torque and power of automobile engine

The maximum power output in horse power / kilowatts etc... is the most efficient point at which the engine will operate (I think, might be wrong?) and is also the point at which maximum power is available, surprise, surprise!!

The figure for maximum torque is usually at lower rpm and is the figure where...... I dunno!

I had better shutup and let someone else answer this one

So I guess I have to admit that I am completely ignorant about torgue... I always thought of torque as something like centripetal force. If you have a piece of metal and it spins at X RPMs, it has Y torque, and if you spin it faster than X you have more torque. If this was the case, then the faster the motor turned, the more torque it would produce.

Where am I going wrong here???

I always think of torque as, 'that which one applies to a nut using a spanner' (wrench) e.g. Two and a half foot-grunts for a wheel nut .

So for a vehicle/engine I think of it pulling up a hill or pulling a load away from stationary....torque as more a static thing whereas power would be dynamic on the flat..

I daresay ths is more of an analogly than a real explanation...(but hey ..even Ken used the term Grunt ...so I s'pose it's ok)

_{(BTW...I did get my golf in yesterday...well 11 holes at least , but no birdies)}

Del

Are your balls still the same color?

Torque relates to rotation as force relates to translation.

That's all.

Well that's cleared that up then

Where am I going wrong here???

On the definition of torque. Torque is twisting force. It is measured in units of distance and force, so a force of one pound at the end of a lever 2 feet long, produces a torque of 2 lb ft, 10 newtons at one meter is 10 newton meters, etc.

In tightening a bolt, if you apply apply 20 lb of force at the end of a one foot wrench to a 1/4" bolt... you snap the durn thing right off!!! How could you be so careless! The typical 1/4" bolt is only good for about 10 lb ft of torque.

If an engine is twisting hard it produces a lot of torque. The BMEP (brake mean effective pressure) in the cylinder (the average working pressure in a cylinder during the power stroke) determines how hard a particular engine twists.

At any RPM beyond the peak torque speed, an engine produces less torque as is turns faster.

In dc motors, torque varies directly with current, speed varies directly with voltage.

I know I'm being somewhat thick here, and it's not your obligation to teach me, but it would seem to me that the maximum twisting force would occur exactly where the horsepower was the highest... Although, I know this not to be true.

If you scan down to post number 11 in this thread from a Subaru enthusiast site, you can see the torque and HP curves for a modified and unmodified normally aspirated (i.e., not turbo) Subaru. As you can see in both cases, the curves cross at 5250 rpm, because HP= torque x RPM/5250. (This would be the case in any engine capable of over 5250 rpm, and in which the graph is drawn so the numeric values for torque and HP are the same at each horizontal graph line.)

If you took the torque values alone, and did the math at any RPM, you'd find that your answer would lie on the HP curve: that's actually how the HP curve is generated -- typically a dyno measures torque directly, RPM directly, and HP is calculated from those.

For the reasons I mentioned in my forst post above, and others, torque falls off on either side of the torque peak (there's too much aerodynamic friction in the intake and exhaust tracts at high rpm (flow rates) and not enough gas inertia at low speeds). As you can see, especially in the curves for the modified engine, even though torque is falling off at 6000 rpm, HP is still rising. Eventually, at about 6700 RPM, torque has dropped off enough that HP is also dropping with higher RPM (the math just doesn't work out favorably).

As you can imagine, if you installed a cam that took greater advantage of inertia effects (which occur at higher RPM) you could maintain torque at a higher RPM. (Such a "high duration" cam would allow the engine to produce more HP, but drivability would suffer, because at low RPM, the engine would produce less torque. In these dyno traces you can see that if the dyno run started at a lower rpm, the torque curves for modified vs unmodified would cross at 2750 RPM, so at RPM below that, the stock engine would have an advantage. The torque curves for both engines are very broad and relatively flat, and even the modified engine would be easy to drive around town.

(The nearly vertical line at the left end of the curve is has nothing to do with the engine, and everything to do with the type of dyno used, and the way in which the test was run: the test was started at 2300 for the modified engine and 2750 for the stock engine. You could do a straight line extrapolation down toward about 1000 rpm and be pretty close to actual figures at lower RPM for the two engines.)

