Ihab Saad – Loading and Hauling resistances, speeds, and cycle times
AI: Summary ©
The speakers discuss the importance of loading and hauling in construction equipment, including factors affecting cycle time, productivity, and cost. They also discuss the importance of resistance and tire flex in the process of loading truck, as well as the factors affecting the speed and efficiency of the machine. The speakers provide examples of various types of rock, including resistance, tire penetration, and soil conditions. They also discuss the factors affecting the speed, efficiency, and maximum speed of the tractor, including weight, traction, and resistance. They provide examples of performance curves and explain how to measure the performance of the machine and estimate the maximum speed. They also discuss the factors affecting the total production of the equipment and the importance of derating factors and preventing acceleration and deceleration during driving conditions.
AI: Summary ©
Steve, hello, into another lecture of construction equipment, and
today we're going to be talking about loading and hauling. So
primarily in this lecture, we're going to learn about how to
calculate the cycle time, and what are the elements affecting the
cycle time, primarily, what kind of resistance is the equipment
going to be subject to that can affect its performance. It can
affect the duration of the cycle time and the number of cycles per
hour.
So the equipment productivity is affected by several things. Speed
affects the cycle time. Speed of each cycle. The Cycle Time affects
the production. Because, again, the production is determined by
the number of cycles per hour times the production per cycle.
And the number of cycles per hour is determined by the length of the
cycle time, and production determines cost. So primarily, we
can say that the speed of the performing the operation for each
cycle affects the project cost for that piece of equipment.
Each piece of equipment requires a certain amount of power to
overcome the resistance that is going to be facing. So the
required power is the power needed to overcome resisting forces and
cause machine motion. So for example, imagine that you are
driving in a muddy condition. There's going to be a certain
resistance to the motion of the tires, so the equipment has to be
able to overcome that resistance in order to move forward.
The magnitude of the resisting forces determines the power
required, the minimum amount of power required to overcome this
resistance and to be able to move the vehicle or the equipment
forward. The
equipment cycle time. The Cycle Time for a piece of equipment is
the time it takes to perform one cycle of its planned job. We
mentioned in last in the last class, that one cycle is basically
going to be, for example, for a loader is to position itself
instead of the area to be excavated, or in front, in front
of the soil to be removed, and then to load that soil in the
bucket, to turn around and to move, to dump that soil, and then
to come back, position itself and get ready for a new cycle. All of
this forms one cycle, so loading, hauling, excavating, lifting, etc.
All of these are parts of the cycle time for the equipment. And
each cycle consists of two components. One of them is called
fixed cycle part, or fixed cycle time, or fixed component of the
cycle time, and the other one is the variable component of the
cycle time. The fixed time is the part of the cycle other than the
travel time includes spotting, loading, maneuvering and dumping.
So mostly the fixed time is done while the equipment is in its
place. It's not moving. Loading takes place while the equipment is
standing in place. Dumping is the same thing, whereas traveling back
and forth, this is the hauling part of the cycle time, which is
going to be part of the variable time. So the variable time
represents the travel time from origin to destination and back,
and it depends on equipment characteristics, like the weight
and the power, the engine, power of that equipment, the road
conditions, whether it's flat or bumpy, whether it's uphill or
downhill, what kind of soil is it trolling on grade and altitude?
Again, grade, which is the slope of the road, again, downhill or
uphill altitude, as we have discussed in class, the higher
from the sea level you're working, the thinner the air is going to
be, which might affect the efficiency of the engine and the
distance traveled. Of course, the farther the distance, the longer
it's going to take to get to and from there, the longer the
variable part of the cycle time.
The loading time is a function of the capacity and cycle time of the
loading equipment,
the capacity of the truck or hauler and the skill of the
loading operator. If you remember what we discussed before about
excellent job conditions, above average, average, below average,
etc. It had something to do with the angle of swing. For example,
of the equipment, the larger that angle of swing, the longer it's
going to take, which means it's going to take the cycle time is
going to be longer, whether it's dumping on the ground or it's
dumping in a truck, whether that truck is a large truck or a small
truck, how skilled the truck driver is? How big is the bucket
for the loading equipment, and how big is the bed of the truck where
you're going to be dumping the soil? All of these are elements
that affect the loading time. The dumping time is affected by.
