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Homework Help: Mass moment of inertia of an airplane about its main gear

  1. Nov 7, 2017 #1
    1. The problem statement, all variables and given/known data
    Okay, so this isn't so much a problem as a design issue while following an example given in a book.
    (Relevant material is here: http://www.aero.us.es/adesign/Slide... 12. Desig of Control Surfaces (Elevator).pdf )

    Now, the calculations require a mass moment of inertia about the main gear to be calculated in order to be factored into equations. Sure enough this is already provided in the example they give, but in order to make sure that while applying it to the design I'm making I decided to check my formula against the info given in the PDF...

    The details of the airplane in the PDF are a total weight of 20 tons, Iyy 150,000 kg m^2, and this drawing of the plane:

    2. Relevant equations
    What formula must I use for such an object.
    For a slightly different design (instead of using a tubular body, instead using a flying wing-style cabin) is there a general formula that can be used - such as the one for a rod rotated at its end, etc.

    3. The attempt at a solution
    The formula I used was that for a rod being rotated at its end (ML^2)/3 ... However that didn't come out right, no matter how I tried it. As a distance I tried using 1.1, 1.7, 0.6, 2.02 (hypotenuse that links CG and the main gear wheel), and several values between 8 and 12 (since the main gear seems to be right about the center, figured the distance to the nose would be the same)... Nothing comes out right - the only value that would fit is about 4m if my memory serves me right.

    I then proceeded to try several formulas. The one that finally got me in that ballpark of 150 tons per square meter was the one for a rectangular plane rotating about an axis at its end ( m/12 * (4h^2 * w^2)) where I fed it 1.1 as height and 2 as width as that is the rough radius of a private jet like this. (to be noted, the design I'm personally working on does NOT use a round fuselage, it is more of a flying wing cabin design for two pilots).

    I thought that this was finally the proper solution, but it didn't feel right... Also attempting this on other problems from this book, it still didn't fit quite right.
  2. jcsd
  3. Nov 7, 2017 #2

    Andrew Mason

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    Hi yavorh. If I understand the problem correctly, you have the moment of inertia of the plane about a vertical axis through the centre of mass of the plane and you want to calculate the moment of inertia about a vertical axis through the main gear. If that is the issue, it should be a simple matter of applying the parallel axis theorem. But why does the plane have two centres of gravity (cgfor and cgaft)? Do they represent the centres of gravity with gear up and gear down?

  4. Nov 7, 2017 #3


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    Typically if you wan't to work out the MOI of something complicated you...

    1) Divide it up into regular shapes for which the MOI is known. (Examples here: https://en.wikipedia.org/wiki/List_of_moments_of_inertia and elsewhere on the web).

    2) The tables will typically give the MOI about the centre of mass or an edge and you may need the MOI about a different point (eg the undercarriage or the CoG). So you "correct" the MOI of each part using the parallel axis theorem to give you the MOI about the required point. https://en.wikipedia.org/wiki/Parallel_axis_theorem

    3) Add up the corrected MOIs to give the total MOI.
    Last edited: Nov 7, 2017
  5. Nov 7, 2017 #4


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    They are usually certified to fly with a CoG within a range of values. The exact position can change depending on passenger distribution, baggage, fuel burn etc.
  6. Nov 7, 2017 #5
    Thanks for the info thus far - I am trying to find the Mass MOI in kg/square meter about the main gear - the idea is to factor it into the calculation of how much force the elevator needs to produce in order to rotate the plane during take off.

    In this case I'm not sure exactly how to calculate each objects MOI... Perhaps it'd be best to take each shape's CoG and calculate its MOI by using the formula for a point-weight (pendulum) rotating about the axis of the main wheel.
  7. Nov 7, 2017 #6

    Andrew Mason

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    In that case, the pivot axis is a line between the points of contact of the wheels with the ground. So you have to work out the moment of inertia about a horizontal axis through the centre of mass of the plane and then apply the parallel axis theorem to find the MOI about the pivot axis.

