Wing Tip Skid/Rear Spar Fitting Load

Discussion in 'Aircraft Design / Aerodynamics / New Technology' started by proppastie, Apr 22, 2019.

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  1. Apr 22, 2019 #1

    proppastie

    proppastie

    proppastie

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    Here is a spread sheet.....Never having calculated angular momentum I am hopping this makes sense to someone......If it is reasonable my assumption is that the mass of the glider must cancel. Anyway these are the dimensions of the Aluminum Dragon and the one calculation gives me P as 207 which I assume is in pounds.
     

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  2. Apr 22, 2019 #2

    wsimpso1

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    Something is wrong someplace... F in the example is 150 pounds. b/2 in the example is not given, but it is a length in inches. L in the example is not given but it too is a length in inches. P is the reaction at the tail wheel or skid, in pounds. What they appear to be doing is calculating a moment due to the tip dragging (F*b/2) and then dividing the moment by the length to the tail skid to get lateral force at the tail skid. So, P at the tail skid is P = F*b/(2*L) The balancing reaction at the spar mount is also P.

    My big question is "what is the wingtip load on your airplane?" I do not understand if they are giving you a conservative 150 pounds for all gliders or if you have to find your own number based upon your glider landing mass and wing geometry...

    Billski
     
  3. Apr 22, 2019 #3

    proppastie

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    looking to the left, of the picture from the book:

    L = 191.5 in
    b = 528 in
    b/2 =264 in

    these are the lengths from the Aluminum Dragon.

    I believe 150# is the load specification for all gliders, and I have to calculate the P based on the lengths for my glider.

    I understand the picture, but I do not know if this makes sense were one to calculate "instead" the angular momentum for the given 150# load.

    I hope that clarifies the post.
     
    Last edited: Apr 22, 2019
  4. Apr 22, 2019 #4

    wsimpso1

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    I am having trouble here. Where does angular momentum come into this? During landing and before wingtip touch, there should be zero yaw rotation and thus zero angular momentum. Once the wing tip touches and produces this magic number of 150 pounds of drag at the wing tip, there will be a moment attempting to drive yaw rotation. The rotation will be resisted by the mass moment of inertia of the sailplane. The forces can be computed and resolved for every place on the airframe.

    The authors of the guide you are using appear to have determined that if you use their magic number and the equation presented, and then design the spar mounts to suit the magic number and equation, you will have an adequately sturdy airplane. Without seeing their methodology, I can not know how appropriate their simplified method is to your airframe.

    One thing you can check is this: Get some idea for the percent contribution to total airframe mass and mass moment of inertia from the wings and loaded fuselage of the typical sailplane in their guide. Get the same info for your airplane. If they are similar, you might be OK with the simplified criteria. If your sailplane has mass and mass moment of inertia greatly different from the typical sailplane, the rigorous analysis might be more appropriate...

    Maybe you can provide a pick to the document. I will look around in it and see what presents itself.

    Billski
     
  5. Apr 22, 2019 #5

    wsimpso1

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    Found the criteria... OK, I looked over the document. The page you extracted the math we were discussing cites two methods for determining design loads for the wing mounts, with one being the equation we were discussing and the other considering the angular inertia of the glider. Angular inertia and angular momentum are two very different things...

    Angular inertia and mass moment of inertia are the same thing - it is mass resisting rotation. I would run an estimate of forces at wing mounts and other spots as they are induced by wing tip touch down and drag for several reasons. The biggest is that your airframe masses are quite a bit lower than those assumed in the Criteria. The next biggest is that the distribution of inertia in your airframe may be quite different from those assumed in the criteria.

    I will take this as far as I can remember off the top of my head, and get the rest later. You will need an estimate of the rearward force and the mass moment of inertia of your airplane so you can compute an estimate of the rotational acceleration from the wing tip being grabbed by the ground. From that we can compute the loads that generates at any place in the airframe...

    I would do a piecewise estimate of masses every foot of the span and half foot of the fuselage, then you can calculate MMOI of the whole plane about its CG. If you already have the mass distribution in other units that are pretty small compared to the span, go for it. I would also get some sort of idea what the biggest drag load on your wingtip would be.

