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11072012, 08:26 PM #1Member
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Why horsepower/torque "changes", dynos "read low", peak torque is not when you accelerate hardest, and gearing matters.
To me  as an engineer, it is very frustrating when this information is taken lightly. We all enjoy our cars  and a good understanding of the physics involved helps everyone understand a lot. This is BY FAR the best article I have ever read on the subject. The gearing subject that recently came up is found in this article. It explains why it matters A LOT when getting your car dynoed what gear the car is in  and why the measured "engine torque" (difference between this and wheel) on a dyno does not change from gear to gear. Another thing that I have read is that people "short shift" their cars  specifically because there is more torque lower in the RPM range after the gear change. This article has a perfect example of why this is in fact almost never true.
Anyways, this isn't my article  but thought some people might enjoy the read. It's not theory  it's physics.
Hope you guys like it. I know it's long  but again, it's the best I have seen on this subject, and the physics involved in measuring power, gearing, acceleration, etc...
Cheers.

Horsepower and Torque
This page describes horsepower and torque and how they relate to each other. (Not all of it is my words, but I can't even remember whose work I plagiarised now. Sorry!)

Horsepower and Torque

Everyone knows race engines make big horsepower and torque, but what do those terms mean?
It doesn't take an engineer to know these engines are powerful. To measure the power in a scientific way, the engineers use the dynamometer to quantify the two measurements that rate an engine's ability. The dyno gives the engine shop the critical info needed to gauge the ability of their engines in the form of torque and horsepower.
First, let's look at torque. Torque is a "turning or twisting force." In an internal combustion engine, torque means the amount of combustion pressure the engine creates. This force turns the rear wheel and propels the bike forward. Torque is measured in foot/pounds or kilogram/metres or newton/metres. A simplified way of looking at torque is to say it is the amount of force from the engine that turns the rear wheel in a turning motion.
After a dyno run is completed, the resulting readout shows the torque available throughout the rev range. This is called a torque curve. The Holy Grail for an engine engineer is a perfectly flat torque curve. If it existed, it would give useable and predictable power throughout the rev range. But port efficiency, timing, carburetion, and exhaust limitations keep the perfectly flat torque curve from existing. The engineers use the intake, exhaust, timing and carburetion systems to eliminate dips in the torque curve.
Since the torque curve isn't a straight line, the point at which peak torque occurs is important. A Gold Wing or Harley cruiser will have its peak lower in the rev range. This gives a feel of good acceleration or useable power in the lower rpm registers, and make the machine easy to ride. A stout Superbike will have its peak torque somewhere near the beginning of the last third of the rev range.
But torque is only half the story. While torque is the force created, it doesn't account for the importance of revs. Imagine trying to remove a wheel from a car with an inadequate tyre tool and all the torque you could produce couldn't loosen the rusty lug nuts on the studs. While you applied lots of force, i.e. torque, you couldn't generate any rpm. Therefore nothing was accomplished, despite your cursing, crying and kicking.

Without rpm, torque is useless.
Two engines may make 75 foot/pounds of torque, but if one is turning at 5,000 rpm and another is turning 10,000 rpm, the latter is doing more work than the former. Remember, torque measures force, but it doesn't measure actual power produced.
To measure that total power output, we have horsepower. Horsepower is torque times rpm divided by 5252 (Torque x RPM / 5252). Through this formula, we can calculate torque and horsepower and see that both are linked.
In the same example above, the engine running at 5,000 rpm and producing 75 feet/pounds of torque is making 71.4 horsepower. The engine turning at 10,000 rpm makes 142.8 horsepower. The force (torque) is the same, but since the latter engine is turning twice as fast, it makes twice the horsepower.
Which is more important, revs or torque? Neither, really. In the real world, high performance engines need both.
Dremeltype tools are handy devices. Say one is advertised to turn in the neighborhood of 22,000 rpm, a pretty astounding figure. However, it has little torque, so for some jobs you'll need a drill that turns at a fraction of that speed, but has more turning force. On the other hand, the drill is illsuited for other tasks like polishing because it can't rev fast enough.
The dyno gives engineers a horsepower curve in addition to the torque curve. They will always be different shapes, and peak rpm will always occur later in the rev range than peak torque. Although the torque curve will decline slightly after peaking, the revs are still increasing, and thus the horsepower curve increases as well. This difference in the curves is very important to how the engines perform on the track.
If we look at both curves, we can see where the power band lies. The rider wants to keep the engine revving between peak torque and peak rpm. The gear ratios will be selected to keep the bike in this "happy zone" when it accelerates.
Sometimes the engineers succumb to a disease called "dyno blindness". To get the numbers, they will change the timing, port size and shape, exhaust and/or carburation to get a higher horsepower, and in turn narrow the power band, or create dips in the horsepower or torque curves. This can a big mistake. Sometimes more isn't necessarily better.
Tuners should first develop a good torque curve and horsepower, then increase the acceleration rates. The acceleration rate is how fast the engine can run through the rev range.
To beat the best on a consistent basis, you need all three elements, torque, horsepower, and acceleration. Then all three combine to make a useable and effective power band. There are no substitutes.

