Power to Weight versus Torque to Weight

[Paul Ranson]

Surely on an incline, or on the flat, with the same gearing, mass and wheel diameter, the vehicle with the larger force vector acting up the hill will accelerate faster (F=MA, A=F/M).

Of course.

 

So with more torque and hence greater thrust at wheel contact point, the diesel will out drag the petrol up the incline?

You are making an assumption that the diesel has more thrust at the wheel contact point. The deciding factor is the power available at a given road speed.

 

Paul

 

Edited by - Paul Ranson on 17 Sep 2002 23:12:51

[Bilbo]

Hi Highlander

 

Well as I posted I thought I was missing something that was the importance of the differences in gearing between the cars.

 

Posting phrases like "larger force vector" leaves me rather brain dead at this time of night without a force vector diagram

 

Properly be in the same state if you did send me one

 

 

1988 200 bhp, 146 ft lbs, 1700cc Cosworth BD? with Brooklands and Clamshell wings, Freestyle Motorsport suspension. Q 979 CGY

 

[Highlander]

No Bilbo I think your on the ball here. Turning effort is what an engine produces and power is the rate that it does this (by definition power is Joules/S).

[Highlander]

Quote

 

You are making an assumption that the diesel has more thrust at the wheel contact point. The deciding factor is the power available at a given road speed.

 

Yes, all other thing being equal, road speed, gearing, mass, rolling circumference the vehicle with more thrust produces most acceleration, therefore the one with the most TORQUE, not power wins.

I'm not talking about engine peak power or torque. If given the above mass, gearing, incline conditions you have two engine turning at 3000 rpm, one is producing 160 ibft torque (the petrol) and one producing 250 lbft of torque (the diesel), which one accelerates faster? The diesel.

[Paul Ranson]

all other thing being equal, road speed, gearing, mass, rolling circumference

But they're not all equal. Read what I said and take it literally.

 

Paul

[Highlander]

Sorry, but I dont see what you're getting at.

[Peter Carmichael]

Sorry but I don't see what you are getting at

 

Quite.

 

This started off by someone writing to Evo saying that "torque to weight" tells you whether a car is fast. By elementary physics I have demonstrated that this is not the case.

 

People like you haven't had the penny drop yet and are testing my patience because you have not been able to disprove the validity of anything I have written. Instead you have remained in the comfort zone of what you believe and have refused to consider that there might be some new knowledge out there for you to acquire.

 

Torque at the flywheel

 

Torque at the wheels

 

Thrust at the contact patch.

 

These are things that I, like you, am interested in. I have been telling you that there is a really simple way to calculate this sort of stuff without any of this "holding all other things equal nonsense" because WHAT IS THE POINT OF COMPARING CAR PERFORMANCE IF YOU HAVE TO HOLD ALL THINGS EQUAL, including (as you list) gear ratio and tyre size.

 

The following is my method:

A car is travelling at 51.5mph (23.02m/s). It has an engine that delivers 200bhp(149.5kW). It has a gearbox, so you can select the gear that gets you closest to operating at peak power. What is the thust at the contact patch? Answer: 149.5/23.02 = 6.49kN.

---

 

This is how much pain you have to go through if your hidebound brain can't get beyond torque:

The same car with a few more details. It produces 200bhp@6000rpm. It is a Caterham fitted with a 6 speed gearbox. It is in second gear which has a ratio of 2.01:1. It has a 3.62 differential. It is fitted with Avon AVB10 tyres in 6x21-13. It is still going at 51.5mph. What is the thrust at the contact patch?

 

What revs at 51.5mph in second gear?

51.5*1.609/3600 = 23.02 m/s

The tyre circumference is:

21*25.4*pi = 1.675m

The wheel rotational speed is:

23.02/1.675 = 13.74 revs per second

The crankshaft speed is:

13.74 * 3.62* 2.01= 100 revs per second

Which is:

100 * 60 = 6000 rpm

 

Ok. The engine is operating at nearasdammit power peak. What torque does it produce?

