Torque and Horse Power vs Car Acceleration

I think we see this type of question about every three months. It's difficult to answer accurately unless you are very specific with the question. It helps if you spell out every detail and include as many equations as you are using.

Are we talking straight-line, level travel?

Who is measuring the velocity of the car? The acceleration? The power?

Are we permitted to assume an infinitely variable transmission or do we use a standard set of ratios?

And, it is not exactly correct to say that torque divided by mass gives linear acceleration. I think I know what you mean, but the answer requires some guesswork that might be different for each of us.

So, post back with some details, please.

From a general point of view, torque dictates acceleration.

You are off base when you talk about kenetic energy and velocity, but a good primer on torque vs. horsepower is here.

Sorry Anon. I did a search but probably didn't use the correct criteria.

Background: Readers of Hot Rod Magazine and Popular Hot Rod ask about torque vs HP. As in 'Why do I care about torque? I want HP.' I'm trying to understand the implications of torque vs HP on acceleration so that I can attemp to write something that will put the debate to rest (HA!).

I know that HP is just torque*RPM/5252 and that having the torque at a higher RPM allows either higher gearing or higher speed in a gear without shifting. I also understand that torque is limited by BMEP (<210) and displacement so you can only get so much torque from a particular engine.

What I'm trying to figure out is if some of the power developed by an engine goes to increasing the kinetic energy of the vehicle and therefore takes away from the torque that can be used for acceleration. This affect would increase as the speed increases if it does. Or does only f=ma matter?

The experiment would be for a vehicle, with infinite traction and no drag, to accelerate on a very long, flat, and straight drag strip.

Hello msm98lw:

I agree with tpv45 it's a difficult question to answer. In fact I wish I had the answer so I will be following the thread. It's been my experience that in the real world were continuously variable transmissions are hard to come by for the dragstrip, it is sometimes advantageous to sacrifice some peak horsepower in favor of a broader torque curve. As far as relating either to the kinetic energy of the vehicle, since both can influence the trap speed I really don't see any difference, both are important.

Which is more important depends largely on the choice the vehicle and configuration. A small block Chevy turning 9000 rpm to make 400 hp that is installed in a 4000 pound car with a two speed automatic with a 3.23 gear ratio will make for a very boring ride. Install a big block Chrysler making the same 400 hp at 5000 rpm and there will be a dramatic difference in the quarter-mile times and trap speeds assuming traction is optimized for both.

Of course un-street able high stall converters and optimizing the gearing will close the gap considerably but not eliminated it. Ultimately a high torque big block is very tolerant of less than optimal gearing, so for a dual purpose machine the old adage of nothing beats cubic inches except rectangular dollars holds true.

What I'm trying to figure out is if some of the power developed by an engine goes to increasing the kinetic energy of the vehicle and therefore takes away from the torque that can be used for acceleration.

I'm not sure what you meant to say here, but what you are saying is circuitous. Increasing the kinetic energy of the vehicle is acceleration. All of the torque available (beyond that to maintain a given speed) goes to acceleration. That part in parentheses is critical though, because at some speed, all of the available hp is used to maintain speed (the vehicle's top speed).

That was a GA Ken.

Grazie!

You said:you need area under the horsepower curve.

Well put, I wish I had your talent phrasing concepts, that was what I was alluding to when I was saying it was advantageous at times to sacrifice some peak horsepower in favor of a broader torque curve. I woke up this morning and was going to add an old term we used, broad range horsepower. However it was too late so I had to give you another good answer.

Reading the thread did bring an old question to mind, how do you equate thrusts to horsepower?. I always wondered this when I was watching the hydrogen peroxide rockets or jet cards run. How much horsepower do they have? Hopefully you have a formula for that, I've come up with guesstimate formulas but never anything definitive.

PS. The origin of this question predates the Internet so I am going to google it and see if I come up with an answer, but would still be interested in yours.

It's probably much easier (and has more in common with measuring engine hp) than you might think. I just checked to see I what I put in my spreadsheet to get the instantaneous hp requirement for any condition of acceleration, speed and grade. I might have calculated backwards from tractive effort to torque, and then used the common torquexrpm/5252 formula. But if you already have tractive effort, then you have thrust -- push forward. So then, the hp calculation is simply the thrust x speed/550 (all in feet and lbs and seconds). (33,000 lb/minute -- the definition of one hp is 550 ft lb per second.) That's the calcuation I use in the spreadsheet.

So, the same applies to a jet engine, a propeller on a boat, etc. When a jet is standing still on the runway doing a runup, it is producing no useful work (from the pilot's perspective), so hp equivalents are done at various speeds. (In fact, even when the engine is standing still on a runup, the engine is doing work [just ask a bug that gets caught in the intake stream.] Flight hp is a little like brake hp. If you put an auto trans car on a chassis dyno, and lock the dyno rollers, and then apply some accelerator, the engine is producing hp (as evidenced by the converter temperature) but is producing no "brake" hp.