The following is a great read that can be found on the Micapeak FJR1300 site here. A a site for FJR1300 enthusiasts. The document is reproduced here in full with the kind permission of the Author Tom Barber.
Torque and Power
Home
Five Popular Urban Myths about Torque and Power Debunked
by Tom Barber
Anyone who spends much time on motorcycle and car related Web forums knows that there is a fair amount of debate on whether the torque or the power of an engine is the true indicator of the vehicle's maximum acceleration. Numerous Web sites offer genuine information on the subject, but there are also a few that offer misinformation and that propagate a few popular myths. This article endeavors to clear up some of the misunderstandings and debunk a few of the myths. Along the way, this article will give a usefully complete theoretical treatment of torque, work, and power, and will offer insight into the metrics of torque, work and power. It will present a theoretical perspective of how dynamometers work and will explain how the two basic types differ. It will even explain precisely how torque and power each relate to acceleration.
Myth #1: Dynamometers only actually measure torque. Power is an abstract quantity that can't be measured, and must be calculated from the torque.
This belief is false and it will not be hard to show that. However, it involves a discussion of how dynamometers actually work, and before we can go there, we have to begin with a brief review of some fundamental physics. I'll try to avoid equations as much as possible, and instead I'll try to instill an intuitive understanding of the basic concepts, which is something that classroom lectures and textbooks often fail to accomplish for many people.
Isaac Newton taught us that any object having a mass M, has an intrinsic resistance to changes in its present momentum when external force is applied. The acceleration of the object is proportional to the strength of the external force, and is inversely proportional to the mass of the object. These observations are summed up succinctly by the infamous equation: F = M x A. The true essence of force, mass and acceleration is embodied in this equation, and this equation tells us how force or mass can be detected and measured. When an object's momentum is not constant, this indicates the existence of an external force, and the strength and direction of that force can be determined by investigating the acceleration. Conversely, if an object is accelerated under the influence of a wellknown force, its mass can be determined.
Whenever the aggregate force acting on an object is not directed through the object's center of mass, the object is caused to rotate, and this rotation will continue to accelerate as long as this condition persists. The laws governing rotational motion parallel the laws governing straightline motion. Torque replaces force, the moment of inertia replaces the mass, and the angular acceleration replaces straightline acceleration. Whereas velocity is measured in terms of absolute distance in straightline motion, in angular motion we measure velocity in terms of degrees of rotation per unit of time, which we may call angular distance. Angular velocity is the instantaneous rate at which angular distance is covered. Angular acceleration is the instantaneous rate at which the angular velocity is changing, and is proportional to the applied torque, and inversely proportional to the moment of inertia. The torque that is associated with an applied force is equal to the force (magnitude) multiplied by the distance from the center of mass of the object to the line of force (measured at right angles to the line of force). For an object that is constrained to rotate about a fixed axis of rotation, i.e., a wheel, the distance is measured from the axis of rotation to the line of force. Two objects with identical mass will not in general have the same moment of inertia. For example, if two perfect spheres of identical mass are made of different materials having different densities such that they are not the same size, the larger one will have the greater moment of inertia. In the case of objects having fixed axis of rotation, if two such objects have the same overall diameter and density but differ in the radial distribution of the mass, the one with the mass concentrated closer to the axis of rotation will have a lower moment of inertia.
Work, in the case of linear motion, is the force multiplied by the distance through which the force was exerted and the object was moved. Work and energy are equivalent; it takes a specific quantity of energy to move an object a given distance in opposition to a given resistive force, and that amount of energy is the same no matter how quickly the movement is performed. Power is simply the instantaneous rate at which work is performed (or energy is expended), and is therefore equal to the product of the force and the velocity. If you measure the force required to move an object, the distance that it moves, and the time interval required to move it that distance, then you can calculate the average power that was applied throughout that motion and over that interval of time. Note, though, that average power is no more equivalent to power than average speed is equivalent to speed.
In the case of rotational motion, work is the torque multiplied by the angular distance through which the object is rotated. The parallel with linear motion is again applicable: it takes a specific quantity of energy to rotate an object a given angular distance, and that quantity of energy is the same no matter how quickly the rotation is performed. Each time an object undergoes one complete rotation, a fixed amount of work will have been performed. It follows that the amount of work performed in a given interval of time will depend on the angular distance covered in that time interval, and the rate at which the work is performed will therefore depend on the angular velocity. In rotational motion, power is equal to the product of the torque and the angular velocity, which may be measured in revolutions per minute or rpm.
Let's reiterate the important rules discussed thus far:
Straightline motion:
•Force = Mass x Acceleration
•Work = Force x Distance
•Power = Force x Velocity
Rotational motion:
Torque is measured by taking the rightangle distance from the line of force to the axis of rotation, and multiplying the force by that distance.
•Torque = Moment of Inertia x Angular Acceleration
•Work = Torque x Angular Distance
•Power = Torque x Angular Velocity
We should note that these equations are of the type of equation known as "identities". As opposed to stating conditions on the value of a variable, they simply state that the quantities on the two sides of the equal sign are identical quantities.
Torque Multiplication
There is one more important theoretical concept that we have to discuss, that being an important consequence of the relationship among power, torque, and angular velocity as given in the last equation above.
If you use a gearbox to achieve a different rotational rate at the wheel as compared to the engine, the torque at the wheel will not be the same as the torque at the engine crankshaft, even if there are no frictional losses. The reason that this is so is that the rate at which work is done must be constant throughout the system (excepting only for the loss of energy within the system due to friction), and because power is the product of the torque and the angular velocity.
