### Garden Math

Gardening is an activity which occasionally require

s the use of math, such as when you are

computing how much fertilizer to use or how much co

mpost to buy. Luckily, the math involved is

simple and easily applied.

All that is probably necessary for most readers is

a review of the basic principles and a step-by-step

description of their application. The best advice

is to start slowly and carefully, draw simple diagr

ams

to help you keep track of things, and use a calcula

tor to avoid arithmetic mistakes. It will become

much easier after a little practice.

This document is organized into the following secti

ons:

Squares …………………………. Page 2

Rectangles ………………………

Page 3

Triangles ………………………… Page 4

Circles …………………………… Page 6

Irregular Shapes ……………….. Page 8

Volumes …………………………

Page 10

Converting Measurements ……. Page 11

Typical Garden Situations …….. Page 12

Michael Spencer, WSU Master Gardener

© Grays Harbor & Pacific Counties Master Gardener F

oundation, Washington.

March, 2007. All rights reserved.

2

Squares

Squares are four-sided shapes where all sides are t

he same length and all angles are 90°.

Some Basic Definitions:

Area

: The two-dimensional space covered by a shape or o

bject. It is usually expressed as the

square of some linear measurement, such as square f

oot (ft

2

) or square meter (m

2

). Terms

such as acre and hectare are also used to describe

areas.

Perimeter

: The one-dimensional or linear distance along the

boundary of a shape or object.

Gardeners often use this to determine the length of

garden borders.

12 ft

12 ft

12 ft

12 ft

Square Example

Side =

12 ft

Square Area (Formula 1) =

Side

2

Area =

12 ft x 12 ft

Area =

144 ft

2

Perimeter (Formula 2) =

Side x 4

Perimeter =

12 ft x 4

Perimeter =

48 ft

Formula 1:

Square Area = Side

2

Formula 2:

Square Perimeter = Side x 4

3

Rectangles

Rectangles are four-sided shapes, with two opposite

sides of one length, and the remaining two

opposite sides of another length. All of the angle

s are right angles (90°).

18 ft

12 ft

12

ft

18 ft

Rectangle Example

Longer Sides =

18 ft

Shorter Sides =

12 ft

Rectangle Area (Formula 3) =

Longer Side x Shorter Side

Area =

18 ft x 12 ft

Area =

216 ft

2

Rectangle Perimeter (Formula 4) =

(Longer Side x 2) + (Shorter Side x 2)

Perimeter =

(18 ft x 2) + (12 ft x 2)

Perimeter =

36 ft + 24 ft

Perimeter =

60 ft

Formula 3:

Rectangle Area = Longer Side x Shorter Side

Formula 4:

Rectangle Perimeter = (Longer Side x 2) + (Shorter

Side x 2)

4

Triangles

Triangles are shapes with three sides.

Triangle definitions:

Right Triangle

: A triangle that has one angle equal to exactly 90

°.

Base

: Any one of the triangle sides (it is often rotat

ed for convenience so that it is horizontal

across the page, but this is not necessary). The b

ase is used to compute the area of the

triangle.

Height

: The perpendicular distance from the base to the a

ngle created by the other two sides.

The height is used to compute the area of the trian

gle.

Height

Base

Base

(Note: While they have significantly different sha

pes, the two triangles above have equal

bases and heights, and from Formula 5 below we see

that they have the same area.

However, they do not have the same perimeter. The

triangle on the right is a right triangle.)

Formula 5:

Triangle Area = Base x Height / 2

Formula 6:

Triangle Perimeter = Side A + Side B + Side C

5

Side A = 4 ft

Side B = 5 ft

Side C = 3 ft

Triangle Example

Side A =

4 ft

Side B =

5 ft

Side C =

3 ft

Base =

3 ft (Side C in this example)

Height =

4 ft (Side A in this example)

Triangle Area (Formula 5) =

Base x Height / 2

Area =

3 ft x 4 ft / 2

Area =

12 ft

2

/ 2

Area =

6 ft

2

Triangle Perimeter (Formula 6) =

Side A + Side B + Side C

Perimeter =

4 ft + 5 ft + 3 ft

Perimeter =

12 ft

(Note: A triangle whose sides have a ratio of 3:4:5

, like the one in the example above, is very

useful. If you lay out such a triangle with stakes

and string in your garden, the angle formed

between the sides with lengths of 3 and 4 is a righ

t angle of exactly 90°. The units of measure

are not important: a 3:4:5 triangle measured off in

yards has the same proportional shape, and

the same perfect 90° angle, as a 3:4:5 triangle mea

sured off in either inches or meters.)

