Explain xkcd: It's 'cause you're dumb.
Title text: With reasonable assumptions about latitude and body shape, how much time might she gain them? Note: whatever the answer, sunrise always comes too soon. (Also, is it worth it if she throws up?)
Angular momentum is the force upon an object having a certain velocity while spinning. You may remember the certain strain when a spinning yoyo returned into your hand, giving it that much "slip" to discomfort you. The energy of that momentum does that. Angular momentum is also forced upon the Earth, as it is spinning 24 hours a day, 7 days a week. This 24/7 rotation enables us to have a clock. We say that the Earth is running "clockwise". But viewed from the north of the ecliptic, the Earth in fact turns counter-clockwise.
Nevertheless, Megan tries to work against this (massive) energy. She is spinning counter-clockwise, trying to generate energy to reduce the Earth's moving. Her move will force the Earth to slow down clockwise, but a stick figure just does not have the mass for a measurable effect. So, it's obvious that she will not be able to alter the turn of the Earth. But the romance part is also obvious, who wouldn't want to be longer with the one they truly love?
Randall states the obvious in the title text: While not being able to reverse time, enjoy your night time. Sunrise always comes too early.
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- [Cueball sits on his bed, looking at Megan who is spinning. It is night.]
- Cueball: What are you doing?
- Megan: Spinning counterclockwise
- Megan: Each turn robs the planet of angular momentum
- Megan: Slowing its spin by the tiniest bit
- Megan: Lengthening the night, pushing back the dawn
- Megan: Giving me a little more time here
- Megan: WITH YOU
The issue date is not given, as i don't have a clue about it. Could someone fix this? Rikthoff (talk) 19:30, 3 August 2012 (EDT)
- When the page was updated to the new comic template by User:Bpothier he fixed the date. lcarsos (talk) 20:48, 28 August 2012 (UTC)
That actually is a neat physics puzzle, which has probably (i.e. certainly) been addressed somewhere on the net. I may incorporate that some day. --Quicksilver (talk) 05:58, 24 August 2013 (UTC)
I tried to calculate the change in Earth's period, assuming that she was standing in the north pole (latitude = 90º N), where her spinning would have more effect. I either did something wrong, or my TI-84 Plus is not capable of detecting the very small effect her spinning would have on the Earth's rotation. I assumed the Earth had a period of exactly 24 hours, and got the same value to the second, even if she was spinning at 1000 turns per second, which seems like a lot.
Here's the formula:
L_Earth_i = L_Earth_f + L_spinner <=>
I_Earth * (2*PI)/T_Earth_i = I_Earth * (2*PI)/T_Earth_f + I_spinner* (2*PI) * f_spinner <=>
(1/T_Earth_f) = (1/T_Earth_i) - (I_spinner/I_Earth)*f_spinner <=>
T_Earth_f = 1/((1/T_Earth_i) - (I_spinner/I_Earth)*f_spinner)
Where the variables have names in the format:
[variable name]_[object it refers to]_[situation (i or f stand for initial and final)]
L = Angular Moment
I = Moment of Inertia
T = Period of rotation about one's axis
f = frequency
I used as values:
T_Earth_i = 86400 seconds (24 hours exactly)
I_spinner = 62,04 Kg.m^2 (Found on Wolfram|Alpha, for a 62Kg adult human being)
I_Earth = 8,03e+37 Kg.m^2 (http://scienceworld.wolfram.com/physics/MomentofInertiaEarth.html)
f_spinner = the frequency of the woman's spinning in complete turns per second. 188.8.131.52 (talk) (please sign your comments with ~~~~)
- Taking that a bit further, the relative decrease is:
(T_Earth_f - T_Earth_i)/T_Earth_i = 1 / (I_Earth/(I_spinner*T_Earth_i*f_spinner) - 1)
= 1 / ( 1.5 e+28 - 1) ~= 67 e-30
- Fwiw, the absolute value is 5.767 yocto-seconds. If the entire world population would spin at that 1000 turns per second (and at favourable locations as in your assumptions), the effect will still be a measly 0.041 pico-seconds. So T_Earth_f = 86 399.999 999 999 999 958 ... But the TI-84 only has about 14 digits precision, i believe, so even that won't show up. -- 184.108.40.206 22:46, 30 October 2013 (UTC)
Is it possible for someone to write an equation that factors in latitude (and, if relevant, longitude) that we could plug our locations into and get a value from? That would be awesome. Thanks. 220.127.116.11
02:48, 23 February 2014 (UTC)