Maybe it's interesting to know how the grid snooth could concentrate a lot percentage of the light in a small zone.
When you look inside of a pipe you see only a little round window of what is in front of you. But what do you see around that? ...it depends. Indeed in case the pipe material is cardboard you'll see a black area, but if it's plastic, also if it's black itself, probably you'll see the light reflected by the scene in front of yourself. So the plastic tube transmits more light than the cardboard one.
You certainly are already familiar with reflection and refraction, in this case refracted rays don't pass through the material (as we could have with water) but they're absorbed, in another words we lose them. So we have that all the rays which come at surface of the pastic with an angle greater than the refraction angle of that material are lost, but all the others are reflected. Because the more the grid is dense the more the angle is narrow, also more light would be reflected, and the exit beam would be thinner.
1. schematic of the grid snoot on the flash
2. schematic of what happens inside a snoot without grid, the amplitude of the beam is 51° (only an example to compare with the other cases)
To be right what matters is the ratio between width (or diameter of the holes) and lenght, this should be less than a certain value to obtain a narrow beam, but we assume that snoots have all about the same lenght, indeed a snoot without grid and 1 meter long is not so easy! Now that you are skilled with theory you could choose your favorite snooth to DIY!
3. the case of a grid snoot with medium holes, in this case we have (almost) no lost rays and the beam amplitude is 32°
4. if we imagine a grid with thinner holes the amplitude of the beam would be even lower
5. here you see how the reflection angle could be in a single straw, this angle had to be lower than the refraction angle (maybe this is about 35-40°)