# Pull Force

Pull Force is the force that a truck or prime mover can exert onto a transporter or any type of trailer. It has been a subject of much controversy and misunderstandings as truck manufacturers and end-users often do not talk the same language. Marco J. van Daal explains how to arrive from engine power to the actual theoretical pull force of the truck.

A truck salesman talks about horsepower (HP), kilowatt (kW), or torque (lb-ft or Nm) when the end-user is often just interested in how many tons or pounds the unit can pull. In order to go from HP to pull force, many terms and conversion factors are thrown on the table such as number of driven axles, gear box ratio, rear end ratio, differential, tire size, truck weight and fifth wheel capacity, just to name a few. It gets even worse and more complicated if the buyer wants a custom truck with, for example, an auxiliary gearbox (also called a transfer case) to achieve even lower speeds and higher torque, or if he wants to have one of the rear axles retractable.

Let’s start with a thumb rule. A truck can deliver a pull force equal to some 80 to 90% of the weight carried by the drive axles. The truck in figure 1 weighs 35 tons. The rear 2 axles are drive axles and carry 10 tons each; and the third rear axle and the steering axle carry 7.5 tons each. The maximum pull force that this truck can generate is between 16 tons (35,000 LBS) and 18 tons (40,000 LBS).

Figure 1:

However, this is only if the correct gear is chosen and there is sufficient engine power (HP of kW) available and this power can actually be transferred to the road surface. Once it is known that the above conditions are complied with, this thumb rule is pretty accurate for practical purposes.

What is more important?
The power and torque of a truck is delivered by the engine and is expressed in HP / kW or lb-ft / Nm. A power of 1 HP equates to about 745 W and a torque of 1 lb-ft equates to about 1.36 Nm (Newton-meter). A 400-HP engine with a maximum torque of 1,500 lb-ft can deliver 298 kW and about 2,036 Nm.

To understand the difference between power and torque, it is important to understand that power (HP and kW) is a measure of the rate at which work is delivered; it is measured per unit of time. It may help to know that the definition of a HP is based on the (empiric) assumption that a horse can move 33,000 LBS 1 foot per minute, therefore 1 HP = 33,000 lb-ft/min = 550 lb-ft/sec.

Torque (lb-ft and Nm) is the measure of an object’s tendency to rotate about a point, a twisting force. It is not measured per unit of time but (in the case of engines) per crankshaft revolution.

In terms of rating an engine, it is the application that determines whether power or torque is the more important measurement. When speed or a certain duty cycle is required, the engine power is of importance; as power, like speed, is measured per unit of time.

When pull force is important, such as with trucks and most off-road and earth moving equipment, the engine torque is of importance as this determines if the truck will be able to move the object/trailer. It is a matter of wanting to know “how fast” versus “how much”.

How does a truck work?
The power and torque are delivered to the engine shaft, which is connected to the crank shaft that is rotated by the engine pistons. The speed of this shaft is determined by the engine RPM, which in turn is influenced by the throttle.

The engine power and torque are not required all the time; you may be idling at a traffic light, but you do not want to shut off the engine. Immediately behind the engine shaft is the clutch, which can be engaged or disengaged. When the clutch is disengaged (you press down the clutch pedal), the engine runs but no power or torque is delivered to any mechanical part of the truck. When the clutch is engaged (you slowly release the clutch pedal), the engine power and torque now set into motion an array of gears. See figure 2.

Figure 2:

The first set of gear can be found in the gearbox. Figure 3 shows an open gearbox. The protruding shaft connects to the engine shaft. The large open disk-like shape is the “bellhousing” that bolts to the engine block and covers the flywheel. The flywheel is a large steel disk or wheel that rotates inside the bellhousing. The function of the flywheel is to store kinetic energy. The speed of the flywheel is not easily changed because of its weight and momentum. Because of this, the flywheel helps to keep the shaft rotating at the same speed. This comes in handy as piston engines usually have uneven torque per piston and per firing. The flywheel fixes this problem.

The gears in the gearbox are of different diameters - each set with its own gear ratio. When you open a gearbox, you can calculate the gear ratio from the number of teeth on the gears, but that is a cumbersome way. The manufacturer can supply you with the gear ratio of each gear. The first gear is most important as this is the gear that will set the truck and transporter in motion. The ratio must be large enough to ensure a slow starting speed. Ratios of 16 or 18 or 20 are not unheard of. For comparison, your average car has a first gear ratio of 3.5 or 4.0.

Figure 3 & 4:

As the output of the gearbox is connected to the drive shaft by a U-joint, in practical terms, a gear ratio of 20 means that it takes 20 crank shaft revolutions to make one drive shaft revolution.

So now we have the drive shaft rotating. The drive shaft is connected by a U-joint to the differential. The differential has two functions:

1. It allows the rear wheels to rotate at different speeds in turns and curves. The outside rear wheel needs to rotate faster than the inside rear wheel, the differential allows for this. See Figure 4.

2. The gears in the differential are designed with a certain ratio, similar to the gearbox but with the difference that the differential has one set (designed) gear ratio. This differential gear ratio is called the “rear end ratio” or “final drive ratio.” A differential gear ratio of 10 means that it takes 10 drive shaft revolutions to make one drive axle shaft revolution.

The differential also has a disadvantage. Since it allows differential speeds between the two wheels, in case one wheel loses traction and starts spinning, the entire power and torque goes into that wheel as this is the way of the least resistance. Many trucks nowadays have “differential locks” (or diff-lock) built in. Once the diff-lock is engaged, both the wheels rotate at the same speed no matter if one axle has lost traction. Note: diff-lock can only be used on straight roads as it completely eliminates the possibility of speed differential between left and right axle in turns.

