# Lift Strut and Cabane Strut loads

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#### wsimpso1

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So it seems that I must concern myself not only with distributed loads on the fwd and aft spars at high AOA, but also all the way down to the AOA that produces Max G at Vd ... So the question this logic would create is how to determine what AOA is going to create Max G at Vd. I suspect this is a complicated answer that involves things like Coefficient of lift and can be better found after TOWS is read and understood. If there is a "rule of thumb"
All true. At Vd and limit g loads, you have worst case for wing spars in shear and bending moment. Va is a close second, but the wing is generally sturdier against bending here. Va is the place where loads forward from lift are maximum and is the big stressor on drag and anti-drag bracing of a fabric covered wing. Put the air moving horizontally, the wing pitched up 18 degrees, and lift is vertical, lift has a significant component towards the front of the wing, which is trying to wrack the wing tips forward. This is why we have some form of diagonal bracing between the spars of fabric covered wings, usually drag and anti-drag wires to hold the wing square. When you go to structural skins, they carry these loads. Anyway, we have long established empirical guidance on sizing of these braces.

This design already has a V-strut and I don’t see any reason to complicate things by changing that to a parallel strut so, to calculate the cabanes needed I need to be able to work with the V-strut arrangement ... I’m not sure the tension and compression capacity of these members is equal to the simple sum of their parts, and if tables exist for the tension and compression capacity of rather thin plywood.
The Design References sticky includes ANC-18 for wooden structure design. Someone has posted links for it. Not my area, so not at my fingertips.

One the topic of strut loads, this is all truss theory. There are a number of good videos online about this and can get you started on how to resolve vertical loads, and then how the loads further increase when the strut is angled.

Please refer to the attached diagram which is supposed to represent a high wing strut braced aircraft ... If this is the case then the I would assume that “rectangular lift distribution” would be like that shown in “A” and does not really exist because there is always (except in a few oddball designs like flying donuts and such) less lift at the tips due to spanwise flow, that being which results in wingtip vortices. This being the case “rectangular lift distribution” is not what is found on a “Hershey bar” wing and really is only a theoretical idea that does not exist on any wing design or planform.
True, but it does allow an easy calculation of an upper bound on wing bending moment curve. Useful for those just starting to calculate such things. Reality is different.

Then you have “linear lift distribution” which is shown in D which also probably does not happen in the real world.
I am accustomed to linear reflecting the plan form of the wing. It too shows too much lift in the outer portion of the span, and demands not enough of the inner portion. Not real, drives excess structure and may underestimate stall speed.

Now ellipses are not all the same (as I recall it has to do with the spacing of the foci, among other factors) so you can have more than one lift distribution that would still be called elliptical – say by comparing B to C.
All ellipses are similar. The line is vertical at the tip and horizontal at the centerline. When you hold span constant (you are flying an airplane with wings not changeable in flight) and you unload the plane, the lift distribution curve scales down , pull back and load up the plane, the lift distribution scales up. All still elliptical.

As to our typically using the upper half of an ellipse instead of a whole ellipse, this is just convenience. If you use the whole ellipse, you still get the same shape, just scaled differently. And the quarter- or half-ellipse is more intuitive to calculate and synthesize from.

Billski

#### wsimpso1

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Your input was very helpful Dana. The above referenced quote is "making some lights come on" but I need to think a bit more about it to get a full grasp.

By the "quarter chord" do you mean 25% of the wing chord?
So I get that this is a "tool" to simplify managing the chordwise lift distribution and a gross oversimplification is good at this point. I can refine it once I grasp the basic concept. One thing that confuses me is if the spar actually WERE at the quarter chord and all of the lift forces were seen to act there then the front spar would be loaded with the total lift created by the wing. Then if there were a pitching moment (which would exist if the center of lift were not actually at the quarter chord, which you indicated it probably would not be, so there would be attributed a pitching moment, that this moment would then put some load on the aft spars making the total forces on the spars greater than the total force generated by the wing. I mean if the fwd spar was assumed to carry all of the lift force and the rear spar also had a force on it the sum of the spar loads would not be equal to the total wing force generated. So I'm missing something in my understanding.

