Forestry Problem Index
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Forestry #1 *
Forestry #2 *
Forestry #3 *
Forestry #4 *
Forestry #5 *
Forestry #6 *
Forestry #7 *
Forestry #8 *
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Forestry #9 *
Forestry #10 *
Forestry #11 *
More #1 *
More #2 *
More #3 *
More #4 *
More #5 *
Go To Right-of-Way or Aquatic Problems Index *
Forestry Problem Index (continued)
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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
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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 *
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Right of way #a *
Aquatics #1 *
Aquatics #2 *
Aquatics #3 *
Aquatics #4 *
Aquatics #5 *
Aquatics #6 *
Aquatics #7 *
Go To Forestry Problem Index *
Forestry problem #1
To prepare 3 gallons of a 3 percent
Accord solution requires how many
ounces of Accord?
FORESTRY PROB #1
How much Accord is needed?
3 gallons
3% solution
FORESTRY
PROB #1 1/2
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
GO TO FORESTRY PROBLEM INDEX
FORESTRY
PROB #1 2/2
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
400 gallon tank
20 gal/ac rate
3 lb ai/ac
70% ai product
FORESTRY
PROB #2 1/4
Treated acres per tank
400 gal / tank / 20 gal / ac =
20 ac / tank
FORESTRY
PROB #2 2/4
Treated acres per tank
20 ac / tank
Pounds active ingredient needed
20 ac x 3 lb ai / ac =
60 lb ai
FORESTRY
PROB #2 3/4
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
PROB #2 4/4
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
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
FORESTRY
PROB #3 2/4
How much area sprayed?
.475 ac
How many gpa?
7.5 gal / 0.475 ac =
15.78 gpa
FORESTRY
PROB #3 3/4
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
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
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
FORESTRY
PROB #4 2/5
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
FORESTRY
PROB #4 4/5
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
GO TO FORESTRY PROBLEM INDEX
FORESTRY
PROB #4 5/5
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
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
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)
FORESTRY
PROB #5 3/5
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
Final Solution in tank
600 gallons total mixture
as
37.5 gal 2,4-DP
24 oz retardant
562.25 gal of water
GO TO FORESTRY PROBLEM INDEX
FORESTRY
PROB #5 5/5
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
60 acres to be treated
3 qt Garlon 4 +
1 qt Accord / ac
FORESTRY
PROB #6 1/2
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
GO TO FORESTRY PROBLEM INDEX
FORESTRY
PROB #6 2/2
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
40 gal / ac output
at 4 mph
new speed 8 mph
FORESTRY
PROB #7 1/2
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
GO TO FORESTRY PROBLEM INDEX
FORESTRY
PROB #7 2/2
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
Calibrated to deliver 25 gpa
at 15 psi
pressure increased to 60 psi
FORESTRY
PROB #8 1/3
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
New rate?
New pressure is 4 x old pressure
4 x pressure = 2 x the rate
25 gal x 2 =
50 gpa
GO TO FORESTRY PROBLEM INDEX
FORESTRY
PROB #8 3/3
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
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
== 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
FORESTRY
PROB #9 3/9
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
Graphically this is:
7”
7in dbh x 3.1416 (-- “pi”) = 21.98 in
circumference
FORESTRY
PROB #9 5/9
x
x
x
x
x
x
x
x
x
x
2 in
x
22 in/stem / 2 in/cut = 11 cuts / stem
FORESTRY
PROB #9 6/9
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
FORESTRY
PROB #9 8/9
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
GO TO FORESTRY PROBLEM INDEX
FORESTRY
PROB #9 9/9
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
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
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
FORESTRY
PROB #10 3/9
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
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 -- LD50 s
Bluegill = >1,000 mg/L
Channel catfish = >1,000 mg/L
FORESTRY
PROB #10 6/9
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
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
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
GO TO FORESTRY PROBLEM INDEX
FORESTRY
PROB #10 9/9
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
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
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
FORESTRY
PROB #11 3/5
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
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
GO TO FORESTRY PROBLEM INDEX
FORESTRY
PROB #11 5/5
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
3 gal backpack
0.4 liq oz Arsenal / gal
2 liq oz Bullseye / gal
0.5% Cide-Kick
FORESTRY PROB
#More 1 1/2
0.4 liq oz Arsenal / gal x 3 gal = 1.2
liq oz Arsenal
2 liq oz Bullseye / gal x 3 gal =
liq oz Bullseye
6
0.005 Cide-Kick x 3 gal x 128 liq oz
/ gal = (1.92) = 2 liq oz Cide-Kick
GO TO FORESTRY PROBLEM INDEX
FORESTRY PROB
#More 1 2/2
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 #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
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
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
GO TO FORESTRY PROBLEM INDEX
FORESTRY PROB
#More 2 2/2
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
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
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
GO TO FORESTRY PROBLEM INDEX
FORESTRY PROB
#More 3 2/2
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
Four nozzle boom
8003 Spraying Systems flat fan
nozzles
40 psi is being used through the
boom
FORESTRY PROB
#More 4 1/2
8003 nozzle applies 0.3 gal/min at 40
psi at an 80o angle
4 nozzles x 0.3 gpm / nozzle =
1.2 gpm
GO TO FORESTRY PROBLEM INDEX
FORESTRY PROB
#More 4 2/2
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
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
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
