When quantifying how far a light can reach, the usual metric used is ANSI distance, which is defined as distance till the illuminance falls to 0.25 lux, assuming no losses to the atmosphere.
However, we all know that the usable range of a light is usually less than the ANSI specification, with the main reason being that 0.25 lux is often not enough to see detail, particularly on darker surfaces like vegetation. Furthermore, particulate matter in the air (or even air molecules!) scatter light and prevent some portion of the light from reaching the target; this effect is particularly pronounced when the air quality is poor, which has been the case for much of North America over the past few months due to wildfire smoke.
For the above reasons, I’ve decided to make a calculator that takes both factors into account.
Input:
Illuminance threshold for visibility (lux). If you believe, say, that 4 lux of illuminance is enough for your use, then enter 4. Entering 0.25 recovers ANSI throw.
Visibility (km). Read that off the weather app; the functional definition is given toward the end of this derivation.
Output:
Usable distance (km). This is the distance where, even after atmospheric losses, the apparent illuminance of the target object equals the selected threshold.
Note that this calculator does not account for backscatter of the beam, which lowers the signal-to-noise ratio and makes things difficult to see even when well-illuminated. However, this is strongly mitigated by holding the light far from your eyes, so that your line of vision is less aligned with the beam.
The math behind the calculator is as follows: the illuminance reaching a certain distance is penalized both by (i) inverse-square decay due to divergence and (ii) exponential decay due to atmospheric losses. Since light needs to travel both ways to reach the observer’s eye, the atmospheric losses are roughly doubled. Setting the resulting “observed illuminance” to the selected threshold yields the usable distance.
Really interesting question! I think my answer might be “mostly yes” for some other types of loss/scattering, but “mostly no” for diffusers like frosted glass, assuming that diffusion can be modeled by blurring, which is a fundamentally different process from scattering.
About diffusers: while they do decrease intensity, the amount of intensity loss (over any distance) is dependent on the light, rather than an intrinsic property of the diffuser itself (which is the case for scattering in air). A uniform emitter like a light bulb doesn’t lose any intensity to a diffuser, while a laser can lose over 99% of its intensity to the same diffuser if placed directly over the light.
Additionally, the placement of the diffuser matters. The closer to the source, the more intensity is lost; the closer to the target, the less intensity is lost. Because the diffuser is a thin sheet rather than a large mass of material that spans the distance between emitter and target, there’s no real notion of “exponential loss over distance”. Also, the way diffusers work is very unlike scattering. Objects through a diffuser look blurred, but even on a heavily polluted day when you can stare directly at sunset, the sun’s contour still looks perfectly sharp.
For other types of loss/scattering like absorption (say underwater, by the slightly blue tint of water), the model still works with minor modifications–the underlying Beer-Lambert law is remarkably general. In particular, the weather app’s notion of “visibility” is defined as a sort of “signal to noise ratio” and only makes sense for scattering; with absorption both signal and noise drop uniformly and the ratio doesn’t change, so even a heavily colored solution could have very long visibility. This is easily fixed by using the absorption coefficient directly, rather than inferring it from visibility.
This is an interesting consideration. For what reason do you think that further distances require more illuminance?
I don’t disagree, I’m just curious why you think this is the case. My guess is that it has less to do with illuminance and more to do with the resolution of the human eye–farther objects appear smaller (and thus more blurry) in the field of view and therefore require more light to tell apart, especially for people with myopia. I might argue that this particular mechanism is mitigated by binoculars, and otherwise depends too much on individual differences in eyesight, to include in a model about lights and atmospheric conditions.
Well, for one, the futher away you are from a lightsource, the less of the light hits you from said source. Since light will spread in every direction after hitting a surface unless it’s a mirror, so being further away means much less photons traveling back to you.
This is true, but does not relate to how well-lit an object appears to the human eye. Recall that illuminance is not bulk output, but output per area, and as an object gets farther away, its apparent area also shrinks.
To see why “total amount of light coming back at you” is not the right quantity to look at, let’s pretend, for a moment, that how well-lit an object appears is proportional to how much light from it hits your eye. Consider a piece of printer paper 1m away, lit by direct sun and reflecting light at you with 0.1lm per mm^2.
Now suppose that the same piece of paper is 1km away from you, and you want to see it equally well-lit. In order for your eyes to receive the same amount of light from it, it needs to be brighter by a factor of (1km/1m)^2 = 1,000,000, putting its intensity at 100,000 lm per mm^2, which is more intense than staring directly at an equivalently-sized patch of the sun! A piece of paper with this much emission would also incinerate everything in its vicinity.
Surely you would agree with me that it doesn’t take being more intense than the surface of the sun for a piece of paper to be well-visible at 1km.
I’m not advocating that only total lumens matter, but how much light from a certain area. Our eyes affect this with their apparent resolution of course, as well as the size and intensity of the spot you are looking at.
The biggest question is how much does each factor impact the result, and or do we even need to account for everything to get a decent estimation at all.
I don’t think we discern a normal sized (A4 for instance) piece of paper at 1km range at all, unless it’s a significant source of light.
Although I don’t know how our angular resolution interacts with point sources of light at all tbh. All I know is most stars is way way smaller than we should be able to discern, yet we can clearly see them.
Also keep in mind the nature of the iris…not sure if a study has been done between the correlation of the contraction versus the intensity of the light being perceived. If we could fix the aperture of the eye (with those dilation drops or whatever) would it be possible that the eye does perceive things linearly?
In that case, the decay of light reaching your eye over distance does not affect the illuminance of the object: the apparent size of the object shrinks with the amount of light reaching your eye, and these two effects cancel to yield the same level of perceived illumination.
