Your contribution does not warrant comment to be honest. "Torque doesn't even have its own units ".... well of course it does. But....
Because you comment on vehicles with 500bhp vs. 500 ft lb torque is laughable because: (a) you don't give information on revs, or even rev/torque curves, and (b) you quote torque with a unit, lb ft in your example - so contradict you initial comment. Doh.
Assume this was a joke comment but for those in doubt or confused..... Select a car with similar high power and torque figures and then no need to worry. I can recommend the Bugatti Veyron 16.4 with 987bhp at 6000rpm but only 922 lb ft torque at 2200rpm :-) Of course a little slow 0-62mph (2.5s) but not so bad 0-124mph (7.4s). You'll have to live with only 253mph top speed until the higher spec version capable of 270mph comes out. Surprisingly small so quite easy to park at your local Tesco :-)
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">Your contribution does not warrant comment to be honest.<"
But I'm sure you will anyway.
">Because you comment on vehicles with 500bhp vs. 500 ft lb torque is laughable because: (a) you don't give information on revs, or even rev/torque curves, <"
Exaclty, we don't need info about revs if we are dealing with 500bhp. 500 bhp is 500 bhp is 500bhp. If we know weight (or even mass) then we can comment about accel. We don't need to know about torque.
I can build a pushbike that will generate 922 ft lb of torque, it won't be very fast but it will have 922 ft lb of torque. If I could build a pushbike with 987bhp it would be very quick. In this context, torque is meaningless.
lb ft is a combination of two units, one for weight and one for distance.
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Power is based on torque and revs. Period. Look into it maybe rather than ignore physics? You seem to ignore simple definitions of power and torque. Not having a go.
"I can build a pushbike that will generate 922 ft lb of torque"
I'd like to see the size of muscles needed to drive a pushbike potentially capable of 922 lb ft torque. Do you even realise what 922 lb ft torque equates to? Probably not. And yes you can get high bhp with high revs even for low torque.... F1 cars actually have quite low torque but can rev to high rpm.
And you have not responded to saying torque has no units and yet you give an example with Imperial units. And the Metric is Newton Metres.
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">Power is based on torque and revs.<"
Ah yes, Power (bhp) = (Torque lb.ft x Angular Speed rpm) / 5252
This equation is responsible for all sorts of misunderstandings and is frequently trotted out by those who have no real grasp of the relationship between torque and power ;-). From the same equation, we can state that rpm is based on power and torque, and torque is based on power and rpm. And so we go round in circles. Now, are we all convinced that the equation applies to all rpm conditions for an engine with forced induction?
">I'd like to see the size of muscles needed to drive a pushbike potentially capable of 922 lb ft torque<"
Pedal crank = 5ft long, me = 200lbs, I stand on pedal and the max torque applied at the bottom bracket axle = 1000ftlbs, no muscle needed, blubber would be sufficient. I can apply 200ftlbs with my humble torque wrench, again we can see how confusing torque is.
">Do you even realise what 922 lb ft torque equates to? <"
Mmmm, that's a difficult one. It might equate to 922lbs hanging on the end of a lever length 1ft, or even 1lb hanging on the end of a lever length 922ft. Other combinations are available.
">F1 cars actually have quite low torque but can rev to high rpm.<"
Which confirms that a car can accelerate rather quickly even though the engine produces low torque, although the other torques that appear at the same time elsewhere in the power train may be higher, or even lower. And it demonstrates why torque creates a smoke screen of mystery. A 600kg car with 300ftlbs of torque might achieve values of accel, on the other hand it might not. A 600kg car with 900bhp will achieve high values of accel.
">And you have not responded to saying torque has no units <"
I think I wrote that torque doesn't even have its own units, it is a product of two different measurements, weight and distance.
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From
sromagazine.com/paris/dictionary/sro-dict.htm
"torque: [1] Turning or twisting force such as the force imparted on the drive line by the engine. Usually measured in lb-ft. It differs from work or power in that torque does not necessarily produce motion. Basically, the magnitude of a torque acting on a body is the product of the magnitude of a force and its force arm (perpendicular distance from the axis of rotation of the body to the line of action of the force). This product is called the moment of the torque about the axis or the torque. Also see self-aligning torque. [2] To tighten a nut or bolt with a torque wrench "
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If we wish, then to achieve maximum acceleration over the whole speed range of the car, the overall gear ratio between the engine and the road wheels must be varied in such a way that the engine is always running at the speed which gives the maximum torque at the road wheels. * This is the ideal -----
*At first sight, one would expect this to be the speed giving maximum engine torque. The torque curve falls off slowly, however, and by using a bigger overall speed reduction ratio, more torque is given at the road wheels at higher engine speeds, and in fact, for a given road speed, the greatest torque is given to the road wheels at maximum power engine speed.
