We have some unbelievable discussions regarding the acceleration of race cars. At first glance, many just look at what they remember, F=ma and go on long rants regarding engines that have high levels of torque. However, even when you rearrange the term to a=F/m, its still a Force at the tires that causes acceleration, not at the engine before the gear box. Many different types of race cars have all sorts of engines. some with high torque and some with low torque, even some of those might have equal HP output. Since HP is a rate of doing work, a rate of change of kinetic energy, if a two cars were to be compared and both had the same HP, yet one had half the torque output, if their HP curve shapes were the same, the rate of acceleration at any vehicle speed would be the same. Correct? This rate of acceleration is proportional to Power at any vehicle speed. (all other things being equal, such as the car's weight, speed) Acceleration=Power/(mass x velocity) Can you guys here enlighten the folks in the racing community regarding equations for acceleration of a race car and the relationship of thier engines torque output to the torque and force found a the wheels to accelerate it at any vehicle speed, anywhere on the track. I posted a set of engine HP/torque curves that showed that even with lower engine torque values, one engine can produce more rear wheel torque and force at any speed, because its HP curve was boader. Not the rule, but the exception that a good percentage of the racing community cannot accept. Here is the graph that was posted based on the original question, "What would be better on the race track if the two engines in question, both had the same HP but one had more torque than the other. " From the data provided from this person posing the question, I put it in a graph and looked like as follows:
There is a reason I don't participate in these types of discussions on the automotive forums that I used to, or still do frequent. It's a waste of time. The people who understand have no interest in participating. The ones who are uneducated and believe the answers can be found in a handful of numbers are drawn like moths to a flame, and for the most part will resist anything different. Yes, but to make your comparison (the graph later) you have to ignore any mechanical advantage. The correct comparison for what you want is chassis dyno info (it makes no difference really from your engine graph, but it seems silly to discuss forces at the tire with flywheel data), which is done in a 1:1 gear ratio. You demonstrate later in your post that you know peak numbers don't mean much, so why do you allow yourself to use peak numbers to describe entire curves here? This mistake is the same one that you are arguing against with your car buddies. What you would be concerned with as far as HP curves for equal acceleration would be the area under the curves being equivalent. This could be done either by having the exact same curve shifted along the rpm axis (accounting for differences in peak torque), or with a different curve that has the same area below it (allowing for more narrow or broad rpm ranges). I understand what you mean, but again your wording is poor. The difference in torque at the flywheel and the wheels (in 1:1 ratio) is simply a difference of drivetrain losses and will not change. HP is a function of torque and rpm and therefore you can supplement lower torque with higher rpms. Again, the important thing will be area under the curve. That's all I have at 4:30 AM :zzz: . I'm trying not to be overly critial of your post because you recognized an important concept about the curves whereas most never will. Try not to concern yourself with what others on the internet don't/can't understand.
All you really need to do with engine torque is multiply it by the gear ratio to get the torque at the wheels, then divide by wheel diameter to get the force.
Thanks for the response. Actually, I could have been more clear. the curve I posted was set from the guidlines of one of the posters on that discussion. IT IS REAR WHEEL HP AND TORQUE values. It would open an entire set of variables if we start talking gear box efficiencies, etc. I used the peak torque and HP, because those were the only requirements, making it easy for me to debunk the idea that engine torque matters at all as a judging value. am I correct in saying that the HP at any vehicle speed will determine the rear wheel torque and thus forces if the tires are the same diameter? I think so and its been the the foundation of the argument. In otherwords, it doesnt really matter what the engine torque peak or value is, if the HP is the same at any vehicle speed. (meaning the shape of the HP curve is the same for both) then, both vehicles will acceleratae at the same rate at any vehicle speed. I used the flywheel data, as measured at the rear wheels by a dyno. we have to assume that the gear rations of the comparison for the two engines of equal HP, will be adjusted to have