Analogies between mechanical systems and electrical systems are often insightful, so I'll ask you to indulge me for a moment while I digress and talk about electrical transformers. The voltage transformer is the electrical equivalent of the mechanical gearbox. A voltage transformer (which, by the way, only works for alternating current) takes advantage of the inductive coupling between the primary and secondary windings to transform the voltage. In electricity, power is the product of the voltage and the current. Because the rate at which work is performed must be constant throughout the system, the product of the voltage and the current in the secondary must be the same as the product of the voltage and the current in the primary ? at least in the case of an ideal transformer where there is no loss to heat. If the voltage is reduced or increased by a factor of N, the current will be inversely changed in that same proportion.
Okay, back to mechanics. Because power is the product of the torque and the rotational speed, and because the power must be constant throughout the system, the transmission increases the torque in the same proportion by which it reduces the rotational speed. If you measure the torque at the rear wheel, you have to use the overall reduction ratio to calculate the engine torque. At a given engine speed, the rearwheel torque will depend on what gear is selected. Any dynamometer chart that shows torque must be engine torque or else it would be applicable only to a specific gear and the chart would have to specify the gear, which would not be especially useful. Note, however, that if the torque is measured at the rear wheel and then the overall reduction ratio is used to calculate the engine torque, the result will not be the same as the result that you would get if you rigged the dynamometer directly to the crankshaft, because the measured rearwheel torque is subject to power train losses. Nevertheless, any dynamometer chart that shows torque and that does not specify the gear is most definitely the engine torque, albeit adjusted for drive train losses.
Metrics of torque and power; dynamometer theory
Okay, at last we're ready to talk about the metrics of torque and power, and about how dynamometers work. Contrary to what the urban myth says, there is no fundamental physical reason by which it is impossible to measure work or power directly on a dynamometer, i.e., without deriving the power through the torque measurement. Furthermore, even if it were true that the only way to measure power is by derivation through the measured torque, that fact would have no bearing on the significance of power.
In fact, measuring power on an inertial dynamometer is no more difficult than measuring torque, and is not predicated on the measurement of torque. Whenever the work performed goes to increase the kinetic energy of an object, power is the instantaneous rate of change of the kinetic energy. It is a trivial matter to calculate the kinetic energy of a moving drum if its rotational speed and moment of inertia are known. The rate of change of the kinetic energy can be determined in a manner essentially identical to the traditional method used to determine the angular acceleration of the drum, which is needed in order to calculate the torque.
In an inertial dynamometer, the engine is allowed to accelerate an inertial drum as quickly as it is able. Newer inertial dynamometers use an inertial accelerometer to give the measure of the angular acceleration continuously. With the traditional method, however, each time the drum rotates through a fixed number of degrees, an electrical pulse is generated which triggers the recording of the elapsed time. >From those data points, the average angular velocity can be computed for each of the individual time intervals between adjacent points, by dividing the fixed angular distance by each of the time intervals. Then an average acceleration is calculated for each adjacent pair of average angular velocity values, by dividing the difference in the adjacent average angular velocity values by the corresponding time difference. For each average acceleration value, the average torque over the corresponding time interval is then found by multiplying the average acceleration by the moment of inertia of the drum.
On an inertial dynamometer, the average rate of change of kinetic energy for each time interval can be determined using a method that is essentially identical to the traditional method used to determine the acceleration. The average kinetic energy for a given time interval can be calculated from the average angular velocity over the interval, using the formula: K.E. = ½M x V². (It is necessary to factor in a standard correction to account for the fact that the square of an average is not the same as the average of the squares.) If the difference between the average kinetic energies for two adjacent time intervals is divided by the corresponding time difference, the result will be the average power for that time interval. I want to emphasize that this is not in any way an impractical, farfetched approach. It is for all intents and purposes identical to the ubiquitous, traditional method used to determine the acceleration and the torque on an inertial dynamometer.
It is also possible to measure power independently of torque on an inertial dynamometer that is equipped with an accelerometer. A computational technique known as numerical integration can be used to derive the angular velocity at many closely spaced points, from the accelerometer readings. The kinetic energy can then be calculated at each of those points. The average rate of change of the kinetic energy between each of those points can then be calculated the same as with a traditional inertial dynamometer, by dividing adjacent pairs of kinetic energy values by the corresponding time differences.
Inertial dynamometers vs. brake dynamometers
With an inertial dynamometer, the engine is allowed to spin up as quickly as it can, accelerating the drum as quickly as possible. Consequently, the measured results reflect the engine's ability to increase its work output rapidly, which ability is more greatly influenced by the engine's own internal inertia than is a measurement of steadystate work output. This factor is treated less rigorously than it could be: there is generally no attempt to quantify the effect that the engine's reluctance to rapidly increase its output has on the measured results, and the measured results are not carefully distinguished from a measurement of steadystate output such as would be obtained on a brake dynamometer. Be that as it may, an inertial dynamometer is a more realistic test of an engine's ability to accelerate a vehicle, whereas a brake dynamometer is at least as realistic when the question at hand is an engine's ability to climb a steep hill at steady speed. Turbine engines have a very high thrust to weight ratio, but due to the combined moments of inertia of all of the turbine blades, they are slow to speed up and slow to slow down. Jay Leno's turbinepowered motorcycle would fare much better on a brake dynamometer than it would on an inertial dynamometer.
With a brake dynamometer, the engine speed is increased in small steps and held steady at each step in order that the braking torque that is required to hold the drum at a steady rotational rate ("dynamic equilibrium") can be read. The throttle is kept fully open, but closedloop feedback using the drum's motion sensors is used to regulate the braking force in order to hold the drum's angular velocity constant. Several different types of brakes are used, but regardless the principle of using feedback to hold the drum velocity and the engine rpm constant is the same. Depending on the type of brake used, transducers will be used to take measurements of force, or hydraulic pressure, or hydraulic flow, or electric voltage, or current, etc. Calibration factors and formulae will be applied, all in order to determine the opposing torque that is applied via the brake to the drum in order to maintain the drum at dynamic equilibrium.
