The Tale of the Tape
Measuring Pond Surface Area and
Volume for Stocking and Chemical
Applications
Billy Higginbotham
Professor and Extension Wildlife and
Fisheries Specialist
Texas Cooperative Extension
How Big is Your Pond?

“It’s a 2,000 yard pond.”

“My pond’s pretty small.”

It’s about an acre, I guess.”

“My pond’s real big.”
Why Measure?

Surface area
– Fish stocking
– Carrying capacity
– A few chemical applications

Volume - most chemical applications
What Do I Measure?

Surface area in acres

Volume in acre-feet
So What is an Acre?

1 surface acre = 43,560 square feet
Equivalent to: a square 209’ x 209’
a circle 235’ across
How Do I Measure Surface Area?

Pace it off

Tape measure

Range finders
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Aerial photos

Geographic Information Systems (GIS)
Determining Surface Area
Three basic pond shapes

Square or rectangular

Triangular

Circular

(or a combination of 2 of the 3)
Area of a Square or Rectangle
Length
Width
An Acre-foot of Water is:

1 surface acre 1 foot deep

43,560 cubic feet

326,000 gallons
Determining Pond Volume
Pond Volume = surface area (acres) x
average depth (feet)
How Do I Measure Depth?

Guess!

Depth finder

Calibrated rope and anchor

Calibrated pole
Measuring Pond Depth

Take at least 2 perpendicular transects

Begin at one bank with “zero” and end with “zero”

The more transects and depth soundings taken,
the better your depth estimate becomes

Average depth is always LESS than you
suspected BM (Before Measuring)!
Measuring Depth
x
x
x
x
x
x
x x x x x x x x x x x x x x x x x x x x x x x x x x x
x
x
x
x
x
x
Determining Average Depth
Average depth = sum of soundings
(feet)  number of soundings
Transect 1 - 0,2,3,6,6,8,7,5,3,1,0
Transect 2 - 0,1,1,1,3,4,8,6,6,4,1,0
76 feet  23 = 3.3 feet
Determining Average Depth
If you do not include zeros:
Transect 1 - 2,3,6,6,8,7,5,3,1
Transect 2 - 1,1,3,4,8,6,6,4,1
76 feet  19 = 4.0 feet
Determining Surface Area in Acres
Square or Rectangular Pond
Length (ft.) x Width (ft.)  43,560=Acres
Determining Surface Area in Acres
Square or Rectangular Pond
Length (ft.) x Width (ft.)  43,560=Acres
Example: Pond is 200’ x 350’ = 70,000 square feet
70,000  43,560 =1.6 surface acres
Determining Volume in Acre-Feet
Square or Rectangular Pond
Example: pond is 200’ x 350’ = 70,000 square feet
70,000  43,560 = 1.6 acres
Average depth = 3.2 feet
1.6 acres x 3.2 feet = 5.1 acre-feet of water
Determining Surface Area in Acres
Triangular Pond
1/2 [base (ft.) x height (ft.)]  43,560 = acres
Area of a Triangle
Base
Height
Determining Surface Area in Acres
Triangular Pond
1/2 [base (ft.) x height (ft.)]  43,560 = acres
Example: pond is 200’ along the dam x 500’ to upper end
1/2 (200 x 500)  43,560 = 1.1 surface acres
Measuring Depth
x
x
x x x x x x x x
x
x
x x x x x
x
x
x
Determining Volume in Acre-Feet
Triangular Pond
Example: pond averages 4.2 feet deep and is 200’
along the dam x 500’ to upper end
1/2 (200 x 500) = 50,000  43,560
1.1 surface acres
1.1 surface acres x 4.2 feet
4.6 acre-feet of water
Determining Surface Area in Acres
Circular Pond
 x (radius)2  43,560 = acres
 is the ratio of the pond circumference to the pond diameter
and approximates 3.14
The pond radius is 1/2 of the pond diameter
Determining Surface Area in Acres
Circular Pond
 x (radius)2  43,560 = acres
Example: Pond is 150’ across the middle (diameter)
 = 3.14 and radius = 75’, so
3.14 x (75)2  43,560 = 0.41 acres
Area of a Circle
Radius
Area of a Circle
Diameter
Measuring Depth
x
x
x
x
x x x x x x x x x x x x x x
x
x
x
x
x
x
Determining Volume in Acre-Feet
Circular Pond
Example: pond averages 3.9 feet deep and is 150’ across the middle
(diameter)
3.14 x (75)2  43,560
17,662.5  43,560 = 0.41 acres
0.41 acres x 3.9 feet = 1.6 acre-feet of water
Pond Math

