07-31-2010, 06:43 AM

Every so often, in the car magazines, you see a question to the technical

editor that reads something like "Should I build my engine for torque or

horsepower?" While the tech editors often respond with sound advice, they

rarely (never?) take the time to define their terms. This only serves to

perpetuate the torque versus horsepower myth. Torque is no more a low rpm

phenomenon than horsepower is a high rpm phenomenon. Both concepts apply

over the entire rpm range, as any decent dyno sheet will show. As a general

service to the list, I have taken it upon myself to explode this myth once

and for all.

To begin, we'll need several boring, but essential, definitions. Work is a

measurement that describes the effect of a force applied on an object over

some distance. If an object is moved one foot by applying a force of one

pound, one foot-pound of work has been performed. Torque is force applied

over a distance (the moment-arm) so as to produce a rotary motion. A one

pound force on a one foot moment-arm produces one foot-pound of torque.

Note that dimensionally (ft-lbs), work and torque are equivalent. Power

measures the rate at which work is performed. Moving a one pound object

over a one foot distance in one second requires one foot-pound per second of

power. One horsepower is arbitrarily defined as 550 foot-pounds per second,

nominally the power output of one horse (e.g. Mr. Ed).

Since, for an engine, horsepower is the rate of producing torque, we can

convert between these two quantities given the engine rate (RPM):

HP = (TQ*2.0*PI*RPM)/33000.0

TQ = (33000.0*HP)/(2.0*PI*RPM)

where:

TQ = torque in ft-lbs

HP = power in horsepower

RPM = engine speed in revolutions per minute

PI = the mathematical constant PI (approximately 3.141592654)

Note: 33000 = conversion factor (550 ft-lbs/sec * 60 sec/min)

In general, the torque and power peaks do not occur simultaneously (i.e.

they

occur at different rpm's).

To answer the question "Is it horsepower or torque that accelerates an

automobile?", we need to review some basic physics, specifically Newton's

laws of motion. Newton's Second Law of Motion states that the sum of the

external forces acting on a body is equal to the rate of change of momentum

of the body. This can be written in equation form as:

F = d/dt(M*V)

where:

F = sum of all the external forces acting on a body

M = the mass of the body

V = the velocity of the body

d/dt = time derivative

For a constant mass system, this reduces to the more familiar equation:

F = M*A

where:

F = sum of all the external forces acting on a body

M = the mass of the body

A = the resultant acceleration of the body due to the sum of the forces

A simple rearrangement yields:

A = F/M

For an accelerating automobile, the acceleration is equal to the sum of the

external forces, divided by the mass of the car. The external forces

include

the motive force applied by the tires against the ground (via Newton's Third

Law of Motion: For every action there is an equal and opposite re-action)

and

the resistive forces of tire friction (rolling resistance) and air drag

(skin

friction and form drag). One interesting fact to observe from this equation

is that a vehicle will continue to accelerate until the sum of the motive

and resistive forces are zero, so the weight of a vehicle has no bearing

whatsoever on its top speed. Weight is only a factor in how quickly

a vehicle will accelerate to its top speed.

In our case, an automobile engine provides the necessary motive force for

acceleration in the form of rotary torque at the crankshaft. Given the

transmission and final drive ratios, the flywheel torque can be translated

to

the axles. Note that not all of the engine torque gets transmitted to the

rear axles. Along the way, some of it gets absorbed (and converted to heat)

by friction, so we need a value for the frictional losses:

ATQ = FWTQ * CEFFGR * TRGR * FDGR - DLOSS

where:

ATQ = axle torque

FWTQ = flywheel (or flexplate) torque

CEFFGR = torque converter effective torque multiplication (=1 for

manual)

TRGR = transmission gear ratio (e.g. 3 for a 3:1 ratio)

FDGR = final drive gear ratio

DLOSS = drivetrain torque losses (due to friction in transmission, rear

end, wheel bearings, torque converter slippage, etc.)