Finally, in the quarter chord plus moment method I assume the calculated moment must change with AOA changes to properly reflect the change in lift distribution?
There is a lot of thought training and then practice doing the work in passing classes on calculus, chemistry, physics, linear algebra, statics, dynamics, fluid mechanics, aerodynamics, vibration, mechanics of solids, and mechanical design. They are all intended to produce individuals capable of first doing the calcs and then being able to synthesize new designs over a wide range of possible topics. Drinking from a fire hose is how much of engineering school seems to many students with this stuff spread over four years. It is a big thing, and you are trying to take a corner of it in one big gulp. Please be realistic in your expectations on yourself and on those of us trying to help you and other interested folks following along.

Net forces on unaccelerated flight are zero. In the vertical direction, we have lift and weight, equaling zero. In the horizontal axis we have thrust and drag, summing to zero. This continues on down into component sets too. Sum of forces in each direction must zero out or the thing you are looking at is accelerating off in some direction. No accelerating away in that direction and you must have forces in that direction summed to zero. You have six possible directions that all must normally null out and they apply to pieces as well as the whole airplane:
• Three are forces in linear directions mutually orthogonal - perpendicular to each other. Easy way is fore-aft, up-down, and port-starboard, and coincides with FS, WL, and BL axes of the airplane;
• Three are rotation and moments along those three axes of the airplane - the familiar Pitch, Yaw, and Roll.
The wing making lift is usually seen as airflow over the wing. Three of the six forces are significant here. We get lift, drag and pitching moment. To keep the wing from blowing away aft (from drag), flying out of sight vertically (from lift), and tumbling nose down (from pitching moment), we must have balancing forces. These forces originate in q (rho/2*v^2), S (wing area), c (chord), Cd, Cl, and Cm.

So to react these forces and make the wing stay put while the air blows over our wing, we anchor the main spar and the drag spar. If the main spar is at .25c, we can get to the loads pretty easily. Let's tackle just the lift and pitching moment: Lift and pitching moment are reacted by the anchors attached to the main spar. This includes the mounts at the root and the struts. Total vertical forces and moments calculated about any datum point will sum to being equal and opposite to the lift and pitching moment of the wing. Calculating all of these forces is taught within a semester long course call Statics, and it produces a lot of puzzled students for a while.

Let's break it down so we can understand it all better. If we put the forward spar and main strut right underneath each other at 0.25c, we will get the lift computed for the wing reacted by the strut and the root fitting on that spar. Got that? Some lift at the root, some at the strut. How much at the strut? Well, we have to figure out how much bending moment the lift produces about the root fittings. Bending moment and pitching moment are not related. Bending moment - imagine the lift along the span is broken into a bunch of local lift. Each local lift times the distance from it to the line of the root mount is the moment for that little local lift, do all of them, sum them up and you have the bending moment about root mounts. Divide that moment by the distance between root mount and strut mount and that is the vertical reaction at the strut. The root then carries the difference between total lift of that wing and the vertical reaction at the strut.

Now let's look at the drag spar and its strut. For the moment, let's make the strut directly below the drag spar. We have pitching moment from aerodynamics, and it produces torsion in the wing that must be reacted off. Fabric covered wings are really soft as torsion boxes. The root fitting reacts some of the torque, and the strut reacts the rest of the torque. The total torque is the wing pitching moment. How it is divided is related to the stiffness of the whole wing. A reasonable guess is that all of the pitching moment from outboard of the strut and half of the pitching moment from the region between root and strut are applied to the strut, while the balance finds its way to the drag spar root mount. Divide the lift by the longitudinal distance between main and drag spars, and you get the vertical force at the mount.

There is also drag and forward component of lift which can be significant at high AOA. Drag and forward component of lift (anti-drag) is distributed along the span, which requires forward or aft forces at the root fittings. Since drag and anti-drag is centered out on the wing, there is a moment about a vertical axis that is reacted by an lateral forces at the main and drag spar mounts.