HP, 33,000 ft lb per minute, does not inherently involve torque. It is a cruel twist of fate that ft lb (a measure of work) and lb ft (a measure of torque) have the same units. They are entirely different concepts, with one requiring motion, and the other not. In the metric system, the units make more sense: you'd never confuse a joule with a newton meter. (Neverthless, I persist in using the US system much of the time, because it offers some convenient shortcuts.)

(If you know a car "weighs" 2200 lb, then you know the rolling resistance, if the Crr is say .010, instantly: 22 lb. In the metric system, the car has a mass of 1000 kg. To get friction, you need a force, so you need to calculate the force [in Newtons] against the ground. On the other hand, doing aerodynamic calculations without using "slugs" in the US system is difficult.)

Thanks for the info, as soon as I get time I will try working a couple of problems and see what I come up with. You might be surprised at the different answers I've come up with through the years. Both from aerospace engineers, and commercial pilots. I've included a link that you might find interesting it containes a rather detailed explanation on the subject.

As you pointed out in your post many times the answer I had previously received involved speed, and altitude of the aircraft, which of course relates to drag. One interesting things I found in the article that may have been a source of some confusion is that a jet engine produces no measurable torque since torque is a measurement of twisting force on an output shaft hence it is far more practical to rate jet and rocket engines in thrust, once a jet engine is mated to an output shaft as in a turboprop it can then be rated in horsepower.

I'm glad to know somebody else uses both English and metric when doing calculations, I'm sure many would cringe if they got hold of one of my worksheets that I flipped back and forth on using whatever was easier. Fortunately we seem to be in good company, as the aerospace industry still seems to use both systems, if you recall the NASA debacle a few years back that involved a failure to properly convert units, I would've hated to be the one responsible as a multimillion dollar satellite missed the planet due to a simple math error.

http://www.aerospaceweb.org/question/propulsion/q0195.shtml

horsepower curve is nonlinear. so different work point gets different situation

my god, ken you use so long words to describle the issue. I admire you, only Sparkstation can match you at it. he wrote llong article, too, for those who ask question.l

in the electrical world, "For a lot of power, do I need a lot of voltage or a lot of amperage?"

my answer is Yes, of cause. either increase v or increase I all both.

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

The spreadsheet has several fun examples. It was written by a friend as a teaching aid. Enjoy.

The problem with all these debates is that, in the real world, torque and gearing will tell you how fast a vehicle will accelerate. Horsepower will not, unless you have a continuously variable transmission, or use some conversion from power to force. Power at the rear wheels is identically engine power (less losses) regardless of gearing. Torque changes with gear ratio.

In all my long years no one has been able to show a formula that uses power to calculate acceleration without:

- converting power to force (e.g., acceleration = power / mass * velocity) and so actually using force to calculate acceleration

- averaging methods of some sort (like changes in kinetic energy).

i basically made the same spreadsheet showing the relation between Power, velocity, displacement and acceleration. What I found and what I assumed was true to begin with, is that as velocity increases, acceleration decreases. I was using my model to see how much CONSTANT power it would take to get a mass of 3120lbm down a 1/4 mile in 11.19s from a dead stop.

If the car did have a continuously variable transmission and could run the engine at peak power the entire run, then the calculations showed that it would require around 450HP at the wheels. This didnt account for losses due to friction, sound and drag, nor did it include stored engergy in the already spinning engine when lauching the car. But it seems to be a safe estimate that it would take far greater than 300WHP to move the car that far in that amount of time.

In all my long years no one has been able to show a formula that uses power to calculate acceleration without:

- converting power to force (e.g., acceleration = power / mass * velocity) and so actually using force to calculate acceleration

Of course, everyone who cares about this stuff converts to tractive force so that F=MA can be applied. If you want to be able to see the effect of a change in one ratio of a gear set, or the effect of induced drag from a down-force wing, you really have to think in terms of F=MA, and you really have to think incrementally (because engine power curves, among other things, do not fit neat equations). Typically, you will need to work from actual dyno curves to get discrete points.

But I think you've missed the point. The classic argument among hot rodders was whether engine torque or engine HP was the key to a fast quarter mile. When this argument was heard more than now, was in the days of three speed transmissions. With these, a highly tuned engine could "fall off the cam" between gears, so some people found that torquier engines (lower HP engines with more torque at low rpm) were faster in the quarter. In recent history however, with more ratios in the gearbox and with variable valve timing, the the car with the greater HP will win, all else being equal. Likewise, in the old days (1970's) a 250 cc Yamaha GP bike with 6 speed box would trounce a Harley Sportster with its "torquey" engine and four speed, because of the better hp to weight ratio and the ability to hold the engine near its hp peak at all times.

