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```					                                                                                                 MINING AND
EARTHMOVING

22
CONTENTS                                                                    ELEMENTS OF PRODUCTION
Elements of Production . . . . . . . . . . . . . . . . . .22-1                 Production is the hourly rate at which material
Volume Measure . . . . . . . . . . . . . . . . . . . . . .22-2            is moved. Production can be expressed in various
Swell . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .22-2   units:
Load Factor . . . . . . . . . . . . . . . . . . . . . . . . . .22-2
Material Density . . . . . . . . . . . . . . . . . . . . . .22-2          Metric
Fill Factor . . . . . . . . . . . . . . . . . . . . . . . . . . .22-3        Bank Cubic Meters — BCM — bank m3
Soil Density Tests . . . . . . . . . . . . . . . . . . . . .22-3             Loose Cubic Meters — LCM — loose m3
Figuring Production On-the-Job . . . . . . . . . . .22-4                       Compacted Cubic
Load Weighing . . . . . . . . . . . . . . . . . . . . . . . .22-4              Meters             — CCM — compacted m3
Time Studies . . . . . . . . . . . . . . . . . . . . . . . . .22-4           Tonnes
English Example . . . . . . . . . . . . . . . . . . . . . .22-4           English
Metric Example . . . . . . . . . . . . . . . . . . . . . . .22-5
Estimating Production Off-the-Job . . . . . . . . .22-5                        Bank Cubic Yards — BCY — bank yd3
Rolling Resistance . . . . . . . . . . . . . . . . . . . . .22-5             Loose Cubic Yards — LCY — loose yd3
Grade Resistance . . . . . . . . . . . . . . . . . . . . . .22-6             Compacted Cubic
Total Resistance . . . . . . . . . . . . . . . . . . . . . .22-6               Yards              — CCY — compacted yd3
Traction . . . . . . . . . . . . . . . . . . . . . . . . . . . . .22-6       Tons
Altitude . . . . . . . . . . . . . . . . . . . . . . . . . . . . .22-7       For most earthmoving and material handling
Job Efficiency . . . . . . . . . . . . . . . . . . . . . . . .22-8        applications, production is calculated by multiply-
English Example . . . . . . . . . . . . . . . . . . . . . .22-8           ing the quantity of material (load) moved per cycle
Metric Example . . . . . . . . . . . . . . . . . . . . . .22-10           by the number of cycles per hour.
Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .22-13       Production = Load/cycle     cycles/hour
Economic Haul Distances . . . . . . . . . . . . . .22-13
The load can be determined by
Production Estimating . . . . . . . . . . . . . . . . . .22-14
Loading Match . . . . . . . . . . . . . . . . . . . . . . .22-14
2) load estimating based on machine rating
Fuel Consumption and Productivity . . . . . . .22-14
3) surveyed volume divided by load count
Formulas and Rules of Thumb . . . . . . . . . . . .22-15
Generally, earthmoving and overburden removal
INTRODUCTION                                                                for coal mines are calculated by volume (bank cubic
This section explains the earthmoving principles                          meters or bank cubic yards). Metal mines and aggre-
used to determine machine productivity. It shows                            gate producers usually work in weight (tons or tonnes).
how to calculate production on-the-job or estimate
production off-the-job.
Machine performance is usually measured on an
hourly basis in terms of machine productivity and
machine owning and operating cost. Optimum
machine performance can be expressed as follows:
Lowest Possible
Hourly Costs
___________________
Lowest cost per ton =
Highest Possible
Hourly Productivity

Edition 40   22-1
Mining and             Elements of Production
Earthmoving             ● Volume Measure ● Swell
● Load Factor ● Material Density

Volume Measure — Material volume is defined               Bank = Loose ÷ (1 + Swell) =
according to its state in the earthmoving process.                           1000 LCM ÷ (1 + 0.25) = 800 BCM
The three measures of volume are:                                            1308 LCY ÷ (1 + 0.25) = 1046 BCY
BCM (BCY) — one cubic meter (yard) of material               Load Factor — Assume one bank cubic yard of
as it lies in the natural bank state.   material weighs 3000 lb. Because of material char-
LCM (LCY) — one cubic meter (yard) of material            acteristics, this bank cubic yard swells 30% to 1.3 loose
which has been disturbed and has        cubic yards when loaded, with no change in weight.
swelled as a result of movement.        If this 1.0 bank cubic yard or 1.3 loose cubic yards
CCM (CCY) — one cubic meter (yard) of material            is compacted, its volume may be reduced to 0.8 com-
which has been compacted and has        pacted cubic yard, and the weight is still 3000 lb.
become more dense as a result of           Instead of dividing by 1 + Swell to determine bank
compaction.                             volume, the loose volume can be multiplied by the
In order to estimate production, the relationships        load factor.
between bank measure, loose measure, and com-                  If the percent of material swell is known, the load
pacted measure must be known.                               factor (L.F.) may be obtained by using the following
Swell — Swell is the percentage of original volume        relationship:
(cubic meters or cubic yards) that a material increases
100%
when it is removed from the natural state. When                           L.F. = ______________
excavated, the material breaks up into different size                            100% + % swell
particles that do not fit together, causing air pockets        Load factors for various materials are listed in
or voids to reduce the weight per volume. For exam-         the Tables Section of this handbook.
ple to hold the same weight of one cubic unit of bank          To estimate the machine payload in bank cubic
material it takes 30% more volume (1.3 times) after         yards, the volume in loose cubic yards is multiplied
excavation. (Swell is 30%.)                                 by the load factor:
for a given weight
1 + Swell = ______________________                      The ratio between compacted measure and bank
Bank cubic volume for                  measure is called shrinkage factor (S.F.):
the same given weight
Compacted cubic yards (CCY)
Loose                                                   S.F. = ____________________________
Bank = __________                                                               Bank cubic yards (BCY)
(1 + Swell)
Shrinkage factor is either estimated or obtained
Loose = Bank       (1 + Swell)                              from job plans or specifications which show the con-
version from compacted measure to bank measure.
Example Problem:
Shrinkage factor should not be confused with per-
If a material swells 20%, how many loose cubic              centage compaction (used for specifying embank-
meters (loose cubic yards) will it take to move 1000        ment density, such as Modified Proctor or California
bank cubic meters (1308 bank cubic yards)?                  Bearing Ratio [CBR]).
Loose = Bank      (1 + Swell) =                                Material Density — Density is the weight per
1000 BCM       (1 + 0.2) = 1200 LCM          unit volume of a material. Materials have various
1308 BCY      (1 + 0.2) = 1570 LCY           densities depending on particle size, moisture con-
tent and variations in the material. The denser the
How many bank cubic meters (yards) were moved
material the more weight there is per unit of equal
if a total of 1000 loose cubic meters (1308 yards)
volume. Density estimates are provided in the
have been moved? Swell is 25%.
Tables Section of this handbook.
Weight     kg (lb)
Density = ________ = ________
Volume    m3 (yd3)
Weight = Volume     Density