During our previous aerodynamics discussion, one of the list members

mentioned

that aerodynamic drag is the reason cars accelerate slower as speed

increases,

implying that, in a vacuum, a car would continue to rapidly accelerate.

This

is only true for vehicles like rockets. Unlike rockets, cars have finite

rpm

limits and rely upon gearing to provide torque multiplication so gearing

plays

a major role. In first gear, TRGR may have a value of 3.35 but in top gear

it

may be only 0.70. By the above formula, we can see this has a big effect on

the axle torque generated. So, even in a vacuum, a car will accelerate

slower

as speed increases, because you would lose torque multiplication as you went

up through the gears.

The rotary axle torque is converted to a linear motive force by the tires:

LTF = ATQ / TRADIUS

where:

TRADIUS = tire radius (ft)

ATQ = axle torque (ft-lbs)

LTF = linear tire force (lbs)

What this all boils down to is, as far as maximum automobile acceleration is

concerned, all that really matters is the maximum torque imparted to the

ground by the tires (assuming adequate traction). At first glance it might

seem that, given two engines of different torque output, the engine that

produces the greater torque will be the engine that provides the greatest

acceleration. This is incorrect and it's also where horsepower figures into

the discussion. Earlier, I noted that the torque and horsepower peaks of an

engine do not necessarily occur simultaneously. Considering only the torque

peak neglects the potential torque multiplication offered by the

transmission,

final drive ratio, and tire diameter. It's the torque applied by the tires

to

the ground that actually accelerates a car, not the torque generated by the

engine. Horsepower, being the rate at which torque is produced, is an

indicator of how much *potential* torque multiplication is available. In

other words, horsepower describes how much engine rpm can be traded for tire

torque. The word "potential" is important here. If a car is not geared

properly, it will be unable to take full advantage of the engine's

horsepower.

Ideally, a continuously variable transmission which holds rpm at an engine's

horsepower peak, would yield the best possible acceleration. Unfortunately,

most cars are forced to live with finitely spaced fixed gearing. Even

assuming fixed transmission ratios, most cars are not equipped with optimal

final drive gearing, because things like durability, noise, and fuel

consumption take precedence to absolute acceleration.

This explains why large displacement, high torque, low horsepower, engines

are better suited to towing heavy loads than smaller displacement engines.

These engines produce large amounts of torque at low rpm and so can pull a

load at a nice, relaxed, low rpm. A 300 hp, 300 ft-lb, 302 cubic inch

engine

can out-pull a 220 hp, 375 ft-lb, 460 cubic engine, but only if it is geared

accordingly. Even if it was, you'd have to tow with the engine spinning at

high rpm to realize the potential (tire) torque.

As far as the original question ("Should I build my engine for torque or

horsepower?") goes, it should be rephrased to something like "What rpm

range and gear ratio should I build my car to?". Pick an rpm range that

is consistent with your goals and match your components to this rpm range.

So far I've only mentioned peak values which will provide peak instantaneous

acceleration. Generally, we are concerned about the average acceleration

over

some distance. In a drag or road race, the average acceleration between

shifts is most important. This is why gear spacing is important. A peaky

engine (i.e. one that makes its best power over a narrow rpm) needs to be

matched with a gearbox with narrowly spaced ratios to produce its best

acceleration. Some Formula 1 cars (approximately 800 hp from 3 liters,

normally aspirated, 17,000+ rpm) use seven speed gearboxes.

Knowing the basic physics outlined above (and realizing that acceleration

can

be integrated over time to yield velocity, which can then be integrated to

yield position), it would be relatively easy to write a simulation program

which would output time, speed, and acceleration over a given distance. The

inputs required would include a curve of engine torque (or horsepower)

versus

rpm, vehicle weight, transmission gear ratios, final drive ratio, tire

diameter and estimates of rolling resistance and aerodynamic drag. The last

two inputs could be estimated from coast down measurements or taken from

published tests. Optimization loops could be added to minimize elapsed

time,

providing optimal shift points, final drive ratio, and/or gear spacing.