The type and condition of material, how easy it is it going
to flow from the bed or from the bucket of that equipment,
whether that material is going to be wet or dry, the Method of
dumping or spreading, whether it's going to be end dump or bottom
dump, or whether you're going to be dumping into a pile or you're
going to be spreading it around, and the type and maneuverability
of the truck or the piece of equipment, basically.
So the haul and return time are a function of the haul road profile,
including the great resistance, rolling resistance and distance to
be traveled. We're going to talk about each one of these in more
detail in a second, the altitude of the project site and the
performance characteristics of the hodding equipment, which is a
characteristic of the vehicle itself, or the equipment itself,
number of loader cycles to load a truck. So if you have, let's say,
a loader with a bucket size of two cubic yards, and you have a truck
with a capacity of 15 cubic yards, how many cycles is it going to be
needed to load the truck? The volume capacity of the truck
divided by the volume capacity of the loader? So in this case, going
to be 15 divided by two, which is seven and a half, which means it's
going to take eight cycles. Now,
the number of loader cycles required times the loader cycle
time is going to determine the loading cycle. So to fully load
that truck, you're going to need eight cycles of that loader. If
each cycle takes, let's say, 35 seconds, then it's going to be
eight times 35 seconds, that's going to be the duration of
loading that truck.
And here, for example, are some examples in this table about the
job conditions, whether it's favorable, average or unfavorable,
for the turn and done time and for the spotting time, which are
basically all of these are part of the fixed time of the equipment.
This is not the variable time. This is the fixed part. So in
under favorable conditions, in end dump, it's going to take about one
minute.
The spotting time is going to be about 15 seconds. For the bottom
dump is going to be less than that, only point four minutes, and
the spotting time is going to be pretty much the same. So under
favorable conditions, the fixed time for that piece of equipment
is going to be for an end up is going to be a minute and 15
seconds. Under unfavorable conditions, as you can see, that
number is going to be much larger, one and a half to two minutes, and
point eight minutes, so almost twice as much as the favorable
conditions. Now,
talking about the resistance, what kind of resistance is the
equipment going to be facing while it's trying to move and to perform
its function? Rolling resistance is the first type, and it's a
measure of the force. The force is going to be expressed in pounds
per ton
that must be overcome to rotate a wheel over the surface on which it
makes contact. So this is got friction is going to be part of
that equation, one of the types of resistance and equipment faces
while moving. So we have two types of resistance, the equipment Scott
has to overcome. One of them is going to be running resistance,
and the other one is going to be the grade resistance. It can be
expressed either in power as a pounds per ton or of equipment
weight, or just in pounds. So you can say either the rolling
resistance going to be 20 pounds per ton of the equipment weight.
So if the equipment weight is 20 tons, then the total resistance,
running resistance going to be 20 pounds per tons, time per ton
times 20, which is the weight of the equipment. That's 400 pounds.
Or you can just express it as number of pounds again, as we did,
by multiplying that factor in pounds per ton by the total weight
of the equipment. Now that weight of the equipment is going to
change because whether the equipment is empty or is loaded.
So when it's going when it's loading and going to dump, it's
going to be moving forward full. So it's going to be heavier,
therefore the resistance going to be much more. When it's coming
back is going to be empty, therefore the resistance is going
to be less than the previous case, it is caused by internal friction
and tire flexing. Tire flexing is very important, so as you know, to
improve your fuel efficiency for your vehicle, for your car,
properly inflated tires are going to yield better fuel consumption.
Same thing here. It's going to increase that rolling resistance
going to increase by about 30 pounds per ton for each inch of
tire penetration. Tire penetration into that soil.
If it's if the tire is penetrating two inches, then that's going to
add 60 pounds per ton of rolling resistance. Properly inflated
tires reduce rolling resistance. What if the equipment is.
Running over asphalt or concrete. Now, in this case, you're not
going to have any tyre penetration, so you're not going
to have much
additional rolling resistance due to tire flexing.
If the tyre penetration is known, then the rolling resistance is
equal to, which is RR, that's the running resistance. It's 40 pounds
per ton of the equipment weight, plus 30 pounds per ton per inch
times inches of tire penetration. So again, if you're running on
concrete or asphalt, where you're not going to have any tire
penetration, that second part of the equation is going to be equal
to zero. You're only going to have the 40 pounds per ton,
the running resistance force. So this is the running resistance,
but the running resistance force for the whole equipment is going
to be the running resistance in pounds per ton, which we
calculated from here the RR times the total weight of the equipment
in tons. That's going to be a force
represented by or expressed in pounds. If tyre penetration is not
known, then the running resistance can be estimated from tables. So
you don't know exactly. You can't measure the type penetration, but
you know what type of soil that equipment is going to be working
on. We can use that table. So for concrete or asphalt, the running
resistance factor in pounds per ton is going to be 40 pounds per
ton, which is basically this one, with the second component of the
equation being equal to zero.