6

Circles

Circles are round shapes where every point on the c

ircle is the same distance from a single point in

the center.

Circle definitions:

Circumference

: The distance around a circle. This is the same a

s the perimeter of the circle.

Diameter

: The straight-line distance across a circle at its

widest point. One half of the diameter

of a circle is the radius of that circle. The diam

eter always passes through the center of a

circle.

Radius

: The straight-line distance from the center of a c

ircle to the edge of the circle. Twice

the radius of a circle is the diameter of that circ

le.

pi (π)

: pi is the ratio between the diameter of a circle

and the circumference of that circle. In

other words, if you take the circumference of a cir

cle and divide it by the diameter of that circle,

you get pi. pi is the same for all circles regardl

ess of their size, and for our purposes you can

consider it to always equal 3.1416.

Radi

us Diameter

Formula 7a:

Circle Area = π x Radius

2

Formula 7b:

Circle Area = π x Diameter

2

/ 4

Formula 8a:

Circle Circumference = π x Radius x 2

Formula 8b:

Circle Circumference = π x Diameter

7

Radius = 8 ft

Circle Example 1

π =

3.1416

Radius =

8 ft

Circle Area (Formula 7a) =

π x Radius

2

Area =

3.1416 x (8 ft x 8 ft)

Area =

3.1416 x 64 ft

2

Area =

201.06 ft

2

Circle Perimeter (Formula 8a) =

π x Radius x 2

Perimeter =

3.1416 x 8 ft x 2

Perimeter =

50.27 ft

Diameter = 16 ft

Circle Example 2

π =

3.1416

Diameter =

16 ft

Circle Area (Formula 7b) =

π x Diameter

2

/ 4

Area =

3.1416 x (16 ft x 16 ft) / 4

Area =

3.1416 x 256 ft

2

/ 4

Area =

201.06 ft

2

Circle Perimeter (Formula 8b) =

π x Diameter

Perimeter =

3.1416 x 16 ft

Perimeter =

50.27 ft

8

Irregular Shapes

Life would be easy if all we had to deal with in th

e garden were simple geometric shapes like squares,

rectangles, triangles, and circles. When dealing w

ith irregular shapes in the garden, there are two

relatively simple (but not the only) approaches: 1)

using a grid overlay to estimate the area, and 2)

calculating the area and perimeter by “exploding” a

n irregular shape into simpler geometric shapes.

Using A Grid Overlay

.

This method involves placing a grid of known scale

over a drawing of the irregular shape. It doesn’t

really matter what grid scale is used as long as it

is uniform. This method will give you an estimate

of

the area of an irregular shape, but will not give y

ou the perimeter measurement.

Scale: Grid Cells = 1 ft

2

To estimate the area of the irregular shape above,

look at each one of the grid cells and see how

much of it is occupied (or covered) by the shape.

You will assign a value to each of the grid cells:

if

all of a cell is occupied by the shape it has a val

ue of 1, if half of the cell is occupied the cell h

as a

value of ½ or 0.5, if only one quarter of the cell

is occupied it has a value of ¼ or 0.25, and so on.

Cells that are not occupied by any part of the shap

e have a value of 0 and can be ignored.

When you have established a value for each of the c

ells occupied by the shape, add them all together

to get an estimate of the irregular shape’s area (d

on’t forget to attach the scale units). In the irr

egular

shape above the estimated area is approximately 19.

75 ft

2

. Do you get the same area estimate?

To determine the perimeter of this shape, one metho

d would be to run a string along the shape

outline, and then remove the string and measure its

length.

9

“Exploding” An Irregular Shape Into Simpler Shapes.

This method involves breaking up a single irregular

shape into two or more simpler geometric shapes.

The following example is relatively complex:

Original Shape

“Exploded” Shape

12 ft

12 ft

8 ft

8 ft

A

B

8 ft

8 ft

C

D

8 ft

8 ft

Shape

Initial Area Calculations

Adjustments

Result

Rectangle A

Use Formula 3:

Area = 8 ft x 12 ft = 96 ft

2

None. 96.0 ft

2

Quarter-circle B

Use Formula 7a:

Area = 3.1416 x (8 ft)

2

=

201.1 ft

2

Only ¼ of the calculated area is

part of the total area.

50.3 ft

2

Square C

Use Formula 1:

Area = 8 ft x 8 ft = 64 ft

2

None. 64.0 ft

2

Semi-circle D

Use Formula 7b:

Area = 3.1416 x (8 ft)

2

/ 4 = 50.3 ft

2

Only ½ of the calculated area is

part of the total area.

