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Posers and Puzzles

Posers and Puzzles

  1. Subscriber sonhouse
    Fast and Curious
    06 Mar '08 01:10 / 1 edit
    Carl Sagan is standing under a 1000 watt output streetlight on a pole 56 feet high. Carl is 6 feet high. Being an astronomer, he wonders how far away from the sun you would have to be for the sun to have the same illumination as Carl receives from that lamp.
    Who can come up with the easiest way to figure it out?
  2. 06 Mar '08 01:43
    Originally posted by sonhouse
    Carl Sagan is standing under a 1000 watt output streetlight on a pole 56 feet high. Carl is 6 feet high. Being an astronomer, he wonders how far away from the sun you would have to be for the sun to have the same illumination as Carl receives from that lamp.
    Who can come up with the easiest way to figure it out?
    by energy flux.
    Energy flux from the sun in Earth's surface is 1300 W/m^2 . And sun is 1.5e8 Km from Earth.
    Find the flux from the lamp (with solid angles -> a sphere has an angle of 4*Pi) at the given distance, and voila. (the flux coming from the lamp and the flux of a sphere with radius = 50 feet is the same - just find out the proportion between areas, and compare to the sun)
  3. Subscriber sonhouse
    Fast and Curious
    06 Mar '08 02:01
    Originally posted by serigado
    by energy flux.
    Energy flux from the sun in Earth's surface is 1300 W/m^2 . And sun is 1.5e8 Km from Earth.
    Find the flux from the lamp (with solid angles -> a sphere has an angle of 4*Pi) at the given distance, and voila. (the flux coming from the lamp and the flux of a sphere with radius = 50 feet is the same - just find out the proportion between areas, and compare to the sun)
    So give me an answer. BTW, here is a jennings constant: suppose you don't remember how to calculate the surface area of a sphere, take the surface area of a cube the same diameter as the sphere and multiply it by 0.52333333.
  4. 06 Mar '08 02:35 / 1 edit
    Originally posted by sonhouse
    So give me an answer. BTW, here is a jennings constant: suppose you don't remember how to calculate the surface area of a sphere, take the surface area of a cube the same diameter as the sphere and multiply it by 0.52333333.
    the flux is power / area
    for the lamp 1000 / (4*Pi*r^2)
    for the sun 3.9e26 / 4*Pi*r'^2)
    (sun is like a 3.9e26 W lamp - if you multiply 1300W by 4*Pi*1.5e11 squared, you get the result - approximately)
    Now... how far do you have to be from the sun the get the same flux per square meter of a lamp 50 feet away?
  5. 06 Mar '08 02:38
    Originally posted by sonhouse
    So give me an answer. BTW, here is a jennings constant: suppose you don't remember how to calculate the surface area of a sphere, take the surface area of a cube the same diameter as the sphere and multiply it by 0.52333333.
    it's easier to know the area of a sphere is 4*Pi*r^2 then memorize that strange number.
  6. 06 Mar '08 03:42
    Originally posted by sonhouse
    ...take the surface area of a cube the same diameter as the sphere ...
    Cubes don't have diameters.
  7. Subscriber sonhouse
    Fast and Curious
    06 Mar '08 10:06
    Originally posted by Gastel
    Cubes don't have diameters.
    Side length then. The same kind of thing can be used in 2 dimensions, like you don't want to do PI R squared,etc for the area of a circle, do a square the same size and multiply by .78, gives the area of a circle. Anyway, nobody has given an answer yet!
    The idea of the surface area of a sphere was the way I did it, that and knowing a couple of facts about light propagation makes it a pretty easy problem.
  8. 06 Mar '08 12:24
    Originally posted by sonhouse
    Side length then. The same kind of thing can be used in 2 dimensions, like you don't want to do PI R squared,etc for the area of a circle, do a square the same size and multiply by .78, gives the area of a circle. Anyway, nobody has given an answer yet!
    The idea of the surface area of a sphere was the way I did it, that and knowing a couple of facts about light propagation makes it a pretty easy problem.
    The answer is just solve the ultra simple system I gave you. Doing by head the calculations, you should be around Mars to get the same energy as the guy under the lamp.
    Unfortunately I didn't have the patience to investigate how many meters 50 feet are.
  9. Standard member wolfgang59
    Infidel
    06 Mar '08 13:20
    50 feet = 15 metres is a pretty accurate approximation.
  10. 06 Mar '08 13:26
    Originally posted by wolfgang59
    50 feet = 15 metres is a pretty accurate approximation.
    either way, the answer is 6.3 e11 * R, where R is the distance to the lamp ( if I got the calculations right)
  11. Subscriber sonhouse
    Fast and Curious
    06 Mar '08 22:05
    Originally posted by serigado
    either way, the answer is 6.3 e11 * R, where R is the distance to the lamp ( if I got the calculations right)
    How I did it: 1000 watts, 50 feet or 15 meters, in feet, assign 50 to r, then 4 PI r^2 or 12.566 * 2500 = 31,415.92654 square feet, the surface area of a sphere 50 feet in diameter. Divide by 1000 you get 31.4 and change square feet per watt, invert that to get .0318 watts per square foot. 126 watts per square foot of solar energy hits the earth. That is about 3958 times the illumination from that streetlight per square foot. Since the power goes down as the square of the distance, just take the square root of 3958 which is about 62.9. That represents the AU you have to be away from the sun for solar radiation to = 0.031 watts per square foot. 62.9 times 93E6 miles is about 5.85 billion miles away from the sun. Times 1.6= ~9.36 billion kilometers away from the sun. Billions and Billions indeed!