For concrete is going to be 40. For asphalt is going to be 30.
Firm, smooth, flexing slightly under load. So we're not talking
about the paved road. We're going to talk about a compacted dirt
road. For example, that's going to be up to 64 pounds per ton, rotted
dirt roadway. One to two inches of penetration is going to be 100
which is basically the 40 plus 30 pounds per ton per inch times two
inches, which is going to be 60. So 40 plus 60, that's 100 pounds,
pounds per ton, soft, rotted dirt, three to four inch penetration,
about 150
loose sand or gravel is going to be up to 200 pounds per ton, soft,
muddy, deeply rooted road conditions is going to be anywhere
between 304 100 pound pounds per ton. You can see the big
difference between a paved road, concrete or asphalt, 40 and soft,
muddy, deeply rotted. 300 to 400 which is 10 times the rolling
resistance. Therefore, these are things that you can control as a
project manager. For example, you can build a temporary access road,
or you can have compacted soil or crushed stone or gravel or
something like that, to improve the working conditions of the
equipment, therefore reducing the rolling resistance, which means
you're going to get better cycle Time, which means higher
productivity and lower cost
the second type of resistance is the grade resistance, which is the
component of a vehicle's weight, which acts parallel to an inclined
surface. It can be positive when moving uphill. Contrary to our
intuition, you might think that positive is something good that
helps, and negative is something that impedes. Here we're going to
use the opposite sign convention. It's positive when moving uphill,
so you're going to add that resistance when when moving
uphill, is going to be working against the equipment, which is
adverse conditions, and negative when moving downhill, which is a
favorable condition, also called grade assistance, not resistance.
In this case, grade assistance, which means it can help reducing
the rolling resistance, and it can be calculated exactly the same way
as we did with the rolling resistance. The grade resistance
can be expressed in pounds per ton, which is equal to 20 pounds
per ton for each 1% slope times the percent slope. So if the grade
is 5% moving uphill by 5% is going to be 20 pounds per ton for each
1% which is 20 pounds per ton times five which means we're going
to have a grade resistance of 100 pounds per ton. If the weight of
the equipment is 20 tons. Again, 200 pounds, 100 pounds per ton,
which is 20 times five times 20 that's going to give you the total
resistance, the total grade resistance force. So the grade
resistance force is equal to grade resistance, the factor that we
calculated from here, times the total weight of the equipment in
tons. Again, in this case, the issue of whether the equipment is
loaded or unloaded is going to make a big difference.
Can we express both resistances into one so? Can we express both.
Is rolling resistance and grade resistance as a common number.
Effective grade is the grade resistance equivalent to the total
resistance encountered by the vehicle. So you can say, for
example, that a vehicle that's moving on a flat surface, but the
road is rotted and is going to exert a lot of rolling resistance.
That is as if the equipment is moving uphill on a certain slope.
It's also called equivalent grade, or percent total resistance, and
can be calculated as the effective grade percentage is equal to the
actual grade if it's moving uphill at 5% so that's going to be five
plus rolling resistance divided by 20. So if the running resistance,
for example, is, let's say, 100 pounds per ton, then we divide
that by 20. That's going to be 100 pounds per ton is going to be
equivalent to moving uphill at the 5% slope, which is the 100 divided
by 20. So in this case, if you already have a 5% slope plus
running resistance of 100 pounds per ton, that's equivalent to
moving to up to a 10% slope uphill.
Of course, that also is affected by the type of soil that you're
running on. Imagine if you're running on ice, you're not going
to have any traction with that ice, so the tires or the wheels
can turn without the equipment moving forward. So we're going to
deal with something called a coefficient of traction, the power
available to move the vehicle and its load can be expressed as
either, if that equipment is running on wheels. It's going to
be called the rim pull. RIM pull, which is the pull available at the
rim of the driving wheels under rated conditions. The driving
wheels, some equipment are four wheel drive. Some of them are two
wheel drive. Some of them have more than two moving axles. So an
equipment might have, might have three axles. For example, two of
them are moving axles. So in this case, we're going to calculate
that pull available a dream of the driving wheels, which is the
moving axles. If that equipment runs on tracks,
then it's going to be called draw bar. So we we're talking about
dream pull in case of wheels, draw bar in case of tracks, which is
the power available at the hitch of the crawler tractor operating
under standard conditions, how much pull can it exert? How much
weight can it pull? The traction depends on the coefficient of
traction and the weight on the drivers. So the maximum usable
pull.
