Jul 30, 2010

Marathon run

(reduced to figures)

1896 Marathon start
Despite the unrealistic belief of Pheidippides route from Marathon battlefield to Athens in 490BC that supports its legend, marathon run has written its own history. Neither the first modern Olympic games in 1896 were actually the first, as they had already been held since 1859. But the marathon race event did start in 1896 in Greece. Though the current distance of 42.195 Km was agreed in memory of Dorando Pietri's race in 1908 Olympics, that first one was originally of 40 Km, and then Spiridon Louis its first ever finisher, not Pheidippides, in 2h58'50".

Your performance in such a race is determined by concepts like VO2max or the lactate threshold.

1. VO2max is the aerobic (moderate intensity for long intervals) capacity, the maximum amount of oxygen you can consume, in ml/Kg body weight/min. That is, as you make more effort, your body increases its need of oxygen, but it just can hold up to an upper limit.

After training your body regenerates muscles and the number of cells increase, thus raising your rate of oxygen uptake level. For instance interval training.

You can estimate your VO2max peak according to Jack Daniels and Jimmy Gilbert formula (Oxygen Power. Performance Tables for Distance Runners, 1979):

VO2max = (-4.60 + 0.182258 * velocity + 0.000104 * velocity^2) / (0.8 + 0.1894393 * e^(-0.012778 * time) + 0.2989558 * e^(-0.1932605 * time))

where velocity is in meters/minute, and time in minutes as well. For instance, Spiridon Louis VO2max, according to his 40 Km marathon finish time, would be at 50.5 ml/Kg/min.
In 1979, Jack Daniels and Jimmy Gilbert published "Oxygen Power. Performance Tables for Distance Runners". This series of tables predicted all-out racing times for virtually every racing distance. Each performance time in the table is related to a VO2max index, called VDOT. The tables were generated using two regression equations: (1) relating oxygen consumption with velocity, and (2) predicting the amount of time one can run at a given percentage of VO2max. By combining these two equations, substituting VDOT indices, and looking for convergence for Newton-Raphson curve fitting analysis, one can then mathematically match up a predictable racing time expected at a given distance for someone having a particular VDOT index. The validity of these tables is strongly supported by looking at the known VO2max scores of some world record holders and their respective record times.

Inversely, from a known VO2max, racing and training paces can be infered.

2. Lactate threshold is the anaerobic (high intensity in a short interval) level at which it (lactate, or lactic acid) is faster produced by muscles than metabolized, thus starts accumulating in the blood. It is usually between 90-95% of your maximum heart rate. It can also be increased with the appropriate workout, like farleks.

Better training will improve your limits. And that also means better programmed, plus a balanced diet.

Slower run paces (i.e. at 75% of your speed in VO2max) metabolize fat better than faster ones. Quick efforts need quicker energy availability, that is glycogens. Glycogen produces glucose, which reacting with oxygen produces carbon dioxide plus water, and energy. When no more carbohydrates are available, fat is metabolized instead. A slower process, driving to decreasing performance.

Only a small proportion of your workout should be run at fast paces. Anaerobic does not use oxigen, therefore it is less efficient, and performance decreases faster.

Panathinaikon stadium

Jul 2, 2010

Greenhouse effect

A rough numerical estimation

If we consider the Total Solar Irradiance (radiation intensity) reaching Earth as an averaged 1365.5 [Watts/m2] (known as the Solar constant), according to the Stefan-Boltzmann law (that is infered from Planck's law of spectral radiance), a "black-bodied" (absorbs all incoming radiation) Sun's emittance (energy flowing per unit surface area and time) depends on its temperature at a epsilon·rho·T^4 rate, where epsilon=1 for black bodies, and rho states for the Stefan constant of proportionality, whose value is 5.67·10^-8 [W/m2·K4].

Thus, the total energy radiated (radiation power) by the Sun is Wt = Ws · (4·pi·radius^2), being Ws the Stefan-Boltzmann law applied on the surface of Sun. As an isotropic radiation (does not depend on direction), at Earth's distance Wt = We · (4·pi·distance^2), being We the solar irradiance measured on Earth (decreases with distance).