Taken from The Sports Car, Its design and performance, Fourth Edition. Written by Colin Campbell M.Sc., C.Eng., M.I.Mech.E.
I, for one don?t doubt the gentleman.
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>>I, for one don?t doubt the gentleman.
Nor do I mjm. He is describing how you would set up an ideal CVT to acheive maximum acceleration, and correctly says that you would hold the engine at maximum power, taking advantage of the optimum gear ratio and torque multiplication to give the maximum tractive effort.
I agree with your post and Mr Campbell's writing 100%.
However, if you have an ordinary gearbox, there are two scenarios to consider.
1) If you stay in any particular gear, and accelerate through the rev range, you will get the most acceleration in that gear at the engine speed for maximum torque. There is a brief description of how this works in both the available power and the tractive effort performance prediction techniques in the diesel overtaking thread, where I show that they are equivalent.
2) You may choose to change down to get better acceleration, and, providing you don't rev much beyond maximum power, you will gain more by the improvement in torque multiplication than you lose by being away from the torque peak. Typically, mechanical and acoustic considerations prevent you from revving much faster than the power peak of most engines! There is a clear link with the CVT logic described above in this post, and by Mr Campbells wise words in mjm's post.
Number_Cruncher
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Engine torque is multiplied by the transmission reduction ratio to give torque at the wheel. With an infinitely low first gear, infinite torque at the wheel can be acheived.
Engine power isn't affected by transmission ratio, apart from small power losses.
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>>With an infinitely low first gear, infinite torque at the wheel can be acheived.
Indeed, this has been a problem with some types of CVT which give a geared "neutral". In doing so, they enter a region where the driveshafts downstream of the gearbox have to withstand very high stresses. Practically, wheel slip and compliance within the driveline prevent a truly infinite torque condition occuring, but it is, nevertheless, a problem.
Number_Cruncher
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">Engine power isn't affected by transmission ratio, apart from small power losses.<"
Hurrah! Which is why power is the most effective way to define the point in the rpm range at which a car has (potentially) greatest acceleration. In this context, forget torque, it confuses the issue.
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Torque torque torque... whyntcha all shaddap? You remind me of me...
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>>power is the most effective way to define the point in the rpm range at which a car has (potentially) greatest acceleration
This is only true if you have an ideal CVT.
Number_Cruncher
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">1) If you stay in any particular gear, and accelerate through the rev range, you will get the most acceleration in that gear at the engine speed for maximum torque. <"
Max torque where? At the crankshaft? At the gearbox output shaft? At the final drive shafts?
">You may choose to change down to get better acceleration, and, providing you don't rev much beyond maximum power,<"
Which is where gear ratio selection is so important. With the correct gear ratios and a reasonable spread of power at the peak of the power curve, it is possible to over-rev past peak bhp but still have reasonable power available, change up and still be close to peak bhp. And repeat. A good example of this is the 1981 KTM 495, an oversquare engine which could rev way, way past peak bhp. This allowed it outrun later 'bikes with more bhp because the '81 could remain in one gear where mortals on lesser 'bikes would be changing up. However, on a long, uphill fire road, the '81 bike would eventually be caught by a 'bike with more bhp and long gearing, so there's no beating bhp ... eventually.
But - interestingly enough (?) - the '81 495 still holds the fastest verifiable top speed for a MXer ..... OK, there are other factors (smaller frontal area etc), but the ability to continue to rev past the powerpeak was crucial.
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I went away for the weekend. Glad to see discussion has taken place.
">1) If you stay in any particular gear, and accelerate through the rev range, you will get the most acceleration in that gear at the engine speed for maximum torque. <" Max torque where? At the crankshaft? At the gearbox output shaft? At the final drive shafts?
We can all agree that maximum acceleration takes place where maximum torque is aplied to the wheels. Ignoring transmission losses and a constant gearing the torque at the wheels must be proportional but may not be equal to the torque at the engine.
