Page 110 - Calculating Agriculture Cover 20191124 STUDENT - A

9-26 Land, Soils and Fertilizers CH 9]

Solution:
Silt loam: 1 ft. x 1.25-2.00 = 1.25 - 2.00 in.
Silty clay loam: 4 ft. x 2.25-3.00 = 9.00 - 12.00 in.
5 ft Total 10.25 - 14.00 in.

Example B: If 0.2 inch per day of water is evaporated from the soil and transpired through the
plant, how much water would be left in the preceding soil profile if no rain occurred in
July? Refer to Figure 9.14.

Solution: (10.25-14.00 in., in 5 ft.) − (31 days x 0.2 in.)

(10.25-14.00 in., in 5 ft.) − (6.2 in.)

10.25 to 14.00 in., in 5 ft.
— 6.20 to 6.20 in., in 5 ft.

= 4.05 to 7.8 in. left in soil

Example C: If corn needs 20 inches of available water, what portion (%) of its water needs are
stored in the saturated soil in the preceding example? Refer to Figure 9.14.

Solution:
10.25 to 14.00 in. available water
———————————— x 100% = 51.25% to 70% of the needs
20 in. needed

PLANT NUTRIENTS and FERTILIZERS

Soil Nutrient Content and Uptake for Crops
Besides getting moisture from the soil, plants also take much of their nutrient supply from the soil.
Soil with its organic and inorganic makeup contains nutrients that plants use. In analyzing the nutrient
content of a soil, the nutrients are of primary importance and include: Nitrogen (N), Phosphoric acid
(P 2O 5), and Potash (as Potassium Oxide — K 2O). These elements are measurable in soil.
From a soil analysis for nutrients and understanding the crop to be grown on a field along with the
crops nutrient requirements it is possible to calculate the amount of nutrients to be added to optimize
crop growth to harvest. Typically, the analysis considers acreage, volume weight, and the weight per
cubic foot of the soil. With the given information, the pounds of a nutrient can be calculated. The
equation is:

(Ft/ac x soil depth) (soil wt/cu.ft) (volume wt) (% element) = element wt.

Example: A chemical analysis of the upper 9 in. of a soil showed that it contained 0.183% nitrogen,
0.244% phosphate, and 0.414% potash. How many pounds of nitrogen would be
contained in one acre of this soil if it had a volume weight of 1.3? (Convert %’s to
decimals by dividing by 100%)

Solution: Note: 9 inches is (9in ÷12 in/ft =) 0.75 ft
(Ft/ac x soil depth) (soil wt/cu.ft) (volume wt) (% element) = element wt./acre
(43560 sqft/1 ac x 0.75ft) (62.3#/cu.ft) (1.3) (0.00183) = # Nitrogen / acre
4,842.08 # = # Nitrogen/ acre

The solution process for all elements of N, P, and K, is the same to arrive at how much of each
element is in the soil to the stated depth per acre.

The hand calculator key strokes are:

43560 sqft/1 ac 0.75ft 62.3#/cu.ft 1.3 0.00183 = # Nitrogen / acre
CL 43560 X 9 ÷ 12 X 62.3 X 1.3 X .00183 = 4,842.076239

4,842.08 # Nitrogen / ac

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