# 44”/12” = 3.67’ 3.67’ x 20 = 73.4’ 6 x 30” = 180” 180”/12” = 15’ 73.4’ x 15’ = 1101 sq ft - PowerPoint PPT Presentation

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#2. A 6-row planter has 44” circumference wheels and is set at 30” centers. ½ lb. Of seed is dropped with 20 revolutions. What is the seeding rate?. 44”/12” = 3.67’ 3.67’ x 20 = 73.4’ 6 x 30” = 180” 180”/12” = 15’ 73.4’ x 15’ = 1101 sq ft 43560/1101 = 39.56

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44”/12” = 3.67’ 3.67’ x 20 = 73.4’ 6 x 30” = 180” 180”/12” = 15’ 73.4’ x 15’ = 1101 sq ft

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#### Presentation Transcript

#2. A 6-row planter has 44” circumference wheels and is set at 30” centers. ½ lb. Of seed is dropped with 20 revolutions. What is the seeding rate?

44”/12” = 3.67’

• 3.67’ x 20 = 73.4’

• 6 x 30” = 180”

• 180”/12” = 15’

• 73.4’ x 15’ = 1101 sq ft

• 43560/1101 = 39.56

• .5 lb x 39.56 = 19.78 lbs/ac

#1. You have a 12’ drill with 91 “ circumference wheels. In calibrating it with 20 revolutions, 3 ¼ lbs. Of seed are dropped. What is the seeding rate at this setting?

• 91”/12” = 7.58’

• 7.58’ x 20 x 12’ = 1819.2 sq. ft

• 43560/1819.2 = 23.93

• 3.25 lbs x 23.93 = 77.77 lbs/ac

#2. A grower wants to apply 80 lbs. Of actual nitrogen per acre using anhydrous ammonia. How many pounds of anhydrous ammonia will be applied per acre? How many pounds are needed for 32 acres?

• 80/.82 = 97.56 lbs./ac

• 97.56 x 32 = 3121.92 lbs.

#1 A grower wants to apply 125 lbs. Of actual nitrogen per acre to a crop using ammonium sulfate. How many lbs. How many pounds fertilizer will he apply per acre? How many tons are needed for 65 acres?

• 125/.21 = 595.24 lbs/ac

• 595.24 x 65 = 38690.6

• 38690.6/2000 = 19.35 tons