">You may choose to change down to get better acceleration, and, providing you don't rev much beyond maximum power,<" Which is where gear ratio selection is so important. With the correct gear ratios and a reasonable spread of power at the peak of the power curve, it is possible to over-rev past peak bhp but still have reasonable power available, change up and still be close to peak bhp. And repeat. A good example of this is the 1981 KTM 495, an oversquare engine which could rev way, way past peak bhp. This allowed it outrun later 'bikes with more bhp because the '81 could remain in one gear where mortals on lesser 'bikes would be changing up. However, on a long, uphill fire road, the '81 bike would eventually be caught by a 'bike with more bhp and long gearing, so there's no beating bhp ... eventually. But - interestingly enough (?) - the '81 495 still holds the fastest verifiable top speed for a MXer ..... OK, there are other factors (smaller frontal area etc), but the ability to continue to rev past the powerpeak was crucial.
The maximum acceleration occures at max. power (and max. torque in a given gear, see my coments above). For simplicity lets say you have a gear box with 2 speed. to get maximum acceleration you must rev beyond max. torque and power in first gear before you engage second gear.
To demonstate lets say in 2nd the engine revs half as fast for a given speed as in first. That is second gear is twice as high as first gear (later sentence may not make technical sense but you get my point?). To get max. acceleration you must rev the engine beyond is max. power and torque in 1st so that at the time of changing gear the torque at the wheels does not change. For instance if max torque occures at 4000rpm, max power at 7000rpm, and half of max torque occures at 8000 rpm, you would change from 1st to second gear when the engine reaches 8000rpm in 1st. At that speed the acceleration offered by 1st and 2nd gears must be equal. If you rev higher in 1st gear you would be better off in 2nd, if you change up before 8000rpm to get to second you would be beter of keeping in 1st.
This of course assumes torque characteristics acting 'normally'. If I am fundermentally wrong in my assertion I would be interested to know why. (I hope I have not left much room for miss interpretation, other than the bit about the torque curves acting 'normaly'.
It stands to reason with what I have said above, max. acceleration in this simple model will occure if the engine is alowed to rev beyond max power.
And Lud, it does not all come down to torque. We could equally take torque as given, but discuss rpm. We just find it intuitivly easier to talk about torque. If all you can disscuss is torque without reference to rpm you will never understand acceleration, power or kinetic engergy (which is proportional to speed*speed). All acceleration is, is the build up of kinetic energy. This energy is supplied throughout time by power (another Rumsfeld moment for me?? - I know what I mean)
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Torque means nothing without RPM
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Torque multiplication by the gearbox is a constant and is not directly related to the engine's torque or power characteristics, i.e. the engine characteristics are not actually changed by the use of different gear ratios rather a certain combination of ratios will be specified to complement certain engine characteristics. So the gear ratio itself is not relevant to the torque v power question except that to say that the closeness of adjacent ratios will help the engine to be kept in it?s effective "power band". Or otherwise wider ratios will reduce the need to change gear regularly by utilising a flexible engine?s wide ?power band?.
So how to define that "power band", in general it can be said to be the area of the rpm range between max torque and max power. Ok, so max torque is where each individual cycle of the engine is at it's most effective and max power is where the optimum balance between the decreasing cyclic efficiency (not cycling proficiency !) as engine speed increases and the multiplication provided by the increase in engine speed.
An engine may develop a very sharp increase in torque over a small rev range, and / or experience a very sharp drop off in power if the engine speed increases much beyond the point of max power, and otherwise may have a very flat torque curve and produce perhaps a high % of it?s max torque as well below the engine speed at which peak torque occurs and / or continue to produce a substantial proportion of it?s maximum power as the engine speed increases well beyond the point at which maximum power occurs, in such cases the useable "power band" stretches to either side or both sides of the max torque to max power range. However the engines within the latter category do not always have a wider "power band" because the main criteria is the spread between max torque and max power. Max torque and max power can be close together, as in a peaky racing engine or wide apart as in a turbo diesel so even though the racing engine continues to produce power at engine speeds well past the point where it produces max power and the TDs power drops off soon after it reaches the engine speed where it produces max power, the TD will have a wider power band.
To further address the relevance of torque and power in determining an engine?s characteristics, it is true to say that you can quote torque alone without mentioning power and the reader will be able to determine the power characteristics.