Optimal gearing for top speed could be determined. Appropriate delays for

shifts and loss of traction could be added. Parametrics of the effects of

changes in power, drag, weight, gearing ratios, tire diameter, etc. could be

calculated. If you wanted to get fancy, you could take into account the

effects of the rotating and reciprocating inertia (pistons, flywheels,

driveshafts, tires, etc.). Relativistic effects (mass and length variation

as you approach the speed of light) would be easy to account for, as well,

though I don't drive quite that fast.

Please put this in perspective for me, using this example:

Two almost identical Ford pickups:

1. 300ci six, five spd man---145 hp@3400rpm----265ft-lbs torque rpm

2. 302ciV8, five spd man----205 hp@4000rpm----275ft-lbs torque rpm

Conditions: Both weigh 3500#, both have 3.55 gears, both are pulling a

5000#

boat/trailer. Both are going to the lake north of town via FWY. There is a

very steep grade on the way. They hit the bottom of the grade side by side

at 55mph. What will happen and why? This theoretical situation has

fascinated

>me, so maybe one of the experts can solutionize me forever.

In short, the V8 wins because it has more horsepower to trade for rear wheel

torque, using transmission gear multiplication. What really accelerates a

vehicle is rear wheel torque, which is the product of engine torque and the

gearing provided by the transmission, rear end, and tires. Horsepower is

simply a measure of how much rear wheel torque you can potentially gain from

gearing.

My previous posting provides all the necessary equations to answer this

question, but we need a few more inputs (tire size, transmission gear

ratios,

etc.) and assumptions. I'll fill in the details as we go along. To do this

properly would require a torque (or horsepower) curve versus rpm, but for

illustration purposes, let's just assume the torque curve of the I6 is

greater than that of the V8 up to 2500 rpm, after which the V8 takes over.

Using the horsepower and torque equations, we can fill in a few points.

300 I6 302 V8

RPM Tq Hp Tq Hp

--- ------- --------

4000 269 205

3400 224 145

3000 275 157

2000 265 100

where:

TQ = torque in ft-lbs

HP = power in horsepower

RPM = engine speed in revolutions per

minute

Assume both trucks have 225/60/15 tires (approximately 25.6 inches in

diameter) and transmission ratios of:

Gear Ratio RPM @ 55 MPH

---- ----- ------------

1st 2.95 7554

2nd 1.52 3892

3rd 1.32 3386

4th 1.00 2560

5th 0.70 1792

I determined engine rpm using:

K1 = 0.03937

K2 = 12.*5280./60.

PI = 3.141592654

TD = (K1*WIDTH*AR*2.+WD)

TC = TD*PI

TRPM = K2*MPH/TC

OGR = FDGR*TRGR

ERPM = OGR*TRPM

where:

K1 = conversion factor (millimeters to inches)

K2 = conversion factor (mph to inches)

WIDTH = tire width in millimeters

AR = fractional tire aspect ratio (e.g. 0.6 for a 60 series tire)

WD = wheel diameter in inches

TC = tire circumference in inches

TD = tire diameter in inches

MPH = vehicle speed in mph for which engine rpm is desired

TRGR = transmission gear ratio (e.g. 3 for a 3:1 ratio)

FDGR = final drive gear ratio

OGR = overall gear ratio (transmission gear ratio * final drive ratio)

TRPM = tire RPM

ERPM = engine RPM

In fifth gear, both trucks are at 1792 rpm (55 mph) as they approach the

hill. Running side-by-side, the drivers then floor their accelerators.

Since the I6 makes greater torque below 2500 rpm, it will begin to pull

ahead. The V8 driver, having read my earlier posting, drops all the way down

to second gear, putting his engine near its 4000 rpm power peak.

Responding,the I6 driver drops to third gear which also puts his engine near its power

peak (3400 rpm). The race has begun.

