Can't I find it out with mathematics?
Say, fill a bucket with 10 liters of tap water. Add 10 gram of dichlor. Dissolve. Take a sample of 10 ml. Dilute with distilled water (1:10 ratio?) and then take the CYA test.
If I can go as high as 10 and my FC is 6, according to the pool calculator, I should add 72 grams of dichlor.
I'm not sure if there's some confusion floating in this thread about the amount of CYA in dichlor -- I think someone asked if a percentage of CYA was listed on your particular product label.
Anyway, the dichlor is a chemical compound, so that inherently sets ratios of the materials we're interested in for pool chemistry (the chlorine and the cyanuric acid). It's not like a mix of salt and sugar, for example, where you could have 10% salt and 90% sugar, or 80% salt and 20% sugar, or... any other ratio you like. It's more analogous to the ratio of hydrogen and oxygen in water -- that ratio is fixed by the chemistry of the molecule.
I don't remember for sure if there are different forms of dichlor (I'll post a link if I come across something). You can search the forum yourself, and perhaps someone more knowledgeable about chemistry will chime in also.
edited: below is a quote from
chem geek, including the info. about the two forms of dichlor (dihydrate and anhydrous):
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It doesn't work out that way for several reasons, but mostly it's because Dichlor has TWO chlorine atoms attached to the ring so it's one part CYA and two parts chlorine. Trichlor has THREE chlorine atoms attached to it so it's one part CYA and 3 parts chlorine. But the "ppm" values don't work out exactly because the ppm for chlorine is for the weight of chlorine gas which has TWO chlorine atoms. Another factor is that the most common form of Dichlor has two water molecules attached to it (dihydrate).
Dichlor•2H2O is 255.97766 g/mole
CYA is 129.075 g/mole
Chlorine gas is 70.906 g/mole
Trichlor is 232.4103 g/mole
Dichlor anhydrous is 219.9471 g/mole
So, Dichlor (dihydrate) is 129.075/255.97766 = 50.4% CYA and 2*70.906/255.97766 = 55.4% "available" chlorine.
Trichlor is 129.075/232.4103 = 55.5% CYA and 3*70.906/232.4103 = 91.5% "available" chlorine.
Dichlor anhydrous is 129.075/219.9471 = 58.7% CYA and 2*70.906/219.9471 = 64.5% "available" chlorine
129.075 / (2*70.906) = 0.91
129.075 / (3*70.906) = 0.61
Anyway, the bottom line is that for every 1 ppm FC added by Dichlor, you get 0.9 ppm CYA while for every 1 ppm FC added by Trichlor you get 0.6 ppm CYA.