In other words: if you had a giant sheet of paper out during the day, it looks equally white 1m away or 1km away–being farther away doesn’t make it less lit (equivalent, look grey).
I agree with this. Factoring this into the model would require accounting for differences in visual acuity for every user, which seems infeasible. I’m very myopic myself, and removing my glasses would cap the usable range of every single light I own to 5m, which would not be a very useful data point for folks with healthier eyes.
Very interesting observation! The limit of human eye resolution is often said to be 1/60 of a degree, equivalently, about 1/30 the angular diameter of the sun/moon. Any objects smaller than this would appear blurred, with an apparent size greater than this threshold. All stars other than the sun are certainly smaller than this threshold.
I believe that @Fairlight is highlighting the difference between an emissive object vs a reflective object. With the objects at the same surface intensity, an observer may have less trouble seeing the emissive object because there may be no level of atmospheric backscattering, a.k.a. beam coherency. This might seem obvious, but A super bright floody light may effectively REDUCE the perceivable throw, especially in a smoggy environment, where the smog also becomes a reflective object and overwhelms/interferes with the brightness of the intended object. Like driving with high beams in heavy fog. Others can see where the high beams originate, but the originator can’t see those being illuminated.
Is there a universal viewing angle where one could always tell what object is being illuminated by ANSI standards?
An emissive object still experiences atmospheric scattering, since light from the object needs to travel through the atmosphere to reach the observer’s eyes. The scattering is reduced (compared to a reflective object being illuminated by a source at the observer’s location), since light no longer needs to travel from the emitting source to the object.
I see, you meant to describe the backscatter of the beam, which doesn’t only cause the useful light to decay but actively adds noise. This is not taken into account by the model, with the following justification:
Could you clarify a bit more? What do you mean by “viewing angle”, and what are the relevant ANSI standards? What if the object has fine detail that cannot be resolved regardless of how well-lit it is?
Fair point…viewing angle as an arbitrary observer, independent of the emissive source; i.e. putting a light down and creating that distance you mentioned by walking away from it to depart from the centerline of illuminance **while maintaining visibility of the target. ANSI would be the 0.25 lux on target, which **could become amoot point if the differential backscattering at any point somehow equates or exceeds the target…
This in an interesting point, but still begs the question…If we sent super illuminator visible light beacons to planets, say Mars, how bright would they need to be to exceed/cut through the atmospheric reflectance from the sun?
Sorry, still not making sense yet–to be more explicit, whenever I see the word “angle”, I ask myself, “what’s the vertex and what are the two rays spanning the angle”, or “angle subtended by what object, from what vertex point?” I don’t yet see an answer to either question.
One may as well ask the same question on Earth! My answer would be that it is not sensible for the model to account for other sources of illumination. During the day it probably takes thousands of lux to illuminate an object well enough to compete against the sun or scattered sunlight, but such a consideration is not relevant to the situations where a light is actually needed.
I think we kinda agree on my point tbh, but I’m not even sure exactly what point I tried to make anymore .
I had a clear idea when I made the first comment, but now I’m not so sure lol.
Visibility of subject being dependent on reflected light…Just because a beam can touch an object to a surface illumance of 0.25 lux, doesn’t mean any of that 0.25 lux would be reflected back to the viewer at longer distances. 0.25 lux viewed at 5 meters vs 0.25 lux viewed at 500 meters.
**Special relativity still messes with my brain too, you’re in good company, lol.
***Per AI lookup, starlight on a dark night is abouy 0.0001 lux, with a single star producing about .000002lux. Dunno what that could do for this convo…
Thank you very much, this diagram makes a ton of sense, and you raise a very interesting point! Here I think it is necessary to consider different modes of reflection.
If you’re considering diffuse reflection, which is the case for 99% of flashlight use, the apparent intensity of the target does not depend on viewing angle, similarly to how Lambertian emitters have the same luminance in every direction.
If you consider retroreflection (e.g., reading road signs from a distance), then the viewing angle definitely matters. In this case: if your line of sight is aligned with the beam, even a very dim light (when held next to the eye) can produce a bright retroreflection. Otherwise, even a very bright light may fail to produce a noticeable reflection.
If you consider specular reflection (as in a mirror), then what matters is not only viewing angle but also the angle of incidence on the target. Unless the target surface is exactly perpendicular to the beam axis, you won’t pick up any reflection at all!
I don’t work in this field, but that sounds like a sensible concern!
THIS! I agree that in an ideal scenario this is true. However, if beam coherency is introduced for a non-zero beam angle/divergence and atmospheric conditions are not ideal, say, australia during spider mating season, it would be tough to pick out a small subject where closer particulate would be more apparently bright… how far out would you have to establish a viewing angle where backscatter would be the least obstructive?
That’s both an important and interesting question!
Backscatter can be roughly thought about as a region of air (the beam) that is simultaneously emissive and translucent, which is quite an unusual combination of properties. Due to translucency, the apparent intensity of backscatter is roughly proportional to the thickness of the backscatter along your line of sight. If you’re perfectly on-axis, then that thickness equals distance to target, and you see a lot of backscatter. If you’re off-axis by a good amount, then your line of sight to the target only intersects the light beam much later on, which makes the backscatter much less intense.
Of course, beam shape also comes into play here. If the beam is a wide cone rather than a thin cylinder, then even for a large offset, your line of sight will intersect the cone very soon and thus see a lot of backscatter. This is a good reason to prefer narrow-beam, low-power setups for throwers.
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