Fallacy of inverse-square law

For photography, the inverse-square law is applicable to point sources: a theoretical construct, but close enough for bare bulbs and the sun. For larger light sources, especially double-diffused softboxes and matte reflector panels, you are better served by a slightly more elaborate trigonometric function. Select values can be tabulated for easy reference. The surface must be Lambertian (uniform emission for all angles), which is a lot more realistic than any point-source assumption about a softbox at customary distances. C Code:

#include <math.h>
#include <stdio.h>

/*
* Given an object of height 'h', return ratio of its apparent size as
* subtended to observers at distance 'd1' versus 'd2'.
*/
float
L1_angle_factor( float d1, float d2, float h)
{

return (atan( h/d2) / atan( h/d1));

}

main( )
{

int i;

float H = 1000; // softbox height
float d = 1; // initial distance of subject
/*
* repeatedly double subject distance from softbox
*/
for (i=0; i<20; ++i) {

float r = L1_angle_factor( d, d*2, H);
printf( "H = %.f, d = %06.f, ratio = %f\n", H, d, pow( r, 2));
d *= 2;

}

}

Results from the program:

H = 1000, d = 000001, ratio = 0.998726 (Case 1)
H = 1000, d = 000002, ratio = 0.997452
H = 1000, d = 000004, ratio = 0.994901
H = 1000, d = 000008, ratio = 0.989790
H = 1000, d = 000016, ratio = 0.979537
H = 1000, d = 000032, ratio = 0.958939
H = 1000, d = 000064, ratio = 0.917629
H = 1000, d = 000128, ratio = 0.836448
H = 1000, d = 000256, ratio = 0.691224
H = 1000, d = 000512, ratio = 0.496687
H = 1000, d = 001024, ratio = 0.344812
H = 1000, d = 002048, ratio = 0.277910 (Case 2)
H = 1000, d = 004096, ratio = 0.257324
H = 1000, d = 008192, ratio = 0.251855
H = 1000, d = 016384, ratio = 0.250465
H = 1000, d = 032768, ratio = 0.250116
H = 1000, d = 065536, ratio = 0.250029
H = 1000, d = 131072, ratio = 0.250007
H = 1000, d = 262144, ratio = 0.250002
H = 1000, d = 524288, ratio = 0.250000

In plain English, if subject at 1' from a 1000' softbox steps back to 2' away, then illumination drops to 99.87% (Case 1). So much for inverse-square! The "law" doesn't start to kick in until you are 2m from a 1m source (Case 2).

and the point of this is?

It's in reply to https://photography-on-the.net …showthread.php?​p=17928013, but also of general interest. Impress people at cocktail parties.

Ok, thanks. Though I think it would be more impressive to actually light a photo with a 1000' softbox 1' from the subject.

Most folks here don't bother with simple division, which is all that it takes to do Guide Number arithmetic!

Next most difficult is Inverse Square; and an f/stop 1.4x larger admits TWICE as much light

You want folks to get what you wrote, when they can't cope with the first two?!

Bassat wrote in post #17930089
Ok, thanks. Though I think it would be more impressive to actually light a photo with a 1000' softbox 1' from the subject.

It would be even more impressive to see the demonstration of how the inverse square law applies to the sun in relation to photography. Anyone up for a quick trip to mars?

absplastic wrote in post #17930115
It would be even more impressive to see the demonstration of how the inverse square law applies to the sun in relation to photography. Anyone up for a quick trip to mars?

Too hard to verify the instantaneous distance between Mars and the Sun, due to the orbital path, so I'll pass on the opportunity.

Wilt wrote in post #17930122
Too hard to verify the instantaneous distance between Mars and the Sun, due to the orbital path, so I'll pass on the opportunity.

So you're volunteering instead to be the one who gathers the data points going towards the sun?

Wilt wrote in post #17930122
Too hard to verify the instantaneous distance between Mars and the Sun, due to the orbital path, so I'll pass on the opportunity.

C'mon Wilt! If you know relative positions, velocity and trajectory, instantaneous distance is a simple calculation. If Copernicus can do it, you can, too!

absplastic wrote in post #17930124
So you're volunteering instead to be the one who gathers the data points going towards the sun?