(BTW, the Dynojet dynos are inertia type, where a heavy weight is spun up, while the engine is held at full throttle. The rotational acceleration of the weight is measured, and and the dyno does the math to translate that into torque. In a more typical dyno, a brake is applied to the input rollers (this can be friction, hydraulic, eddy current, etc -- anything which will provide resistance while permitting the engine to continue to turn. As you can imagine, an ordinary break would soon be glowing red hot, but there are still dynos with mechanical brakes. The reaction force on the brake arm can be measured to measure torque -- although usually, the torque is measured a little less directly via pressure in hydraulic fluid, or current in an eddy current brake or generator.)

V,

Ken's on the money in 16, however he's omitted to tell you that it's BMEP over the mean effective LEVER. Torque is force times lever. The crankshaft is a lever, just like the lug-nut wrench is, so's you can change a wheel. Pistons push down on the lever, just like your foot does on the wrench.(Are you too young to have hand-started a Flivver? That'll teach you about torque.) Peak torque occurs in an engine when all the dynamics of the combustion process are at their most effective. To further illustrate this, consider that big diesels, truck engines, etc, have their peak torques at relatively low revs, typically 1200 - 1800 rpm. It's to do with the the 'speed' of the rise in cyl. pressure as the fuel burns. Like the examples given.

When an engine is dyno'd to find it's peak power, the measured force is that of Torque. Horsepower is then calculated from that, with time being the next consideration. No, we don't sit down with a calc. and do it, the dynamometer computer is programmed to read it out as the test is running.

Hope this also helps.

Cheers,

Stu.

idustrial engines are designed to have nearly flat torque curves. They have increasing torque to 75% of rated RPM and then flat to 105% of rpm, then decreasing torque. Horsepower = torque * RPM / (about 5350 a conversion factor based on ft-lb torque)

What this does is allw us to couple the engine to a compressors that have linear horsepower curves, so at 75% RPM, their HP requirements are 75% that of 100% RPM. To do this, the engine efficency is low at 75% rated speed and is maximum at rated speed.

Good answer Ken...

_{That's just what I was going to say in my post # 1}

Basic to the design of every engine is bore and stroke. An "oversquare" engine, where the stroke is greater than the piston diameter, is good for low RPM applications because it produces a lot of torque at low speed. Short stroke engines produce less low speed torque but produce their maximum torque at higher speed.

Valve timing, computer settings, etc. will do a lot to determine the exact torque curve, but the bore/stroke ratio will always provide the starting point.

Yes, in the UK 1960's we had a motorbike called the Panther it was a single 600 cc 4 stroke, it had horrible HP but it had enormous torque at very low revs...

It was a common saying that in top gear doing 60 mph the cylinder fired about once per lamp post!!!

And that it could pull a car out of a ditch - whether that is true or not is debatable!!

Nostalgia fix.... here

Del

Ahhhh yes..... I remember that well...

Brought back the smell of Castrol R exhaust fumes - nothing like it!!!

Your definition of "oversquare" is reversed, but otherwise what you say is more or less true, or at least has been.

Nowadays, these ratios are still key engine design parameters, but are no longer so much "in control" of the engine dynamics as they used to be. Larger bore (oversquare) engines tend to breathe better, and generally have more valve area, per unit of displacement. Therefore even at low speeds they can have higher BMEP which can more than compensate for the greater leverage provided by a long stroke (but doesn't always).

The Honda S2000 has one of the peakiest, highest-revving engines in any passenger car. The 2008 model produces max HP at 7800 rpm, and produces a modest 162 lb ft of torque at 6800 rpm. This undersquare design (87mm bore, 90.7mm stroke) has the usual characteristics of an oversquare design. But, supporting your general point, the earlier years of this model, had an even more extreme engine (max torque at 7500 rpm, max HP at 8300) and it was oversquare. So going from oversquare to undersquare was one of the things Honda did to make the engine a little less peaky, and easier to drive without having to pay so much attention to always being in the optimum gear.

My own Honda, also somewhat peaky and lacking torque at low RPM, is an undersquare (long stroke) design -- again, not the expected characteristic. In the motorcycle world, the XR 600 Honda is a "thumper" but is quite oversquare (100mm bore, 82 mm stroke).