Get your head wrapped around all that, and then we can shift the main spar away from 0.25c and start sharing forces between the main and drag spars.

Billski

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#### Protech Racing

##### Well-Known Member
This plane swings the wings from the rear strut anchor, the front strut is anchored at the rear strut anchor . What is the change in loading from a front strut that is in line with the loads? Thanks,MM

#### Dana

##### Super Moderator
Staff member
This plane swings the wings from the rear strut anchor, the front strut is anchored at the rear strut anchor . What is the change in loading from a front strut that is in line with the loads? Thanks,MM
Tension in the front strut will pull the wingtips aft, adding to drag loads.

#### Protech Racing

##### Well-Known Member
Tension in the front strut will pull the wingtips aft, adding to drag loads.
How do you tell how much? My lift loads are 1000 per wing , with V strut mounted the same way . The rear strut is 24 in behind the front strut .

#### Dana

##### Super Moderator
Staff member
Analyze it like any other simple truss. But you gotta know the geometry and lift distribution...

##### Well-Known Member
I think this might be in the resources list, but I would suggest Richard Hiscocks' book Design of Light Aircraft. It is primarily focused on explaining how to determine the loads on an aircraft. It does it in a clear, accessible manner. It's a good stepping stone to structures books like Bruhn, Peery, etc.

#### Protech Racing

##### Well-Known Member
Ihave the books .None show the loads on an offset strut. I can easily add a front cable to even up the pull on the wing. I would like to avoid that if possible . I will post up the strut sizes/ angles in the next day or so . Thank you.

#### Fenix

##### Well-Known Member
Yes, "quarter chord" is 25% of the wing chord.

A "moment" is also called a "couple". Think of it as two equal but opposite forces some distance apart. The sum force in any direction is zero, but the twisting force still exists.

So, for example, assume a wing with two spars, one at the quarter chord point and one three feet farther back. Let's say the actual wing lift (Cl * q * wing area) is 100#, and that the center of lift is one foot behind the front spar. You now have a negative pitching moment of 100 ft-lb, which works out to 33.33# on the rear spar (100 ft-lb divided by 3 ft). That 33.33# is subtracted from the 100# on the front spar, so the front spar carries 66.67#.

A force at any point is equivalent to an equal force at any other point plus a moment around that other point. If there is no acceleration, the vector sum of all forces and moments is zero.

This is covered in "statics", a first or second year engineering class (before you get into aerodynamics and structures).

Ahhh, now the light is on. Thanks Dana for the excellent description. I think I've got it:

While the chordwise lift distribution of an airfoil could be shown as a “line graph” sort of like the Schrenk line but chordwise instead of spanwise (and maybe not elliptical or uniform), and such a depiction is probably useful in designing wing ribs (another topic for a later thread) the characteristics of an airfoil can be/are defined simply as the moment around c/4. And this changes with AOA so a given airfoil would have a "table of data" relating to it.

From this “torque” you can calculate the force at any given distance, that being the rear spar location (or any spar location). Once the force of one spar is known the balance of the force applies to the other spar. Or both spars, or 3 spars if so constructed, could have their loads independently derived using the “torque and distance formula”.

What a clever way of dealing with this. OK, and now I see this does not only apply to spars in airfoils but to any loaded “beam??” application. Ahhh, its broad applicability is why it is part of a fundamentals course (apparently called statics). Maybe I need these books first instead of AE books…..

But the light is spreading and I can see some things coming into view in the distance suddenly. I’m now suspecting that this, or something quite similar, is going to be used when I attempt to determine the amount of down force on the stab/elevator so I can apply these loads to analyze the aft fuselage truss structure, as I’ve been studying truss loads (to analyze the struts assy) on You Tube. This is all making sense, one of the moments that keeps me digging around in the books and forums I don’t understand much of, just waiting to find a new “nugget” that then helps to understand other previously confusing elements.

