected and the numerical difference between 2.5 g and 3.5 g aircraft is 140 percent (if the subtraction is ignored) and 146,4 percent, when it is not ignored. Sizing the final constant in the middle of aircraft of interest, it is thus possible to have less than 3 percent error in estimates of turn time.

The final equation is therefore:

t ~ P^0.5 * Vmax^(-0.5) * b * W^(-1)

The constant for sustained turn time in 1000 meters height and metric units is 8.80 (see the first equation). For sea level turn it comes as 9.48.

For those using imperial units, 1000 meters constant is 4.69 and sea level 5.05 (P in horses, Vmax in statute miles per hour, b in feet and W in pounds).

Example of use:

Fw-190 A-3, 1560 HP @ SL, 540 km/h @ SL, span 10.5m, 3850 kg.

t = (3850/10.5) * (540/1560)^0.5 * (1/8.80) = 24.5 seconds

Possible errors: Oswald span efficiency and propeller efficiency varies from aircraft to aircraft. Though it is possible to compute them, they vary so slightly, that there should be no significant errors. Above example yields the same result with exact formulas for 0.78 propeller efficiency and 0.868 oswald. If the propeller efficiency was 0.858 (+10 percent), turn time would be 23.2 seconds and with Oswald 0.7812 (-10 percent) turn time rises to 26.1 seconds. Thus +- 10 percent differences produce about 5-6 percent difference in results. With variable pitch propellers of second world war I don't expect much that much effective difference, while the differences in Oswald span efficiency are very low between these high aspect ratio aircraft (for AR between 5 and 7, Raymer computes 0.9 to 0.84 Oswald efficiency). Thus I don't expect these factors to be significant.

Other than this, the power in 1000 meters is different than at sea level. Using sea level power is however convenient, and it introduces usually only small error. E.g. Yak-1 has about 1020 hp at sea level and 1050 at 2000 meters. The error in square root of 1035 versus 1020 is 0.7 percent. In weight terms it is for 2900 kg fighter equal to 20 kg (less than average ammunition load).

The last known problem is data. If you have no sea level speed and power data, you can for first orientation use data at known altitude and correct them for the different air density. Quick and dirty approach would be to replace the speed in above equation with (Valt*exp(-alt/28.0)) and use appropriate horsepower (i.e. power at that altitude); Valt is speed in altitude (km/h), alt is the altitude (m). But remember, garbage in, garbage out. Try not to abuse the god of physics and take care to avoid this approach as much as possible. Some engines are especially unsuitable for this approach, which essentially supposes, that the engine power remains the same with altitude.

Example:

Spitfire Mk VB, 1470 HP @ 6100m, 597 km/h @ 6100m, span 11.23m, 2950 kg.

t = [(exp(-6.1/28.0) * 597)/1470]^0.5 * (2950/11.23) * (1/8.80) = 17.1 seconds

I apologize for such a long post and hope you find the equations useful. If you have any feedback, find any errors or have questions, don't hesitate to ask.

Zdenek,
badger 45

P.S. note that the span loading and speed-to-power coefficients are after all the only things that affect the aircrafts turning rate. Thus it seems that the turn time was really more or less unsubstantial characteristic in WWII design, since every aircraft designer in a quest for performance sought to maximize these and thus inevitably worsened turning characteristics of aircrafts.

 

[drgondog]

"Cd0 can be estimated numerically, using formula Cd0 = Cd - Cl^2/K, but for the purpose of further simplification it can be assumed as equal (in max speed level flight) to total coefficient of drag. Error from this assumption is negligible (on the order of less than 1% accuracy in final time to turn computation). This is because of the 90 percent share of Cd0 in total drag (as long as you take max speed level flight conditions). Total coefficient of drag is proportional to P * Vmax^(-3) * S^(-1), where the Vmax denotes sea level max speed.

First comment - the fighter is NEVER in max speed condition for turns As you fall away from top speed, even the individual components within the summation of all other drag components are non-proportional to each other. If you look at CD0 and CDi at cruise speed for example they are equal to each other at the bottom of the drag polar. I suspect you know this but the "90 percent share of Cd0 in total drag" confused me with respect to your thesis.

You know that to obtain any performance profile you need to model from a free body diagram starting from T=D and solve for that equation throughout the curvelinear path? And that the Drag is equal to the Parasite Drag plus the Vortex Drag plus the Lift-dependent viscous Drag plus Compressibility Drag (and you CAN assume that is not a factor for a Mustang or Yak in a turning fight)? and that the inviscid (vortex) drag is only the zero lift term due to twist and lift dependent parts that that depend on twist and planform? You calculated it as Profile Drag plus Induced Drag and derived all your profile drag terms by assuming that zero lift drag at max speed and power is all that remained of parasite drag plus all the remaining vortex drag componebts at increasing angles of attack (like a turn vs level flght).

Not so in low to moderate speed ranges at high AoA.

The remaining portion of 'induced drag" is the viscous part (increase due to increase of both friction drag with increasing angles of attack - which is NON trivial). Additionally empirically-estimated terms arise from fuselage vortex drag, changes in trim drag with angle of attack and a change in drag due to engine power effects (incremental values of drag affected by the change in mass flow rate through the prop disk, votex driven pressure distributions as well as drag due to exhaust stacks and radiator inlets).

In addition to your 'Cd0' calculated term at top speed, you have to introduce many different and meaningful drag deltas when you roll the airplane in a bank angle at high angle of attack, including more drag due to greater span wise flow distributions, added trim drag of rudder/ailerons/elevator to keep the a/c in a bank, added prop/disk drag, asymmetric vortex flow around the fuselage, increased non-linear (and uncalculated) adverse pressure distributions as the bird nears stall in the turn.

So, while you may be bored with 'aero' complexities, if you wish to semi-accurately predict the aircraft state during each segment of the entire curvilinear path, you have to solve for each drag sum for the physical aspects of the fighter, the Thrust required - Thrust available, the velocity profile and all the terms that are angle of attack dependent in the banked turn. If you are 'comfortable' that any such profile may be maintained perfectly in optimal mode - then find that velocity and bank angle and thrust and drag - and apply those as constants...for that single altitude and power condition.