To summarize, with an inertial dynamometer, the drum is allowed to accelerate as rapidly as the engine can make it accelerate, and the torque that the wheel exerts on the drum is determined using the data from the drum's motion sensors and simple mathematical methods to obtain average rates of change over small time intervals. With a brake dynamometer, the torque that the wheel applies to the drum is found more directly by applying a regulated countertorque, using transducers to measure the braking force, which must be calibrated and which can be a significant source of error if not done properly with a good understanding of metric processes.
Myth #2: Engine torque is the same as the dynamometer drum torque.
Peculiar though it is that anyone with any real knowledge of the subject could be this misinformed, it is apparent that many of the same people who believe that dynamometers can't measure power, believe this as well. Regardless, the true nature of the relationship between engine torque and dynamometer drum torque is sufficiently important such that I would be remiss if I were to omit that subject from this article.
Regardless of which type of dynamometer is used, the dynamometer measures the torque that is applied to the drum by the wheel, which is applied by way of a longitudinal, frictional force between the wheel and the drum. That longitudinal force can be found by dividing the measured drum torque by the drum's radius. The rear wheel torque can then be found by multiplying the longitudinal force by the wheel's radius. Of course, the longitudinal force is not usually of interest unless perhaps you wish to calculate the theoretical acceleration of your vehicle. To get the rear wheel torque more directly, you divide the drum torque by the drum radius and multiply by the wheel radius. To convert the rear wheel torque to engine torque, you divide the rear wheel torque by the overall reduction ratio. The overall reduction ratio is found by multiplying together the primary reduction ratio between the crankshaft and the transmission's input shaft, the final reduction ratio between the transmission's output shaft and the rear wheel, and the transmission ratio that depends on the specific gear that was used during the dynamometer run.
Myth #3: Acceleration will be greatest when the engine speed matches the engine speed where the torque peak occurs.
Myth #4: Power is an abstract, derived quantity that is meaningless insofar as concerns the goal of selecting shift points that yield the greatest acceleration.
In accordance with the well known relationship among force, mass and acceleration given by Newton's wellknown F = M x A, the acceleration of the vehicle is proportional to the longitudinal force at the tire contact patch. (We can ignore the fact that some of the force goes to angular acceleration of the wheels; this will not alter the validity of the analysis or the conclusions.) If the engine torque (adjusted for power train losses) is known for a given engine speed, the corresponding longitudinal force at the contact patch can be calculated by multiplying that torque by the overall reduction ratio and then dividing by the rear wheel radius. So long as the reduction ratio remains fixed at a given gear, the wheel torque, the longitudinal force and the acceleration will all have their maximum values at the engine speed where the torque has its maximum value. That is fine and good, and obvious, but it bears little on the question of how the maximum rear wheel torque and acceleration are determined and maximized when you can use gear selection to alter the engine speed at a given wheel speed.
When you use gear selection to change the relationship between the engine speed and the wheel speed, two things happen. First, the change to the engine speed generally alters the engine torque. Second, the change to the overall reduction ratio changes the torque multiplication between the engine and the rear wheel.
The vehicle will have greatest acceleration at a given wheel speed when the gear selected results in the greatest rear wheel torque. If a gear is selected that puts the engine speed somewhat higher than the engine speed at which the engine torque peak occurs, that numerically lower gear will result in greater torque multiplication and the rear wheel torque will be greater even though the engine torque will be somewhat less than its maximum value. This will be true so long as the torque curve remains reasonably flat above its peak value. Even with engines that have a pronounced peak in the torque curve, the torque curve will be essentially flat for some distance near the peak. If the gear ratios are properly matched to that torque curve and the wheel speed is within the normal operating range, it will always be true that the acceleration will be greatest when the engine speed is higher than the engine speed at which the torque peak occurs.
This begs the question of how the acceleration is related to the power. That relationship is slightly complicated by the coupling of several facts. Power is the rate of change of the kinetic energy, kinetic energy depends on the square of the velocity, and it is not intuitively obvious whether the velocity should change most rapidly at the same engine speed where the square of the velocity changes most rapidly. However, by combining two of the identities (equations) that were presented near the beginning of this article, we can derive an identity that describes acceleration as a function of power, velocity and mass.
Combining the two identities: Power = Force x Velocity; Force = Mass x Acceleration, we get another useful identity: Power = Mass x Acceleration x Velocity. Rearranging the terms to isolate acceleration, we get an identity that describes acceleration as a function of power, velocity and mass:
•Acceleration = Power / (Mass x Velocity)
This formula provides us with a couple of useful facts. For one, it tells us that for a given power and mass, the acceleration decreases as the velocity increases, which is consistent with the fact that kinetic energy increases as the square of the velocity.
Probably of greater interest to most readers is the fact that for a given velocity and mass, the acceleration is directly proportional to the power. There are two distinct and meaningful consequences of the proportionality between acceleration and power. First, at a given speed, the acceleration will be greatest when the gear selected is such that the power at the associated engine speed is the greatest among all the gears. Second, given any two vehicles with identical mass (to include the mass of the rider), the one with the more powerful engine will exhibit the greatest maximum acceleration, regardless of which one produces the most torque. To be sure, the power to weight ratio determines the vehicle's maximum acceleration, which of course is why that ratio is frequently quoted.
The simplest way to assess what the acceleration can be at any given wheel speed, is to convert that wheel speed to the equivalent engine speed for each gear, and then look at the power curve to find which of those engine speeds yields the most power. You can also answer this question from the standpoint of rear wheel torque, but then after looking up the engine torque for each of the engine speeds, you have to turn back around and multiply those engine torque values by their corresponding reduction ratios in order to find the rear wheel torque for each gear. You'll get the same result either way; if you don't, then at least one of the two graphs is in error. However, the torque method requires more computational work as compared to simply looking at the power curve, so why would anyone want to do that?
