# 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 (N_{2}) 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/m^{3} and ρ_{Earth} = 1.2 kg/m^{3} (note the measured surface density of air on Earth is 1.2 kg/m^{3} at Earth's mean temperature even without the simplifying assumption of 100% N_{2}).

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/m^{3} 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 L_{0} and you weigh W_{0}, on Titan you'd have a lift of 4.4 L_{0} (Cueball calculated 1.5 L_{0}) due to the greater air density. Your weight would only be 0.14 W_{0}, due to Titan's lower surface gravity. If lift balances weight, you would be able to fly on Titan, that is if 4.4 L_{0} = 0.14 W_{0}. That means to fly on Titan you need a lift on Earth of L_{0} = 0.03 W_{0}, that is 3% of your Earth weight. Substituting Cueball's Titan density you would get the critical value from the comic: L_{0} = 0.14 W_{0}/(1.5) = 9% W_{0}.

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)