# Lift Strut and Cabane Strut loads

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#### Dana

##### Super Moderator
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Different curves for different Reynolds numbers, yes. Remember Reynolds number changes not only with size but also velocity (airspeed).

$$Re = \frac{\rho V L}{\mu}$$ where ρ is air density, V is airspeed, L is length (wing chord in this case), and μ is the dynamic viscosity.

#### wsimpso1

##### Super Moderator
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Log Member
+1 on the comments above. As BBerson points process might be helpful. TOWS, in typical engineering textbook fashion, just jumps right into "this is how to calculate this", sometimes with only a modest basis for the work. Hmm.

Data in appendix IV of TOWS is based upon wind tunnel work. By the time the data cited in TOWS was taken, a whole bunch of stuff had become known and was taken into account in doing the work. Knowledge included:
• Reynolds number matters;
• Turbulence in the air matters;
• Moment changes very little if you capture forces at 0.25c;
• Boundary layer of the model matters;
• Wing tip effects matter;
• Boundary layer on the walls matter;
• The tunnel is restricted by the presence of the model;
• Forces vary linearly with projected wing area S and with dynamic pressure q. Moment varies with S, q, and chord;
So, they set out to control for these issues in order to define the characteristics of the foil, not the characteristics of the tunnel, the model, the air temp, density, velocity, and turbulence of the air. So what did they do about them?
• Reynolds number - is a dimensionless ratio of dynamic to viscous forces in the air. Dana cited it for foils above. The researchers found that most subsonic airplanes of the day used Re between about 1M and 6M, and that behaviour was fairly smooth between these numbers, so those were the Re they tested and reported;
• Turbulence - you get the wind in a wind tunnel by running a big propellor, and that can cause all sorts of burbles in the air that can poison the data. So much effort was spent to smooth flows and reduce turbulence;
• Moment changes very little if you capture forces at 0.25c - so they attached the foil to the precise balances at 0.25c;
• Boundary layer of the model matters - Very smooth models were generated with very low chordwise waviness so that turbuence in the boundary layer would be due to the airfoil shape and not the quality of the model. Then to show effects of surfaces less than perfect, they had runs with artificially roughened surfaces;
• Wing tip effects matter - in a real airplane, the wings end, and air rushes around the tips under the effects of lowered pressure on one side and raised pressure on the other side. Also, we know that the lift vs alpha curve is steepest at high aspect ratio, and flatter as aspect ratio gets smaller. To beat this, the model is constant chord and runs from one wall to the other wall;
• Boundary layer on the walls matter - a layer of air sticks to any solid in the tunnel, with air moving zero speed at the surface and speeding up towards free stream speed some distance away. This is boundary layer, and it can become quite thick. So they apply suction along the walls just upstream of the model. Boundary layer must form again, and it is just getting started at the model, minimizing this effect;
• The tunnel is restricted by the presence of the model - the model reduces the tunnel cross section, slightly choking the tunnel at this point. There are rules of thumb for how much reduction is allowed, computational approaches to correcting data for the choke, etc.
Then they deliberately run from well into stall at negative alpha through stall at positive alpha, at a variety of Re and model surface roughness, with and without split flaps. Lift, drag, and pitching moment are all carefully measured during these runs. Most runs are repeated multiple times, data reviewed for anomolies, and averaged. Then Cl, Cd, and Cm is solved for out of the well known equations of the form F = q*S*C. For lift, C is Cl, for drag, C is Cd, and for moment C is c*Cm. Data is reviewed and plotted.

Standard roughness was selected at a way overdone level. It is about the texture of anti-skid surfaces you will find on a boat deck or wing walk or maybe in your shower. This is way over the top compared to any bug accretion on leading edges, irregularities from rain drops, and really only represents the wing walk surfaces. But it does give you an upper bound on how bad the perfect shape but rough surface can be.