Even if you have a Drag polar (unlikely) for all the aircraft of interest for all the reasons noted above, a banked aircraft in a high G turn striving to a.) stay in the air, and b.) minimize the turn turn time is in a different physical state and drag profile than the level wing/top speed assumptions you started with.

In other words, I simply disagree that accurate modelling can be performed using the elegantly constructed series of assumptions that you used?


This is not that much surprising. Speed for best load sustained turn (at SL) occurs at constant percentage of max speed for that altitude, usually 64 percent. Variations in actual aircraft seem to be under 1%, when the result was computed exactly with all coefficients. The second equation (after further substitution for K ~ (b^2)/S) yields:

Curiosity compels me to ask your sources for the above statement?

 

[drgondog]

There is a reason you don't get a BS degree in Aero following Algebra II. Aircraft performance discussions are non-trivial exercises in both assumptions and the math surrounding them. Having said this, when trying to estimate one fighter versus another in a performance calculation can be ordered in original assumptions by considering a.) wing loading, b.) airframe/wing relative drag, c.) power available, and d.) Clmax

you can make a lot of educated guesses with just those four factors - assuming a perfect pilot!

 

[badger45]

Thank you for your reply!

First, I would like to clarify one thing, which is probable source of misunderstanding. The equations above are to be used as quick estimates, not fine analysis. It is obvious, that the best method would be to do finite element analysis, or even better, to build 1:1 replica and test it in tunnel. I am aware of the limitations and I am willing sometimes to sacrifice the precision for speed.

Second, it is neither from lack of my aerodynamics education, that I assume these relationships, nor from finding aerodynamics boring. May be I made a mistake in choosing the word bothersome - English is not my native language. But believe me, it was not meant to offend. Just to point out, that with a few percents of error we can arrive at the result with much simpler approach. And because the Cd0 and K contributions are quite large, I was not willing to sacrifice that much accuracy by neglecting them (e.g. aircraft like I-16 would become too slow turning if you omit Cd0 representations).

Most of your post deals with changing and undstable status of Cd0. I am aware of the fact. In case somebody else is interested in the matter, he can look at e.g. Torenbeek, Synthesis of Subsonic Airplane Design, or Roskam, Methods for Estimating Drag Polars of Subsonic Airplanes. But I think, that for the performance calculation you can with some degree of accuracy actually employ a constant Cd0. And I am not the one, who says it. See for example Hale, Aircraft Performance, Selection and Design, Raymer - Aircraft Design: A Conceptual Approach. Both use the simple unchanging Cd0 for estimate of sustained turn performance. How much problems in terms of numerical errors it brings I yet have to see. Perhaps you may supply some data. But given the fact, that overall contribution of Cd0 is of 1/6 power, it should not be of much concern (for quick estimation, again). Especially bearing in mind, that we are interested only in difference in sum of all the drag additions and subtractions. If the basic Cd0 is 0.025 for average fighter in max speed and typical increase in drag compared to max speed flying condition is 0.003, then if some examined fighter due to its e.g. high vortex and trim drags has 0.006 of additional drag (double!), but the same basic Cd0 in max speed, it would introduce around 1.7 percent of error. This is due to the fact, that the numerical constant "swallows" typical increase in parasite drag (in other words, calibration process can include actual aircraft and thus already some typical proportion of the increased zero lift drag coefficient). I hope it's clear.

For the simplified representation of induced drag holds essentialy the same. The source are the authors mentioned above. And they give the same warning about nature of estimates as I give. There are some other things, you mention, but don't make much sense, at least to me. Like mentioning departure of parabolic drag relationship near stall (i.e. regular invalidness of induced drag equation near max Cl). Sustained turn performance is nowhere near that high angles of attack, so it is largely inconsequential.

Simplification is normal scientific method. In the absence of detailed data, less detailed data are normally used (with the possible error analysis). One example of this is Performance Analysis and Tactics of Fighter Aircraft from WWI by Scott Eberhardt, the other is Quest for Performance by Laurence K. Loftin. I intended to supply another material in this line and nothing more.

To sum up - next time I will use bold type for the word estimate

Your second post somewhat surprised me.

First, the method you propose is largely unusable, as you can see from many pointless discussions around the web. I wished to propose alternative to just this "educated guessing", since many times it involves much greater (and unquantifiable) errors.

Second, why do you include Clmax to sustained turn performance estimation? I see only a theoretical possibility, that the Clmax does not allow turning at best load sustained turn. Do you know such WWII fighter? I don't.

That's all. Thanks again for your reply. I hope I don't appear hostile, as it was not my intention.

Cheers,

Zdenek,
badger45

 

[drgondog]

Thank you for your reply!

First, I would like to clarify one thing, which is probable source of misunderstanding. The equations above are to be used as quick estimates, not fine analysis. It is obvious, that the best method would be to do finite element analysis, or even better, to build 1:1 replica and test it in tunnel. I am aware of the limitations and I am willing sometimes to sacrifice the precision for speed.

I got that. Having said that, predictions for high G curvilinear flight introduces many non-linear components that make even the most sophisticated Navier Stokes model solutions 'questionable' until flight test instrumentation and results are obtained for comparison.

Second, it is neither from lack of my aerodynamics education, that I assume these relationships, nor from finding aerodynamics boring. May be I made a mistake in choosing the word bothersome - English is not my native language. But believe me, it was not meant to offend.

Your command of the english language is excellent -

Most of your post deals with changing and undstable status of Cd0. But I think, that for the performance calculation you can with some degree of accuracy actually employ a constant Cd0. And I am not the one, who says it. See for example Hale, Aircraft Performance, Selection and Design, Raymer - Aircraft Design: A Conceptual Approach. Both use the simple unchanging Cd0 for estimate of sustained turn performance.