Since the engines are now in different gears, we must figure in the effects

of the gear ratios to determine which vehicle has the greater rear wheel

torque and thus the greater acceleration. We can determine axle torque

from:

ATQ = FWTQ * CEFFGR * TRGR * FDGR - DLOSS

where:

ATQ = axle torque

FWTQ = flywheel (or flexplate) torque

CEFFGR = torque converter effective torque multiplication (=1 for

manual)

TRGR = transmission gear ratio (e.g. 3 for a 3:1 ratio)

FDGR = final drive gear ratio

DLOSS = drivetrain torque losses (due to friction in transmission, rear

end, wheel bearings, torque converter slippage, etc.)

Assuming there are no friction losses, the equation reduces to:

ATQ = FWTQ*TRGR*FDGR

= 269*1.52*3.55 = 1452 ft-lbs for the V8 at 4000 rpm

= 224*1.32*3.55 = 1050 ft-lbs for the I6 at 3400 rpm

Since the V8 now makes considerably more rear axle torque, it will easily

pull away from the I6. Falling behind, the I6 driver might shift down a

gear to take advantage of second gear's greater torque multiplication. He

will still lose the contest because his I6 engine, now operating at close to

4000 rpm, is making less torque than the V8. If he shifts up to a gear that

places his engine at its maximum torque output, he will lose the torque

multiplication of the lower gear ratio and fall even farther behind.

Note that I picked the gear ratios so both engines can operate near their

respective horsepower peaks at 55 mph by shifting to a lower gear (second

gear for the V8 and third gear for the I6). This was necessary to make the

contest equal. I could have manipulated the gear ratios to favor one engine

or the other, but that would not have been a fair comparison. In any case

where both engines are optimally geared, the V8 will win because it simply

has more horsepower to trade for rear wheel torque.

P.S. Since we know the weights and the tire diameter, we can convert this

rotary torque to a linear tire force and, given the angle of the hill,

compute the linear accelerations of the two trucks using F=MA. This

computation is left as an exercise for the reader.

editor that reads something like "Should I build my engine for torque or

horsepower?" While the tech editors often respond with sound advice, they

rarely (never?) take the time to define their terms. This only serves to

perpetuate the torque versus horsepower myth. Torque is no more a low rpm

phenomenon than horsepower is a high rpm phenomenon. Both concepts apply

over the entire rpm range, as any decent dyno sheet will show. As a general

service to the list, I have taken it upon myself to explode this myth once

and for all.

To begin, we'll need several boring, but essential, definitions. Work is a

measurement that describes the effect of a force applied on an object over

some distance. If an object is moved one foot by applying a force of one

pound, one foot-pound of work has been performed. Torque is force applied

over a distance (the moment-arm) so as to produce a rotary motion. A one

pound force on a one foot moment-arm produces one foot-pound of torque.

Note that dimensionally (ft-lbs), work and torque are equivalent. Power

measures the rate at which work is performed. Moving a one pound object

over a one foot distance in one second requires one foot-pound per second of

power. One horsepower is arbitrarily defined as 550 foot-pounds per second,

nominally the power output of one horse (e.g. Mr. Ed).

Since, for an engine, horsepower is the rate of producing torque, we can

convert between these two quantities given the engine rate (RPM):

HP = (TQ*2.0*PI*RPM)/33000.0

TQ = (33000.0*HP)/(2.0*PI*RPM)

where:

TQ = torque in ft-lbs

HP = power in horsepower

RPM = engine speed in revolutions per minute

PI = the mathematical constant PI (approximately 3.141592654)

Note: 33000 = conversion factor (550 ft-lbs/sec * 60 sec/min)

In general, the torque and power peaks do not occur simultaneously (i.e.

they

occur at different rpm's).

