The Bugwood Network

Practice Problems

Forestry Problem Index
(click on appropriate number below)

Forestry #1 Forestry #2 Forestry #3 Forestry #4
Forestry #5 Forestry #6 Forestry #7 Forestry #8
Forestry #9 Forestry #10 Forestry #11 More #1
More #2 More #3 More #4 More #5
More #6 More #7 More #8 More #9
More #10 And more #19 And more #20 And more #21
And more #22 And more #23 Extra #a Extra #b


ROW & Aquatic Problem Index
(Click on desired problem)

Right of way #1 Right of way #2 Right of way #3 Right of way #4
Right of way #5 Right of way #6 Right of way #7 Right of way #8
Right of way #a Aquatics #1 Aquatics #2 Aquatics #3
Aquatics #4 Aquatics #5 Aquatics #6 Aquatics #7


Forestry problem #1

To prepare 3 gallons of a 3 percent Accord solution requires how many ounces of Accord?.

How much Accord is needed?

  • 3 gallons
  • 3% solution

3 gallons x 0.03 = .09 gallon
However it is easier to measure this as liquid ounces-- so…
0.09 gal x 128 liq oz / gal = 12 liq oz


Forestry Problem #2

A spray unit with a 400 gallon tank is calibrated to apply 20 gallons per acre. Instructions call for 3 pounds of active ingredient per acre of a 70 percent wettable powder. How many pounds of the product should be added to the tank?

  • 400 gallon tank
  • 20 gal/ac rate
  • 3 lb ai/ac
  • 70% ai product

Treated acres per tank
400 gal / tank / 20 gal / ac = 20 ac / tank

Treated acres per tank 20 ac / tank
Pounds active ingredient needed
20 ac x 3 lb ai / ac = 60 lb ai

Treated acres per tank
20 ac / tank
Pounds active ingredient needed
60 lb ai
Pounds of product needed
60 lb ai / 0 .7 lb ai / lb WP = 86 lb WP


Forestry Problem #3

A contractor arrives, and you request a calibration check. The contract requires an application rate of 16 gpa. The check run is made at the planned speed, expending 7 1/2 gallons. You measure the area sprayed, and it is 30 feet by 690 feet. Do you accept the application rate? (In most instances, 5 percent above or below the allowed rate is acceptable.)

  • Calibration check
  • Require 16 gpa
  • 5% tolerance
  • Sprayed 30’ x 690’
  • Expended 7.5 gals

How much area sprayed?

30 ft. x 690 ft. / 43,560 sq. ft / ac = 0.475 ac

How many gpa?

7.5 gal / 0.475 ac = 15.78 gpa

Rate O.K.?

15.78 gpa / 16 gpa = 98.6% of desired

Rate is O.K. (only 1.4% off)


Forestry Problem #4

You have a site preparation project. The spray unit tank holds 300 gallons and you want to apply 30 gallons of the mixture per acre. The prescription calls for 1.5 gallons of Garlon 4 per acre, 1/2 percent adjuvant, and 4 ounces of drift control agent per 100 gallons of spray mixture. How much of each would you add per tank?

  • 300 gal tank
  • 30 gal mix per ac
  • 1.5 gal Garlon 4 / ac
  • 0.5% adjuvant
  • 4 liq oz drift control / 100 gal mix

Acres per tank?

300 gal / tank / 30 gal / ac = 10 ac / tank

Gallons of Garlon 4?

10 ac x 1.5 gal / ac = 15 gal

Adjuvant?

300 gal x 0.005 = 1.5 gal

Drift control agent?

300 gal x 4 liq oz / 100 gal = 12 liq oz


Forestry Problem #5

You have a 30 acre area to be sprayed with 5 quarts of herbicide per acre in a 20 gallon/acre water spray mixture. The mixture is also to include 4 ounces of drift control agent per 100 gallons. How much of each chemical will you need to do the job, including water?

