A) There is plenty of empirical evidence that the concept is hard to grasp B) There is plenty of empirical evidence that many never achieve an understanding Correct I accept this statement since: P/V = m*a :which says: at any given speed (V constant); P is proportional to m*a. The counterintuitive notion is that power (force * distance) is acting like a force at constant speed.
the torque is indicative of the acceleration in a given gear, while the max HP figure gives you an ide of how quickly the car can accelerate when in a lower (i.e. numerically better for the speed) gear. huge hp, with little torque in a 1 spd car would give you acceleration akin to the torque curve (fairly weak) but for a loooooong time, if its a high revver. Given a second gear, the torque would be further multipled by in effect creating a "half" first gear. THis is how HP can help. Hope this helps.
This subject is right up there with politics and religion. I like to think of HP as what keeps you going at a certain speed, like up a hill or it takes x amount of HP to go 200mph. And torque is what gets you going. The turning effect. Also they are related so there is no perfect laymans explanation, sorry for the bad spelling no dictionary here.
This thread has really made my day! Its quite funny when you read it all through. I love the way my 355 revs high to get its power, makes a riduculously loud noise, and thrashes nearly all cars (in the UK) at high speed - and its 10 yrs old!!! Although a lot of heavily modded (and illegal) cars here in UK can beat my 355 doing 0-62mph (in 4.7 seconds), I'm really please a 10 year old car like mine can still compete/beat the large majority of legal cars made today. If the 355 was modded in the same way as some of these street cars....... nothing would have a chance. I say - increase the revs all the way.
I will add this empirical anecdote..... If merely a torque value makes a vehicle "fast", then why dont all the performance cars have turbo diesels in them? One can make a 1/4 hp electric motor lift a million pounds if they liked (with the proper gearing). But, no matter what gearing is used, theyll never be able to lift that million pounds during any time period that could be considered even remotely "quick". Larry Tampa Bay, FL 2002 MB C320 Sport Wagon 1996 Ford Ranger 1991 Dodge Spirit R/T - 13.0@115
Throw my .02 in here. Ferrari, to me, means style and finesse. Ferrari has never been (in their street cars) about making the just pure raw 0-100 monsters. They have been about finding a middle ground with engines that share F1 technology, and hand crafted bodies/interiors. When I think of a Ferrari, I think of a car winding through the mountains at full wail... screaming out it's exhaust note at 7K RPM or higher. I don't think of stop-light races and the like. Ferraris can have more power or torque, but they aren't going to win the knock-down brutality test for a street car. They aren't about that. Vipers and the Ford GT rule the torque department, no doubt. a SRT-10 is a hellova car (but ugly as sin) in a strait line and brakeing, but it doesn't have the "feel" of a finely crafted item like a 360 does. It's not supposed to..it's job is to provide gobs of tire-roasting power on demand, under 5500RPM. I hope ferrari does increase the power levels, but power is not what I go by 100% when looking at ownership. If it was all about the torque numbers, then most f-cars wouldn't even be in the running. Lotus has never been about high HP numbers. Chapman was all about power to weight, which is where Ferrari has always been as well. A Esprit won't out accelerate a GT2 or a Viper, but, with the right driver, can carve up the corners like you would not belive! So, Power is good...but it's not everything. James
The 200" telescope on mount Palomar has been driven by the same 1/4 HP motor for the entirety of its use as a large professional telescope (over 50 years, every clear night) The moving telescope weights in over 1 million pounds (530 tons).
That sure is interesting. But Im sure it moves slowly! If they needed it to move more quickly, they would need a larger motor.
Although I voted for the 8500 redline I would prefer a 10K+ redline. After all, the F1 cars can do much higher....
During normal tracking the 1/4 HP motor is just loafing along. Horizon to horizon is just over 3 minutes for 530 tons, but more importantly, the telescope tracks to better than 1 arc-second continuously (an accuracy of 5 PPM), and is supported by high pressure oil (300 PSI) and a large bearing surface area. The oil pump was replaced twice in 50 years--if I remember correctly.