Dear friend

Power is sth and torque is sth else .

power = t * angular velocity. torque is the result of combustion and gear box and the other things.

At heavy machines for example in TRACTORS the torque is high but in a common car the torque is low with respect to the former.then by a given power the more the torque the less the rpm(heavy gear or tractors and...).generally , power and torque in comparison with fuel consumption will be efficient in deciding.

regards

Go back to basic definitions to intuit your answer. Power includes a "per-time-interval" factor; torque does not (is not dependent on how long force is applied). Operation (for example) at such-and-such RPM will (say) convert the most tire tread to smoke versus other engine speed rates. Reaching another RPM rate will coincide with the instant of quickest acceleration (or resistance to deceleration) within a current gear ratio range.

Reaching another RPM rate will coincide with the instant of quickest acceleration (or resistance to deceleration) within a current gear ratio range.

True. That particular RPM rate at which which the engine produces the greatest acceleration for a particular gear is the point at which maximum tractive force (for that gear) is reached, which is in turn the RPM at which peak torque is reached.

Further rambling: By happy coincidence, the tires on smallish cars are very close to 24" diameter, which means the lever arm from axle to contact point is very close to one foot. Therefore, if an engine produces 150 lb ft torque at let's say 2000 rpm, and first gear is 3:1 and the final drive ratio is 4:1, the tractive force at the rear wheels would be 1800 lb. If the coefficient of tractive friction is .8 and half the weight is on the drive wheels, then if the car is less than 2880 lb, the tires will, as CowAnon mentions, smoke.

Ken,

You must be deep into racing to have such abstruse knowledge of auto engines.

Reading what you have written, and the limiting factors that mechanically (Cam) driven valves introduce I could not help but wonder if anybody had thought to cut valve operations loose from the rotating parts of the engine and instead operate them electrically with a computer doing calculation based on engine sensors and data.

We do indeed inject fuel in almost all of today's engines (I am not clear if any engines inject directly into the cylinders as opposed to an intake manifold, but (And I am extrapolating from what I do know about engines and am a fair self service mechanic) the volumes of air that the intake requires and the volumes of exhaust gases would seem to require mechanical valves of some size.

Could we drive such valve systems electrically and if so what kind of edge would we get on engine efficiency?

Even as I write this my mind suggests, even if we stick with a cam, a cam driven by an electric motor whose speed therefore could be varied as to the best conditions for car and motor at any point.

Am I crazy or are these possibilities if not now actualities?

Of course the best alternative is to dump these archaic machines going for free mass transport over long distances and free use of small electrics at either end of such a system. I wonder if anybody has calculated the costs of such a system against our current gross transportation costs.

j.

j.

Jack, direct fuel injection is being used today on some new vehicles, I think mainly diesels.

No body has mentioned hydraulic operated valves? these are once again new and used on some performance vehicles... The advantage over a spring is obvious, no chattering or speed limits as the valve is opened AND closed hydraulically.

John.

Actually I thought of hydraulic but thought to narrow the consideration of valving freed from group reliance on the same rotating appliance, i.e., a cam.

j.

Electroman, Jack, Ken, I've been watching this conversation; perhaps I can add something worthy of consideration.

In some quarters the reference would be to hydraulic lifters, which, indeed, are commonplace in engines in which valve timing is chain driven, or otherwise by resort to rocker arms (i.e., not overhead cams).

As to advantages respecting solid vs hydraulic, that would certainly be a debate which is neither new nor recent, with each system having it own virtues (non-chattering not being one of them; and response time favoring solid, insofar as I've heard respecting not-unduly-worn or not-ill-maintained engines). The most typical rationales in favor of solid lifters seem (to me at least) to center on engine performance (in particular, high performance) as well as ease of maintenance/service, including lash adjustment service. Other people, and especially manufacturers, on the other hand, might see compensating virtue with hydraulic lifters in that, even though more costly & less easily serviced, they are dynamically self-adjusting for most of, if not all of, engine life cycle...not a small virtue given the precise fuel metering required to meet emissions standards...and even more critical since the imposition of (USA) 5/50 emissions system warranties upon all passenger vehicle manufacturers.

