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**Plato and Socrates Discuss Torque, Power and Acceleration (Revised 03/25/2007)**

By Thomas Barber

**Plato:**Dude, nice toga. Say, Ive just

been reading up on torque and power. Torque, it seems, is the rotational equivalent of force in straight-line

motion.

**Socrates:**To fully appreciate what that means and get us off on a solid footing,

lets start with a quick look at the familiar equation: F = M x A. This equation tells us that whenever an

objects present velocity is changing, the acceleration is given by the ratio

of force to mass. Manifestly, the

greater an objects mass, the greater the force needed to yield a specified

amount of acceleration. Mass also

determines how much kinetic energy an object contains when it is moving at a

given velocity. In rotational motion,

that familiar equation is replaced by a similar equation: Torque =

Moment-of-Inertia x Angular Acceleration. The moment of inertia determines how much torque is needed to yield a

given amount of angular acceleration, as well as how much kinetic energy an

object contains when spinning at a given angular velocity. When an ice skater in a spin brings his or

her arms in closer to the torso, there is no loss of kinetic energy, and the

observed increase in angular velocity reveals that the moment of inertia has

been made smaller.

**Plato:**To find the torque associated with a straight-line force, you multiply

the force by the length of the lever arm, which is always measured along a line

of direction that is square to the direction of the force. Whenever I tighten a bolt, if I vary the

length of the lever arm, the force that I sense in my hand and arm will change,

yet the torque doesnt change unless the amount of friction in the threads

changes. Whenever I think about

overcoming friction, I think about work and power.

**Socrates:**To find the work associated with any steady force, you multiply the

force by the distance covered. Whereas

work is cumulative over time and distance, power is the measure of how quickly

work is being performed, instantaneously in time. Work and energy are truly the same concept, so the measure of how

quickly work is being performed, is also the measure of how quickly energy is

being spent. If you turn the crank of a

well to lift a 1-lb bucket at a steady rate of 1 ft/min, you are doing work at

the steady rate of 1 ft-lb/min, which is simply the product of the force and

the velocity. If the radius of the

spool is 1 ft, then the torque applied to the crank by the 1-lb bucket, will be

1 lb-ft. In each complete rotation of

the crank, the bucket will move a distance equal to the radius of the spool

multiplied by twice pi. It follows that

the work associated with a specific amount of torque, for one complete

rotation, may be found by multiplying the torque by twice pi. The calculation of power from torque and

rotational speed is similar. The

following chart summarizes:

straight-line motion | rotational motion | |

work | force x distance | torque x number of rotations x 2 x pi |

power | force x velocity | torque x angular velocity x 2 x pi |

The

expression for power in rotational motion reveals that you always multiply by

the same constant value (2 x pi) to calculate power from the product of torque

and angular velocity. Note, though,

that this assumes that angular velocity is measured in complete rotations per

unit of time. You could just as easily

measure angular velocity using a smaller angular distance, such that you would

have to multiply that smaller angular distance by twice pi in order to yield

one complete rotation. If you measured

angular velocity using that smaller angular distance (which is known as a

radian), the expression for power would be simply the product of torque and

angular velocity, i.e., you would not multiply by twice pi to calculate power. Hence, the business of multiplying by twice

pi is equivalent in effect to converting from one unit of measure to another,

and it is correct to say that power is simply equal to the product of torque

and angular velocity.

**Plato:**If the bucket is raised at steady velocity, its kinetic

energy will be steady. Only its

potential energy will be changing, and the power will be simply the static

weight of the bucket multiplied by the steady velocity. It is easy enough to measure instantaneous

velocity when the velocity is steady, but in real-world scenarios, doesnt it

get more complicated?

**Socrates:**A common approach to measuring the power of an engine is to use a

regulated brake to hold the engine steady at the desired speed. You have to measure the force that resists the

pull of the engine on the brake, so that you can deduce the engine torque from

that force and from the lever arm, which you also have to figure out. And, of course, you have to measure the

engine speed. Dynamometers of this sort

are known as brake dynamometers. Conceptually, you could implement a brake dynamometer of sorts by

applying the engine to the task of lifting an elevator car, using a

continuously variable transmission in the coupling. The CVT would allow you to stabilize the speed of both the engine

and the elevator car at any desired engine speed. To deduce power, you would multiply the elevators steady

velocity by its weight, and as with brake dynamometers in general, those

measurements would be unaffected by the engines inertial moment. The other common approach is to hitch the

engine to a massive drum that spins freely. As long as the increase in the kinetic energy of the drum is the only

energy sink, the power will be given by the instantaneous rate of increase of

the drums kinetic energy, which can be deduced from the drums moment of

inertia and its instantaneous angular acceleration. Dynamometers of this type are known as inertial

dynamometers. The angular acceleration

can be measured with the help of an accelerometer, or deduced from closely

spaced measurements of time and angular distance. The drums moment of inertia can be measured separately, or

calculated from its dimensions and the density of its substance. The increase in the kinetic energy of the

engine itself is an energy sink. The

measurements are influenced to a degree by the engines moment of inertia, and

they reveal, to a degree, the ability of the engine to quickly increase its

work output. As such, measurements

taken on an inertial dynamometer give a more realistic picture of an engines

actual performance on the road. For

purposes of ordinary performance tuning on a test bench, that sort of accuracy

isnt particularly beneficial, whereas the ability to keep the engine running

steadily for extended periods can be beneficial.