I am aware of Raymer, still have in my library Corning's Supersonic and Subsonic Design and Hoerner's Fluid Dynamic Drag as two volumes I use to refresh my failing memory..I disagree Raymer and others for using the simplistic approach to calculating parasite drag - particularly from top speed at high altitude (and presumably Hale) for the reasons stated. My academic advisor in graduate school was Dr. Jack Fairchild who headed up LTV's Flight mechanics group - he forced me to work along the entire drag polar when crafting performance models and showed me that material differences were shown at various points in the dynamic model at different velocity points for different attitudes and loads. This is the basis of my discussion with you. Having said this CD0 calculated from low altitude is close.

How much problems in terms of numerical errors it brings I yet have to see. Perhaps you may supply some data.

I don't have any of my works after 40 years of throwing trash out for moves.. but I suggest you do what was forced down my throat - namely find the Corner Speed for best turn rate using the initial assumptions you presented (including 'CD0' picked from top speed flight test charts).

For the Free Body Diagram and constant solution for T=D (and be able to find Power Required versus Power Available, acceleration available, etc) then:

Test for CL at the first iteration for velocity and the bank angle and load for max rate of turn. Pick an AoA (approximate) for that condition from the appropriate Airfoil charts. Look to the drag polar for that velocity (and remember it is for level wing constant altitude flight).Pick the value of Lift related Drag at that value, and 'All other Drag total' at that value. Now calculate two additional components 1.) effective reduction in AR due to increased tip vortex at high G/High Lift, and 2.) increase in Trim drag due to control surface deflections to maintain constant altitude.

Now go back into your equations for Vmax and re-iterate and see what the difference is in the result.

But given the fact, that overall contribution of Cd0 is of 1/6 power, it should not be of much concern (for quick estimation, again). If the basic Cd0 is 0.025 for average fighter in max speed and typical increase in drag compared to max speed flying condition is 0.003, then if some examined fighter due to its e.g. high vortex and trim drags has 0.006 of additional drag (double!), but the same basic Cd0 in max speed, it would introduce around 1.7 percent of error. This is due to the fact, that the numerical constant "swallows" typical increase in parasite drag (in other words, calibration process can include actual aircraft and thus already some typical proportion of the increased zero lift drag coefficient). I hope it's clear.

Yes it is clear why you assumed what you presented as I was clear on your thesis earlier. What should be equally clear is that Vmax for max turn rate is far lower that Vmax at top speed level flight. Accordingly the components for Lift related Drag and 'all other drag' are closer to 'equal' - than in the flight regime at top speed, which has low angle of attack, low trim drag, low induced drag, lower effective reductions to AR due to tip effects.

I may be missing something in my old age and long years away from the math but two things are clear to me.
1. The Vmax and load for sustained max turn rate is far less than the highest velocity attainable at max power/level flight. That fact alone reduces the weight of total parasite Drag from top speed and places it at a value much closer to the total Lift related Drag.

2. The AoA and associated, (and incrementally more important) drag components which are a function of increased AoA are much more important than at top speed level flight. That fact raises the contribution of trim drag, adverse pressure gradients, etc which must be considered and calculated for Vmax at max turn rate.


There are some other things, you mention, but don't make much sense, at least to me. Like mentioning departure of parabolic drag relationship near stall (i.e. regular invalidness of induced drag equation near max Cl). Sustained turn performance is nowhere near that high angles of attack, so it is largely inconsequential.

What is not inconsequential - but very difficult to quantify - is the combination of turbulent flow asymetrically applied by the prop vortex tube (in banked High G flight attitude, the contribution of increased spanwise flow along the lifting line plus other AoA related Drag components have the effect of adversely increasing pressure gradients causing earlier separation of boundary layer. These combinations push the CL point toward a higher 'effective AoA' and significantly increase lift related drag.

That is why I suggested that you iterate to determine a closer value for CL at Vmax. The CL in a steeply banked turn in high G while maintaining constant altitude is far greater than for level flight (as you know). For the same reasons asymmetrical control deflections are required to keep the thrust line above the plane of the turn

I don't recall saying anything about 'departure of parabolic drag relationship near stall" but in the real world Lift is affected by many things at increasing hig angles of attack - not the least of which is both increased spanwise flow component realized in steep banks/sideslips and adverse pressure gradiants behind the boundary layer 'attach region'


To sum up - next time I will use bold type for the word estimate

Your second post somewhat surprised me.

First, the method you propose is largely unusable, as you can see from many pointless discussions around the web. I wished to propose alternative to just this "educated guessing", since many times it involves much greater (and unquantifiable) errors.

See above for the approach I would propose that you consider

Second, why do you include Clmax to sustained turn performance estimation? I see only a theoretical possibility, that the Clmax does not allow turning at best load sustained turn. Do you know such WWII fighter? I don't.

That's all. Thanks again for your reply. I hope I don't appear hostile, as it was not my intention.

You didn't sound hostile to me and I did not mean to convey that I was dismissing your approach.

You understand the principles - but what we disagreed is the selection of Drag components for parasite drag at peak velocity when the actual turning speeds will be at a point in the Drag Polar where Induced and Parasite Drag values will be closer to each other... say the values for each near max economical cruise speed versus top speed

Cheers,

Zdenek,
badger45

Zdenek - about the second post.. and CLmax as a factor of interest.- it is nice to know the RELATIVE CLmax between to two ships you wish to compare - as it is essentially the approximate boundary at the stall point. For example a Me .109 has about 20% higher value than a P--51.

Secondly, the resultant CL in a banked level turn is higher than for a 'wing level' flight - as the vertical lift vector(opposite weight) is less than the total lift vector required (normal to aircraft axis)- it is directly related to the angle of Bank. Accordingly the relative AoA is higher on that wing in a curvilinear/banked flight path and becomes greater (assuming constant velocity) as the bank angle increases.

Summary - in a high G, turning flight near best turn rate, the airfoil is much closer to, or at, CLmax than the CL at the same speed in level flight... and in a much higher relative AoA...

and to maintain a constant altitude turn in these lower velocities/higher G manuevers the rudder and elevator trim drag components to drag become more of a factor in computing drag. While induced Drag is a small compared to the sum of all other Drag components for top speed/level flight, this is not the case at or around Corner Speed.

What we can say is that we agree to disagree for all the reasons presented.

 

[badger45]

Thank you for your reply!

