MASS

Since we said our main goal is to fight gravity, let's measure how strong it's going to be by making an estimate of the weight.

• Battery: 43g
• Servos: 10g (4)
• Receiver: 16g
• Expected airframe weight: 200g

This gives a total mass of: 399g

This seems a bit light, but I'm going to leave it like that for now.

SPEEDS, HEIGHTS and Gs

Now I'm going to calculate some speeds and expected heights from the throw. The basic technique is to grab the plane by its wing tip, twirl once to build some speed up in the plane, and then let go.

Reading up a bit, I should expect to launch this thing to a height of around 40m. That means that the release velocity, not considering air friction, has to be:

Let’s say the wing will be about 1.2m long (I'm starting with a rough estimate based on the size I want) which gives it 0.6m from the body to the wing tip. Since the plane's centre of mass will rotate around me, I need to take the length of my arm into account too (measured at about 0.7m). This means for a mass centred around the point M, and me and the plane twirling like a pretty ballerina around point O, the total centripetal force exerted on the mass through the wing is calculated below.

This diagram lays out the situation for us

This diagram lays out the situation for us

This means right at the point of release, I will spin around twice in about a seccond. But this still doesn't tell us what kind of pulling force will be exerted on the wing, so the following equation will tell us:

180N. That's quite some force! That's like hanging 9 bottles of coke from the wingtip (vertically of course).

Even though this was a simplification of the situation (compitition throwing speed conditions), it was a conservative one so if we design for this we are still safe.

This only accounts for the force exerted on the wing along its length. It's actually the easiest force to design for and not the one I expect to break the wing.

We need to figure out what the largest possible lift is that the wing will produce (and therefore the largest bending moment) since I definitely think that will be the one to cause the most trouble.

Unless I dive straight down to terminal velocity and pull up, I supsect the largest load on the wings will be right after I release the plane and pull up for the climb. The release velocity may actually be higher than terminal velocity and just for interest sake I'll check for that afterwards. I looked around and also watched some videos on the typical steep banking radii of these planes and it seems to be about 10m. Let me ilustrate:

This means the force needed to turn the plane around that circle is given by the same equation we used before:

24N doesn't seem so bad, it's like hanging 8 planes from this one while it flies.

There are a lot more forces going to act on the plane, but I'm not building a passenger plane and I don't have to compete for cost or efficiency. I also don't see the tailplane I build breaking under any normal flight conditions so I won't concentrate much effort in proving its strength.

Questions to Address:

• How big do the wings need to be? How wide? How long?

Provide lift for 10m radius turn at 28m/s.

The length I am going to choose based on what I can comforatbly swing.

More refined constraints on drag and lift will provide the final rules for the wings.

• What is airfoil?

Airfoil design is right next to antenna design on the B̵͓̭̰̗̼̺͇̗͈̄̓̃͠ͅL̶̟̤͔͎̘͔͔͉͐̂̄̎̀͛͂̇͘͘͜ͅA̴̦̮͗̾͘͝C̶̢̨̫̞̼͓͙͂̽́͊͂̐̐͑̐͜ͅḴ̸̡̡̫̳̘̱̝̖̥͒͊̉̌̽͗͑̋͜ ̶̧̡̢̨̰̬̞̋̿̓̈̈̃̄͌͘͝͝ͅM̸̛͇̓̏̆͌͆̾̿A̵̦͖̱̿̽̀G̷̨̻͔̮̘͙̣͚̯͑̂̀͊͐̋̑͘͝͝͠Ī̷̧̢̤̼̝̮͍́͊̌͑͆̾͒̎͊ͅͅͅC̴̗͇̄̆͋͗͛̍̆͐͝ shelf.

Many good airfoils exist to choose from, created by wizzards from shady government projects about the same time they exprimented with LSD in the 70s section X will show how.

• How thick does the wing need to be to carry the forces we worked out above?

Thick enough so that the material we make it out of is far from its strength limits, but ultimately the thickness will be determined by the airfoil. I will have to ensure the strength requirements are met under those dimensions.