    Mass mount of inertia - On a plan view of your airplane, establish the x and y position of every increment of your span and fuselage, and establish what the weight of that chunk of the airplane weighs. We need everything, wing structures, fuselage structures, landing gear, instruments, pilot, engine. A seated human's CG is about their belly button.

    Here is a hint: even experienced degreed ME's can get tangled up in their undershorts trying to do inertia in Brit units. I use SI units and avoid a lot of trouble.
    • Set up a column in Excel for X and Y dimensions in inches and weights in pounds. Put in the positions and weight of everything. Do yourself a favor - start at the wingtip you are putting the braking load on, and work your way down the table by first going in the wing that is hitting, then the fusealge, then the other wing;
    • Set up columns for X and Y in meters and mass in kg. Divide weights in pounds by 2.2 to get kg. Divide distances in inches by 39.37 to get distance in meters;
    • Next two columns are first moments - X*m and Y*m. At the bottom sum the mass, X*m, and Y*m columns;
    • Divide Sum X*m and Sum Y*m by Sum m and those are your CG. I put those on the summation line under the X's and Y's;
    • You need a new set of X and Y columns for how far the X or Y is from the CG.
    • Now on a last column, you put in m*(X^2 + Y^2) for each line, using the m in kg and X and Y in meters from the CG. That is your mass moment of inertia.
    Next step is to compute moment trying to rotate your bird about its CG. That force you think is conservative for rearward force at the wing tip (might be 150 pounds, might be less) times the lateral distance to the CG is the moment. But we gotta put it in SI units. Force in pounds time 4.45 gives force in Newtons. Distance in inches divided by 39.37 is distance in meters. Multiply your Newtons times your meters and you have a moment in N*m.

    F = m*a and T = MMOI*alpha. Solving for alpha = T/MMOI. Units are radians/s^2.

    Next we want to look at the wing root, so we need distance from the tip load to the wing root (call that L). alpha*L is the acceleration at the wing root.

    This is where I have research to do to make sure we will compute the right values. Get started and I will get back to this in a day or two...

    Billski
     
    Last edited: Apr 22, 2019
  6. Apr 22, 2019 #6

    proppastie

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    Well if anyone would understand this....... given your work with transmissions it would be you.......

    I can live with 150# number....(my gust loading on the tail is 243 lb), and my P calculated is 207 lb. My calculated resultant load in in the span direction on the rear fitting is not large.

    The question revolves around any load, any mass, and typical mass distribution of an aircraft, does the mass cancel out in the calculation for angular inertia, and does this formula look reasonable for a typical aircraft.......what might be the proper equation for angular inertia where this approximation makes sense?
     
  7. Apr 23, 2019 #7

    wsimpso1

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    In the purest sense, the glider has mass and mass moment of inertia. Mass resists changes in linear speed while MMOI resists change in rotational speed. A rearward force aligned with the CG would simply accelerate the glider rearward, slowing it down. A rearward force that is not aligned with the CG will both accelerate the glider rearward AND accelerate the glider in rotation.

    The amount of linear acceleration at the CG is the force divided by the the total mass of the glider. The amount of rotational acceleration about the CG is the MMOI divided my the moment of force times arm from force to CG. They both matter, and neither cancels out anything.

    How the authors got the relationship they did, I can only guess at. I do know that In the typical sailplane, where much of the weight and almost all of the MMOI is the wing, the calc shown may be a pretty good representation of the forces at the wing mounts. In a typical power plane where the wing is a much smaller part of the airplane weight and a much smaller part of the MMOI, well, I suspect that the full up analysis might be a better idea for sizing the wing mounts.

    In your airplane, where you will weigh more than the airframe and you will have an engine and fuel too, well, I would go for a first run through and see what the loads look like. If you want to, i will check my approach and tell you how it is done. If not, I will spend that effort on the next pieces of my homebuilt...

    Billski
     
  8. Apr 23, 2019 #8

    proppastie

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    make parts, let me play with equations for a while, I might be able to work it out with what you have mounted so far.
     
  9. Apr 23, 2019 #9

    wsimpso1

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    Once you have the linear and rotational accelerations, the whole airplane is doing that. If you want the forces at the wing root, the linear force is the mass of everything minus the mass of the one wing times the linear acceleration. That force is putting axial loads on the wing root fittings. The torque at the wing root is the inertia of the whole airplane minus the Inertia from the wing that hit ground times the rotational acceleration. Take that torque and divide by the distance between the mounts and you have the forces, one out, one in, on the wing mounts. Make sense?