One thing I want to make a bit more clear, as I realize, it may confuse other people. The equations above are meant for best load sustained turn time, not for the fastest or tightest sustained turn. Thus it is possible, the aircraft in question will turn faster in sustained turn near the stall speed. On the other hand, stall speed turning equations require knowledge of maximum lift coefficient and other subtleties involving high angles of attack and slow speed regularly encountered during such flight conditions. Whether and on what conditions it is possible to parametrize these aircraft characteristics to make quick estimates comparable to the above approach for best load sustained turn time, I do not know.

Solutions above can be mentally grasped as quick estimate of turning capability at about 60-65 percent of flying speed, where the best load sustained turn usually appears. Such turns are typically less than Cl=1 even for high wing loading and although flown much faster than stall speed, they still show very good turn rates (and thus are near best sustained turn rate).

Put into perspective in another way, best sustained turn is for WWII fighters usually an intersection between stall turn curve and power turn curve (my terminology, I forgot the official one). Above certain speed, a fighter does not have enough power to turn faster without loss of energy, below that speed he may not have enough lift. An intersection of these two is usually the true fastest turn. The equations above were derived from expressions related to the power turn curve and serve to determine time to turn when the g-load on the power turn curve is greatest. Fastest turn rate is usually higher and occurs at lower speeds. Please see the attachment with generic curves (x - speed, y - turn time). May be I should have emphasized this earlier, since it is not that much clear from my first post. But nobody's perfect...

Nevertheless I find the results quite telling about the real combat turning performance of selected aircraft, since I think that turns at higher speeds were more often than the stall fights. But that would be another topic altogether

Regards,

Zdenek
badger45

 

[drgondog]

Thank you for your reply!

One thing I want to make a bit more clear, as I realize, it may confuse other people. The equations above are meant for best load sustained turn time, not for the fastest or tightest sustained turn. Thus it is possible, the aircraft in question will turn faster in sustained turn near the stall speed. On the other hand, stall speed turning equations require knowledge of maximum lift coefficient and other subtleties involving high angles of attack and slow speed regularly encountered during such flight conditions. Whether and on what conditions it is possible to parametrize these aircraft characteristics to make quick estimates comparable to the above approach for best load sustained turn time, I do not know.

Here is an article which I believe will help clarify my point of view..

The Bootstrap Approach to Aircraft Performance<br>(Part Three — Maneuvering)

You are overlooking several critical factors in both your simplified calculations and assumptions.

First and foremost is overlooking CL vs bank angle. the Lift coefficient required to sustain a constant altitude turn increases inversely to the square of cosine of the bank angle - as does the load factor.

Increased Induced drag (plus all lift associated drag with increased AoA) and changes (increases) to Stall speed are functions of bank angle.

Further the stall speed increases as bank angle increases as the angle of attack on the wing must aslo be increased (compared to level flight) to provide the necessary total lift to balance the weight for level flight..

Additionally the result of increasing CL by the inverse square of cosine bank angle which results in the tremendous increase in Induced drag - then directly affects the amount of power(Thrust) required to offset Total Drag and therefore the corresponding decrease of (T-D)/W ratio and the crossover point at which the aircraft may no longer climb or even hold its constant altitude.


If you don't think that angle of attack increase in a banked turn, reflect on the fact that you have to pull the stick back increasingly as you increase bank angle.


Solutions above can be mentally grasped as quick estimate of turning capability at about 60-65 percent of flying speed, where the best load sustained turn usually appears. Such turns are typically less than Cl=1 even for high wing loading and although flown much faster than stall speed, they still show very good turn rates (and thus are near best sustained turn rate).

That is a statement which as yet has zero factual basis.. the best turn RATE in degrees per minute is usually close to stall, although stall is appreciably higher in a 6G turn than for level flight. The consequence of operating very close to stall is that there is no power reserve due to the significantly increased CL (and AoA to get there) to sustain slight climb rate or even level flight. There are no data to support a thesis for 60% (or any constant factor) of top speed.

Put into perspective in another way, best sustained turn is for WWII fighters usually an intersection between stall turn curve and power turn curve (my terminology, I forgot the official one). Above certain speed, a fighter does not have enough power to turn faster without loss of energy, below that speed he may not have enough lift. An intersection of these two is usually the true fastest turn. The equations above were derived from expressions related to the power turn curve and serve to determine time to turn when the g-load on the power turn curve is greatest. Fastest turn rate is usually higher and occurs at lower speeds. Please see the attachment with generic curves (x - speed, y - turn time). May be I should have emphasized this earlier, since it is not that much clear from my first post. But nobody's perfect...

Unless I missed something, you do not account for the various bank angles and associated CL to maintain a specific rate of turn - these factors have to be accounted for to plot Speed vs N and then extract rate of turn and radius. Without determining the CL required as a function of bank angle, the increase in Induced drag due to earlier approach to CLmax necessary to sustain a high G turn, you cannot predict the point at which level flight (and approach to stall) is no longer possibe.

Nevertheless I find the results quite telling about the real combat turning performance of selected aircraft, since I think that turns at higher speeds were more often than the stall fights. But that would be another topic altogether

Regards,

Zdenek
badger45

I will summarize the key points;

1. Maximum turn radius for given speed occurs at minimum bank angle and load factor

2. For constant altitude, Maximum turn rate (time to travel 360 degrees) occurs at or near stall, at the highest G load sustainable for level flight and max angle of attack, at the point where available thrust closely approaches total drag.

3. To model the mechanics of flight for that aircraft, one must know the CLmax and AoA for CLmax, gross weight, the AR (and assumed Oswald efficiencies), the Power Available at the altitudes you wish to solve for, incremental thrust due to exhaust or Meridith effect, the density of the air at that altiude, and strictly speaking - the other Lift related important drag factors (i.e Trim Drag and Pressure Drag at high angles of attack) as well as other important Parasite Drag factors (fricton drag, surface imperfections, cavities, protrusions, interference between fuse and wing). As the high g max turn rate manuever is at a much lower air speed than at Max speed level flight, you have to check eta for Thrust equation for piston powerplant.