You might have a lot of power for the equipment. You might have a
lot of RIM pull, lot of drawbar but especially lot of RIM pull in
this case, but you're working on a very slippery soil, so the wheels
turn in place, therefore it's not all translated into motion. So
here we have a something called the coefficient of traction. For
concrete that's dry, it's going to be point nine, which is 90% of the
power is going to be translated into motion, whereas for tracks,
it's only 45%
concrete that's wet, 80% and 45% respectively. And you keep going
down until we reach for example, ice is only 10% 90% of the power
of the equipment is wasted. And in case of tracks, it's going to be
85% of that power that's going to be wasted.
So the maximum usable pull is the coefficient of traction, which we
can obtain from such a table, depending on the soil conditions,
times the weight on drivers, not the weight of the whole equipment,
but the weight on the moving axles. That's going to be what
affects the motion of the equipment. That's why, if you
remember in the last lecture, when we were talking about that
coefficient of traction, in case you are driving uphill on an icy
road or where you have snow, it might, especially in a rear wheel
drive car, putting a heavier weight on the rear axle, like
having sandbags in your trunk, for example, in the trunk of Your car,
might help overcoming that kind of resistance,
the equipment available power is the engine, horsepower and
operating the engine horsepower and operating gear are the primary
factors in determining the power available at the drive wheels or
the draw bar of a machine
drawbar, in case of tracks, RIM pulled in case of wheels.
Horsepower involves the rate of doing work, and one horsepower is
equivalent to 33,000 foot pound per minute. Therefore the
traveling speed of the machine should be considered when
calculating the amount of.
Poll since we're talking about
per minute, so we're talking about speed, which is going to be a
factor.
Performance charts are provided by equipment manufacturers to enable
the calculation of the estimated machine speed. So with each piece
of equipment, you're going to have a manual that has some performance
charts that tells you, under first gear, what's the maximum
attainable speed on second gear, third gear, fourth gear, under
different loading conditions, whether the equipment is fully
loaded or it's empty, it's gross weight or net weight and so on and
so forth. The charts relate trim pull or draw bar pull to gross
vehicle weight, speed and total resistance as a percent, which is
the effective grade.
Here's an example of a performance curve. It shows both in kilograms
or in pounds, so metric and imperial, and that's the drawbar
pull.
And this is the speed that the equipment can reach. And it tells
you here under such and such
speed under such and such gear. So under the first gear, your maximum
speed is going to be about two and a half miles per hour. Under the
second gear, your maximum speed is going to be about 4.25
miles per hour, and the maximum overall maximum speed of that
equipment is going to be less than seven miles per
hour. So knowing the drawbar pull here, for example, 25,000 pounds
is going we go horizontally, is going to interact, is going to
intersect with two different gears. So under the first gear,
with that 25,000 pounds of available drawbar, available pull,
we're going to have about 1.4 miles per hour, and under the
second gear, we're going to have only about one mile per hour.
Here's another example of these sets of
of performance curves. It shows at the top the gross weight of the
vehicle, and it shows whether when it's empty, that's the weight, and
this is when it's loaded. So we can do the same thing, the gross
weight, the rim pool, the total resistance, which is grade plus
rolling resistance. We converted the rolling resistance into grade
by dividing it by 20
and each that's going to be represented as percentage points.
So here, for example, if that way, vehicles weight is
about 100,000 pounds, and it's running over a surface with an
Effective grade of 6%
then
in the fourth gear,
it's going to give us
a speed of about 14 miles per hour, and that's got the available
rim pole is going to be or the available the required
power To generate is going to be around 6000 pounds.
Here we have something called a retarder curve. If that equipment
is moving downhill, downhill again, the same piece of
equipment, here's the weight and it's moving the effective grade is
a favorable grade moving downhill 20%
in the second gear is going to give us a speed of about maybe
seven miles per hour.
And if it's empty, this is when it's loaded. If it's empty, same
equipment and moving again on the same slope, is going to give us
also
in the second gear, it's going to give us about the same speed,
which is about 7% seven miles per hour.
