Garden Math

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
.