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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 *. Forestry Problem Index (click on appropriate number below).

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Forestry problem index click on appropriate number below l.jpg

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 *

Forestry Problem Index(click on appropriate number below)

Go To Right-of-Way or Aquatic Problems Index *


Forestry problem index continued click on the appropriate number l.jpg

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 *

Forestry Problem Index (continued)Click on the appropriate number


Row aquatic problem index click on desired problem l.jpg

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 *

ROW & Aquatic Problem Index(Click on desired problem)

Go To Forestry Problem Index *


Forestry problem 1 l.jpg
Forestry problem #1

To prepare 3 gallons of a 3 percent Accord solution requires how manyounces of Accord?


Forestry prob 1 l.jpg

FORESTRY PROB #1

How much Accord is needed?

3 gallons

3% solution

FORESTRY PROB #1 1/2


Go to forestry problem index l.jpg

GO TO FORESTRY PROBLEM INDEX

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 PROB #1 2/2


Forestry problem 2 l.jpg
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?


Forestry prob 2 l.jpg

FORESTRY PROB #2

400 gallon tank

20 gal/ac rate

3 lb ai/ac

70% ai product

FORESTRY PROB #2 1/4


Forestry prob 2 2 4 l.jpg

FORESTRY PROB #2 2/4

Treated acres per tank

400 gal / tank / 20 gal / ac =

20 ac / tank


Forestry prob 2 3 4 l.jpg

FORESTRY PROB #2 3/4

Treated acres per tank

20 ac / tank

Pounds active ingredient needed

20 ac x 3 lb ai / ac =

60 lb ai


Forestry prob 2 4 4 l.jpg

FORESTRY PROB #2 4/4

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

GO TO FORESTRY PROBLEM INDEX


Forestry problem 3 l.jpg
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.)


Forestry prob 3 l.jpg

FORESTRY PROB #3

Calibration check

Require 16 gpa

5% tolerance

Sprayed 30’ x 690’

Expended 7.5 gals

FORESTRY PROB #3 1/4


How much area sprayed 30 ft x 690 ft 43 560 sq ft ac 0 475 ac l.jpg

How much area sprayed?

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

0.475 ac

FORESTRY PROB #3 2/4


How much area sprayed 475 ac how many gpa 7 5 gal 0 475 ac 15 78 gpa l.jpg

How much area sprayed?

.475 ac

How many gpa?

7.5 gal / 0.475 ac =

15.78 gpa

FORESTRY PROB #3 3/4


Slide16 l.jpg

How much area sprayed?

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)

GO TO FORESTRY PROBLEM INDEX

FORESTRY PROB #3 4/4


Forestry problem 4 l.jpg
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?


Forestry prob 4 l.jpg

FORESTRY PROB #4

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

FORESTRY PROB #4 1/5


Acres per tank 300 gal tank 30 gal ac 10 ac tank l.jpg

Acres per tank?

300 gal / tank / 30 gal / ac =

10 ac / tank

FORESTRY PROB #4 2/5


Acres per tank 10 ac tank gallons of garlon 4 10 ac x 1 5 gal ac 15 gal l.jpg

Acres per tank?

10 ac / tank

Gallons of Garlon 4?

10 ac x 1.5 gal / ac =

15 gal

FORESTRY PROB #4 3/5


Acres per tank 10 ac tank gallons of garlon 4 15 gal adjuvant 300 gal x 0 005 1 5 gal l.jpg

Acres per tank?

10 ac / tank

Gallons of Garlon 4?

15 gal

Adjuvant?

300 gal x 0.005 =

1.5 gal

FORESTRY PROB #4 4/5


Go to forestry problem index22 l.jpg

GO TO FORESTRY PROBLEM INDEX

Acres per tank?

10 ac / tank

Gallons of Garlon 4?

15 gal

Adjuvant?

1.5 gal

Drift control agent?