the same final vehicle speed at each maximum rpm point on the engine, AND the same % rpm drop for each shift, to keep the argument centered around what the accelerative potential is of the two engines. sure, we need to look at the optimal use of HP to create the maximum amount of HP available to the rear wheels, using gear box ratios. I dont think this is done by looking at the area under the curve, because wouldnt the time spent at the different HP levels be more accurate? So, I start thinking in terms of "HP-seconds", unit measures of work, because at the higher rpm levels, near maximum engine speeds, the acceleration rates will be less and more time will be spent at those levels. I know the concepts are simple, but the application is relatively complex, correct. you have variable power on the curve, non-constant jerk, (is that a term?) and a host of other variables we are not even talking about regarding the acceleration of a vehicle on a race track. Getting back to your last paragraph about my wording. maybe i wasnt making my point well enough. my point was that the engine power determines rear wheel torque as it is multiplied through the gear box at any vehicle speed. sure engine torque will be multiplied through the gear box to get rear wheel torque based on the reduction ratios for a given speed. So, as an example, if both cars are at max hp, the rear wheel torque will be the same, even if the engine torque is 1/2 for one vs the other. (because the rpm are traded off for gear reduction). of course it is consistant with the equation, Acceleration = power/(Mass x velocity) where the power determines the acceleration rate at any vehicle speed, not the engine torque numerical value. Thanks for the replies Mk
Yeah, I think they all get this, but some how cant get their arms around the power being the driving factor. The main point here is that torque is a factor of power. power determines what torque you are going to have available at any vehicle speed. Seems most get it backward, after all power is the rate of doing work. heck, force is just a factor of work. What is the formula for acceleration of a mass over a speed range based on power? since power is the rate of change of kinetic energy, Is there a formula that ties the two together in one neat equation? Thanks,
Power in watts = torque (in Newton-meters) times RPM x 2 pi/60 power in HP = power in watts/746 So a torque of 300 Newton meters at 6000 RPM is 188,500 watts or 253 HP. Getting the right torque to the wheels is selecting the right gearbox ratios. If torque is in pound feet, multiply by 1.356 to get Newton meters: So a torque of 300 pound feet is 407 Newton meters 407 Newton meters x 6000 RPM x 2 pi/60 = 255,700 watts = 343 HP
Im not sure what your point is below. You just gave he difference in power output using the same rpm but different amounts of torque (i.e 300ft-lbs is different than 300NM) The point of all this is that with the same power output, the rear wheel torque at any speed on the road will be the same. true? Power dictates power to the rear wheels at any vehicle speed. Obviously, engine torque is going to change inversely with RPM needed to acheieve the same vehicle speed, if both cars have the same power output at any same vehicle speed. For example, if we have a total gear ratio of 8:1 for a high torque engine (most first gears of most cars) and have the same size diameter tires, at 300hp, and 300ftlbs of torque at 4000rpm, i would have 2,400ft-lbs of torque at the tires. A same hp 300hp car with 150ft-lbs of engine torque and 8000rpm would have 2,400ftlbs of torque because it would be running a 16:1 gear ratio in that same gear. All other gears would follow suit, because the RPM drop % would be the same (very common) and the HP curve shape is the same. In some cases, the curve might be more optimal if broader for a lower torque engine, so it would have more accelerative forces at any vehicle speed. this is the crux of the argument I'm trying to show those with a little less basic physics 101 knowedge than I have :), how acceleration is determined by HP available, not engine torque at any same speed. With distilling out a couple of newtonian identities, a=F/M and P=Fv, I use a=p/(mv) to show that acceleration is proportional to power at any same vehicle speed. (and same car ). This isnt going over well with folks that think and say "Torque gets you out of turns and hp gets you down the straight". :) Thanks for the comments!
You are confusing and misusing terms, as I tried to be polite about in my first post. HP is a function of torque and rpm, period. There is no way for two equivalent HP values at different rpms to have the same torque. If you are considering your graphs to be at the wheels, then you already know exacty what the torque is at the wheels. I know what you are trying to say, but you are doing a poor job of it and as is it is technically incorrect. I was trying not to focus on that, as it is not what is really important at this point.