Only engine HP can tell you how fast you can do a quarter mile or how quickly you can climb a hill, because gearing simply trades torque for speed (so you can only get high acceleration in low gear). (The argument is not about rear wheel torque, because then there can be no argument -- obviously greater rear wheel torque means greater tractive force, which means greater acceleration. This "argument," is obviously not an argument for engineers who are engaged in doing this stuff, because we can agree on calculation to within a tenth of a second or so.) Engine torque alone cannot tell you how fast you can accelerate, because one engine producing a peak of 100 lb-ft torque at 2500 rpm has half the horsepower of one peaking at 100 lb-ft at 5000 rpm. The engine with the higher horsepower will have the better quarter mile time. Torque does not require any motion at all, so without rpm, it is a useless for acceleration -- and as soon as you put RPM into the equation you have HP.

CVTs are not necessary in the rationale (i.e., it is hair splitting.) Cars with CVTs vs manuals and manumatics with 5-8 speeds are all comparable in 1/4 mile times (unless the engine is unusually peaky) Audis are available all three ways and the differences are slight.

The problem with all these debates is that, in the real world, torque and gearing will tell you how fast a vehicle will accelerate.

Torque and gearing cannot tell you either a rate of acceleration or a maximum speed up a grade. As soon as you mention speed, or a change in speed (acceleration) then you have HP whether you like it or not. Getting back to your CVT, a car will always accelerate faster if the CVT is set for the HP peak rather than for the torque peak.

If you want to get reasonably close, you need to look at resistance (aero drag, rolling resistance, transmission drag) at each speed increment, calculate the tractive force available at that engine speed and gear ratio, subtract the resistances from the available tractive force to get a net tractive force, and then use F=MA for a suitably small increment. If the increment is as large as 5 mph, you can come within a tenth of a second or two tenths of real world performance, for the portion of acceleration during which the drive wheels are not slipping. Once the tires spin, then tire size, type, and brand, driver technique, etc enter in.

Without HP the car stands still.

yes, basically they are arguing that a static force, torque, will win a dynamic race. Anyone who makes this argument hasnt taken statics and dynamics. Since power is derived from torque and RPM, in engine terms, when you have RPM and you have an applied toque, you get power. So the argument should be that area under the torque curve wins races, which of course is POWER, so POWER wins races.

Of course more torque from the get go allows you to get to your higher RPMs, which is your higher power band faster, so a flatter torque curve will provide faster rotational acceleration which in turn leads to greater linear acceleration.

Take the Tesla Roadster for example, the torque curve is flat due to the electric motors from 0-6000RPM so the power curve is linear. Even though the engine makes close to 300lb-ft of torque from 0-6000RPM it only makes 288 peak HP at 6000RPM. This translates to an incredibly fast 0-60 of 3.7s but leaves its 1/4 mile in a "still pretty good, but not great" 12.7s.

Now, if the power were constantly held at 288HP by spinning the engine at 6000RPM and used a continusouly variable tranmission then, just like any engine doing the same, it would accelerate faster because it has more power the entire run.

I think this is the relationship that people miss out on. More power will ALWAYS win IF that power can be used more of the time. More power translates to more energy which translates to more force which translates to more acceleration.

Given two engines, say a 19000RPM F1 engine that makes a peak power of 700BHP but only 200 lb-ft of torque vs a 462ci BBC that makes 550HP at 6300RPM and 550lb ft of torque: Which one will win? At this point it all depends on the transmission. If the F1 engine has a continuous variable tranmission that will allow it to stay at its peak power of and RPM it will win. However, since most CVTs cant handle that kind of power, a more traditional 7speed would probably need to be used and even with the close gear ratios the engine would still have to start at a relatively slow RPM and would not have enough torque to spool up fast enough to reach its max RPM and stay in that high RPM band. On the other hand, the 550HP BBC would also benefit from a CVT to get its max power output the entire run, but would loose to the F1 engine if the transmission were the same. However, because you probably wont find a CVT for either of these applications, and standard 7 speed would be the best choice and the higher torque of the BBC would allow it to get to its peak power and RPM faster, thus giving it faster acceleration and the win.

oops forgot to login on that last comment.

But in conclusion, transmissions and properly selected gear ratios win races. A high HP high RPM engine with low torque but with the proper transmission will beat a high torque, low power engine.

Well put.

Actually, the math is quite simple, but involves three variables, not two. Next to torque and horsepower, you need to take the third variable in account, being speed.

Power = Torque * 2 * pi * rotational speed output shaft of the engine

In general terms, more HP means higher top speed, more torque yields higher acceleration properties.

HP and torque however are no constants for a given engine. They vary in function of speed. Therefore car manufacturers give torque and hp graphs in their catalogue, so you can figure out how you car could perform. A higher rotational speeds, the engine could come short of "breath", thus lowering engine power output. When those losses become substantial, depends on the motor characteristics, intake and outlet design.