200*5252/6000= 175.07 lbft which in sensible units is 237.94Nm

What torque appears at the wheels?

237.94*2.01*3.62 = 1731.29 Nm

What thrust appears at the contact patch?

1731.29 / 21*2 /25.4 *1000= 6491N = 6.49kN

---

 

So yes, using a torque focussed view you can come up with the same answer as the power focussed view, but which one was easier? Which one was more real world? Which one required the least amount of additional information?

 

Two cars side by side at 50 miles an hour. Which one can accelerate quickest? The one with the higher power to weight ratio. QED.

[tom7]

Bravo!!!!

I have spent the last couple of days following this thread. I've not got involved as I don't have the necessary knowledge of mechanics or physics.

But, I have to say the conclusion that Peter has just written is one of the best demonstrations of "proving your point" I have ever read. It has brought I smile to my face 😬

 

Tom

 

Edited by - tom7 on 18 Sep 2002 12:50:01

 

Edited by - tom7 on 18 Sep 2002 12:54:33

[Highlander]

Calm down Peter, no need to get personal!

 

1) My brain is not hidebound and I'm more than willing to assimilate new information.

2) I wasn't trying to disprove the validity of your comments, rather try to answer another point that was a little off thread. Perhaps badly as it was nearly midnight.

3) I wasn't trying to define 'car performance' by torque.

 

I agree with all the maths you have stated, in the long hand version is clearer to people as you don't have to know that you are assuming that in the first caluculation the road speed corresponds to peak power in the selected gear. Correct?

 

As I say, I agree with all you say, but I'd just add that it can be difficult for people to relate power/weight as its connection to resultant thrust as it's not immediately obvious.

 

Out of interest Peter, where does you calculation of torque from power come from (200*5252/6000 = 175.07 lbft)

 

 

[Peter Carmichael]

[calming down]

 

200/6000*5252 is a conversion related to the units of horsepower and lbft. In SI it is simpler:

 

power (W)/rotational vel (rad/s)= torque (Nm)

 

[/calming down]

 

I agree with all the maths you have stated, in the long hand version is clearer to people as you don't have to know that you are assuming that in the first caluculation the road speed corresponds to peak power in the selected gear. Correct?

---

 

No. Absolutely not. This is the point that you have consistently failed to learn. The engine can produce 200 horsepower. IT HAS A GEARBOX. It could be at 18,000rpm or 2rpm, but as long as the gearing allows it to make 200 horsepower at a road speed of 51.5 mph, the resulting thrust force will be 6.49kN. If it isn't making exactly 200 horsepower it isn't far off.

 

What power does an engine produce when it is not at peak? When you are driving, you have the choice of gears. A gearbox allows you to keep an engine on the boil and with appropriate close ratios that means at or above ~90% of peak power output. So a 200 horsepower engine can *always* be operated at 180 horsepower or above. So the thrust value can always be determined at 90% or above the 6.49kN figure I gave. This doesn't require any knowledge of gearing other than that you have a gearbox.

 

Addressing your point:

If given the above mass, gearing, incline conditions you have two engine turning at 3000 rpm, one is producing 160 ibft torque (the petrol) and one producing 250 lbft of torque (the diesel), which one accelerates faster? The diesel.

 

At 3000rpm and 160lbft, the petrol engine is producing 91 Horsepower.

 

At 3000rpm and 250lbft, the diesel engine is producing 142.8 Horsepower.

 

So power wins but it appeared that torque had won because you presented the problem in that way.

 

The petrol engine driver has the option to change down a gear or two because he has an extended rev range. If he selects an appropriate gear that allows the petrol engine to develop more than 142.8 horsepower, then the petrol engined car will accelerate faster. The diesel driver does not have this choice. Power wins.

 

The question you need to consider as a Seven driver is whether you are more interested in what happens when:

A: you lazily extend a toe and can't be arsed to change gear, or

B: you rev the car to the redline and change gears to keep on the boil

 

If you drive your Seven on a racetrack, answer A is never considered.