Combining both, the Sun's temperature is derived from Ws = rho·Ts^4 = We · (distance/radius)^2, then Ts = (We/rho)^0.25 · (distance/radius)^0.5 = (1365.5 [W/m2] /5.67·10^-8 [W/m2·K4])^0.25 · (149.5978707M [Km] / 696K [Km])^0.5 = 5775.44 [Kelvin] (actually 5780 K, or 5507º Centigrades), where the Sun-Earth distance is approx. 150 million Kms (that is 1 AU, astronomical unit), and Sun's radius is around 700 thousand Kms.

Wien's displacement law denotes that temperature as a function of the shorter wavelength (lambda) at which radiation is mostly emitted: lambda · T = 0.2898 [cm·K]. The Sun radiates its peak at lambda close to 500 nm (10E-9 m), that corresponds to the yellow. Graphically:

In order to deduce Earth's temperature, we must take into account its albedo, or rate of reflected radiation (what makes Earth visible from space), due to atmosphere (mostly clouds 23%, and air 4%), and Earth's surface (snow, sea, 4%), that may average to 30% as a whole.

Furthermore, Earth's surface average insolation (incoming solar radiation) is approximately Ws/4, averaged. This can be explained by the projection of Earth's "spheric" surface (4·pi·radius^2) as a circle of area pi·radius^2. Because any unit of area, due to Earth's rotation, is not always being insolated. At any time, half the Earth is on the dark side. And the rest depends on the direction of the incident radiation: that is, when they are perpendicular, you get the whole radiation; in any angle, you just get the cosine component (Lambertian cosine law).

The solid angle is then integrated as a function of sin(theta)·d(theta) from 0 to pi/2 (half insolated sphere); do not forget to multiply it by cos(theta), and by d(phi) from 0 to 2·pi (the whole circumference). The latter is just 2·pi; and the former (considering u=sin(theta) and du=cos(theta)·d(theta)) sums 1/2; thus totals 2·pi/2 = pi. So finally get the same 1/4 ratio, as the whole sphere subtends a 4·pi angle.

Then, Earth's temperature might be deducted from the former black body relationship Te = ((1-albedo)·(Ws/4) / rho)^0.25 = ((1-30%)·(1365.5/4) [W/m2] / 5.67·10^-8 [W/m2·K4])^0.25 = 254.79 [Kelvin] = -18.21º [C]. But that's obviously untrue, because of Greenhouse effect.

Incoming solar radiation energy is 95% in the 300 to 3000 nm wavelength range: 10% in the UV (ultraviolet, 300-400 nm); 40% in the visible (400-700 nm), and 50% in the infrared (700-3000 nm). Along with the albedo, not all incoming radiation is transmitted through atmosphere to the ground because of:

- Clouds (3%) and atmosphere (21%) partially absorb it, then converted into heat and causing the emission of their own radiation in all directions. Basically shortwaves are by H2O vapour, CO2 carbon dioxide, methane CH4, nitrous oxide N2O; and ultraviolets mostly by ozone O3, and nitrogen N2 and oxygen O2; also aerosols CFCs contribute. These are the main Greenhouse gases.

- Scattering takes place when small particles and gas molecules diffuse part of the incoming solar radiation in random directions, even back to space; this causes the sky bluish color (Rayleigh's dependance on wavelength^-4) and white clouds (Mie's on wavelength^-1).

Earth emittance is 90% in the range from 3 to 30 microm, with its peak around 10 microm. It corresponds to a black body temperature around 300 K, close to the 288 K average surface temperature.

One step further, if Earth emits energy at a T'e=288 K temperature, then yields an emissivity epsilon = (1-albedo)·(Ws/4) / rho·T'e^4 = (1-30%)·(1365.5/4) [W/m2] / 5.67·10^-8 [W/m2·K4] · 288^4 [K4] = 0.61.

Over half of Earth’s emittance lies in the infrared range from 8 to 14 microm. Without this window, the Earth would become too warm to support life.

Let's build a very simplified model to outline this temperature increase.

If we consider Earth as a black body (thus ideal emissivity = 1), and we close that infrared window, then when thermodynamical equilibrium is again reached, power emitted per unit surface outside is all but the selected window: Pem = Integral[ Radiance(frequency, Temperature) dfreq ] avoiding the infrared range.

