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The Tale of the Tape

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.”

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The Tale of the Tape

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  1. The Tale of the Tape Measuring Pond Surface Area and Volume for Stocking and Chemical ApplicationsBilly Higginbotham Professor and Extension Wildlife and Fisheries Specialist Texas Cooperative Extension

  2. 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.”

  3. Why Measure? • Surface area • Fish stocking • Carrying capacity • A few chemical applications • Volume - most chemical applications

  4. What Do I Measure? • Surface area in acres • Volume in acre-feet

  5. So What is an Acre? • 1 surface acre = 43,560 square feetEquivalent to: a square 209’ x 209’ a circle 235’ across

  6. How Do I Measure Surface Area? • Pace it off • Tape measure • Range finders • Aerial photos • Geographic Information Systems (GIS)

  7. Determining Surface Area Three basic pond shapes • Square or rectangular • Triangular • Circular • (or a combination of 2 of the 3)

  8. Area of a Square or Rectangle Length Width

  9. An Acre-foot of Water is: • 1 surface acre 1 foot deep • 43,560 cubic feet • 326,000 gallons

  10. Determining Pond Volume Pond Volume = surface area (acres) x average depth (feet)

  11. How Do I Measure Depth? • Guess! • Depth finder • Calibrated rope and anchor • Calibrated pole

  12. 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)!

  13. 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

  14. 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

  15. 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

  16. Determining Surface Area in Acres Square or Rectangular Pond Length (ft.) x Width (ft.)  43,560=Acres

  17. 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

  18. 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

  19. Determining Surface Area in Acres Triangular Pond1/2 [base (ft.) x height (ft.)]  43,560 = acres

  20. Area of a Triangle Base Height

  21. Determining Surface Area in Acres Triangular Pond1/2 [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

  22. Measuring Depth x x x x x x x x x x x x x x x x x x x x

  23. 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

  24. 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 3.14The pond radius is 1/2 of the pond diameter

  25. 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

  26. Area of a Circle Radius

  27. Area of a Circle Diameter

  28. 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

  29. 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

  30. 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?

  31. 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 acre-foot = 20 lbs of Rotenone needed

  32. 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?

  33. 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

  34. 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?

  35. 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 acre-feet = 4.2 lbs of copper sulfate

  36. 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 acre-feet = 0.54 lbs of copper sulfate

  37. Spot Treatments • When fish are important pond resources • When vegetation covers more than 50% of thepond • When treatment is required in the summer months • To spot-treat, applyherbicide to no morethan 20% of the totalpond surface at weeklyintervals

  38. 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)?

  39. 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

  40. Restricted Use Pesticides for Aquatic Applications • Rotenone • 2,4-D products

  41. The End

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