Because they do not rev, and thus you (like trucks) need 40 gears to maintain that acceleration. When you are beside that 70 tonne truck driving to work and you easily blow it away and think fncken trucks are so slow ... have you ever thought if your so called powerful Ferrari could accelerate 70 tonnes as fast!!! ... ofcourse not as it takes serious torque to accelerate that much mass that fast. As I said in the other thread, I have shared a race track with a racing truck (I do not know if America has this sort of racing but in England, NZ and Australia tractor units race, but they are limited to a 100 mph top speed ... it really is quite pathetic but many spectators like it). Anyway I was in my classic racing Alfa Sud with 170 - 200 hp and this little car only weighing less than 900 kg could NOT keep up with a Kenworth tractor unit as it accelerated down the backstraight ... until it hit the speed limiter. Naturally my lap time was something like 30 seconds faster but that was caused by braking and cornering performance NOT acceleration. If they could make a turbo diesel pull serious revs AND still make the huge torque then it would be the idea performance motor! Returning to: and Mitches comments indicating the formula P/V = m*a, and thus proving that P is proportional to m*a at a constant speed. All this formula does is confirm that P is proportional to a at a constant velocity and mass. It does not confirm that peak acceleration at ANY given speed occurs at peak horsepower. Why is that?, because at many different SPEEDS in a particular gear you are NOT at peak horsepower, and thus acceleration will NOT be peak for that gear. In the end like Brian reiterated it all comes down to maximising the force that pushes via the driving wheels the vehicle down the road. Maximise that force in a particular gear and you will maximise your change in velocity relative to time, or acceleration. Anyway when I workout the maximum load on say a driving wheel wishbone, or axle I use torque, and gear ratios, etc. ... not interested in power. And we also have to remember that ALL forms of dynos measure torque not power as torque is what an engine produces as a measureable force, and that same force is what moves the car. Pete
What I haven't seen talked about here yet is the fact that the kinetic energy of a moving vehicle is directly related to the SQUARE of its speed. Thus, a cars acceleration graph will show lower and lower acceleration rates as speed increases given a constant horsepower input. The faster a car goes, the higher the energy input (horsepower) required to maintain a constant acceleration. Of course there are resistances (wind, rolling, bearings, thrashing fluids, etc.) to deal with but forget about them for a minute. Look at an acceleration rate vs speed graph for any car, the rate falls as the car's speed increases (regardless of the resistances mentioned above). This is due to the fact that the kinetic energy being added to the car is increasing as the SQUARE of the speed. In other words; you put a lot more energy in the car going from 50-60 mph than you do from, say 20-30mph. The way we increase the kinetic energy in the car is through doing work, work in this case is done by the engine, the maximum rate of work that can be done is determined by the engines HORSEPOWER (not it's torque: not now, not ever, never will be). You do a LOT MORE work on a car to get it from 50-60 than from 20-30, so even though the horsepower remains constant it takes longer to get the car from 50-60. The great OZ behind the curtain is the fact that the accleration rate in a given gear in a standard transmission car is greatest at peak engine torque, not peak engine horsepower. BUT, this is because peak torque occurs at a lower SPEED than peak horsepower in any given gear. The greatest energy input into the car is done at peak engine horsepower, not peak engine torque. This is where most people get confused. They equate greatest acceleration in a given gear with the stupid saying "torque rules". This is because peak torque occurs at a lower rpm than peak hp, allowing for a greater acceleration rate at a lower speed (thus less kinetic energy being added to the car), but not a greater amount of work being done on the car. If you consider a constantly variable ratio transmission (Audi's is coming along well) and ask the engineer's designing it where peak acceleration occurs. All the ones that haven't been fired yet will tell you.....AT PEAK HORSEPOWER. This is not their opinion, not how "they feel", not because they wrote a hundred different equations upside down and backwards all over the whiteboard. It is incontravertible FACT. Just like the earth orbits the sun, horsepower rules acceleration. I'm sure if Galileo had proffered that "horsepower rules", the Catholic church would have made sure he spent even more of his life incarcerated. Of course the church, at the behest of physicists and reality, would have had to forgive him of that sin too, just like they forgave him for not believing the sun revolved around the earth.
