Talk:620: Wings

Cueball's physics has a mistake on this one (or at least assumes we've managed to heat the atmosphere of Titan to Earth's temperature). The temperature of Titan is roughly 1/3 the temperature of Earth on an absolute scale. Starting with the Ideal Gas Law, PV = NkT (k is Boltzmann's constant, N is # of molecules, P is pressure, V is volume, T is temperature), its easy to define the density of a gas, ρ as:

ρ = m/V = (m P)/(N k T) = P (m/N) / (k T)

Titan's atmosphere is 98.4% molecular nitrogen (N₂) and on Earth only 78.1% molecular nitrogen (by volume), but for simplicity we'll assume 100% for both. The weight of one molecule of Nitrogen is (m/N) ~ 2 × 14 × 1.67x10^-27 (kg/molecule) (there are 28 nucleons per molecule with a mass of about 1.67x10^-27 kg.

The pressure on Titan is P_Titan=146.7 kPa, and T_Titan = 93.7 K, while on Earth P_Earth=101.3 kPa and T_Earth = 287 K.

Plugging in numbers, we get ρ_Titan = 5.3 kg/m³ and ρ_Earth = 1.2 kg/m³ (note the measured surface density of air on Earth is 1.2 kg/m³ at Earth's mean temperature even without the simplifying assumption of 100% N₂).

Hence Titan's atmosphere is 4.4 = (5.3/1.2) times denser than Earth's (or 340% denser); not 50% denser as stated in the comic.

You will get the 50% denser if you assume the same planetary temperature on Titan as on Earth. Titan at 287 K would have a density of ρ_{Titan at 287K} ~ 1.73 kg/m³ which is about 50% greater than Earth's.

For the second calculation (panel 2), note lift is proportional to the density of air. If your action on Earth creates a lift of L₀ and you weigh W₀, on Titan you'd have a lift of 4.4 L₀ (Cueball calculated 1.5 L₀) due to the greater air density. Your weight would only be 0.14 W₀, due to Titan's lower surface gravity. If lift balances weight, you would be able to fly on Titan, that is if 4.4 L₀ = 0.14 W₀. That means to fly on Titan you need a lift on Earth of L₀ = 0.03 W₀, that is 3% of your Earth weight. Substituting Cueball's Titan density you would get the critical value from the comic: L₀ = 0.14 W₀/(1.5) = 9% W₀.

PS: I largely adapted this my writeup on xkcd forums from 2009 when the comic was made. Jimbob (talk) 05:44, 8 June 2013 (UTC)

That was the whole point of Blackhat's presence. He was there to make sure Rob (AKA Cueball) wasn't hurt.

Fortunately, Blackhat couldn't care less about the outcome. So he's got that going for him, which is nice. I used Google News BEFORE it was clickbait (talk) 01:21, 29 January 2015 (UTC)

Why is this not a thing you can do in amusement parks or places like that?141.101.104.44 17:00, 8 September 2016 (UTC)

I think that the bridge to which Cueball attached his wings is inspiration for the bridge in Click and Drag.--Obscure xkcd reference (talk) 13:47, 13 November 2021 (UTC)

Okay so maybe I just don't know enough about the physics, but... shouldn't the bricks be dangling? In the illustration, it looks like the rope should have a little slack in it. With nothing keeping it taught, I don't see how it would cancel out any of a person's weight. It also seems like you'd have to account for the friction of the pulleys, which I think would both increase the force required to get off the ground and slow the fall back down. Idk maybe I'm just overthinking this. 172.70.111.34 16:18, 5 October 2024 (UTC)

I agree completely. Conjuncts (talk) 04:03, 24 July 2025 (UTC)

It occurred to me that you could probably recreate this safely using rock climbing/belay equipment. Clip the flyer into the pulley system; on the other end, set up a harness weighing 91% that person's body weight. Also clip the harness to an active belayer. The belayer's job is to keep the rope only slightly slack, in case of sudden falls. However, you'll need to find a pulley with sufficiently low friction (although that problem is also present in Randall's original setup.) Conjuncts (talk) 04:03, 24 July 2025 (UTC)