Now, to make things a bit more complicated. All cars have multi-ratio gearboxes, enabling you to shift peak power from shaft rotation speed to high or low wheel speeds, thus also manipulating torque values. The amount of power is more or less the same (only gearbox and engine speed losses will have an influence on that). So more power in higher gear, will enable less acceleration than the same engine in a lower gear (unless of course your engine runs at very high speeds where engine speed losses are too large). Max. speed will be less in lower gear than in higher gear for the same reason. So that might look paradoxal, but given the engine speed losses, it's very logical.

So indeed, it depends on what you're after. A surplus on HP at a given speed also gives more torque and thus better acceleration values at that given engine speed (mind the equation above). But you have to recalculate engine speed from your vehicle speed and mind the manufacturers graphs (unless you tinkered with engine tuning, but that's another story).

There are some pretty good answers here. Just skimmed them so I may have missed it. But if we are talking about performance of a car then are we not talking wheel torque? It does equate into linear force of the vehicle but vehicle wheel torque is the output of the drive system.

HP of a rated engine has to go through a driveline with all of it's losses including contact with the ground. The variables to get to the actual work done (linear force) are too many beyond that. So the output torque is the only known. And this is determined by the losses. Lowering losses increases output.

Mass, resistance and efficient applicable torque curve are a different story.

you said that --This assumes no drag or friction.

is quite wrong ! as we all know no friction the car will not move at all, its no other than this friction make car motion.

from this spot, we can deduce their relationship. several steps can be divided into.

from start, accelaeration moment,

acceleration moment,

into constant speed moment,

ok, at the first step, frictin force is variable force from zero to enough make car motion. i.e maximum static friction force. at this moment, air resistance approximate to zero.

the second, the speed is increasing whick is caused by dynamic friction force. and air resistance is added according to speed ( from linear to square nd cube and etc. here we consider its maximum is square term)

when speed is constant, that mean, frichion force is contant and air is constant. the acceleration is zero.

now we can calculate further more their reltion.

how much the friction foce? = mss x friction coefficient.

what will balance thes forces? the anser is torque which is added on the wheels, front or rear. generally black. or both, this doesnt matter.

futher more, where is the torque from? ok, engine give out energy. we can nme it Hp or other power units.

thats all.

in short, we can simply write as this,

friction force + air resitance ) x speed = power -----unit, horsepower,

or

torque x wheel turn sppeed = power.

if any one out there need more detail equation about the car motion , write to me I can show you basic differential equation at different station.

Here are some web sites that I found interesting when I was researching my question:l

http://www.vettenet.org/torquehp.html

http://www.epi-eng.com/piston_engine_technology/engine_technology_contents.htm

From the answers I've gotten so far it sounds like the increase in kinetic energy doesn't take away for the power available for acceleration. It's just f=ma.

When I do my calculations I use 24" tires to make the conversion from torque to force easy.

Each of these sites is at least potentially misleading in several different ways.

http://fis.cie.uma.es/old/docencia/2003-04/C105/links/uwinnipeg/mod_tech/node30.html - sorry, link no longer available

This one takes a physics teacher's view of the situation. Calculating power from energy consumed over time is pretty convoluted, even with the (invalid) simplifying assumptions. If one wanted to calculate a real acceleration curve, then rolling resistance and aerodynamic drag must be taken into account (it is mainly aerodynamic drag that causes acceleration eventually to go to zero). That means that to be consistent in this approach, one would have to measure the energy change in the tires (change in heat, taking into account the radiation, conduction and convection that occurs as the car moves) as well as the energy change in all of the air moved out of the way of the car. This becomes quite convoluted.

http://www.vettenet.org/torquehp.html

This one perpetuates the myth that there is a torque vs hp debate, and even presents "The Case for Torque" and "The Case for Horsepower." Just as you cannot have electrical power without both voltage and current, or hydraulic power without both flow and pressure, you cannot have hp without both torque and rpm. The author eventually gets around to saying the the higher HP LT1 will out accelerate the lower hp L98.

What he says re the Integra (The Integra GS-R, for instance, is faster than the garden variety Integra, not because it pulls particularly harder (it doesn't), but because it pulls *longer*. It doesn't feel particularly faster, but it is.) is simply wrong. The GS-R is faster (in top speed) and quicker (in acceleration) and from looking at the acceleration curves, it clearly pulls harder, not just longer (in other words, the peak accelerations in each gear are greater than in the garden variety Integra.) It also feels faster.

http://www.epi-eng.com/piston_engine_technology/engine_technology_contents.htm

This is better than the others, in my view. It says:

It often seems that people are confused about the relationship between POWER and TORQUE. For example, we have heard engine builders, camshaft consultants, and other technical experts ask customers:

"Do you want your engine to make HORSEPOWER or TORQUE?"

And the question is posed in a tone which strongly suggests that these experts believe power and torque are somehow mutually exclusive.

In fact, the opposite is true, ...

I just scanned this article, but didn't see anything to quibble with.