• What will be the best way of building the wing?

Based on the forces encountered and my available resources, a composite spar wing with thin balsa ribs.

• How long the tail?

Long enough so that the tail plane can turn the plane using as little force as possible, but not too long as to unnecessarily increase the weight or the rotational inertia. Another potential issue is if the tail starts to wobble or vibrate, as the carbon tube acts as a nice energy-storing spring. If it's short and stiff, we don't have to worry about that.

• How big do the fins and control surfaces need to be?

There are general good practice guidelines for sizing up the tailplane. Tailplane needs to stabilize the aircraft like a dart's fletching.

A larger tailplane will mean the aircraft will be faster to point its nose true to the incoming airflow and much faster to respond to controls but also provides more drag.

The horizontal stabilizer also needs to apply a force to counter the pitching moment caused by the wings as they produce lift, which I will explain below.

The rest of the aspects in this section will be guesstimates made from observation and a little experience in folding a paper plane before.

For simplicity's sake I'm going to choose the general shape of my wing now since I've been making wings before I could cook eggs

My wing has the following profile:

Where the thickest part of the wing runs straight and the width decreses from w to half of w:

There's three reasons a wing tapers to the tips

1] Reducing tip drag. A lot of complicated theories and effort go into reducing tip drag, and you can read up on it, but basically the fatter your tip is, the more drag you get.

2] Reducing moment of inertia, concentrate the lift and mass as close to the CG as possible.

3] The bending moment on the wing structure increases with distance from the body.

There's a fourth secret reason, because fat wing-tips look retarded, probably because we internally know it's a bad idea

You can have all sorts of tapering paterns, like the oval of the spitfire wings, the segmented gull wings etc.. but I chose linear because our upcoming calculations are already complicated.

Wing Lift

Let's look at the equation we are going to use to predict the lift the wing will produce:

This equation sumarises the dark mocking wizzardry of earonautics into 4 innocent looking terms, but the MONSTER of insanity is nicely captured like a pokemon in the Cl Term. It basically describes how good an object is at creating lift, and for given aerofoils the dark wizzards of the 70s have tamed these into little numbers on graphs for us.

We now know that the ammount of lift an object produces as air flows over it is preportional to its top down surface area. So if you imagine the wing being chopped up into little slices and we calculated the lift each of those slices produced, we would see that it decreases as we go towards the tip like this:

Now, mathematically we can chop this wing up into infinitely thin slices and add those infinitely many infinitely thin slices up to get a smooth graph that descibes the lift from the root of the wing all the way to the tip, calculus is beatiful.

This is why I chose a straight lined outline for the wing because it means the graph for the lift will also be a straight line.

But this only looks at lift in two dimensions, and this linear simplification is not how lift is really distributed across a wing, but I feel it's close enough for my hobby purposes anyway. The true lift distrubution is very close to elyptical, where the vortices created by the wing tip and disturances by the body at the wing root is taken into account.

I am going to make a good guess at the size of wing I need. It will be based mostly on what I'd feel most comforatble swinging around myself and what size I want to carry around. We don't know yet what Fr or Ft is, but we know the one will be half the other.

Now, since we chose the tip to be half the width of the root, we can easily see that the amount of force a slither of wing at the tip will produce (let's call it dFt) is half the ammount of force a slither of wing will produce at the root (let's call that force dFr).

So, if we know now what lift each little slither of wing will produce, that meens if we add them all up we will get the total lift produced by the wing. What we also know is that we need to produce at least 24N of lift to pull the aircraft into that tight turn.

Before we go further please note: We might get a very weird answer from the following method I will try, and if the answer is nothing what I expected then I'm going to try another method. I say this because if the only criterion we're taking into account at the moment is at the wing's most extreme state of flight we may end up with a wing that's only good at meeting those criteria and not very good at normal flight. There are other aspects we will take into account after this.

As I was saying, we need the wings to give us 24N of lift traveling at 28m/s. One important thing to note about wings are that they can give different amounts of lift depending on how much you turn their tip up against theincomming air. This is known as it's Angle of Attack (since it's the angle the air attacks it at).

