Answer to the Question 03/96

The question was:
What is the maximal distance from which you can see (with your naked eye) a burning candle in the middle of the night?

I admit: it was a difficult question... It can be solved in one of two ways:

1. Eric M. Bram suggested a solution which seems to set the true maximal distance from which the candle can be practically seen. He assumed that we can see the stars of 6th magnitude, and by comparing their illuminance with that of a candle he arrived to the conclusion that the candle can be seen from 10 km distance. (See details of his solution below.) Later, assuming that even 8th magnitude stars are visible he revised his estimate to 60 km.

2. There is also a different approach: Since the luminous flux of candle is about S=0.03 Watts (see discussion of the problem), it emits approximately N=0.03 Watts/(3*10^{-20})Joules=10^{17} photons/second. (The number by which it was divided is the energy of a photon of yellow light.) From distance R the number of photons per second which reach the eye is N*A/(4{pi}R^2), where A is the area of the pupil (A is about 40 mm^2 for the dark-adapted eye). In order to see a continuous spot of light we need about 20 photons/second entering our eye (this is the "memory" of an eye; think about the speed of the movies!). Thus, N*A/(4{pi}R^2)=20, leading to R of about 140 km. (Some estimates in this solution were approximate and their modification may increase the number by, maybe, factor 2 or 3.)

Comment: The second solution assumed absolute darkness (no noise from surrounding light sources) and no atmospheric interference. We should probably conclude that away from a city on a really dark night the maximal distance is about 10 km.

Here is what Eric M. Bram actually wrote regarding the solution in part No. 1:

From C. W. Allen, "Astrophysical Quantities" (3d Ed, 1973) I see that a magnitude 0 star gives (outside the Earth's atmosphere) an illuminance (light received per unit surface) of 2.54(10 Exp-10) phot ==> [1 phot = 1 lumen/cm^2].
The faintest stars one is supposed to be able to see with the unaided eye is 6'th magnitude (of course we're inside the earth's atmosphere, so this might throw off the calculation, but forget that). Since the difference in magnitude m1-m2 between two stars with fluxes S1 and S2 is -2.5(log(S1/S2)), the ratio S1/S1 = 10 Exp ((-0.4)(m1-m2)).
The six magnitude difference between magnitude 6 and 0 is therefore 10 Exp ((-0.4)(6)) or 10 Exp -2.4, so a magnitude 6 star (assumed to be the dimmest that can be seen) gives an illuminance of 2.54(10 Exp-10)(10 Exp -2.4) phot = 2.54(10 Exp -12.4) phot.
Since 1 meter-candle (which I think I recall is the illuminance given by a candle at 1 meter) is 1 lux = 1 lumen/m^2 = 10 Exp -4 phot, the candle at 1 meter looks (10 Exp -4)/((2.54)(10 Exp -12.4)) = (10 Exp 8.4)/(2.54) as bright as the faintest star that can be seen.
Brightness varying inversely as distance squared, of course, in order to be as faint as that faintest star that can be seen with the unaided eye, the candle would therefore have to be Sqrt((10 Exp 8.4)/(2.54)) = 9944.5 meters away.
HOWEVER: 6'th magnitude is the traditional limit among amateur astronomers for "naked eye" objects, but I and most experienced observers can see 7'th magnitude stars in a really good sky. And you're right, the night sky is never totally dark with only one star in it, so that should depress the "practical" value even more. Also, the initial illuminence value for the 0 magnitude star I used is for outside the Earth's atmosphere. Without the atmosphere the stars would be quite brilliant, and steady! So the faintest then might be more like 8'th magnitude. And every magnitude fainter the eye can see multiplies the "candle distance" answer by a factor of about 2.5. So if 8'th magnitude is the actual "naked eye star" limit with absolutely dark sky and no atmospheric absorption, perfect seeing, etc., the answer would be 6.31 times as large.

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