To answer the question "Is it horsepower or torque that accelerates an

automobile?", we need to review some basic physics, specifically Newton's

laws of motion. Newton's Second Law of Motion states that the sum of the

external forces acting on a body is equal to the rate of change of momentum

of the body. This can be written in equation form as:

F = d/dt(M*V)

where:

F = sum of all the external forces acting on a body

M = the mass of the body

V = the velocity of the body

d/dt = time derivative

For a constant mass system, this reduces to the more familiar equation:

F = M*A

where:

F = sum of all the external forces acting on a body

M = the mass of the body

A = the resultant acceleration of the body due to the sum of the forces

A simple rearrangement yields:

A = F/M

For an accelerating automobile, the acceleration is equal to the sum of the

external forces, divided by the mass of the car. The external forces

include

the motive force applied by the tires against the ground (via Newton's Third

Law of Motion: For every action there is an equal and opposite re-action)

and

the resistive forces of tire friction (rolling resistance) and air drag

(skin

friction and form drag). One interesting fact to observe from this equation

is that a vehicle will continue to accelerate until the sum of the motive

and resistive forces are zero, so the weight of a vehicle has no bearing

whatsoever on its top speed. Weight is only a factor in how quickly

a vehicle will accelerate to its top speed.

In our case, an automobile engine provides the necessary motive force for

acceleration in the form of rotary torque at the crankshaft. Given the

transmission and final drive ratios, the flywheel torque can be translated

to

the axles. Note that not all of the engine torque gets transmitted to the

rear axles. Along the way, some of it gets absorbed (and converted to heat)

by friction, so we need a value for the frictional losses:

ATQ = FWTQ * CEFFGR * TRGR * FDGR - DLOSS

where:

ATQ = axle torque

FWTQ = flywheel (or flexplate) torque

CEFFGR = torque converter effective torque multiplication (=1 for

manual)

TRGR = transmission gear ratio (e.g. 3 for a 3:1 ratio)

FDGR = final drive gear ratio

DLOSS = drivetrain torque losses (due to friction in transmission, rear

end, wheel bearings, torque converter slippage, etc.)

During our previous aerodynamics discussion, one of the list members

mentioned

that aerodynamic drag is the reason cars accelerate slower as speed

increases,

implying that, in a vacuum, a car would continue to rapidly accelerate.

This

is only true for vehicles like rockets. Unlike rockets, cars have finite

rpm

limits and rely upon gearing to provide torque multiplication so gearing

plays

a major role. In first gear, TRGR may have a value of 3.35 but in top gear

it

may be only 0.70. By the above formula, we can see this has a big effect on

the axle torque generated. So, even in a vacuum, a car will accelerate

slower

as speed increases, because you would lose torque multiplication as you went

up through the gears.

The rotary axle torque is converted to a linear motive force by the tires:

LTF = ATQ / TRADIUS

where:

TRADIUS = tire radius (ft)

ATQ = axle torque (ft-lbs)

LTF = linear tire force (lbs)

What this all boils down to is, as far as maximum automobile acceleration is

concerned, all that really matters is the maximum torque imparted to the

ground by the tires (assuming adequate traction). At first glance it might

seem that, given two engines of different torque output, the engine that

produces the greater torque will be the engine that provides the greatest

acceleration. This is incorrect and it's also where horsepower figures into

the discussion. Earlier, I noted that the torque and horsepower peaks of an

engine do not necessarily occur simultaneously. Considering only the torque

peak neglects the potential torque multiplication offered by the

transmission,

final drive ratio, and tire diameter. It's the torque applied by the tires

to

the ground that actually accelerates a car, not the torque generated by the

engine. Horsepower, being the rate at which torque is produced, is an

indicator of how much *potential* torque multiplication is available. In

other words, horsepower describes how much engine rpm can be traded for tire

torque. The word "potential" is important here. If a car is not geared

properly, it will be unable to take full advantage of the engine's

horsepower.

Ideally, a continuously variable transmission which holds rpm at an engine's

horsepower peak, would yield the best possible acceleration. Unfortunately,

most cars are forced to live with finitely spaced fixed gearing. Even

assuming fixed transmission ratios, most cars are not equipped with optimal

final drive gearing, because things like durability, noise, and fuel

consumption take precedence to absolute acceleration.