 

Would you drive a fireblade engine at 3000rpm and expect it to be performing at its best?

 

but I'd just add that it can be difficult for people to relate power/weight as its connection to resultant thrust as it's not immediately obvious.

 

It bloody well would be obvious if people stopped quibbling with really basic physics. Power is THE answer. It is physically exact. Its units exactly relate to the problem of accelerating a moving vehicle. It is the definitive tool for the job. Take POWER and divide by SPEED and divide by WEIGHT. That is the ACCELERATION. Are we done yet? How much less difficult can I make this?

 

Edited by - Peter Carmichael on 18 Sep 2002 14:25:14

[Paul Ranson]

but I'd just add that it can be difficult for people to relate power/weight as its connection to resultant thrust as it's not immediately obvious.

 

Thrust at a power and speed is power/speed.

Acceleration at a power, weight and speed is power/(speed*weight), power to weight divided by speed.

 

Paul

 

Edited by - Paul Ranson on 18 Sep 2002 14:37:15

[Peter Carmichael]

Is this discussion ready to handle "losses" yet 🤔

[Highlander]

Peter,

 

The 5252 bit, yes when I was posting I thought is was the furlongs/fortnight to SI fiddle factor.

 

Yes I do know what a gearbox is. It's that thing you don't need in a diesel cause of all the torque.

 

OK maybe 'assume' was the wrong word. What I meant was that in your initial calculation P/V =F you had chosen a road speed that corresponded to peak power in your chosen gear. I realise that it doesn't matter where an engine makes it's power as long as you select the right gear to utilise that power.

 

As to the two cars petrol & diesel. Yes it 'appears' that torque wins when it's power, which as you state, (and I fully understand), is THE motive 'force' but as they are inextricable linked P=C*Torque*RPM, (C being the SI fiddle factor), then we are both 'correct'.

 

Fireblade? Of course I wouldn't drive it at 3k as all it's useful power is at the top end, conversely would you drive a truck at 10k rpm? No. They have different power delivery characteristics. Apart from the fact that it'd have exploded by then anyway.

 

Losses? What are we talking here, frictional, aerodynamic? Enron's stock price?

 

Bring it On.

[AMMO]

As previously stated in my little story about the schoolboy there are very few things you actually need to know. The schoolboy's question goes beyond all the horsepower and torque debate and actually goes to the crux of the question. How fast will it go? The more horsepower you have the faster it will go.

 

I'm basically a simpleton. But is fairly obvious even to me that no power = no speed. When I try to drive my car fast I change gears around or beyond the point of maximum horsepower. Why is that?

 

I once bought a Peugeot 205. Before parting with my money the dealer let me road test it. To see if the engine was in good nick I got it flat out on a Motorway. If it did not acheive it's published maximum speed I would not have bought it. Later I had it dynoed and the power was spot on (in fact it had about 3 bhp more).

 

The only reason we are interested in power is because it is a means of acheiving acceleration and speed. This is what makes us grin from ear to ear.

 

😬

 

What about losses Peter?

 

AMMO

[Peter Carmichael]

Does anybody have a good turbo-diesel torque or power curve to hand. I would love to overlay it with some petrol ones to show that the nominally impressive "torque" does not in fact give as good flexibility as the extended rev range of a good petrol engine.

 

I am interested in parity of comparison.

 

If someone was able to demonstrate that a diesel outperformed (answering Ammo's 8 year old's question) any petrol engine I was considering, I would not hesitate in commissioning a racing diesel.

 

Highlander,

Fireblade? Of course I wouldn't drive it at 3k as all it's useful power is at the top end,

---

 

Yet you set up the petrol vs. diesel question as a less extreme example of the same phenomenon.

 

THE motive 'force' but as they are inextricable linked P=C*Torque*RPM, (C being the SI fiddle factor), then we are both 'correct'.

 

Only because you fiddled the problem by disallowing the petrol car driver from selecting an appropriate gear. If you remove the constraint then RPM is not necessarily consistent between the two cases and TORQUE LOSES.

 

Always.