You stated you wished a 'simplified' approach, but if you assume that all WWII fighters were essentially the same with respect to the other drag factors you would be wrong to some or greater degree - and the individual summations in the low to middle speed ranges are important between say a Me 109 and a P-51. For the latter you have to consider either additional thrust due to Meridith effect, or conversely reduced drag of radiator cowl/inlet geometry.

4. Even for a simplified approach (i.e eliminate all drag components other than Induced Drag strictly for thw wing, you should plot variations of required CL for given speeds as a function of bank angle and constantly check for Total Drag, AoA Stall points and G (N) loads. Hopefully the article and the charts help explain the approach.

 

[drgondog]

duplicate

 

[badger45]

You still don't understand, what my formula calulates. It DOES NOT calculate best sustained turn time. It estimates turn conditions for the peak load in sustained turn. See attached page from Hale, Aircraft Performance. My formula estimates the nmax point. As you can see from the graph, minimum turn radius and max turn rate occur at different point, i.e. at intersection with stall speed.

Thus far for most of your arguments. What concerns me, however, is your talk about me overlooking induced drag and so on. Have you looked to the way the formula is derived? If you are unable to follow the algebra, I can try to explain better. If you have followed the algebraic simplification, you have to see that induced drag is included, and then I don't understand your remarks.

In case you are from some reason unable to understand, that formulas above DO NOT estimate best turn rates, I can give you a simplified formula for SL conditions, that DOES.

This time I will limit myself just to the formula, assumptions and example, since the algebra was first time thrown out of window, as it seems.

Best sustained turn rate = [153 * V^0.25] / [ (W/S)^0.5 * (W/P)^0.25 * (W/b^2)^0.25] in degrees per second

W/S is wing loading (kg/m2), W/P power loading @ SL (kg/HP), b is span in meters (w/b^2 is weight over span squared), V is max speed @ SL.

I can be put up in simpler way as (153 * V^0.25 * P^0.25 * S^0.5 * b^0.5) / W.

Assumptions: Clmax = 1.2 (others regarding Oswald number and propeller efficiency as above).

Example:
W = 3170 kg, b = 12 m, S = 20 m2, P = 1100 HP, V = 500 km/h (Ki-61 I)
turn rate = 153 * 4.73 * 5.76 * 2.11 * 1.86 / 3170 = 20.4 deg/sec
turn time = 17.7 sec

Analysis:
Aspect ratio A = 7.2
Oswald efficiency (Cavallo) = 1.78*(1-0.045*A^0.680)-0.64 = 0.833
Induced drag coefficient k = 1/(pi*A*e) = 0.053
Dynamic pressure (max speed) = 0.5 * ro * (V^2/12.96) = 11825
Dyn. pressure * S (ref area) = q *S = 236497
Coefficient of lift (max speed) = 9.81 * 3170 / 236497 = 0.131493
Induced drag (max speed) = k*Cl^2 = 0.0009
Total drag (max speed) for prop efficiency 0.78 = (1100*0.78*735.5) / (500/3.6) = 4544
CD (max speed) = 4544 / 236497 = 0.0192
Zero lif drag (max speed) Cd0 = 0.0192 - 0.0009 = 0.0183

That much for gaining all essential parameters. Now that we have drag equation CD = 0.0183 + 0.053 Cl^2, we can start iterating for stall speed conditions at given Clmax (I will assume 1.2). I actually don't iterate manually anymore, since spreadsheet can do it for me. The result is 17.48 seconds for best sustained turn time, equal to turn rate 20.59 degrees per second.

As I have not given up hope, that my first two formulas will be understood properly, let's have a look at what they yield for SL conditions: turn time 18.8 seconds, speed (0.64 * V max) = 320 km/h. This is the BEST LOAD sustained turn. As this one needs not iteration, we can calculate it exactly (within limits of standard drag model as above) according to my first post (formula is actually from Hale). The result is 19.0 seconds at 320.46 km/h. Simple calculation can show, that at this condition g-load is 3.17, while for the best sustained turn time the g-load was only 3.13.

From this you can see, I hope, actual use of my formulas. Those from first post calculate conditions of maximized load in sustained turn (this time the estimate was optimistic by 1.1%), the one from this post can be used for best time to complete sustained turn (this one was pessimistic by 1.3%). Overall they give splendid estimates, by my standards. Of course the model can be further complicated e.g. by increase in zero drag in lower Reynolds numbers, but there is good reason, it is not. But on this level of modeling, my equations should give quick estimate within few percents of modeled results (that are otherwise quite pain to calculate, as I have shown).

I hope this helps.

Zdenek
badger45

P.S. I know Clmax is actually not 1.2 for all WWII aircraft. But the formula can be easily applied to any given Clmax if you square the difference betweeen Clmax desired and 1.2.

 

[badger45]

One further thing comes to my mind: what is the simplified approach good for and why and when to use it?

Basically I use them in two different circumstances. First, when I read about some less well known aircraft, perhaps a prototype or even just a design, the formulas enable me to to quickly evaluate its turning performance. There are many books dealing with the historical development of aircraft full of wonderful prototypes (like popular series by Yefim Gordon) and many relatively unknown aircraft as well. Most of the time you won't find any information about their characteristics, except for a few basic parameters. Then my simplified approach comes to play.

Second, if you like to create what-if scenarios or are interesting recreating design process of WWII aircraft, the formulas give a basic understanding of what can be traded for what. E.g. time to time people speculate about performance of aircraft with other than historical engines and so on. Creating detailed analysis is desirable only when you have some idea, what the results might be. E.g. imagine change of water-cooled 1100 HP engine for same weight air cooled 1300 HP engine and let's say, that because of the increased drag, the newer version of aircraft has the same top SL speed. Is the air cooled engine aircraft turning better? According to my formulas, the air cooled project will have better turning time by 4 percent (at stall, incalculable without further data) and will be able to pull 8 percent higher sustained g (above stall at 320 km/h).

Of course there should follow detailed analysis and after all, only actual aircraft can prove or disprove any design concept. But for preliminary analysis it is still very useful. At least it seems to me.