Now what if that equipment? Imagine if the road is divided
into different segments, part of it is uphill, part of it is flat,
and part of it is downhill,
depending on the length of the segment. If the equipment is going
to move in very short segments, it's not going to gain enough
speed to move faster, so the longer the segment,
the better the factor that we're going to use here, and we're going
to see how to use that factor in a minute. If the length of the
segment is only 10 100 feet, it has to start and stop in 100 feet.
It hasn't gained enough speed but, but if it's moving 5000 feet, is
going to give all.
Almost gain 96% of its maximum speed. Here is going to have only
about 45% of its maximum speed.
And then we're going to have to compare whether it's coming from a
stop and going uphill or coming from a stop moving downhill. Is it
increasing or decreasing speed? We're going to see all of these in
the in a problem in a moment.
If the truck stops at both ends of a segment, divide the segment
length into two parts and determine the speed factor for
each part. So if, for example, we have
a flat road segment of, let's say, 700 feet, the equipment is going
to travel the the truck is going to travel 700 feet,
but it's gone. It's going to be standing still at the beginning to
be loaded, and it's going to stop at the end of these 700 feet to
dump that load. In this case, we're going to divide that 700 by
two, so as if the segment's length is only 350
and we're going to use that factor only once, which is we can
interpolate between these two numbers.
Effect of altitude. If the equipment operates at a higher
altitude where the air is less dense, the air is thinner, the
engine may perform at a reduced power output, the engine power is
going to be decreasing approximately 3% for each 1000
feet above sea level. So in increments of 1000 feet,
for each increment of 1000 feet, you lose 3% of the engine power.
Turbocharged engine are more efficient at higher altitudes
because they are not affected by that equation. So if you have in a
problem, or if you have in real life that you're using
turbocharged equipment, this does not apply. You don't have to worry
about the effect of altitude
a the rating factor is used to reduce the engine production based
on the altitude. For from this equation, the rating factor as a
percentage is equal to three times altitude minus 3000 divided by
1000 3000 is going to be our benchmark. So working 3000 feet
above sea level is going to be where we're going to measure the
equipment performance.
If you are working at 4000 feet. So in this case, the altitude is
4000 minus 3000 that's 1000
times three that's 3000 divided by 1000 so that the rating factor is
3%
if you're working at 5000 feet. So five minus three, 5000 minus 3000
that's 2000 times three 6000 divided by 1000 so the rating
factor is going to be 6% and so on. Now
let's look at an example that can illustrate all of the things that
we've talked about so far.
Using the performance curve determine the maximum speed of the
vehicle, if its gross weight is 150,000 pounds, the total
resistance, which is rolling resistance, plus grade resistance,
both of them combined and translated into effective grade 4%
altitude. The rating factor is point two, 520, 5%
altitude. The rating factor is 25% so obviously this equipment is
working at a relatively high altitude.
So we're going to look at this performance table here. The first
thing that we can detect is 150,000
pounds. So the weight of the equipment is going to be under
50,000 pounds, and we have an effective grade of 4% so we're
going to look at 150,000
and the intersection with the 4%
here's the 150,000
the interaction with 4% it means that for this equipment to
overcome the resistance, it needs 6000 pounds of rimple, 6000
pounds. But remember that at this higher altitude we're not going to
be able all to use all of that 6000 pounds. So to overcome that
resistance at the higher altitude, we need actually more than 6000
6000 1000 that was not taken into consideration the effect of
altitude. So we're going to derate that. So we're going to decrease
that by 25%
which means dividing by one minus 25%
divide by one minus the derating factor, which is 25% so as if we
are dividing 6000 6000 divided by point seven, five, which is one
minus point two five, which gives a required rainfall of 8000
pounds. So at a rainfall of 8000 pounds,
we're going to check third gear is going to give us a speed of about
10 miles per hour. So this equipment can operate.
In the third gear, at this altitude, with a speed of 10 miles
per hour.
So basically, that's what we're looking for determine the maximum
speed of the vehicle. So we determine the maximum speed is
going to be 10 miles per hour. Let's
look at another example, a four wheel drive, wheel tracker. Four
wheel drive, which means that both axles are going to be moving.
It's wheeled. It's not on tracks.
It weighs 41,000 pounds and produces a maximum rim pull of
40,000 pounds.
