How many lux or lumens is the Sun?

Just wondering.

Anyone knows?

Easy

Once we know how bright the sun is in lux directly overhead, we can use that number to figure the lumens (since the sun radiates more or less equally in all directions.

Digging around, I found that bright sunlight is 50K-100K lux. Let's go halvies on that, and say 75K lux.

That answers one of your questions.

Now, lux is defined as lumens per square meter. So, we are measuring 75,000 lumens per square meter at the Earth's surface. All we have to do is find the number of square meters on the surface of a sphere with a radius equal to the Earth's distance from the sun.

The Earth is about 150 million km from the sun (150 billion meters). The formula for the area of a sphere of radius r is 4(pi)r^2.

Thus, the sphere has a surface area of 4(pi)(150x10^9)^2 = 2.83*10^23 square meters.

Multiply that by the lux value from before (75,000 lux = 75,000 lumens/m^2) and we get 2.12x10^28 lumens.

me not understand /ubbthreads/images/graemlins/thinking.gif /ubbthreads/images/graemlins/confused.gif /ubbthreads/images/graemlins/icon3.gif
/ubbthreads/images/graemlins/bowdown.gif /ubbthreads/images/graemlins/bowdown.gif /ubbthreads/images/graemlins/bowdown.gif

LOL!

OK. How about lumens at the surface of the sun!!!

Easy enough...

The sun's diameter is 1.4 million km, so the radius is 700,000 KM (or 700 million meters). Taking the above formula for surface of a sphere, and the sun's surface area is 6.2x10^18 meters.

Dividing the above luminous output by the surface area gives 3.4 billion lumens per square meter.

... and still not taking into account the losses in earth's atmosphere ... /ubbthreads/images/graemlins/grin.gif

For those who don't know... lol

2.12x10^28 lumens

this is scientific notation... /ubbthreads/images/graemlins/smile.gif an easier way to express really large or really small numbers..

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Negeltu said:
For those who don't know... lol

2.12x10^28 lumens

this is scientific notation... /ubbthreads/images/graemlins/smile.gif an easier way to express really large or really small numbers..

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2.12x10^28 lumens aka 21,200,000,000,000,000,000,000,000,000 lumens! /ubbthreads/images/graemlins/faint.gif

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Kiessling said:
... and still not taking into account the losses in earth's atmosphere ... /ubbthreads/images/graemlins/grin.gif

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True - how much loss would you say it accounts for? I'm sure that my calculation is well within one order of magnitude.

absolutely no idea, I only wanted to add a smart-*** reply /ubbthreads/images/graemlins/grin.gif /ubbthreads/images/graemlins/grin.gif
anyway, it is pretty impressing, this little yellow thing that hangs in the sky... /ubbthreads/images/graemlins/smile.gif ... a fusion lamp ...
cool math evan! /ubbthreads/images/graemlins/thumbsup.gif
bernhard

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evan9162 said:
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Kiessling said:
... and still not taking into account the losses in earth's atmosphere ... /ubbthreads/images/graemlins/grin.gif

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True - how much loss would you say it accounts for? I'm sure that my calculation is well within one order of magnitude.

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I would say *well* within an order of magnitude. I have a referance that addresses the atmospheric absorbtion question. As you would expect, it varies a bit across the spectrum but it appears that the incoming flux is about 55% higher above the atmosphere than at sea level with the sun overhead.

Yep! impressive /ubbthreads/images/graemlins/thumbsup.gif

Thanks for the info, Doug.

So taking into account atmospheric losses, we can round and say around 3x10^28 lumens.

Zero Lux, but only at night!

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Doug S said:
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evan9162 said:
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Kiessling said:
... and still not taking into account the losses in earth's atmosphere ... /ubbthreads/images/graemlins/grin.gif

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True - how much loss would you say it accounts for? I'm sure that my calculation is well within one order of magnitude.

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I would say *well* within an order of magnitude. I have a referance that addresses the atmospheric absorbtion question. As you would expect, it varies a bit across the spectrum but it appears that the incoming flux is about 55% higher above the atmosphere than at sea level with the sun overhead.

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As I recall from some photoelectric work, at sea level, when the sun is directly overhead, you get about 1 kilowatt per square meter of irradiation. In space, as far away from the sun as earth orbits you get about 1.7KW. So 55% is certainly close enough. Obviously, this varies with clouds, smoke, smog, fog, dust, humidity and don't forget ALTITUDE.

To get an accurate lumen value, one would have to construct an integrating sphere in space around the sun (a Dyson sphere with a pure flat white interior), taking into account the planets and other objects in orbit around the sun. Until that happens though, one would have to extrapolate the lumen value based upon a very small area of flat illuminated surface from a known distance from the sun.

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ESD said:
As I recall from some photoelectric work, at sea level, when the sun is directly overhead, you get about 1 kilowatt per square meter of irradiation. In space, as far away from the sun as earth orbits you get about 1.7KW. So 55% is certainly close enough. Obviously, this varies with clouds, smoke, smog, fog, dust, humidity and don't forget ALTITUDE.

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I believe that your 1KW and 1.7KW figures apply to total solar radiation. Since the original question was about lux and lumens which apply only to light, not total radiation, I looked at only the 400-700nm portion of the solar spectrum to get that 55% figure.

For the Lux measurement, you want to know the value about which distance? At one meter /ubbthreads/images/graemlins/grinser2.gif
or such as read by a luxmeter on earth /ubbthreads/images/graemlins/yellowlaugh.gif

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evan9162 said:

Easy

Once we know how bright the sun is in lux directly overhead, we can use that number to figure the lumens (since the sun radiates more or less equally in all directions.

Digging around, I found that bright sunlight is 50K-100K lux. Let's go halvies on that, and say 75K lux.

That answers one of your questions.

Now, lux is defined as lumens per square meter. So, we are measuring 75,000 lumens per square meter at the Earth's surface. All we have to do is find the number of square meters on the surface of a sphere with a radius equal to the Earth's distance from the sun.

The Earth is about 150 million km from the sun (150 billion meters). The formula for the area of a sphere of radius r is 4(pi)r^2.

Thus, the sphere has a surface area of 4(pi)(150x10^9)^2 = 2.83*10^23 square meters.

Multiply that by the lux value from before (75,000 lux = 75,000 lumens/m^2) and we get 2.12x10^28 lumens.

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Yes....but,

Digging around some more, for what is more applicable to us down on earth, and more specifically how we rate incident light from the flashlights we use, for comparison; noon daylight reaching the surface of the earth at the equator(relatively constant throughout the year), after being filtered by the atmosphere we get around only 10,000-12,000 fc(lumens/f-sq'd) and full moon only about 0.02 fc. So turning your Aurora or simliar very high output lights, and shining directly into your eye's is like looking at the sun, you'll go blind!

Keep these 'toys', like a loaded gun, away from children.


PDF link: Units of measure(pdf)

HTML conversion link:
Units of measure HTML(usual number of formatting errors in conversion) /ubbthreads/images/graemlins/eek.gif

how many lumens does the sun put out???

so how many lumens does the sun put out?
any way to calculate that?