So I suspect that the “c/4 moment” data is probably published somewhere on a great many airfoils. Where can this be found? TOWS? (I wonder if the PIetenpol is a published airfoil).

Dana, from your inclusion of (Cl * q * wing area) at the right time when I was in the correct frame of reference suddenly I think I understand Coefficient of lift. It is a term encountered “all over the place” but suddenly, from what must be happening in the equation, it makes sense. I think it is simply the ability or effectiveness an airfoil (or body) has to turn q into lift. Hmmm, and does an airfoil (or body) also have a Coefficient of drag that is the (usually undesirable) ability of being able to turn q into a force in the same direction as q? Which makes it seem that lift is a term pilots use but that “purists” consider it simply a force 90 degrees to the direction of q. And what is the amount of this force we are dealing with? Suddenly the term “mass flow” which I’ve encountered (even my truck has a mass flow sensor – whatever that does…) seems like a very relevant term to describe the amount of force that is being converted into lift and drag. So from a Google search it appears mass flow and q are closely related or the same thing. Gosh, suddenly this is easy. I apologize to the contributors and “viewers” who were suffering “is this brick ever going to get anything”.

And these co-efficients of lift and drag and such are probably determined from testing (and as of late in some cases computer modeling) and published? Maybe in TOWS?

#### Fenix

##### Well-Known Member
At Vd and limit g loads, you have worst case for wing spars in shear and bending moment. Va is a close second, but the wing is generally sturdier against bending here. Va is the place where loads forward from lift are maximum and is the big stressor on drag and anti-drag bracing of a fabric covered wing. Put the air moving horizontally, the wing pitched up 18 degrees, and lift is vertical, lift has a significant component towards the front of the wing, which is trying to wrack the wing tips forward. This is why we have some form of diagonal bracing between the spars of fabric covered wings, usually drag and anti-drag wires to hold the wing square. When you go to structural skins, they carry these loads. Anyway, we have long established empirical guidance on sizing of these braces.

Billski,

I’ve been thinking about your description of how Max G and Va is usually not the critical scenario but rather that it occurs at Vd. You were describing the distance between the spars and horizontal and vertical forces. Is it like when you place a ladder across two saw horses and stand on it (the thin way) it can support a given load. This is like the wing at low AOA where the vertical loads are acting on its “thin dimension”. But if you stand the ladder “on edge” and then stand on it (yes it will teeter and you’ll likely fall off) it can support a lot more weight because the vertical load is now acting on its “thick axis” (and in the case of the wing the drag/anti-drag truss is going to get loaded up as you said). Of course the latter analogy is an AOA of 90 degrees which would not happen (and still create lift) but is this the concept of why the risk is higher at low AOA such as Vd?

It being the case that the high G at lower AOA is more critical I guess what I need to obtain is the “c/4 moment” data for the Pietenpol wing at the AOA which gives Max G at Vd. That is probably a long answer all unto itself that has to do with coefficient of lift which is starting to make some sense.

#### Dana

##### Super Moderator
Staff member
And these co-efficients of lift and drag and such are probably determined from testing (and as of late in some cases computer modeling) and published? Maybe in TOWS?
Starting in the 1920s NACA (the predecessor of NASA) tested many airfoil sections and published the data in graphs of Cl, CD, and Cm vs. AOA. Many of those are reproduced in TOWS.

#### Fenix

##### Well-Known Member
There is a lot of thought training and then practice doing the work in passing classes on calculus, chemistry, physics, linear algebra, statics, dynamics, fluid mechanics, aerodynamics, vibration, mechanics of solids, and mechanical design. They are all intended to produce individuals capable of first doing the calcs and then being able to synthesize new designs over a wide range of possible topics. Drinking from a fire hose is how much of engineering school seems to many students with this stuff spread over four years. It is a big thing, and you are trying to take a corner of it in one big gulp. Please be realistic in your expectations on yourself and on those of us trying to help you and other interested folks following along.