It is working at an altitude of 8000 feet on wet earth. Wet earth
means coefficient of traction is going to decrease the performance
of the equipment. 8000 feet means that we're going to have a
derating factor.
Operating conditions require a pull of 20,000 pounds to move the
tractor and its load. Can the tractor perform under these
conditions? That's the question. Now, we need a force or a power of
20,000 pounds. Are we going to be able to generate that power, or is
that equipment not going to be able to do that?
So first of all, we look at the derating factor. We have 8000 feet
minus 3000 times three, so five times three, which gives 15%
so the percent rated power available is going to be the power
that was given here, which is
40,000 pounds times 85%
therefore the available power is 40,000 times point eight five,
which is 34,000 pounds. Now the coefficient of traction based on
wet earth. Wet earth, we're going to look here,
wet earth for rubber times is going to give us a factor of point
four, five,
so the available power is going to be the maximum usable pull.
Remember, the maximum usable pull is point four, five,
which is what we got from the table times the weight, because,
again,
we use the whole weight in this case, because
you have two moving axles. If it were only a two wheel drive, we
would look at the weight on the moving axle, which might be less
than that. So it's equivalent to 18,450
pounds.
The usable power
is less than the required pull.
Less than the required pull. The required pull was the 20
the 20,000
pounds. So basically, the usable pool is less than the required
pool. Therefore the tractor cannot perform under these conditions in
order to work, increase the weight or the coefficient of traction. So
again, improve the soil conditions, or increase increase
the weight, because if you increase the weight times the low
traction, factor of traction, or coefficient of traction, is going
to give you something higher. So if that will, weight were to
increase by about maybe 3000 pounds or 4000 pounds, when you
multiply it by this is going to give you something more than the
25 20,000 which would be able to pull the load behind the
equipment. To estimate the travel time, we have to account for
acceleration and deceleration, and not only for the maximum speed of
the vehicle, as we just mentioned a couple minutes ago, the longer
the stretch of the road, the longer the part where you're going
to be moving at maximum speed, because you have to accelerate at
the beginning and you have to decelerate at the end. If you're
moving in a very short distance, by the time you accelerate, you
have to decelerate. You haven't reached the maximum speed of the
equipment. But the longer the stretch of the road, the longer
part with a maximum seed you're going to be achieving. Using the
average speed factor from tables, converts the maximum speed to the
average speed. Remember that point nine, 5.96 etc. Let's go back
here.
So here, for example, under if you were moving only 100 feet,
you're going to be moving at only 45% of the maximum speed. So the
maximum speed that we calculate the from the performance tables,
you're going to multiply that times point four, five, that's
going to be the operating speed. Whereas if you are moving at a
length of 5000 feet, you're going to be operating at 96% of the
maximum speed, which shows a big difference
the travel time.
Is obtained by dividing the treble distance by the average speed.
Travel distance by the average speed. The average speed factor
applies twice, if starting from rest and ending at stop. So as we
said, if you're going to be moving only 100 feet, you're going to
divide that by two. And as if you're only moving 50 feet, and
you apply that factor, or you apply the factor twice, because
you're going to be starting from standstill and ending at the
standstill as well at the end. Let's look at an example which
might again explain this idea and illustrate the ideas.
We are using a caterpillar, D, 6r standard.
It weighs 39,800
pounds,
with a coefficient of traction of point six.
What is the maximum speed when up when moving up an 8% slope.
So the available pull is going to be the weight on the driving axles
times the coefficient of traction here, by the way, we use the whole
weight, which means that it's a four wheel drive.
So 39,800
times point six, which is equivalent to 23,880
pounds. That's the available pool,
the grade resistance.
We are moving at 8% slope, which is 20
times the weight in tons,
times eight, which is the percent 20. That's a constant times 20
pounds per ton times the weight in tons, which is 20 tons, which is
this one, the 39,800
almost 40,000 pounds, which is 20 tons. So 20 pounds per ton times
20 tons times 8% the slope, which gives 3200 pounds. That's the
grade resistance, 3800 pound. So the net drawbar pull that we want
is going to be 23,880
the available minus 3200 which is the resistance which gives 20,680
pounds. That's the available drawbar pull. So we're gonna go
here and look at
this. By the way, either is a four wheel or it's on tracks. In this
case, I believe it's on tracks because we're using drawbar pull
and not trim pull. So the net drawbar pull is 20,680
we're gonna go along this axle axis here, 2680
which is about here, 20,006 80 that's going to intersect with the
first gear and the second gear almost at the same point, which
gives a speed of about two miles per hour.