Just for grins, let's consider an actual example. The March, 2003 issue of Motorcycle Consumer News has a dynamometer chart for the 2003 FJR1300. The Yamaha shop manual gives the primary, secondary, and gearspecific reduction ratios for the five gears. I want to find the optimal road speed for shifting from 1st gear to 2nd gear. I worked out the value of the multiplier for converting the road speed in mph to engine speed in rpm. That multiplier is 60, and the primary and secondary reduction ratios are already factored in to that, but I still have to multiply by the gearspecific reduction ratio. At 50 mph, the engine speed will be 7590 rpm in 1st gear (more or less depending on the accuracy of my measurement of the wheel radius), and looking at the dynamometer chart, I read about 120 hp for that engine speed. That's very close to the peak power, but the peak is located a little higher, just shy of 8000 rpm. Therefore, I expect that I should go beyond 50 mph in 1st gear before shifting to 2nd gear.
At 60 mph, the engine speed will be 9104 rpm in 1st gear, which is just off the chart because when you cross 9000 rpm you're into the red zone, but I can visually extrapolate the curve and estimate 110 hp, well below the maximum hp. (The error in my measurement of the wheel diameter might be the reason for the engine speed being in the red zone at 60 mph in 1st gear, but that error won't effect the essential comparative result, so I will ignore the red zone.) I can't know whether I should shift to 2nd before reaching 60 mph, until I check what the power will be in 2nd gear at 60 mph and confirm that it is more than 110 hp. At 60 mph in 2nd gear, the engine speed will be 6380 rpm, and the chart indicates about 105 hp, which is slightly less than the 110 hp that I'll get at 60 mph if I remain in 1st gear. Therefore, if this chart is correct and subject to the accuracy of my measurement of the rear wheel diameter, the implication is that for purposes of maximum acceleration, I should wait until I reach a speed slightly higher than 60 mph before shifting from 1st to 2nd.
But what happens if I compare the engine torques? At 60 mph in 1st gear, at 9104 rpm, the engine torque has dropped to about 60 lb. ft., a value well below the peak value of about 87 lb. ft. At 60 mph in 2nd gear, at 6380 rpm, the engine torque is about 85 lb. ft., which is very close to the peak. If I were to believe that the acceleration is greatest when the engine torque is greatest, then I would conclude that I should shift well before I reach 60 mph, probably somewhere in the neighborhood of 50 mph. But let's see what happens when we convert those engine torque values to rear wheel torque. When 60 lb. ft. is multiplied by the overall reduction ratio for 1st gear, the rear wheel torque at 60 mph in 1st gear is found to be about 658 lb. ft. When 85 lb. ft. is multiplied by the overall reduction ratio for 2nd gear, the rear wheel torque is found to be about 653 lb. ft. In other words, I have slightly more rear wheel torque at 60 mph in 1st gear than I do in 2nd gear, which suggests that I should wait until I reach a speed slightly higher than 60 mph before shifting from 1st to 2nd, which agrees exactly with the result that I got when I compared the power!
Myth #5: The reason that longstroke, undersquare engines produce more torque at low rpm than shortstroke, oversquare engines, is because the longer crank throws are more distant from the axis of rotation, and that causes the torque to be greater.
Let's consider the question of what determines the engine torque, and in doing so, let's be careful to distinguish the instantaneous torque from the average torque through a full rotation of the crank. The instantaneous torque varies considerably throughout that rotation, even within the ½ crank rotation corresponding to the power stroke. The quantity of work done during one complete rotation of the crank is fully determined by the integral of the instantaneous torque over that complete rotation. The average torque over the rotation is the integral of the instantaneous torque divided by the angular distance, so it follows that the average torque over the crank rotation effectively determines the quantity of work done over that crank rotation. (I have of course simplified matters by adopting an engine that has but a single cylinder.) Now, if it were possible to increase the average torque over a crank rotation simply by lengthening the stroke and the crank throw, then it would be possible to arbitrarily increase the work, the power, and the acceleration simply by lengthening the stroke! Talk about your free lunch!
The work performed by an engine during the movement of the piston through a single power stroke is determined by the quantity of energy released by the combustion of fuel and by the compression ratio (the compression ratio determines the thermal efficiency). Any desired compression ratio can be achieved for a given stroke, by shortening the space between the piston face and the cylinder head. Since the length of the stroke does not fundamentally determine either the compression ratio or the amount of energy released in the combustion, the quantity of work performed during a single crank rotation must be independent of the length of the stroke. Since the average torque over the rotation likewise determines the quantity of work performed, it follows that the average torque over the rotation must also be independent of the length of the stroke.
The torque applied to the crank depends on the force as well as the crank throw distance. The force that the gas exerts on the piston face is proportional to both the gas pressure and the area of the piston face. As the stroke is make longer, for a given displacement, the piston face area is made proportionally smaller, and the force exerted on the piston face is made proportionally smaller. Thus, the effect of increasing the stroke length is cancelled by the coupled effect of reducing the piston face area. Here, we see that even the maximum instantaneous torque applied to the crank will be independent of the stroke if the stroke variation is subject to a constraint on the displacement.
To the extent that a long stroke engine happens to produce more engine torque at low engine speed as compared to a short stroke engine, the reason for this is at best indirectly related to the length of the stroke. Rather, it can only be due to a difference in volumetric efficiency at that lower engine speed, i.e., design characteristics such as valve lift/duration and the rate of volumetric expansion of the combustion chamber on the intake stroke. Instead of the long stroke causing the engine to produce more torque at low engine speed, the truth of the matter is that the long stroke and its associated greater piston speed and piston acceleration limit the engine speed. As such, it only makes sense that the design characteristics such as the valve lift and duration be optimized for greatest volumetric efficiency at the lower engine speeds where that engine will always be operated. Because of that specific optimization, you would expect that such an engine should be capable of producing more torque at those low engine speeds than an engine that is not similarly optimized for low engine speed.
Whether or not an engine that by design produces its maximum torque at a low rpm is a Good Thing™, is subjective. The power and acceleration will come on a little sooner off the line, and this sort of engine will be able to increase its output from low output to maximum output more quickly, because the rpm range through which it must be accelerated to reach its peak power will be smaller. However, power determines acceleration, and power depends on the engine speed, so an engine of this sort is inherently incapable of achieving the same power or acceleration as compared to a high rpm engine of similar displacement, at any road speed. Even "off the line", if a short stroke, high rpm engine is mated to a gear box with 1st gear set adequately low, the short stroke, high rpm engine will outaccelerate the long stroke, low rpm engine, anytime, anywhere.
Copyright © 2003, by Tom Barber. All rights reserved.