Regards,

Zdenek,
badger45

 

[drgondog]

You still don't understand, what my formula calulates. It DOES NOT calculate best sustained turn time. It estimates turn conditions for the peak load in sustained turn. See attached page from Hale, Aircraft Performance. My formula estimates the nmax point. As you can see from the graph, minimum turn radius and max turn rate occur at different point, i.e. at intersection with stall speed.

It doesn't calculate anything correctly.. you continue to use 1 G straight and level dash speed characteristics when trying to apply to mid speed turning manuever.

Thus far for most of your arguments. What concerns me, however, is your talk about me overlooking induced drag and so on. Have you looked to the way the formula is derived? If you are unable to follow the algebra, I can try to explain better. If you have followed the algebraic simplification, you have to see that induced drag is included, and then I don't understand your remarks.

In case you are from some reason unable to understand, that formulas above DO NOT estimate best turn rates, I can give you a simplified formula for SL conditions, that DOES.

This time I will limit myself just to the formula, assumptions and example, since the algebra was first time thrown out of window, as it seems.

Best sustained turn rate = [153 * V^0.25] / [ (W/S)^0.5 * (W/P)^0.25 * (W/b^2)^0.25] in degrees per second

W/S is wing loading (kg/m2), W/P power loading @ SL (kg/HP), b is span in meters (w/b^2 is weight over span squared), V is max speed @ SL.

I can be put up in simpler way as (153 * V^0.25 * P^0.25 * S^0.5 * b^0.5) / W.

Assumptions: Clmax = 1.2 (others regarding Oswald number and propeller efficiency as above).

Example:
W = 3170 kg, b = 12 m, S = 20 m2, P = 1100 HP, V = 500 km/h (Ki-61 I)
turn rate = 153 * 4.73 * 5.76 * 2.11 * 1.86 / 3170 = 20.4 deg/sec
turn time = 17.7 sec

Analysis:
Aspect ratio A = 7.2
Oswald efficiency (Cavallo) = 1.78*(1-0.045*A^0.680)-0.64 = 0.833
Induced drag coefficient k = 1/(pi*A*e) = 0.053
Dynamic pressure (max speed) = 0.5 * ro * (V^2/12.96) = 11825
Dyn. pressure * S (ref area) = q *S = 236497
Coefficient of lift (max speed) = 9.81 * 3170 / 236497 = 0.131493
Induced drag (max speed) = k*Cl^2 = 0.0009
Total drag (max speed) for prop efficiency 0.78 = (1100*0.78*735.5) / (500/3.6) = 4544
CD (max speed) = 4544 / 236497 = 0.0192
Zero lif drag (max speed) Cd0 = 0.0192 - 0.0009 = 0.0183

That much for gaining all essential parameters. Now that we have drag equation CD = 0.0183 + 0.053 Cl^2, we can start iterating for stall speed conditions at given Clmax (I will assume 1.2).

Coaching tip - you DON'T have all the essential parameters. Thrust will be a crucial variable and you haven't even estimated exhaust thrust which is 10% or more of the total thrust generated by the propeller/engine system.

Second coaching tip. If you do not run a series of calculations comparing bank angle to Lift (CL) required to balance weight, extract the velocity required to generate that resultant Lift and generate the Nz factors required to sustain a constant altitude - you have no hope of proceeding to the next step... You will be constantly calculating to obtain lift Load, velocity, AoA/CL required, Thrust Available and thrust required.

Third coaching tip. As you iterate on bank angle versus Velocity (and CL vs AoA to obtain the lift to offset the aircarft weight) required to maintain constant altitude you must test Thust available to Thrust required. The last test is a 'killer' for your theories as you must once again (and many more times) re-calculate DRAG.

Fourth Coaching tip. In the 180-300 kts/1G to say 6 G envelope - you will be constantly checking to make sure the airplane stays in the air at a certain bank angle, AoA and velocity. This is where the 'tedious' aero stuff comes in and extends beyond simpleton assumptions and 'algebra'. At the velocity near best sustained turn rate you are far closer to the region where induced drag approaches parasite drag because of the high angles of attack

In these velocity ranges the Induced and Angle of attack/Lift related components are very important. Conversely the Parasite Drag components reduce in weighting relative to Max level flight speed - but important nevertheless.

Its time to re calculate the various Paraiste Drag components, test lift related components like trim drag, and re-solve for

T=D

It is equally true that the flight regimes of high bank angle/high G flight exceed the ability of ailerons and elevators to sustain level flight once Total lift in 'Z' direction fails to offset W in opposite direction. You need to know where they are. You need to know what the structural capability is but most WWII fighters fall out of the sky (either stall or too great of a bank angle and not enough CL) .

When your fighter of interest reaches the braek point at max power, the stick (and resultant control deflections of elevator and rudder) location in the aft rear quadrant has reached its limit - the bird is going down - either by stall or descending spiral.

I actually don't iterate manually anymore, since spreadsheet can do it for me. The result is 17.48 seconds for best sustained turn time, equal to turn rate 20.59 degrees per second.

As I have not given up hope, that my first two formulas will be understood properly, let's have a look at what they yield for SL conditions:

Lets for the moment that we ignore your supercillious, condescending and boring contempt for aero engineers 'complicating things?

turn time 18.8 seconds, speed (0.64 * V max) = 320 km/h. This is the BEST LOAD sustained turn. As this one needs not iteration, we can calculate it exactly (within limits of standard drag model as above) according to my first post (formula is actually from Hale). The result is 19.0 seconds at 320.46 km/h. Simple calculation can show, that at this condition g-load is 3.17, while for the best sustained turn time the g-load was only 3.1

If you mean by 'Best Load', the maximum sustained turn rate as a funtion of N (acceleration to center of turn radius) you have to solve for N as a function of CL, V, and AoA. N is a function of Theta (bank angle) and symbolized as L tan theta where theta is bank angle.

From this you can see, I hope, actual use of my formulas. Those from first post calculate conditions of maximized load in sustained turn (this time the estimate was optimistic by 1.1%), the one from this post can be used for best time to complete sustained turn (this one was pessimistic by 1.3%).

Based on what? what is your refernce point and sources for the values you compared against?