Now this is going to be the maximum speed, looking at the
distance is going to be traveling, and the conditions of the road,
we're going to multiply that by the factor that's going to reduce
that to the actual speed, rather than the maximum speed.
Looking at the third example here a contractor is to use a
caterpillar, d7, G, crawler, tractor, crawler on tracks, the
whole weight is going to be used and not part of the weight with
the power shift transmission to excavate 1500 bank cubic yards for
the foundation of a large house. The swell of the excavated
material is estimated to be 25%
remember now this problem, each word here has a meaning, so we
have to break it down later on to look at the meaning of each one of
this piece of information, the tractor must push the excavated
material up a 12% slope where it will be stockpiled for later
removal.
The contractor has measured the pile of excavated material in
front of the tractor, universal blade, just before spinach occurs,
and has determined the pie. Length,
12.6 feet. Width, 8.1 and height, four feet, if you remember the
equation that we used in the previous lecture, point 375,
wlh, this is something that we're going to be using here.
The tractor weighs 44,400
pounds, and the coefficient of traction is estimated to be point
seven.
The average haul distance is estimated to be 300 feet. What is
the estimated productivity of the tractor? If the contractor plans
to average 50 minutes of operation per hour, lots of information.
Let's break it down to see what is given and what is needed. How are
we use what's given into what's needed? What kind of equations are
we going to use? So we're going to break that down into steps and
look at each step is going to give us certain deliverable. We're
going.
Use that deliverable in the following steps. We're going to
process it until we reach the finance
so step number one, the volume of material that can be removed moved
during one operation cycle, which is point 375, WHL. Remember that
equation, but also remember that this equation gives us a volume in
cubic feet in loose cubic feet. So we need to convert that into cubic
yards. Therefore the number that we got 153.09
cubic feet. We divide that by 27 which is equivalent to 5.67
again, this soil, remember, is loose. So it's in loose cubic
yards. We need to convert that into bank. Do we have the swell?
Yes, we do. So to step number two is to convert the volume to bank
cubic yards. The volume bank is equal to volume loose divided by
one plus the swell factor, which gives us 4.54
bank cubic yards. That's the volume of soil that's going to be
moved into one cycle.
Now we're going to look at the resistance for that crawler
tractor. We do not have
rolling resistance. We don't have any tire flexing. Therefore, all
the resistance for the crawler tractor is going to be only the
grade resistance, which is going to be 20 pounds per ton, per
percent slope
times 12%
times the weight of the equipment, in tons, which was given here as
44,400
which is 22.2 tons. You know, 22,000 pounds per ton. So the
total resistance is equivalent to 50 328 pounds.
The usable power
is going to be the maximum usable drawbar pull is the coefficient of
traction point seven times the weight of the draw on the driving
wheels or tracks. In case of tracks, we use the whole weight of
the equipment. So 44,400
which gives a maximum usable draw bar of 31,080
pounds.
The usable power that's available is greater than the required
drawbar pull. Required drawbar pull is to overcome the
resistance, which is the 5328
so we have much more than that. Therefore, the
equipment is going to be able to move without slip. So that's the
first step. Yes, this equipment can move forward.
Next we're going to look at the speed. At what speed can this
equipment move? So to determine the maximum speed the track will
operate in first gear, we're going to look here
at
the available drawbar pull,
which is three, 3000
something, and
that's going to give us, in the first gear, a speed of 2.1
about 2.1 miles per Hour. So the maximum speed to overcome the
resistance, which is
5328
not 3053 28 pounds is going to be 2.1 miles per hour.
Now, when returning,
the tractor will return empty, going downhill, therefore we have
no rolling resistance,
and we have no great resistance either. It's actually going to be
great assistance. So instead of using a negative number, we're
just going to assume a zero, because we don't have any slopes
here on this performance curve. So we're going to assume zero
resistance.
So the maximum speed available in second gear is going to be four
miles per hour.
So moving uphill, moving forward, loaded, we're going to operate at
2.1 miles per hour, moving downhill backward. We're going to
operate at four miles per hour,
so the tractor will return empty, going downhill, no rolling
resistance. Step number six is going to be determined the cycle
time. Cycle time is going to be equal to fixed time plus haul time
plus return time, fixed time. If you remember, here, it mentioned
something about
shift transmission, power, shift transmission, if you remember,
from the previous lecture, that was equivalent to.