You wrote:

From the answers I've gotten so far it sounds like the increase in kinetic energy doesn't take away for the power available for acceleration. It's just f=ma.

It's not just F=MA. It is Net F=MA. If F (tractive force) is equal to 200 lb, and aerodynamic drag is equal to 130 lb, and rolling resistance is equal to 20 lb, then Net F is equal to 50 lb. Therefore, acceleration (at that speed) will be one quarter the value predicted if the actual forces at work are ignored.

Yes, 24" tires make sense, although it also makes sense to do these calculations in a spread sheet, so you can change such things and see the effects.

if I would had been aware of your needing such basic formula, I would not have given out the abstruse answer. hehe

Thanks Ken. You seem to have a good grasp of this.

BTW what is that vehicle?

Lew

Hi Lew,

It's the test mule for my Gaia Transport MC2, an ultra high efficiency plug-in hybrid commuter vehicle. (I've had lots of tother things going on recently, so the MC2 progress has been slow, but I hope to step up the pace soon, so that I am ready for the Progressive Automotive X Prize.)

Ken

Hey, I know this thread is old, but I found this looking for some info to back up a discussion I was having with some people on a very interesting subject. A guy on a board had posted some numbers for his car and he would need to make around 450WHP (continuously, and not taking into effect drag or friction losses) to get a 3200lb car to do an 11.2s 1/4 mile. he claims that he can do it with far less, to the tune of just over 300whp.

I have told them that acceleration is exponentially proportional to power if the distance and mass remain the same.

These proportions assume CONSTANT HP, no spooling up of the engine, basically a continuously variable transmission could do this.

P=k*a^(2/3) - when distance and mass are constant

and acceleration is inversely and exponentially proportional to distance.

a=k*d^(-1/3) - when power and distance are constant

these I have solved for from the equation:

P=W/t = F*d/t = m*a*d/t = 2*m*d^2/t^3

I have also used a spreadsheet to solve for acceleration with the P=m*d^2/t^3 equation when m and d, are constant and power is from 100-8000HP and the trendline is the same as the P=k*a^(2/3) equation.

When I solve for acceleration with the P=m*d^2/t^3 equation, in a spreadsheet where power is now constant, as well as mass, but distance is changed from 100-8000ft, I get the same function of a=k*d(-1/3) from the trendline.

So it seems that I can safely say that these proportionalities are correct.

This means that for an engine with a given amount of power, its acceleration will decrease the farther, and faster it goes.

So now I use the simple equation I have stated before to solve for an AVERAGE power required at the wheels to make this run, and I come up with around 450HP, which is far different from the 300HP he is claiming he makes at the wheels. Also the run he did has some interesting 60', 660' and 1320' times where the times follow the exact curve that is predicted by a car with the weight of 3200lb and CONTINUOUS 450HP when solving for time at these distances. He has basically told me how his tranmission is set up to run so that it basically slips the cluth so that he can run the engine at maximum power and RPM for the whole run, thus making a CONTINOUS amount of power to the wheels, no spooling of the engine since it runs at max power the whole time.

He attributes his low HP numbers to the gearing, where the gearing allows him to accelerate faster. But as I understand it, gearing CAN NOT increase the amount of POWER to the ground, just how fast the engine spools up, but in this case, doesnt matter because he is constantly able to run the engine at maximum power.

Am I right to conclude that he is making far greater power than he is leading on or can gearing account for the appearent massive discrepency in power required to do what he is saying?

Hello PIF:

As an educated guess I would say you are right. 11.2 second quarter miles on a 3200 pound car that is hooking up should give him a trap speed of about 120 miles an hour ,so a trap speed would be helpful for your calculations.ET'S vary dramatically with how well the car hooks up, quarter-mile speeds are a much better indicator of horsepower.

My avatar ran in the 11 70s at 120 miles an hour and weighs about 4100 pounds with a non-lightweight driver like me. The old direct connection handbook had a chart to take weight and trap speed and calculate horsepower, by their chart I came out to about 560 hp. I just thought I'd give you some numbers to play with, obviously I needed more tire.

It is quite common nowadays in high horsepower cars (professional cars) to adjust the lockup rate of the clutch, as I understand it this primarily to tune the car to the track for traction purposes. (Excluding top fuel cars) if you think about it slipping the clutch all the way down the track doesn't seem to make much sense, you're either putting the power to the ground or turning it to heat, believe me I know you can turn a flywheel blue in a heartbeat. Launched in third gear one time.

A continuously variable transmission could put the power to the ground without that loss. So my guess is in that weight car he's probably closer to 400 HP than 300 hp at the flywheel.

data from his run

60ft: 1.47

⅛ ET: 7.09

⅛ MPH: 97.7

¼ ET: 11.19

¼ MPH: 119.63

I will see it I can find my old direct connection manual lying around here. I went online looking and found something you might be interested in

http://www.allpar.com/eek/hp-vs-torque.html

From his numbers I would say he is right where a well set up car with optimal traction should be, so I will still guess there's more than 400 hp at the flywheel.