There is no attempt at showing you the effects of rivets, seams, and other irregularities. In so-called turbulent flow airfoils, these effects are modest, mostly adding a couple thousandths to Cd. In airfoils designed as laminar flow foils, the flow over the forward portion of the foils stays laminar at low alphas, and the drag has a low region through there called the laminar bucket. In these foils, waviness, bugs, water drops, seams, paint stripes, trip the boundary layer to turbulent. In your mind's eye, just pull the parabola looking curve from higher and lower Cl across the laminar bucket to get your new Cd...

What about the other appendices in TOWS?

Appendix I is theoretical velocity ratio of air flowing over airfoil thickness distributions known to be useful in airplanes at zero Cl and at some non-zero Cl.

Appendix II is the theoretical pressure ratio of air flowing over airfoil camber curves. We can assemble thickness distributions and camber curves to make a huge variety of airfoils. Likewise we can assemble the velocity profile of any of those foils from this information. Through Bernoulli's well known equation, velocity over a spot predicts pressure at that spot, so you can do detail resolution of forces and thus predict loads in structures.

Appendix III is the unitless coordinates of the airfoils discussed in the book.

Billski

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#### wsimpso1

##### Super Moderator
Staff member
Log Member
OK, a graph. Even better than a table!

Why are there multiple lines for the same airfoil, most noticeable in the drag section? At the bottom it appears to say "R".
Is that Reynolds number and would that then mean the curves change based on the size of the airfoil? For example it may be on an airliner, a GA plane or an RC plane. The same airfoil, but very different in size.
If you like graphs, you will LOVE TOWS. More than half of the book is appendices and most are plotted graphs.

You do get to learn by looking at the plots. R is the authors' abbreviation for Reynolds number. Most of us use Re for this. I mentioned it above. Behaviour of stuff with fluids flowing over it varies with Re, with surface roughness, and with flap deployment, all of which are shown on these plots.

Billski

#### Fenix

##### Well-Known Member
HBA Supporter
+1 on the comments above. As BBerson points process might be helpful. TOWS, in typical engineering textbook fashion, just jumps right into "this is how to calculate this", sometimes with only a modest basis for the work. Hmm.

Data in appendix IV of TOWS is based upon wind tunnel work. By the time the data cited in TOWS was taken, a whole bunch of stuff had become known and was taken into account in doing the work. Knowledge included:
• Reynolds number matters;
• Turbulence in the air matters;
• Moment changes very little if you capture forces at 0.25c;
• Boundary layer of the model matters;
• Wing tip effects matter;
• Boundary layer on the walls matter;
• The tunnel is restricted by the presence of the model;
• Forces vary linearly with projected wing area S and with dynamic pressure q. Moment varies with S, q, and chord;
So, they set out to control for these issues in order to define the characteristics of the foil, not the characteristics of the tunnel, the model, the air temp, density, velocity, and turbulence of the air. So what did they do about them?
• Reynolds number - is a dimensionless ratio of dynamic to viscous forces in the air. Dana cited it for foils above. The researchers found that most subsonic airplanes of the day used Re between about 1M and 6M, and that behaviour was fairly smooth between these numbers, so those were the Re they tested and reported;
• Turbulence - you get the wind in a wind tunnel by running a big propellor, and that can cause all sorts of burbles in the air that can poison the data. So much effort was spent to smooth flows and reduce turbulence;
• Moment changes very little if you capture forces at 0.25c - so they attached the foil to the precise balances at 0.25c;
• Boundary layer of the model matters - Very smooth models were generated with very low chordwise waviness so that turbuence in the boundary layer would be due to the airfoil shape and not the quality of the model. Then to show effects of surfaces less than perfect, they had runs with artificially roughened surfaces;
• Wing tip effects matter - in a real airplane, the wings end, and air rushes around the tips under the effects of lowered pressure on one side and raised pressure on the other side. Also, we know that the lift vs alpha curve is steepest at high aspect ratio, and flatter as aspect ratio gets smaller. To beat this, the model is constant chord and runs from one wall to the other wall;
• Boundary layer on the walls matter - a layer of air sticks to any solid in the tunnel, with air moving zero speed at the surface and speeding up towards free stream speed some distance away. This is boundary layer, and it can become quite thick. So they apply suction along the walls just upstream of the model. Boundary layer must form again, and it is just getting started at the model, minimizing this effect;
• The tunnel is restricted by the presence of the model - the model reduces the tunnel cross section, slightly choking the tunnel at this point. There are rules of thumb for how much reduction is allowed, computational approaches to correcting data for the choke, etc.
Then they deliberately run from well into stall at negative alpha through stall at positive alpha, at a variety of Re and model surface roughness, with and without split flaps. Lift, drag, and pitching moment are all carefully measured during these runs. Most runs are repeated multiple times, data reviewed for anomolies, and averaged. Then Cl, Cd, and Cm is solved for out of the well known equations of the form F = q*S*C. For lift, C is Cl, for drag, C is Cd, and for moment C is c*Cm. Data is reviewed and plotted.