As to the other, valve train controlled, dynamic fuel intake issue, no less that three points have (I would say) always militated, and would always militate, against Jack's brainstorm notion regarding direct, non-manifold injection:

1. Uncertainty (at best) as to how, not only fuel compression, but also nozzle integrity and function could be durably and reliably achieved with repeated, direct exposure of nozzles to combustion chamber environments.
2. The great simplicity and very reliable dynamic precision that comes from mechanically linking fuel intake directly with all conditions of engine output demand.
3. The formidable difficulty and prohibitive cost—possibly insurmountable—of (somehow) devising a computer (or computations) which would indirectly (and unfailingly), both, calculate and control instantaneous multi-point injection while simultaneously controlling all other timing requirements...and would do so under all feedback conditions without possibility of internal conflict.

As to the other, valve train controlled, dynamic fuel intake issue,...

I think Jack was think of two separate issues. One is mechanical control of valves via camshaft vs more flexible methods. The mechanical system has many limitations that could be addressed by solenoid or hydraulic actuation, so that there is no longer a fixed mechanical link but instead a software controlled link that could adapt to different engine conditions. There are already variable valve timing engines on the market, but the degree of variability is limited, because they rely on advancing of retarding the cam relative to the crankshaft, giving control over timing but not duration. Sturman Industries is promoting fully electro-hydraulic control to provide great flexibility in all aspects of valve control. Current ECUs control fuel injection timing and duration, and an analogous system could control valve open timing and duration. Adding the degree of control shown in their diagrams (variable deceleration for closing, for example) would add additional processing speed requirements beyond the current fuel injection processing requirements. However, it has been 8 years since they drove a truck up Pikes Peak with their system, so processing speeds may now be approaching the point where the system could be up to the task for higher rpm engines.

There are now both "collision" and "non-collision" engines in which a broken timing belt or chain either causes damage (from valve-piston contact) or not. I'd think engineers would be reluctant to use a fully electrohydraulic system with a collision engine. If a fuel injector fires at the wrong time 1/100,000 times, it's no big deal. If a valve collided with a piston 1 out of 100,000 strokes, we'd have engine failures every few minutes.

The other issue Jack raised was direct fuel injection. This is currently widely used in diesels and in marine two-stroke engines. For marine two-strokes, direct injection enables emissions to be adequately controlled to meet the more rigid standards that have led to four-stroke engines becoming very common for outboard use.

Ken,

I would have thought they would have learned their lesson about "collision" engines by now for all the fools that fail to get those drive belts changed before they come apart.

j.

I guess it ends up being a philosophical decision. The best combustion chamber shape approximates a hemisphere, in the interests of minimum surface area, to keep as much heat in the gases and out of the cylinder head. However, to achieve high compression, the piston must be domed, intruding into the head portion of the combustion chamber. The valves need to be as large as possible, in the interests of good breathing. Normally, this means that clearances must be cut into the top of the piston to permit valve overlap (where both valves are slightly open at TDC between exhaust and intake strokes) even without any cushion for valve float. These valve pockets produce sharp edges, restrictions to flow, reduced compression ratio, etc, so are in themselves a compromise. So, in general, very high performance engines will put the valves very close to the pistons. In simple terms, the things done to provide enough clearance that collision is impossible, reduce performance.

In racing engines, (pre electronic rev-limiter days) it was not too uncommon for valve float (during inadvertent excursions to beyond redline, as in a "missed shift") to cause piston/valve collision. Often, this would cause the valve head to bend slightly relative to the stem, causing poor sealing and poor conduction of heat from valve head to cylinder head. If the racer was unaware of the damage, the stem could weaken from overheating, and the head of the valve would drop into the cylinder, fairly well destroying much of the engine.

But there are always several ways to skin a cat, and in engines a notch down from racing engines, it is possible to get good performance while still implementing the compromises needed to get valve clearance no matter what the valve and piston positions. Turbo engines are typically low compression, so these can be easily made non-collision (non-interference). But for high-performance normally aspirated engines, the designer has to weigh to cost of a chain or gear drive, vs cheaper, lighter belt drive, the likelihood that the owner will follow the replacement guidelines, etc.