**Plato:**I read somewhere that to measure power, you measure torque and then you

deduce power from torque. That

supposedly demonstrates that power is just an abstraction of torque.

**Socrates:**Clearly, there are various ways to measure

power independently of torque. Moreover, the notion, that power is less real than torque, has no

meaning or interpretation that is capable of being confirmed

experimentally. As far as the orthodoxy

and methodology of empirical science is concerned, notions of that sort are

meaningless.

**Plato:**I should be able to measure the power of my mare, by

measuring how quickly she is able to lift a large bucket of water from my

well. If I adjust the amount of water

such that the velocity is steady, the actual force will be equal to the weight

of the bucket. That way, I wont have

to measure the actual strain in the rope, and of course, it will be easier to

measure the velocity.

**Socrates:**In the future, a fellow by the name of James Watt will determine that

his horse is able to perform work at an ongoing, instantaneous rate of 33,000

foot-pounds of work per minute. If you

measure torque in lb-ft and rotational speed in rpm, and you want to express

the power in hp, you can use the conversion factor: 1 hp = 33,000 ft-lb/min. The value of twice pi is 6.283, and that

divided by 33,000 is about 1/5252. So,

as long as torque is measured in lb-ft, rotational speed is measured in rpm,

and you want to express the power in hp, you can take a short cut and divide

the product of torque and rotational speed by 5252.

**Plato:**Does that mean that torque and power are equivalent at 5252 rpm?

**Socrates:**Nope. Torque and power are

distinct properties, with each being analytically related to acceleration in

its own special way. The value 5252 is

merely an artifact of the English system of measure, and that value is not the

least bit special if another system of measure is used. In most of the world, torque is expressed in

Newton-meters (N-m), and power is expressed in Watts or kilowatts (kW), which

we use for electrical power. The engine

speed where torque in lb-ft and power in hp coincidentally take on the same

numerical value, happens to fall within the operating range of most engines, so

on dynamometer plots, it is convenient to use a single number scale for both

torque in lb-ft and power in hp. When

that is done, the two curves will cross at 5252 rpm.

**Plato:**But, it seems that torque should determine acceleration, so why does

power matter?

**Socrates:**Power matters because at any point in time, acceleration is proportional

to the rate at which the engine is performing work. Engine torque tells you how much work is performed over any

specific interval of crankshaft rotation, but does not tell you how quickly the

work is being performed. It is of

course possible to deduce acceleration from the engine torque using other

information such as the overall gear ratio and wheel diameter, but that doesnt

change the pertinent and useful fact that at any point in time, acceleration is

proportional to power. Recall that

power is equal to the product of force and velocity. If you turn that around, it says that force is equal to power

divided by the (non-zero) velocity. If

you substitute that expression for force into the familiar equation that

relates force, mass, and acceleration, you get this:

*acceleration = power / (mass x velocity) =>*

*acceleration = engine_torque x 2 x pi x engine_speed / (velocity x mass)*

Hence,

*given*

the vehicular velocity that is applicable to some point in time, the

the vehicular velocity that is applicable to some point in time

acceleration that you get, for a given amount of engine torque and a given

mass, depends on the engine speed. Of

course, if the ratio of engine speed to vehicle speed is given, as it

effectively is while the gear ratio is held constant, acceleration will then

vary according to the engine torque. (Note that if you plug a set of values into that equation to calculate

acceleration, in order to get proper units of measure for acceleration, you need

to use lbf instead of lb for the force component of the torque. 1 lbf is the force of gravity on 1 lb of

mass: 1 lbf = 1 lb x 32.2 ft/s^2 = 32.2 ft-lb/s^2 = 4.45 N.)

**Plato:**But, if acceleration is proportional to power, why does acceleration

track with the engine torque curve as the engine speed and the vehicle speed

increase in a given gear?

**Socrates:**The perception of a contradiction, between the fact that wheel torque

tracks with the engine torque while the gear ratio remains fixed, and the fact

that acceleration is proportional to power, is a false perception. The equations reveal that the

proportionality between acceleration and power is different at different

vehicle speeds. The acceleration that you

get for a given amount of power decreases as the vehicle speed increases, yet,

at any point in time, acceleration is proportional to power, and depends as

much on engine speed as on engine torque.

**Plato:**What does this mean from the perspective of gear selection strategy?

**Socrates:**Whenever you change gears, as long as you are quick to avoid any

significant loss in vehicle speed during the up-shift, the proportionality

between power and acceleration will be steady across the up-shift. Hence, in order for acceleration to be

steady across the up-shift, power must be steady across the up-shift, which

means that the engine torque must increase to compensate for the drop in engine

speed. If the throttle is held open so

that actual power follows the engines power curve, the engine speed must transition

between two equal-power points on opposite sides of the power peak. Note that shifting such that power will be

steady across the up-shift, and shifting such that you are always using the

gear that yields the greatest power, are two different ways to describe the

same optimal strategy.