Overall they give splendid estimates, by my standards. Of course the model can be further complicated e.g. by increase in zero drag in lower Reynolds numbers, but there is good reason, it is not. But on this level of modeling, my equations should give quick estimate within few percents of modeled results (that are otherwise quite pain to calculate, as I have shown).

Zdenek - yes they are a pain - no you have not demonstrated the 'elligance' of your method

I hope this helps.

Zdenek
badger45

P.S. I know Clmax is actually not 1.2 for all WWII aircraft. But the formula can be easily applied to any given Clmax if you square the difference betweeen Clmax desired and 1.2.

Every airfoil has a different Cl/CD profile as a function of AoA - as well as different CD0's for that airfoil. For example, If you picked a Mustang (or Ta 152 or Do335 or P-47N/M) at top speeds at ~440-460 mph at 25K altitude you enter the region where compressibility drag begins and will affect CD=CDi+CDo+CDm

Different aircraft have dramatically parasite drag and even thrust components. Different aircraft will be flow at different prop pitch although max thrust will be at max power/fine pitch

Your approach presumes no effect on bank angle to required AoA to increase lift as a function of velocity -

Aside from that, well done on the algebra side of the discussion. Boundary conditions and physical assumptions for the model? - not so much.

 

[badger45]

I have only one coaching advice for you: stop coaching me. I don't like it. I told you once, now it is the second time. I am not some teenage enthusiast who read one book on airfoils, so please lay aside your arrogance or quit discussion with me. I will not repeat myself for the third time...

Now to your post.

1] You will have to prove that my formula calculates anything wrong. As of now you have proved nothing. Just enumerating possible sources of systematic error is entirely useless. They may or may not make a major contribution. So please don't hesitate to make a definite proof supported by some numbers or academic evidence.

2] Formula CD=Cd0 + K*Cl^2 is normally used for estimation of Cd0. If you are uncomfortable with that, it's entirely your problem, not mine. See for example Torenbeek, Synthesis of Subsonic Airplane Design (p. 153, section 5.3.4 Retracing a drag polar from performance figures).

3] This approach to drag is also regularly used in performance prediction. If you don't agree, don't blame me. See for example Ojha, Flight Performance of Aircraft, (p. 363 and following).

4] Exceeding aileron capacity, structural strenght limit... You ARE kidding this time, aren't you? Are we still speaking about sustained turn performance?

5] Start to read thoroughly. I attempted to explain to you the term 'best load' so many times, I can only recommend re-reading previous posts and looking at the pdf attachments. The same holds for the percentage comparisons. You are asking what is their basis. Well the answer is right there. It is comparison of standard drag model calculation of turning performance versus my simplified formulas.

6] Last thing. Iteration is boring. It is boring. No matter what I think of aerodynamics, iterative and numerical methods are always boring. I have not said, that aerodynamics is boring. This is not the first time you offended me with such assumptions, giving me the thoughts that YOU misinterpret from my sentences. Stop doing that and speak to the matter or don't speak at all.

Zdenek,
badger 45

 

[drgondog]

I have only one coaching advice for you: stop coaching me. I don't like it. I told you once, now it is the second time. I am not some teenage enthusiast who read one book on airfoils, so please lay aside your arrogance or quit discussion with me. I will not repeat myself for the third time...

Go ahead and repeat yourself - I am consumed by indifference. As to 'lecturing' perhaps you might go back and review your condescending remarks to me to gain some insight?

Now to your post.

1] You will have to prove that my formula calculates anything wrong. As of now you have proved nothing. Just enumerating possible sources of systematic error is entirely useless. They may or may not make a major contribution. So please don't hesitate to make a definite proof supported by some numbers or academic evidence.

What I have done is meticulously point out what you don't do in concert with applied aerodynamics and flight mechanics. I have offered what I consider to be excellent examples of such applications, presented by aerodynamicists and published on the internet. You have neither responded or acknowledged the perspectives or the physics behind their presntations...

2] Formula CD=Cd0 + K*Cl^2 is normally used for estimation of Cd0. If you are uncomfortable with that, it's entirely your problem, not mine. See for example Torenbeek, Synthesis of Subsonic Airplane Design (p. 153, section 5.3.4 Retracing a drag polar from performance figures).

Yes - the 'gamers' are fond of misusing CD0 and apply the term to all drag components except Induced Drag. CD0 has a very specific definition and it is solely the zero lift drag of that SPECIFIC airfoil at the AoA for ZERO Lift. It IS a COMPONENT of Total Drag, IT IS a Component of Profile Drag. It Is the Same value at Stall as at top speed and the only Drag component that remains constant througout the flight profile. It is NOT total Drag less low AoA Induced Drag.

Further, those boneheads will lecture from total ignorance on a stream of assumptions that come form a place 'where the sun doesn't shine'. Nearly all seem to have passed an Algebra II course, have read different articles from texts and think they have acquired the requisite knowledge to offer 'simple' solutions with amazing facility.

You seem to fit neatly in that category. If you are uncomfortable with that, its entirely your problem, not mine.

3] This approach to drag is also regularly used in performance prediction. If you don't agree, don't blame me. See for example Ojha, Flight Performance of Aircraft, (p. 363 and following).

I stated earlier that I DON'Tagree, and stated the reasons why 'your approach', which presumably follows 'Ojha, is a crock. If you care to post the pages I will look at them and respond - I must confess curiosity regarding how well you applied his approach and to what degree he framed his assumptions. I will tell you this - if he ignores the extremely high drag factors equivalent in magnitude to raw Induced Drag equation - while in a much lower banked angle speed range - he doesn't know what he is talking about.

4] Exceeding aileron capacity, structural strenght limit... You ARE kidding this time, aren't you? Are we still speaking about sustained turn performance?