I haven't looked at the national records in quite awhile, but I doubt they've changed much as far as the relationship between mile an hour and ET. Assuming optimal traction, personally my mile an hour was almost constant between street tires and drag slicks, but the ET varied by over a second between the two sets of tires.

The 425 hp RT challenger , depending on what test you read has a trap speed of about 105 miles an hour and weighs about 4100 pounds a rough rule of thumb is a change of 1 mile an hour for every hundred pounds so of Chrysler's not lying, if you lightened the challenger up by 1000 pounds it still wouldn't do better than 115 miles an hour through the quarter with 425 net installed horsepower.

http://www.edmunds.com/insideline/do/News/articleId=130809

Cool, thanks for the info.

I figured that the car must be putting down close to 450hp at the wheels to do this with my equations. It also seems conceptually valid because the equations show that for a given HP, i.e. the engine is running at maximum HP and with a CVT, acceleration will decrease as velocity increases according to a specific curve, that can be theoretically estimated with a spreadsheet. The curve shows an exponential decay in the acceleration as distance increases, and since velocity is increasing as well, this means acceleration decreases as velocity increases.

In my assumptions, I assume an AVERAGE HP and try to solve for that, based on an AVERAGE acceleration, even though I know that acceleration is not average for an IC engine on a car with a 6 speed transmission. But am I right to assume that an engine running at this calculated AVERAGE HP with a CVT, so that it would be capeable of running at this AVERAGE HP, is going to be the MINIMUM HP needed to accomplish such a feat? Since the engine with a 6 speed will at some points operate below this AVERAGE HP, it is safe to say that at some points, it will need to operate above the average, in order to get to the average.

So, does this make sence, or am i making a fatal flaw somewhere?

oops, I forgot to login, that last comment was from me.

thanks

I think you're on the right track, ken has posted a spreadsheet that you might be interested in if you haven't tried using it yet. I'm sure if you put enough variables you can accurately predict horsepower versus performance.

I don't think any of the spreadsheets take all the factors into account but they should get close. I'm a little weak on spreadsheets but I've always said empirical data outweighs theory.

Factors that have to be considered that may not be in spreadsheets besides the ones that you've already addressed that skew the results. Vehicle weight.

An extreme example would be a motorcycle acceleration curve versus a heavy car, a 500 pound motorcycle, for the moment assuming all other factors are equal i.e. horsepower under the curve ,will accelerate faster 4000 pound car with an equal horsepower per weight ratio. I'm sure somebody will disagree but it's a fact.

Another factor obviously it's air drag off the top of my head it increases at the cube of the velocity, real life it doesn't mean much below 60 miles an hour to drag car but you know you're moving some air at 120 miles an hour, so obviously that has to be factored in to any acceleration curve obviously again, this is less of a factor with more aerodynamic vehicle .

Before the advent of constant variable transmissions many times it was advantageous to to select the gear ratios in the transmission there were less than optimal for the first to second and second to third gear changes on faster cars, the reason being as you previously referred to the car would be accelerating faster, with less air resistance at the lower speeds, and it was more important to keep the engine and its peak performance range at higher vehicle speeds.

Reducing rotating mass, all else being equal , using the lightest rotating assembly i.e. crankshaft rods Pistons will make a significant difference in acceleration in a drag racing vehicle. And virtually none at high and relatively constant engine rpm such as in a NASCAR vehicle. I don't have any accurate figures, but as an educated guess 25 to 50 hp more might be required to produce the same acceleration rate from the more robustly built engine, than the bonsai drag racing engine. I should express as a percentage, but I'm referring to the engines I am familiar with in the 550 to 600 hp range.

Hm, good info. I was wondering, do you know why it is that the motorcycle accelerates faster than the car? Im not disagreeing or second guessing, i just dont have any knowledge as to why that is and would be cool to know.

I guess for me it really comes down to being able to express this, not so much with math, which I think i have done, but conceptually. If a 200HP motor can lift an object vertically at a constant velocity, then it is producing a constant amount of power. So the only way for the motor to get the object up quicker, would be to decrease the amount of acceleration it is overcoming, thus allowing it to raise the object faster, if say a constant external force was applied below it to help raise it. So in my example, as the cars velocity increases, its acceleration must decrease if the motor is producing a constant amount of power. this seems conceptually and mathematically correct.

These may help:

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

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

The reality is that horsepower per se has little to do with acceleration. The spreadsheet at the first link explains why I say that. I've never seen a formula that calculates acceleration directly from horsepower. The ones that have been presented are either estimation methods (like changes in kinetic energy) or have a conversion from power to force [as in (acceleration = power / (mass * velocity)]

The Sportrider article is pretty good for showing the differences in ideal shift points from bike to bike: some benefit by winding to redline, some do not. This, of course, depends upon the shape of the torque curve and the gear ratio spacing. With the RC51 you wouldn't need to think at all about up-shifting -- wind it to redline in each gear. But for the R6, each change has a different optimum point (with the range being 1500 rpm). (If I were racing an R6, I'd want a shift light programmed to come on at the right speed in each gear).