11072012, 09:49 PM #2
@bobS pay attention to this.
Good article, good stuff. Lot's to soak in though he makes it pretty easy to understand. Don't like when he does the electricity analogy as that complicates it a bit but he tried to keep it fairly simple.Stage 2 or 2.5 E9X M3 S65 V8 supercharger kit for sale: http://www.boostaddict.com/showthrea...rkitforsale

11072012, 10:02 PM #3Member
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Good post op, I saw this @Sticky and honestly what I'm trying to say is I prefer the drivability of a fbo n54 over a stock s65, I saw that you commented on the thread about the AA supercharger and the ESS, where the comparison was made between the e39 m5 and the e9x m3, even though they have similiar hp, the tq curve is much different, I prefer the flatter tq curve of the e39 m5 in that comparison as well. It's the drivability I prefer when it comes to tq in the fbo n54
ESS 6XX kit

11072012, 10:05 PM #4Stage 2 or 2.5 E9X M3 S65 V8 supercharger kit for sale: http://www.boostaddict.com/showthrea...rkitforsale

11072012, 10:11 PM #5Member
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I would be fine with a supercharged s65, and might still end up with one some day, depending how the new m3 turns out...just not a stock s65
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11072012, 10:43 PM #6Member
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@Sticky, yeah  I hear you. Although he has a ton of good information  he sometimes goes all over the place and says things that aren't necessary. I think he is trying to keep it simple by doing this  but agree that it makes it even more difficult to read. Plus it's long.
@bobS, I know what you mean with the flexibility. It certainly feels a lot different than the M3, no doubt about this. I think it's having that push (boost) from practically right off idle that gives this flexibility. I like the 335 and it's engine. I just have a hard time when people say the S65 is torqueless  it's got ~260 ftlbs after drivetrain losses  and it's curve is as flat as a countertop all the way out to 8400 RPM. With gearing reduction, it has much more torque at the wheels in most gears than the 335i  but certainly doesn't have the same feel off the line. Not at all...
They aren't even comparable cars/engines in that regard. Both are fun, but I prefer the high strung/predicable/quick nature of the M. I do wish I had boost though  or just some NA offering that put me at a similar power level though.