  • 30 acres to be sprayed
  • 20 gal spray mix / ac (gpa)
  • 5 qt prod / ac
  • 4 liq oz drift retardant / 100 gal mix

Total mixture needed

30 ac x 20 gal /ac = 600 gal

2,4-DP needed

5 qt / ac x 30 ac = 150 qt (37.5 gal)

Drift retardant needed

4 liq oz / 100 gal x 600 gal = 24 liq oz

Final Solution in tank – 600 gallons total mixture as

  • 37.5 gal 2,4-DP
  • 24 oz retardant
  • 562.25 gal of water

Forestry Problem #6

You plan to treat a 60 acre site preparation area with 3 quarts of Garlon 4 and 1 quart Accord per acre. How many gallons of each chemical do you need for the job?

  • 60 acres to be treated
  • 3 qt Garlon 4 +
  • 1 qt Accord / ac

Gallons Garlon 4?

60 ac x 3 qt / ac = 180 qt / 4 qt / gal = 45 gallons

Gallons Accord?

60 ac x 1 qt / ac = 60 qt / 4 qt / gal = 15 gal


Forestry Problem #7

A ground spray unit is calibrated to apply 40 gallons per acre at 4 mph. The spray unit speeds up to 8 mph. What is the new application rate?

  • 40 gal / ac output at 4 mph new speed 8 mph

Rule of thumb

Double the speed = 1/2 the coverage

40 gal / ac at 4 mph -- 40/2 gal / ac or 20 gal / ac at 8 mph


Forestry Problem #8

Another spray unit is calibrated to apply 25 gallons per acre at 4 mph. The spray pressure is increased from 15 psi to 60 psi. What is the new application rate?

Calibrated to deliver 25 gpa at 15 psi pressure increased to 60 psi.

New rate?

Rule of thumb
Rate increases as a function of the square root of the pressure increase.

60 psi / 15 psi = 4

New pressure is 4x old pressure

4 x pressure = 2 x the rate

25 gal x 2 = 50 gpa


Forestry Problem #9

You plan to inject a 40 acre stand with Accord. The Accord is to be diluted with two parts water to one part herbicide, and 1 ml of the diluted mixture is to be injected into each cut, with 2 inches between centers of the injector cuts. There are 400 stems per acre to be treated, with an average stem diameter of 7 inches. How many gallons of undiluted Accord is needed for the project?

  • 40 acres to be treated
  • Accord used at 1:2 in water
  • Inject 1 ml dilution / cut
  • Injection on 2 in. centers
  • 400 stems per acre
  • Average dbh is 7 in.
  • == Reorganize data ==
  • 40 acres to be treated
  • 400 stems per acre
  • Average dbh is 7 in.
  • Injection on 2 in. centers
  • Inject 1 ml dilution / cut
  • Accord used at 1:2 in water

Number of stems to treat

40 ac x 400 stems / ac = 16,000 stems

7 in. x 3.1416 = 21.98 in. circumference
22 in. / 2 in. center = 11 cuts per stem

Graphically this is:

7 in. dbh x 3.1416 (-- "pi") = 21.98 in. circumference

22 in. / 32 in. = 11 cuts / stem

16,000 stems

Number of cuts per stem

11 cuts per stem

Total number of cuts

16,000 stems x 11 cuts / stem = 176,000 cuts

Total ml of dilution needed

176,000 cuts x 1 ml / cut = 176,000 ml

176,000 ml = 176 L

Total Accord needed

176 L / 3.785 L/gal = 58.66 gal
58.66 gal / 3 [2 water :1 Accord] = 15.5 gals Accord


Forestry Problem #10

You have a pond which averages 20 feet wide, 100 feet long, and 5 feet deep, with a population of bluegills and rainbow trout. A five gallon container of Accord is accidentally spilled into the pond. Would you expect a fish kill? If so, what could be done to reduce or prevent the kill? (1 cubic foot of water = 28.32 liters; 1 pound = 453.6 grams.)

