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
>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 @2000 rpm
>2. 302ciV8, five spd man----205 hp@4000rpm----275ft-lbs torque @3000 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.
Q.E.D.
Dan Jones
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.
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