Hi,

JoeReal gave very detailed info on nutrient calculation, thanks;

JoeReal wrote:

300 ppm N would translate to 300 mg N dissolved in 1 liter of water. in terms of ammonium sulfate simply 300 mg/ 21% gives 1.428571 grams ammonium sulfate per liter of water is the same as 300 ppm N. There are various types of ammonium sulfate powder and thus could vary in densities. Pure crystals are about 1.7 gram per ml, but the powder or prilled form that I buy from the stores has a density of about 0.73 gram per ml. Converting density to volume, that would be 1.428571/0.73 to give roughly 1.956946575 ml ammonium sulfate powder dissolved in 1 liter of water. And 1 tsp is 4.92892161 ml, and that would translate to 0.4 tsp ammonium sulfate powder per liter of water: so roughly slightly less than half a tsp of ammonium sulfate powder per liter of water for the UK guys. And about 1 and a half tsp of ammonium sulfate powder per US gallon for the US folks. But beware also that various powder size could affect density of the ammonium sulfate, plus that in humid countries, it readily absorbs water. (My previous statement of dissolving 1 tsp ammonium sulfate per US gallon is only actually 200 ppm N of the current powder form that I use. This is because I assumed that the density of ammonium sulfate powder is 1 gram per ml, while in fact it is lower, approx 0.73 gram/ml. You should look it up for actual densities and refine the calcs. Nonetheless, it was erred on the safer side of not burning the roots, I still use that rule of thumb not to exceed 1 tsp fertilizer per gallon for my potted plants because I usually apply a mixture, and the N ratios could be higher plus all the other nutrients along with N come into play, so I don't exceed the 1 tsp per gallon).

I have a question.Since we will follow the 5-1-3 ratio, so we need to calculate other elements, right.
In above example, we used 300mg N, calculate and found that we need to add 1.42gr Ammonium Sulfate per liter of water.

So, for 300mg N we need to add 60mg P and 180mg K per liter of water, do I get it right?

We need to find a K source and calcuate how much do we need to add from that source, according how many percent K it includes, right?

Thanks,

Your logic is right on-- just multiply 60 mg x % P/100 and you have the mg of whatever material you need--divide that by 1000 and you have the value in grams.

Thanks Skeeter

So, in rough estimate that will be a total of 2gr. or more fertilizer per liter.
Is it safe to give that much fertilizer with every watering? Say, we give water, at least, 2 times a week, won't it be too much?

Thanks,