  • Pond size - average
  • 20 ft. wide - 5 ft. deep - 100 ft. long
  • Spill 5 gal Accord
  • Kill bluegill?
  • Kill channel catfish?

Volume of pond?

10,000 cu ft

Convert to metric

10,000 cu ft x 28.32 L / cu ft = 283,200 L

How much Accord in pond in metric?

  • 8.3 lb/gal water x 1.23 (spec. grav. of Accord) = 10.2 lb/gal Accord
  • 5 gal x 10.2 lb /gal x 453.6 gm / lb = 23,133.6 gm x 1,000 mg / gm = 23,133,600 mg

Volume of pond in metric

283,200 L

How much Accord in pond?

23,133,600 mg

Average Accord in pond?

23,133,600 mg / 283,200 L = 81.7 mg / L (= ppm) = 81.7 ppm

  • From MSDS -- LD50 s
  • Bluegill = >1,000 mg/L
  • Channel catfish = >1,000 mg/L

Average Accord in pond?

81.7 ppm

Neither bluegill nor catfish are expected to be seriously affected by this spill.

Had Roundup been spilled in the pond at the same 81.7 ppm

  • From MSDS -- LD50 s are
  • Bluegill = 5.8 mg/L
  • Channel catfish = 16 mg/L

Had Roundup been spilled in the pond at the same 81.7 ppm

Both would be expected to be seriously affected bluegill more than catfish.


Forestry Problem #11

You have another pond which averages 25 feet wide, 100 feet long, and 4 feet deep; it contains a population of bluegills and is regularly used by mallard ducks. A one gallon container of Garlon 3A is accidentally spilled into the pond. Would you expect a fish kill? Would you expect the ducks to die from eating fish, minnows, plants, and seeds from the pond? (1 cubic foot of water = 62.4 pounds; 1 gallon of water = 8.3 lbs.).

  • Pond size
    25 ft. wide x 4 ft. deep x 100 ft. long
  • 1 gal Garlon 3A spill
  • Bluegill and ducks are of concern

Volume of the pond

25 ft. x 4 ft. x 100 ft. = 10,000 cu. ft

1 cu ft = 7.48 gals

10,000 cu ft x 7.48 gal/cu ft = 74,800 gal

Concentration in the Pond

gal Garlon / gal pond x 1,000,000 = ppm Garlon in pond

1 / 74,800 x 1000000 = 13.4 ppm

At the concentration there is little risk to either fish (bluegill = 471 ppm) or ducks (>10,000 ppm) from this spill.


Forestry Problem More #1

You are supervising a crew performing pine release by directed foliar spray, using backpack sprayers. The herbicide tank mix calls for 0.4 ounces of Arsenal and 2 ounces of Bullseye Spray Pattern Indicator per gallon in water, plus 1/2% Cide-Kick by volume. Your backpacks hold three gallons apiece. How much of each chemical do you mix with water in each backpack?

  • 3 gal backpack
  • 0.4 liq oz Arsenal / gal
  • 2 liq oz Bullseye / gal
  • 0.5% Cide-Kick
  • 0.4 liq oz Arsenal / gal x 3 gal = 1.2 liq oz Arsenal
  • 2 liq oz Bullseye / gal x 3 gal = 6 liq oz Bullseye
  • 0.005 Cide-Kick x 3 gal x 128 liq oz / gal = (1.92) = 2 liq oz Cide-Kick

Forestry Problem More #2

Your contractor arrives at the work center with a 200-gallon nurse tank in his crew truck to pick up herbicide mix for site preparation. You are aware that DOT regulations prohibit transportation of more than 1,000 pounds of herbicide (total weight of material plus packaging) in a vehicle unless the driver has a commercial drivers licence with hazardous materials certification, and the contractor does not have such a license. -- Cont’d.

