How does grid depth determine beam angle?

I just got a 28" Westcott Apollo softbox. They don't sell a grid/egg crate for it, so I'd like to create my own, similar to this.

My understanding is that the deeper the grid, the narrower the angle of light spill, which means a smaller area lit and thus more control over lighting. What I'd like to know is how do you determine the depth/angle ratio, besides trial-and-error.

Also, I wouldn't mind any advice on what the most useful grid beam angles are.

• Also, if there's a better term than "lighting angle", please feel free to post it and/or edit into my question. Mar 3 '11 at 22:00
• I submitted an edit to change it to "beam angle" which is what the spread is more commonly called in lighting. Mar 3 '11 at 22:29
• @Cabbey Is there an authoritative reference to "beam angle" you could share? It might help resolve the question concerning how to calculate (or measure) it. Replies in this thread differ by a factor of two--which seems like a pretty big difference to me--but I wonder whether that might just be a matter of definition. Mar 5 '11 at 20:01
• I'm sure we have one in the stage lighting world, let me dig one up. Note that a 2x or 1/2 difference can easily be attributed to someone looking at the radius instead of the diameter of the beam. Mar 5 '11 at 21:48
• le-us.com/stagemath.html and en.wikipedia.org/wiki/Stage_lighting_instrument#Field_angle are the closest I see in a few minutes search. Otherwise I'd have to quote from my copy of amazon.com/Backstage-Handbook-Illustrated-Technical-Information/… that's on my shelf. Mar 5 '11 at 22:02

Consider a 2D cross section ABCD straight through a cell of the grid, parallel to (and containing) the lighting axis. AD = BC is the depth of the cell and AB = CD is the length of the opening (horizontally, vertically, or even at an angle). In this diagram light can come anywhere from the left in any direction (created by your softbox or otherwise). The illuminated subject is represented abstractly as the line JL. Three of the possible light rays passing completely through the cell are shown: BL, AJ, and HK (a ray in a "generic" position). Evidently all rays emanating from the cell (without any intermediate reflection) must land between J and L on the subject. (This is obvious if you start at the subject and trace the light path back through the cell: only by starting between J and L will you be able to find some line that makes it back through the cell to the light source.) The angle subtended by the lit portion of the subject is the angle JGL--the left tip of the yellow triangle--which is identical to the angle CGD. You can compute it trigonometrically if you like: the tangent of half this angle equals (CD/2) / (AD/2) = CD/AD. But it may be good enough to note that the extreme rays, BL and AC, intersect in the center of the cross-sectional rectangle at G. That gives you an effective way to visualize the angle of the beam and also shows that it's twice the angles you would measure across the cell at CBD or CAD. In short, the beam angle is (at most) what would be observed were a tiny light source placed exactly in the (3D) center of each cell of the grid and it's (approximately) twice the angle you would estimate by going from any single point on the back of the cell through the cell's opposite opening. This justifies your understanding--as the cell gets deeper, the angle at G must get smaller--and also quantifies it.

This reasoning is enough to recover the entire 3D angle by considering different possible orientations of cross sections along the axis of the cell (the lighting axis).

That's not the whole story. The quality of the light depends slightly on the quality and extent of the source. Most importantly it won't be uniform: even when the source is uniform and diffuse, the emitted light falls off substantially towards the edges (approximately linearly). That should not be noticeable (except at the very edges of the total illumination) because the actual light is the composite of beams from all the grid cells, not just from one of them. And the source won't always be uniform, either. Lack of uniformity will tighten the beam angles, especially among the grid cells furthest (off-axis) from the light.

• Nice explanation! Mar 5 '11 at 7:18

Assuming square grid bins, the dimensions of each grid bin are WxWxD, where D is the depth of the grid and W is the square edge length. Then, using trigonometry, we know that:

tan(A) = W / D

where A is the beam angle (from center line - axis - to one side). But, when considering rays passing through the square corners, there are two more angles to consider:

tan(A') = W / D' = W / sqrt(D^2 + W^2)

tan(A") = W' / D = sqrt(2) * W / D

It can be seen that A" > A and A > A', and thus A" > A'. A" is the largest angle and should be considered the beam angle.

UPDATE: To clarify, by convention, the angle I calculate above is measured from the beam axis to its edge. Since the beam is symmetrical, then the spread is in both direction, and one should consider double this value when calculating the lit area. • This correctly computes the maximum angle created by light rays emanating from any single point at the back of the "bin." However, (a) very few of the rays will be separated by that angle yet (b) it underestimates the angle of spread from an extensive (i.e., non-point) light source. Maybe we need to clarify what "beam angle" really means. Mar 4 '11 at 20:43
• @whuber - I agree with (a). Obviously, the light intensity is not uniform across the beam cross section. I think that, per section, there is a (smaller) square across which the intensity is maximal. Outer of that square, the intensity decreases as you approach the edge of the section. As of (b), I don't see how the analysis underestimates the fact that the source is not a point source.
– ysap
Mar 5 '11 at 0:06
• @ysap I provide an analysis of (b) in my reply to this thread. Your analysis considers the spread from a single point, as if all the light were emanating from one corner of your bin. That's not how the setup works: there's usually a fairly broad light source behind the grid. You are correct about (a); the falloff can be computed as a convolution of two squares: that makes a middle square maximally bright with linear decrease in intensity outwards from it. Mar 5 '11 at 1:22
• @whuber - I don't think my analysis limits the result to a single point source. It just presumes that the maximal angle is obtained from the far opposite corners of the bin. Any other ray from any other point in the source will be limited to a smaller angle. Note that this is not a strict stereometric proof, but rather an explanation in which I take freedom to not mention the obvious.
– ysap
Mar 5 '11 at 5:05
• @ysap Geogebra (geogebra.org/cms ) It's kind of a grown-up computer-assisted ruler&compass system. The interface is a little funky but simple and quick to learn. You can create interactive (Java) Web pages with it, too. To make my diagram I had to specify only seven points: ABCD, H, and two (invisible) points on line JL. Everything else was constructed from them. If they ever make a 3D version it will awesome :-). Mar 5 '11 at 20:16

To complete the answer of whuber, the opening angle is α = tan⁻¹(2×diameter/length). My most often used grid is made of straws with a diameter of 5 mm and a length of 3 cm = 30 mm, resulting in an opening angle of approximately 20°, or a beam that gets wider by about 33 cm after each meter (imho that's an easier way to imagine the opening angle). Latter is calculated by: 1 m × 2 × diameter / length.

An interesting fact about grids by the way: The shape it throws on the wall is defined by the shape of the single elements. If you take a grid of squares, you (more or less) get a square pattern. With round straw the result is a circle.

I’ve written a Tutorial about building a DIY grid with an online calculator for the beam width some time ago, perhaps this helps as well :) (It’s for small flashes though.)

• +1 Great illustrations! The shape on the wall, by the way, is the dilation (convolution; Minkowski sum) of the outlet cross-section via the inlet cross-section. As you say, when both are squares the shape is square and when both are circles the shape is circular. And yes, I left my explanation in terms of tangents because they correspond exactly to how you are thinking about the angle: as amount of horizontal spread for each unit distance outward from the grid. I think most people would agree that's more intuitive than computing an inverse tangent :-). Mar 5 '11 at 19:10
• Whew, I need to bookmark this and put the Minkowski sum on the to-read-list for calmer times :) And, thanks! Mar 7 '11 at 20:52