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.