22-2   Edition 40
Elements of Production               Mining and
● Fill Factor             Earthmoving
● Soil Density Tests

A given material’s density changes between bank                        CCY      10,000
and loose. One cubic unit of loose material has less        a) BCY = _____ = _______ = 12,500 BCY
S.F.       0.80
weight than one cubic unit of bank material due to
air pockets and voids. To correct between bank and          b) Load (BCY) = Capacity (LCY)
loose use the following equations.                                    Load factor (L.F.) = 20    0.81
kg/BCM         lb/BCY
1 + Swell = ________ or ________                                 (L.F. of 0.81 from Tables)
kg/LCM         lb/LCY                     Number of            12,500 BCY
(1 + Swell)
lb/BCY = lb/LCY       (1 + Swell)                                                    ●●●
Fill Factor — The percentage of an available vol-           Soil Density Tests — There are a number of
ume in a body, bucket, or bowl that is actually used        acceptable methods that can be used to determine
is expressed as the fill factor. A fill factor of 87% for   soil density. Some that are currently in use are:
a hauler body means that 13% of the rated volume                      Nuclear density moisture gauge
is not being used to carry material. Buckets often                    Sand cone method
have fill factors over 100%.                                          Oil method
Balloon method
Example Problem:
Cylinder method
A 14 cubic yard (heaped 2:1) bucket has a 105% fill            All these except the nuclear method use the fol-
factor when operating in a shot sandstone (4125 lb/         lowing procedure:
BCY and a 35% swell).
a) What is the loose density of the material?               1. Remove a soil sample from bank state.
b) What is the usable volume of the bucket?                 2. Determine the volume of the hole.
c) What is the bucket payload per pass in BCY?              3. Weigh the soil sample.
d) What is the bucket payload per pass in tons?             4. Calculate the bank density kg/BCM (lb/BCY).
a) lb/LCY = lb/BCY ÷ (1 + Swell) = 4125 ÷ (1.35) =             The nuclear density moisture gauge is one of the
3056 lb/LCY                                              most modern instruments for measuring soil den-
b) LCY = rated LCY fill factor = 14 1.05 = 14.7 LCY         sity and moisture. A common radiation channel
c) lb/pass = volume     density lb/LCY = 14.7               emits either neutrons or gamma rays into the soil.
3056 = 44,923 lb                                         In determining soil density, the number of gamma
BCY/pass = weight ÷ density lb/BCY = 44,923 ÷            rays absorbed and back scattered by soil particles
4125 = 10.9 BCY                                          is indirectly proportional to the soil density. When
or bucket LCY from part b ÷ (1 + Swell) = 14.7 ÷         measuring moisture content, the number of moder-
1.35 = 10.9 BCY                                          ated neutrons reflected back to the detector after col-
d) tons/pass = lb ÷ 2000 lb/ton = 44,923 ÷ 2000 =           liding with hydrogen particles in the soil is directly
22.5 tons                                                proportional to the soil’s moisture content.
All these methods are satisfactory and will provide
Example Problem:                                            accurate densities when performed correctly. Several
Construct a 10,000 compacted cubic yard (CCY)               repetitions are necessary to obtain an average.
bridge approach of dry clay with a shrinkage factor         NOTE: Several newer methods have been success-
(S.F.) of 0.80. Haul unit is rated 14 loose cubic yards              fully applied, along with weigh scales to deter-
struck and 20 loose cubic yards heaped.                              mine volume and loose density of material
a) How many bank yards are needed?                                   moved in hauler bodies. These measurements
b) How many loads are required?                                      include photogrammatic and laser scanning
technologies.

Edition 40    22-3
Mining and                        Figuring Production On-the-Job
● Time Studies
● Example (English)

FIGURING PRODUCTION ON-THE-JOB                                      Time is any time, other than wait time, when a
machine is not performing in the work cycle (scraper
Load Weighing — The most accurate method of
determining the actual load carried is by weighing.
To determine trips-per-hour at 100% efficiency,
This is normally done by weighing the haul unit one
divide 60 minutes by the average cycle time less all
wheel or axle at a time with portable scales. Any
wait and delay time. Cycle time may or may not
scales of adequate capacity and accuracy can be used.
include wait and/or delay time. Therefore, it is pos-
While weighing, the machine should be level to reduce
sible to figure different kinds of production: meas-
error caused by weight transfer. Enough loads should
ured production, production without wait or delay,
be weighed to provide a good average. Machine weight
maximum production, etc. For example:
is the sum of the individual wheel or axle weights.
Actual Production: includes all wait and delay time.
The weight of the load can be determined using
Normal Production (without delays): includes wait
the empty and loaded weight of the unit.
time that is considered normal, but no delay time.
Weight of
Maximum Production: to figure maximum (or opti-
load = gross machine weight – empty weight
mum) production, both wait time and delay time
To determine the bank cubic measure carried by
are eliminated. The cycle time may be further
a machine, the load weight is divided by the bank-
altered by using an optimum load time.
state density of the material being hauled.
BCY = _____________
Bank density                                 A job study of a Wheel Tractor-Scraper might yield
Times Studies — To estimate production, the                      the following information:
number of complete trips a unit makes per hour must                   Average wait time        = 0.28 minute
be determined. First obtain the unit’s cycle time with                Average load time        = 0.65
the help of a stop watch. Time several complete cycles                Average delay time       = 0.25
to arrive at an average cycle time. By allowing the                   Average haul time        = 4.26
watch to run continuously, different segments such                    Average dump time        = 0.50
as load time, wait time, etc. can be recorded for each                Average return time      = 2.09
______
cycle. Knowing the individual time segments affords                   Average total cycle
a good opportunity to evaluate the balance of the
= 8.03
______ minutes
spread and job efficiency. The following is an exam-                  Less wait & delay time = 0.53
ple of a scraper load time study form. Numbers in                     Average cycle 100% eff. = 7.50 minutes
the white columns are stop watch readings; numbers                  Weight of haul unit empty — 48,650 lb
in the shaded columns are calculated:                               Weights of haul unit loaded —
Weighing unit #1 — 93,420 lb
Total                                                                Weighing unit #2 — 89,770 lb
Cycle                                                                Weighing unit #3 — 88,760 lb
Times                                                                                      __________
(less Arrive Wait Begin Load End Begin Delay End                                          271,950 lb;
delays) Cut Time Load Time Load Delay Time Delay                                             average = 90,650 lb
0.00    0.30     0.30    0.60 0.90                       1. Average load weight = 90,650 lb – 48,650 lb =
3.50     3.50    0.30     3.80    0.65 4.45                          42,000 lb
4.00     7.50    0.35     7.85    0.70 8.55 9.95   1.00   10.95   2. Bank density = 3125 lb/BCY
4.00    12.50    0.42    12.92    0.68 13.60                                 Weight of load
NOTE: All numbers are in minutes                                    3. Load = _____________
Bank density
42,000 lb
This may be easily extended to include other seg-                3. Load = ____________ = 13.4 BCY
ments of the cycle such as haul time, dump time, etc.                           3125 lb/BCY
Haul roads may be further segmented to more accu-                   4. Cycles/hr =
rately define performance, including measured                               60 min/hr       60 min/hr
__________ = _____________ = 80 cycles/hr
speed traps. Similar forms can be made for push-                            Cycle time 7.50 min/cycle
ers, loaders, dozers, etc. Wait Time is the time a unit             5. Production = Load/cycle       cycles/hr
must wait for another unit so that the two can func-                   (less delays) = 13.4 BCY/cycle     8.0 cycles/hr
tion together (haul unit waiting for pusher). Delay                                  = 107.2 BCY/hr