300 gal x 4 liq oz / 100 gal =

12 liq oz

FORESTRY PROB #4 5/5


Forestry problem 5 l.jpg
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?


Forestry prob 5 l.jpg

FORESTRY PROB #5

30 acres to be sprayed

20 gal spray mix / ac (gpa)

5 qt prod / ac

4 liq oz drift retardant / 100 gal mix

FORESTRY PROB #5 1/5


Total mixture needed 30 ac x 20 gal ac 600 gal l.jpg

Total mixture needed

30 ac x 20 gal /ac =

600 gal

FORESTRY PROB #5 2/5


Total mixture needed 600 gal 2 4 dp needed 5 qt ac x 30 ac 150 qt 37 5 gal l.jpg

Total mixture needed

600 gal

2,4-DP needed

5 qt / ac x 30 ac=

150 qt (37.5 gal)

FORESTRY PROB #5 3/5


Slide27 l.jpg

Total mixture needed

600 gal

2,4-DP needed

37.5 gal

Drift retardant needed

4 liq oz / 100 gal x 600 gal =

24 liq oz

FORESTRY PROB #5 4/5


Go to forestry problem index28 l.jpg

GO TO FORESTRY PROBLEM INDEX

Final Solution in tank

600 gallons total mixture

as

37.5 gal 2,4-DP

24 oz retardant

562.25 gal of water

FORESTRY PROB #5 5/5


Forestry problem 6 l.jpg
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?


Forestry prob 6 l.jpg

FORESTRY PROB #6

60 acres to be treated

3 qt Garlon 4 +

1 qt Accord / ac

FORESTRY PROB #6 1/2


Go to forestry problem index31 l.jpg

GO TO FORESTRY PROBLEM INDEX

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 PROB #6 2/2


Forestry problem 7 l.jpg
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?


Forestry prob 7 l.jpg

FORESTRY PROB #7

40 gal / ac output

at 4 mph

new speed 8 mph

FORESTRY PROB #7 1/2


Go to forestry problem index34 l.jpg

GO TO FORESTRY PROBLEM INDEX

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 PROB #7 2/2


Forestry problem 8 l.jpg
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?


Forestry prob 8 l.jpg

FORESTRY PROB #8

Calibrated to deliver 25 gpa

at 15 psi

pressure increased to 60 psi

FORESTRY PROB #8 1/3


Slide37 l.jpg

New rate?

Rule of thumb

Rate increases as a function of the square root of the pressure increase

60 psi / 15 psi =

4

FORESTRY PROB #8 2/3


Go to forestry problem index38 l.jpg

GO TO FORESTRY PROBLEM INDEX

New rate?

New pressure is 4 x old pressure

4 x pressure = 2 x the rate

25 gal x 2 =

50 gpa

FORESTRY PROB #8 3/3


Forestry problem 9 l.jpg
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?


Forestry prob 9 l.jpg

FORESTRY PROB #9

40 acres to be treated

Accord used at 1:2 in water

Inject 1 ml dilution / cut

Injection on 2” centers

400 stems per acre

Average dbh is 7”

FORESTRY PROB #9 1/9


Slide41 l.jpg

== Reorganize data ==

40 acres to be treated

400 stems per acre

Average dbh is 7”

Injection on 2” centers

Inject 1 ml dilution / cut

Accord used at 1:2 in water

FORESTRY PROB #9 2/9


Number of stems to treat 40 ac x 400 stems ac 16 000 stems l.jpg

Number of stems to treat

40 ac x 400 stems / ac =

16,000 stems

FORESTRY PROB #9 3/9


Slide43 l.jpg

Number of stems to treat

16,000 stems

Number of cuts per stem

7 in x 3.1416 = 21.98 in circumference

22 in/stem / 2 in/cut =

11 cuts per stem

FORESTRY PROB #9 4/9


7in dbh x 3 1416 pi 21 98 in circumference l.jpg
7in dbh x 3.1416 (-- “pi”) = 21.98 in circumference