    To find the moment of inertia of the plane about a horizontal lateral axis through its centre of mass I would treat the plane basically as a non-uniform cylinder. Divide the plane up into short cylindrical sections along the long axis of the plane and find the mass per unit length for each section. From that work out the MOI of each section (each is a thick-walled cylinder) about a horizontal lateral axis through the section's centre of mass. Then apply the parallel axis theorem to find the MOI of the section about the parallel axis through the centre of mass of the plane. Then add them up to find the MOI of the entire plane about the horizontal lateral axis through the centre of mass . Finally, then apply the theorem to determine the MOI of the entire plane through the parallel pivot axis.

    Just a quick note about units for the MOI, the units should be kg m2 not kg/m2.

  8. Nov 7, 2017 #7
    Awesome! I just managed to find some similar info through this one book, but it wasn't as detailed on finding the CG MOI as you explained it. Thank you very much!

    Since I was using this given above example as a base on which to test if the formulas I'm using on my own design are accurate, I'll probably have to treat the shapes as cuboids, since it's the simplest approach:
  9. Nov 11, 2017 #8
    Had posted my results to verify validity, buuuut nevermind, forgot to do parallel axis through CG
  10. Nov 11, 2017 #9
    Wings MOI = ((45wingshell+15ribs)/12)*(2.5^2+.55^2) = 32.7kg/m2 *2 = 66kg m2

    CoM is right ontop of CG, rounding up to 70kgm2 to account for minor differences

    2x Wing Ibeams (6/12)*(.08^2+.005^2)= 0.003213

    Per wing, one of the beams is on CG, the other is .6m back, the ones who are back have a MMOI of 2.52

    Thus, total 5.5kg m2 for a safety margin.

    Battery packs (4 total) =(75/12) *(.15^2 + .2^2) = .39 * 4 = 1.5kg m2 They are ontop of CG

    TOTAL = 77 kg m2

    Cabin MOI = ((+ 50cabin equip)/12)*(3^2+1.2^2) = 448 kgm2

    Pilots – 100kg each, avg dimensions from the side (180cm by 200cm) = 27.5 *2 = 55kgm2 CG seated right over plane Cg, so minimal change – chalking it up to 60kgm2 just in case

    Motor+ inverter = 210kg 100cm by 40cm = 20.3 kgm2. CoM is 90cm from CG so MMOI is 190.4, rounded to 200kgm2 for safety margin.

    Cain+equipment and luggage etc spaced evenly about it = 150kg. Cabin dimensions are 300cm by 150cm = 140.625kgm2. it will be optimized to have CoM over CG, but to account for possible shifts due to luggage spacing and the such, will give it up to .5m of variation, thus 178.125kgm2. Rounding to 200kgm2 for safety margin.

    TOTAL = 460 kg m2

    Tail MOI

    HTail = (5/12)*(1.75^2 + 0.21^2) = 1.019kg kg m2 ; CoM is 5m from CG, thus 126.019, rounded to 130 for SM.

    HTail supports = 4kg, 80x5mm bar, after parallel axis, comes up to 1kgm2

    VTail = (2/12)*(1.5^2*0.6^2) = 1.08 kg m2; CoM is 5m from CG, thus 51.08, there being two of them = 105kgm2 (+SM)

    TOTAL = 236 kgm2

    Main support = Total weight is 30kg. Since it’s an Ibeam:

    Flat (bottom and top) = (10/12)*(0.008^2 +6.125^2) = 31.26 = parallel 58.8 * 2= 120kgm2

    Middle = (10/12)*(0.08^2 +6.125^2) = 31.27 = 60 parallel

    Total = 180 kgm2


    Since the total about CG is 953 and the distance from the CG to the main gear is 1.31244, assuming a max take off weight of 1250kg, then the MMOI about the Main gear MG in reference to the Y axis is 3106.123442
  11. Nov 24, 2017 #10
    I'd call that a conventional configuration with a blended wing-fuselage, not a flying wing.

    Small propeller-driven singles have a short nose chord, and all the big masses are clustered together, so the moment of inertia is low. How fast the elevator can change the pitch angle is not usually a concern in that case. (With large turbine multis, the passengers and payload are spread out along the fuselage, and sometimes the engines are in the rear, so you can have big masses far from the CG.)
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