Net forces on unaccelerated flight are zero. In the vertical direction, we have lift and weight, equaling zero. In the horizontal axis we have thrust and drag, summing to zero. This continues on down into component sets too. Sum of forces in each direction must zero out or the thing you are looking at is accelerating off in some direction. No accelerating away in that direction and you must have forces in that direction summed to zero. You have six possible directions that all must normally null out and they apply to pieces as well as the whole airplane:
• Three are forces in linear directions mutually orthogonal - perpendicular to each other. Easy way is fore-aft, up-down, and port-starboard, and coincides with FS, WL, and BL axes of the airplane;
• Three are rotation and moments along those three axes of the airplane - the familiar Pitch, Yaw, and Roll.
The wing making lift is usually seen as airflow over the wing. Three of the six forces are significant here. We get lift, drag and pitching moment. To keep the wing from blowing away aft (from drag), flying out of sight vertically (from lift), and tumbling nose down (from pitching moment), we must have balancing forces. These forces originate in q (rho/2*v^2), S (wing area), c (chord), Cd, Cl, and Cm.

So to react these forces and make the wing stay put while the air blows over our wing, we anchor the main spar and the drag spar. If the main spar is at .25c, we can get to the loads pretty easily. Let's tackle just the lift and pitching moment: Lift and pitching moment are reacted by the anchors attached to the main spar. This includes the mounts at the root and the struts. Total vertical forces and moments calculated about any datum point will sum to being equal and opposite to the lift and pitching moment of the wing. Calculating all of these forces is taught within a semester long course call Statics, and it produces a lot of puzzled students for a while.

Let's break it down so we can understand it all better. If we put the forward spar and main strut right underneath each other at 0.25c, we will get the lift computed for the wing reacted by the strut and the root fitting on that spar. Got that? Some lift at the root, some at the strut. How much at the strut? Well, we have to figure out how much bending moment the lift produces about the root fittings. Bending moment and pitching moment are not related. Bending moment - imagine the lift along the span is broken into a bunch of local lift. Each local lift times the distance from it to the line of the root mount is the moment for that little local lift, do all of them, sum them up and you have the bending moment about root mounts. Divide that moment by the distance between root mount and strut mount and that is the vertical reaction at the strut. The root then carries the difference between total lift of that wing and the vertical reaction at the strut.

Now let's look at the drag spar and its strut. For the moment, let's make the strut directly below the drag spar. We have pitching moment from aerodynamics, and it produces torsion in the wing that must be reacted off. Fabric covered wings are really soft as torsion boxes. The root fitting reacts some of the torque, and the strut reacts the rest of the torque. The total torque is the wing pitching moment. How it is divided is related to the stiffness of the whole wing. A reasonable guess is that all of the pitching moment from outboard of the strut and half of the pitching moment from the region between root and strut are applied to the strut, while the balance finds its way to the drag spar root mount. Divide the lift by the longitudinal distance between main and drag spars, and you get the vertical force at the mount.

There is also drag and forward component of lift which can be significant at high AOA. Drag and forward component of lift (anti-drag) is distributed along the span, which requires forward or aft forces at the root fittings. Since drag and anti-drag is centered out on the wing, there is a moment about a vertical axis that is reacted by an lateral forces at the main and drag spar mounts.

Get you head wrapped around all that, and then we can shift the main spar away from 0.25c and start sharing forces between the main and drag spars.

Billski
Billski,

You spent considerable time in your replies today. Thank you, I appreciate that and feel the need to also say that I realize you cannot devote such time to answering my posts on an ongoing basis. If you, or any of the other contributors, feel I expect it or feel obliged to do so you will quickly tire of my inquiries. So while I appreciate it from all of you, please know I don’t expect anyone to contribute unless they have a particular interest in replying to that particular thread on that particular day.