22-4     Edition 40
Figuring Production On-the-Job               Mining and
● Example (Metric)              Earthmoving
Estimating Production Off-the-Job
● Rolling Resistance

Example (Metric)                                      ESTIMATING PRODUCTION OFF-THE-JOB
A job study of a Wheel Tractor-Scraper might yield      It is often necessary to estimate production of
the following information:                            earthmoving machines which will be selected for a
job. As a guide, the remainder of the section is
Average wait time         = 0.28 minute
devoted to discussions of various factors that may
Average load time         = 0.65                                                                                22
affect production. Some of the figures have been
Average delay time        = 0.25
rounded for easier calculation.
Average haul time         = 4.26
Rolling Resistance (RR) is a measure of the
Average dump time         = 0.50
force that must be overcome to roll or pull a wheel
Average return time       = 2.09
______                    over the ground. It is affected by ground conditions
Average total cycle       = 8.03
______ minutes            and load — the deeper a wheel sinks into the ground,
Less wait & delay time = 0.53                       the higher the rolling resistance. Internal friction
Average cycle 100% eff. = 7.50 minutes              and tire flexing also contribute to rolling resistance.
Weight of haul unit empty — 22 070 kg                 Experience has shown that minimum resistance is
Weights of haul unit loaded —                         1%-1.5% (see Typical Rolling Resistance Factors in
Weighing unit #1 — 42 375 kg                        Tables section) of the gross machine weight (on tires).
Weighing unit #2 — 40 720 kg                        A 2% base resistance is quite often used for estimat-
Weighing unit #3 —__________
40 260 kg                    ing. Resistance due to tire penetration is approxi-
mately 1.5% of the gross machine weight for each
123 355 kg;                    inch of tire penetration (0.6% for each cm of tire pen-
average = 41 120 kg         etration). Thus rolling resistance can be calculated
1. Average load weight = 41 120 kg – 22 070 kg =      using these relationships in the following manner:
19 050 kg                                            RR = 2% of GMW + 0.6% of GMW per cm tire
2. Bank density = 1854 kg/BCM                                     penetration
Weight of load                              RR = 2% of GMW + 1.5% of GMW per inch tire
Bank density
It’s not necessary for the tires to actually penetrate
19 050 kg
3. Load = ____________ = 10.3 BCM                     the road surface for rolling resistance to increase
1854 kg/BCM                                above the minimum. If the road surface flexes under
4. Cycles/hr =                                        load, the effect is nearly the same — the tire is always
60 min/hr       60 min/hr
__________ = _____________ = 80 cycles/hr     running “uphill”. Only on very hard, smooth sur-
Cycle time 7.50 min/cycle                     faces with a well compacted base will the rolling
5. Production = Load/cycle       cycles/hr            resistance approach the minimum.
(less delays) = 10.3 BCM/cycle     8.0 cycles/hr     When actual penetration takes place, some vari-
= 82 BCM/hr                          ation in rolling resistance can be noted with various
●●●                             inflation pressures and tread patterns.
NOTE: The Cat Cycle Timer Program software uses       NOTE: When figuring “pull” requirements for track-
laptop computers in place of stopwatches,                type tractors, rolling resistance applies only
organizes the data, and allows study results             to the trailed unit’s weight on wheels. Since
to be printed.                                           tracktype tractors utilize steel wheels mov-
ing on steel “roads”, a tractor’s rolling resis-
tance is relatively constant and is accounted
for in the Drawbar Pull rating.

Edition 40    22-5
Mining and             Estimating Production Off-the-Job
● Total Resistance
● Traction

