# Ihab Saad – Loading and Hauling resistances, speeds, and cycle times

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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.