The power absorbed per unit surface Pab will be the sum of the known total black body radiance rho·Te^4, plus the Integral[ Radiance(frequency, Temperature) dfreq ] over the window range.

Given that in thermal equilibrium net energy flux must be null, Pem = Pab :

Int[ Rad(freq,Te') dfreq ]{freq2..infinite} + Int[ Rad(freq,Te') dfreq ]{0..freq1} = rho·Te^4 + Int[ Rad(freq,Te') dfreq ]{freq1..freq2}

where Te' is the new equilibrium temperature. We can express this in terms of total radiance :

Int[ Rad(freq,Te') dfreq ]{0..infinite} - Int[ Rad(freq,Te') dfreq ]{freq1..freq2} = rho·Te'^4 - Int[ Rad(freq,Te') dfreq ]{freq1..freq2} = rho·Te^4 + Int[ Rad(freq,Te') dfreq ]{freq1..freq2}

So :

rho·Te'^4 - rho·Te^4 = 2 · Int[ Rad(freq,Te') dfreq ]{freq1..freq2}

Solving the integral :

Int[ Rad(freq,Te') dfreq ]{freq1..freq2} = Int[ (c/4) · (8·pi·freq^2·Kb·Te' / c^3) dfreq ]{freq1..freq2}

here Kb refers to the Boltzmann constant 1.38·10^-23 [J/K].

We are using the Rayleigh-Jeans expression instead of Planck's to afford easier calculations. But take into account that this approximation is only valid for long wavelengths, due to this curve tends to infinite as wavelengths become shorter. Planck's, instead, falls exponentially at short wavelengths.

The c/4 factor is because Planck's (and Rayleigh-Jeans) law stands for density (unit volume) of radiant energy, but what is actually measured is radiant emittance or spectral radiance (per unit surface), which depends as well on speed of outgoing radiation.

Finally :

Int[ (c/4) · (8·pi·freq^2·Kb·Te' / c^3) dfreq ]{freq1..freq2} = 2·pi·(freq2-freq1)^3·Kb·Te' / 3·c^2

Thus :

rho·Te'^4 - rho·Te^4 = 2 · [ 2·pi·(freq2-freq1)^3·Kb·Te' / 3·c^2 ] = 4·pi·(freq2-freq1)^3·Kb·Te' / 3·c^2

Rearranging the latter expression after dividing it by rho·Te^4 :

(Te'/Te)^4 = 1 + 4·pi·(freq2-freq1)^3·Kb·Te' / 3·c^2·rho·Te^4 = 1 + ( (4·pi·(freq2-freq1)^3·Kb / 3·c^2·rho·Te^3) · (Te'/Te) ) = 1 + Constant · (Te'/Te)

because Constant is adimensional, and its value is :

Constant = 4·pi·(freq2-freq1)^3·Kb / 3·c^2·rho·Te^3 = 4·pi·((c/8 [microm])-(c/14 [microm]))^Kb·Te / 3·c^2·((1-albedo)·(Ws/4)) = 4·pi·(3.75·10^13-2.14·10^13)^3·1.38·10^-23 [J/K]·254.79 [K] / 3·(299792458 [m/s])^2·238.9625 [W/m2] = 2.841

Constant's value is "almost" small, so we "can" approximate :

(Te'/Te) = (1 + Constant·(Te'/Te))^(1/4) ~ 1 + (1/4)·Constant·(Te'/Te)

Then :

(1 - (1/4)·Constant) · (Te'/Te) ~ 1 ; Te' ~ (1 - (1/4)·Constant)^-1 · Te = (1 / (1 - (1/4)·Constant)) · Te

Here we will make a new approximation, due to the mentioned trend to infinite for short wavelengths. A geometric series can be  expressed as Sum[ c·r^i ]{i=0..infinite} = c / (1-r), being r the ratio. In our case c = 1 and r = (1/4)·Constant. We take just the first terms of the series and discard from the quadratic term onwards, and :

Te' ~ (1 + (1/4)·Constant) · Te = 435.75 [Kelvin]

That way, Earth's temperature would approximately be increased up to 162.75º Centigrades.