GO TO FORESTRY PROBLEM INDEX
FORESTRY PROB
#More 5 2/2
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
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
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)
GO TO FORESTRY PROBLEM INDEX
FORESTRY PROB
#More 6 2/2
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
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
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
GO TO FORESTRY PROBLEM INDEX
FORESTRY PROB
#More 7 2/2
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
2600 stems / ac to be treated
135 ac
3 ml Garlon 4 per stem
FORESTRY PROB
#More 8 1/2
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
GO TO FORESTRY PROBLEM INDEX
FORESTRY PROB
#More 8 2/2
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
50 gal nurse tank
Ratio of 1 : 5 (Garlon 4 : JLB Oil
Plus)
FORESTRY PROB
#More 9 1/2
Garlon 4
1/ 6 x 50 gal = 8.3 gal
JLB Plus
5/6 of mix x 50 gal = 41.7 gal
GO TO FORESTRY PROBLEM INDEX
FORESTRY PROB
#More 9 2/2
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
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
15 gal / jug x 8 jug / pallet =
gal / pallet
120
5/6 x 280 gal = 233.33 gal needed
233 gal / 120 gal / pallet = 2 pallet
GO TO FORESTRY PROBLEM INDEX
FORESTRY PROB
#More 10 2/2
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
FORESTRY More #19
15 gpa at 4 mph
Reduce to 2 mph
Half speed = double rate, or
2 x 15 gpa = 30 gpa
GO TO FORESTRY PROBLEM INDEX
FORESTRY PROB
#More 19 1/1
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
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
GO TO FORESTRY PROBLEM INDEX
FORESTRY PROB
#More 20 1/1
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
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
GO TO FORESTRY PROBLEM INDEX
FORESTRY PROB
#More 21 1/1
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?
FORESTRY More #22
1.5 gal / ac
As a 25 gal mix in water
30 ac
FORESTRY PROB
#More 22 1/2
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
GO TO FORESTRY PROBLEM INDEX
FORESTRY PROB
#More 22 2/2
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?
FORESTRY More #23
3 gal / min
200 ft / 2 min
40 ft wide swath
FORESTRY PROB
#More 23 1/2
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
GO TO FORESTRY PROBLEM INDEX
FORESTRY PROB
#More 23 2/2
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?
Forestry Problem #a
40 ac
3000 stem / ac
2 ml / stem
FORESTRY PROB
#More a 1/2
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
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FORESTRY PROB
#More a 2/2
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?
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
5”
5 in dbh x 3.1416 (-- “pi”) = 15.7 in
circumference
FORESTRY PROB
#More b 2/4
x
x
x
3 inches
x
x
15.7 in/stem / 3 in/cut = 5 cuts/ stem
FORESTRY PROB
#More b 3/4
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)
GO TO FORESTRY PROBLEM INDEX
FORESTRY PROB
#More b 4/4
R-O-W Problem #1
If you treat 15 feet on both sides of a road
for 10 miles, how many acres were
treated?
RIGHT-OF-WAY PROB #1
10 mi of R-O-W
15’ each side of road
R-O-W PROB # 1 1/3
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
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
GO TO R-O-W & AQUATIC PROBLEM INDEX
R-O-W PROB # 1 3/3
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?
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
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
GO TO R-O-W & AQUATIC PROBLEM INDEX
R-O-W PROB # 2 2/2
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?
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
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?
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
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 2,4D 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?
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
400 gal / tank / 20 gal / ac =
ac / tank
20
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
GO TO R-O-W & AQUATIC PROBLEM INDEX
R-O-W PROB # 5 2/2
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.)
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
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%)
GO TO R-O-W & AQUATIC PROBLEM INDEX
R-O-W PROB # 6 2/2
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.
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
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
GO TO R-O-W & AQUATIC PROBLEM INDEX
R-O-W PROB # 7 2/2
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
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
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
GO TO R-O-W & AQUATIC PROBLEM INDEX
R-O-W PROB # 8 2/2
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?
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
R-O-W PROB # a 2/4
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
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
GO TO R-O-W & AQUATIC PROBLEM INDEX
R-O-W PROB # a 4/4
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?
AQUATICS PROB #1
20 surface acres
Average 2’ deep
20 ac x 2 ft = 40 ac ft
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AQUATIC PROB # 1 1/1
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?
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
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?
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
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
GO TO R-O-W & AQUATIC PROBLEM INDEX
AQUATIC PROB # 3 2/2
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?
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
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?
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
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!]
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
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.
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
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
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
GO TO R-O-W & AQUATIC PROBLEM INDEX
AQUATIC PROB # 7 1/2