Horsepower vs. Torque, More than you wanted to know... Options

[jgreening]

Saw this on another forum, thought you folks might enjoy it:

Torque Versus Horsepower - More Than You Really Wanted to Know
by Dan Jones

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.

Later,
Dan Jones