What about the other appendices in TOWS?

Appendix I is theoretical velocity ratio of air flowing over airfoil thickness distributions known to be useful in airplanes at zero Cl and at some non-zero Cl.

Appendix II is the theoretical pressure ratio of air flowing over airfoil camber curves. We can assemble thickness distributions and camber curves to make a huge variety of airfoils. Likewise we can assemble the velocity profile of any of those foils from this information. Through Bernoulli's well known equation, velocity over a spot predicts pressure at that spot, so you can do detail resolution of forces and thus predict loads in structures.

Appendix III is the unitless coordinates of the airfoils discussed in the book.

Billski
Wow, I was already thinking about how thankful we should be for the volumes of data contained in the graphs of so many airfoils.

And that was before I read Billski's account of all the difficulties.

I'm sure glad they shared their results!

HBA Supporter

#### Fenix

##### Well-Known Member
HBA Supporter
We paid for it.

BJC
Yes I know, but we don't always get what we pay for.

BJC

#### Fenix

##### Well-Known Member
HBA Supporter
Thanks to all of the above help and guidance I now know how to (at least in simple cases) determine the load distribution between the fwd and aft spar and how to calculate the lift applied at the strut fitting as a result of spanwise lift distribution.

This lift at the strut fitting will create tension in the lift struts (I'll just talk positive G's herein) that is greater than the lifting force applied at this point. This is trig and truss theory I've been studying on you tube. This tension in the lift strut will be reacted by compression (again calculated with truss theory) in the spar section from the root to the lift strut attach point. Where the lift struts attaches to the fuselage it will create tension across the fuselage which is reacted by the opposing wing strut fittings so of course the fuselage must have members capable of transferring these loads across the fuselage at these points (assume parallel lift struts for now). The connection of the lift strut at the fuselage will also create a compression force upward between the strut attach to the fuse and the wing root fitting. These forces will be carried by the side of the fuselage and the cabane struts (in the case of a parasol). Let's just look at the cabane strut here. This cabane will be subject to the above described compressive forces. But if there is any of the lift distributed to the root fitting (because the lift strut attaches far enough outboard that it does not carry all of the lift acting on its spar) then this cabane will also be subjected to some tension as this lift is transferred down to the fuselage and its associated mass.

Question 1:
Is the net force on this cabane the compressive forces minus the tension forces? It seems it must be but then things are not always as they intuitively appear in physics.....

Now let's change the above parallel lift struts to a V type strut which attaches to the fuse below the fwd spar but the cabanes remain vertical in a parallel "4 posted" type arrangement.

The aft lift strut which is now angled fwd will (among other things) create a compressive force between the front spar and rear spar where the struts attach to the spars. This will require adequate compression members (in addition to any that already exist there in the drag/anti drag truss) to react this compression.