Dont worry, im conferring with you guys because I want it to be technically correct. so, I appreciate the input. If you read what i wrote again, you can see that I said that two cars with equal Hp, one having more "Engine" torque than the other, will still produce the same rear wheel torque or rear wheel Force (as measured after the gear reductions) at the same vehicle speed. (since we use 24" diameter tires, the force would be equal to the torque found at the rear wheel input shaft :) ) Maybe its confusing because there are two types of rear wheel torque. Rear wheel torque that is calcuated by the dyno, relating to flywheel torque, and the rear wheel torque that is responsible for the rear wheel force on the pavment. I'm trying to be clear with the two, and is why I say, "Rear wheel torque, as multiplied through the gear box", when im speaking about the end force that is developed at any same engine HP level. So, when I say, "Engine HP determines rear wheel torque" I'm saying that Engine HP detrermines Rear wheel Forces at any vehicle speed. Again, the terms are: Engine HP= obvious flywheel HP= obvious Rear wheel Hp= that which is measured at the dyno Rear wheel torque= that is which is measured at the dyno,but relates to torque at the flywheel as measured and calculated at the rear wheels using engine RPM Rear wheel torque/force= that which is found at the rear wheels as acting on the pavement. This value is engine torque multiplied by gear ratios at any given vehicle speed. Maybe this should be called "Wheel Force" and forget about torque as it can be confused with rear wheel torque that the dyno calculates. So, with your first objection, I'm talking about the same HP of two engines making the same rear wheel torque. (the 2,400ft-lbs value) made up from either 300ft-lbs at the engine at 4000rpm or the 150ft-lbs made by the other engine at 8,000rpm. Both engines have the same HP output at that vehicles same speed, and thus will have the same rear wheel forces acting on the pavement. If there is a better way to explain this, I surely would like to hear it. Its all in an effort to show that power of the engine determines acceleration at any same vehicle speed. Is ok, I have a thick skin. I want to be techically correct and explain it in a way that is not confusing. Thanks
Going back to your first post: From the equation that relates torque and horsepower, getting half the torque at the same horsepower requires double the rpm. You tell me - does a car that requires double the rpm to generate the same horsepower accelerate slower or faster? Or perhaps more telling: if these two cars get the same horsepower, torque, and acceleration at a given speed (their hp to the wheels is the same) and to do that, one has to operate at twice the rpm, what does that tell us about the horsepower at the engine? And unless one car redlines at twice the rpm as the other, what does that tell us about their max hp and max speed?
WE need Mike in here. With regard to one of your questions in the OP. "What would be better on the race track if the two engines in question, both had the same HP but one had more torque than the other. " The one with more torque, torque adds drivability to an engine. Thats an easy question to answer. But what if the question was "What would be better on the race track a turbo 4pot 500HP with 200Nm (avg) torque, or somewithing like a v8 with 450HP and 350Nm (avg) torque (these are figures I just plucked out of the air)" The answer would depend on, whats the track, whats the gearing of both cars, how good is the driver etc etcetc. In this case its not as simple as the first, as the car with more torque would accelerate harder in any given gear lower down, but has less ultimate grunt, the engine would feel 'torquey'. Its also importent to remember that the higher torque gives an 80% max power reading across a wider rev range, increasing drivability. With the other engine, all the power and acceleration comes at he top end. However this isnt as bad as it seems for acceleration as some of this low down loss can be eiminated with clever selection of gears. So what would you want here. If the track had long open straights, and relatively few slow corner to fast straight, you'd be better with the HP of the turbo 4pot. If the track had lots of slow corners mid length strights you'd want the extra torque of the v8 for inital acceleration. I'll try to find a nice link that explains this better.
In the way you're asking about this, it doesn't really matter. Assume that the horsepower versus RPM curves look the same, with the only difference being the rpm scale at the bottom of the chart. Say the chart for the higher torque engine shows rpm from 0 to 5000 rpm, spaced 1000 rpm per inch across, while the one for the lower torque engine shows rpm from 0 to 10000 rpm, spaced 2000 rpm per inch across. If the curves appear to be the same, then the performance will be the same. If you then overlay these charts with torque versus rpm, then the first torque curve will be twice as high as the second torque curve, but as you stated, the actual torque curve doesn't matter, it's the power curve. The shape of the torque curve matters, but this would show up in the power versus rpm curve.
Eh... Jeff thats not really comprehensive. Are you saying if the power curves look the same it'll go the same? Edit: also people dont take into account the fact that its the gearbox thats the critial factor of how the engine power gets put down. All people seem to talk about is single value readings, this is a horrible way of determining performance.