Graphically this is:

7”

FORESTRY PROB #9 5/9


22 in stem 2 in cut 11 cuts stem l.jpg
22 in/stem / 2 in/cut = 11 cuts / stem

x

x

x

x

x

x

2 in

x

x

x

x

x

FORESTRY PROB #9 6/9


Slide46 l.jpg

Number of stems to treat

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

FORESTRY PROB #9 7/9


Total number of cuts 176 000 cuts total ml of dilution needed 176 000 cuts x 1 ml cut 176 000 ml l.jpg

Total number of cuts

176,000 cuts

Total ml of dilution needed

176,000 cuts x 1 ml / cut =

176,000 ml

FORESTRY PROB #9 8/9


Go to forestry problem index48 l.jpg

GO TO FORESTRY PROBLEM INDEX

Total number of cuts

176,000 cuts

Total ml of dilution needed

176,000 ml = 176 L

Total Accord needed

176 L / 3.785 L/gal = 46.5 gal

46.5 gal / 3 [2 water :1 Accord] = 1.5 15.5 gals Accord

FORESTRY PROB #9 9/9


Forestry problem 10 l.jpg
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.)


Forestry prob 10 l.jpg

FORESTRY PROB #10

Pond size - average

20’ wide - 5’ deep - 100’ long

Spill 5 gal Accord

--

Kill bluegill?

Kill rainbow trout?

FORESTRY PROB #10 1/9


Volume of pond 20 ft x 5 ft x 100 ft 10 000 cu ft l.jpg

Volume of pond?

20 ft x 5 ft x 100 ft =

10,000 cu ft

FORESTRY PROB #10 2/9


Volume of pond 10 000 cu ft convert to metric 10 000 cu ft x 28 32 l cu ft 283 200 l l.jpg

Volume of pond?

10,000 cu ft

Convert to metric

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

283,200 L

FORESTRY PROB #10 3/9


Slide53 l.jpg

Volume of pond in metric

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

FORESTRY PROB #10 4/9


Slide54 l.jpg

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)

FORESTRY PROB #10 5/9


Average accord in pond 81 7 ppm from msds ld 50 s bluegill 1 000 mg l channel catfish 1 000 mg l l.jpg

Average Accord in pond?

81.7 ppm

From MSDS -- LD50 s

Bluegill = >1,000 mg/L

Channel catfish = >1,000 mg/L

FORESTRY PROB #10 6/9


Slide56 l.jpg

Average Accord in pond?

81.7 ppm

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

FORESTRY PROB #10 7/9


Slide57 l.jpg

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

FORESTRY PROB #10 8/9


Go to forestry problem index58 l.jpg

GO TO FORESTRY PROBLEM INDEX

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 PROB #10 9/9


Forestry problem 11 l.jpg
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 plants, and seeds from the pond (assme they are contaminated at the same rate as the water)?


Forestry prob 11 l.jpg

FORESTRY PROB #11

Pond size

25’ wide x 4’ deep x 100’ long

1 gal Garlon 3A spill

Bluegill and ducks are of concern

FORESTRY PROB #11 1/5


Volume of the pond 25 ft x 4 ft x 100 ft 10 000 cu ft l.jpg

Volume of the pond

25 ft x 4 ft x 100 ft =

10,000 cu. ft

FORESTRY PROB #11 2/5


Volume of the pond 10 000 cu ft 1 cu ft 7 48 gals 10 000 cu ft x 7 48 gal cu ft 74 800 gal l.jpg

Volume of the pond

10,000 cu. ft

1 cu ft = 7.48 gals

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

74,800 gal

FORESTRY PROB #11 3/5


Slide63 l.jpg

Volume of the pond

74,800 gal

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

ppm Garlon in pond

1 / 74,800 x 1000000 =

13.4 ppm

FORESTRY PROB #11 4/5


Go to forestry problem index64 l.jpg

GO TO FORESTRY PROBLEM INDEX

Concentration in the pond

13.4 ppm

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

FORESTRY PROB #11 5/5


Forestry problem more 1 l.jpg
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?


