There’s very few things to like about How To Train Your Dragon: The Hidden World, but that doesn’t mean they exist. In fact, the lift at the end of the movie is legitimately my favourite thing about it (and I feel a bit bored today). This, of course, means we’re gonna analyze it a bit.
Do you even lift?
We’re gonna pretend that we’re in high school again and that things like ‘friction’ and ‘drag’ don’t exist where we don’t need them to simplify our calculations.
The first question is: how much power do those windmills produce? To start things off, let’s re-cap some already known calculations that should put us somewhere in the ballpark.
I’m not gonna go into this level of detail today, I’ll just do the eyeballing. If you take the right lowermost blade of the big windmill and measure its length, you’ll get about 110px. Using same conversion factor as we did for the viking, we get about 6.6m for the long side of the blade on average. The width of the blade seems to be about 30px, or about viking-sized — and we’re running off the 180cm estimate for that. That gives us 11.88 m² for each blade. Since there’s 8 of them, that brings our total surface to about 95 m².
But surface area is only one of the things to see how much we could lift by the power generated by said windmill: the other two are wind speed and density. For density, we’ll just take the density at 2000m from engineering toolbox, which comes out at 1 kg/m³.
Wind speed is trickier. In the movie, wind speed doesn’t exist at all, so we’ll have to use our own. 5 m/s seems to be fairly reasonable constant speed. 10 m/s feels slightly generous, but given that the island sticks two to three and a half kilometers out of the ocean … we’ll allow it. 15 m/s would already get into risky conditions, probably and is almost ridiculous. We’ll do all of these three.
Fortunately for us, some people have already made online calculators that calculate wind power based on our parameters. We’ll take this one because it has units that we want. Results are out: ~6 kW for 5m/s, 47.5 kW for 10 m/s, ~160 kW for 15 m/s.
Boy these numbers surely rise up fast, don’t they? But while we’re cheering at those very high numbers, Albert Betz is standing in the corner with a baseball bat, ready to beat some harsh reality in our results. Turns out that you can only turn 59% of that wind energy into something else, with modern turbines being only ~40% efficient. Berk isn’t a modern wind turbine, though — it’s more like those Dutch windmills. Internet says ~15% or less for those, so we’ll just operate with 10% just because the Dutch know their shit better than Berk.
That leaves us with … 600W, 4.75 kW and 16 kW.
But that’s just the biggest windmills. What about the second and the third one?
The second windmill (#2) probably doesn’t produce anything even under full load. The blades are parallel to the wind. The third one is just a smaller version of the first. Eyeballs say blades are ~8 m² (6×2), and there’s 6 of them. This gives us a surface area of ~48 m², and power outputs of 360W, 2.9 kW and 9.7 kW.
Combined outputs are 960W, 7.65 kW and 25.7kw.
If we factored losses from pulleys that run the lift, rather than just the losses in the windmill portion of the lift, this amount of power probably couldn’t lift shit.
Speed of the lift
See this speeds?
It’s nowhere this fast.
To brush up on high school physics, everything that you lift has a potential energy. The amount of potential energy depends on the mass of the object, how far above the ground it is and the acceleration of the free fall:
U = m*g*h
Free fall acceleration is 9.8 m/s², height is 2-3.35 km (note: ‘2 km’ number came up after the linked post was written, but I haven’t written the third revision quite yet). We’ll be using 2 km and 3.2 km for the calculations.
And now we get to the weight. We’ll assume that the lift has to lift a standard viking warship, about ten warriors and about their weight in cargo. For weight of a viking ship, we’re surely in luck. After all, Draken Harald Hårfagre — a modern reconstruction of a viking ship, constructed using traditional materials and somewhat traditional methods — is a thing that exists. With displacement of 95 tons (presumably empty), it’s pretty heavy.
