Independent of aerodynamic drag, at what rpm does an rx8, or any car for that matter, actually accelerate the car the hardest? Peak torque seems like a viable choice, but then so does peak hp...or anything in between...

This is a tough question that I have pondered for years and I am curious what others say.



oh and obviously if you say "first gear" you missed the point.

 

Peak torque. Max speed happens at peak hp

 

Peak tourque or HP at what RPM, dyno the car but different dyno's will conclude different results.

Hey Carbon I've ran into you twice on the beltway going up 270, I was in the Nordic Green few months back. You were in the red rx8 right, just trying to figure out who was who???

Anyways back to the OP's topic.....

 

Are we talking the perfect CVT? Or a car with actual gears? If you have a car where 30 mph is at the peak HP in second and peak torque in third, the car will accelerate fastest in second. However, if you go into a specific gear of fixed ratio, the acceleration of the car will follow the torque curve.

If you are driving at peak horsepower, the gear ratio you are using is putting the most torque possible to the pavement. You can't drive a car at constant RPMs without a perfect CVT though.

 

I have a problem with the "peak torque" or "follows the torque curve" statements. They seem logical, but I am not so sure. Here is why: If you look at dyno charts uploaded onto this forum, you will notice that a typical stock 8 shows a torque curve with figures that are VERY close when comparing the 3500-4000 range to the 8000-8500 range. So if the statements are true, then the car would pull just as hard at 3500 rpms as it does at 8500 rpm. BUT IT DOESNT. At 3500 rpms in say, 2nd or 3rd gear, the car feels pretty weak, but in contrast, at 8500 rpm, the car feels much peppier and certainly seems to pull much stronger.

Explain...?

 

Take the torque delivered by the engine and multiply by your gear ratio, diff ratio, tire radius - napkin tire force on the ground. F=MA. So if you shift into a gear at low RPMs, you should feel acceleration that matches the torque curve. Given that losses generally mount with speed, I should think at 8k rpms, your acceleration is actually lower at 8000 rpms than at 3500. However, it would be impossible to get the car to run at 3500 rpms and go that speed without reducing your gearing advantage. The "feel" of acceleration can be entirely psychological. Minus an actual dyno chart to reference, I'm just taking your torque curve description at face value.

 

Someone take an iphone, download a g-meter app, and go in a straight line. That would help remove any misconceptions based on sight and feel.

My 2-cents: Torque is the force, therefore peak torque is the highest mass times acceleration possible.

Now, what you are saying is "pull", which might not ride entirely on the peak torque, but also how steady the torque is after or before the peak.

 

did u just say an rx8 pulls harder at 3500 rpm vs 8000 rpm? That might be a bit hard for me to swallow...

at 3500, in 2nd for example, the car bogs when floored and doesnt really pick up till about 4500 or so. At 8000 rpm in 2nd, hell, you can power slide the thing. The motor is definitely pulling harder at 8000, despite having equal torque as 3500-4000

 

OMG that's a good f'ing idea. ok someone do this please cuz i really want to know and I don't have an i phone. Just do this: get in a nice healthy stock 8, get going like 10 mph in 2nd gear, and floor it to redline. Have someone sitting next to you or whatever record results.

link for app info: http://iphone.wareseeker.com/trackpe...app/42ef28df87

 

So you're power sliding an Rx8 at 55 miles per hour? Get an accelerometer. This is all according to physics. Are you sure your car makes the same tq at 3500 and 8000? Powersliding its self is irrelevant as the speed at which you are cornering factors in.

 

Well, don't confuse torque at the flywheel with the amount of force hitting the payment. A car may make less fw tq at HP peak, but it has a higher gearing advantage than if you were to "upshift" the gearing to bring the car to its tq peak. I'm pretty sure it's mathematically impossible to put more force on the pavement than what a car exerts when at its hp peak. It has the best combination of tq and speed at that point (hence HP).

