You cannot calculate drive wheel thrust from engine torque without .. its RPM and car speed (or even RPM and transmission ratio.. same you introduce power). If you have to know both, actually you calculate it from power.

Don't care. You're right. Honor's safe.

How do you know your engine will produce 100 lbs ? Because you should know the speed of the wheel and then the RPM of the engine and then its power. You could use a different engine with different torque/rpm couple and a different gear, but same torqueXrpm (as known as ...power) you will have the same force. So we don't care the engine torque, the prevailing factor for force is well the power. It's just that complex.

Thank you.

No, not at all. You do not need power or rpm to calculate wheel thrust at any point. Ben showed you how. I showed you how. For example: If there is x torque at the flywheel and the total effective gear ratio is 3:1, there will be 3x torque at the drive axle -- regardless of the power, rpm or speed at which it occurs. Thrust at the drive wheels is at all times proportional to torque at the flywheel.

Gear reduction cannot multiply power. It can only multiply torque, and can only do that by dividing speed. In first gear there is several times more thrust at the drive wheels than in top gear because there is several times more torque multiplication. But at any given engine rpm in first gear or top gear, power at the drive wheels remains the same. Gearing cannot change it. It can only exchange torque for speed or vice versa.

To calculate the acceleration of a car in a given state you need to know either power and speed or rpm/torque/effective gear ratio. The information is completely equivalent. rpm/torque is power and rpm/effective gear ratio is speed.

The 'power sells engines/torque wins races' aphorism comes (IMO) from the days when engines had two valves/cylinder, push rods, mechnical ignition timing and carbs. Increasing power while retaining 'driveability' is very difficult. An engine that compromises peak power is probably faster than one that concentrates on the rpm range near peak power and won't run cleanly lower down.

With four or more valves per cylinder, overhead cams and engine management the rules shift. If you watch an F1 onboard with the data overlay you can see that they run the engine down to 10000rpm or less in slow corners, run cleanly round to 19000 then shift with an rpm drop to 17500. The engine has enough power at 10000 to trigger the traction control coming off a slow corner, and can still pull cleanly and smoothly up to max power. And we have 7 gears to play with.

So everybody is right. It just depends on your engine restrictions. But ideally we go for absolute max power, because that gets us down the straights in the shortest time.

Paul

What he said.

do you have copyright on your single formula you really gave to me ?


What it's wonderfull with you is that the car doesn't seem to roll. So let me know your formula. I'm quite desesperated to read it. But it's more easy to copy/paste Ben. By the way, Ben used RPM to and gear ratio is in his equation. Your examples of torque at wheel did have no applications, no variables. They're ridiculous. Their utility is null.

You determine the torque of the engine without its RPM of even the speed of the car and the gear ratio, great !

What is YOUR formula of the torque at wheel function of the torque engine (torque curve provided) ?

The simplest, "mine", is still Force = Power / Speed. But you surely already know it before i wrote it.

But before to "publish" it here it was impossible, according to you, to exprime a force from a power.

Try to find a happy end for you alone. I don't care.

"You're wrong
I'm right".

That is the first law of Mc Guire in physics.

You cannot miss it. Each message recalls it at beginning.

The compromise has never changed. Virtually any engine can be made to produce more high end power. It's simple to do: greater port volumes, more valve timing. But that increase will come at the expense of torque everywhere else in the range. Top end power is of little use if the engine doesn't have enough torque to pull it there.

As you say, modern F1 engines have impressive output curves. Well, they were not developed by going for the "absolute max power" to which you refer. That is not how engine development works.

I admire your tenacity and perseverance.

As if you don't have enough opinions, torque is about two things. Cylinder filling at a certain rpm, and cylinder pressure.
You can increase cylinder filling by having sufficient size valves and ports for a certain rpm. At the same time, you want air speed going into and out of the combustion chamber as high as possible because of the ram affect of moving gases. Rod ratio affects this, as well header size and length, intake length, cam profile, and lots of other stuff.

(I did not write this myself, but I think it's a good eksplenation)

How can torque push a engine up to speed? you nead horcepower to do that.
Torque is a none moving fysical size.
angular velocity x Nm = kW

The below is a exsample to bring it to a easy understandabal level.
1kW=1kNm/s=1kJ/s
If a car produce 200Nm it sound like a nice torque but what if we made it into the same as kW?
it would be 0,2kNm and the engine produce maybe 120kW at the same RPM.
All the suden the torque look less inpresive.
it would represent a mass off only 20,4kg to compare it to something that is easely understod.
I can lift that with one hand, but I would not be able to acelerate a motorbike up to more then 15km/h

If the torque is 200Nm at 4900RPM and power is 130kW at 5500RPM I would defently belive the gearbox will deliver more force if the input shaft spinn at 5500RPM than at 4900RPM if the ratio is so that the output shaft will spin at the same speed.



