# Horsepower vs Torque

## Understanding Horsepower and Torque in Diesel and Gas Engines

The differences between horsepower and torque are not nearly as importance as the relationship between the two concepts. Peak horsepower and torque ratings are often used to identify performance characteristics in internal combustion engines. In passenger cars and light vehicles, horsepower seems to be a primary selling point. On the contrary, torque is often the sought after characteristic in diesel powered trucks. In this article, we'll examine the relationship, in addition to the differences between horsepower and torque and how the two engine ratings apply to various situations.

### Horsepower vs Torque

#### What is Torque

Torque is the measure of an objects tendency to rotate about a point. It is a measure of work, or the magnitude of a force acting through a distance. In terms of engine output, it can be described simply as the net twisting force by the engine about the crankshaft. The greater the torque output, the greater the load that the engine is capable of overcoming.

It's important to understand that torque is not a time dependent variable. Imperial units of torque include in-lbs (lb-in) and ft-lbs (lb-ft) while metric units of torque are typically given in N-m (Newton meters). Notice that these units include a force and a distance.

#### What is Horsepower

Horsepower is a measurement of, no surprise, power; the measure of the rate at which work is performed or, more importantly, how much energy is consumed during a process. Horsepower is dependent on time and torque as it is the force generated through a distance per a unit of time. Therefore, power identifies the total work done over a given time interval.

Horsepower is a fictitious concept based on the assumption that a horse can move a 33,000 pound object 1 foot per minute (33,000 lb-ft/min). Horsepower itself is not actually measurable by any instrument; it is calculated by measuring the torque at a particular angular velocity (speed). In the case of engines, this would refer to rotations per minute (RPM) of the crankshaft. By measuring the torque and RPM, horsepower is then calculated using the following formula:

Horsepower = (torque x RPM)/5,252

The constant "5,252" is derived from the fact that a circle 1 foot in diameter has a circumference of 6.2832 feet. Dividing 33,000 by 6.2832 results in 5,252 and resolves the units for the equation. Interestingly, but not coincidently, horsepower and torque curves will always cross paths at exactly 5,252 RPM (horsepower and torque will be equal).

#### Engine Horsepower/Torque vs Drive Wheel Horsepower/Torque

Automakers and engine manufactures always advertise peak rated engine horsepower and torque, whereas a vehicle dynamometer measures actual drive wheel horsepower and torque (often referred to as rear wheel horsepower and rear wheel torque). Engine horsepower/torque will never equal drive wheel horsepower/torque, i.e. that 500 horsepower engine will never put 500 horsepower to the pavement. This is primarily do to parasitic losses from friction, heat, and the mass of rotating objects; energy that is consumed in transmitting power through the drivetrain.

There's also torque multiplication through the drivetrain that must be considered. For example, an engine producing 600 lb-ft of torque at a particular RPM through a 2:1 transmission ratio and 3.73:1 rear differential ratio actually produces, disregarding all parasitic losses, 4,476 lb-ft of torque at the drive wheels (600 x 2 x 3.73). Up-shifting the transmission to a 1:1 ratio reduces the torque multiplication and our vehicle now only experiences 2,238 lb-ft of torque at the drive wheels (600 x 1 x 3.73).

Torque multiplication through the drivetrain is a form of mechanical advantage that allows the engine to overcome greater resistive forces, such as those experienced when trying to bring a relatively heavy load up to speed. An important use of a vehicle dynamometer is to identify engine performance characteristics while taking into consideration drivetrain losses. Thus, correction factors are used in order to negate all torque multiplication through the drivetrain and deliver real-world engine horsepower and torque figures.

#### What Does Torque Actually Tell us About an Engine?

In Newtonian physics, torque is directly related to acceleration. Newton's second law of motion ascertains that the force produced by an object is equal to its mass times its acceleration (F=M*A). Since torque is the measurement of a force through a distance, we can isolate the force and use this equation to calculate the rate of acceleration of a moving vehicle at a given instance (A=F/M). As a result, a great deal of torque is advantageous in applications where rapid acceleration of a vehicle is desirable.