 

Here's a couple links to two rx8's which show how the torque bulges in the middle, peaking around 5500-6000, and both ends go down. You can see how 8000-8500 is typically about the same in terms of torque for 3500-4000 rpm. These are just two, but you can find many many others just by looking on the rx8 club forums.

http://www.dragtimes.com/images_dyno...-RX-8-Dyno.jpg

http://www.rx7.com/RPGReddyRX8exhaus...ffler_dyno.gif

 

Wait, if you are putting the most force down to the pavement at peak hp, then how would the car pull equal to or even harder way down at 3500 like you said? Remember, I am talking about a single gear full rpm range pull, so don't worry about shifting and all this confusing stuff. also, as already stated, i am talking independent of wind resistance as a car speeds up.

 

I'll stipulate to the torque curve. Even if it wasn't true, torque generally has a curve to it on these engines so you will see high rpm points and low rpm points with similar torques.

Don't forget about speed. Accelerating at 30 mph and accelerating at 10-15 are not the same HP wise.

Coming up with an example.

 

Say your engine puts out a flat 150 ft/lbs. First gear is a pure 4:1 and second is 2:1. At 9000 rpms, you're doing 2250 length/time in first. The same speed occurs at 4500 rpm in second. However, the "pavement force" is 300 in second. It's 600 in first. So in first you accelerate twice as hard as in second at that speed.

Now if a car is pulling X hard at 3500 rpm, if you can find a gear lower than that, it can pull even harder if you get near the HP peak. Hence downshifting.

 

ok so with the engine you described, le'ts imagine you just keep it in one gear for the entire rev range. Now because this mythical motor has the exact same torque at all revs, (maybe it's electric...), would the actual acceleration G's be the same for entire rev range in that gear? Or would it gradually increase as rpms increase, as bhp increases, since bhp = work over time thus more work getting done faster?

Something tells me it would pull the hardest at high revs, but alas, I can't explain why, so I still don't really know.

 

I'll chime in before RotaryGod sees this and throws a shitfit...

Peak horsepower is when the engine will pull the hardest.

Torque is in instantaneous force, and tells you nothing about how fast work is being done. It is only one variable in the equation that actually tells you how fast your car is going to be:

HP=(Torque*RPM)/5252

Horsepower tells you how fast your car is using it's available torque by applying engine speed to it. If you want to know how much stuff your car can tow, the torque number is what to look at, but if you want to know how FAST it's going to use that torque you need to look at horsepower.

 

According to my argument, it would pull exactly the same in a frictionless vacuum and probably slightly less the faster it went in actual application.

 

The only way that theoretical engine would pull exactly the same throughout the RPM range is if the torque curve was negative. That is, peak torque is at idle, and it gets lower as the RPMs increase in such a manner that Horsepower remains constant.

 

So you are saying torque is not directly proportional to acceleration?

 

The problem is that most people seem to think of torque as the ONLY variable that matters, but in reality it's the COMBINATION of torque and RPM that gives you acceleration. If either one increases, your rate of acceleration will increase.

So in that theoretical engine, if you maintain constant torque, but increase RPM, your rate of acceleration will increase throughout the RPM range. At the same time, if you somehow increase torque while maintaining constant RPM your acceleration will increase. Both are important in figuring out how fast a car really is.

 

So just use HP and call it a day...

 

Yes so it looks like some are saying peak bhp, some say peak torque. I would tend to lean towards saying peak bhp, but I just can't seem to get the following fact out of my head:

any motor at peak torque requires the most force to equal it and stop it from increasing rpm, so why would the car not pull the hardest at peak torque?


so far i have seen some interesting and thought provoking discussion, but nowhere near a conclusion.

 

Whatever car - that is making more HP at a given speed (MPH) will accelerate faster.... assuming the same weight.

 

Exactly.

Think of it this way: Which is faster, a Formula 1 car or a Semi Truck (with no trailer)?

Obviously it's the F1 car, even though the Semi produces more torque. This is because the F1 car make more HP. It compensates for the low torque by having extremely high RPMs, thus giving it a high HP number.

The truck's redline is so low that it doesn't have enough RPMs to give it a very high HP number. Some Semi's may produce 1,500+lb/ft of torque, but only ~400HP. The F1 car may produce 700HP @22,000 RPM while only having 167 lb/ft of torque. It's the 700HP that makes it fast, not the 167lb/ft of torque.