Forestry more 1 l.jpg

FORESTRY More #1

3 gal backpack

0.4 liq oz Arsenal / gal

2 liq oz Bullseye / gal

0.5% Cide-Kick

FORESTRY PROB #More 1 1/2


Go to forestry problem index67 l.jpg

GO TO FORESTRY PROBLEM INDEX

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 PROB #More 1 2/2


Forestry problem more 2 l.jpg
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


Forestry problem more 269 l.jpg
Forestry Problem More #2

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?


Forestry more 2 l.jpg

FORESTRY More #2

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

FORESTRY PROB #More 2 1/2


Go to forestry problem index71 l.jpg

GO TO FORESTRY PROBLEM INDEX

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 PROB #More 2 2/2


Forestry problem more 3 l.jpg
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?


Forestry more 3 l.jpg

FORESTRY More #3

30 gal tank

3 liq oz Garlon 4 / gal

0.4 liq oz Arsenal / gal

0.5% Cide-Kick

FORESTRY PROB #More 3 1/2


Go to forestry problem index74 l.jpg

GO TO FORESTRY PROBLEM INDEX

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 PROB #More 3 2/2


Forestry problem more 4 l.jpg
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?


Forestry more 4 l.jpg

FORESTRY More #4

Four nozzle boom

8003 Spraying Systems flat fan nozzles

40 psi is being used through the boom

FORESTRY PROB #More 4 1/2


Go to forestry problem index77 l.jpg

GO TO FORESTRY PROBLEM INDEX

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 PROB #More 4 2/2


Forestry problem more l.jpg
Forestry Problem More #

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? 


Forestry more 5 l.jpg

FORESTRY More #5

Spray covers an 8 foot wide swath

Puts out 1.2 gpm

Desire 5 gal / ac of the tank mix

FORESTRY PROB #More 5 1/2


Go to forestry problem index80 l.jpg

GO TO FORESTRY PROBLEM INDEX

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 PROB #More 5 2/2


Forestry problem more 6 l.jpg
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?


Forestry more 6 l.jpg

FORESTRY More #6

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)

FORESTRY PROB #More 6 1/2


Go to forestry problem index83 l.jpg

GO TO FORESTRY PROBLEM INDEX

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 PROB #More 6 2/2


Forestry problem more 7 l.jpg
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?


Forestry more 7 l.jpg

FORESTRY More #7

5 pt / ac Velpar L

Spot apply 2 ml per spot

Determine appropriate spacing for the spots (spot grid application)

FORESTRY PROB #More 7 1/2


Go to forestry problem index86 l.jpg

GO TO FORESTRY PROBLEM INDEX

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 PROB #More 7 2/2


Forestry problem more 8 l.jpg
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?


Forestry more 8 l.jpg

FORESTRY More #8

2600 stems / ac to be treated

135 ac

3 ml Garlon 4 per stem

FORESTRY PROB #More 8 1/2


Go to forestry problem index89 l.jpg

GO TO FORESTRY PROBLEM INDEX

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 PROB #More 8 2/2


Forestry problem more 9 l.jpg
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?


Forestry more 9 l.jpg

FORESTRY More #9

50 gal nurse tank

Ratio of 1 : 5 (Garlon 4 : JLB Oil Plus)

FORESTRY PROB #More 9 1/2


Go to forestry problem index92 l.jpg

GO TO FORESTRY PROBLEM INDEX

Garlon 4

1/ 6 x 50 gal = 8.3 gal

JLB Plus

5/6 of mix x 50 gal = 41.7 gal

FORESTRY PROB #More 9 2/2


Forestry problem more 10 l.jpg
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?