As an example, think of holding your hand out a car window and tilting it so it gets pushed up by the wind.

The problem is that as you increase the angle, you not only increase the lift, but the drag goes up too. SO, you need to be carful to not design a wing only at the maximum lift it can produce (small highly loaded wing), because the plane will slow down very quickly. Here are two simple graphs showing how the two properties are related:

Now that we know that, I'm going to add that when we pull the nose of the aircraft up after the throw we will most likely be opperating near the top of that lift graph's highest possible point. The drag will also be pretty high but we won't worry about that for now since the turn is going to be very quick.

So let's proceed with a typical Lift Coefficient value you can expect from a glider airfoil at that point. Most research I did put that at around 1.2 Now we can go on to find the wing size we need at those conditions.

To find the sum of all the little lifting forces produced by the slithers of wing, we just have to integrate along the wing length with and equation that describes the distribution of lift. Even more simply, we just need to find the area under the graph. The following drawing will show how:

Note how Ft only represents the ammount of force produced, not the actual width of the wing. I will now manipulate a few equations to give me an estimate of the tip size I need:

The lift equation with the formual for the area where

Area = (w*l) + Half w * l

Substituting in my wanted length Ammount of force onee wing should produce, the density of air, lift coefficient, and velocity This puts the wing tip at about 0.12 m long which is a very reasonable estimate. This also means the root airfoil length should be 0.24m

Now that we have decent wing size for managing our sharp banking, let's see how well it will perform in normal gliding flight.

What we need for a good gliding flight is enough force to counter gravity and as little drag as possible. Remember how you can make a small wing generate lots of lift by increasing the angle of attack but at the same time its drag goes exponentially higher?

Well what we want to see is what kind of drag we will get at a certain lift coefficient of the wing, and if it seems horrible, we may need to incrase the size of the wing, but I have a feeling we will be fine

The tool I will use to see what kind of drag to expect from my wing is a program that a lot of model aircraft builders use called XFLR-5 and was introduced to me by a brilliant aerodynamycist I had the privilge of working under during some vacation training I did. It's a program that takes in an airfoil at certain conditions, and then does a whole bunch of math at different angles of attack. One of the graphs it gives you back is one that shows lift vs drag, and that will be the one I'm going to look at here.

What we'll need to give the program is the follwing:

1 ] Reynold's Number of air flow over our wing. (Look it up under the principle list)

2 ] An airfoil we want to evaluate.

Getting some advice from one of the deciples of the order of the foil (who happens to be a close family friend) I have two very popular airfoils that people use for DLG sail planes. The one is the HM-51 that looks like this:

And the other is the AG38

Just looking at these two wing profiles you can barely see a difference, but every subtle change in curvature or difference in angle changes the way air and flows and follows the surface drastically.

These two diagrams are runes to be etched into materials endowing them with flight

I chose to use the HM-51 airfoil because it's slightly thicker and since I plan on building the plane out of balsa and other squishy stuff I want all the strength I can get.

What we need now is the Reynold's Number for the flow of air over our wing. I will use the average air speed and average chord length (distance between leading edge of wing and trailing edge) of the wing and use a value for the viscosity of air I got off the internet:

Note how the Reynold's number has no unit, it's purely a number that tells us how turbulent the flow is in these conditions.

These aerodynamisists have left us the equivalent of a tome in the form of a small computer program that can roughly calculate a lift-curve based on the angle of attack the program is called XFLR-5

Lift coefficient at different angles of attack (alpha)

Lift coefficient (Cl) vs Drag Coefficient (Cl)

This is all very nice and pretty, but how do we use this? Well first let's see what kind of Lift Coefficient we need to keep the plane in the air at 10m/s and then we can find that lift coefficient on the graph and therefore the corresponding drag coefficient.

So to maintain level flight we need a coefficient of lift of about 0.247

Looking at our Cl vs Cd graph, this corresponds to a coefficient of drag of about 0.012

What does this mean? Well this will help us figure out the ammount of drag the aircraft will experience and give us an estimate for how fast it will loose speed and therefore altitude (since it has to dive to keep the speed up).

