That is not Rex's formula nor is it correct.
I'm not using anyone's formula here. I am using "unit cancellation," though not all the steps are shown. You can learn more about using unit cancellation methods by watching the following slide show. It may seem kind of kiddy with the cheesy clip art, but the step by step animation is really really good:
http://www.dmacc.org/medmath1/UNITCANCEL/Unitcanhtml/sld001.htm
Assuming zero light loss, what is the lux value of 1 2340 40 watt 48" 9325K bulb above 864 square inches of tank?
Here is how you "do the math" using unit cancellation for this particular questions. (please ignore all the periods, they are place holders):
1 lumen = 1 lux, (aka "1 lumen per square meter= 1 lux))
....m2
Using your lumen and surface area values:
2340 lumens *
1550 in2 = 4198 lumens/m2, aka 4198 lux
864 in2
...............1 m2
4198 lux = 4.86 lux per square inch
864 in2
Making the same assumption, what is the lux per square inch of 18720 initial lumens?
Here is it calculated with the light you assume you will need (ignore all the periods, they are place holders so you can see the units):
18720 lumens *
1550 in2 = 33583 lumens/m2, aka 33583 lux
864 in2
..................1 m2
4198 lux = 38.86 lux per square inch
864 in2
How does a smaller surface area change the lux value?
The basic way to get lux is here:
x lumens *
1550 in2 = z lumens/m2, aka z lux
y in2
..................1 m2
Keeping lumen values constant, if you have a smaller surface area you will have a bigger lux value. If you have a larger surface area you will have a smaller lux value.