Grade Resistance is a measure of the force that            Total resistance can also be represented as consist-
must be overcome to move a machine over unfavor-           ing completely of grade resistance expressed in per-
of the force that assists machine movement on              component is viewed as a corresponding quantity of
Grades are generally measured in percent slope,          approach, total resistance can then be considered in
which is the ratio between vertical rise or fall and       terms of percent grade.
the horizontal distance in which the rise or fall            This can be done by converting the contribution of
occurs. For example, a 1% grade is equivalent to a         rolling resistance into a corresponding percentage of
1 m (ft) rise or fall for every 100 m (ft) of horizontal   grade resistance. Since 1% of adverse grade offers a
distance; a rise of 4.6 m (15 ft) in 53.3 m (175 ft)       resistance of 10 kg (20 lb) for each metric or (U.S.)
equals an 8.6% grade.                                      ton of machine weight, then each 10 kg (20 lb) of
4.6 m (rise)                                resistance per ton of machine weight can be repre-
53.3 m (horizontal distance)                          resistance in percent grade and grade resistance in
15 ft (rise)
__________________________ = 8.6% grade               percent grade can then be summed to give Total
175 ft (horizontal distance)                         Resistance in percent or Effective Grade. The fol-
lowing formulas are useful in arriving at Effective
Grade resistance is usually expressed as a positive        Rolling Resistance (%) = 2% + 0.6% per cm tire
(+) percentage and grade assistance is expressed as                                      penetration
a negative (–) percentage.                                                           = 2% + 1.5% per inch tire
It has been found that for each 1% increment of                                        penetration
tance must be overcome for each metric (U.S.) ton of          Effective Grade (%) = RR (%) + GR (%)
machine weight. This relationship is the basis for           Effective grade is a useful concept when working
expressed in kg/metric ton (lb/U.S. ton):                  curves, Brake Performance curves, and Travel Time
= 20 lb/U.S. ton   % grade      Traction — is the driving force developed by a
wheel or track as it acts upon a surface. It is expressed
Grade resistance (assistance) is then obtained by
as usable Drawbar Pull or Rimpull. The following
multiplying the Grade Resistance Factor by the
factors affect traction: weight on the driving wheel
machine weight (GMW) in metric (U.S.) tons.
or tracks, gripping action of the wheel or track, and
Grade Resistance = GR Factor           GMW in metric       ground conditions. The coefficient of traction (for any
(U.S.) tons                     roadway) is the ratio of the maximum pull devel-
Grade resistance may also be calculated using per-       oped by the machine to the total weight on the drivers.
centage of gross weight. This method is based on the                                              Pull
relationship that grade resistance is approximately                  Coeff. of traction = _________________
weight on drivers
equal to 1% of the gross machine weight for 1% of
grade.                                                       Therefore, to find the usable pull for a given
machine:
Usable pull = Coeff. of traction weight on drivers
Grade resistance (assistance) affects both wheel
and track-type machines.                                   Example: Track-Type Tractor
Total Resistance is the combined effect of rolling       What usable drawbar pull (DBP) can a 26 800 kg
resistance (wheel vehicles) and grade resistance. It       (59,100 lb) Track-type Tractor exert while working
can be computed by summing the values of rolling           on firm earth? on loose earth? (See table section for
resistance and grade resistance to give a resistance       coefficient of traction.)
in kilogram (pounds) force.
Total Resistance = Rolling Resistance +

22-6   Edition 40
Estimating Production Off-the-Job                Mining and
● Altitude                Earthmoving

Answer:                                                 power deration will be reflected in the machine’s
Firm earth — Usable DBP =                                  gradeability and in the load, travel, and dump and
(0.90 59,100 lb = 53,190 lb)      machine itself). Altitude may also reduce retarding
Loose earth — Usable DBP =                                 performance. Consult a Caterpillar representative
0.60 26 800 kg = 16 080 kg        to determine if deration is applicable. Fuel grade                        22
(0.60 59,100 lb = 35,460 lb)      (heat content) can have a similar effect of derating
If a load required 21 800 kg (48,000 lb) pull to move   engine performance.
it, this tractor could move the load on firm earth.           The example job problem that follows indicates one
However, if the earth were loose, the tracks would         method of accounting for altitude deration: by increas-
spin.                                                      ing the appropriate components of the total cycle
NOTE: D8R through D11R Tractors may attain                 time by a percentage equal to the percent of horse-
higher coefficients of traction due to their     power deration due to altitude. (i.e., if the travel time
suspended undercarriage.                         of a hauling unit is determined to be 1.00 minute at
full HP, the time for the same machine derated to 90%
Example: Wheel Tractor-Scraper                             of full HP will be 1.10 min.) This is an approximate
What usable rimpull can a 621F size machine exert          method that yields reasonably accurate estimates
while working on firm earth? on loose earth? The           up to 3000 m (10,000 feet) elevation.
total loaded weight distribution of this unit is:             Travel time for hauling units derated more than
10% should be calculated as follows using Rimpull-
Drive unit                 Scraper unit                    Speed-Gradeability charts.
wheels: 23 600 kg          wheels: 21 800 kg                  1) Determine total resistance (grade plus rolling)
(52,000 lb)               (48,000 lb)             in percent.
Remember, use weight on drivers only.
GROSS MACHINE WEIGHT (GMW)
Firm earth — 0.55      23 600 kg = 12 980 kg
(0.55   52,000 lb = 28,600 lb)
Loose earth — 0.45     23 600 kg = 10 620 kg
(0.45   52,000 lb = 23,400 lb)

TOTAL RESISTANCE
On firm earth this unit can exert up to 12 980 kg
RIMPULL

(28,600 lb) rimpull without excessive slipping. How-
ever, on loose earth the drivers would slip if more
than 10 620 kg (23,400 lb) rimpull were developed.
●●●

Altitude — Specification sheets show how much
pull a machine can produce for a given gear and
speed when the engine is operating at rated horse-                                  SPEED
power. When a standard machine is operated in high
altitudes, the engine may require derating to main-           2) Beginning at point A on the chart follow the
tain normal engine life. This engine deration will         total resistance line diagonally to its intersection,
produce less drawbar pull or rimpull.                      B, with the vertical line corresponding to the appro-
The Tables Section gives the altitude deration in       priate gross machine weight. (Rated loaded and
percent of flywheel horsepower for current machines.       empty GMW lines are shown dotted.)
It should be noted that some turbocharged engines             3) Using a straight-edge, establish a horizontal
can operate up to 4570 m (15,000 ft) before they           line to the left from point B to point C on the rim-
require derating. Most machines are engineered to          pull scale.
operate up to 1500-2290 m (5000-7500 ft) before               4) Divide the value of point C as read on the rim-
they require deration.                                     pull scale by the percent of total horsepower avail-
The horsepower deration due to altitude must be         able after altitude deration from the Tables Section.
considered in any job estimating. The amount of            This yields rimpull value D higher than point C.