**Plato:**What would happen if the engine torque were to be held steady

across the up-shift, i.e., you kept the engine speed within the flat region of

the engine torque curve?

**Socrates:**The acceleration would drop abruptly at the up-shift, matching the drop

in engine speed. Lets look at it

another way, and lets take a quick side trip that may help to put the

significance of power into better perspective. In an electrical transformer, any increase in voltage between the

primary and the secondary windings, must be accompanied by a compensating

decrease in current. Power is equal to

the product of voltage and current, and as the saying goes, power in is power

out. That saying applies as well to

the physics of mechanical motion. Except for the energy losses due to friction, the product of torque and

rotational speed will be the same at the wheel as it is at the engine, and as

it is anywhere else that you measure it along the drive train. You want the wheel torque to be steady

across the up-shift, and since the wheel speed will also be steady at the

up-shift, the product of torque and rotational speed will be steady at the

up-shift, not only at the wheel, but at the engine as well. That, of course, means that the engine

torque must increase to compensate for the drop in engine speed.

**Plato:**Okay, but given two vehicles that are identical except for the

engines, the one with the greater peak engine torque will still exhibit greater

peak acceleration in each gear, right?

**Socrates:**If the vehicle with greater peak power is allowed to use a different

final drive ratio, then by shifting its engine torque peak to lower vehicle

speed, the corresponding wheel torque will increase. Thus, the vehicle with greater peak power may exhibit greater

peak acceleration in each gear, even if its peak engine torque is less than

that of the other vehicle.

**Plato:**Well, there are still certain benefits to emphasizing torque in lieu of

power, arent there?

**Socrates:**Certain effects, such as improved acceleration from a full stop and less

frequent shifting, are the result of a comparatively flat, uniform spread of

engine output, starting at comparatively low engine speed. It makes perfect sense to attribute such

effects to a de-emphasis on peak power. However, logically speaking, torque and power are not opposites, and it

does not follow from the fact that you have de-emphasized peak power, that you

have emphasized torque. Of course, if

there exists some other justification for the practice of equating the engines

low-speed performance to torque, that will also constitute justification for

equating a de-emphasis on peak power to an emphasis on torque, never mind that

torque and power are not opposites. At

the wheel, the affinity between low rotational speed and torque is quite

genuine, owing to the fact that the transmission is used to exchange rotational

speed for torque. But this effect does

not apply to the engine. The practice,

of equating engine performance at low and moderate engine speed exclusively to

torque, seems to derive essentially from the fact that the peak engine torque

occurs at a lower engine speed than does the peak power. This seems a weak justification when you

consider that the peak engine torque reveals the engine performance accurately

at only a single engine speed. That

engine speed is often above the midpoint of the engines operating range, and

no matter how low the actual engine speed, the actual performance depends

partly

*on*the engine speed, and is fully revealed by the actual power.

**Plato:**What else?

**Socrates:**Many people seem to believe that the full explanation, for why longer

stroke generally means improved low-end performance, is simply that by

increasing the effective lever arm (the crank throw distance is one-half of the

stroke distance), you increase the torque. For whatever reason, they dont realize that if it were that simple, the

improvement in engine torque would be uniform over the operating range, which

would not explain why the performance improvement is specific to low engine

speed. They have somehow gotten the

idea that any change, that directly improves engine torque, will automatically

favor lower engine speed. Clearly, it

isnt that simple. If you increase the

stroke while keeping the volume displacement constant, the piston surface area

will decrease, which will nullify the effect of the increased lever arm, since

the force depends on the surface area of the piston face. Engine torque corresponds to the amount of

energy spent over any specific interval of crankshaft rotation, and that amount

of energy depends on the amount of oxygen used. It follows that the variation in engine torque with engine speed

reveals the variation in the amount of air captured per individual intake

stroke. Cylinder shape interacts with

the duration of the intake stroke to influence the amount of air that is

captured on the intake stroke. When the

cylinder is made long and skinny, the effect is to increase the amount of air

captured at low engine speed, and to decrease the amount of air captured at

high engine speed. Note also that if

the relationship between stroke and torque were as direct as the naïve

explanation suggests, you could get free energy just by making the cylinder

long and skinny.

**Plato:**I need to go get measured for a new toga, but before I run along, Id

like to know what you think about the various claims that engine torque is the

true indicator of engine performance.

**Socrates:**Those sorts of claims have to be interpreted to mean that you are always

supposed to get the same acceleration for a given amount of engine torque, no

matter the engine speed at which that much engine torque is delivered. There simply is no other meaningful,

tangible interpretation of those claims. Yet, as we have already seen, wheel torque depends just as much on

engine speed as it does on engine torque. Anyone who is not convinced of that, need only discover for themselves

that at any of the various vehicle speeds where the transmission will permit

you to choose between two equal-torque points on opposite sides of the torque

peak, the acceleration will be dramatically greater in the lower of the two

gears. It is logically dubious to

infer, from the fact that the peak power does a poor job of revealing the

engines performance at low and moderate engine speeds, that torque is the true

indicator of engine performance.