This explains why large displacement, high torque, low horsepower, engines

are better suited to towing heavy loads than smaller displacement engines.

These engines produce large amounts of torque at low rpm and so can pull a

load at a nice, relaxed, low rpm. A 300 hp, 300 ft-lb, 302 cubic inch

engine

can out-pull a 220 hp, 375 ft-lb, 460 cubic engine, but only if it is geared

accordingly. Even if it was, you'd have to tow with the engine spinning at

high rpm to realize the potential (tire) torque.

As far as the original question ("Should I build my engine for torque or

horsepower?") goes, it should be rephrased to something like "What rpm

range and gear ratio should I build my car to?". Pick an rpm range that

is consistent with your goals and match your components to this rpm range.

So far I've only mentioned peak values which will provide peak instantaneous

acceleration. Generally, we are concerned about the average acceleration

over

some distance. In a drag or road race, the average acceleration between

shifts is most important. This is why gear spacing is important. A peaky

engine (i.e. one that makes its best power over a narrow rpm) needs to be

matched with a gearbox with narrowly spaced ratios to produce its best

acceleration. Some Formula 1 cars (approximately 800 hp from 3 liters,

normally aspirated, 17,000+ rpm) use seven speed gearboxes.

Knowing the basic physics outlined above (and realizing that acceleration

can

be integrated over time to yield velocity, which can then be integrated to

yield position), it would be relatively easy to write a simulation program

which would output time, speed, and acceleration over a given distance. The

inputs required would include a curve of engine torque (or horsepower)

versus

rpm, vehicle weight, transmission gear ratios, final drive ratio, tire

diameter and estimates of rolling resistance and aerodynamic drag. The last

two inputs could be estimated from coast down measurements or taken from

published tests. Optimization loops could be added to minimize elapsed

time,

providing optimal shift points, final drive ratio, and/or gear spacing.

Optimal gearing for top speed could be determined. Appropriate delays for

shifts and loss of traction could be added. Parametrics of the effects of

changes in power, drag, weight, gearing ratios, tire diameter, etc. could be

calculated. If you wanted to get fancy, you could take into account the

effects of the rotating and reciprocating inertia (pistons, flywheels,

driveshafts, tires, etc.). Relativistic effects (mass and length variation

as you approach the speed of light) would be easy to account for, as well,

though I don't drive quite that fast.

Please put this in perspective for me, using this example:

Two almost identical Ford pickups:

1. 300ci six, five spd man---145 hp@3400rpm----265ft-lbs torque

2. 302ciV8, five spd man----205 hp@4000rpm----275ft-lbs torque

Conditions: Both weigh 3500#, both have 3.55 gears, both are pulling a

5000#

boat/trailer. Both are going to the lake north of town via FWY. There is a

very steep grade on the way. They hit the bottom of the grade side by side

at 55mph. What will happen and why? This theoretical situation has

fascinated

>me, so maybe one of the experts can solutionize me forever.

In short, the V8 wins because it has more horsepower to trade for rear wheel

torque, using transmission gear multiplication. What really accelerates a

vehicle is rear wheel torque, which is the product of engine torque and the

gearing provided by the transmission, rear end, and tires. Horsepower is

simply a measure of how much rear wheel torque you can potentially gain from

gearing.

My previous posting provides all the necessary equations to answer this

question, but we need a few more inputs (tire size, transmission gear

ratios,

etc.) and assumptions. I'll fill in the details as we go along. To do this

properly would require a torque (or horsepower) curve versus rpm, but for

illustration purposes, let's just assume the torque curve of the I6 is

greater than that of the V8 up to 2500 rpm, after which the V8 takes over.

Using the horsepower and torque equations, we can fill in a few points.