For instance, take a Yamaha R1 engine in 1250rpm increments towards the top of its rev range, the R1 produces 72lb ft 8750rpm, 82lb ft at 10,000rpm, 78 lb ft at 11250rpm, 76lb ft at 12500 and around 63lb ft at 13,750rpm redline, that is all you need to know because simple maths tells you that this equates to 120bhp, 156bhp, 167bhp, 180bhp and 164bhp respectively. This also serves to illustrate how the torque being sustained at high revs enables the phenomenal power peak, torque at 12500rpm is little higher than at 8750rpm though power is 50% greater due to the multiplication provided by the increase in engine speed, also although each individual cycle of the engine is 10% less effective at 12500 rpm compared to 10,000 rpm, the fact that there are 25% more cycles means that more power is produced.
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Thank you Cheddar for an informative post, I have only one question remaining and that is, where does the figure of 5050 come from in the formula shown below? It is obviously a constant and must in some way relate lb/ft to rate of work, but I do not know its origin, can you explain please?
Power (bhp) = (Torque lb.ft x Angular Speed rpm) / 5252
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5252 is simply a convertion factor, if you quote torque in Newton meters then it is 7183, i.e:
Power (bhp) = (Torque Nm x Angular Speed rpm) / 7183.
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To clarify, it might be 7180, IIRC:
Power (bhp) = (Torque Nm x Angular Speed rpm) / 7180.
Power (PS) = (Torque Nm x Angular Speed rpm) / 7124.
PS is effectively a metric hp, and is an often used alternative unit for power.
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So, Cheddy old bean, do these formulae apply to forced induction engines at all rpm?
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do these formulae apply to forced induction engines at all rpm?
Yup, all engines, steam, rotary, 2 stroke etc.
Take two different Fords:
2.2 TDCi turbo charged diesel:
(400nm x 1800rpm) / 7124 = 101 PS (power at max torque rpm)
(155PS x 7124) / 3500rpm = 315nm (torque at max power rpm)
ST220 n/a V6 petrol
(285nm x 4900rpm) / 7124 = 196 PS (power at max torque rpm)
(226 PS x 7124) / 6250rpm = 257nm (torque at max power rpm)
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Diesels are the torque meisters however for a petrol the 335i is phenomenal, look how it compares with a 335d, and it does 29mpg to the 335d?s 37mpg.
335d turbo charged diesel:
(580nm x 1750rpm) / 7124 = 142 PS (power at max torque rpm)
(286PS x 7124) / 4400rpm = 463nm (torque at max power rpm)
Over 460nm between 1750 and 4400 rpm.
335i turbo charged petrol
(400nm x 1300rpm) / 7124 = 73PS (power at max torque rpm)
(306 PS x 7124) / 5800rpm = 375nm (torque at max power rpm)
Over 375nm between 1300 and 5800 rpm!
So if it is pulling 30mph / 1000 rpm then 1300 rpm would equate to 39mph, from which it would acclelerate hard, and 5800 would equal 174mph, perhaps it would do it if not restricted to 155mph.
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It's torque curves like these that confuse me:
tinyurl.com/ym5d7w
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OK, take the 225, the graph is approx though it works out:
78 hp = (205lb ft x 2000rpm) / 5252
155 hp = (205lb ft x 4000rpm) / 5252
225 hp = (196lb ft x 6000rpm) / 5252
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If you choose what are technically** the most common units in European use;
Power (W) = Torque (Nm) * Angular Velocity (radians per second)
Then there are no troublesome constants to remember.
Although radians aren't much used outside technical usage, they are an excellent unit of angle - one radian is where the length of an arc is equal to the radius used to describe it - there are 2 * pi of them in a full circle. Among the many beauties of this measure are that it is dimensionless - it is simply the ratio of two lengths, and for small angles, many trig functions become much simpler. e.g. for small angles sin(theta) is nearly equal to theta itself, if you use radians!
Of course, for public consumption, old units of Bhp and lbf ft are used, or even horrendous mixtures of Bhp and Nm!!
** By this I mean used internally by manufacturer's and research labs, not necessarily by the rolling road in the industrial unit round the corner, and not that published in blurb to punters in showrooms.
Number_Cruncher
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>>It's torque curves like these that confuse me:
Do you mean the unusually flat torque curve for a petrol?
If so, it's because a basic engine even a basic turbo-charged engine gives maximum torque at the engine speed where the volumetric efficiency is greatest. This is a function of how all the slugs of gas in the manifolds interact, and how pressure waves bounce around the system, all conspiring to cram pack the cylinder full of air and fuel just before the intake valve shuts. So you end up with a typical old fashioned torque curve with a peak.