The empty nurse tank weighs 60 pounds, and the herbicide mix you are using is an emulsion of 5 ounces of Garlon 4 per gallon in water. The specific gravity of Garlon shown in the MSDS is 1.08 (1.08 times the weight of water), so you correctly assume that the Garlon will not significantly increase the weight of the mixture above the weight of plain water. Given that water weighs 8.3 pounds per gallon, how many gallons of herbicide mix should you put into the tank?

  • 200 gal nurse tank
  • Weighing 60 lb
  • 5 liq oz Garlon 4 / gal water
  • Spec grav of triclopyr = 1.08
  • Water = 8.3 lb / gal

At a specific gravity of 1.08 the effect is negligible on the total mixture weight so use 8.3 for entire mix.

1,000 lbs max weight – 60 lb tank = 940 lb allowable mixture / 8.3 lb/gal = 113.25 gal allowable mixture.

Probably wise to keep the load to 100 gal.


Forestry Problem More #3

You are preparing to do a site preparation job on a pine regeneration area, using an electric spray rig mounted on a four-wheeler. The sprayer has a 30-gallon tank, and your tank mix is to be 3 ounces of Garlon 4 and 0.4 ounces of Arsenal per gallon of mix, in water, plus 0.5% Cide-Kick. How much of each chemical do you need per tankful?

  • 30 gal tank
  • 3 liq oz Garlon 4 / gal
  • 0.4 liq oz Arsenal / gal
  • 0.5% Cide-Kick
  • 3 liq oz Garlon 4 / gal x 30 gal = 90 liq oz Garlon 4
  • 0.4 liq oz Arsenal / gal x 30 gal = 12 liq oz Arsenal
  • 0.005 Cide-Kick x 30 gal x 128 liq oz / gal = 19.2 liq oz Cide-Kick

Forestry Problem More #4

The spray rig in problem #3 has a boom equipped with four Spraying Systems #8003 flat fan nozzles. At its normal operating pressure of 40 PSI, what is the sprayer's application rate in gallons per minute?

  • Four nozzle boom
  • 8003 Spraying Systems flat fan nozzles
  • 40 psi is being used through the boom

8003 nozzle applies 0.3 gal/min at 40 psi at an 80o angle.

4 nozzles x 0.3 gpm / nozzle = 1.2 gpm.


Forestry Problem More #5

The spray rig above has an effective swath width of eight feet. You want to apply five gallons of tank mix per acre. At what speed (in feet per second) must you operate to achieve this application rate? What is this in miles per hour?

  • Spray covers an 8 foot wide swath
  • Puts out 1.2 gpm
  • Desire 5 gal / ac of the tank mix
  • 43,560 sq ft / 8 ft = 5445 ft
  • 5 gal / 1.2 gal / min = 4.17 min
  • 5445 ft / 4.17 min = 1306.8 ft / min
  • 1306.8 ft / min / 5280 ft / mi = .2475 mi / min x 60 min / hr = 14.85 mi / hr

Forestry Problem More #6

Maximum safe speed for the four-wheeler with the spray rig above in your site preparation area is 5 MPH. How could you modify the sprayer to apply herbicide at the desired rate without exceeding this speed?

Maximum safe speed is 5 mph.

From previous problem we know that we should be going 14.8 mph to get the desired output ( = 3 times the safe speed).

If we reduce speed to 5 mph we triple the rate of application, so we must reduce flow of product.

Changing the spray nozzles from 8003 to 8001 nozzles will reduce the output per nozzle from 0.3 gpm to 0.1 gpm which reduces the output the desired amount.

(What you would not do is adjust pressure in the system).


Forestry Problem More #7

You are going to do a site preparation project with Velpar L herbicide, using backpacks and gunjets equipped to apply 2 milliliters per "spot." Based on your soil type, you want to apply 5 pints per acre. What spacing should you use for your spot application grid?