In less technical terms, torque can be thought of as the extent to which an engine can overcome a load. A common example would be an engine's ability to begin to move a trailer at rest and bring it up to speed. The heavier the trailer, the more torque required to move it and bring it up to cruising speed. The relationship between torque and acceleration can be misleading as torque alone does not necessarily define the performance characteristics of an engine in its entirety. A typical torque curve will peak, flat line, then taper off as RPM increases. When torque peaks, how long it remains constant, and when/how rapidly it begins to fall off is just as important as the peak torque figure.

#### What Does Horsepower Actually Tell us About an Engine?

Unlike torque, horsepower builds in a comparatively linear fashion and only begins to drop in the higher RPM ranges (relatively speaking). Low engine speeds produce low horsepower, while higher engine speeds produce higher horsepower. High horsepower engines are typically associated with racing applications and thus many assume that more horsepower denotes a faster vehicle. There may be some truth to this statement, however it is extremely misleading. As previously outlined, torque is directly related to the rate of acceleration of a vehicle. However, one must not forget the relationship between horsepower and torque. Horsepower is the rate at which torque is being produced, and it provides some useful insight into the characteristics (and manners) of an engine, including operating range.

To some extent horsepower can be used to compensate for an engine's relatively low torque output. Likewise, torque can be used to compensate for an engine's relatively low horsepower rating. If this seems confusing, recall that an engine is only one piece of an otherwise complex drivetrain system and the mechanical advantage provided by gear reduction in the transmission and differential can also compensate for both low engine torque and horsepower ratings.

#### Practical Applications (with Math!)

If you're still over-complicating this whole horsepower vs torque dilemma, follow along with the practical applications below. We'll break the mathematics down into nitty-gritty details so you won't have to. This may help you understand how horsepower and torque ratings can be applied in different scenarios.

##### Example 1 - Moving a heavy object from rest

How much drive wheel torque is required to move from rest a 10,000 object if the friction coefficient between the object and the road is 0.90 and the vehicle has 30 inch tires? (neglect the weight of the vehicle).

Recalling that ∑F=MA (F = force, M = mass, A = acceleration)...

∑F=MA
A = 0 since the object is at rest, thus:

∑Fx = 0 -----> Fdrive = Ffr
Friction force Ffr= μ*Fn (Fn = normal force, μ = coefficient of friction)
μ = 0.90
Fn = 10,000

Ffr = μ*Fn = (0.90)*(10,000 lbs) = 9,000 lbs

A drive force of 9,000 lbs is required to overcome the friction between the object and the road. Knowing that drive force = drive wheel torque/tire radius:

Tdrive = Fdrive*tire radius = (9,000 lbs)*(15 inches) = (9,000 lbs)*(1.25 ft) = 11,250 ft-lbs

11,250 lb-ft of torque will be required at the drive wheels in order to move the object from rest. To better put this in perspective, lets imagine the transmission ratio is 6:1 and the rear axle ratio is 3.73:1. The total gear reduction is therefore 6*3.73 = 22.38. Under these conditions, the required engine torque (assuming no torque converter or clutch slippage) would be:

Tdrive = Teng*total gear reduction -----> Teng = Tdrive/total gear reduction = (11,250 ft-lbs)/(22.38) = 503 ft-lbs

It's important to note that more torque is required to move an object from rest than to keep it in motion, i.e. once the object starts moving it will have a lower coefficient of friction (static coefficient of friction > dynamic coefficient of friction). Also, the object in the example above is NOT a trailer, which would 1) require less torque to move and 2) have a more complex set of calculations. However, this example provides insight into the importance of vehicle torque; we could not necessarily apply the same logic to horsepower.

##### Example 2 - Maintaining speed up an incline

How much horsepower is required to maintain a speed of 55 mph up a 10° incline assuming the following information:

• 5,000 lb vehicle
• 5% loss do to friction between tire and road
• 10% loss do to wind resistance (drag force)

∑F=MA
A = 0 since the vehicle is at moving at constant velocity, thus:

∑F = 0 -----> ∑Fx = 0

Fx - (5000 lbs)*sin(10°)
Fx = 868.24 lbs

The vehicle experiences a total loss of 15% do to wind resistance and friction, so we'll multiply 868.24 lbs by 1.15 in our power equation below.

