Soil Lab – Taking samples and fertilizer application rates ANSC 110 – Pasture Management October 12, 2010

Taking the samples  The top 2-3 inches is the most important to the plants Root location Pasture fertilization practices  Use of a soil probe (auger or tube)  Sample in the fall (or spring if not able to do fall)  Sample every 2-3 years

Soil Testing  Take at least sample cores per pasture or paddock If you have a large pasture and it has different soil characteristics, sample those separately No sample should represent more than acres  Transverse each field in zig-zag pattern  Sample parts of field that appear different in soil or crop  Avoid sampling areas that were recently fertilized, obvious dung/urine areas, or areas adjacent to roads

PASTUREslope water

Soil Testing  Collect soil in clean bucket  Break up lumps  Mix well, take out a subsample (about ½-pound)  Place in soil sample box  Label the box!!! Manure application? Legumes in pasture? Crop to be fertilized (i.e., grass type) Field information

Soil Testing Labs  A & L Eastern Agricultural Labs – Richmond, VA  Agri Analysis, Inc. – Leola, PA  Brookside Labs, Inc. – New Knoxville, OH  PA Agricultural Analytical Serv. – Penn State  Spectrum Analytic, Inc. – Washington Ct. House, OH  Univ. Delaware Soil Testing – Newark, DE  Waters Agricultural Labs, Inc. – Camille, GA

Calculating Fertilizer Rates  Fertilizers are listed as N- P 2 O 5 - K 2 O  Recommendations are given as: N P 2 O 5 K 2 O

Fertilization Examples  Multiply rate of desired nutrient by 100 and divide by the percentage in the fertilizer  Recommended rate of N is 80 lb/acre Using (Urea) Equation:  (80 x 100) / 46 = 174 lb/acre

Fertilization Examples  Recommended rate of P 2 O 5 is 40 lb/acre Using (Monammonium Phosphate) Equation:  (40 x 100) / 51 = 78 lb/acre

Fertilization Examples  Recommended rate of K 2 O is 15 lb/acre Using (Potassium Chloride - crystalline) Equation:  (15 x 100) / 60 = 25 lb/acre

Fertilization Example – Price Per Unit  Cost of fertilizer is related to plant nutrient content  Compare prices on cost per pound of “actual” nutrient, not price per ton of fertilizer  Divide percentage of nutrient by 100 and multiply by 2,000 lb. (1 ton)

Fertilization Example – Price Per Unit  Example: Urea (46-0-0) costs $253/ton, what is the cost per pound of nitrogen? Equation:  Nitrogen in 1 ton of Urea (46-0-0)  (46/100) x 2,000 = 920 lb.  Cost per lb. of Nitrogen  $253 / 920 lbs = $0.275

Fertilization Example – Price Per Unit  Example (Based on at $403/ton) What is the cost of P 2 O 5 in ? Calculate nitrogen cost first  (12/100) x 2,000 = 240lbs.  240lbs. X $0.275 = $66 (value of nitrogen)

Fertilization Example – Price Per Unit  Example (Based on at $403/ton) Cost of phosphate  $403-$66 = $337 Phosphate in one ton of  (51/100) x 2,000 = 1020 lb. Cost per lb. of phosphate  $337 / 1020 lbs = $0.33