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"CAN TOOTHLESS FLY?: AN ATTEMPT AT AERODYNAMICS Saturday, July 19, 2014

9:45 PM

 (A .pdf version of this process in the form of a question prompt and answer sheet can be downloaded  here )

I have to admit, aerodynamics is not a subject I have much authority on. I know the basic concepts, but I don’t know multivariable calculus, which is required to get a more accurate answer. Therefore, I have relied on mostly elementary concepts and simplifications that do not deviate heavily from the correct method, so that we can get a pretty good approximation that will suffice for what we’re looking for.

The reason why I focused on Toothless is that 1) we know his wingspan and 2) he’s one of the more aerodynamically sound dragons. Meatlug obviously cannot fly, for example.

And with those things out of the way, let’s get started.

First of all, we only have one given that is specific: Toothless has a wingspan of 48 feet. This is a crucial information that we’ll use to derive numerical data from what we see.

Our primary reference resource is the opening sequence from  Riders of Berk  as it contains one scene that shows Toothless from a top-down perspective, flying across the ocean:

This picture is a composite of three consecutive frames from the sequence that shows Toothless flying at pretty much constant speed. Let’s see what we can dig up from this.

My goal is three-fold: The video is in 24 frames per second, so if three frames are taken: It is obvious that this composite shows Toothless over the span of  1/12 seconds. We have the time, but we’ll need the distance that Toothless has traveled over this period of time to calculate his average velocity. How do we get that from pixels?
 * 1) Examine if the velocity of Toothless in this sequence is reasonable for both the dragon and the rider’s safety
 * 2) Calculate the lift generated at this velocity (investigating how to calculate lift from limited information)
 * 3) Find the minimum velocity that will allow Toothless to cruise horizontally without falling into the sea
 * first frame
 * (1/24 sec. interval)
 * second frame
 * (1/24 sec. interval)
 * third frame

We use conversion factors, of course. In the picture, you’ll see Toothless’s wingspan in the final frame (where his wings are fully expanded) is 270 pixels. Pixel is a perfectly consistent unit of length in images, so we can use this to get a conversion factor to convert pixels into real-world length:

48 ft x 0.3048 m/1 ft x 1/270 px =  0.05419 m/px

Now we are ready to calculate, after we get the distance traveled.

<p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555">The distance is tricky here. Toothless is moving in the camera’s perspective, but the camera is also moving along with Toothless: we have a moving reference point, in other words. You can see the stationary rock (thank goodness it’s there) has moved 185 pixels  downwards on screen. This would mean we have to  add those 185 pixels to the 43 pixels, the distance that Toothless has traveled from the camera’s perspective, to get absolute distance.

<p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555">We have the distance and time, let’s calculate:

<p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555">v = d/t = [(185 px + 43 px) x (0.05419 m/px)] / (1/12 sec) = 148.255 m/s = 533.9 km/hr

<p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555">I did the m/s to km/hr conversion to show you how insane this speed really is. According to the numbers, Toothless is flying at nearly twice the speed of an average Formula One vehicle, or a bit more than half the speed of a Boeing 767. I don’t know about you, but I would  not  want to be on that Night Fury if I was riding. Otherwise, there would be about five trillion bugs in my teeth.

<p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555">That is one down, two more to go.

<p style="margin:0in;line-height:18pt;font-family:Calibri;font-size:11.0pt">

<p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555">This is a bit more of an exercise, but we should calculate the lift that is generated at this speed. It gives me time to explain the formulae, my simplifications, and some concepts that even I struggle to fully grasp. SKIP THIS SECTION IF YOU WANNA JUST GET TO THE IMPORTANT STUFF.

<p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555">Here are the formulas: <p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555">where <p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555">Our goal is to calculate  L. This might take a while.
 * 1) L  = 0.5  ρ  x  v ^2 x  A  x  C_L
 * 2) l  = 0.5  ρ  x  v ^2 x  c  x  c_l
 * 3) c_l ≈ 2 πα
 * L =  lift generated
 * l  = section lift (lift generated by a cross section of the wing)
 * ρ  = density of air
 * v  = velocity/true airspeed
 * A  = area of the planform (wings)
 * c  = chord length
 * C_L ,  c_l  = lift coefficient, section lift coefficient
 * α  = angle of attack
 * ( n  = wingspan. This will appear later)

<p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555">But let’s go through this. The difference between lift and section lift is that section lift only looks at a cross section. If you take the units into account, you’ll see that lift is in Newtons while section lift, due to the missing “m” in chord length compared to area, is technically in N/m. Where did that “m” come from?

<p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555">Since the only thing that significantly varies in the formula is the area vs. chord length, let’s examine these two. Chord length (shown “39 px” in the picture) measures the wing from front to back in the most basic terms; area measures, well, the area of the wings.

<p style="margin:0in;font-size:12.0pt;color:#555555"> But if chord length is front to back, couldn’t we multiply chord length by wingspan, which measures the wings left to right, to approximate the area of the wings? Or, isn’t  A   ≈   cn ?