25.1 ft

2

Total Area of the Original Shape:

235.4 ft

2

Shape

Initial Perimeter Calculations

Adjustments

Result

Rectangle A

Use Formula 4:

Perimeter = (12 ft x 2) + (8 ft X 2)

= 40 ft

One of the short (8 ft) sides of

the rectangle is not part of the

perimeter.

32.0 ft

Quarter-circle B

Use Formula 8a:

Circumference = 3.1416 x 8 ft x 2

= 50.3 ft

Only ¼ of the calculated

circumference is part of the

perimeter.

12.6 ft

Square C

Use Formula 2:

Perimeter = 8 ft x 4 = 32 ft

Only two sides of the square are

included in the perimeter.

16.0 ft

Semi-circle D

Use Formula 8b:

Circumference = 3.1416 x 8 ft

= 25.1 ft

Only ½ of the calculated

circumference is part of the

perimeter.

12.6 ft

Total Perimeter of the Original Shape:

73.2 ft

10

Volumes

We have seen how to compute one-dimensional or line

ar garden distances, such as the perimeter of

geometric shapes. We have also worked with areas,

the two-dimensional shapes that have both

width and length. Volumes are three-dimensional ob

jects, where one considers not only the physical

area covered, but also the height or thickness of t

he coverage.

A volume is the space contained within a three-dime

nsional object or space. It is usually expressed

as the cube of some linear measurement, such as cub

ic yard (yd

3

) or cubic centimeter (cc or cm

3

),

but terms such as gallon and liter are also used to

describe volumes.

The gardener most often deals with volumes when he

or she is adding mulch, compost, or a bulk soil

amendment to a bed. The instructions will often sa

y to cover the soil with 3 inches or 6 inches of

some material. Bulk materials vary greatly in weig

ht, so most are sold by volume, such as the cubic

foot (ft

3

) or cubic yard (yd

3

). In order to purchase the correct amount of bulk

material, you need to

determine what volume of material you need.

Follow these steps to determine the volume of mater

ial needed to cover an area to a specified depth:

1. Calculate the area of the space you wish to cove

r

. Use the formulas and examples presented

above to calculate or estimate the area to be cover

ed. For simplicity, whenever possible do

your measurements and calculations in the units use

d to supply the bulk material. In other

words, if the bulk material is sold by the cubic fo

ot, do your calculations in feet rather than

yards so there is no need to convert yards to feet

later.

2. After you have calculated the area to be covered

, multiply the area by the depth desired (using

the same units of measure)

. This is where you turn a two-dimensional shape (

area) into a

three-dimensional volume.

Warning

: Make sure you are using the same basic units of

measure for both the area

and the depth of the material! If you measured the

area in ft

2

, then you must use feet (or

a fraction of a foot) when you do this step or your

answer will be incorrect

.

For example, if you want to cover a 100 ft

2

area to a depth of 3 inches, multiply the 100 ft

2

area

by 0.25 ft (because 3 inches is 25% of a foot), giv

ing you the correct answer that 25 ft

3

of

material will be needed. If you were to multiply t

he 100 ft

2

directly by 3, you would get the

incorrect answer of 300 ft

3

of material, which is 12 times more material than

you need!

3. Consider how the material you wish to purchase i

s supplied

. Using the 25 ft

3

example above:

Volume

Needed

How Supplied

Calculations

What To Purchase

25 ft

3

Bulk Delivery in yd

3

. 1 yd

3

= 27 ft

3

Pick up or order delivery of 1

yd

3

(2 ft

3

will be left over).

25 ft

3

2/3 yd

3

bags.

1 yd

3

= 27 ft

3

2/3 yd

3

= 18 ft

3

25 ft

3

/ 18 ft

3

= 1.4 bags

Purchase 2 of the 2/3 yd

3

bags (11 ft

3

will be left over).

11

Converting Measurements

There are instances when you will need to convert f

rom one measurement unit to another. Examples

are when you need the number of square feet in a sq

uare yard, or to convert pounds to kilograms.

The table below gives some useful equivalents for g

ardeners.

Some examples:

– To convert 13.5 L to gal: multiply 13.5 L x 0.26

4 to get 3.56 gal.

– To convert 6 m

2

to yd

2

: multiply 6 m

2

by 1.196 to get 7.18 yd

2

.

– To convert 100 lbs to Kg: first multiply 100 lbs

by 453.6 to get 45360 g, then divide 45360 g

by 1000 to get the final answer of 45.36 Kg.

Weight

1 ounce (oz) avdp.