We are not talking about energy required, simply acceleration. 100% agree but again we are not interested in work done, but maximum acceleration. No not just because it is a lower speed but because the force acting on the rear wheels are at maximum at maximum engine torque, not power. Also acceleration rate does not care what speed you are going it is a measure of the delta velocity over time. Yes it is harder to accelerate at faster speeds due to all the resistence etc, but if you do theoretical calculations and ignore all the real world resistences (such as rolling, wind, etc.) you will still obtain a lower acceleration figure at peak horsepower than at peak torque ... add on top the real world resistences and peak power looses out even more, as you quite rightly pointed out that the resistences will be larger at the higher speed. Also a cars acceleration tapers off as you go up the gears and increase speed because the torque being applied to the rear wheels reduces due to gearing ... thus while there is more resistence to over come, there also is less force via torque to push the car and maintain the same acceleration rate. Agree but not interested in work done or energy, simply how fast the car will increase speed. Different things. Again not interested in work done, simply how fast the mass accelerates, and is not just because it is at a lower speed, but because the force rotating the rear wheels is maximum at maximum engine torque and as long as you have perfect traction you have maximum acceration. I'm not sure on this one ... but I believe that we only currently rev higher than peak torque so that when we change up gear we end up atleast at peak torque. Again not interested in work done or energy. Peak torque is not just a fluke it is a characteristic of the engine design. Physics is fact too, and again we do not care a rats arse about energy or work done ... this is where EFWUN got confused. Anyway I have work to do or else I will get accelerated out of my job, and then their will not be much WORK done by this expanding (as I get older) mass Pete
Thats because torque is merely a COMPONENT of horsepower. Lots of torque and high rpm equals lots of HORSEPOWER. [/QUOTE] ... not interested in power. Pete[/QUOTE] Good, you stick a 5 foot breaker bar on the lug nut of your car wheel and stand on it, you'll be developing about 1,000 lbs ft of torque! WOW!. Without RPM you have no HP your'e going nowhere. I'll race you in my measly 400 hp Modena for pink slips. Torque without rpm is meaningless. Torque combined with rpm is POWER. A massively TORQUEY truck that develops 300 hp will not accelerate as hard as it would with a "peaky" 400 hp Modena engine with the correct gearing. Physics doesn't lie, but humans sure get confused.
Let the hand brake go and that car will accelerate at whatever rate a 1,000 lbs ft of torque will provide. You will win as I drive a measly er, probably less than 100 hp Toyota family wagon . Also are Australian as you know about our pink slips? ... I live in Sydney. Again do not agree. Most big trucks only develop around the 300 to 350 hp, but they accelerate that 70 tonne a hell of a lot faster than your Modena engine ever will not matter what gearing. Remember the gearbox is a torque multiplier NOT a power multiplier. Thus if you start off with more torque in your truck engine you will end up with MORE torque after the gearbox multiples it. Agree. Horsepower is very important as it is a measure of an engines ability to maintain torque and thus push the vehicle to higher speeds in that same gear ... and thus relates to top speeds in that gear. Again everything measures torque, so isn't power a factor of torque not the other way around. Pete ps: My father used to own a Jensen Interceptor that was a very heavy car. It had (supposedly) 330 hp, but a **** load or torque from its 383 Chrysler motor ... nothing could stay with that car from a standing start, NOTHING! I also remember watching a classic race where a guy entered his Jensen Interceptor. He was racing against 308 Ferraris, E-type Jags (modified, etc.) and many other cars. Which car was first to the first corner? ... the Jensen Interceptor ... infact he lead the first half of the first lap ... until the powerful cars overhauled him, and then his brakes overheated trying to slow such a heavy vehicle. Real life example of torque providing massive acceleration. Note Dad's Jensen used to top out at only 130 mph ... because it did not have enough POWER to push the increasing resistences.