That is different than the issue asked about by msm, though : "Which determines the acceleration at speed torque or horse power?" He provided background in post #3, "Readers of Hot Rod Magazine and Popular Hot Rod ask about torque vs HP. As in 'Why do I care about torque? I want HP.'

Obviously, interpreting torque to mean "rear wheel torque" begs the question. Rear wheel torque relates directly to tractive force which relates at any instant to acceleration.

While you are correct that everyone calculates acceleration by F=MA, and that the F is net tractive force (which must be calculated from engine torque), the only way that you can have a net tractive force is if you have more horsepower available than required to maintain a constant speed. We all agree that the top speed of a vehicle is determined by horsepower. A vehicle's top speed is simply that point where acceleration ceases because the hp consumed matches the sum of the drags (a force) times a speed. (Force x speed = power)

In the motorcycle world, there are bikes that answer the classic argument convincingly. The Buell 1200 has 84 lb-ft torque and weighs 395 lb. The GSXR 1000 has 80 lb-ft torque, and weighs just a few pound less. The GSXR, however is far quicker in a quarter mile and essentially out of sight in a half mile. The difference is that the GSXR has about 75% more horsepower, because its torque peak is at a much higher RPM (8500 vs 6000) and usable torque is maintained to much higher rpm. The 600 cc crotch rockets also out accelerate a Buell, despite having far lower torque, but only slightly less weight.

My 50cc motor scooter has the same torque (5 lb-ft @ 5500 rpm) as an old Kreidler racing 50cc motorcycle (5 lb-ft @ 22,000 rpm) The Kreidler produces 4 times the horsepower (20 vs 5) and accelerates very quickly through 60 (on its way to over 100 mph) whereas my scooter has leveled of by 48mph. Same engine torque, dramatically different acceleration rates. To accelerate faster from 48 mph (faster than zero g) my scooter needs more hp, not more torque -- because the Kriedler shows me that its 5 lb-ft is plenty for good acceleration at 48 mph.

This problem is partly in the question, and in linguistics. As soon as an engine turns you have HP. An analogy is in electrics (and I may be repeating something that I or someone else has said) to do any work with a motor you need both voltage and current. HP is torque x rpm, so the two cannot exist independently.

The Google spread sheet, incidentally, is misleading in its commentary. For example it says: "The engine in the second case would have to turn just 13,558 RPM to match the top speed of the first case. With a 15,000 RPM redline the gear ratios could be even lower (higher numerically) yielding more acceleration with less HP and torque but still attaining the same road speeds."

This is completely incorrect. Maximum road speed depends only on HP. This should be obvious, because maximum road speed is a direct expression of HP: it is a force (that required to overcome all the drags) times a speed, therefore a power. Redline is independent of the whole mess, and is sometimes close to the hp peak, and sometimes quite a ways beyond it. If only all you had to do to make a car go faster was to change the redline -- you could make a killing in selling stronger valve springs.

The Google spreadsheet is also completely wrong in its calculation of accelerations, because it ignores the fact that at higher speeds more and more hp is consumed in overcoming ever increasing drag.

A 100 hp engine with a 10,000 redline (with relatively low peak torque) will produce exactly the same acceleration as a 100 hp engine with a 5000 rpm redline (with twice the peak torque) provided each is appropriately geared (e.g., each with CVTs).

To elaborate, the two will experience all the same drags so we will ignore them. We know the tops speeds will be the same. That means (if we assume peak power at redline in each case) that the primary drive ratio leading to the CVT will be twice as high (numerically) for the 10,000 engine as for the 5000 rpm engine. The input to the CVTs (torques and speeds) will therefore be identical. Thus the acceleration performance will be identical, because the HP is the same, even though the engine torque is different.

Put most simply, from only an engine power to vehicle weight ratio you can determine 1.4 mile times (while making assumption for the many expected things). From an engine torque to vehicle weight ratio you cannot make this determination.

Hello Blink:

My initial intention was to supply some old world trivia to the discussion, along the way I've managed to confuse myself.

I had previously read an article stating that the air drag increased at the cube of the velocity, I am now reading that it increases at the square of the velocity, the square is probably right but it sure feels like the cube of the velocity. So that part would be a question?

My previous experiences would be comparing like vehicles i.e. Cars to cars, pickups to pickups, motorcycles to motorcycles there would be a negligible difference in the sub 100 mph range as far as your spreadsheet calculations. After 100 miles an hour things change quickly.