  • 5 pt / ac Velpar L
  • Spot apply 2 ml per spot
  • Determine appropriate spacing for the spots (spot grid application)
  • 5 pt / ac x 0.125 gal / pt x 3,785 ml / gal = 2366 ml / ac
  • 2366 ml / ac / 2 ml / spot = 1183 spot / ac
  • 43,560 sq ft / ac / 1183 spot / ac = 36.8 sq ft / spot
  • 38.6 sq ft = 6.066 ft or a 6’ x 6’ grid

Forestry Problem More #8

You are planning a "streamline" application of Garlon 4 to release oak regeneration from hardwood competition in a group of stands following shelterwood cutting. The total area to be treated is 135 acres, and a preliminary check indicates that there are an average of 2,600 stems per acre to be treated. Experience tells you that you will use an average of 3 milliliters of herbicide mix per stem. How much of the herbicide mix will you need for the project?

  • 2600 stems / ac to be treated
  • 135 ac
  • 3 ml Garlon 4 per stem
  • 135 ac x 2,600 stems / ac x 3 ml / stem = 1,053,000 ml
  • 1,053,000 ml = 1,053 Liters
  • 1,053 L / 3.785 L / gal = (278.2 gal) or 280 gals

Forestry Problem More #9

For the project in problem #8, you are going to use a solution of Garlon 4 in JLB Oil Plus "Improved." In the past, you have ordered JLB Oil Plus in 30-gallon drums containing 25 gallons of oil, and prepared the mix by adding 5 gallons of Garlon to each drum. This year, you want to order the carrier oil in JLB's new 15-gallon plastic, returnable containers, and mix the herbicide in a 50-gallon nurse tank. To duplicate the mixture you formerly used, how much Garlon and how much oil would you put in each tankful?

  • 50 gal nurse tank
  • Ratio of 1 : 5 (Garlon 4 : JLB Oil Plus)

Garlon 4

1/ 6 x 50 gal = 8.3 gal.

JLB Plus

5/6 of mix x 50 gal = 41.7 gal.


Forestry Problem More #10

To get JLB Oil-Plus in the 15-gallon returnable containers, you have to order them by the pallet, which is 8 containers. How many pallets will you need to order for the project in problem #8?

  • 15 gal containers of JLB Oil Plus
  • Packaged by 8 containers / pallet
  • Need 280 gal of mix
  • 5/6 of mix is JLB Oil Plus
  • 15 gal / jug x 8 jug / pallet = 120 gal / pallet
  • 5/6 x 280 gal = 233.33 gal needed
  • 233 gal / 120 gal / pallet = 2 pallet

Forestry Problem More #19

A sprayer is calibrated to apply 15 gallons per acre (gpa) at a speed of 4 miles per hour (mph). What would the application rate be if the speed were slowed to 2 mph?

  • A. 7.5 gpa
  • B. 30 gpa
  • C. 20 gpa
  • D. 22.5 gpa
  • 15 gpa at 4 mph
  • Reduce to 2 mph
  • Half speed = double rate, or 2 x 15 gpa = 30 gpa

Forestry Problem More #20

A sprayer is calibrated to apply 15 gpa at a pressure of 20 psi. What is the pressure required to increase the output to 30 gpa without a change in speed, or a change in nozzles?

  • A. 40 psi
  • B. 10 psi
  • C. 80 psi
  • D. 60 psi
  • 15 gpa at 20 psi
  • Increase pressure to deliver 30 gpa
  • (30 gpa / 15 gpa)2 x 20 psi = 4 x 20 psi = 80 psi

Forestry Problem More #21

Prescription requires 8 gallons of herbicide in a 200 gallon spray solution. If you only need 50 gallons of solution, how many gallons of herbicide will you need?

  • A. 2 gals
  • B. 3 gals
  • C. 4 gals
  • D. 6 gals
  • Prescription: 8 gal herbicide / 200 gal mix
  • Require only 50 gal of mix
  • 50 gal / 200 gal x 8 gal = 2 gal

Forestry Problem More #22

You want to apply 1 1/2 (1.5) gals of pesticide per acre in 25 gallons of a mixture with water. How many gallons of pesticide would you need for 30 acres?