 

Oh, by the way, following up on your reply to Bilbo's contributions:

So with more torque and hence greater thrust at wheel contact point, the diesel will out drag the petrol up the incline?

 

No. Greater thrust at the wheel contact point is achieved by generating more power, not more torque, so the petrol will win if the driver changes down a gear. The concern is that operating an engine at consistently close to its rpm limits may be inappropriate and lead to reduced wear and excessive noise in operation - because of this, in traction applications, it is preferable to run an engine that generates sufficient power at low rpm rather than optimum power at high rpm. Good turbo-diesels can achieve this, but would you want that in your Seven 🤔

[Highlander]

Turbo Diesel power curves? unfortunately I cant help you there, at the moment, but if I find one I'll e-mail it to you.

 

OK I think you've mad your point that absolute power determines a vehicles absolute performance. But I still maintain that a modern diesel is more 'flexible', as you put it, without giving more ultimate performance.

You mention a petrol engine with an extended rev range. Surely as you push the power peak (assuming the same absolute power figure) up the rev range, you effectively narrow the useable power band and make it less flexible.

 

Example. I had a Integra Type R 187hp @ 7900 rpm. As comparison take a BMW 330D with almost identical 186hp but at 4200rpm. Are you trying to say that the Type R is more flexible than the 330?

I know which one I'd rather tow my 7 to the track with. BTW both top out at around 145 ish. That's kph, aledgedly, if the Five O are listening.

 

As to commisioning a racing Diesel, to late VW have been rallying a Mk3 Gold TDI in the GB Rally series for some time. With reasonable results if memory serves.

 

Do I want a diesel in the 7 ? Only if I can have VWs forthcoming V10 with 350bhp (ish) and 500+ lb-ft!

[AMMO]

What is all this talk about flexibility? Flexibility is for people who can't drive. Are we talking performance here or what you use to go down the shops.

 

I may not be the most experienced person in the world but good racers carry good corner speed and usually have higher gearing than the other guys they are racing against.

 

The only time the engine sees 5,000 rpm is on the start line.

 

AMMO

[Highlander]

AMMO

 

Yeah but carrying higher corner speed has little to do with power,but rather driver ability don't you think?

[DanB]

Surely flexibility is determined not by the peak torque but by how wide a power spread the engine has. I hope that by now we've establish that power is what makes a car accelerate and keep it at any fixed speed and the amount of power determines how fast it can accelerate - anything else can be ignored if the gearing of the car is correct. i.e. it doesn't matter whether the car is producing 200bhp at 1rpm or 200bhp at 10,000rpm so long as the gearing allows it to use that power at the appropriate road speed.

 

For the sake of argument, let's assume that a car's usable power spread is that part of the engine's power curve where it's producing at least 90% of its maximum power. Flexibility will be determined by the proportion of the rev range in which the engine can produce at least this power.

 

As an example, we have 2 cars - one revs to 10k, one revs to 5k. Both are in a gear which allows a top speed of 80mph - car 1 will do 100mph at 10k, car 2 will do 100mph at 5k.

 

If car 1's power spread (over 90% of max power) is between 5k and 9k, that's 40% of its rev range. So that gear is suitable (albeit not necessarily optimal) for speeds between 50 and 90mph.

 

If car 2's power spread (over 90% of max power) is between 3k and 4k, that's 20% of its rev range. So that gear is suitable (albeit not necessarily optimal) for speeds between 60 and 80mph.

 

Therefore car 1 has a more 'flexible' engine, despite the fact that it has a lower maximum torque figure.

 

But let's assume that, in fact, car 2's power spread is from 2k to 4k and car 1's power spread is between 6k and 8k. The results are reversed and car 2 is more 'flexible'.

 

However, this ignores the fact that an engine of the same capacity will generally, if it produces its power at higher revs, will produce more power. Without forced induction, the level is quickly reached where VE cannot be increased and so torque cannot be increased. Therefore the only way to increase power is to increase revs. It follows therefore that limit revs means limited power.

