Denver, CO February 5 8. Landscape Math. Mike Fedison Pickens Technical College Horticulture Instructor

Transcription

1 Denver, CO February 5 8 Landscape Math Mike Fedison Pickens Technical College Horticulture Instructor

2 Landscape Math Preparation for The CLT Certification Tests

3 The use of scales: Architect Scale

4 General Landscape Math Three steps Perimeter Outside border usually measured in feet. Measurements added together M + M + M + M= Total measurement If you add feet to feet the units do not change = Answer is in feet w a l P= 2l +2w C=2 π r b P= 2b +2a

5 Area Area measurements are defined by square feet or square yards Area is calculated by multiplying two measurement. You can only multiply like units. By multiplying unit by unit you get square units. (ft², sq. ft., yd²) Use the following as examples: h r w h l A= l w A= π r² b A=b h 2 b A= b h

6 Volume Volume is referred to as cubic feet or cubic yards and is found by multiplying length (l) x width (w) x height (h) The three types of volumes that you will work with are prisms, cylinders and cones. Prisms: V= Ab x h or volume = area of the base x height h 4 A= 20 x 50 l 50 w 20 A= 1000 sq. ft. V= 1000 sq. ft. x.333 V= 330 cu. Ft. H= =.333ft

7 Cubic feet to cubic yards Cubic yards (cu. yd.) = cu. ft. 27 Cu. yd. = 330 cu. ft. 27 Cu. Yd. = 12.2 yd³

8 Cylinders V= Ab x h r = radius 2 H = 6 A= pi x r² (area of a circle) A= 3.14 x (2 x 2) A= sq. ft. V= (Ab) x 6 (h) V= cu. Ft.

9 Cones: V= 1/3 (Ab x h) A= pi r² H=6 A= 3.14 x (2 x 2) A= sq. ft. R = 2 V= x 6 V= V= cu. ft.

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11 Other Math Problems Associated to CLT Production Times Production based on time it takes to do a job. You install 25 plants at a rate of 20 minutes per plant. 25 x 20 = 500 minutes 60 = 8.33 hours One person can lay 400 sq. ft. of sod in one hour. How long will it take to lay 2700 sq. ft. of sod? = 6.75 hours

12 Instrument/Leveling (Hardscape) There several ways to describe the steepness of a slope Ratio : Horizontal distance to change in elevation Gradient : Change in elevation to the change in distance Percent : Change in elevation to the change in distance expressed as a %. Angle : Slope expressed as an angle Calculations: G =D/L L = D/G D = GL G= Gradient (as a decimal) D= Difference in elevation between two points L= Horizontal distance between two points

13 Using the formulas: The slope between two points when the difference in elevation is known. and the horizontal distance is known. G = D/L Point a = 94.3 ft Point b = ft The two points are 250 ft apart G = ( ) 250 G = = To convert to a % slope x 100 = 4.64% slope

14 The difference in elevation between points when the gradient and distance are known. D = GL A ball field is 350 ft long and requires a 2% slope for drainage. The elevation of the lower end of the field is 35.7 ft. What is the elevation at the other end of the field? D =.02 x 350 = 7 D = = 42.7 ft.

15 The horizontal distance between points when the gradient and difference in elevation is known. L= D/G On a 6% slope, how far away are two points when an elevation difference of 2.5 ft. L= = ft L = ft. 6% =.06 as a decimal

16 Another example: Rod reading at point A = 2.32 Rod reading at point B = 7.84 ( since the rod reading decreases as elevation increases, point A is higher than point B) A 4% slope is required for drainage, with the ground sloping from point A to point B. What is the difference in elevation between point A and B to yield a 4% slope? D = B-A = = 5.52 The horizontal Distance (L) is 50.1 ft The existing slope between A and B is G = D/L = x 100 = 11.02% slope (current) 11.2% (current) 4% (desired) = 7.2% or.072 Depth of fill at point B = x 50.1 = 3.61 ft of fill needed (D=GL) Rod reading for 4% slope Reading B = = 4.23

17 Fertilizer Calculations Ounces of Nitrogen / 1000 sq. ft. Simple formula = What you want What you got You have a fertilizer and need to apply 1 pound of nitrogen per 1000 sq. ft. How much fertilizer would you apply? 1 pound.24 = 4.17 pounds of applied to 1000 sq. ft sq ft of turf / 1000 = sq ft pieces x 4.17 = 148lbs It would take 148 pounds of to apply 1 pound of N over sq ft A tree fertilizer is applied at 1 tablet per ½ in caliper. If you have a 6 caliper tree how many tablets would you apply? There are 12 -½ s in 6 = 12 tablets

18 Other sample problems 1. Conversion of yards to tons: 1 cu. yd. weight 1.5 tons 12 cu. yd. converts to 18 tons 12 cu. yd x 1.5 = 18 tons or 1 ton is = to 1.5 cu. yd. 12 cu. yd. converts to 8 tons cu. yd. 1.5 = 8 tons If you have 8 cu. yd. of soil and you spread it 6 deep, how many sq. ft. will it cover? 8 cu. yd. x 27 = 216 cu. Ft. X sq. ft x.5 = 216 cu. Ft. x = 216 cu. ft..5 ft = 432 ft²

19 3. Precipitation rates x Total GPM Total Area 1. 8 spray heads at 1.65 GPM and an area of 60 by x 1.65 = 13.2 gpm 60 x 12 = 720 ft² x 13.2 = ft² = 1.76 inches per hour

20 4. If you need 3 ounces of a chemical in 6 gallons of water, then you will need ounces in 3 gallons. 3 oz = x oz 6 gal 3 gal 3oz x 3 gal 6 gal = x so so 3 oz x 3 gal = x oz x 3 gal 6 gal 1 3 gal 1 9 oz = 1.5 oz 6

21 5. What is the total length of steel edging needed to edge a tree ring that is 7 feet in diameter? Cir = 2 pi r 2 x 3.14 x radius 2 x 3.14 x feet of steel edging Edging Tre e Steel edging

22 Denver, CO February 5 8

23 Denver, CO February 5 8 Thank you! Insert speaker contact info