Let's see if we can expect a decent value for a glide slope (ratio of foreward movement to drop in altitude). I expect it to be fairly ok since a wing thats about 20cm wide looks like a standard size on these models.

For this we need to work out the total frontal area of the aircraft and use the drag equation to calculate the drag force on the plane. The frontal fuselage area I'm going to calculate as a semi square the size of my expected fuselage. The wings will have dimensions specified by their airfoils which for the HM-51 having a thickness of about 8% of its length. This gives us a root wing thickness of 18mm.

Now we can calculate the drag. I will use the coefficient of drag that the graph predicted for the wings and one I found on the internet I think most clsely matches the shape of the body I'll build:

To keep the plane moving forward, we need to tilt the force vector of the lift slightly over till its horizontal component is equal to the magnitude of the drag force. This diagram shows what I mean:

Wing Strength and Structure

Since the wing will carry a significant amount of force, I want to ensure the current plan will hold up. To do this, let's first review how the force is distributed through the wing and how we expect it to transfer that force into the body to lift the plane:

Lift Force Graph

Here’s the lift force graph we calculated:

Shear Stress Distribution

By integrating the lift graph (simply summing up all the force generated by each small part of the wing), we can determine the shear stress distribution:

Moment Distribution

From the shear stress, we can integrate to find the moment distribution the wing will experience:

For reference, 11.2 Nm of torque is roughly equivalent to the effort required to twist your hand while holding a box of milk hanging from a pole one meter away.

Wing Load Analysis and Spar Design

These graphs help us visualize how forces and stresses build up in the wing. On the shear force diagram, note that the total force added from the tip to the root (from 0.6m to 0m on the graph) equals 12N. With a wing on either side, this totals 24N, which is sufficient to accelerate the plane through its tightest turn.

We can see that the wing can have a thin, weak tip since there isn’t much force acting on it yet. However, as we move toward the root, the wing must become thicker and stronger to carry the accumulated load of the lift until it can finally dissipate into the body and counter the weight of the plane.

If you recall, we already chose the dimensions for the wing root and tip, as well as the airfoil. This gives us a linearly tapering wing that goes from 18mm thick at the root to 9mm thick at the tip.

Since I am building the plane out of balsa, the simplest structure to handle these forces consists of a few balsa ribs covered with a film to shape the wing’s airfoil and a strong spar running through the middle. The force—caused by pressure differences on the top and bottom of the wing—will transfer through the ribs from the film into the spar, which will carry it to the body.

With this setup, I won’t worry much about the ribs and will focus on designing the spar instead. I believe the primary challenge will be managing the moment in the spar. Although the shear force maxes out at 12N, which is significant over a small surface, we should examine the stress caused by the moment at its maximum point (the root).

First, let’s consider a spar made from a uniform material (a solid piece of wood, for example). The stress on the material varies linearly from a maximum at the top and bottom to zero at the center. This is because the most stretching or compression occurs furthest from the center, where the maximum stress is.

From this graph, we can infer that improving the material properties is most effective at the surface of the top and bottom where the stress is highest. Using a stiffer material at these points allows it to bear most of the load, as it resists stretching, leaving less stress for the core to handle. For example, here’s the stress distribution of a bar with very stiff top and bottom pieces glued to a softer core:

An important detail about our spar design is that the balsa core will have its grain running perpendicular to the spar's length. The reason for this is quite interesting. With a stiff upper portion of the spar under compression and a stiff lower portion under tension, there must be an interface where these forces cancel each other out. If the spar were composed of unglued slabs, they would slide past each other when bent. However, gluing them together forces them to push and pull on each other, translating the force effectively. The following image illustrates this concept more clearly:

So, we now have details about the forces in the wing and the geometry of its construction. I will now perform some simple strength calculations and see what I get. First and easiest, a simple calculation to see if the wing can handle the vertical shear force of 12N. I will assume the wood doesn't resist this force at all, and it's only the carbon strips that take it.