    Billski
     
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  10. Apr 23, 2019 #10

    proppastie

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    It does make sense.....but I thought I would try something else.....I never was afraid to make a mistake so why should I change now.

    Looking at just the mass moments to compare with the given formula this is what I got......105 lb load P instead of 207.
    edit:....I am thinking ignoring angular acceleration at different value of radius is wrong.


    tipskidcalc..jpg
     
    Last edited: Apr 23, 2019
  11. Apr 23, 2019 #11

    wsimpso1

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    I wish you would just attach the Excel file...That way I can see your equations, understand where you got numbers, Ctrol-C your data for my calcs, etc. I shall lead by example below...

    You appeared to be doing first moments (weight times distance from some reference... which was also unclear) which are useful for computing CG position only. You really need to keep track of X (BL) and y (FS) positions of masses to do rotational stuff.

    Enclosed is my spreadsheet with my interpretation of:
    Your masses are (A1-H10),
    Listing of weights and positions in Brit units (A20-D85),
    Listing of weights and positions in SI units (F20-H85),
    First moments for CG Calcs (J20-K88),
    Mass and CG (E87-K88),
    MMOI calc (M20-Q88),
    Then back to A12-G18 for calculation of accels;
    Then K1-Q14 for forces at the wing root.

    You can go in and fix data in A1-H12 and then adjust weights if you can for the piece wise sections of the wing and fuselage. MMOI will go down if your wings and fuselage tend to have weight more centrally located than the uniform distribution I have. I guessed at the 3 feet between forward and aft mounts, you can fix that too.

    I continue to question the 150 pound force at the tip. My guess is that is good for a standard sized trainer with two people on board. Scaling the 150 pounds for your all up weight might make more sense. On the other hand, that may be chosen for the forces the wing walker can be expected to give...

    Nonetheless, I would not get all freaked out about having to stand 646 pounds, one outward, one inward. I suspect an AN3 bolt or similar sized pin will carry that in single shear with ease, and in double shear, well, it will be stouter to this load than to lift while pulling g's on bumpy ridge...

    Billski
     

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  12. Apr 23, 2019 #12

    proppastie

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    had to massage the W&B data.....used the wing centroid for that arm.

    Edit: delete bad spread sheet
     
    Last edited: Apr 24, 2019
  13. Apr 23, 2019 #13

    wsimpso1

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    You know, the more I look at the page from the Criteria, the more I think they are just finding out how much lateral force will be applied to tailwheel if it is stuck in a rut while a wing walker grabs the wing tip and applies fore and aft force to scoop the tailwheel around for the next tow.

    But what I did in my spreadsheet was figure out how much force is in the wing mountings to do that. Now I will get into your spreadsheet and see what you were doing...

    Billski
     
  14. Apr 24, 2019 #14

    proppastie

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    revised spread sheet

    rear fitting loads agree
     

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  15. Apr 26, 2019 #15

    proppastie

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    cs-22 tip skid.jpg

    Found this......Do we assume the Europeans know how to design gliders?....Do we have a FOS of 2 for Glass? With a FOS of 2 the number is 242 lb. For my ultra lite glider I like the smaller number of 121 lb.....Airplane design is fun.
     

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    Last edited: Apr 26, 2019
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  16. Apr 27, 2019 #16

    wsimpso1

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    I agree that the Europeons probably understand gliders and fiberglass, and their rules may make more sense for your airplane.

    Minimum FOS: Metal structures are usually 1.5, composite structures are usually 2.0, bolts and the joints they go into are typically more like 4.0, but I do not recall where I got that from. So the FOS you have to use depends on what pieces you are working on. Be careful to not stack FOS. If the landing gear reactions are 121 and it is a metal structure it is hung on, the metal structure has to be designed to stand an ultimate load of 182 pounds at the wheel, but the bolted joints should be designed to stand an ultimate load of about 484 pounds at the skid for the same yawing resistance portrayed in the drawing. DO NOT take 182 pounds at the landing gear and multiply that by 4, as you will then be designing to 6.0 and that can lead to excess of weight.

    Billski
     
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