No, the trip takes too long. The minimum distance from the Earth to Mars is about 54.6 million kilometers so getting to Mars (at the right time) involves a much shorter trip (about 1/4 the time). Also the radation from cosmic rays and solar flares would be much less intense going to Mars, compared to going to the Sun, and I am not a tanning bed fanatic.

You want me to take a picture of you? Hold on, let me just run my inverse square-law program to make sure the falloff is mathematically correct

Even wikipedia says if the light source is smaller than a fifth of the distance, the inverse square law holds to 1% error.

RicoTudor wrote in post #17930076
For photography, the inverse-square law is applicable to point sources: a theoretical construct, but close enough for bare bulbs and the sun. For larger light sources, especially double-diffused softboxes and matte reflector panels, you are better served by a slightly more elaborate trigonometric function. Select values can be tabulated for easy reference. The surface must be Lambertian (uniform emission for all angles), which is a lot more realistic than any point-source assumption about a softbox at customary distances. C Code:

#include <math.h>
#include <stdio.h>

/*
* Given an object of height 'h', return ratio of its apparent size as
* subtended to observers at distance 'd1' versus 'd2'.
*/
float
L1_angle_factor( float d1, float d2, float h)
{

return (atan( h/d2) / atan( h/d1));

}

main( )
{

int i;

float H = 1000; // softbox height
float d = 1; // initial distance of subject
/*
* repeatedly double subject distance from softbox
*/
for (i=0; i<20; ++i) {

float r = L1_angle_factor( d, d*2, H);
printf( "H = %.f, d = %06.f, ratio = %f\n", H, d, pow( r, 2));
d *= 2;

}

}

Results from the program:

H = 1000, d = 000001, ratio = 0.998726 (Case 1)
H = 1000, d = 000002, ratio = 0.997452
H = 1000, d = 000004, ratio = 0.994901
H = 1000, d = 000008, ratio = 0.989790
H = 1000, d = 000016, ratio = 0.979537
H = 1000, d = 000032, ratio = 0.958939
H = 1000, d = 000064, ratio = 0.917629
H = 1000, d = 000128, ratio = 0.836448
H = 1000, d = 000256, ratio = 0.691224
H = 1000, d = 000512, ratio = 0.496687
H = 1000, d = 001024, ratio = 0.344812
H = 1000, d = 002048, ratio = 0.277910 (Case 2)
H = 1000, d = 004096, ratio = 0.257324
H = 1000, d = 008192, ratio = 0.251855
H = 1000, d = 016384, ratio = 0.250465
H = 1000, d = 032768, ratio = 0.250116
H = 1000, d = 065536, ratio = 0.250029
H = 1000, d = 131072, ratio = 0.250007
H = 1000, d = 262144, ratio = 0.250002
H = 1000, d = 524288, ratio = 0.250000

In plain English, if subject at 1' from a 1000' softbox steps back to 2' away, then illumination drops to 99.87% (Case 1). So much for inverse-square! The "law" doesn't start to kick in until you are 2m from a 1m source (Case 2).

Oh, Jeez. Have you never done a manual flash calculation in your head. Inverse square law may be complicated in theory. In practice it is 3rd grade arithmetic.

Benitoite wrote in post #17930145
Even wikipedia says if the light source is smaller than a fifth of the distance, the inverse square law holds to 1% error.

By my calculation it's about a seventh but, either way, softboxes are commonly used at a far closer distance: 1x the diagonal for a nice wraparound effect. Personally, I prefer the hard stuff.

Bassat wrote in post #17930146
Oh, Jeez. Have you never done a manual flash calculation in your head. Inverse square law may be complicated in theory. In practice it is 3rd grade arithmetic.

If you think you can do it in your head, tell me then if you meter f/8 at a distance of 2 meters from a bare bulb light source, what aperture reading should you get if you if you move the lightmeter to 5 meters from the bulb? Or, alternately, how many stops do you have to increase the light to keep the meter reading f/8?

Most adults I know couldn't even do the logarithms with a calculator, let alone in their heads. I doubt many 3rd graders can.