Three seconds, point oh, five
so point o5, minutes. That's the fixed time. The hull time is going
to be the distance, which is 300 feet, divided by the speed, which
is 2.1 miles per hour. Now this is in feet, and this is in miles per
hour. To convert that, we divide by a factor. Remember that that's
going to be a constant to convert from feet to miles per hour. We
divide by 288,
so 88 feet per minute, per miles per hour,
the return
part of the cycle time,
which is again a variable time, 300 feet, divided by the speed,
four miles per hour. Again, we're going to divide by the same factor
the 88 to convert from feet into miles per hour. So the total cycle
time is going to be point oh, five. That's three three seconds
plus 1.62 minutes, which is the number that we get from here, plus
point eight, five minutes, which is the number that we get from
here. So the total cycle time is 2.52 minutes to perform one cycle.
It's 2.52 minutes.
Step number seven, the productivity in each cycle,
each cycle is going to take 2.52 minutes. And we have 50 minutes of
operation per hour. So how many cycles in the 50 minutes? 50
divided by 2.52
almost 20 cycles per hour. Each cycle is going to be 4.54 bank
cubic yards. So almost 20 times 4.54
is going to give us a total of 91.3
bank cubic yards per hour. That's going to be the total production
of that piece of equipment. Another
example. As you can see in the previous example, we broke it down
into steps, processed the information for each step, and it
led to additional elements of the problem. We combine all of these
elements at the end to get the final answer. A wheel tractor is
being operated on a soft roadway. Wheel operate wheel tractor so it
has wheels soft roadway, which means we're going to have tire
penetration. The tire penetration is five inches. The tractor weighs
five tons. What is the total resistance and the effective
grade? If the tractor is going uphill, ascending a slope of 4% or
if the tractor is descending a slope of 6%
let's look at it this way. Now we have rolling resistance because of
the time penetration, and we have a grade resistance that's coming
from that slope that we're talking about. So we need to combine these
to calculate the total resistance, which is the effective grade,
total resistance, as in pounds per ton, and translate that into an
effective grade as a percentage.
So the running resistance factor is going to be 40 plus. That's the
constant 40, as if you're moving on asphalt plus 30 pounds per ton
times the tire penetration. So 40 plus 30 times five, that's 190
pounds per ton. That's the rolling resistance factor. The total
rolling resistance is the total resistance factor times the weight
of the equipment. The weight is 20 tons. So 190 times 20, which gives
a total of 3800 pounds.
The grade resistance going uphill is equivalent to point oh four.
That's the slope times 20, which is 20 pounds per ton, per 1% of
slope times
to convert that into times 2000 pounds per ton. 20,000 that's the
weight of equipment, 2000 pounds per ton for the
to convert the total into pounds. So the total resistance is
equivalent to 1600 pounds. Descending is going to be
negative. Remember that going uphill is positive, downhill is
negative. So the resistance going downhill is negative, 2400
pounds.
So the total resistance if you're moving uphill, is rolling
resistance plus grade resistance, 3800 plus 1600 that's the total of
5400 pounds
going in the opposite direction, descending downhill is going to be
the same 3800 minus, in this case, 2400 which gives only 1400 pounds.
Now to convert this resistance into effective grade, we're going
to divide by the weight of the equipment and 20 pounds per ton
for each percent of slope. So 5400 divided by 20.
Times 20, that gives an effective grade of 13.5%
uphill. So converting the rolling resistance into
a
a grade resistance is almost nine and a half percent, which is the
difference between 13 and a half and four and for the downhill
trip, is going to be that 1400
divided by 20 by 20. So it reduced the
the
the total to this is going to be 3.5%
remember, we had negative 3.5 so it's exactly the same 9.5 if you
notice here, 9.5 is the rolling resistance. When added to the 4%
gave a total of 13 and a half. When you subtract the six, it
gives three and a half. So nine and a half plus 413, and a half,
nine and a half minus six. That's the three and a half
when descending.
This is basically an introduction about how to calculate the slopes,
how to calculate the resistances, how to calculate the speeds and
how to calculate the cycle, the cycle time. We have posted online
a more comprehensive example that shows different segments of the
road uphill, downhill, and it follows the same procedure to
calculate the different segments into different steps, and then we
combine that at the end to get the total cycle time, and therefore
the total production of that equipment. I'll see you in the
next lecture so.