SportsTurf provides current, practical and technical content on issues relevant to sports turf managers, including facilities managers. Most readers are athletic field managers from the professional level through parks and recreation, universities.
Issue link: http://read.dmtmag.com/i/369217
ft 2 . The extreme values in this case are the ones on the outside of the equation and the means are those near the = sign. And 218 fl oz x 57,600 ft 2 does equal 43,560 ft 2 x 288 fl oz. (If you multiply these out there is a small discrepancy due to rounding.) One of the most import calculations a turfgrass manager makes is the determination of how much of an input to apply to an area of turfgrass. Every calculation of this type comes down to the same prin- ciple applying an amount of a product over an area. Most discussions of turfgrass mathematics spend some time discussing the determi- nation of areas of different shapes and even how to determine the areas of oddly shaped features such as golf greens and sand bunkers. When I think of athletic fields I don't see many of these odd shapes but mostly rectangles and quarter circle arcs of baseball and softball diamonds. For the rectangular shapes the areas are easily determined by multiplying the length by the width. Most field managers know the length and widths of their fields because at some point they have pulled a tape and measured them. With today's technology it is also fairly easy to determine areas of fields using smart phone apps such as Measure My Land, Planimeter, or Google Earth. Nearly every product we apply to turfgrass is not in a pure form so we must determine application rates to allow for this. For example, if we are applying 21-0-0 fertilizer it only contains 21% N so even though we have applied 100 lbs. of product we have only applied 21 lbs. of N. Two applications that apply a product in a pure form are topdressing and irrigation. We will begin with some examples of those applications. Example: How much sand topdressing is required to apply 1/4th inch of topdressing to an area of soccer fields that is 250 yards long and 75 yards wide? A topdressing layer can be visualized as long, wide, and thin box; in this case 250 yards long, 75 yards wide and 1/4th inch thick. We have units of yards and inches so we need to convert the inches into yards. So we will need 130 cubic yards of sand for this application. Our sand supplier sells sand by the ton and a cubic yard of dry sand weighs 2700 lbs. Now the sand is probably not totally dry depending on the weather. If I were buying sand for this application I would buy 15% extra to allow for this water. So for this application I would order 202 tons of sand. Example: How much water is needed to apply 1 inch of irrigation to a football field with the dimensions 130 yds by 70 yds? In this case it may be easier to work in cubic feet. Or we could do an internet search and ask "How many gallons are in an acre inch of water?" Answer 27,152 gallons. The problem now is an equation of ratios. Solving for x we get 51,000 gallons. Sand and water are some of the commodities that are "pure" in that the contain 100% of their ingredient. ChemiCal appliCations With the exception of fertilizer, all of our chemical applications that are sprayed or spread on turfgrass almost always come with a label that gives of a rate of product to use per area of turfgrass. Example: We have an adult softball 4-plex with full skin infields, 65 ft bases and 275 ft to centerfield. Each field has 61,450 ft 2 of grass area. Each field is grassed with MS-Pride bermudagrass and we need to apply Primo-MAXX to tighten up our canopy and cut down on our mowing. We are treating 61,450 ft 2 x 4 = 245,800 ft 2 of grass area. The rate of Primo Maxx for athletic field height (1/2 inch) hybrid bermudagrass is 11 oz per acre. Fertilizer appliCations Fertilizers are a bit different than other chemical applications in that they are made in response to a soil test or fertility plan, and we must take into account the percent element of interest (usually N) in the fertilizer. For example we have 233,000 ft 2 of bermudagrass soccer fields on soils modified with shallow sand cap. Our fertility plan calls for 1.5 lbs of N per 1000 ft 2 for the months of June, July, August and September. Our soil test also indicates we need to apply some potas- sium per our soil test so we choose a 20-0-20 fertilizer. How much 20-0-20 do we need to purchase? www.stma.org September 2014 | SportsTurf 21 1 1 yd 4 36 inches inch X = 0.00694 yds 250 yds x 75 yds x 0.00694 yds = 130 cubic yds 2700 lbs 1 ton yd 3 2000 lbs 130 yd 3 x x = 176 tons 176 tons + (15 % x 176 tons) = 202 tons of sand 11 fl oz x fl oz 11 fl oz x 245,800 ft 2 43,560 ft 2 245,800 ft 2 43,560 ft 2 3 ft 3 ft ft yd yd 12 in. 130 yd s x x 70 yds x x 1 in. x = 6,800 ft 3 x 7.5 = 51,100 gallons gallons ft 3 3 ft 3 ft yd yd 130 yd s x x 70 yds x = 81,900 ft 2 27,152 gal x gal 43,560 ft 2 81,900 ft 2 = so = 62 fl oz Primo Maxx = =