If we now look at the total gear reduction between the gearbox (reduction 20) and the differential (reduction 10) we can conclude that it takes 20 crankshaft revolutions to make one drive shaft revolution and 10 drive shaft revolutions to make one rear axle shaft revolution. Therefore, it takes 10 * 20 = 200 crankshaft revolutions to make 1 rear axle shaft revolution.

At the end of each drive axle shaft we find the wheels. The diameter of the wheels or tires has an influence on the velocity of the vehicle. Larger diameter tires gain speed much quicker than smaller diameter tires. In addition, larger diameter tires generally have a lower rolling resistance, particularly in off-road condition.

It is important to understand the difference between drive axles, non-driven axles, and steering axles. To determine pull force, we are solely interested in drive axles as these are the axles that convert the engine power into traction on the road surface. In general, it can be stated that the more weight a drive axle carries, the more traction it can develop. Keeping in mind, however, that engine power and axle capacity can be limiting factors.

The truck in Figure 5 has a Gross Vehicle Weight (GVW) of 56 tons. Of the five axles, the rear three axles are drive axles and each carry 12 tons. The front two axles are steering axles; these carry 10 tons each. Since the front two axles are not driven axles, these do not contribute to the development of the pull force.

Figure 5 & 6:

As stated, the thumb rule for determining pull force is that a truck can deliver a pull force equal to about 80-90% of the weight carried by the drive axles. The drive axles of the truck in Figure 17-6 carry a total weight of 3 * 12 ton = 36 tons.

The maximum pull force that this truck can generate is between 28.8 tons (63,436 LBS) and 32.4 tons (71,366 LBS). We arrive at these numbers by performing the following calculations:

Pull force at 80% = 80% * (3 * 12 ton) = 28.8 ton
Pull force at 90% = 90% * (3 * 12 ton) = 32.4 ton
Once the pull force is known, this can be used to determine how much weight can actually be set in motion by this truck. For this, it is necessary to know how much the rolling resistance is of the transport vehicle. The rolling resistance is generally expressed as a percentage of the GVW. On dry asphalt or tarmac the rolling resistance is around 2-3%. This means that the truck in Figure 5, with a GVW of 56 tons, requires between 1.12 tons (2% rolling resistance) and 1.67 tons (3% rolling resistance) of its pull force to set itself in motion.

Rolling resistance is a number that is empirically (by tests and observations) determined. When the surface is hard and solid, the rolling resistance is often in the 2-3% range. However, when performing transports on sand or dirt roads, this percentage is generally higher.

Figure 6 shows a transport in operation; it consists of a truck that is pulling a combination of a single wide 12-axle line transporter in the front and a double wide 6-axle line transporter in the rear. The rear double wide transporter provides the required stability. Both transporters are equipped with turn tables.

Calculation to determine if this truck is strong enough to pull this transport combination:

Overview of the various weights;

Truck 40 ton of which 32 ton on the drive axles
Vessel 466 ton
Turn table 5 ton each
Transporter 3.5 ton per axle line

The GVW is the total weight of the entire transport combination;

GVW = truck + vessel + turn tables + transporter = 40 ton + 466 ton + (5 ton * 2) + (3.5 ton * 24 axles)
= 40 ton + 466 ton + 10 ton + 84 ton
= 600 ton

For the rolling resistance in this example, a conservative 3% is taken. The required pull force is therefore 3% of the GVW. Required pull force = 3% * GVW = 0.03 * 600 ton = 18.0 ton

The truck can generate a pull force between 25.6 ton (80% of 32 ton) and 28.8 ton (90% of 32 ton).

The conclusion of this exercise is that the truck is strong enough to pull this transport combination.

Now let’s assume that in the route to the final destination there is a 2% incline. Is this truck still strong enough to perform this transport? The GVW does not change; it remains at 600 tons. The truck still generates a pull force between 25.6 and 28.8 tons. The difference here lies in the required pull force, which is the sum of the rolling resistance + incline.

Required pull force = (3% + 2%) * GVW = (0.03 + 0.02) * 600 ton = 30 ton.

It can now be concluded that this truck is adequate for pulling this transport on a flat road surface, but it is not strong enough to pull this transport up a 2% incline. A stronger truck will be required.

How do you really find out the thru power (pulling capacity) of a truck so that you do not have any surprises when it needs to deliver this force?
The only real test is to put it to the test. This is not a test recommended by the manufacturers, but it is a test that will tell you the thru power of the truck. Connect the truck via a wire rope sling and a load cell to an immovable object (I have used a large crawler crane and it worked just fine for this application). Let the truck pull in its lowest gear, slowly increasing the RPM. When the engine stalls or dies, you have reached the maximum for this truck. When the drive axles spin, you could add more counterweight and perform the test again, provided that the fifth wheel capacity, axle, tires is sufficient.

Three cautionary notes here:
• Never stand between the truck and the immovable object
• When performing this test, be aware that in the chain of components between the engine and the rear axles, the drive shaft is often the weakest link (the cheapest component and the easiest to replace). The standard drive shaft may not be strong enough for this test and can snap/break so you may need a stronger drive shaft version for heavy pulling
• This test needs to be performed only once in the lifetime of a truck, unless an engine or gearbox is replaced.
About the author: Marco J. van Daal has been in the heavy lift & transport industry since 1993 starting with Mammoet Transport from the Netherlands and later with Fagioli PSC from Italy, both leading authorities in the industry. His 20 years+ experience extends to 5 continents and over 55 countries and has resulted in a best selling book “The Art of Heavy Transport” which is available at www.the-works-int.com.
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