Question 2:
Will this new "triangle" (created by the V struts and the compression members just described between the fwd and aft spars) also create a compression load in the fwd lift strut? If so I assume the net force in the fwd lift strut will be the same as in the answer to Question 1 above?

Also now both the fwd and rear lift struts are going to apply the combined tension across the fuselage to the opposite strut fitting all at this one strut attach point (as opposed to the two that existed in the parallel strut assy). This will make a greater tension load across the fuselage at this point than existed in the parallel strut arrangement. Correct? If so, similarly:

Question 3:
Wont the compression load transmitted up to the wing root also be greater here putting more compressive forces on the fwd cabane strut and virtually eliminating compressive forces in the rear cabane strut?

#### Fenix

##### Well-Known Member
HBA Supporter
Also in considering the angle of the strut attachment to the spar the matter of dihedral comes up. So in thinking about this is it not that case that:

A plane that weighs 2,000 LBS in level flight is creating 2000 LBS of vertical force (lift) but if it has dihedral not all of the force which acts 90 degrees to the wing panel is vertical. So it seems to get 1,000 lbs (one panel or semi span) of vertical force with say, 3 degrees of dihedral, the total force of the wing would need to be the hypotenuse of a rt triangle with a 3 degree angle and an adjacent side of 1,000 (the vertical component). So the relevant function is the COS of 3 degrees or .9986 making the total force of the panel 1,001.4 LBS of which 1,000 acts vertically.

Is this how it works? Or maybe not????

If so then the difference between vertical force (lift) and total wing force in any reasonable dihedral would be minimal - and may usually be ignored??
But is it technically correct that in the above situation where each semi span needs 1,000 LBS of lift the bending loads of the spar and the calculated loads applied to the struts and root attach points would be based off of 1,00.4 and not 1,000?

#### wsimpso1

##### Super Moderator
Staff member
Log Member
The difference in lift required between zero dihedral and in modest dihedral typically is tiny. If it were large, all airplanes would be high wings.

#### BobDaly

##### Well-Known Member
This is an excel spreadsheet I did using Hiscocks' "Design of Light Aircraft". One can toggle between Minimax and Himax parameters. With it one can examine symmetric positive loading and with some assumptions maybe negative and asymmetric loads. Minimax wing load spreadsheet

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#### Fenix

##### Well-Known Member
HBA Supporter
This is an excel spreadsheet I did using Hiscocks' "Design of Light Aircraft". One can toggle between Minimax and Himax parameters. With it one can examine symmetric positive loading and with some assumptions maybe negative and asymmetric loads. Minimax wing load spreadsheet
That is a lot of the info I was looking for. Much of it is advanced beyond my current level of understanding but more of it will become clear as I work my way through it. What did you mean by "toggling between the MMax and HiMax parameters"?

#### BobDaly

##### Well-Known Member
There is a tool called "scenario manager" in Excell that allows for substituting several cell contents with alternate values. The spread sheet has sets of values for both airplanes.

#### wsimpso1

##### Super Moderator
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Log Member
Hmm. I have looked at the spreadsheet above. Substantial piece of work, but it includes a number of shortcuts which will generally overestimate some loads and underestimate others - I suspect that it is conservative (safe) but may drive unneeded weight. See my comments:

#### Fenix

##### Well-Known Member
HBA Supporter
Thanks to all of the above help and guidance I now know how to (at least in simple cases) determine the load distribution between the fwd and aft spar and how to calculate the lift applied at the strut fitting as a result of spanwise lift distribution.