How are you planing to get more high end power if the torque drops anywhere else in the range???

angular velocety x torque = power

Good luck making that hapend.

If you want to increase power the torque has to increase.

There is an upper bound on torque governed by the fuel and the effective displacement. IIRC anything more than 90lbft/litre is unlikely on pump fuel.

Paul

McGuire,
About two years ago the following was posted by you:
quote:
--------------------------------------------------------------------------------
Originally posted by WPT
McGuire said , "In any gear, maximum acceleration occurs at the rpm of peak torque. "

True, but at any speed maximum acceleration occurs in the gear that delivers the most hp.
WPT
--------------------------------------------------------------------------------




Sorry, I don't know quite what you mean. Gearing cannot multiply horsepower. Gearing can only multiply torque. If you are stating that the top of each gear must be selected to match max hp rpm in order to achieve top speed for each gear, that is quite correct. But that also means that +/-100% of the acceleration in each gear occurs BELOW the rpm of peak hp.

Meanwhile, max acceleration in each gear occurs at the rpm of peak torque (subject to the apparent exception previously discussed). That is when the maximum force per rotation of the crankshaft is being exerted. Yes, we can make more power by turning the crankshaft faster than peak torque rpm (to get more more revs per unit of time) but maximum force per crankshaft rotation is exerted at the rpm of peak torque, by definition. And again, gearing cannot multiply horsepower, only torque.

Do you still hold to this reply? WPT

I appreciate your magnanimity.

I'm not sure I follow you. We appear to be saying the same thing for the most part, and I can't possibly have that right.

If you are also saying that a proportional increase in torque is necessary to get a max power increase, not necessarily. Beyond the torque peak, torque is already falling by definition. With port volume you can get more gross breathing capacity, extend operating rpm and squeak out a power gain, at least on the dyno, but it will kill the flow velocities and thus the output curves. Or, how to build a light-switch engine.

Yes I do, though I suspect you have some heavy-duty lawyering in mind. The four graphs I posted here earlier explain it completely as far as I am concerned. Do you have a problem with them?

But clearly this stage isn't often reached. Current 2.4l F1 isn't 'light switch', for example and they're probably at an extreme of power/litre for naturally aspirated on petrol.

Torque is bounded by the type of fuel and the engine's displacement. You might get similar max values for very different power outputs and rpm ranges.

Power is bounded by how much air the engine can pump, which crudely depends on valve perimeter, displacement and rpm.

If we were running simulations of engine characteristics around a circuit the winner would probably be the one that burned the most fuel, generated the most energy, where the integral of power against time was maximised for the speeds and gears available. Direct considerations of torque aren't too helpful in grasping this.

The whole compromise of what makes a winning race engine is too big to fit on this page. Which is why we have arguments about 'torque' even among people who understand the basic physics. Those who don't understand the basics are in trouble.

Paul

If the torque stay the same and you increase the RPM the power will just droop after you reatch the previously peek hp.
If a peek torque is increased or lets say you have peek torque at 200Nm at 5000RPM and 190Nm at 6000RPM. If you get a 195Nm at 6000RPM and the RPM at 5000RPM will drop to 190 you will have a increase in power. You lost some torque on the previously peek, but you where able to move the peek from 5000RPM to 6000RPM. The peek torque is lover, but it's higher at a higher RPM.

Doos a honda CBR600RR have more torque than a HD with 1200?
If you add weight to the Honda it will lose a aceleration test?
HD 1200 79.0 ft. lbs. @ 3500 rpm VS Honda CBR600RR 49lb.ft @ 11,000rpm

Acording to you the honda would never have the torque to push it to max hp at 117bhp @13,000rpm, but the HD would.

Wow, Stian, it's really hard to understand what you're getting at there.

This is all a matter of degrees. If you have an engine that performs well with modest port sizing and then decide to 'make it really good' by putting drain-pipe sized ports in it, well, you will run into the exact problem(s) that Mac is talking about. Of course, that's not the only way. There are other situations where relatively small changes to valve seats, cam timing, intake trumpet length, or exhaust configuration will not effect the torque peak, allow a power gain, and not introduce any unwanted qualities to the engine. It's harder to do, but it does happen. If you are in this situation, you should understand that it is a gain. No, you have not changed the peak efficiency of the engine, but you have increased it's total output. However you want to measure it, you have made your racecar faster. It's nothing to gloss over.

