Sustained Turn Performance Of Piston Engine Aircraft

Discussion in 'Flight Test Data' started by badger45, Oct 26, 2010.

  1. badger45

    badger45 New Member

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    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.
     
  2. drgondog

    drgondog Well-Known Member

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    #2 drgondog, Oct 26, 2010
    Last edited: Nov 27, 2010
    "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?
     
  3. drgondog

    drgondog Well-Known Member

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    #3 drgondog, Oct 26, 2010
    Last edited: Oct 27, 2010
    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!
     
  4. badger45

    badger45 New Member

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    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
     
  5. drgondog

    drgondog Well-Known Member

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    #5 drgondog, Oct 28, 2010
    Last edited: Dec 13, 2010
    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.
     
  6. badger45

    badger45 New Member

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    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
     

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  7. drgondog

    drgondog Well-Known Member

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    #7 drgondog, Nov 3, 2010
    Last edited: Dec 13, 2010
    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.
     

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  8. drgondog

    drgondog Well-Known Member

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    #8 drgondog, Nov 3, 2010
    Last edited: Nov 27, 2010
    duplicate
     
  9. badger45

    badger45 New Member

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    #9 badger45, Nov 9, 2010
    Last edited: Nov 9, 2010
    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.
     

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  10. badger45

    badger45 New Member

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    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
     
  11. drgondog

    drgondog Well-Known Member

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    #11 drgondog, Nov 10, 2010
    Last edited: Dec 13, 2010
    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.
     
  12. badger45

    badger45 New Member

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    #12 badger45, Nov 11, 2010
    Last edited: Nov 11, 2010
    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
     
  13. drgondog

    drgondog Well-Known Member

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    #13 drgondog, Nov 11, 2010
    Last edited: Nov 11, 2010
    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.
     
  14. drgondog

    drgondog Well-Known Member

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    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.
     
  15. badger45

    badger45 New Member

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    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
     
  16. drgondog

    drgondog Well-Known Member

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    Oh, well... Bye.
     
  17. Timppa

    Timppa Active Member

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    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.
     
  18. Juha

    Juha Well-Known Member

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    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
     
  19. badger45

    badger45 New Member

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    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
     
  20. badger45

    badger45 New Member

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    Ta-152H, [email protected], [email protected], 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 [email protected] with MW-1 boost). I am short of time to browse my books for more reliable data.

    Regards,
    badger45
     
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