Not really kidding but trying to convey to you what conditions exist in a cockpit when you actually approach a max G sustained turn in a high performance fighter. Simply, no WWII fighter could ever approach 8+ Gs in a sustained turn - not because it will fall apart, but because the performance of the airframe will not enable it even if the pilot could. On the other hand when applying the physics of flight mechanics to say a Cessna 172, whose limit load factor is around 3g - IT IS A CONSIDERATION -

Having said that, those ARE factors 'normally calculated' by the practicing engineer when crafting plots of turn radius, velocity and N. Real world has the ship falling out of the sky in a deep stall long before that. Further, the ability of the rudders and elevators to constantly maintain a level altitude and the necessary angle of attack in a sustained turn is a key factor. 'Real world' analysis back in WWII examined the change in stick force for a desirable 'load factor' (for you, load factor in this case is 1/cos theta where theta is the bank angle.

5] Start to read thoroughly. I attempted to explain to you the term 'best load' so many times, I can only recommend re-reading previous posts and looking at the pdf attachments. The same holds for the percentage comparisons. You are asking what is their basis. Well the answer is right there. It is comparison of standard drag model calculation of turning performance versus my simplified formulas.

Yes you attempted to explain... reading your posts and your simplified math doesn't really nail what YOU mean and how that either applies to 'load factor' or 'lift loading' and their relationship to a sustained turn. I know the relationship of the latter and I know how to calculate the required load factor, CL, V and radius. What is clear is that you don't as you have continuously ignored the flight mechanics of banked flight. There is a 'max load factor' that every aircraft will reach based on the thrust available, the max bank angle and velocity achievable, the drag which at that point equal the thrust, and the CLmax - beyond which any increase in AoA (driven by the aforementioned 'elevator and rudder') will cause the aircraft to stall.

6] Last thing. Iteration is boring. It is boring. No matter what I think of aerodynamics, iterative and numerical methods are always boring. I have not said, that aerodynamics is boring. This is not the first time you offended me with such assumptions, giving me the thoughts that YOU misinterpret from my sentences. Stop doing that and speak to the matter or don't speak at all.

Or what? Are you going to tell your mommy?

Zdenek,
badger 45

I am profoundly and deeply sorry that you are offended. Having said that - get over it.

You may be 'bored' with iterative solutions but absent some pretty high powered aerodynamic models (which you will not easily understand and certainly do not have access to), you are stuck with iterative processes to generate tables - as neither YOUR equations nor your boundary conditions nor your assumptions nor your knowledge of the aerodynamics nor your source data - are up to the task.

Your model and your knowledge are probably fine for discussion with your gaming friends and perhaps you are an acknowleged thought leader in your inner circle. You may rest comfortably with that testimonial. Perhaps we may cease with the insults and each go our own way? At any rate I have spent far too much time with you.

 

[drgondog]

Btw - While i was not able to see details in his book, the table of contents stresses that Ojha discusses 'Load Factor", Maximum Bank Angle, Maximum Turn Rate and Minimum Turn Radius.

I suspect that your understanding of 'Best Load' does not relate to anything Ojha presents in his book.

 

[badger45]

Still no proofs and only more offenses? Well then our conversation is over. I have too little time to waste it on idiots.

Regards,
badger45

 

[drgondog]

Oh, well... Bye.

 

[Timppa]

Hi All,

since I noticed that there are not only many fans of WWII aircraft, but also many people actually interested in calculation of basic performance indicators, I would like to post some useful equations. I have found the exact approach of aerodynamics engineers quite bothersome to use and tried (successfully) to reduce the equations as much as possible, while retaining sufficient degree of accuracy.

The simplest approach to calculate time to perform sustained turn at 1000 meters height:

t = (W/b) * (Vmax/P)^0,5 * (1/8,80)

where W is weight in kg, b is span in meters, V is max sea level speed in km/h and P is power at the maximum sea level speed (measured in metric horspower, i.e. 735.5 W).

Thanks for your post. I'll make a spreadsheet. The extreme cases, like the Ta-152H can be interesting..)
FYI, there is the Ignore List option in this forum.

 

[Juha]

Hello turning times according to Soviet tests at 1000m
FW 190A-4 22sec when turning to left, 23 sec to right
Spitfire Mk VB 18,8 sec

 

[badger45]

Hello turning times according to Soviet tests at 1000m
FW 190A-4 22sec when turning to left, 23 sec to right
Spitfire Mk VB 18,8 sec

For best time sustained turn you have to use the second set of formulas and because of the 1000 meters height (instead of SL), there must be a slightly higher constant: instead of 2.35 (for SL), use 2.47 (1000 meters). Input can be taken form SL figures (SL max speed and SL power), as these parameters usually change only slightly and they are under one-fourth power, so the difference should be negligible (and is calibrated in the constant - 2.47 is average from actual interval of about 2.45 to 2.5).

Using 2.47:

1560HP, 540km/h, 18.3m2, 10.5m, 3850kg for Fw-190A-4: 22.6 seconds
1030HP, 470km/h, 22.48m2, 11.23m, 2950kg for Spitfire VB: 17.4 seconds

The Russian Spitfire seems to be turning a bit slower than expected. Unfortunately I am quite lost in Spitfire variants and I don't know, which one was used in the test. The data above are for Merlin 45 and unclipped wing.

8 percent difference worse real performance may be explained by clipped wing, but it may well be something other. Aircraft in many tests were run in suboptimal regime of engine or excessive weights, for different reasons. I have ran standard detailed computations, which show best turn time 17.23 seconds for above configuration. To make it up to 18.8 seconds would mean counting either around 20 percent less power (similar to usable power and speed of Spitfire I) or Clmax less than 1 (not probable) or weight increase to 3185 kg. Or some combination of these factors

Regards,
badger45

 

[badger45]

Thanks for your post. I'll make a spreadsheet. The extreme cases, like the Ta-152H can be interesting..)
FYI, there is the Ignore List option in this forum.

Ta-152H, 1750HP@SL, 530km/h@SL, 14.44m, 23.5m2, 4625kg: 19.98 seconds (formula) and 20.17 seconds (full model) for best sustained turn time and 20.03 seconds (formula) and 20.45 seconds (full model) for sustained turn at maximum load factor.

Thanks for idea for extreme example! Data used are quick jump from chronically unreliable Wiki (speed is rough estimate for 1750HP derived as cube root from cited 560km/h@2050HP with MW-1 boost). I am short of time to browse my books for more reliable data.

Regards,
badger45