Shear Force in Carbon Strips

The shear force in the carbon strips is 6.667 MPa.

• Maximum Shear Strength: Carbon can withstand around 90 MPa (assuming this is across the grain of the fibers).
• Safety Factor: This gives us an impressive safety factor of over 60—great news!

Normal Stresses on the Spar

To ensure the spar holds up, we now calculate the normal stresses at the most critical points: the top and bottom surfaces.

• Definition: Normal stress acts along the spar’s length, causing compression and tension that can crush or pull apart the fibers.
• Assumption: Only the carbon layers take up the force, while the balsa contribution is negligible. To calculate this, we determine the second moment of area (how well the shape resists bending).

Material Strength

The tensile strength of carbon laminate (aligned with the fiber orientation) is approximately 600 MPa, according to a textbook and an online source.

• Safety Factor: This results in a safety factor of 3.4, even without considering the small contributions of balsa or glass reinforcements.

Wing looks like it will hold up just fine.

This diagram shows the load the tab will be under:

Project Update

Current Thoughts

Right now, I don’t see any immediate areas for improvement. Any further refinements would require a lot of iteration and calculations, and significant improvements would only come by combining several of them. I want to start building, as I’ll most likely encounter problems along the way and work around them.

Construction Begins

Body

• Photocopied the design and scaled it by two to create paper templates for parts.
• Traced one side of the fuselage onto a 1.5mm balsa sheet.
• Tacked the sides to the base with superglue and shaped the nose block using glued balsa pieces.
• Reinforced the inside of the fuselage with two strips of carbon and used a resin-carbon mat mix as a strong, lightweight filler to secure the boom (half a fishing rod).

Tailplane

• Cut out the tailplane pieces and added a thin carbon layer to the top and bottom of the horizontal stabilizer.
• Created a stiff elevator linkage by wrapping carbon tow around a small dowel, clamped it, and let it cure.

I placed everything I had so far into and onto the fuselage to see the progress. I also finished covering the tailplane pieces, but I still need to use a heat gun to properly shrink the plastic.

WANGS

• Decided on a pull-pull system for the tail since a push rod system won’t fit due to the receiver’s position near the tail boom entrance.
• Printed and scaled airfoil templates, then traced them onto balsa:
  • Three inner ribs from 1.5mm balsa.
  • Outer ribs from 1mm balsa.
• Cut and sanded ribs to match the airfoil lines exactly.

A rack of ribs


• Added vertically grained strips of wood between ribs as shear webs and used clamps while the resin cured.
• Used Smooth-On laminating resin for spar construction, wrapping thread around the spar to hold everything in place while applying clamps.
• Assembled and glued the two wing halves together, leaving reinforcement with glass and carbon strips for later.
• Added servo wire extensions and made small mounts for the wing servos, gluing them into place.

SPARS

Here’s a close-up of the spar construction with thread wrapped around it for stability during curing. The two wing halves are glued together, reinforced with glass and carbon strips, and fitted with servo mounts. Next, I’ll finish covering the wings and work on attaching the finger tab.

Covering the Wings

I decided on a clear and thin orange covering film to showcase the wood underneath. After tracing a rough outline on the film, I spent an hour struggling to make it stick. Eventually, I realized there was a thin protective sheet over the adhesive layer. Once removed, I could proceed with covering the wing.

There were several mistakes during this process:

• Wrinkles appeared in some areas due to improper heating, even resulting in holes in the film.
• Surfaces weren't entirely smooth, which caused bumps under the film.
• The film didn't adhere well to glass laminate, leaving wrinkles at the wing's center.

Despite these challenges, I managed to patch holes and smooth out the film where possible.

Final Steps

• Created a finger tab for throwing the plane by layering carbon and compressing it under textbooks while it cured. The tab was strong and glued directly to the spar with reinforcement.
• Extended the triangular balsa ailerons by cutting a V slot and key for a better bond.
• Calculated the optimal thickness and reinforcement needed for the finger tab, considering carbon’s tensile strength of 600 MPa and a safety factor of 1.6.

Thar She Be