Forestry more 10 l.jpg

FORESTRY More #10

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

FORESTRY PROB #More 10 1/2


Go to forestry problem index95 l.jpg

GO TO FORESTRY PROBLEM INDEX

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 PROB #More 10 2/2


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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 C. 20 gpa

B. 30 gpa D. 22.5 gpa


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FORESTRY More #19

15 gpa at 4 mph

Reduce to 2 mph

Half speed = double rate, or

2 x 15 gpa = 30 gpa

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FORESTRY PROB #More 19 1/1


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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 C. 80 psi

B. 10 psi D. 60 psi


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FORESTRY More #20

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

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FORESTRY PROB #More 20 1/1


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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 C. 4 gals

B. 3 gals D. 6 gals


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FORESTRY More #21

Prescription: 8 gal herbicide / 200 gal mix

Require only 50 gal of mix

50 gal / 200 gal x 8 gal = 2 gal

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FORESTRY PROB #More 21 1/1


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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?


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FORESTRY More #22

1.5 gal / ac

As a 25 gal mix in water

30 ac

FORESTRY PROB #More 22 1/2


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GO TO FORESTRY PROBLEM INDEX

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 PROB #More 22 2/2


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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?


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FORESTRY More #23

3 gal / min

200 ft / 2 min

40 ft wide swath

FORESTRY PROB #More 23 1/2


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GO TO FORESTRY PROBLEM INDEX

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 PROB #More 23 2/2


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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?


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Forestry Problem #a40 ac3000 stem / ac2 ml / stem

FORESTRY PROB #More a 1/2


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40 acx 3000 stem / acx 2 ml / stem = 240,000 ml240,000 ml / 1,000 ml / L = 240 L= approximately 60 gallons

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FORESTRY PROB #More a 2/2


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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?


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Forestry Problem #b

  • 40 ac

  • 200 stem / ac

  • stem average 5” dbh

  • injected on 3” centers

  • 1 ml solution / cut

FORESTRY PROB #More b 1/4


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5 in dbh x 3.1416 (-- “pi”) = 15.7 in circumference

5”

FORESTRY PROB #More b 2/4


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

x

x

x

3 inches

x

x

FORESTRY PROB #More b 3/4


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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)

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FORESTRY PROB #More b 4/4


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R-O-W Problem #1

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


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RIGHT-OF-WAY PROB #1

10 mi of R-O-W

15’ each side of road

R-O-W PROB # 1 1/3


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How many sq ft

15 ft x 2 x 5280 ft / mi x 10 mi =

1,584,000 sq ft

R-O-W PROB # 1 2/3


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GO TO R-O-W & AQUATIC PROBLEM INDEX

How many sq ft

1,584,000 sq ft

How many acres?

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

36.4 ac

R-O-W PROB # 1 3/3


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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?


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RIGHT-OF-WAY PROB #2

15 feet on both sides

5 mi R-O-W

2 gal / ac

R-O-W PROB # 2 1/2


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GO TO R-O-W & AQUATIC PROBLEM INDEX

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 PROB # 2 2/2


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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?


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RIGHT-OF-WAY PROB #3

25 gpa at 8 mph

Speed reduced to 4 mph

25 gpa x 8 mph / 4 mph = 50 gpa

GO TO R-O-W & AQUATIC PROBLEM INDEX

R-O-W PROB # 3 1/1


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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?


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RIGHT-OF-WAY PROB #4

25 gpa at 6 mph

Pressure reduced from 50 psi to 12.5 psi

20 gpa / 50 psi / 12.5 psi = 10 gpa

GO TO R-O-W & AQUATIC PROBLEM INDEX

R-O-W PROB # 4 1/1


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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?


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RIGHT-OF-WAY PROB #5

400 gal tank

20 gal / ac

1.25 gal 2,4-D / ac

3 liq oz drift control agent / 100 gal

1% adjuvant

R-O-W PROB # 5 1/2


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GO TO R-O-W & AQUATIC PROBLEM INDEX

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 PROB # 5 2/2


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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.)