How much total mixture will you require?

How many gallons of water will you need?

  • 1.5 gal / ac
  • As a 25 gal mix in water
  • 30 ac
  • 1.5 gal / ac x 30 ac = 45 gals herbicide
  • 30 ac x 25 gal / ac = 750 gal total
  • 750 gal – 45 gal = 705 gal water

Forestry Problem More #23

The spray rate is 3 gpm, spray width is 40 feet, and the sprayer travels at a rate of 200 feet every 2 minutes. What is the spray rate in gallons per acre?

  • 3 gal / min
  • 200 ft / 2 min
  • 40 ft wide swath
  • 200 ft / 2 min x 40 ft = 4,000 sq ft / min
  • 43,560 sq ft / ac / 4,000 sq ft / min = 10.89 min / ac
  • 10.89 min / ac x 3 gal / min = 32.67 gpa

Forestry Problem #a

You must treat a 40 acre site by injection. There are an average of 3,000 stems per acre to be injected with 2 ml of herbicide. How much herbicide is needed?

  • 40 ac
  • 3000 stem / ac
  • 2 ml / stem
  • 40 ac x 3000 stem / ac x 2 ml / stem = 240,000 ml
  • 240,000 ml / 1,000 ml / L = 240 L = approximately 60 gallons

Forestry Problem #b

You must treat a 40 acre site by injection. There are an average of 200 stems per acre to be injected with 1 ml of herbicide per cut. Hack is on a 3” center and the average stem diameter is 5”. How much herbicide is needed?

  • 40 ac
  • 200 stem / ac
  • stem average 5 in. dbh
  • injected on 3 in. centers
  • 1 ml solution / cut

5 in. dbh x 3.1416 (-- "pi") = 15.7 in. circumference.

15.7 in. / 3 in. = 5 cuts/ stem.

5 cut / stem x 1 ml / cut x 200 stem / ac x 40 ac = approximately 40,000 ml = 40 L or 10 gallons.

Note that this number is a little low since the average stem required 5.2 cuts (= 10.5 gal solution).


Right-Of-Way (R-O-W) Problem #1

If you treat 15 feet on both sides of a road for 10 miles, how many acres were treated?

  • 10 mi of R-O-W
  • 15 ft. each side of road

How many sq ft

15 ft. x 2 x 5280 ft. / mi x 10 mi = 1,584,000 sq ft.

How many acres?

1,584,000 sq ft / 43,560 sq ft / ac = 36.4 ac.


R-O-W Problem #2

How many gallons of a herbicide is needed to treat 15 feet on both sides of a road for 5 miles at 2 gallons per acre?

  • 15 feet on both sides
  • 5 mi R-O-W
  • 2 gal / ac
  • 15 ft x 2 x 5 mi x 5280 ft / mi = 792,000 sq ft
  • 792,000 sq ft / 43560 sq ft / ac = 18.2 ac
  • 18.2 ac x 2 gal / ac = 36.4 gal

R-O-W Problem #3

A road right-of-way spray unit is calibrated to apply 25 gpa at 8 mph. The speed is reduced to 4 mph. What is the resulting application rate?

  • 25 gpa at 8 mph
  • Speed reduced to 4 mph
  • 25 gpa x 8 mph / 4 mph = 50 gpa

R-O-W Problem #4

Another spray unit is calibrated to apply 20 gpa at 6 mph. The spray pressure is reduced from 50 psi to 12.5 psi. What is the application rate now?

  • 25 gpa at 6 mph
  • Pressure reduced from 50 psi to 12.5 psi
  • 20 gpa / 50 psi / 12.5 psi = 10 gpa

R-O-W Problem #5

You have a spray tank which holds 400 gallons, and you want to apply 20 gallons of the mix per acre. The prescription calls for 1 ¼ gallons of 2,4-D per acre, 3 ozs of drift control agent per 100 gallons of mix, and 1% adjuvant. How much of each would you add per tank?