I think that nowadays, most people are aware of the importance of replacing timing belts at the recommended time, and I would guess that interference damage is fairly rare because there have been so many horror stories of serious engine damage.

I'm sure there is a newer list, but this list highlights interference engines with an asterisk. Most engines are interference type. As you'd expect, the choice is fairly consistent for a given manufacturer. No Subaru engine is an interference engine, and few Fords are. Almost all Hondas (and Acuras), Nissans, BMWs, Porshes, Fiats and Volvos interfere. Toyota is schizophrenic.

OK. So in the top-fuel eliminator class of NHRA drag racing, not only do they run a toxic fuel - Nitro-benzene (was nitro-methane?), they still use Hemi's, and how much horsepower do they produce?

...and how much horsepower do they produce?

The common term for the quantity of HP produced is "gobs". In some university reports, the more descriptive term, "sh*tloads" is used.

When I was just a dust speck, I saw the 200 mph barrier broken. And now they're turning three second quarter-miles! Frigging amazing!!!

On early BMW's the problem was not so much the chain breaking, but loss of valve timing caused by overheating. The chain would jump one link over the sprocket causing the valve interference problem in spite of a chain tensioner used to keep out slack from the tension side of the chain.

Some felt the change from chain to belt was responsible for the interference problems. This change (around 1984) demanded belt replacement every 30k miles. Even before this change the OEM recommended chain replacement at this same mileage interval.

Your discussion on hemispherical combusion chambers is spot on.

To prevent valve float, increase ther red-line using stiffer valve springs and after- market cam shafts.

Another problem with the hemi was intake valve deposits. Solution: Cheveron gas additive (Techron, sp?) every other tank full or valve jobs every 40k. This was back in the 80's.

So, I agree that the decision to build a performance engine is a philosophical one.

You get a GA!!

Am I crazy or are these possibilities if not now actualities?

I think it's not an either/or situation: You are crazy and these are possibilities.

There are several patents for solenoid operated valves. The forces involved in overcoming inertia of the valves are surprisingly high, so the solenoids involved would need to be pretty powerful. (Valve springs can have installed on-seat loads of 100 - 200 lbs, which is required to keep the cam follower in contact with the cam when the valve is closing at high speed.) To eliminate the springs, you'd want solenoid control in both directions.

Ducati has promoted desmodromic valve actuation, in which the valve is both opened and closed by cams (and linkages). This eliminates valve float, at the cost of some complexity.

A huge advantage of electronic valve actuation (whether via big solenoids, or small solenoids controlling hydraulics or pneumatics) is that valve timing can be controlled (along with injection and ignition) by the ECU, so one second you can have a fuel economy "cam" and the next a power "cam".

The definition of torque as the twisting force similar to that on a nut by a wrench is correct, but whereas the screw is stationary that the nut being tightened on, the torque applied by an engine is applied with continuous motion. Horsepower is equal to torque in foot pounds at 5250 RPM and is proportional to speed so at 10500 RPM the horsepower is 2 x torque in foot pounds.

If you have a high-torque engine in a car, it will be capable of pulling easily at low revs - think of the old Triumph TR-6 that had 150 ft-lb of torque but only 104 hp. You could shift into overdrive while pulling onto a highway because it did not need to rev to keep up with traffic. More typical would be the first-year (1993) Subaru Impreza where the torque was 117 ft-lb and the power was 110 hp. You had to use the gears because peak torque came in at 4000 rpm, whereas the Triumph would not require much shifting. The other extreme would be the Honda as mentioned above, where there was plenty of power available, but not until the engine was turning fast.

The problem with any discussion on torque vs. horsepower is that horsepower is a construct. It was designed (by James Watt) for a specific purpose, and was defined very carefully. For example, horsepower doesn't change going through a gearbox (neglecting mechanical losses). In a vehicle rear wheel horsepower always = engine horsepower regardless of gearing. Another example, horsepower cannot be used to directly calculate acceleration. All formulas that I've seen either are estimates (typically change in kinetic energy formulas) or have a conversion from power back to force (as in acceleration = power / mass * velocity). Here are two links that may help understand the relation between force and power.

http://spreadsheets.google.com/ccc?key=0ApPRT4wdrOMvcEJDSDJDbHptcnBsSFVORmdya1Q3SEE&hl=en

and

http://www.sportrider.com/tech/146_0402_art/index.html

I don't mean to pick on you, but your first spreadsheet is largely incorrect or misleading, for reasons mentioned in this post from another thread.