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RIGHT-OF-WAY PROB #6

25 gal / ac required

6 gal expended on

20’ x 545’ area

+/- 5% tolerance

R-O-W PROB # 6 1/2


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GO TO R-O-W & AQUATIC PROBLEM INDEX

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 PROB # 6 2/2


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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.


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RIGHT-OF-WAY PROB #7

25 gpa

20 mi

15 ft both sides

1.5 gal 2,4-D / ac

3 liq oz Poly Control / 100 gal mix

R-O-W PROB # 7 1/2


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GO TO R-O-W & AQUATIC PROBLEM INDEX

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 PROB # 7 2/2


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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


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RIGHT-OF-WAY PROB #8

5 gal Garlon 4 spilled into a

200’ x 320’ pond with average depth of 6’

Bluegill 96 hr LC50 is 0.87 mg/L

R-O-W PROB # 8 1/2


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GO TO R-O-W & AQUATIC PROBLEM INDEX

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 PROB # 8 2/2


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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?


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RIGHT-OF-WAY PROB #a

6 mi of R-O-W

12’ each side of road

2 gal Krenite / ac

R-O-W PROB # a 1/4


How many sq ft 12 x 2 x 5280 mi x 6 mi 760 320 sq ft l.jpg

How many sq. ft

12’ x 2 x 5280’ / mi x 6 mi =

760,320 sq ft

R-O-W PROB # a 2/4


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How many sq ft

760,320 sq ft

How many acres?

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

17.45 ac

R-O-W PROB # a 3/4


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GO TO R-O-W & AQUATIC PROBLEM INDEX

How many acres?

760,320 sq ft

How many acres?

17.45 ac

How much Krenite?

17.5 ac x 2 gal / ac =

35 gal

R-O-W PROB # a 4/4


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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?


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AQUATICS PROB #1

20 surface acres

Average 2’ deep

20 ac x 2 ft = 40 ac ft

GO TO R-O-W & AQUATIC PROBLEM INDEX

AQUATIC PROB # 1 1/1


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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?


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AQUATICS PROB #2

Pond: 100’ long x 500’ wide x 4’ average depth

100 ft x 500 ft x 4 ft / 43560 cu ft / ac ft = 4.6 ac ft

GO TO R-O-W & AQUATIC PROBLEM INDEX

AQUATIC PROB # 2 1/1


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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?


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AQUATICS PROB #3

30 lb chemical / ac ft

Using a 50% granule

300’ long x 10’ wide with average depth of 10’

AQUATIC PROB # 3 1/2


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GO TO R-O-W & AQUATIC PROBLEM INDEX

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

AQUATIC PROB # 3 2/2


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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?


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AQUATICS PROB #4

1 gal / 2 ac ft

420’ long x 105’ wide by average 4’ deep

420 ft x 105 ft x 4 ft / 43,560 cu ft / ac ft x 1 gal / 2 ac ft = 2 gal

GO TO R-O-W & AQUATIC PROBLEM INDEX

AQUATIC PROB # 4 1/1


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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?


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AQUATICS PROB #5

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

GO TO R-O-W & AQUATIC PROBLEM INDEX

AQUATIC PROB # 5 1/1


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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!]


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AQUATICS PROB #6

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

GO TO R-O-W & AQUATIC PROBLEM INDEX

AQUATIC PROB # 6 1/1


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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.


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Aquatics Problem #7

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


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AQUATICS PROB #7

10 gal spill

2 lb a.i. / gal – toxicity based entirely on the a.i.

2 ac x 4’ deep

1 ppm may result in fish kill

AQUATIC PROB # 7 1/2


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GO TO R-O-W & AQUATIC PROBLEM INDEX

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

AQUATIC PROB # 7 1/2


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