  • 400 gal tank
  • 20 gal / ac
  • 1.25 gal 2,4-D / ac
  • 3 liq oz drift control agent / 100 gal
  • 1% adjuvant
  • 400 gal / tank / 20 gal / ac = 20 ac / tank
  • 20 ac x 1.25 gal / ac = 25 gal 2,4-D
  • 400 gal x 3 liq oz / 100 gal = 12 liq oz drift control agent
  • 400 gal x 0.01 = 4 gal adjuvant

R-O-W Problem #6

You require a contractor to make a calibration check. The contract requires 25 gpa of mix to be applied. The contractor makes a calibration run at the planned speed expending 6 gallons. You measure the treatment area and it is 20 x 545 feet. Do you accept this application rate? (5% above or below the specified rate is considered acceptable.)

  • 25 gal / ac required
  • 6 gal expended on
  • 20 ft. x 545 ft. area
  • +/- 5% tolerance
  • 20 ft x 545 ft / 43560 sq ft / ac = 0.25 ac
  • 6 gal / 0.25 ac = 24 gpa
  • 24 gpa / 25 gpa = .96
  • 96% rate is acceptable (range is 95 – 105%)

R-O-W Problem #7

You plan to apply a 25 gallon per acre treatment of herbicide to 20 miles of right-of-way, spraying 15 feet on both sides of the road. The prescription calls for 1 ½ gallons of 2,4-D ester per acre plus 3 ounces of Poly Control (drift retardant) per 100 gallons of spray solution. Calculate the gallons of 2,4-D, ounces of drift retardant, and the minimum gallons of water needed for the project.

  • 25 gpa
  • 20 mi
  • 15 ft both sides
  • 1.5 gal 2,4-D / ac
  • 3 liq oz Poly Control / 100 gal mix
  • 20 mi x 5,280 ft / mi x 2 x 15 ft 43,560 sq ft / ac = 72.7 ac
  • 72.7 ac x 1.5 gal / ac = 109 gal 2,4-D
  • 72.7 ac x 25 gal / ac = 1817.5 gal of mix – 109 gal 2,4-D = 1,708.5 gal water
  • 1817.5 gal x 3 liq oz / 100 gal = 54 liq oz Poly Control

R-O-W Problem #8

You are transporting 20 gallons of Garlon 4 to a road right-of-way spray site. The transport truck overturns and one 5 gallon can burst spilling into a pond containing Bluegill (fish). You measure the pond and it is 200 feet wide, 320 feet long, and an average of 6 feet deep. Would you expect a fish kill? Notes:1 cu. ft. water = 28.32 liters; 1 lb = 453.6 gms.

  • 5 gal Garlon 4 spilled into a
  • 200 ft. x 320 ft. pond with average depth of 6’
  • Bluegill 96 hr LC50 is 0.87 mg/L
  • 200 ft x 320 ft x 6 ft x 28.32 L /cu ft = 10,874,880 L
  • 8.33 lb / gal (water) x 1.08 (spec grav of G4) x 5 gal (G4) x 453.6 gm / lb x 1000 mg / gm = 20,389,300 mg in 5 gal G4
  • 20,389,200 mg / 10,874,880 L = 1.88 mg (G4) / L (pond water)
  • 1.88 mg/L > 0.87 mg/L – fish kill expected

R-O-W Problem #a

You are planning to treat a 12’ swath on each side of a road R-O-W for 6 miles. How many acres are to be treated? How much Krenite will you nee to treat the R-O-W with 2 gpa of Krenite?

  • 6 mi of R-O-W
  • 12 ft. each side of road
  • 2 gal Krenite / ac

How many sq. ft

12 ft. x 2 x 5280 ft. / mi x 6 mi = 760,320 sq ft

How many acres?

