Supplementary material

Supplementary material for: 

Comment by user "filtherton", from here:

    "Okay, so assuming the person is in the middle of nowhere, naked, and that 3K is the blackbody temperature of the surrounding universe, with some simplifications one could apply the following equation:

    Q = S*A*E_s(T_b^4 - T_s^4)


    S is the Stephan Boltzmann constant, 5.67x10^-8 W/m^2-K^4

    A is the surface area of the person's naked body, 1.8m^2, roughly

    E_s is the emmisivity of human skin, 0.97

    T_b is the temperature of the body in K, here ~310K (I have assumed that the body temp is uniform)

    T_s is the temperature of space in K, 3K

    Q is about 900 Watts. The typical metabolic heat generation for a sleeping person with 1.8m^s worth of skin is 40W/m^2 is 108 W. This means that your net heat loss would be ~800 W.

    Assuming the the specific heat of the human body is approximately the same as the specific heat of water, (C_p = 4187 J/kg-K), a 200 lb person (mass = 90 kg), and that we're only concerned with temperatures above 0 C, you could figure out how fast your temperature would be dropping using the following equation:

    dT/dt = Q/(mass*C_p)

    dT/dt is about 2 thousandths of a degree Celcius per second. So if your body was uniformly 310K, initially you'd be losing a degree every 8 minutes. It would take about 40 minutes of naked stillness to reach stage 3 hypothermia.

    There are probably errors in this analysis. And even if error free, it is pretty approximate. If you were behind the earth you'd be getting some heat from the earth, so you'd take a little longer to freeze to death."


Assumed resting body heat is 108W.   With rigorous exercise this can go up to 1000 W.  Mass = 150kg, area of astronaut = 3m2,  heat capacity = 4187, emissivity = 0.9,  start temp = 37C, temp of space = 3K, Stefan-Boltzmann = 5.67x10^-8, watts output = 1414, watts input (body heat) = 108.  Assumes perfect conduction of internal heat to outside of suit.


Assuming half the watts from the plate is received by the shaded plate, using 0.5^(1/4) = 84%.  If the plate absorbs EMR and re-emits it on both sides of the plate equally, the shaded plate gets roughly half the energy emitted by the sunlit plate.

Using S-Boltzmann T^4, take 4th root of a ratio of 1/2 and you end up with 84%.  84% of the sunlit's plate's temperature to equals half the watts output.


Index for Fallacy series:


Index for space reentry vehicles:

 space reentry vehicles part 1.
  • ablation and the detached shock wave
space reentry vehicles part 2.

  • aerodynamics

 space reentry vehicles part 3.
  • ICBMs and the cold war
space reentry vehicles part 4.

  • more on ablation

 space reentry vehicles part 5.

  • more on the detached shock wave

space reentry vehicles part 6.

  • control

space reentry vehicles part 7.
  • pressure

Harries et al 2001

Increases in greenhouse forcing
inferred from the outgoing longwave
radiation spectra of the Earth in
1970 and 1997

link mirror

Harries Brindley 2003

Observations of the Infrared Outgoing Spectrum of the Earth from Space:
The Effects of Temporal and Spatial Sampling

link mirror

Griggs, J. A., & Harries, J. E. (2004)

Comparison of spectrally resolved
outgoing longwave data between 1970 and present.

link mirror

Griggs, J. A., & Harries, J. E. (2005)

Comparison of spectrally resolved outgoing longwave
radiation between 1970 and 2003: The ν4 band of methane

link mirror

Griggs, Harries 2007

Comparison of Spectrally Resolved Outgoing Longwave Radiation over the Tropical
Pacific between 1970 and 2003 Using IRIS, IMG, and AIRS

link mirror

Chen, Harries et al 2007

Spectral signatures of climate change in the Earth’s infrared
spectrum between 1970 and 2006

link mirror

Harries et al 2001 corrected graphs:

No comments:

Post a Comment