=

28.35 g 1 gram (g)

=

0.0353 oz avdp.

1 pound (lb)

=

453.6 g 1 kilogram (kg)

=

2.205 lb

1 short ton

=

0.9078 metric ton 1 metric ton

=

1.1016 short tons

Volume

1 cubic inch (in

3

)

=

16.39 ml 1 milliliter (ml)

=

0.0610 in

3

28.32 L 1 liter (L)

=

61.02 in

3

1 cubic foot (ft

3

)

=

7.48 gal (US)

264.2 gal (US)

27 ft

3

35.3 ft

3

1 cubic yard (yd

3

)

=

0.765 m

3

1 cubic meter (m

3

)

=

1.308 yd

3

Liquid Measure

1 teaspoon (tsp)

=

5 ml

1 ml

=

0.034 fl oz

1 Tablespoon (Tbsp)

=

3 teaspoons (tsp) 1.057 qt (US)

1 fluid ounce (fl oz)

=

29.6 ml

1 L

=

0.264 gal (US)

1 quart (qt) (US)

=

946.3 ml

1 gallon (gal) (US)

=

3.785 L

Area

1 square inch (in

2

)

=

6.452 cm

2

1 square cm (cm

2

)

=

0.155 in

2

1 square foot (ft

2

)

=

929.09 cm

2

10.76 ft

2

43,560 ft

2

1 sq meter (m

2

)

=

1.196 yd

2

1 acre

=

0.4047 ha

2.471 acres

1 hectare (ha)

=

1,000 m

2

Length

1 inch (in)

=

2.54 cm 1 cm

=

0.394 in

1 foot (ft)

=

30.48 cm

39.37 in

91.44 cm

3.281 ft

1 yard (yd)

=

0.9144 m

1 m

=

1.094 yd

Temperature

°F

=

(°C x 9/5) + 32 °C

=

(°F – 32) x 5/9

12

Typical Garden Situations

Note: Fertilizer and soil amendment application in

structions are usually given as a rate: a specific

amount of product per specified area. An example w

ould be an application rate of 3 lbs of

product per 100 ft

2

of area to be covered.

Example #1.

The soil pH of your established blueberry patch is

too high. You have been advised by the soil

testing laboratory to lower the soil pH by applying

a topdressing of elemental sulfur at a rate of

10 lbs of sulfur per 1000 ft

2

. Your blueberry patch is 25 ft wide by 50 ft long

. How much sulfur

do you need to purchase and use?

a. Determine the size of the area to be treated

. You want to treat a 25 ft x 50 ft rectangle,

so use Formula 3: area = 25 ft x 50 ft = 1250 ft

2

.

b. Compare your area with the application rate area

. The recommended amount of sulfur

to be used is based on a 1000 ft

2

area. 1250 ft

2

/ 1000 ft

2

= 1.25. Your area is 1.25

times larger.

c. Determine the amount of product to apply

. Since your area is 1.25 times larger than the

application rate area, you will need 1.25 times the

specified amount of sulfur:

10 lbs sulfur x 1.25 = 12.5 lbs sulfur.

Answer

: You will need to purchase 12.5 lbs of sulfur to c

over your 1250 ft

2

blueberry patch at a

rate of 10 lbs of sulfur per 1000 ft

2

of soil.

Example #2.

The soil testing laboratory has indicated you need

to raise the soil pH around your fruit tree

with an application of 7 cups of ground limestone p

er 100 ft

2

of soil being treated. You want to

amend the soil in a 15 ft circle out from your tree

. How much ground limestone do you need?

a. Determine the size of the area to be treated

. In this example the tree trunk represents

the center point of a circle with a 15 ft radius.

Use Formula 7a to find the area of this

circle: area = 3.1416 x (15 ft)

2

= 707 ft

2

.

b. Compare your area with the application rate area

. The recommended amount of

limestone to be used is based on a 100 ft

2

area. 707 ft

2

/ 100 ft

2

= 7.07. Your area is

7.07 times larger.

c. Determine the amount of product to apply

. Since your area is 7.07 times larger than the

application rate area, you will need 7.07 times the

specified amount of limestone:

7 cups limestone x 7.07 = 49.5 cups limestone.

Answer

: You will need to purchase 49.5 cups of limestone

to cover a 15 ft radius circle around

your tree at a rate of 7 cups of limestone per 100

ft

2

of soil.

13

Example #3.