The problem with any discussion on torque vs. horsepower is that horsepower is a construct.

There is no "problem" in discussing torque or horsepower. The problem only occurs when you create a problem by trying to incorrectly equate or compare the metrics. Power, force, work, energy, are all constructs with which thousands of people work every day. Horsepower is just one of many units for power, and all are easily and simply converted.

It was designed (by James Watt) for a specific purpose, and was defined very carefully*. For example, horsepower doesn't change going through a gearbox (neglecting mechanical losses).

You say "For example..." as if your next words will be an example of how horsepower was defined very carefully. But then you say that it does not change going through a gearbox. Of course it does not change! Why should it? (The work done on one end of a lever is the same as the work done on the other. Does that mean there is a problem with the concept of leverage?)

As another example (of something) you say that horsepower cannot be used to directly calculate acceleration. But in fact, it is the best measure for assessing the acceleration ability of a vehicle, and if all the various drags are taken into account, will produce a perfect times for any acceleration. From weight-to-power ratio we can assess the acceleration capability of a car but cannot do the same thing from weight-to-torque.

If there are problems with horsepower, are there also problems with kilowatts?

Because power in any rotary device is torque x rotational speed (times a constant) you can derive torque from power and rpm... and can derive power from torque and rpm. What could be easier? I don't see the problem. The problem is only created if one thinks the two terms can be used for the same purpose, which is simply and fundamentally wrong -- bad physics. They are no more alike than voltage and current (which when multiplied also produce a figure for power).

*Many would argue with this -- the carefully part. It is only a rough approximation of the power of a horse (an animal which varies widely in power output).

If there are errors in the spreadsheet, please state what they are. All the formulas are visible. In what way is it misleading?

There is no problem working with power as Watt or as anyone else defined it. You've misread or misunderstood what I said.

Units are convertible, but some units must be converted before certain operations can be done. The fact is that there are no formulas that calculate acceleration directly from power of any kind. They are estimates (like using changes in kinetic energy) or contain a conversion back to force (usually reduced out, but there).

If you look at the spreadsheet it demonstrates clearly that horsepower is not a good way to assess acceleration potential. The best assessment is by using the rear wheel torque curve. The rear wheel torque curve is always mathematically similar to the acceleration curve. The horsepower curve is not.

Hi devil,

If there are errors in the spreadsheet, please state what they are. All the formulas are visible. In what way is it misleading?

In the most fundamental way, it makes assumptions that are unrealistic. You cannot ignore air resistance, rolling resistance, etc in computing acceleration because you can only accelerate if you have available more hp than required to sustain a particular speed. The F in F=MA is only the net force, (left after the drag forces are subtracted) not the total force (tractive force).

If the spreadsheet is intended to be instructional for high school students, then it would be far better in SI units. As it is, it uses ft-lb for torque and ft-lb for work. For many students, this leads to loads of confusion. The convention in US units has been and should remain lb-ft for torque and ft-lb for work. Although the math would appear to work either way, the distance units are shorthand for different concepts -- in one case a displacement (how far something has been moved), in the other a lever arm length (with no movement required).

There are several other issues, especially with terminology, in the spreadsheet, but we can get to them later. First, read through this post, and see if it makes sense to you.

Then try this exercise:

1. Calculate the 0-60 time of a 2500 lb car with an engine that produces 50 lb-ft. (Good luck... to score full points you need to come up with the same figure I calculated.)

2. Calculate the 0-60 time of a 2500lb car, with a 95hp engine that has a flat torque curve at 50 lb-ft from 0 rpm to its 10,000 rpm redline.

3. Calculate the 0-60 time of a 2500lb car with a 47.5 hp engine that has a flat torque curve at 50 lb-ft from 0 rpm to its 5000 rpm redline.

For each, use a single gear which produces a top speed of 60 mph. Per your reasoning, the times should be the same, it seems.

Knowing torque alone (force times distance) leaves out a critical unit needed to determine acceleration, time. Power is a rate, torque is not.

