Lux translated into metres for a layman (like me)

I've read much reviews here and the unit used to compare the throw of a light is lux. That I get it and the more lux, the further the light throws! But what would lux translate in metres??

1000 lux = xx metres, 5000 lux = xx metres etc.
The conversion might not exist (google failed ), but based on your experiences, I'd like an estimate in terms of metres from you experts

Something, like you looking for extists. It's pure mathematical approach, but some manufacturers follows it, claiming their "range" distance, based on lux.

This formula assumes, that the minimum lux number on some surface, to see anything, is "0,25". Of course the longer distance, the harder to see details of object, when only 0,25 lux hits its surface. This formula assumes also that the air is crystal clear and doesn't absorp any light. In real life, the air is not clear, and some light is absorbed ...and when it finally hits the surface, this "reflected" 0,25 lux must return to your eyes through the same way. So, the longer the distance is, the more lux is absorbed by the air.

This simple forumla looks following:

distance [m] = sqrt (intensity [lux/1m] / 0,25)

For example, we have a small edc light that gives 625 lux/1m. So:

dist=sqrt(625/0,25)
dist=sqrt(2500)
dist=50m

For a real thrower that puts out 50K lux/1m:

dist=sqrt(50000/0,25)
dist=sqrt(200000)
dist=447m

...More or less it looks like this. Of course, in real life those distances are little exaggerated.

Kestrel said:

This really doesn't belong in /LED/, so I'm moving it to /General/ ...

Click to expand...

Im sorry for wrong posting. Ill try to be more cautious henceforth.

@coolperl: Thanks for a great, informative and comprehensive answer man!

The ANSI spec uses 'light equivalent to a full moon' which work out to 0.25 lux.
or
On a moonless night you will see a dot of light.
When the moon is out moonlight will wash out the dot of light.
Pretty useless but official.
ANSI distance [m] = sqrt (intensity [lux/1m] / 0.25) as coolperl pointed out.
-
Within CPF the ad hoc standard is 1 lux.
So 1 lux at distance [m] = sqrt (intensity [lux/1m]) is more common here. You will see that in the charts in selfbuilt's reviews and at http://www.flashlightreviews.com/ which was operated by CPF member Quickbeam.
(You can simply half the ANSI distance to get 1 lux at distance or double the 1 lux at distance to get the ANSi distance.)

Recognized standards use the range at wich the object reveives 1 lux illumination. Thats why CPF uses 1 lux

ANSI, using 0.25 lux is a bunch of manufacturerers (read marketeers)
Illuminating Engineering Society (more independent) uses 1 lux, IES-range calculates with losses between lightsource and object, I.e. due to air quality etc.
Deutschen Instituts für Normung (German Institute for Standrads) DIN 5037 Lichttechnische Bewertung von Scheinwerfern uses the 1 lux as standard, but without losses between lightsource and object I think.

Some info I found that might be helpfull: Download Lighting Engineering information (pdf, 147 KB) , at the site of Wiska manufacterer of searchlights at http://www.wiska.de/eng/116,wiska-light.html .

Anyone more information about the content or definition of range in those standards?

Too technical for me.
Thanks for sharing though. Im sure other members will understand a bit more than my limited comprehension.

SuLyMaN said:

Im sorry for wrong posting. Ill try to be more cautious henceforth.

Click to expand...

No problem. Your topic fits perfect here in the General Flashlight Discussion forum.

Bill

It looks like this has been answered pretty thoroughly. I will add just one more thought, which is that the basis for this is the inverse square law, which states: intensity = 1 / distance^2 (actually n / distance ^ 2, as the two units are apparently not equally scaled). So to find lux at another distance, you solve the equation (intensity1 * distance1 ^ 2 = intensity2 * distance2 ^ 2). Similarly, to find the throw distance, you plug in whatever lux value you think is the minimum usable intensity of illumination for intensity2 and solve for distance2. Plugging in various numbers yields the various simplifications stated above.

The short answer is distance = sqrt(lux) ... the conditions of which have been detailed above.

I'm a bit confused with this.

Using coolperl's first scenario:

For example, we have a small edc light that gives 625 lux/1m. So:

dist=sqrt(625/0,25)
dist=sqrt(2500)
dist=50m

Click to expand...

Plus the moonlight ~ 0.25 lux thing, does this mean that if I shine this flashlight onto a white barn wall at 50m distance on a night of full moon, I won't be able to discern the flashlight's beam on the wall?

When we talk about "throw" do we mean the absolute, theoretical limit, or the practical limit? Or in other words, might a theoretical 50m throw only equate to a (say) 35m practical throw?

Or am I totally left field here LOL?

—Epi

Light will travel forever if nothing gets in its way.

The ability of a flashlight put a useful amount of light on a distant object will depend on several factors, including the ambient light level and the reflectivity of the target. This is why I have never been a fan of describing the throw of a flashlight in terms of distance.

Think of your light as a shotgun. The closer you are, the closer together the pellets will be.

As you get further out, the pellets get further out and the distance between them increase exponentially as you get further out.

Imagine a circle at distance X made by your flashlight. Its area at 2X distance is 4 times what it was at distance X due to the Pi-R-squared formula for area.


So what they are saying is that lux, just like shotgun pellets decreases over distance. How do you define the 'range' of the shotgun? Its tricky because you have probabilities of a pellet hitting an X sized target at Y distance.

This depends heavily on the Choke and pattern. A tight choke is like a high throw flashlight with a narrow beam. A floody light is like a sawed-off shotgun.
Hence it is difficult to really nail down a definition if you asked for the range of a shotgun. It kind of depends.

Light is a bit easier since it is uniform, but you have to agree on exactly how much light it takes to consider something as 'illuminated'. They say .25 lux but that varies based on light absorption by the air, reflectivity of the target, and other factors that the above poster mentioned. So...it depends.


Clear as mud?

In short, same formula ordered in three ways:

Candela = Lux x distance2

Lux = Candela / distance2

Distance2 = Candela / Lux

(Where 'lux' is the lux-measurement at 'distance' .)

joe1512 said:

As you get further out, the pellets get further out and the distance between them increase exponentially as you get further out.

Imagine a circle at distance X made by your flashlight. Its area at 2X distance is 4 times what it was at distance X due to the Pi-R-squared formula for area.

Click to expand...

That could confuse people - "exponentially" has a specific mathematical meaning and is different from the square-law relationship that applies to things travelling in a straight line.

The beam-width or pellet-separation increases linearly -
- double the distance = double the beamwidth.

The beam area or pellet pattern size increases by square-law, because area is proportional to the square of the diameter.