Edition 40   22-7
Mining and           Estimating Production Off-the-Job
Earthmoving           ● Job Efficiency
● Example Problem (English)

5) Establish a horizontal line right from point D.     Job Efficiency is one of the most complex ele-
The farthest right intersection of this line with a    ments of estimating production since it is influenced
curved speed range line is point E.                    by factors such as operator skill, minor repairs and
6) A vertical line down from point E determines      adjustments, personnel delays, and delays caused by
point F on the speed scale.                            job layout. An approximation of efficiency, if no job
7) Multiply speed in kmh by 16.7 (mph by 88) to      data is available, is given below.
obtain speed in m/min (ft/min). Travel time in min-                                             Efficiency
utes for a given distance in feet is determined by     Operation         Working Hour             Factor
the formula:                                             Day                50 min/hr               0.83
Distance in m (ft)              Night              45 min/hr               0.75
Time (min) = ______________________
Speed in m/min (ft/min)          These factors do not account for delays due to weather
The Travel Time Graphs in sections on Wheel          or machine downtime for maintenance and repairs.
Tractor-Scrapers and Construction & Mining Trucks      You must account for such factors based on experi-
can be used as an alternative method of calculating    ence and local conditions.
haul and/or return times.
●●●●●●●●●●●●●●●
The following example provides a method to manu-       1. Estimate Payload:
ally estimate production and cost. Today, computer     Est. load (LCY)     L.F.   Bank Density = payload
programs, such as Caterpillar’s Fleet Production and   31 LCY     0.80    3000 lb/BCY = 74,400 lb payload
Cost Analysis (FPC), provide a much faster and more    2. Establish Machine Weight:
accurate means to obtain those application results.    Empty Wt. — 102,460 lb or 51.27 tons
Example problem (English)                              Wt. of Load — 74,400 lb or 37.2 tons
Total (GMW) — 176,860 lb or 88.4 tons
A contractor is planning to put the following spread
3. Calculate Usable Pull (traction limitation):
on a dam job. What is the estimated production and
Loaded: (weight on driving wheels = 54%) (GMW)
cost/BCY?
Traction Factor     Wt. on driving wheels =
Equipment:                                                  0.50   176,860 lb    54% = 47,628 lb
11 — 631G Wheel Tractor-Scrapers                      Empty: (weight on driving wheels = 69%) (GMW)
2 — D9T Tractors with C-dozers                         Traction Factor     Wt. on driving wheels =
2 — 12H Motor Graders                                     0.50   102,460 lb    69% = 35,394 lb
1 — 825G Tamping Foot Compactor                      4. Derate for Altitude:
Check power available at 7500 ft from altitude
Material:                                              deration table in the Tables Section.
Description — Sandy clay; damp, natural bed             631G — 100%                 12H — 83%
Bank Density — 3000 lb/BCY                              D9T — 100%                  825G —100%
Shrinkage Factor — 0.85
Traction Factor — 0.50
Altitude — 7500 ft
Job Layout — Haul and Return:                                                         0% Grade
4% G                Sec. D — Fill 400'
'
Haul            RR = 200 lb/ton
C — /ton
Sec. A — Cut 400'         Sec. B — Haul 1500'        Sec. 80 lb               Eff. Grade = 10%
%
RR = 200 lb/ton           RR = 80 lb/ton             RR = rade = 8
Eff. G

Total Effective Grade = RR (%) ± GR (%)
Sec. A: Total Effective Grade = 10% + 0% = 10%
Sec. B: Total Effective Grade = 4% + 0% = 4%
Sec. C: Total Effective Grade = 4% + 4% = 8%
Sec. D: Total Effective Grade = 10% + 0% = 10%

22-8   Edition 40
Estimating Production Off-the-Job            Mining and
● Example Problem (English)              Earthmoving

Then adjust if necessary:                                Rolling Resistance —
Load Time — controlled by D9T, at 100% power, no           RR = RR Factor       Empty Wt (tons)
change.                                                    Sec. D: = 200 lb/ton 51.2 tons = 10,240 lb
Travel, Maneuver and Spread time — 631G, no                  Sec. C: = 80 lb/ton 51.2 tons = 1,4091 lb
change.                                                    Sec. B: = 80 lb/ton 51.2 tons = 1,4091 lb
5. Compare Total Resistance to Tractive Effort               Sec. A: = 200 lb/ton 51.2 tons = 10,240 lb           22
on haul:                                                Total Resistance —
Grade Resistance —                                         TR = RR – GA
GR = lb/ton       tons    adverse grade in percent           Sec. D: = 10,240 lb –     0    = 10,240 lb
Sec. C: = 20 lb/ton       88.4 tons    4% grade =          Sec. C: = 4096 lb – 4096 lb = 0
7072 lb                                      Sec. B: = 4096 lb –       0    = 1,4096 lb
Rolling Resistance —                                         Sec. A: = 10,240 lb –     0    = 10,240 lb
RR = RR Factor (lb/ton)         GMW (tons)                   Check usable pounds pull against maximum
Sec. A: = 200 lb/ton 88.4 tons = 17,686 lb               pounds pull required to move the 631G.
Sec. B: = 80 lb/ton        88.4 tons = 1,7072 lb           Pounds pull usable … 35,349 lb empty
Sec. C: = 80 lb/ton 88.4 tons = 14,144 lb                  Pounds pull required … 10,240 lb
Sec. D: = 200 lb/ton       88.4 tons = 17,686 lb           Estimate travel time for return from 631G empty
travel time curve.
Total Resistance —                                           Travel time (from curves):
TR = RR + GR                                                   Sec. D: 0.40 min
Sec. A: = 17,686 lb +        0    = 17,686 lb                Sec. C: 0.55
Sec. B: = ,7072 lb +         0    = 1,7072 lb                Sec. B: 0.80
Sec. C: = ,7072 lb + 6496 lb = 14,144 lb                             0.40
Sec. A: ____
Sec. D: = 17,686 lb +        0    = 17,686 lb
Check usable pounds pull against maximum                             2.15 min
pounds pull required to move the 631G.                     7. Estimate Cycle Time:
Pull usable … 47,628 lb loaded                           Total Travel Time (Haul plus Return) = 5.55 min
Pull required … 17,686 lb maximum total resistance       Adjusted for altitude: 100%      5.55 min = 5.55 min
Estimate travel time for haul from 631G (loaded)         Load Time                                   0.7 min
travel time curve; read travel time from distance          Maneuver and Spread Time                    0.7 min
________
and effective grade.                                       Total Cycle Time                            6.95 min
Travel time (from curves):
Sec. A: 0.60 min                                       8. Check pusher-scraper combinations:
Sec. B: 1.00                                             Pusher cycle time consists of load, boost, return
Sec. C: 1.20                                           and maneuver time. Where actual job data is not
Sec. D: ____
0.60                                         available, the following may be used.
3.40 min                                            Boost time = 0.10 minute
Return time = 40% of load time
NOTE: This is an estimate only; it does not account          Maneuver time = 0.15 minute
for all the acceleration and deceleration time,   Pusher cycle time = 140% of load time + 0.25 minute
therefore it is not as accurate as the infor-     Pusher cycle time = 140% of 0.7 min + 0.25 minute
mation obtained from a computer program.                              = 0.98 + 0.25 = 1.23 minute
6. Compare Total Resistance to Tractive Effort               Scraper cycle time divided by pusher cycle time
on return:                                              indicates the number of scrapers which can be han-
Grade Assistance —                                         dled by each pusher.
GA = 20 lb/ton       tons    negative grade in percent                        6.95 min
Sec. C: = 20 lb/ton      51.2 tons     4% grade =                           ________ = 5.65
4096 lb                                                       1.23 min