Yes I’m getting an appreciation for how big this subject is and trying to find a corner to start nibbling on. I realize now how much “fundamentals” of physics and such (my high school sadly did not have a physics class, only a (somewhat “rendered safe”) biology class and chemistry class offered on alternating years) comes before aero engineering that is not really part of AE. This perhaps explains why so many AE books don’t make much sense to me. I think I need to find some of those “other books” to lay some more foundation.

Having said that: Billski, your content in the above quoted reply was excellent! I understood almost every phrase. Is/are there a book(s) written just like that? You cleared up several things I was wondering but had not gotten to studying yet, one of them being computing the amount of lift that goes to the root mount vs the strut attach. Also the description gave a new level of clarity to a question I had posed a while back about having a “pinned” connection at the fuselage (such as in a 3 piece parasol wing with center section) as opposed to it not being pinned there, as in a one piece parasol wing.

One question in regard to computing the lift of “each tiny little section” of wing along the span: Because the lift is not a “rectangular distribution” you cannot just divide total lift by span to determine how much lift each foot or inch creates, correct? This is where the “schrenk curve” comes into play?? If so do you do something like calculating the area of the section of the schrenk section above that little section of the wing and then “pro rate” the lift accordingly, or some mathematical function roughly equivalent to the “computing the area of each section” method?

I suppose you can also calculate the bending moment of the short (relative) wing section outboard of the strut in the same manner you calculate it for the entire wing around the root fitting? Seems as this is something like a short cantilever wing? And this bending moment “peaks” at the strut attachment (would be root attachment in true cantilever wing)? Previously were you trying to describe to me that some of THIS bending moment gets transferred to the strut and adds tension to the strut or did I misunderstand that?

You left off with shifting the main spar away from .25c and calculate the load sharing between the main and drag spars. Is this not what we were already discussing? Is this not the matter of calculating the pitching moment from .25c and applying it to the drag spar based upon torque and distance and then the balance of the force must apply to the main spar? I think when examining what happens in shifting the main spar away from .25c this will become clear.

#### Fenix

##### Well-Known Member
I see now that Billski already addressed a couple of the questions I just asked above in his initial response to this thread - I just didn't understand what he was saying at the time. In fact I re-read the entire course of this thread and so many things that several of you contributed are already much more meaningful than just a few days ago. Thanks all for your help and contributions.

#### wsimpso1

##### Super Moderator
Staff member
Log Member
I’ve been thinking about your description of how Max G and Va is usually not the critical scenario but rather that it occurs at Vd ... This is like the wing at low AOA where the vertical loads are acting on its “thin dimension”. But if you stand the ladder “on edge” and then stand on it (yes it will teeter and you’ll likely fall off) it can support a lot more weight because the vertical load is now acting on its “thick axis”.
I is smallest when a flat beam (the wing is a flat beam) is laid horizontal, largest when the beam is turned vertical, and varies smoothly between the two extremes with angle. Beams become stiffer (less deflection per unit load) with increasing I and stresses decrease with increasing I/h, where h is height of the beam.

Aero engineering, civil engineering, and mechanical engineering have a lot in common. Structure theory, control theory, basic fluids, thermodynamics, heat transfer, are all absolutely common, many are cross listed between the departments. We have been talking structures, which was already pretty well understood when Lilienthal started his first gliding experiments.

I guess what I need to obtain is the “c/4 moment”
Do yourself a favor. Stick to the nomenclature and taxonomy of each field, you can communicate with other folks and learn from them much more easily than if you start making up your own terms for stuff. Cm as used in TOWS is moment coefficient and M is pitching moment or airfoil moment in pitch. Aerodynamacist's terms. M as used in structures is bending moment. They are different things, but to figure out airplane design, you have to work in both, so we commonly refer to aerodynamic moment as pitching moment or moment in pitch. When we are talking about beams (structures that carry shear, bending, and torsion), we talk bending moments, torsional moments, bending stiffness, torsional stiffness, etc.

Some quality time in statics, mechanics of materials, and TOWS is in order. You don't need the whole course, but relationships between forces and moments, equivalent loadings, etc are useful. There are some great video series out there on these topics. Then truss analysis will take you a long ways.