<p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555">Let’s assume that is “close enough.” We still have to find the relationship between lift coefficient and section lift coefficient to turn section lift formula into one for total lift. Getting lift coefficient is actually not a feasible task to do without either a model of Toothless and a wind tunnel, or multivariable calculus and some serious talent in 3D modeling.

<p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555">So, I simplified it using things I know. Remember how the section lift is in the unit of N/m, the “m” coming from the missing wingspan? This would imply section lift is the  rate of change  of lift as wingspan changes.

<p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555">Rate of change.

<p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555">Hmm, I wonder if we can fit basic integral calculus into this?

<p style="margin:0in;font-size:12.0pt;color:#555555"> L   ≈ integral of  l  from 0 to the given wingspan with respect to the wingspan variable ≈   ln

<p style="margin:0in;font-size:12.0pt;color:#555555"> Since we’re assuming that the cross section of the wing is always the same, section lift is constant no matter the wingspan. Fortunately, since section lift is constant, the integral simplifies down into  ln. If  L   ≈   ln, that would mean that C_L  and  c_l , the only variables that we have seen no correlation between, are pretty much equal. This means that if we can get the section lift coefficient, we should be able to approximate the lift pretty well as we can get density of air from other sources with reasonable accuracy, and we already know the other variables.

<p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555">The third formula I listed approximates section lift coefficient using angle of attack. Since we cannot measure the angle of attack, let’s assume 7.5 degrees (or π/24 radians, which we will use to calculate) which is a reasonable angle based on my observations.

<p style="margin:0in;font-size:12.0pt;color:#555555"> C_L  ≈   c_l ≈   2 π α =   2 π x π/24 =   0.8225

<p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555"> A  =  cn  = 39 px x 270 px x (0.05419 m/px)^2 =  30.922 m^2

<p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555">The density of air has been taken from the internet, with consideration that this is at sea level (1 atm pressure) with temperature probably somewhere around 5 to 10 degrees Celsius:  1.250 kg/m^3.

<p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555"> L  = 0.5  ρ  x  v ^2 x  A  x  C_L =  0.5 x 1.250 kg/m^3 x (148.255 m/s)^2 x 30.922 m^2 x 0.8225 =  3.494 x 10^5 N

<p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555">That’s a lot of lift. (Do you even lift, bro?)

<p style="margin:0in;line-height:18pt;font-family:Calibri;font-size:11.0pt">

<p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555">Now that we’re done tweaking with the nitty gritty stuff of calculating lift, determining minimum cruising speed should be a bit easier. One question needs to be answered though: how much does Toothless weigh?

<p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555">After some digging from the web, a pretty good reference I found: Komodo monitors can grow up to 3 meters long and weigh up to 70 kilograms (or 686.7 N, if you turn mass into force of gravity acting on the thing). I determined from the final frame of the composite image that Toothless is 150 px (or 8.128 m) long. The factor of 8.128/3 represents the ratio of the two reptiles’ lengths, so if we cube it we’ll get an approximate ratio of the two’s volumes. Use this to calculate Toothless’s approximate weight:

<p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555">Weight_toothless = Weight_Komodo x (factor)^3 = 686.7 N x (8.128/3)^3 = 13651.9 N

<p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555">That is Toothless’s weight in Newtons; it’s about 1400 kg, which I think is reasonable.

<p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555">I specified that I wanted to get Toothless’s required cruising speed. When he is horizontally cruising, he is not moving vertically at all, which would mean in classical mechanics that the vertical forces acting on him add up to zero. The only two significant vertical forces on him are lift (upwards) and weight (downwards), so their magnitudes must be equal. This means that while cruising  L =  weight = 13651.9 N. Rearrange the lift formula to solve for velocity:

<p style="margin:0in;font-size:12.0pt;color:#555555"> v  =  √ [(2 L ) / ( ρ  x  A  x  C_L )] =  √ [(2 x 13651.9 N) / (1.250 kg/m^3 x 30.922 m^2 x 0.8225)] = 29.31 m/s =  105.5 km/hr

<p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555">That’s about the same as many U.S. highways’ speed limit. Hiccup is not wearing a helmet, goggles, nada. He won’t even be able to open his eyes.

<p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555">(sorry for inappropriate humor/not sorry)

<p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555">That brings this analysis to its conclusion. Without flapping wings, Toothless himself will be able to fly if he has been given enough thrust before, but Hiccup will be holding onto his dear life. Either that, or Toothless will try to fly slower and slowly plummet since he isn’t getting enough lift.

<p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555">There is still hope, though. Flapping wings provides both thrust and lift, so Toothless might be able to cruise at lower speeds if he flaps his wings. Toothless might have a lower body density (as many birds do) and subsequently might need less lift to stay afloat. I can think of more reasons, but I’ll leave those up to your critical thinking.

<p style="margin:0in;font-family:Calibri;font-size:12.0pt;color:#555555">All in all, this is just an estimation."