300 I6 302 V8

RPM Tq Hp Tq Hp

--- ------- --------

4000 269 205

3400 224 145

3000 275 157

2000 265 100

where:

TQ = torque in ft-lbs

HP = power in horsepower

RPM = engine speed in revolutions per

minute

Assume both trucks have 225/60/15 tires (approximately 25.6 inches in

diameter) and transmission ratios of:

Gear Ratio RPM @ 55 MPH

---- ----- ------------

1st 2.95 7554

2nd 1.52 3892

3rd 1.32 3386

4th 1.00 2560

5th 0.70 1792

I determined engine rpm using:

K1 = 0.03937

K2 = 12.*5280./60.

PI = 3.141592654

TD = (K1*WIDTH*AR*2.+WD)

TC = TD*PI

TRPM = K2*MPH/TC

OGR = FDGR*TRGR

ERPM = OGR*TRPM

where:

K1 = conversion factor (millimeters to inches)

K2 = conversion factor (mph to inches)

WIDTH = tire width in millimeters

AR = fractional tire aspect ratio (e.g. 0.6 for a 60 series tire)

WD = wheel diameter in inches

TC = tire circumference in inches

TD = tire diameter in inches

MPH = vehicle speed in mph for which engine rpm is desired

TRGR = transmission gear ratio (e.g. 3 for a 3:1 ratio)

FDGR = final drive gear ratio

OGR = overall gear ratio (transmission gear ratio * final drive ratio)

TRPM = tire RPM

ERPM = engine RPM

In fifth gear, both trucks are at 1792 rpm (55 mph) as they approach the

hill. Running side-by-side, the drivers then floor their accelerators.

Since the I6 makes greater torque below 2500 rpm, it will begin to pull

ahead. The V8 driver, having read my earlier posting, drops all the way down

to second gear, putting his engine near its 4000 rpm power peak.

Responding,the I6 driver drops to third gear which also puts his engine near its power

peak (3400 rpm). The race has begun.

Since the engines are now in different gears, we must figure in the effects

of the gear ratios to determine which vehicle has the greater rear wheel

torque and thus the greater acceleration. We can determine axle torque

from:

ATQ = FWTQ * CEFFGR * TRGR * FDGR - DLOSS

where:

ATQ = axle torque

FWTQ = flywheel (or flexplate) torque

CEFFGR = torque converter effective torque multiplication (=1 for

manual)

TRGR = transmission gear ratio (e.g. 3 for a 3:1 ratio)

FDGR = final drive gear ratio

DLOSS = drivetrain torque losses (due to friction in transmission, rear

end, wheel bearings, torque converter slippage, etc.)

Assuming there are no friction losses, the equation reduces to:

ATQ = FWTQ*TRGR*FDGR

= 269*1.52*3.55 = 1452 ft-lbs for the V8 at 4000 rpm

= 224*1.32*3.55 = 1050 ft-lbs for the I6 at 3400 rpm

Since the V8 now makes considerably more rear axle torque, it will easily

pull away from the I6. Falling behind, the I6 driver might shift down a

gear to take advantage of second gear's greater torque multiplication. He

will still lose the contest because his I6 engine, now operating at close to

4000 rpm, is making less torque than the V8. If he shifts up to a gear that

places his engine at its maximum torque output, he will lose the torque

multiplication of the lower gear ratio and fall even farther behind.

Note that I picked the gear ratios so both engines can operate near their

respective horsepower peaks at 55 mph by shifting to a lower gear (second

gear for the V8 and third gear for the I6). This was necessary to make the

contest equal. I could have manipulated the gear ratios to favor one engine

or the other, but that would not have been a fair comparison. In any case

where both engines are optimally geared, the V8 will win because it simply

has more horsepower to trade for rear wheel torque.

P.S. Since we know the weights and the tire diameter, we can convert this

rotary torque to a linear tire force and, given the angle of the hill,

compute the linear accelerations of the two trucks using F=MA. This

computation is left as an exercise for the reader.