Furthest distance a light source can be seen

[Insanity]

Trying to determine at what maximum distance a light source of a known lumen output can be seen in complete darkness. i.e. a 60W bulb or a single candle.

I known lux is the term for light intensity perceived by the human eye, but do not know what the lowest lux limit of the naked human eye is. One source I read said that after dark adaptation, the human eye can recognize objects in 0.007 lux¹, but does not state if this is the lowest limit.

Given that 1 watt = 680 lumen² (@ 555 nm) and 1 lx = 1 lm/m², the lux at 1 meter would be.
total surface area = (4/3)*pi*r³ =~ 4.19 m²
lx = 680 lm/4.19 m² = 162 lx

if 0.007 lx is the lowest limit, the distance at which the source would produce this is:
lm/lx = m²
lm/lx = (4/3)*pi*r³
((lm/(4/3)*lx*pi))^(1/3) = r
((680/(4/3)*0.007*pi))^(1/3) = r = 28.5 m

for a 60W bulb, the distance is:
((40,800/(4/3)*0.007*pi))^(1/3) = r = 111.7 m

Is this correct?
Or is this the distance the light source produces enough light to recognize objects, and not necessarily the apparent brightness of the source? Or are these the same basically?

I apologize for the formulas, I don't know latex that well.

¹Encyclopedia of Physics, Second Edition — ed. by Rita G. Lerner and George L. Trigg, VCH Publishers, Inc., 1991, p. 1342.
²American Institute of Physics Handbook, Third Edition, page 6-10

 

[Dickfore]

Perhaps this will be of value:

http://en.wikipedia.org/wiki/Luminosity_function"

 

[Bob S]

See table in http://www.andor.com/learning/digital_cameras/?docid=316 regarding the ultimate human eye sensitivity in photons per second (wavelength-sensitive) and watts and lumens per m. This article claims ~10 photons per sec (doesn't state area) at ~5500 Angstroms.

This experiment found ~90 photons per pulse into a fully-conditioned eye in a very dark room:

http://math.ucr.edu/home/baez/physics/Quantum/see_a_photon.html

Assume a fully dilated eye is ~ 6-mm diameter. So area is ~0.3 cm³, or ~300 photons/cm². Maybe dc continuous rate is ~3000 photons per cm²-sec.

Bob S

 

[Insanity]

Did find another source that mentioned scotopic vision, or low-light ambiance, where the rod cells are active in human eyes is at 0.003 lm/m².

Given this a 60W bulb produces low-light ambiance at ~147 m.

 

[Insanity]

See table in http://www.andor.com/learning/digital_cameras/?docid=316 regarding the ultimate human eye sensitivity in photons per second (wavelength-sensitive) and watts and lumens per m. This article claims ~10 photons per sec (doesn't state area) at ~5500 Angstroms.

This experiment found ~90 photons per pulse into a fully-conditioned eye in a very dark room:

http://math.ucr.edu/home/baez/physics/Quantum/see_a_photon.html

Assume a fully dilated eye is ~ 6-mm diameter. So area is ~0.3 cm³, or ~300 photons/cm². Maybe dc continuous rate is ~3000 photons per cm²-sec.

Bob S

The article does state that the lux of a 6th magnitude star is 1.0E-08 lm/m². I know 6-6.5 apparent magnitude is about the limit for most people.

So then:

((lm/(4/3)*1.0E-08*pi))^(1/3) = r = approx. max distance it can be seen, assuming lm is known.

i.e. a candle, ~13 lm max distance 677m or 0.42 mi.

 

[Insanity]

Found another solution, though it gives completely different results:

Under ideal conditions, about three and a half miles.

For a point source against a dark background, and with a fully dark-adapted human eye, the threshold for human visibility T (in footcandles) can be found by this equation:

ln T = .0828 [K ln(B)]^2 + .194[K ln(b)] - 9.73

... where B is the background brightness in nanoLamberts, and K = .4343. A perfectly dark, clear night sky will have a brightness of about 100 nanoLamberts, and this equation is valid only for backgrounds of 0.1 nanoLamberts or greater, which is essentially totally dark.

Solving this equation for T, the threshold visibility against a background of 0.1 nL gives us 5.3 x 10^-5 footcandles, which is the brightness of 1 candle at 18,792 feet, or 3.6 miles.

Allowing for atmospheric attenuation will reduce that number a bit.
Source(s):
http://www.dioi.org/vols/wc0.pdf

 

[Bob S]

Solving this equation for T, the threshold visibility against a background of 0.1 nL gives us 5.3 x 10^-5 footcandles, which is the brightness of 1 candle at 18,792 feet, or 3.6 miles.

Could it be sqrt(18,792) = 137 feet?

Definition: [A foot-candle is] the amount of illumination produced by a standard candle at a distance of one foot.

Bob S

 

[Insanity]

Could it be sqrt(18,792) = 137 feet?

Definition: [A foot-candle is] the amount of illumination produced by a standard candle at a distance of one foot.

Bob S

Still, if its 137 ft, and I get 677m, big difference.

 

[Insanity]

Just went got back from testing this. Took two tealight candles, drove out to the back country roads around my house, found a good mile stretch that was fairly straight and level. Lit them and walked away until I couldn't see them any more, noted where I was and walked back, noted where the candles were and drove my vehicle back to where I couldn't see them. The odometer read 0.4 miles and I went a bit more to my mark. Odometer only reads in one tenth mile, but my distance was a little over.

I'm guessing my math is more accurate then the other guy's.