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Raghu's Stability Spreadsheet

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  • Raghu's Stability Spreadsheet

    https://docs.zoho.com/sheet/publishe...49361bf72180d6

    In the above stability spreadsheet Raghu calculates the tail contribution as:
    C_tail =L2*J2*K2/(iyy*391*1.46) (cell G5)
    The numerator is the tail volume times the HS lift curve slope. Iyy is the mass moment of inertia. Can someone shed some light on why the two constants 391 and 1,46 are needed?


    PS: I have developed a stand alone version of the spreadsheet which I would make available if Raghu has no objections. It would, of course, contain a link to his original work. Looking forward to your answer Raghu, if you happen to read this.
    Cheers,

    Juergen

    ..Il semble que la perfection soit atteinte..
    ....non quand il n'y a plus rien à ajouter,...
    ...mais quand il n'y a plus rien à retrancher...
    - Antoine de Saint-Exupéry -

  • #2
    It just occured to me that the air density is missing in the formula. Now 1/391 is 0.002558 which is close to the usual value for air density of 0.0023762 slug/ft^3, could that be the solution to the 391 mystery?
    Looking forward to your comments.
    Cheers,

    Juergen

    ..Il semble que la perfection soit atteinte..
    ....non quand il n'y a plus rien à ajouter,...
    ...mais quand il n'y a plus rien à retrancher...
    - Antoine de Saint-Exupéry -

    Comment


    • #3
      The tail contribution is used to calculate Mq, e.g. cell C17, where it is multiplied by the airspeed, which is entered in mph. Now all other values in the formula have feet as dimension for length. Tonight I found that 1.467 is the conversion from mph to fps. Thank you for your support.
      Cheers,

      Juergen

      ..Il semble que la perfection soit atteinte..
      ....non quand il n'y a plus rien à ajouter,...
      ...mais quand il n'y a plus rien à retrancher...
      - Antoine de Saint-Exupéry -

      Comment


      • #4
        Jurgen,
        You are correct about the 1.46 factor. As for 391, if you divide the velocity squared in miles per hour by 391 you get the dynamic pressure in pounds per square feet: 1/(0.5*0.23762*1.467^2)= 391.

        Comment


        • #5
          Jurgen,
          You are correct about the 1.46 factor. As for 391, if you divide the velocity squared in miles per hour by 391 you get the dynamic pressure in pounds per square feet: 1/(0.5*0.23762*1.467^2)= 391.

          Comment


          • #6
            Click image for larger version  Name:	RaghuSpreadSheetSc.jpg Views:	1 Size:	517.1 KB ID:	1142282




            Thank you very much for chiming in on this one Raghu! I would very likely not have been able to figure out what the 391 is...;-( I have opened this thread because I will enter a new phase in my work on rotary wing stability this year. So far I have focused on getting the stability derivatives right. At first I had hoped that building a good model and then numerically calculating the derivatives would be the way to go. It turned out that I got stuck at some point because every time the derivatives were very different (sometimes even in sign) from measured or calculated values I had for comparison, I was lost, because I couldn't figure out where the problem was. I have therefore turned to analytical expressions about a year ago. One of the best worked out examples is the tandem rotor helicopter in naca-6763 "Helicopter Stabiltiy Handbook", that's why I have taken the fairly big leap to add a tandem rotor capability to my program and implement all formulae for the analytical derivatives in 6763. Unfortunately this tandem rotor example does not incorporate a horizontal stabilizer, which is not unusual for a tandem rotor helicopter. I was therefore looking for a way of comparing my stabilizer derivatives somewhere and that is where your stability spreadsheet came in. Currently I am implementing Prouty's formulae for the HS. The new phase I talked about earlier is that I will now try to use the derivatives to assess the stability characteristics of a rotary wing aircraft. I came across two criteria, first the one dating from the earliest flight tests which states that after a pull up the normal acceleration curve should become concave downwards within two seconds. To me it looks as if this one were not very helpful for autogyros, since pull up maneuvers are a bit dangerous since you loose disk loading at the peak of the your pull up. The second one is from Padfields "Theory of Flying Qualities" where he states that the short period mode damping should be larger than 0.35. Unfortunately the damping is very sensitive to the normal speed derivative dZw/dt or Zw dot. The only source for this derivative that can be calculated reliably using analytical expressions is the horizontal stabilizer. Omitting the dZw/dt term will result in a conservative, albeit very crude approximation which is fairly unsatisfactory and I will therefore try to incorporate it.
            In your spreadsheet you seem to have used other critiria to evalute the properties of the short period mode, which I have tried to reverse engineer in the flow diagram below. They incorporate time to half/double (C15/C16) and oscillation period (C17) One point where I am really lost is where you state that if the absolute value of the minimum of the time to half/double values is larger than 1.5 this will result in marginal stability.
            Over the years you have been so kind as to share a lot of your quest for understanding autogyro stability with the forum and I am planning to read through all the old threads, so if you have elaborated on the stability critiria you used elsewhere I would really appreciate a link to the thread.

            Thank you very much for your support!
            Last edited by kolibri282; 02-20-2019, 02:07 PM.
            Cheers,

            Juergen

            ..Il semble que la perfection soit atteinte..
            ....non quand il n'y a plus rien à ajouter,...
            ...mais quand il n'y a plus rien à retrancher...
            - Antoine de Saint-Exupéry -

            Comment

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