 

Car 1 may be peakier, but it ability to rev harder *may* mean that car 1 can produce more power that the 90% power figure for car *2* over a greater proportion of its rev range than car 2 can. So although the power curve will look peaky, it will actually be more 'flexible' than car 2.

 

Fire away!

 

Dan

[ashaughnessy]

Contents of this posting deleted - it was embarrassingly wrong

Anthony

😳

 

Edited by - ashaughnessy on 19 Sep 2002 07:22:17

[ashaughnessy]

Another thing in favour of the power argument. Say car A produces N pounds feet of torque at 5000 rpm and car B produces N pounds feet of torque at 10,000 rpm. At any given road speed, to get the engine producing N lbsft of torque, car B is in a gear with half (double???) the ratio that car A is in (e.g. B is in second whereas A is in fourth, you know what I mean!) which means it has double the thrust available at the wheels (assuming all other things being equal, which I'm sure they won't be).

 

I think I'm assuming that car A's maximum torque is N and car B's maximum is N. I'm also assuming that the torque production drop off reasonably rapidly after the maximum torque revs so that A isn't still producing N at 10,000 rpm.

 

Anthony

[scooby dooby doo]

you'll still get a thrust of 6.5 kN minus the transmission losses giving you 5 kN or something. but they'll also be a force of about 0.7 kN in the other direction due to aerodynamics.

 

(0.7 comes from reckoning on a 140 mph top speed and using a simple squared law for drag. so 5kN is max thrust, that must equal drag at 140mph. so 0.7 ish at 51 mph)

 

The result is therefore a 5 - 0.7 = 4.3 kN thrust pushing the car forward. So we get 4300 N acting on a 645 kg mass (with driver and passenger).

 

(and a torque trying to lift the front end as the toruqe comes fro above the contact patch. and another trying to lift the car as the aerodynamics operate higher than the contact patch in the opposite direction. lets ignore these)

 

so the accelration is 4300 / 645 = 6.7 m s^-2. ie about .7 g. this seems feasible.

hmm... even with some squat I doubt that 70% of the mass is over the rear wheels - so i guess this will give wheelspin as well.

 

a 435 kg car (notice i'm keeping the division easy ❗) would accelerate at 1g. except it would wheelspin instead etc.

 

 

 

HOOPY R706KGU See you at Cadwell

[ashaughnessy]

No, I was wrong and Peter's mathematics was right. I withdraw my "tosh" allegation and apologise to Peter. I've been trying to get my head round this so-called "basic" physics for the last hour and I think I've got it now and I am convinced by Peter's arguments.

Sorry Peter.

Anthony

[scooby dooby doo]

another convert to the faith

 

HOOPY R706KGU See you at Cadwell

[Bilbo]

Yes its me again just listening and staying on the fence

 

Seems there are a few irrelevant creeping in here.

 

AMMO posted "The only time the engine sees 5,000 rpm is on the start line" yes I agree with that but on a good track run on my engine hopefully its the last time its that low [:D}

 

Old adage use to be change gear so that the change up drop in revs hit max torque curve something I seem to have trouble achieving due to the car being over geared at present.

 

DanB posted "Surely flexibility is determined not by the peak torque but by how wide a power spread the engine has"

 

Think you really mean that, as torque only thing that the engine produces gives a power band relative to the overall gearing i.e. Power = revs x torque... performance is relative to load, as in fact you go on to say . Just trying to make the point that an engine only produce torque as an output to the gearing relative of load, its the car overall that we normally measure as power (work done) to a load or the engine relative to external load in case of having you car on a dyno its still against an external brake load as they only measure torque at rpm against the external applied load. Think that follows Peter's argument.

 

So just to bottom this out are we all agreed engines only produce "torque" in fact ANY engine, the fact its in a car engine is irrelevant.

 

Hi Hoopy slow down on the typing your spelling beginning to look if its one of my posts

 

 

1988 200 bhp, 146 ft lbs, 1700cc Cosworth BD? with Brooklands and Clamshell wings, Freestyle Motorsport suspension. Q 979 CGY