I really don't feel that these types of particulars have much place in this conversation, though. I'm approaching this discussion from the standpoint of what is optimum. When we have 'optimum' defined, then we can look at each particular situation and determine what is realistic (generally a little more conservative than optimum). Mac noted earlier that it's easy to outrun a Honda S2000 if it's in the wrong gear. I can outrun one on foot when it's shut off. What does that matter? Let's look at optimum first before throwing a bunch of special situations in the mix.

I think what we've shown here is that you can calculate a car's acceleration by using force method based off the engine torque and overall gear ratio. You can just as well calculate acceleration using power without ever looking at the actual thrust at the contact patch of the tire. It can be argued that the thrust method gives you the advantage of seeing potential wheelspin, but that is (at least) a three dimensional graph with lateral acceleration being the dimension running 'out of the page' with anything I'm intersted in. I don't see that as much of an advantage. Using the power method does allow you to easily calculate the gear that will provide maximum acceleration at any given speed. This can be helpful when choosing gearing.

In terms of shift points and gear selection, the two methods ultimately arrive at the same answer. Mac's graphs are valid and (I'm assuming here) correct. This is not the only method available, though. I think most of the frustration in this thread is Mac's refusal to acknowledge that there are more perspectives to this issue than his. He can acknowledge it or not, at this point, we all know it's true.

Honestly, I think we have this issue pretty much covered.

That's the point.

You don't choose the speed of your car. This is an external parameter. But you can choose the gear. That makes you have to choose a gear that provides from the engine its maximum power. F = P / S

Best Shift points can be resolved according to the torque curves and gears ratio using 2 method approches (general and numeric) . This is mathematics, not a sweet talk.

No, not at all. I've said from the start and all along that the calculations can be performed from either perspective. (You have done your best to suggest otherwise hmm.) However, behind the calculations there are real physical properties at work that are not interchangeable.

OK then. Whew! Glad that's done with.

I didn't remember you writing what Engine Guy quoted, but there it is. What don't we agree on?

McGuire,
You say that for two years now that you have maintained that the thrust force can be calculated by using torque method or power method. Do you agree that the correct equation for the power method is; F=P/S (here keeping same representation as GSX-R, F=thrust force, P=power, and S=speed)? Using this equation, is it not immediately obvious that my statement is true? And, torque method would give the very same values for F, we all agree. So why the argument about gearboxes multiplying torque, but not horsepower? One sees immediately from F=P/S that F is inversely proportional to S. If this was your understanding two years ago, why didn't you just say 'Your right'? Or, do you think I'm still wrong?
After I made clear what I meant, you posted the following:
quote:
--------------------------------------------------------------------------------
Originally posted by WPT
I mean that at any given road speed the max acceleration is given by selecting the gear that delivers the most hp to the drive wheels at that given speed. WPT
--------------------------------------------------------------------------------



No. That is a common misconception, but one that still never fails to surprise me when I hear it stated by knowledgeable auto enthusiasts. Anyone who can drive a stick-shift car knows this is not really true intuitively (if not conciously). Acceleleration is greater at the bottom of each gear, while speed is greater at the top of the gear. When max engine rpm and vehicle speed in a given gear are reached, we know in the seat of our pants it is time to upshift to the next gear. The statement also includes a logical paradox. We know that every gear has a top speed, and that top speed in any gear occurs at the rpm of max horsepower. If the vehicle is already at top speed for that gear, how can it also be increasing in speed at the same time?

In truth, the only way to accelerate the vehicle from that point is to upshift to a higher gear, at which point engine speed can only drop below the rpm of max hp. We have increased the number of wheel rotations per crankshaft rotation in order to increase the vehicle's ground speed relative to engine rpm. How great the rpm drop is a function of engine torque, the number of gears in the box, and the spread between them etc. But this we know: if the engine is truly at max hp the vehicle is not accelerating in that gear. It's done accelerating.

Top speed occurs at the rpm of max hp, while maximum acceleration occurs at the rpm of peak torque. By definition, peak torque rpm is where maximum force at the crankshaft per unit of crank rotation is obtained. This force is multiplied through gear reduction and the tire's loaded radius to produce a quantity of wheel thrust, which accelerates the vehicle in real time. Where does hp fit into this? Horsepower doesn't exist in instantaneous time -- it's mathematical artifact, representing the amount of torque the engine can produce in one minute. Dynos can't measure it either. They can only measure torque.