Edition 40   22-9
Mining and            Estimating Production Off-the-Job
Earthmoving            ● Example Problem (English)
● Example Problem (Metric)

Each push tractor is capable of handling five plus     11. Estimate Total Hourly Cost:
scrapers. Therefore the two pushers can adequately       631G             @ \$65.00/hr 11 units          \$715.00
serve the eleven scrapers.                               D9T              @ \$75.00/hr       2 units      150.00
9. Estimate Production:                                  12H              @ \$15.00/hr       2 units       30.00
Cycles/hour       = 60 min ÷ Total cycle time            825G             @ \$40.00/hr       1 unit        40.00
= 60 min/hr ÷ 6.95 min/cycle           Operators        @ \$20.00/hr 16 men _________   320.00
= 8.6 cycles/hr                        Total Hourly Owning and
Estimated load = Heaped capacity           L.F.            Operating Cost                             \$1,255.00
= 31 LCY      0.80                     12. Calculate Performance:
= 24.8 BCY                                               Total cost/hr
Cost per BCY = _____________
Hourly unit                                                               Production/hr
production = Est. load          cycles/hr                                 \$1,255.00
= 24.8 BCY       8.6 cycles/hr         Cost per BCY = ____________
= 213 BCY/hr                                            1947 BCY/hr
Adjusted                                                 Cost per BCY = 64¢ BCY
production = Efficiency factor          hourly         NOTE: Ton-MPH calculations should be made to
production                                judge the ability of the tractor-scraper tires
= 0.83 (50 min hour)       213 BCY              to operate safely under these conditions.
= 177 BCY/hr                           13. Other Considerations:
Hourly fleet                                               If other equipment such as rippers, water wagons,
production = Unit production            No. of units   discs or other miscellaneous machines are needed
= 177 BCY/hr       11                  for the particular operation, then these machines
= 1947 BCY/hr                          must also be included in the cost per BCY.
10. Estimate Compaction:                                                         ●●●
Compaction
requirement = S.F.        hourly fleet production
= 0.85     1947 BCY/hr                 Example problem (Metric)
= 1655 CCY/hr                          A contractor is planning to put the following spread
Compaction capability (given the following):             on a dam job. What is the estimated production and
Compacting width, 7.4 ft                   (W)         cost/BCM?
Average compacting speed, 6 mph            (S)         Equipment:
Compacted lift thickness, 7 in             (L)           11 — 631G Wheel Tractor-Scrapers
No. of passes required, 3                  (P)            2 — D9T Tractors with C-dozers
825G production =                                           2 — 12H Motor Graders
W     S      L  16.3 (conversion                1 — 825G Tamping Foot Compactor
CCY/hr = __________________ constant)
P                                  Material:
7.4     6     7   16.3                          Description — Sandy clay; damp, natural bed
CCY/hr = __________________                                Bank Density — 1770 kg/BCM
3
CCY/hr = 1688 CCY/hr                                       Shrinkage Factor — 0.85
Given the compaction requirement of 1655 CCY/hr,         Traction Factor — 0.50
the 825G is an adequate compactor match-up for             Altitude — 2300 meters
the rest of the fleet. However, any change to job lay-
out that would increase fleet production would upset
this balance.

22-10   Edition 40
Estimating Production Off-the-Job              Mining and
● Example Problem (Metric)                Earthmoving

Job Layout — Haul and Return:

4% G                 Sec. D — Fill 150 m
300                                         22
Haul             RR = 100 kg/t
C — g/t
Sec. A — Cut 150 m       Sec. B — Haul 450 m            Sec. 40 k                 Eff. Grade = 10%
%
RR = 100 kg/t            RR = 40 kg/t                   RR = rade = 8
Eff. G

Total Effective Grade = RR (%) ± GR (%)                   Rolling Resistance —
Sec. A: Total Effective Grade = 10% + 0% = 10%            RR = RR Factor (kg/mton)         GMW (metric tons)
Sec. B: Total Effective Grade = 4% + 0% = 4%                Sec. A: = 100 kg/metric ton       80.48 metric tons
Sec. C: Total Effective Grade = 4% + 4% = 8%                             = 8048 kg
Sec. D: Total Effective Grade = 10% + 0% = 10%              Sec. B: = 40 kg/metric ton        80.48 metric tons
1. Estimate Payload:                                                     = 3219 kg
Est. load (LCM)     L.F.   Bank Density = payload           Sec. C: = 40 kg/metric ton        80.48 metric tons
24 LCM 0.80 1770 kg/BCM = 34 000 kg payload                              = 3219 kg
Sec. D: = 100 kg/metric ton       80.48 metric tons
2. Machine Weight:                                                       = 8048 kg
Empty Wt. — 46 475 kg or 46.48 metric tons
Wt. of Load — 34 000 kg or 34 metric tons                 Total Resistance —
Total (GMW) — 80 475 kg or 80.48 metric tons              TR = RR + GR
Sec. A: = 8048 kg +        0     = 8048 kg
3. Calculate Usable Pull (traction limitation):             Sec. B: = 3219 kg +        0     = 3219 kg
Loaded: (weight on driving wheels = 54%) (GMW)              Sec. C: = 3219 kg + 3219 kg = 6438 kg
Traction Factor     Wt. on driving wheels =               Sec. D: = 8048 kg +        0     = 8048 kg
0.50   80 475 kg     54% = 21 728 kg                   Check usable kilogram force against maximum
Empty: (weight on driving wheels = 69%) (GMW)             kilogram force required to move the 631G.
Traction Factor     Wt. on driving wheels =               Force usable … 21 728 kg loaded
0.50   46 475 kg     69% = 16 034 kg                   Force required … 8048 kg maximum total resistance
4. Derate for Altitude:                                     Estimate travel time for haul from 631G (loaded)
Check power available at 2300 m from altitude           travel time curve; read travel time from distance
deration table in the Tables Section.                     and effective grade.
631G — 100%        12H — 83%                              Travel time (from curves):
D9T — 100%         825G — 100%                               Sec. A: 0.60 min
Then adjust if necessary:                                    Sec. B: 1.00
Load Time — controlled by D9T, at 100% power, no               Sec. C: 1.20
change.                                                              0.60
Sec. D: ____
Travel, Maneuver and Spread time — 631G, no                            3.40 min
change.                                                 NOTE: This is an estimate only; it does not account
5. Compare Total Resistance to Tractive Effort                    for all the acceleration and deceleration time,
on haul:                                                       therefore it is not as accurate as the infor-
Grade Resistance —                                                mation obtained from a computer program.
GR = 10 kg/metric ton     tons    adverse grade           6. Compare Total Resistance to Tractive Effort
in percent                                           on return:
Sec. C: = 10 kg/metric ton    80.48 metric tons         Grade Assistance —
4% grade = 3219 kg                       GA = 10 kg/mton       metric tons     negative grade
in percent
Sec. C: = 10 kg/metric ton      46.48 metric tons