760,320 sq ft / 43,560 sq ft / ac = 17.45 ac

How much Krenite?

17.5 ac x 2 gal / ac = 35 gal


Aquatics Problem #1

A pond has a 20 surface acres and an average depth of 2 feet. How many acre feet of water does it contain?

  • 20 surface acres
  • Average 2 ft. deep
  • 20 ac x 2 ft = 40 ac ft

Aquatics Problem #2

Another pond is 100 feet long by 500 feet wide with an average depth of 4 feet. How many acre feet of water does it contain?

  • Pond: 100 ft. long x 500 ft. wide x 4 ft. average depth
  • 100 ft x 500 ft x 4 ft / 43560 cu ft / ac ft = 4.6 ac ft

Aquatics Problem #3

You plan to apply 30 pounds of a chemical per acre foot of water. The surface of the water body is 300 feet by 10 feet with an average depth of 10 feet. How many pounds of a 50% granule do you need?

  • 30 lb chemical / ac ft
  • Using a 50% granule
  • 300’ long x 10’ wide with average depth of 10’
  • 300 ft x 10 ft x 10 ft / 43,560 cu ft / ac ft = 0.689 ac ft
  • 0.689 ac ft x 30 lb / ac ft x 2 (or divide by 0.5) = (41.34 lbs) = 42 lbs

Aquatics Problem #4

Pro-Noxfish (Rotenone) is normally applied at the rate of 1 gallon per 2 acre feet being treated. How many gallons of Pro-Noxfish are required to treat a pond 420 ft by 105 ft with an average depth of 4 ft?

  • 1 gal / 2 ac ft
  • 420 ft. long x 105 ft. wide by average 4 ft. deep
  • 420 ft x 105 ft x 4 ft / 43,560 cu ft / ac ft x 1 gal / 2 ac ft = 2 gal

Aquatics Problem #5

You are applying herbicide with a 30 foot spray boom at a rate of 50 feet per minute. If you apply 2 gallons per minute, how many gallons are applied per surface acre?

  • 30 ft spray boom

  • 50 ft / min

  • 2 gal / min

  • 30 ft x 50 ft / min / 43,560 sq ft / ac = 0.034 ac / min
  • 2 gal / min / 0.034 ac / min = 58 gpa

Aquatics Problem #6

Cutrine Plus liquid is labeled for the control of algae. A pond (10 surface acres x 4 ft deep) is treated. How much Cutrine Plus is needed to treat the pond at a rate of 0.4 ppm? [HINT: Check the label!]

  • 10 surface ac x 4 ft
  • Desired rate is 0.4 ppm
  • From the label – 4 ft deep requires 4.8 gal / surface ac
  • 4.8 gal / ac x 10 ac = 48 gal

Aquatics Problem #7

An applicator is planning to apply 0.3 ppm of a herbicide to one surface acre of a 2 acre pond averaging 4 feet deep. While the 10 gallon container of herbicide is being loaded into the boat the bottom ruptures, spilling the entire contents into the lake. The label indicates that 1 ppm in water may result in fish kill. The product contains 2 lbs a.i./gal and its toxicity is based entirely on the a.i. -- Cont’d.

Would you expect a fish kill in part of the pond, all of it, or none of it? What could have been done to prevent the accident? Notes: 1 cu ft of water = 62.4 lbs.

  • 10 gal spill
  • 2 lb a.i. / gal – toxicity based entirely on the a.i
  • 2 ac x 4 ft. deep
  • 1 ppm may result in fish kill
  • 2 ac x 4 ft x 43,560 sq ft / ac x 62.4 lbs / cu ft = 21,745,152 lb of water in lake
  • 2 lb (ai) x 1,000,000 /21,745,152 lb (water) = 0.92 ppm
  • 0.92 ppm < 1 ppm
  • There should not be a significant fish kill

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Last updated on Thursday, November 07, 2002 at 11:53 AM
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