It’s also 35m long, which the ships in HTTYD aren’t. Eyeballing the lift, those ships appear to be more like 15 meters long. Maybe 20m, but even 15 might be a stretch. We’ll assume uniform scale in all three dimensions. Scaling down to 15 m gives us the weight of about 8 tons. The 10 warriors at the average ideal weight of 75kg and their weight in cargo add up another tonne and a half. We’ll round the ship weight to 10 tons.
And then there are the ropes. Given this is middle ages, they’re probably using hempen ropes. And boy do we have data for that. Breaking strengths, safe load factors, weight per meter. Long story short, it quickly turns out that 2-3 km is too far for a single run of rope as its own weight blows the standard safe load factor in under a kilometer. However, even at three kilometers (and two hundred meters) we don’t blow the minimum breaking strength yet. While transfer stations would be preferred, we’ll just use a single rope and a meager safety factor instead.
Two kilometer run of 2″ rope would weight about three tons. 3200m run of rope would weight about 4.9 tons. Load at the top of the rope would be thus ~29.8 kN and ~48 kN, respectively. Minimum breaking strength is 120 kN. We want to maintain a safety factor of a measly 2, which gives us 60 kN. Rope takes out 29.8-48 kN of our 60 kN capacity, which leaves us with 30-12 kN to work with.
10 tons is 98 kN, which means we need 4-10 (It’s actually 9, but we’ll round up to keep rope counts on both ends of the ship the same).
This gives us another 12-49 tons that we have to lift. However, since weight of the rope changes with how far up the lift is, so does the weight we have to lift. In the end, ropes effectively add only 6-24.5 tons to the mass we have to lift.
Potential energy at the top of the lift comes out to be anywhere between ~315 MJ (2km) to ~1.1 GJ (3.2 km).
Energy is also power over time. If we divide the energy we need to put in with the rate at which we’re putting it in (so, power), we get the amount of time for which we need to keep putting the power in.
So let’s see how the wind speed translates into lift speed.
| One lift, windmill #1 | ||
|---|---|---|
| Wind speed | 2 km | 3.2 km |
| 5 m/s | 6 days (14 m/h) | 21d 5h (6.3 m/h) |
| 10 m/s | 18h 24m (1.8 m/min) | 2d 15h (50 m/h) |
| 15 m/s | 5h 30m (6 m/min) | 19h 15m (2.8 m/min) |
| One lift, windmill #3 | ||
| Wind speed | 2 km | 3.2 km |
| 5 m/s | 10 days (8.3 m/h) | 35 days (3.8 m/h) |
| 10 m/s | 1d 6h (66.6 m/h) | 4d 9h (30 m/h) |
| 15 m/s | 9h (3.7 m/min) | 1d 7h 30m (1.69 m/min) |
| One lift, both windmills | ||
| Wind speed | 2 km | 3.2 km |
| 5 m/s | 3d 19h (21 m/h) | 13d 6h 30m (10 m/h) |
| 10 m/s | 11h 25m (2.9 m/min) | 1d 15h 30m (1.35 m/min) |
| 15 m/s | 3h 24m (9.8 m/min) | 12h (4 m/min) |
| Both lifts, both windmills | ||
| Wind speed | 2 km | 3.2 km |
| 5 m/s | 7d 14h (11 m/h) | 26d 13h (5 m/h) |
| 10 m/s | 22h 50m (1.4 m/min) | 3d 7h (40 m/h) |
| 15 m/s | 6h 24m (5.2 m/min) | 1d (2.22 m/min) |
If you convert those numbers to actual speed, you’ll quickly notice that this is borderline snails pace. And mind: those are the average speeds. Speeds are going to change depending on the rope length.
Of course, that assumes they don’t use counterweights (and in the movie, it’s clear that they don’t) and that they don’t have transfer stations (cannot be determined from movie). Just by adding counterweights (about as heavy as the ship), you could drastically reduce the time you need for a trip pretty drastically — and by adding transfer stations (hopefully located near a waterfall for free water power) you could also reduce the weight associated with ropes.
But as things are in the movie … well. Better take the stairs.