Edition 40   22-11
Mining and           Estimating Production Off-the-Job
Earthmoving           ● Example Problem (Metric)

Rolling Resistance —                                     Each push tractor is capable of handling five plus
RR = RR Factor       Empty Wt.                         scrapers. Therefore the two pushers can adequately
Sec. D: = 100 kg/metric ton      46.48 metric tons   serve the eleven scrapers.
= 4648 kg                               9. Estimate Production:
Sec. C: = 40 kg/metric ton       46.48 metric tons   Cycles/hour       = 60 min ÷ Total cycle time
= 1859 kg                                                 = 60 min/hr ÷ 6.95 min/cycle
Sec. B: = 40 kg/metric ton       46.48 metric tons                     = 8.6 cycles/hr
= 1859 kg
Sec. A: = 100 kg/metric ton     46.48 metric tons    Estimated load = Heaped capacity           L.F.
= 4648 kg                                                 = 24 LCM       0.80
= 19.2 BCM
Total Resistance —
TR = RR – GA                                           Hourly unit
Sec. D: = 4648 kg –       0    = 4648 kg               production = Est. load          cycles/hr
Sec. C: = 1859 kg – 1859 kg = 0                                        = 19.2 BCM        8.6 cycles/hr
Sec. B: = 1859 kg –       0    = 1859 kg                               = 165 BCM
Sec. A: = 4648 kg –       0    = 4648 kg             Adjusted
Check usable kilogram force against maximum            production = Efficiency factor           hourly
force required to move the 631G.                                               production
Kilogram force usable … 16 034 kg empty                                = 0.83 (50 min hour)
Kilogram force required … 4645 kg                                            165 BCM
Estimate travel time for return from 631G empty                        = 137 BCM/hour
travel time curve.                                     Hourly fleet
Travel time (from curves):                             production = Unit production
Sec. D: 0.40 min                                                          No. of units
Sec. C: 0.55                                                        = 137 BCM/hr        11 units
Sec. B: 0.80                                                        = 1507 BCM/hr
0.40
Sec. A: ____
10. Estimate Compaction:
2.15 min                                  Compaction
7. Estimate Cycle Time:                                  requirement = S.F.         hourly fleet
Total Travel Time (Haul plus Return) = 5.55 min                                production
Adjusted for altitude: 100%      5.55 min = 5.55 min                     = 0.85      1507 BCM/hr
Load Time                                   0.7 min                      = 1280 CCM/hr
Maneuver and Spread Time                    0.7 min
________   Compaction capability (given the following):
Total Cycle Time                            6.95 min     Compacting width, 2.26 m                   (W)
8. Check pusher-scraper combinations:                    Average compacting speed, 9.6 km/h (S)
Pusher cycle time consists of load, boost, return      Compacted lift thickness, 18 cm            (L)
and maneuver time. Where actual job data is not          No. of passes required, 3                  (P)
available, the following may be used.                  825G production =
W      S     L    10
Boost time = 0.10 minute                        CCM/hr = ________________ (conversion factor)
Return time = 40% of load time                                         P
Maneuver time = 0.15 minute                                       2.26    9.6     18    10
Pusher cycle time = 140% of load time + 0.25 minute    CCM/hr = _____________________
3
Pusher cycle time = 140% of 0.7 min + 0.25 minute
CCM/hr = 1302
= 0.98 + 0.25 = 1.23 minute
Given the compaction requirement of 1280 CCM/h,
Scraper cycle time divided by pusher cycle time
the 825G is an adequate compactor match-up for
indicates the number of scrapers which can be han-
the rest of the fleet. However, any change to job lay-
dled by each pusher.
out that would increase fleet production would upset
6.95 min
________ = 5.65                     this balance.
1.23 min

22-12   Edition 40
Estimating Production Off-the-Job              Mining and
● Example Problem (Metric)                Earthmoving
Systems
● Economic Haul Distances

11. Estimate Total Hourly Cost:                           SYSTEMS
631G             @ \$65.00/hr 11 units         \$715.00
Caterpillar offers a variety of machines for differ-
D9T              @ \$75.00/hr       2 units      150.00
ent applications and jobs. Many of these separate
12H              @ \$15.00/hr       2 units       30.00
machines function together in mining and earth-
825G             @ \$40.00/hr       1 unit        40.00
moving systems.
Operators        @ \$20.00/hr 16 men _________   320.00                                                              22
● Bulldozing with track-type tractors
Total Hourly Owning and
Operating Cost                             \$1,255.00
12. Calculate Performance:                                  configurations, or push-loaded by track-type tractors
Total cost/hr                          ● Articulated trucks loaded by excavators, track
Cost per BCM = _____________
\$1,255.00                            ● Off-highway trucks loaded by shovels, excavators
1507 BCM/hr
Haul System Selection: In selecting a hauling
= 83¢/BCM                                 system for a project, there may seem to be more
NOTE: Ton-km/h calculations should be made to             than one “right” choice. Many systems may meet
judge the ability of the tractor-scraper tires   the distance, ground conditions, grade, material
to operate safely under these conditions.        type, and production rate requirements. After con-
13. Other Considerations:                                 sidering all of the different factors, one hauling sys-
If other equipment such as rippers, water wagons,       tem usually provides better performance and better
discs or other miscellaneous machines are needed          potential for lowest cost per ton or BCY/BCM. This
for the particular operation, then these machines         makes it critical for the dealer and customer to work
must also be included in the cost per BCM.                together to get accurate information for their oper-
ation or project. Caterpillar is committed to provid-
SOFTWARE NOTE: The Cat DOZSIM program                     ing the correct earthmoving system to match the
can provide a valuable tool for production dozing         customer's specific needs.
applications. Motor Grader Calculator can be used                                 ●●●
to determine the number of graders required to main-
tain haul roads, given a set of site parameters.