Wings (and tailplanes and control surfaces) are externally highly specialized shapes that are also beams. So are things like fuselages, wing struts, landing gear legs, etc. The strut supported wing is a truss, as is the steel tube fuselage, retractable landing gear, and so on. Then you have column theory for elements in compression because there is buckling... it does get broad to design airplanes.

Billski

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#### wsimpso1

##### Super Moderator
Staff member
Log Member
You spent considerable time in your replies today. Thank you, I appreciate that and feel the need to also say that I realize you cannot devote such time to answering my posts on an ongoing basis. If you, or any of the other contributors, feel I expect it or feel obliged to do so you will quickly tire of my inquiries. So while I appreciate it from all of you, please know I don’t expect anyone to contribute unless they have a particular interest in replying to that particular thread on that particular day.
Worry not. We are all volunteering our time here. If we feel up to answering, we do, but I am sure most of us feel no obligation. Speaking for myself,I find it motivating to have good students. Get my drift? Additionally, we are not just writing to you on a forum. We are writing to the other folks that are following along and to the folks who come along later and want to know how this works. If you recognize that your writing has "legs", you can be motivated to do it well.

Engineering is applied physics.

Having said that: Billski, your content in the above quoted reply was excellent!

One question in regard to computing the lift of “each tiny little section” of wing along the span: Because the lift is not a “rectangular distribution” you cannot just divide total lift by span to determine how much lift each foot or inch creates, correct? This is where the “schrenk curve” comes into play?? If so do you do something like calculating the area of the section of the schrenk section above that little section of the wing and then “pro rate” the lift accordingly, or some mathematical function roughly equivalent to the “computing the area of each section” method?
First, I do not like Shrenk's Approximation. It was invented while we were all still working with slide rules and needed closed form solutions to do the work. A closer approach to reality was needed than just applying a single wing loading to the area of a wing. Averaging the elliptical and area based (or linear decreasing) models allows closed form hand calculation, but it still over represents shear and bending moment at zero aileron and does not necessarily provide a good estimate of effects from deflected flaps and/or ailerons. We can break the wing into discrete spanwise chunks, each with its own loading, then numerically integrate to get a more accurate and intuitive image of the whole, so why just do that? Excel is your friend...

So, how do we do this?

First we break the semi-span into chunks. Why semi-span? The left and right sides are the same, let's just do one side. Semi-span is b/2. Don't go crazy. 10 chunks might be plenty, but electrons are cheap, I usually go 16 on a cantilever wing. For a strut braced wing I go 12 from root to strut mount, and another 10 from strut mount to tip. Now in the middle of each chunk, you know x and can calculate w(x) for it too. Let's do that...

Total lift from a wing that has lift elliptically applied spanwise is:

L = PI()/4*wr*b

This might look familiar, as it is the area of an ellipse. If wr=b it is the area of a circle. These terms all come right out of TOWS and I am writing the equation in Excel. L is lift needed in pounds. PI() is Pi, the ratio of circumference to diameter of a circle. Excel has it as a function, thus the PI(). wr is spanwise wing loading at the root, in lb/ft or lb/in or Newtons/furlong or whatever units you can live with. b is span of this wing in the same length units. You know how much force you want the wing (or tailplane or rocket fin or...) to generate, you know the span you are thinking about, solve for wr:

wr = (4*L)/(PI()*b)

Now what? Well, we get local spanwise wing loading from this equation for an ellipse:

1 = (x/x1)^2 + (y/y1)^2

x1 and y1 are the primary radii of the ellipse. x is spanwise dimension starting from the centerline, an y is local wing loading w(x). First let's use algebra to get the height (y) of the ellipse at any spot x:

y = y1*sqrt(1-(x/x1)^2)

Now let's put in our already established terms for things - use b/2 for x1, and wr for y1, x is still position along the semi-span, and w(x) is wing spanwise load at x:

w(x) = wr * sqrt(1 - (2*x/b)^2)

Now your spreadsheet has some input for span, and lift and shape, then it has several columns. First column is BL dimension from root to tip of the wing, broken into your desired chunks. Then there will be a column for chord, then w(x), then L(x). Lift for each chunk is just the spanwise width of the chunk times the spanwise lift per unit length. Make sure they use the same units. Mixing up feet and inches will surely mess up the numbers.

