Talk:620: Wings

Explain xkcd: It's 'cause you're dumb.
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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 PTitan=146.7 kPa, and TTitan = 93.7 K, while on Earth PEarth=101.3 kPa and TEarth = 287 K.

Plugging in numbers, we get ρTitan = 5.3 kg/m3 and ρEarth = 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 ρTitan 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? 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)