Top speed -- maximum distance covered per unit of time -- occurs at the rpm of maximum horsepower, because that is the crank speed where the engine produces its maximum torque per unit of time. And so we gear the car for max hp rpm, not max torque rpm (and then put a bunch of change speed gears in the box to provide torque multiplication so we can accelerate the vehicle too.) If we gear the engine 2:1 (2 crank rotations to 1 axle rotation) we will double the torque, but not the horsepower because we have also reduced by half the number of axle revolutions per unit of time. Horsepower remains the same. Torque * rpm/ 5252 = HP. Double the torque, cut the rpm in half and see what happens: Gearing can only multiply torque, not horsepower. Horsepower is simply torque over time. (5252 is 33,000/2pi. One hp = 33,000 ft-lb/min. of work.)

McGuire, do you still hold to this post? WPT

Think one second, top speed, (except ideally for last or pre-last gear), doesn't occur at top HP but after RPM of top HP (straight line, no slope). When you drive and change gears optimally to get the max acceleration, you can have to exceed top HP RPM and to never use the max torque RPM. Don't make too many "nodes" at your brain with this subject.

No paradox at all.

Which also magically corresponds to (X) torque *at* the rpm of max horsepower!

So once we know the general area of peak HP @ (X) rpm and use that point as a quick reference of where efficiency is dropping off faster than HP is increasing....why would we need to know the hp anywhere else in the powerband? We have the torque values at each rpm data point....simply fatten the torque numbers including the area around the quick reference point and were good to go.



John

ahahaha how funny

You tell me. You have been arguing about it for three years.

Problem solved by Stian's answer as far as I'm concerned (in the F1 equivalent thread).

Amazing how when you try to be nice to some people they can't help but be rude back.

http://www.cosworth.com/download.php?file=...688fbc1cb7072c5

Interestin reading on the second page off this pdf document.

The link is bad.

Place a bicycle front wheel against the wall. Step on the pedal. Depending on your weight, the crank length, and gears, some amount of torque will be applied to the rear wheel.

In my opinion, over-revving or even changing the engine-speed too much is sometimes foolish.

I was once given an explanation about engines and was told that torque is only effective in a certain range. I am not a technical person, but I was told that it is called the Effective Torque Band. Can any of you propeller-heads give a brief explanation of how the different characteristics of V8 and V10 engines maps onto cornering speed and lap times?

If this question has already been asked, please point me at the relevant thread.

FAT BOY "I did simulations to show that the actual acceleration of the car based on these engines was not going to be a based on the engine torque, but engine HP."
--------------------------------------------------------------------------------

MCGUIRE "Not possible. The acceleration of a wheel-driven vehicle is resultant to the thrust available at the drive wheels, obviously. The relationship between engine torque and hp relative to engine rpm is cast in stone. It is a simple matter to calculate drive wheel thrust from either power or torque. Both will produce the identical answer. If they do not, you are doing your sums wrong."

"Sorry McGuire but you are wrong here. Power and vehicle speed are all you need to calculate traction force available for acceleration. Torque and vehicle speed does NOT allow calculation of traction force UNLESS you also know the engine speed or the gearing (either of which will also allow you to calculate the POWER which is the engine property which determines the acceleration available)"

Wow, talk about bringing a thread back from the dead! Curiously, I was involved in a heated discussion on the same subject on a different forum where I'm active that was raging at about the same time as this one.

In that discussion as in here, most people had a piece of the puzzle to correctly answer the question, but it took pages and pages of back-and-forth to reconcile small differences in semantics. In short, McGuire is correct that torque at the wheels accelerate the vehicle through a=F/m, F=T/r, but referred back through gear multiplication back to the engine, it's the developed power that determines the rate of acceleration at the wheels. Torque by itself, ignoring speed, is meaningless other than to get a vehicle rolling from a standstill. The discussion about torque curve and shape of the powerband are all contributory, but again just pieces to a puzzle.

No you di-int. You so did not bring back this topic. (somebody hit this kid with a bat. We can dispose of the body out near the lake.)

Of course you can calculate vehicle thrust from torque without reference to power, engine rpm, or vehicle speed. Watch this:

drive wheel torque/rolling radius of tire = drive thrust on pavement

It doesn't get any more essential than that. Put a load cell on the axle to measure the twisting force (torque) on the shaft and you will always know how much drive thrust is available at the wheel at any moment, regardless of vehicle speed, engine power or rpm. This force is the only thing propelling the vehicle forward. Power is a mathematical expression of that force as work acting over time.