11072012, 10:52 PM #7Member
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One of the interesting parts of the article to me was where he compared the types of dynamometers (inertial and brake). This explains why a lot of people have issue with the Mustang dyno. It's "virtual inertial" system (it's basically a brake dyno)  coupled with his explanation of this type of dyno explain why people get upset when they read low. It seems that a quickly revving engine is not suited to this type of dyno, but need to understand it a bit more to make an accurate statement. Seems to be due to the weight of the internals of the engine, and the acceleration of those internals  the inertial dyno seems better suited for this type of thing (if I am reading it right). He also mentions how much more complicated they are to operate properly/accurately. I really wonder how many dyno operators ensure all the variables are properly set (correct gear, weather, etc.) and know how to properly operate them... In my opinion, all dynos should read exactly the same, as 1 HP should equal 1 HP and 1 Lbft is 1 Lbft no matter what. In the real world however, this will never be the case.

11082012, 09:58 AM #8Member
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See I'm being a little unfair though, I'm comparing a full bolt on stg 2 335 n54 engine, with the increased hp/tq to a stock s65 in the m3. That's why I'm saying I prefer the feel of the 335. I'm not comparing a stock 335 to a stock m3.
ESS 6XX kit

11082012, 10:50 AM #9
All you need to know is that the torque units don't contain time. They are in the dimension of force times length, there is no time (seconds in there). So torque alone cannot possibly represent any form of acceleration alone, without time. That is why power (horsepower, watts, whatever you want) contains the units time, torque applied per unit time, that causes acceleration, nothing really else matters. Well gearing is just how you experience that power, so that matters.
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11082012, 12:12 PM #10Some dude
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Good read, albeit long as mentioned. Corrected some of my internal thought processes regarding torque and HP WRT shift points, as well.
The S52 engine is ubiquitious in BMW CCA and NASA club racing. It's important to maximize shift points to go with your torque and HP curves. I have instructed several club racers on proper shift points based on stuff I've read a lifetime ago, but this really explained to me again how to correctly calculate optimum shift points. Just giving a shootfromthehip estimation of shift points on the critical 34 and 45 shifts looking at a Dynojet plot, we increased the straightaway top speed of a 225RWHP (restricted) E36 M3 GTS2 car by more than 3MPH. He was wasting way too much time revving above his HP peak.James Muskopf
RRT Racing
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11082012, 12:59 PM #11
Some of you may or may not already know this. This is not directed to anyone specifically, but since James already brought up shift points, things start to get a little interesting.
What is my optimal shift point?
That question has been answered by so many people, so many different answers. But let's look at what is important. Average horsepower.
When you shift your car, lets say on the dyno, if your average horsepower in that gear is maximum, you have found your optimal shift point. Torque doesnt apply, anywhere, in racing. Because racing involves time.
If I took a gas turbine that makes 50 ft lbs of torque at 80,000 RPM and dropped it in my car, guess what, thats a 761 hp engine and I will feel every bit of it if my transmission is geared correctly.
So where should you shift? It depends on your cars power output and gearing. The best way to know exactly where you should shift is by going to a dyno, use it as your tool, and look at your average horsepower in every gear, 1st through 5th and optimize from there.Some people live long, meaningful lives.
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12182012, 02:15 AM #12Member
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It's "virtual inertial" system (it's basically a brake dyno)  coupled with his explanation of this type of dyno explain why people get upset when they read low. It seems that a quickly revving engine is not suited to this type of dyno, but need to understand it a bit more to make an accurate statement. Seems to be due to the weight of the internals of the engine, and the acceleration of those internals  the inertial dyno seems better suited for this type of thing (if I am reading it right). He also mentions how much more complicated they are to operate properly/accurately.
Aden

