Luminance from multiple light sources I have a LED diode:
Given Luminance intensity [I] of $1.85$ cd.
I want to have Luminance [B] from the light source of $75 \ {\rm cd/m^2}$.
From $B=I/S$, where S is light source plane area, I get $S=I/B=0.0246 \ {\rm m^2} \ = 246 \ {\rm cm^2}$.
From this I understand that area of the light source must be $246 \ {\rm cm^2}$ to get $75 \ {\rm cd/m^2}$ out of it. Right? Of course my diode has a much smaller area (for instance $1 \ {\rm cm^2}$). 
I know that Luminance is an additive value from all light sources (at a given point $L=L1 + L2 + \ldots + Ln$)
Question: So do I just arrange them in a area that is $246 {\rm cm^2}$ (for example in $15 \times 16$ matrix) or that is not how it is done? Is there more effective way??
 A: This may or may not answer your question.  I hope that it helps you understand the principles involved, and why I keep asking for the details of the application.
To conceptualize luminance, think of a small element of area somewhere in space.  To specify luminance we need to know the location of the area element, its area,  a direction relative to the normal of the area, and a solid angle centered on that direction.  The quantity of lumens that pass through that area (divided by the area), at that point, in that direction, into that solid angle, is the luminance.
Suppose you have an imaginary LED whose luminous area is 1 cm${}^2$, and has a luminous intensity of 1.85 cd uniformly distributed across the entire area.  We also have to know the direction of the specified intensity; we'll assume, reasonably, that it's perpendicular to the plane of the LED luminous area.  We also need to know the angular spread of the light; we don't, but let's assume that whatever the spread is, it's the same spread required of our final light source.  Then the luminance of this LED is 1.85 cd/ 1 cm${}^2$ or $1.85 \times 10^4$ cd/m${}^2$ perpendicular to the area. For simplicity, let's assume that the LED is in a horizontal plane, and we're interested in the luminance in the vertical direction.  If I now take 246 LEDs and array them in a rectangle, then the  luminance is uniform across the entire array, and its value is still $1.85 \times 10^4$ cd/m${}^2$ .
Suppose you have a more realistic LED whose luminous area is 1 mm${}^2$, and whose luminous intensity is 1.85 cd.  The luminance of this device is $1.85 \times 10^6$ cd/m${}^2$.  If I now arrange these in a $15 \times 16$ grid having a 1 cm spacing, I end up with a source of non-uniform luminance.  Its luminance is $1.85 \times 10^6$ cd/m${}^2$ at the luminous areas, and zero elsewhere.
But you can ask a different question:  What's the luminance at some point in space above this light source?  Consider a tiny imaginary area at the point of interest oriented parallel to the light source.  The illuminance in that tiny area at that point might turn out to be uniform.  But the luminance will not.  The luminance will vary according to the angle being considered.  Remember, to specify luminance you have to specify the direction you are looking.  As the direction varies from pointing at an LED to pointing at empty space, the luminance will vary.  Furthermore, in this case we really do need to know the angular spread of the LED and the distance from the source array to the point we're interested in.  It's possible that light from the LEDs is concentrated in such a small solid angle that no light from certain LEDs reach our point of interest.
A: The key to solving this is to think of the sign as the source of the light. What's really happening is the light energy from the LED is reaching the sign surface and then is reflecting off of it into a hemisphere that subtends a solid angle of 2pi. So the luminance of the sign would be the total power of the light shining on it divided by (2pi * Area) where the area is the area of the sign (approximately). There is a factor for the relectivity of the sign, it's not a perfect reflector so you can't use the full power of the LED, and that depends on the sign construction and color. And it's not perfectly diffuse so some of the light reflects off of it like a mirror and does not get scattered into the hemisphere, so this acts like another loss which would be bigger for gloss paint and less for matte. You could estimate both of these using .80 for example and you would then multiply the power of the LED lamps by .64 to get the total power to use in the calculation but this is of course a guess. There are several systems of photometric units but you can work that out by looking up the definitions, the one you want to use is the one that's based on the response of the human eye, candelas. 
