Cincinnati guy· Registered
I have never seen a formula for calculating the amount of acidifying material needed for lowering pH. Lots of charts (see my post below), but they often differ significantly. There are formulas that could be used to estimate the meq of H+ for a specific pH and you conceivably could then calculate the quantity of a acidifying material needed to theoretically produce that amount of H+ needed for a targeted pH, but it would still be a crap shoot due to all the variables. Even lime application calculations are considered only accurate to within 500 lbs per acre (+ or - 10 lbs/M).The acidification comes from the nitrification process. The acidifying ratio is 2.8:1 lb of AMS to S. Do you disagree that the AMS application alone isn't enough to lower the ph?
Most turf will use K at an N:K ratio somewhere between 2:1 and 1:1 or for every pound of N, the turf will use somewhere between 0.5 lbs of K and 1 lb of K. Once again, this can vary due to a number of variables: turf type, climate, cultivation practices, ET, etc. Spectrum chose to use a ratio of 1lb N to 0.75 lb of K. They split the difference. Only a follow-up soil test will show what the K use is for a specific lawn in a specific location.Regarding K, they are recommending almost a 1:1 ratio. I thought a 2:1 ratio was maintenance. Do you feel differently?