In the past, excluding economy cars, cars typically had a superior power to drag ratio as opposed to motorcycles, motorcycles often had a superior power to weight ratio.

Of course this is changed in recent years however the effects of air drag were dramatically demonstrated in the seventies to a friend of mine who was driving a modified 750 Honda which had a 836 kit installed in it and was probably a high 12 second motorcycle.

He had the misfortune of engaging a Pontiac Bonneville in a series of high speed role on's starting at 70 miles an hour the motorcycle easily thrashed the Pontiac, the last role on was started from the mid 90 mph range and after a brief standoff " I watched the Pontiac float down the road and leave me in the dust"

The handling characteristics of my avatar change dramatically with a 20 mph increase,at 120 miles an hour it is well behaved and completely predictable. The first trip into the mid 140 range resulted in floating down the road like a Cadillac, barely controllable,a sufficient increase in force from the air coming through the hood scoop and grill to raise the hood approximately 3 inches at the center from its normal resting position.

Most recently, though I have frequently cruised in an upright position in 100 mph range on my ZRX1200, I arose from tucked down position while doing a test run at approximately 140 miles an hour to check my mirrors (making sure it wasn't a highway patrol car going the other way) let me just say it will get your attention and you'll grab the handlebars and hang on for dear life the force was incredible. So I'm probably wrong but it sure feels the air drag increases more dramatically than at the square of the velocity.

Any rate besides the useless babble the horsepower to air drag ratios have changed on motorcycles dramatically I found a link that you might be interested in looking at.

http://www.sportrider.com/tech/146_0106_aero/index.html

Motorcycle aero is interesting. "Dustbin" fairings were outlawed years ago, and as a result, road racing fairings are ineffective, because they do almost nothing to deal with the predominate cause of drag (once the front of the bike is reasonably rounded), the wake.

It is possible to get a fully enclosed motorcycle down to a Cd of less than .20. (My proof of concept prototype started off as a fully-enclosed two wheeler, in fact, with a goal of sub .2 Cd. The Aptera started out advertising .06, which is utter nonsense -- their claims have now migrated up to .18, IIRC, about reasonable). The typical crotch rocket has a Cd of around .6, so there is a lot of room for improvement.

In racing, I would experience the effect you noted with your bike, several times per lap. When in a tuck, if you come off the tank just a little, your upper body floats, almost weightless. Then as you sit up for a turn, you get a strong air brake effect.

For some shapes, the change in drag with speed does not vary precisely with the square of speed, (because the Cd changes with Reynolds number, which changes with speed). But for ordinary calculations, the square works well, and if it were not for the fact that the Cd changes, it would work perfectly. For most reasonably-streamlined shapes of car size moving at car speed, the changes in Cd with speed are slight enough to ignore, given all the other variables involved.)

The cube shows up in discussions because the hp consumed goes up the cube: if you double the speed, you not only have 4 times the drag, you are also doing twice as much work per second because of the greater speed (even if the aero drag had not changed).

A "mere" squaring of force can be pretty impressive, though, as you have seen. Your upper body, sitting up, has a Cd of about 1, so you are absorbing a lot of HP -- and horses are big animals. When you crouch, you both reduce your frontal area and also improve your Cd -- so when you sit up, you are getting an impressive double whammy.

Good sport rider article, btw. Thanks.

its been TOO LONG since fluid dynamics!

Well thank you for clearing things up, I had to delay my post until I had a chance to do a practical test. It all makes sense now.

The last time I was in the 140 mph range was in 1982 on a modified GS 1100, which to despite the exaggerated claims of 125 mph 750 Hondas was one of the first production bikes that could be relatively stock and reach those speeds. It also had no fairing.

My ZRX1200 has a small cafe style faring after doing some freeway ad hoc airflow I would guess that the ferrying in a tuck position and the shape of the helmet was probably even putting a little down force on the helmet, big change from having your head into the wind.

So I guess 25 years,( you will probably conservative concerning MY Cd it is probably closer to 1.2) and a extra 60 pounds is what led to my surprise.

You probably already know the rest of the trivia . I've been discussing the situation with a friend of mine that remained more involved involved with high performance motorcycles than I have. At one point he had a turbocharged Hayabusa, leading to an expensive lesson in melted pistons.

The trivia, one of reasons supposedly for going to be under tale exhaust on some models was an attempt to reduce the Cd by filling the void similar to the technique used by some military artillery rounds, with exhaust. Apparently not worth the trouble.

I forget what class of road racing he was referring to that it is a common tactic for the riders to arise from their tuck position when approaching a corner to use aero breaking to help preserve the brakes, he was talking about speeds in the 170 to 180 mph range, I can only imagine the forces they encounter.

You're absolutely correct motorcycle aerodynamics is a fascinating subject. When I was actively drag racing my 900 Kawasaki in the mid seventies if anyone had ever told me that stock motorcycles, in the future could be approaching 200 miles an hour I would've checked to see what they were smoking.