Power is equal to force x velocity (P = F*V), thus:

P = (1.15)*(868.24 lbs)*(55 mph)
Convert miles per hour to ft/sec:

55 mi/hr * 5,280 ft/mi * 1 hr/60 min * 1 min/60 sec = 80.66 ft/sec

P = (1.15)*(868.24 lbs)*(80.66 ft/sec) = 80,537 lb-ft/sec
Convert to horsepower (1 hp = 550 lb-ft sec)

P = 146 horsepower

To maintain a constant speed of 55 mph up a 10° incline the engine must produce 146 horsepower (rear wheel horsepower).

##### Example 3 - Minimum input power

What is the minimum input power required to operate a 14,000 KW generator at full load assuming 10% parasitic losses?

Pout = Pin - Ploss
The power out "Pout" is equal to the power input "Pin " minus the total parasitic loss Ploss, thus substituting 0.10Pin for Ploss yields:

14,000 kW = (0.90)*(Pin)
Convert kilowatts to horsepower (1 hp = 746 kW)

18.77 hp = (0.90)*(Pin)
Isolating and solving for Pin:

18.77 hp/0.90 = Pin

Pin = 20.86 hp

A minimum 20.86 engine horsepower will be required to operate a generator at a peak 14,000 kW output with a 10% power loss do to heat, friction, and resistance.

#### Horsepower and Torque in Diesel Engines

The horsepower vs torque debate surfaces often with regard to diesel engines, which produce a relatively large amount of torque but generally exhibit low horsepower ratings (350 horsepower/750 lb-ft of torque, for example). Recalling that horsepower is time dependent and torque is not, a diesel engine's ratings are primarily limited by engine speed. A typical light duty diesel has a maximum engine speed in the 3,500 rpm range while a typical heavy duty diesel has a maximum engine speed in the 1,800 to 2,200 rpm range. If we were to recalculate the horsepower of these engines at a much higher engine speed, power would be proportionately greater.

Diesel engines are speed limited for several reasons. Torque drops off sharply at higher engine speeds do to frictional losses and the fact that diesel fuel burns at a relatively slow rate. Thus, the combustion process becomes extremely inefficient at high engine speeds as the speed of each power stroke theoretically "out-paces" the rate of combustion (piston returns to BDC without ample time for all energy to be extracted). Furthermore, there is the concern that the high compression ratio and long stroke length of a diesel engine may cause excessive wear at high engine speeds. Diesel engines are therefore not well suited for high rpm applications.

#### Horsepower and Torque in Gasoline Engines

Gasoline engines may exhibit horsepower ratings that are lower, equal to, or greater than the respective peak torque rating. As previously discussed with regard to low horsepower in diesel engines, this is indicative of operating range. It is generally acceptable to assume that a high horsepower, low torque (600 hp/400 lb-ft, for example) rated engine has a high maximum engine speed while an engine that has a similar horsepower and torque rating (500 hp/500 lb-ft, for example) has an average maximum engine speed (in the 5,250 rpm range; recall that horsepower and torque always equal each other at 5,252 rpm). An engine that was limited to a much lower engine speed (such as that found in diesel engines) would have a significantly higher torque rating than horsepower (300 hp/400 lb-ft, for example).

Torque is no greater nor no less important in gasoline engines than in diesel engines, however we typically seem to rank gasoline engines by their horsepower ratings as it provides insight into certain performance characteristics, such as the engine's redline. Higher engine speeds are often desirable in high performance applications because shifting at high rpm allows an engine to hold a lower transmission gear longer, thus [theoretically] producing more drive wheel torque for longer periods of time (recall that torque is multiplied through the transmission and rear axle gear ratios, so with each transmission upshift drive wheel torque is reduced).

### Summary - Horsepower and Torque, NOT Horsepower vs Torque

If you take nothing else from this article, it should be that horsepower and torque are closely related, albeit separate performance metrics. Neither is necessarily more important than the other until you attempt to apply them to real-world logic, at which point horsepower, torque, or the combination of the two are more favorable under the circumstances. Torque is the measurement of a twisting force and horsepower is nothing more than the rate at which torque is produced. Don't get hung up on comparing the two, but focus on the big picture - the relationship between horsepower and torque.