I don't agree, but the difference might be in the testing.... Yes, assuming one can top out at a much higher rpm than the other. But what is being tested to get that curve? Are we talking about a car sitting on a dyno in first gear? Do they usually go through all the gears? Or is this just the engine output curve? -If it is the engine output curve, then the one that operates at lower rpm has lower drive losses and performs better when installed in a car. -If this is a dyno run in 1st gear and the curves produced are identical except for the engine rpm scale at the bottom, drive losses decrease as you go through the gears, so the car that operates at higher engine rpm accelerates faster...and generates more horsepower at the wheels than the other engine, as both accelerate. As with a lot of questions, most of the problem here is with the question being not specific enough. If you make the question specific, the answers are pretty straightforward - these are relatively simple issues.
Normally you'd do a dyno test (all the ones i've done) with as close to 1:1 drive through the gearbox as possible (usually 4th gear). Yes I know about the final drive at the diff, I dont really know why, i'd guess its so it doesnt rev up to the limiter too quickly.
The reason the gear closest to 1:1 ratio (usually 4th gear) is chosen, is that normally the drivetrain losses are less in that gear than other gears. Going back to my point about power versus rpm curves, assume that the dyno power curves (force versus speed) are the same. The car with the lower torque, double the rpm engine, would have a final drive ratio with twice the reduction factor. I think the real issue here is the shape of the power or torque versus rpm curve. Generally more power can be acheived with a peakier curve, at the sacrifice of a smaller rpm range where that high power is acheived. The spacing between gear ratios and shift speed are factors in how narrow the powerband can be. Formula 1 cars have 7 closely spaced gears, with computerized shifts that take 30ms to 50ms, so a fairly narrow power band can be used. A race car with fewer gears and a larger spacing between gear ratios will need a wider power band. Nascar race cars are only allowed 4 gears for example, and they run at a few road courses, like Infineon, each season. In race classes where computerized shifting isn't allowed, shift times become significant, and translate into fewer and wider spaced gearing. On a side note, with improvments in metal alloys and engine design, piston speeds in sport vehicle engines have increased to result in more power, especially motorcycles. The Corvette Z06 has a 7 liter (427.6 in^3) engine that redlines a bit over 7000 rpm, to provide a minimum (SAE) of 505 hp at the flywheel, translating in to 440 hp a the rear wheels (the average is more like 450 rwhp). Note the Z06 engine is liter than the turbocharged Porsche 911 3.6 liter engine (480 hp for turbo 911, 530hp for the GT2), and gets better gas mileage. In the case of motorcycles, a 1 liter road racer replica will have about the same 1/4 mile performance as a 1.3 liter superbike (Hayabusa), but the 1.3 liter bike gets 80% of peak torque for the upper 2/3rds of it's rpm range, while the 1.0 liter bike gets 80% of peak torque only in the upper 1/3rd of it's rpm range, half the powerband width. When crusing on a freeway at around 65mph, in top gear (6th), the liter bike won't be able to accelerate as fast as the superbike (1.3 liter engine) without downshifting to get the rpms up. In the case of Suzuki, the 1 liter engine redlines at 14,000 rpm, while the 1.3 liter engine at 11,000 rpm.
Hmm thinking about it, thats obvious really as the 4th gear locks the input and output shafts. Bit of a brain fade moment :P
Not in the case of a manual transmission. It's just another pair of gears, in this case equally sized gears that generally have a bit less power loss than the other gear combinations. Motorcycles transmissions never bother with this, losses with a tranny and chain drive are about 8%. Strength of the gears is a factor in determining gear ratios. The bigger the difference in gear size, the less contact (fewer gear teeth engaged) between gears. To compensate the gear widths are varied. Back in the 70's and 80's rear end ratios were less than 3.00, and first gear ratio over 3.00, but since the 1990's, this was swapped to reduce gear width in the trannies of the higher powered cars (Z06, Viper) to allow more gears (6 instead of 4) and to be able to handle more engine torque. Using the C6 Z06 as an example, 1st gear ratio is 2.66, while the rear end ratio is 3.42.
In MT it locks the shaft. The layshaft isnt used so it has to. EDIT: most the gearboxes i've seen have run direct drive as its better then running a ratio of 1:1
I forgot that some MT's still lock the shaft. I doubt racing cars transmissions do this though. I don't think the Tremec T56 or T6060 do this either.