GENERAL LOADED HAUL DISTANCES FOR MOBILE SYSTEMS

Track-Type Tractor

Wheel Tractor-Scraper

Articulated Truck

Rear Dump Truck
10 m        100 m    1000 m      10 000 m
33 ft       328 ft   3280 ft     32,800 ft

Edition 40   22-13
Mining and               Production Estimating
Fuel Consumption and Productivity

PRODUCTION ESTIMATING                                     FUEL CONSUMPTION AND PRODUCTIVITY
tion range that varies with material, bucket config-      sumption and machine productivity. It is expressed
uration, target size, operator skill and load area        in units of material moved per volume of fuel con-
conditions. The loader/truck matches given in the         sumed. Common units are cubic meters or tonnes
following table are with the typical number of passes     per liter of fuel (cubic yards or tons/gal). Determin-
and production range.                                     ing fuel efficiency requires measuring both fuel con-
based on your specific conditions.                           Measuring fuel consumption involves tapping into
the vehicle’s fuel supply system — without contami-
nating the fuel. The amount of fuel consumed during
operation is then measured on a weight or volumetric
basis and correlated with the amount of work the
machine has done. Cat machines equipped with
VIMS™ system can record fuel consumed with rela-
tive accuracy, given the engine is performing close
to specifications.

Cat Earthmoving and Mining Systems                        Cat Aggregate Systems
Production/50 Min. Hr.                                    Production/50 Min. Hr.
Tonnes       Tons         Tool     Passes    Target        Tonnes          Tons           Tool          Passes         Target
2270/2450   2500/2700    994F HL      7       793D/F      1530/1710      1700/1900         992K            4-5        777D/777F
2450/2700   2700/3000     994F        5        789C       1450/1630      1600/1800         992K             3            775F
2270/2450   2500/2700    994F HL      6        789C       1090/1270      1200/1400         990H             4            775F
2450/2700   2700/3000     994F        4      785C/785D     910/1180      1000/1300         990H            3-4           773F
1800/2000   2000/2200    993K HL      6      785C/785D
700/900        770/990          988H            4-5           773F
1800/2000   2000/2200     993K        4      777D/777F
800/1000       880/1100         988H             4             772
1530/1710   1700/1900     992K       4-5     777D/777F
540/730        600/800        980H HL            6             772
1180/1360   1300/1500     990H       3-4       773F
700/900        770/990          988H             3             770
800/1000    880/1100      988H       3-4       769D
450/630        500/700        980H HL            5             770
2720/2900   3000/3200   5230B ME*     7       793D/F
1500/1800      1700/2000      5130B FS*           5         777D/777F
2540/2720   2800/3000   5230B FS*     8       793D/F
1270/1450      1400/1600      5130B FS*           4            775F
2630/2810   2900/3100   5230B ME*     6        789C
2450/2630   2700/2900   5230B FS*     6        789C       1180/1360      1300/1500      5130B FS*           3            773F
2540/2720   2800/3000   5230B ME*     5      785C/785D     630/900        700/900       5090B FS*           7            773F
2360/2540   2600/2800   5230B FS*     5      785C/785D     730/910        800/1000      5090B FS*           5             772
1900/2100   2100/2300   5130B ME*     7      785C/785D     630/820        700/900       5090B FS*           4             770
1700/1900   1700/2100   5130B FS*     7      785C/785D   *5000 Series Front Shovels and Mass Excavators are no longer produced. This
information is included for reference only.
1800/2000   2000/2200   5130B ME*     5      777D/777F
1540/1810   1700/2000   5130B FS*     5      777D/777F
910/1090    1000/1200   385 LL ME     7        773F
730/820     800/1000    5090B FS*     7        773F
730/910    800/1000    385 LL ME     5         770
630/820     700/900    5090B FS*     5         770

22-14   Edition 40
Formulas and Rules of Thumb               Mining and
Earthmoving

FORMULAS AND RULES OF THUMB                             Total Resistance
= Rolling Resistance (kg or lb) + Grade Resistance
(kg or lb)
cycles/hr
= Load (BCY)/cycle             Total Effective Grade (%) = RR (%) + GR (%)
cycles/hr                 Usable pull (traction limitation)
22
100%                      = Coeff. of traction    weight on drivers
Load Factor (L.F.)       = _______________                 = Coeff. of traction
100% + % swell
(Total weight      % on drivers)
Load (bank measure) = Loose cubic meters
(LCM)      L.F.           Pull required = Rolling Resistance + Grade
= Loose cubic yards (LCY)                          Resistance
L.F.                   Pull required = Total Resistance
Compacted cubic meters      Total Cycle Time = Fixed time + Variable time
(or yards)
Shrinkage Factor (S.F.) = _______________________       Fixed time: See respective machine production
Bank cubic meters           section.
(or yards)          Variable time = Total haul time + Total return time
Density                  = Weight/Unit Volume                             Distance (m)
Weight of load              Travel Time = ______________
Load (bank measure) = ______________                                     Speed (m/min)
Bank density                                  Distance (ft)
Rolling Resistance Factor                                             = ______________
Speed (fpm)
= 20 kg/t + (6 kg/t/cm     cm)
= 40 lb/ton + (30 lb/ton/inch     inches)                                           60 min/hr
Cycles per hour = _________________________
Rolling Resistance                                                           Total cycle time (min/cycle)
= RR Factor (kg/t)     GMW (tons)                     Adjusted production = Hourly production
= RR Factor (lb/ton)     GMW (tons)                                               Efficiency factor
Rolling Resistance (general estimation)                                           Hourly production required
= 2% of GMW + 0.6% of GMW per cm tire                 No. of units required = _________________________
Unit hourly production
penetration                                       No. of scrapers a       Scraper cycle time
= 2% of GMW + 1.5% of GMW per inch tire                  pusher will load = __________________
penetration                                                                Pusher cycle time
vertical change in elevation (rise)       Pusher cycle time (min) = 1.40 Load time (min) +
% Grade = _______________________________                                            0.25 min
corresponding horizontal
distance (run)                                         GMW (kg)       Total Effective
___________________________
= 20 lb/ton    % grade                                           273.75
Grade Resistance = GR Factor (kg/t) GMW (tons)                                   GMW (lb)      Total Effective
= GR Factor (lb/ton) GMW (tons)                                   Grade     Speed (mph)
___________________________
er                    =

Edition 40   22-15
Notes —

22-16   Edition 40

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