If now, you load the engine up with devices whic can adjust manifold lengths and configurations, if you can alter the flow patterns in the air within the manifolds, if you can alter when the intake valve shuts, you can enable this "sweet-spot" to move along with the engine speed, so you are always running at very good volumetric efficiency and hence very good torque (and therefore power) - in a way, it is a bit like building a whole range of engines all with the same cc, but all with slightly different manifolds, timing, etc, and then taking the envelope of all of their torque curves.
This can't go on forever - eventually, and engine speed is reached when the intake gas flow becomes sonic (technically, if a little confusingly for those from the carburettor era, it is termed a choked flow) all the way across the port - at this point, the maximum mass flow rate of air into the cylinder has been reached, and you cannot get any more air/fuel mixture into the cylinder, so the torque curve eventually must begin to drop off beyond this point. (fixed mass flow rate and less time available owing to higher engine speed -->> less air/fuel mixture in the cylinder, hence less torque output)
Number_Cruncher
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Do you mean the unusually flat torque curve for a petrol?
A flat torque plateau can be engineered in a turbo charged engine by controlling the boost (manifold pressure) and the injection and ignition timing etc. However there will still be an engine speed where optimum volumetric efficiency is achieved, where even though the same amount of torque is produced either side of it more fuel is used per cycle in doing so.
A similar effect is achived on an n/a engine via adjusting intake lengths, VVT etc.
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Thank you Cheddar, but you have not answered my question, which was,'where does the number 5252 come from'? I was asking why is it 5252 and not some other number?
I have done a little research using Google and the answer to my question is as follows:
Power (in brake horse power) = Torque (in lb/ft) x 2π x rotational speed
Since 2π = 6.248 and there are 33,000 ft/lb in 1 brake horsepower, it follows that the conversion from Torque
to Power is given by the formula ...................
Power = Torque (in lb/ft) x rotational speed (in RPM)
5252
and the mysterious figure of 5252 comes from: 33,000 divided by 6.248
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Oops!!!!
The strange "2π" in my previous post was supposed to be the symbol for Pi, which I copied from Word into the post. It looked OK on screen, but obviously the HJ board could not handle it.
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Oh god. 22 over 7. 3.141586 (contd p. 94). Hope you guys are having a good time. You're certainly impressing me with your knowledge of physical theory.
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Thank you Cheddar, but you have not answered my question, >>
I did say that 5252 is simply a conversion factor for lb ft to hp and that it changes depending on the units of measurement used, i.e. hp to Nm is 7180, PS to Nm 7124.
To further clarify power = torque x (2x Pi) x rotational speed however in that simple equation because the rotational speed is measured in time, i.e. rpm, then torque is quoted as lb ft/min or Nm/min. 33,000 is conversion factor for lb ft to hp enabling torque to be quoted as a static force, 33000 / (2 x Pi) = 5252.11, this is rounded to 5252, the conversion factor first quoted much further up this thread.
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This has been a really interesting thread, as has the 'Diesel overtaking' one.
But having learned much I'm not sure I understand more. How can the knowledgeable experts disagree so much? Would someone like to summarise the ansewr to the original poster's question in no more than three brief, uncontentious sentences?
As for the real world on the road, I'm inclined to maintain my view that the best way to extract maximum performance out of a particular engine is to experiment in an intelligent fashion first: you get to feel what gear and engine speed works well enough for a given situation. (A view which would sit equally well in 'Diesel overtaking', IMO.)
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>>Can anyone explain to me in simple English what is the difference between Torque and Power when used to describe the performance of an internal combustion engine?
I know what the units are that the two are measured in, both in the Imperial system and in the SI system, but I have to confess that I do not really understand how they are measured, or how to explain what they are in simple words.>>
Three sentences:
The engine speed that at which maximum torque occurs is where each individual cycle of the engine is at it's most effective.
The engine speed where maximum power occurs is where the optimum balance between the decreasing effectivness of each cycle as engine speed increases and the multiplication provided by the increase in engine speed.
For instance the Yamaha R1 example above, the R1 produces 82lb ft and 156bhp at 10,000rpm and 76lb ft and 180bhp at 12500rpm, so although each individual cycle of the engine is 10% less effective at 12500 rpm compared to 10,000 rpm, the fact that there are 25% more cycles means that more power is produced.
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