This lift at the strut fitting will create tension in the lift struts (I'll just talk positive G's herein) that is greater than the lifting force applied at this point. This is trig and truss theory I've been studying on you tube. This tension in the lift strut will be reacted by compression (again calculated with truss theory) in the spar section from the root to the lift strut attach point. Where the lift struts attaches to the fuselage it will create tension across the fuselage which is reacted by the opposing wing strut fittings so of course the fuselage must have members capable of transferring these loads across the fuselage at these points (assume parallel lift struts for now). The connection of the lift strut at the fuselage will also create a compression force upward between the strut attach to the fuse and the wing root fitting. These forces will be carried by the side of the fuselage and the cabane struts (in the case of a parasol). Let's just look at the cabane strut here. This cabane will be subject to the above described compressive forces. But if there is any of the lift distributed to the root fitting (because the lift strut attaches far enough outboard that it does not carry all of the lift acting on its spar) then this cabane will also be subjected to some tension as this lift is transferred down to the fuselage and its associated mass.

Question 1:
Is the net force on this cabane the compressive forces minus the tension forces? It seems it must be but then things are not always as they intuitively appear in physics.....

Now let's change the above parallel lift struts to a V type strut which attaches to the fuse below the fwd spar but the cabanes remain vertical in a parallel "4 posted" type arrangement.

The aft lift strut which is now angled fwd will (among other things) create a compressive force between the front spar and rear spar where the struts attach to the spars. This will require adequate compression members (in addition to any that already exist there in the drag/anti drag truss) to react this compression.

Question 2:
Will this new "triangle" (created by the V struts and the compression members just described between the fwd and aft spars) also create a compression load in the fwd lift strut? If so I assume the net force in the fwd lift strut will be the same as in the answer to Question 1 above?

Also now both the fwd and rear lift struts are going to apply the combined tension across the fuselage to the opposite strut fitting all at this one strut attach point (as opposed to the two that existed in the parallel strut assy). This will make a greater tension load across the fuselage at this point than existed in the parallel strut arrangement. Correct? If so, similarly:

Question 3:
Wont the compression load transmitted up to the wing root also be greater here putting more compressive forces on the fwd cabane strut and virtually eliminating compressive forces in the rear cabane strut?
So I've been working through "truss calculations" of actually quantifying these values on the various components.
Now I see I made an "error in logic" in the above referenced description.
The above statement: "The connection of the lift strut at the fuselage will also create a compression force upward between the strut attach to the fuse and the wing root fitting" is not correct. This upward acting force on the fuselage bracket where the lift strut attaches does NOT create a compression load between this bracket and the wing spar root to cabane bracket above it. Instead, this upward force on the bracket is reacted by GRAVITY. Duhhh.... This is (in part) where the fuselage is lifted from. This must have been "cringe worthy" to the readers who recognized this error.

I may still have this wrong but this describes the new current state of my understanding on this matter. The attached spreadsheet (with multiple tabs at the bottom) shows the progression of this understanding and what I believe to be the forces on the cabanes and lift struts, which was the point of this inquiry from the start. So I guess, in simple cases, I have arrived at that goal - unless I still have some things wrong.

So to answer my Questions 1,2 and 3 above:

Question 1: This is a "nonsense question" because, as described above, the referenced compression forces do not exist. The cabanes are (in this configuration and under positive G's) only subject to the tension forces created by the portion of lift which occurs at the wing root.

Question 2: It will not be the same as Question 1 because Question 1 was in error. I still think the net forces in the forward lift strut of a V configuration will be as posed in Question 2 and I expect to find a definitive answer to this question when I next analyze the V strut configuration.

Question 3: Again this is a "nonsense question" because the referenced compressive forces do not exist. I still think that the tension load across the fuselage will be greater at the front strut attach because the rear strut attach does not exist in the V strut configuration so all the loads will have to go through the "front tension strap" across the fuselage. When I work through the V Strut "truss math" I expect to definitively find these answers. I have just begun the V strut analysis and can already see there are a LOT more load paths. The V Strut analysis is going to be fun! - and I expect informative...

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#### Fenix

##### Well-Known Member
HBA Supporter
Update: Well in working through the V Strut design I can already see that Question 2 is also a nonsense question because the "resulting compression" loads in the front lift strut I referred to also do not exist. I think I "envisioned" these forces because of a "simple analogy" given to me by the designer of an aircraft I was building from plans about 30 years ago. (I've been bugging people with questions for quite a while now I guess....) His analogy was useful in the context of the question I asked, but was not really "how things work". I guess I carried this "misunderstanding" for a long time - until today.

