Some Fw-190 questions

Discussion in 'Aviation' started by tomo pauk, Sep 19, 2012.

  1. tomo pauk

    tomo pauk Creator of Interesting Threads

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    Hello, people,
    I'd like to find out some details re. Fw-190; 1st two are pasted from the another thread:
    1.- If some kind soul has original German data about Fw-190A's Cd0, it would be neat if it's posted here
    2. -Dr. Tank said that his fighter is the 'Dienstpferd' (service horse), not the Racing horse. He states further that both Spitfire Bf-109 were 'racing horses' - a very large engine on the front of the smallest possible airframe; in each case armament had been added almost as an afterthought. I don't see the Fw-190 being anything but - it was based around the very large engine, that was attached at the smallish airframe, while featuring 4 LMGs as the armament in the 1st iteration. The wing area was at 1st so small that it was upped immediately, even prior the 1st series version; the armament also received prompt increase. So what it was - race or service horse?
    3. - What amount of the engine power was drawn for the fan cooler?

    So far, I've tried this approach to find out the Cd0: found the flat plate drag (from the Lednicer's analysis), 'f', and he states it to be at 5.22 sq ft. Then, I've divided the f with 197 sq ft of the wing area (the formula from the table at US 100 thousand book), and arrived at number of 0,0265, and that number should present the Cd0. Any corrections are more than welcomed, as are other veritable informations.
     
  2. stona

    stona Well-Known Member

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    #2 stona, Sep 19, 2012
    Last edited: Sep 19, 2012
    Tank was wrong about the Spitfire armament being an afterthought. 6 or 8 guns,in the wings,was a requirement of specification 10/35 of April 1935 when the Spitfire only existed as a mock up.
    It had been calculated in the early 1930s that due to higher speeds a pilot would be able to hold his guns on target for about two seconds. In that time 8 guns could theoretically deliver 256 rounds on target. Men like Squadron Leader Sorley and Air Marshall Ludlow-Hewitt (at the Air Ministry) were already demanding an eight gun armament when the Spitfire was still on the drawing board.

    I'll let Tank explain his own horse analogy.

    "[The Messerschmitt 109 and the British Spitfire]...could be likened to racehorses: given the right amount of pampering and an easy course they could outrun almost anything. But the moment the going became tough they were liable to falter.
    I felt sure that a quite different breed of fighter would also have a place in any future conflict: one that could operate from ill prepared front line airfields;one that could be flown or maintained by men who had only a short training; and one that could absorb a reasonable amount of damage and still get back.
    This was the background thinking behind the Focke-Wulf 190;it was not a racehorse but a "Dienstpferd",a cavalry horse."

    Cheers

    Steve
     
  3. bada

    bada Member

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    fw190dragtable1gn-1.jpg
     
  4. davparlr

    davparlr Well-Known Member

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    #4 davparlr, Sep 20, 2012
    Last edited: Sep 20, 2012
    nevermind! errant post.
     
  5. drgondog

    drgondog Well-Known Member

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    Careful about context: D/q=Cd*S; Cd=Dt/q*S

    While Cd is TOTAL drag, where Dt is also = Thrust and q is the dynamic pressure which varies with density and Velocity squared.

    So, Flat Plate drag is not Cd0. It is an expression of Total Drag = Dp + Di +D compressibility (Do + Dvortex +Di +Dcomp) at a particular speed and altitude as a function of wing area.

    Do and Cvortex is the sum of all parasite drag including friction drag and form drag. At Zero Lift the CDp becomes the base CDo fro sub compressibility flight speeds.

    To get to Cdo from your source you have to also know the speed and altitude. Fortunately if you have that, you don't need the Hp becuase in this case Hp has been reduced to Engine Thrust plus Exhaust thrust already.

    From the above equation, assuming speed for your calcs is below ~ .55M, for level flight, and assuming you have a known altitude and gross weight:

    D= Dp + Dinduced (strip compressibility and vortex drag)
    plug rho of the altitude you are working the problem for (i.e rho @ SL = .0023769 lb-sec>>2/ft>>4)
    Solve for q=1/2*rho*v>>2
    D/q= Cd*S = 5.22 (your value) x 197ft>>2 = 1093 lbs. (Presumably this is the thrust of the engine/prop system plus exhaust)
    then D= q * 1093
    Dinduced =CDi*S = Cdi*197 but CDi= (CL)>>2/(AR*pi*e) where CL=L/q*S for the weight and altitude you are analyzing.

    But CD = CDi + CDo, so CD0 = CD-CDi
    That is how you get to CDo when you know Total Drag, Speed and altitude for given wing area.
     
  6. Hop

    Hop Member

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    The RAE gave a figure of 0.0269 based on tests of Faber's A3. Wing area was reported as 203 sq ft.
     
  7. drgondog

    drgondog Well-Known Member

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    Remember Flat Plat Drag is the sum of ALL drag components. At low speed Induced Drag will dominate and Parasite Drag will be very low., at the bottom of the Drag Bucket where L/D is Max, Induced Drag will equal Parasite Drag and at max speed Induced drag will be at a minimum.

    CDo can not be extracted from Flat Plate Drag by simply dividing by wing Area S.
     
  8. tomo pauk

    tomo pauk Creator of Interesting Threads

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    #8 tomo pauk, Feb 9, 2013
    Last edited: Feb 9, 2013
    The gentleman at LEMB arrived at this conclusion, using the table kindly posted by Bada:

    The Cw0*F walue is the one for high speed (schnellflug), for the climb (steigflug) the value is 0,623 (or 0,629?).
     
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