You want to cover the following area with 4 inches

of compost, which is supplied in 2/3 yd

3

bags:

16 ft

4 ft

6 ft

6 ft

a. Determine the total area

. In this example you could break the irregular ar

ea into two

simpler shapes: a rectangle which is 4 ft by 16 ft,

and a square that is 6 ft by 6 ft. The

area of the rectangle can be found using Formula 3:

area = 4 ft x 16 ft = 64 ft

2

. The

area of the square can be found using Formula 1: ar

ea = (6 ft)

2

= 36 ft

2

. Add these two

areas together to get the total area: 64 ft

2

+ 36 ft

2

= 100 ft

2

.

b. Determine the volume of material needed

. Since the area measurements were done in

feet, you should use feet when calculating the volu

me. 4 inches of compost is equal to

0.33 ft of compost (4 inches is 1/3 of a foot). Mu

ltiply the total area by the desired depth

to get the volume of compost needed: 100 ft

2

x 0.33 ft = 33 ft

3

.

c. Determine the amount of product to purchase

. Since 1 yd

3

= 27 ft

3

, each 2/3 yd

3

bag of

compost contains 18 ft

3

of compost (2/3 of 27 is 18). Dividing the volume

you need by

how it is supplied will give you the number of 2/3

yd

3

bags needed: 33 ft

3

/ 18 ft

3

= 1.83

bags.

Answer

: Since partial bags are not sold, you will need to

purchase two 2/3 yd

3

bags of

compost to cover a 100 ft

2

area to a depth of 4 inches (with 3 ft

3

of the compost left over).

14

Note: On packages of commercial fertilizer the per

centage by weight of nitrogen (N), phosphorous

(P), and potassium (K) is shown with the N-P-K labe

l. For example, a 10-6-4 fertilizer contains

10% nitrogen, 6% phosphorous, and 4% potassium by w

eight, whereas a fertilizer with a label

of 42-0-0 (such as urea) contains 42% nitrogen by w

eight but no phosphorous or potassium.

The fact that the N-P-K label lists percentages by

weight is important, for if you have a 100 lb

bag of a fertilizer you can read directly how many

lbs of these nutrients it contains. For

example, a 100 lb bag of 42-0-0 fertilizer contains

42 lbs of nitrogen (100 lbs x 0.42 = 42 lbs).

Example #4.

You want to place 10-6-4 fertilizer on 300 ft

2

of your garden at a rate of 3 lb per 100 ft

2

. How much of

this fertilizer will you need?

a. Compare your area with the application rate area

. Divide your area by the application

rate area: 300 ft

2

/ 100 ft

2

= 3. Your area is 3 times larger.

b. Determine how much of the 10-6-4 fertilizer to u

se.

Since your area is 3 times larger

than the application rate area, you will need 3 tim

es the specified amount of fertilizer:

3 lb x 3 = 9 lbs.

Answer

: You will need 9 lbs of 10-6-4 fertilizer to treat

300 ft

2

of garden at a rate of 3 lbs of

fertilizer per 100 ft

2

.

Example #5.

You have a triangular plot, with a base of 10 ft an

d a height of 8 ft, and you wish to fertilize at a

rate of

1 lb of nitrogen per 1000 ft

2

. The 100 lb bag of fertilizer you have on hand ha

s a label of 12-0-0. How

much of this fertilizer should you use on this area

?

a. Determine the size of the area to be treated

. Use Formula 5 to find the area of this

triangle: area = 10 ft x 8 ft / 2 = 40 ft

2

.

b. Compare your area with the application rate area

. The amount of nitrogen desired is

based on a 1000 ft

2

area. Divide your area by the application rate ar

ea: 40 ft

2

/ 1000 ft

2

= 0.04. Your area is 0.04 (or 4%) of the applicati

on rate area.

c. Determine the amount of nitrogen to apply

. Since you are going to fertilize only 4% of

the application rate area, you only will need 4% of

the 1 lb of nitrogen. Multiply the 1 lb

of nitrogen by your percentage of the application r

ate area: 1 lb nitrogen x 0.04 = 0.04

lbs of nitrogen is needed for your area.

d. Determine how much of the 12-0-0 fertilizer to u

se

. In the 100 lb bag of 12-0-0 fertilizer

there is 12 lbs of nitrogen, or 0.12 lbs of nitroge

n per lb of fertilizer. Divide the desired

amount of nitrogen by the nitrogen in each lb of fe

rtilizer: 0.04 lb / 0.12 lb = .33 lbs.

Answer

: Use 1/3 of a lb of the 12-0-0 fertilizer on the 4

0 ft

2

triangular area to fertilize at a rate of

1 lb of nitrogen per 1000 ft

2

.