You can reliably predict a vehicles 1/4 mile trap speed and et with only two things: HP and weight. You CANNOT do the same with TORQUE and weight.
Try using /quote in the closing brackets, not just quote . I believe in F = m*a ... Pete ps: I'd love to see you put a Ferrari motor in a 70 tonne truck and try and move it as fast as a big torque diesel ... now that would be real funny, probably have to have 400 ratios, and still be slower ... but it would sound good. I do not know what a 'Democrat' is being a simple New Zealander.
And that is exactly what is shown by P/v = m*a, as the velocity goes up, it takes (linearly) more power to provide the same amount of acceleration. In fact, the way you get to P/v = m*a is by differentiating: (assuming constant mass m) E= 1/2*m*v**2 dE/dt = 1/2*m*(dv**2/dt) dv**2/dt = 2*v*dv/dt dv/dt = a dE/dt = P Therefore P = 1/2*2*m*a = m*v*a Therefore P/v = m*a but since f = m*a and P/v = m*a then P/v = f = m*a and we are back to forces ruling acceleration. Unless you can find fault in the above math, there is no justification in using HP to govern the process of acceleration. EFWUN and his college physics professor friend found none--go for it! This was the argument posited by EFWUN in the old thread "tell me about driving around at high RPMs". I really suggest (strongly) that you go and pruse that thread before questioning the math of physics involved here. No, it is because acceleration is more directly related to force than to power. f = m*a; whereas P/V = m*a = f This was exactly EFWUNs argument, go back and read the whole thread--note: EFWUN lost the argument.
However, with TQ, the transmission ratios, differential ratio, rear tire rolling radius, and weight; you can. HP makes the calculation simpler, but does not give you the insight as to what is happening and why it is happening. Car V: 3000 lbs 400 HP 380 lb-ft Car F: 3000 lbs 400 HP 280 lb-ft Your formula iindicates each car is equal, yet on the drag strip, the one with the bigger TQ wins almost all the time. Why?
You are confusing yourself with minutia. If the peak torque values are at the same RPM then there is more area under the hp curve for car V.
And buy your own definition/comment (above) the HP does not the provide acceleration ... Again F = m*a, has been used for many years and a = acceleration. Pete ps: Note: area under the HP curve is NOT HP ...
Too many people here are confusing themselves with gear ratios, weight, traction, etc. Just like a Congressional hearing, the simple facts get clouded overwhelmingly by details and BS. A simple challenge to the other posters on this thread: If anyone here can get a reputable physicist to truthfully deny this statement I will GIVE them my Modena: "Knowing nothing more than the engines maximum horsepower, is a predictor of maximum acceleration of a Ferrari Modena. Knowing nothing more than the engines maximum torque is not." Put up or shut up, as the saying goes. I'm off to the Bikini thread, you guys can have the big Torquey girls.