Don't worry about air drags, rolling resistance, tire and clutch slippage, etc for the above calculations.

There is no problem working with power as Watt or as anyone else defined it. You've misread or misunderstood what I said.

However, you wrote: The problem with any discussion on torque vs. horsepower is that horsepower is a construct.

You can perhaps see how I might have thought that you thought there was a problem.

The fact is that there are no formulas that calculate acceleration directly from power of any kind.

This is not true, but we'll get back to that.

Let's begin at the beginning. The spreadsheet shows potential acceleration rates. Per common practice the examples are made with only the mass involved, not including wind resistance or mechanical drag. Sort of like a high school example problem? That doesn't make it incorrect or misleading.

As to your example problems, all three depend on "on the ground" torque, not engine torque. Acceleration = force / mass, you've got to know the driving force, not the engine force. So for the first problem you would need to know the overall gear ratio between the engine and the ground, including rear wheel radius.

The second and third problems say the engine has a flat torque curve and certain horsepower. With constant torque across the RPM range the horsepower increases continuously across the RPM range. At what RPM are the power numbers? I don't understand the point of those two examples. And they both have the same gear ratio issue as the first problem.

In any case, I am not talking about 0-60 elapsed times. I am talking about rate of acceleration.

Power is indeed a rate, but of what? Acceleration the first derivative of velocity with respect to time, the function is [acceleration=(mass*acceleration)/mass]. Work is the first derivative of force with respect to distance, the function is [work=(mass*acceleration)*distance]. Power is the first derivative of work with respect to time, the function is [power=(work/time)].

(Here's where the break is) the second integral of power with respect to time is not force, the first integral of work with respect to time is not force. It is possible to estimate acceleration from power or work only if boundary conditions are known. That insinuates the conversion to force. Otherwise conversion of power or work back to force is necessary.

Another example, algebraic this time. Look at the commonly quoted formula [acceleration = (power / mass * velocity)]. With one equation we can have only one unknown. With two unknowns on the right side of the equation the result is ambiguous. You can get the same result with different values of velocity and power. Once you know both the velocity and the power, power converts back to force and you have a single value for acceleration.

Non-SI units are where I started with these examples almost five decades ago. I suppose I should update. It does make things easier.

Hope this helps.

Hi Devil,

The second and third problems say the engine has a flat torque curve and certain horsepower. With constant torque across the RPM range the horsepower increases continuously across the RPM range. At what RPM are the power numbers?

Obviously, the power numbers would have to be at the redlines, 5000 and 10,000 rpm. In the real world, this is the situation with a DC permanent magnet motor.

I don't understand the point of those two examples. And they both have the same gear ratio issue as the first problem.

The point of all three is that given only torque you cannot calculate acceleration, but that given only hp you can. Further, given two different hp values but the same torque values you can calculate that acceleration is greater with the greater hp value. Instantaneous or potential instantaneous acceleration has no utility in the real world challenge of getting to the next light faster than the other guy (unless it is part of an integration).

Oops! Left out a step in the calculus paragraph. You should be able to see what it should be and fill it in, right?

Most vehicle engines produce maximum horsepower before maximum RPM.

The first integral of acceleration with respect to time is velocity. Using that you can tell which car is traveling faster at any given point in time. Integrating power gives work which is not time dependent.

I would be interested in seeing your answer to the second problem.

I was looking at your example problems again. I still don't understand the premise of your argument. Why would my reasoning say the times would be the same?

In example #2 the elapsed time would be 1/2 the elapsed time in example #3. If the maximum speed with a single gear is 60 MPH, then the gear reduction of a 10,000 RPM engine can be twice the gear reduction of a 5,000 RPM engine. With the same torque available with each engine, gear reduction = rear wheel torque multiplication = more rear wheel force = shorter elapsed times. Twice the gear reduction with the 10,000 RPM engine, twice the driving force, same weight (so same mass), half the elapsed time (about 11.5 seconds vs. 23 seconds, assuming 1 foot radius drive wheel).

If we gear both vehicles with the reduction needed for the 5,000 RPM engine to reach 60 MPH, then the elapsed times for both would be the same. Same gear reduction = same rear wheel force = same elapsed time (about 23 seconds).