I can agree with most everything you said except "power is the property which determines the acceleration available." Torque is the actual physical property allowing your statement about power to appear true. Force, a vector property, that which causes a displacement. Acceleration, a vector property, rate change in velocity. Power is a scalar quantity, work x time. Speed is a scalar quantity too, dimensionally distance and time. To accelerate a body requires force. The force to which we refer here is torque.

But didn't Williams tested precisely such a fancy CVT in 1993 ?!

Here's the video:

http://www.youtube.com/watch?v=x3UpBKXMRto

must...resist...compulsion...to...respond...prying...fingers...from...keyboard...

[QUOTE]Originally posted by McGuire
[B]

I can agree with most everything you said except "power is the property which determines the acceleration available."

Sorry about the resurrection - couldn't resist. You cannot detrermine acceleration available at a given road speed if you only know ENGINE torque. On the other hand you CAN determine acceleration available at a given road speed knowing only ENGINE power. Engine torque is NOT the determinant of acceleration available at a given speed - you can't even CALCULATE the acceleration without also knowing of either engine speed or gearing.

In short - if you know the torque (and therefore also the power) curves of your engine and you want to maximise acceleration at a particular road speed, you select the gear which allows the engine to produce the most power.

Yes, that's the correct and mathematically iron-clad theory behind it. I'm actually pleasantly surprised that a correct consensus has largely formed on the Internet on this issue, given how common the myth of the peak torque number being meaningful to a driver is among car enthusiasts just knowledgeable enough to be dangerous. Just half a dozen years ago Googling this subject would return you all sorts of tripe, and unfortunately this forum itself was not the exception.

Trivial and circular. Independently of speed, horsepower is meaningless. Independently of torque, horsepower does not exist. In itself, horsepower is not a physical determinant of anything. It's a very useful calculation.

That's a pretty stupid remark, since torque is just a handy calculation as well, the actual physical phenomenon being a force vector which, if infinitely extended in both directions, doesn't intersect rotation axis.

An example: You have a force and a distance from the rotation axis (pictured two dimensional in a crankshaft as the line you can draw between the conrod centerline to the rotation axis, being perpendicular to the conrod centerline), you can keep working with these 2 data points (and convert them to what you need as well as torque) or you can combine those 2 and call the result 'torque'. No need to do so, but it makes life easy.

Now when you want to know how much work can be done with your force you need to introduce a new variable: time. Same story, you can keep both seperate and calculate with time (rpm) and torque or you can yet again introduce a unit called 'power' which is the result of joining torque and time. I seems the concept of power appealed to some people at least so it stuck...

Disagree strongly. Power is a FUNDAMENTAL PROPERTY of all heat engines - torque is not. Some heat engines dont produce torque at all (jet engine, Peltier (and yes electricity is energy of the highest quality - as good as work)). A heat engine is something you put heat energy into and get work out of - plus some surplus heat as waste heat. The rate at which you put heat in (fuel in this case) and the rate you get work out are both expressed in POWER units.

Producing POWER is THE PRIMARY FUNCTION of any heat engine.

You also skipped over a very important point in my post.

If you tell me the engine power available and the vehicle speed, I can tell you the force available at the tyres for acceleration.

If I tell you the engine torque available and the vehicle speed - you CANNOT tell me the force available at the tyres for acceleration.

Mac

Don't give in to your compulsion to respond.

It is a waste of time to remind people that HP is found on a dyno by measuring torque and multiplying it by RPM and dividing by a constant.

Or that motion of the vehicle is being impeded by the aero and friction forces and if the force at the rear wheel/road patch is equal to the aero and friction forces then there is no + or – acceleration.

Or that the +/- acceleration is the balance of engine derived/brake derived forces minus the sum of aero and friction forces divided by the vehicle mass.

Or that the + force at the wheel/road patch derived from the engine torque multiplied by the overall gear ratio working through the radius of the wheel/tyre system.

Keep prying your fingers from the keyboard. Any other alternative is only to keep this dynosaur being resurrected. It was happy as being dead as in the past.

Regards

Sigh... So where on the dyno is the torque sensor exactly? Torque is just as much a calculated value as is HP... HP just factors in one more variable.

No, torque is a real physical property independent of time or speed. What is the force that accelerates the crankshaft?

An example for the honourable doubting members to solve before us would illustrate the point quite nicely. The tyre rolling diameter would need to be given, but gear ratios not.

What is a "real physical property" anyway, and how is it of any consequence to the question at hand? I just want to know which gear to use to go fast, how is knowing what is and isn't a real physical property going to help me?