03152013, 06:25 AM #13Member
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i dont agree with this average at all. you shift at the limiter except when power falls off past peak hp badly enough such that you'd have more power in the next higher gear.
you need to maximise the hp available at any instant because this generates the most tractive effort that propels the vehicle

03182013, 11:24 PM #14Member
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I think you're misunderstanding what average means. Over a given gearspan when accelerating, the car will accelerate the quickest when the power over each instant is the highest possible. Say your average gear runs from 4500rpms to 6200rpms or something... over that span, if you sum the power at each instant and then divide the total amount of data points (which is averaging it); you will find the average horsepower. In effect, this is doing exactly what you're saying "...when power falls off past peak hp badly enough..."
The thing is, your rpm span is set in stone so far as width goes; so your shift point determines where the average power calculation starts.

03192013, 01:51 AM #15Member
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oops i misread i thought he meant average over whole rpm not that used on a gear by gear basis.

03262013, 11:17 PM #16Member
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03272013, 08:43 AM #17Member
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05142013, 06:45 PM #18Member
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Just have to say thanks for this thread, it's made it much easier to explain to people exactly what torque is in relation to horsepower, and help out some confused people with dyno graphs (especially the torque>horsepower equation)
boop

05272013, 02:56 PM #19Member
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Ah what a relief to have this. As many of you who actually understand this stuff will agree, it gets frustrating and frankly time consuming to correct every person who talks about "low torque of the m3" Not sure why it drives me mad. Certainly is not because I own an m3 and want to justify it. Rather stupidity bothers me. Even more, a lack of common sense bothers me. Now I can just copy and paste this bad boy on any thread where the poster is an idiot! Sweet! Thank you.
Almost as ridiculous as the "torque wins" races comments is the guys that claim peak acceleration is at peak engine torque. And people who say you want to shift right after peak torque. It is SO ridiculous if you think about it logically or rather go try it with your car. I mean go take an m3 and shift at 6500 torque peak and race someone revving to 8500 in each gear. Pretty sure you can visualize who will win.
HOwever the 335 and even the new N63 v8 engines, really do fall off so terribly after their torque peaks that it really is probably the closest engine I have seen to something you pretty much do shift not to far after the peak torque

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05272013, 05:24 PM #21Stage 2 or 2.5 E9X M3 S65 V8 supercharger kit for sale: http://www.boostaddict.com/showthrea...rkitforsale

05272013, 05:25 PM #22Stage 2 or 2.5 E9X M3 S65 V8 supercharger kit for sale: http://www.boostaddict.com/showthrea...rkitforsale

05282013, 09:48 PM #23Member
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Even with the post above clearly, some people are just not sharp enough and that is why the world is the way it is. Not everyone can be smart so its not surprising.
Ok go out, take an m3 and mash the throttle in 1st, 2nd, 3rd and onward. Then tell me how often you spend under 6k RPMS. So while its true the m3 has relatively low ENGINE torque, you never see low end if you are trying to make the car go fast. If you are just putsing around then 300 foot pounds is fine.
And just for one last time, 300 foot pounds multiplied by gearing is more than a 335!!!! Low and high.
Some just are dense. We live amongst all types

05282013, 09:57 PM #24Stage 2 or 2.5 E9X M3 S65 V8 supercharger kit for sale: http://www.boostaddict.com/showthrea...rkitforsale

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Interesting article/post.
I much prefer flat flat torque output, from turbocharging, di, and small/responsive turbos. The s65 is a beautifully orchestrated symphony when pegged at 8k+ rpms.
I Agree with itch sticky and lostmarine, however. It's gutless below 4k rpms.
I think the n54 spoils us in regards to lowmid rpm power output. Phenomenal if you ask me.
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