Now once you have all this, you can copy it. Then you can add in a deflected aileron or a deflected flap. Study the chapter in TOWS on High Lift Devices to figure out how much higher the Cl and Cm and Cd is for the part of the wing with a deflected surface, and multiply loading over the area of the wing where the device is by that amount, the rest of the program will calculate the new shear, bending torsion, drag, drag moment, etc.

You left off with shifting the main spar away from .25c and calculate the load sharing between the main and drag spars.
This is statics, relationship between forces and moments, equivalence of one to the other, etc. Not terribly complicated, not much more than lever rules, but you can get tangled up in your undershorts here. Go to your statics references and make sure you really understand it by doing practice exercises and check that your answers are right before going further.

As to a book that describes this well... no, the authors are partly about showing how smart they are, and partly about writing to apply the science as broadly as they can. The topics have broad applicability, and then we in the field apply the knowledge where ever we need too. Some of us like to talk about it and are motivated to teach on more specific topics within.

A friend of mine took airplane design with Professor Lesher and doodled airplanes in the margin. There were hardly any airplanes pictured in the books...

Have fun and come back for more guidance.

Billski

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#### Geo

##### New Member
the center of pressure (lift) of the pietenpol aircamper wing is 33% of the cord

#### Pops

##### Well-Known Member
Log Member
Billski is a master teacher. I read every word.

#### Fenix

##### Well-Known Member
Do yourself a favor. Stick to the nomenclature and taxonomy of each field, you can communicate with other folks and learn from them much more easily than if you start making up your own terms for stuff. Cm as used in TOWS is moment coefficient and M is pitching moment or airfoil moment in pitch
So it seems that Cm IS the moment coefficient around the 25% chord point. Correct? This would be useful for determining chordwise distribution of lift as we've been discussing.
And in regard to M (in aerodynamics) : Is that the moment around the center of lift? That would seem useful for determining the downforce required of the stab/elevator. Or is M something different than this? Certainly M is not the same as Cm or it would not have its own term.

#### Dana

##### Super Moderator
Staff member
the center of pressure (lift) of the pietenpol aircamper wing is 33% of the cord
No. Well, it might be, at some particular AOA, but it moves.

So it seems that Cm IS the moment coefficient around the 25% chord point. Correct? This would be useful for determining chordwise distribution of lift as we've been discussing.
And in regard to M (in aerodynamics) : Is that the moment around the center of lift? That would seem useful for determining the downforce required of the stab/elevator. Or is M something different than this? Certainly M is not the same as Cm or it would not have its own term.
M = Cm × MAC x q x S. (S is wing area), so it's used just like Cl, Cd.

Strictly speaking, a moment is independent of position. It's usually given as Cmc/4, which is the moment if you assume the lift vector is acting through the quarter chord point. Sometimes it's given as Cmac (around the aerodynamic center). Cmac is a constant, that's the definition of aerodynamic center, but the location of the a.c. is different for different airfoils. For most airfoils the a.c. is close to c/4 so it's more convenient to use c/4.

Any force vector (a force acting through a particular point in a particular direction) can be converted to a force of the same magnitude and direction through a different point plus a moment. So the wing lift, acting through the center of lift somewhere aft of c/4, is equivalent to the same lift force through the c/4 point plus a negative (nose down) pitching moment equal to the lift force times the distance between the center of lift and c/4. Divide those forces by q*S and you have the coefficients Cl and Cm.

#### wsimpso1

##### Super Moderator
Staff member
Log Member
the center of pressure (lift) of the pietenpol aircamper wing is 33% of the cord
Does it have a non-zero Cm? If yes, center of lift changes with changing alpha....