**** teak360, you really should not say such things on the internet ... it could be binding ... Even I can calculate the acceleration of a F360 Modena with just the engines torque figure ... infact I think I have already done it on the other thread ... or atleast 90% of the calculation. So who is going to pay for the shipping to Australia? Pete
teak360, From the other thread that we suggested you read. I did not go as far as calculating acceleration ... but it is just a a = F/m away ... Okay for those that need real numbers, I have obtained figures for the 360 Modena: Max Hp - 395.1 Hp @ 8500 rpm Max Tq - 275.0 ft lbs @ 4750 rpm Rear Tyres - 275/40 ZR 18 (thus approx 670mm dia.) Weight - 1390 kg Ratios: 1st 3.29:1 including diff (4.44:1) = 14.6076:1 overall. 2nd 2.16:1 thus 9.5904:1 overall. 3rd 1.61:1 thus 7.1484:1 overall. 4th 1.27:1 thus 5.6388:1 overall. 5th 1.03:1 thus 4.5732:1 overall. 6th 0.85:1 thus 3.774:1 overall. Right torque at peek power: Tq = Hp/(rpm/5252) = 395.1/(8500/5252) = 244.13 ft lbs at engine. Thus lets work out how much force is applied by the rear wheel(s) to the ground at peak torque and peek horsepower per gear: 1st: @ 8500 rpm we have 244.1 ft lbs x 14.6076 = 3566.085 ft lbs. Thus using T = F.d (and it is this simple Tim, no dot products. T at ft lbs or Nm is simply a force in lbs or N times a distance in feet or Metres) Thus F = T/d and d is the radius of the wheel (1.11 ft or 338.6 mm) = 3566.1/1.11 = 3210.1 lbs to the road via wheel. @ 4750 rpm (peek torque) we have 275 ft lbs x 14.6076 = 4017.1 fb lbs. Thus the force applied by the wheel is = 4017.1/1.11 = 3616.1 lbs to the road via the wheel. Thus using a = F/m you would be able to obtain the acceleration. Note that the force is greater ... thus as mass is constant the acceleration WILL be greater 2nd: @ 8500 rpm we have 244.1 ft lbs x 9.5904 = 2341.3 ft lbs. Thus F = T/d = 2341.3/1.11 = 2107.5 lbs to the road via wheel. Thus using a = F/m you would be able to obtain the acceleration. @ 4750 rpm (peek torque) we have 275 ft lbs x 9.5904 = 2637.4 fb lbs. Thus the force applied by the wheel is = 2637.4/1.11 = 2374.1 lbs to the road via the wheel. Thus while maximum force is applied to the road at peek torque, due to the large ratio increase revving to maximum rpm in first does apply equate to maximum acceleration, ie. a = F/m. Lets look at 3rd gear: @ 8500 rpm we have 244.1 ft lbs x 7.1484 = 1745.1 ft lbs. Thus F = T/d = 1745.1/1.11 = 1570.9 lbs to the road via wheel. @ 4750 rpm (peek torque) we have 275 ft lbs x 7.1484 = 1965.8 fb lbs. Thus the force applied by the wheel is = 1965.8/1.11 = 1769.5 lbs to the road via the wheel. Thus again due to the large increase in ratio, maximum rpm in 2nd equates to more force on the road than 3rd at peek torque ... but it is getting closer. 4th gear: @ 8500 rpm we have 244.1 ft lbs x 5.6388 = 1376.6 ft lbs. Thus F = T/d = 1376.6/1.11 = 1239.2 lbs to the road via wheel. @ 4750 rpm (peek torque) we have 275 ft lbs x 5.6388 = 1550.67 fb lbs. Thus the force applied by the wheel is = 1550.67/1.11 = 1395.9 lbs to the road via the wheel. 5th gear: @ 8500 rpm we have 244.1 ft lbs x 4.5732 = 1116.4 ft lbs. Thus F = T/d = 1116.4/1.11 = 1004.9 lbs to the road via wheel. @ 4750 rpm (peek torque) we have 275 ft lbs x 4.5732 = 1257.63 fb lbs. Thus the force applied by the wheel is = 1257.63/1.11 = 1132.1 lbs to the road via the wheel. As you will notice the higher the gear the closer the peek torque of the higher gear is getting to the same force as the lower gear at peek Hp. teak360 ... I do not want your F360 Modena, because this is just a discussion and not worth that sort of deal. Note I would love to own one, but it would not be fair or the right way of obtaining one. I could get a physicist to verify the above calcs if you like ... but I would not want to waste their time. Pete Note: I only included HP in the above calculations to prove a point and thus calculate Torque at peak HP ... and also to make sure that I broke some of the rules of teak360's bet. Also acceleration is an instantaneous measurement that changes over time ...