My pond is 0.5 surface acres and averages
4 feet deep. How much powdered 5%
Rotenone do I need to apply at the
recommended rate of 10 pounds per acrefoot of water to kill my fish so I can restock?
Pond Math

My pond is 0.5 surface acres and averages 4 feet
deep. How much powdered 5% Rotenone do I
need to apply at the recommended rate of 10
pounds per acre-foot of water to kill my fish so I
can re-stock?
0.5 surface acres x 4 feet deep = 2 acre-feet
2 acre-feet of water x 10 lbs Rotenone per acrefoot = 20 lbs of Rotenone needed
Pond Math
Information from Texas Cooperative
Extension states that I can stock 1,000
catfish per surface acre if I feed 5-7 days
per week from April through October. How
many catfish should I order for my triangleshaped pond that is 125 feet across at the
dam and 200 feet long from the dam to the
upper end?
Pond Math
Information from Texas Cooperative Extension states that I
can stock 1,000 catfish per surface acre if I feed 5-7 days
per week from April through October. How many catfish
should I order for my triangle-shaped pond that is 125 feet
across at the dam and 200 feet long from the dam to the
upper end?
1/2 b x h = 1/2 (125 x 200)
12,500  43,560 = 0.29 surface acres
1,000 fingerlings per acre x 0.29 acres = 290
fingerlings
Pond Math

I have a pond that is a rectangle on one side
connected by a culvert under my driveway to a
triangular shaped upper end. The rectangle side is
100 feet by 150 feet and the average depth is 6.2
feet. The triangle is 100 feet along the driveway
and runs up into a neck 75 feet from the driveway
and average depth is only 3.0 feet. How much
copper sulfate do I need to apply at 2 pounds per
acre-foot of water to kill the filamentous algae
covering the entire pond?
Rectangular Side
100 x 150 = 15,000
15,000  43,560 = 0.34 acres
0.34 x 6.2 = 2.1 acre-feet of water
2 lbs of copper sulfate per acre-foot x 2.1 acrefeet = 4.2 lbs of copper sulfate
Triangular Side
1/2 (100 x 75) = 3750
3750  43,560 = 0.09 acres
0.09 acres x 3 feet = 0.27 acre-feet of water
2 lbs of copper sulfate per acre-foot x 0.27 acrefeet = 0.54 lbs of copper sulfate
Spot Treatments




When fish are important pond resources
When vegetation covers more than 50% of the
pond
When treatment is required in the summer
months
To spot-treat, apply
herbicide to no more
than 20% of the total
pond surface at weekly
intervals
Spot Treatment
I have bushy pondweed growing from the pond edge
to about 20 feet out around 250 feet of shoreline.
The water where the weeds are growing averages
2 feet in depth. How much
Aquathol Super K do I
need to buy to treat
the affected area if the
application rate is 3ppm
(13.2 pounds per acre-foot
of water)?
Spot Treatment
250’ x 20’ = 5,000 square feet of area to be treated
5,000  43,560 = 0.11 surface acres
0.11 acres x 2 feet average depth = 0.22 acre-feet
0.22 acre-feet x 13.2 pounds of Aquathol Super K per acre-foot
2.9 pounds of Aquathol Super K
Restricted Use Pesticides for
Aquatic Applications

Rotenone

2,4-D products