So it appears the study of the V Strut has already been informative.

I'll be back when I have a spreadsheet of the V strut forces.........

#### wsimpso1

##### Super Moderator
Staff member
Log Member
This lift at the strut fitting will create tension in the lift struts (I'll just talk positive G's herein) that is greater than the lifting force applied at this point. This is trig and truss theory I've been studying on you tube. This tension in the lift strut will be reacted by compression (again calculated with truss theory) in the spar section from the root to the lift strut attach point. Where the lift struts attaches to the fuselage it will create tension across the fuselage which is reacted by the opposing wing strut fittings so of course the fuselage must have members capable of transferring these loads across the fuselage at these points (assume parallel lift struts for now). The connection of the lift strut at the fuselage will also create a compression force upward between the strut attach to the fuse and the wing root fitting.
Stop there. The strut is in tension aligned with the strut. The fitting the strut is bolted to must generate forces inward and down the strut, with TENSION and downward force between the mounts.

These forces will be carried by the side of the fuselage and the cabane struts (in the case of a parasol). Let's just look at the cabane strut here. This cabane will be subject to the above described compressive forces. But if there is any of the lift distributed to the root fitting (because the lift strut attaches far enough outboard that it does not carry all of the lift acting on its spar) then this cabane will also be subjected to some tension as this lift is transferred down to the fuselage and its associated mass.
Maybe. There is one place along the spar where the strut could be attached that produces zero force in the vertical direction on the cabane. That spot is where the net moment calculated around the strut mount on the wing from lift is zero. Exactly where you calculate this spot to be depends upon if you use elliptical, Shrenk's Approximation, or other lift distribution. If you hang the lift strut further out, the root end tries to rise, further in and the root end tries to fall, with commensurate vertical forces from the cabane to restrain it.

Question 1: Is the net force on this cabane the compressive forces minus the tension forces? It seems it must be but then things are not always as they intuitively appear in physics.....
Uh, see above, do the sum of forces. The two spars will be pushing toward each other with the cabane in the middle. Differences between the two sides should be small, but how you get the forces from one wing to the other will be where you need to worry over magnitudes, as these compression forces are on the order of twice the lift being generated. Thinking pulling g's up near Vne. Up and down forces will be less with sense and magnitude depending upon how far out the lift struts are attached.

Now let's change the above parallel lift struts to a V type strut which attaches to the fuse below the fwd spar but the cabanes remain vertical in a parallel "4 posted" type arrangement.

The aft lift strut which is now angled fwd will (among other things) create a compressive force between the front spar and rear spar where the struts attach to the spars. This will require adequate compression members (in addition to any that already exist there in the drag/anti drag truss) to react this compression.
Yes.

Question 2: Will this new "triangle" (created by the V struts and the compression members just described between the fwd and aft spars) also create a compression load in the fwd lift strut? If so I assume the net force in the fwd lift strut will be the same as in the answer to Question 1 above?
Nope, as long as you are talking parasol or high wing and positive g, the forward strut is in BIG tension. When the lift strut fuselage mount is forward or aft of the spar its other end hangs on, the loads go up a bit more.

Also now both the fwd and rear lift struts are going to apply the combined tension across the fuselage to the opposite strut fitting all at this one strut attach point (as opposed to the two that existed in the parallel strut assy). This will make a greater tension load across the fuselage at this point than existed in the parallel strut arrangement. Correct? If so, similarly:

Nope. The horizontal load across the fuselage between lift strut mounts is unchanged by shifting the mount fore or aft. There is also a vertical load and a longitudenal load. The strut is a truss element akin to a wire when in tension, and the force in it is aligned with its long axis. The force in each of the three axes from it turn out to be based upon the relative direction cosines of the angle. The strut is mostly faced toward centerline from out on the wing, is somewhat faced down, and is a little faced forward or aft. The forces in those three directions sums through vector addition to equal the force in the strut. Fx^2 + Fy^2 + Fz^2 = Ftotal^2 and so the directions go as well...

Question 3:
Wont the compression load transmitted up to the wing root also be greater here putting more compressive forces on the fwd cabane strut and virtually eliminating compressive forces in the rear cabane strut?

#### Fenix

##### Well-Known Member
HBA Supporter
Billksi, you said: The forces in those three directions sums through vector addition to equal the force in the strut. Fx^2 + Fy^2 + Fz^2 = Ftotal^2

I was not aware of this "formula" but it makes sense. It is, of course, a great tool for checking your conclusions. And so far my new conclusions are consistent with this "test".

Yes, as you've explained above, I had several things wrong with my understanding when I began this.

#### wsimpso1

##### Super Moderator
Staff member
Log Member
Also in considering the angle of the strut attachment to the spar the matter of dihedral comes up. So in thinking about this is it not that case that:

A plane that weighs 2,000 LBS in level flight is creating 2000 LBS of vertical force (lift) but if it has dihedral not all of the force which acts 90 degrees to the wing panel is vertical. So it seems to get 1,000 lbs (one panel or semi span) of vertical force with say, 3 degrees of dihedral, the total force of the wing would need to be the hypotenuse of a rt triangle with a 3 degree angle and an adjacent side of 1,000 (the vertical component). So the relevant function is the COS of 3 degrees or .9986 making the total force of the panel 1,001.4 LBS of which 1,000 acts vertically.

Is this how it works? Or maybe not????

If so then the difference between vertical force (lift) and total wing force in any reasonable dihedral would be minimal - and may usually be ignored??
But is it technically correct that in the above situation where each semi span needs 1,000 LBS of lift the bending loads of the spar and the calculated loads applied to the struts and root attach points would be based off of 1,00.4 and not 1,000?
You are doing the vector mechanics correctly. This just vector mechanics, taught in Statics. As you can see the cosine of small angles is vanishingly close to unity, and so a lot of folks get away with neglecting dihedral just fine. In power airplanes the error in wing deflection is similarly small. In a competition sailplane pulling g's and big angles on the wing tips, yeah, your math would have to get more complicated to compute actual lift carried. In both extremes though, the straight wing model for lift distribution, accumulating shear, bending moment, and torsion in the wing proceeds pretty accurately.

One thing to know about the current discussion is that low wing airplanes typically use only 3 to 5 degrees dihedral, and high wing or parasol birds typically need zero. The Vari-Eze, and a number of aerobatic biplanes (upper wing) have anhedral, as does the Harrier, Tornado II, Galaxy, Globemaster.

Another point that has much more to do with actual lift generated is that the tail plane is generally pulling down to balance the airplane in the pitch axis. Since net vertical forces sum to zero, the wing has to carry both the weight of the entire airplane and the negative lift from the tail.

Last point for this post, a lot of programs take the weight of the wing off of the lift that the wing has to carry. Not really correct all the way through. Let's look:
• The wing still has to lift all the weight of the bird plus the downforce of the tail;
• When you are calculating shear and bending moment in the wing, the wing must carry the forces coming in from the fuselage, so you first accumulate shear with the weight of the wings that are outboard of the fuselage is what is subtracted from the forces at the root. The weight of the wing parts internal to the fuselage stay put;
• Then after you have the distributed shear from the weight of the wings times load factor added back into the shear in the beams. From that you can calculate bending moment, angular deflection, and vertical deflection;
• Similar things are done with weight of things that may be hung on the wing - fuel, engines, baggage, external stores, etc. You have to lift it with the wings, but it lowers the bending moment to put it on the wing and thus reduces how much structural weight must be in the wing.
Billski

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