Stupid and unrealistic "Which is faster?" question

Why do deeper gears give better acceleration, generally speaking? Because they provide torque multiplication. If I have 100 lb-ft and put it into a 4:1 rearend, I get 400 lb-ft out.

The 8500 rpm engine makes less torque. It has already used up the gearing advantage in order to make the same axle torque as the 6500 rpm engine at the point where we are doing our comparison.

At that instant, the power output at the axle is equivalent. The crankshaft torque (the input to our rear end) is not equivalent. The torque multiplication of the deeper gears which you feel should give better acceleration is already in use making the lesser torque of the high-rev motor equivalent to the greater torque of the low-rev motor after the rear end.

Gearing affects torque, not power. If an engine is putting out 400hp and 400 lb-ft of torque (at 5252 rpm, obviously), and you run that into a 2:1 reduction, you'll get out 800 lb-ft at 2626 rpm, but still 400 hp. Gearing never gives you free power (or takes it away), it only allows you to change where/how you're using it.

Swank Force One wrote: Why are the only diesel race cars out there the ones that are forced to be diesel by their class rules, then?

Because they're not starting from a predicate of "same peak power at two different revs". Er, is that true? I think of racing diesels in terms of the Audis and Peugeots in LMP1, and I thought that class allowed gas as well, but the diesel guys lobbied for rules making diesel the better choice in part to showcase diesel technology for car sales? But I really don't know, that's a hell of a tangent, and I really haven't a clue...

If you can make the engine live, and keep it breathing as well (tall order), you can make more power by revving higher. Rev it twice as high while making the same torque, and double the power...

There can be an advantage to revving higher, but that's in terms of being able to make more power. There's not an intrinsic advantage to revving more for the same power output.

I think i need to actually run the rear end ratios and come up with what i need for these two scenarios instead of flying seat of the pants. Might come up with a different answer. Going to use a 23" tire.

Oh yeah.... i was WAY off.

Here's some numbers for BOTH cars.

1st 3.483:1
2nd 2.015:1
3rd 1.391:1
4th 1.000:1
5th 0.719:1

This gives Car A (8500rpm peak, 9500rpm rev limit) a 4.5 rear end to reach 200mph.

Car B (6500rpm peak, 7500rpm rev limit) would have a 3.55 rear end.

Does this change things?

Or is the answer truly as simple with the information that we have as "Whichever one actually makes more power in the real world."

[edit] Did the math, looks like it doesn't quite change things, but it brings it a LOT closer. If we're talking torque multiplication, at peak:

Car A = 1668
Car B = 1721

Swank Force One wrote: My problem with this is that HP isn't used to figure torque, rather HP is a synthesized number based on torque/rpm. And what i may be going at this incorrectly is where i see that there's two ways to make the same horsepower in this case. They're both making 600hp, just at different times, in different ways, but the end result is the same, is it not?

Simplified to the point where it can't be simplified any further:

No matter what the engine's torque or gearing or anything, a given amount of horsepower at a given speed WILL have the same amount of torque at the wheels.

If the engine has less torque, to make that horsepower it has to be revving faster, so to be going the same speed it has to have shorter gears, which multiply the torque more.

So. Ignore notions of torque as a number. Horsepower is ALL that matters for acceleration.

I didn't read it all. To compare the original 'A' to 'B', you have to assume a distance is involve. If the distance is in the operating range of 'B' (SHORT), 'B' wins. 'A' would need 4.22:1 to be in the hunt.

BUT

Depends on too many other factors

Consider our Stude

1 mile

3.25:1 gears 168.202 (7230 rpm)

3:00:1 gears 168.561 (6690 rpm) (0.07 Gs / Highest 0.63 Gs at 53.6 mph in 1st gear)

1.5 mile

3:00:1 gears 179.644 (7130 rpm) (0.02 Gs)

2.91:1 gears 178.707 (6880 rpm)

Gear splits are critical to a high speed engine. 1st - 0.63Gs at 53.6 mph, 2nd - 0.48Gs at 64.0 mph, 3rd - 0.39Gs at 83.04 mph, 4th - 0.19Gs at 127.33 mph. Too much gear split in 4th killed the recovery and by the time in cranks back up its pinned down by the air.

Long story short...too much blah, blah, blah to say the original question did not have enough information to answer.

Having just read all 4 pages -- I would say that since the Force only works on the feeble minded, Chewbacca (car B) would take out Darth Vader (car A).

How did bradyzq's post not get any likes before i clicked it? He answered the EXACT question that was being asked!!

Car B is accelerating harder at 6500 than Car A is at 8500RPM. A: 600 5252/8500 = 371 torques @ flywheel * 3.9:1 rear end ratio = 1447 lbft at the wheels B: 6005252/6500 = 485 torques @ flywheel * 3.23:1 rear end ratio = 1566 lbft at the wheels So, Car B is accelerating 8% quicker.

We're looking at the potential acceleration at a given point in time. For this all we need is axle torque [because all other factors are assumed to be the same in both cases]. From there we can easily calculate pounds of thrust - at that moment. ... all we really need to know is axle torque because everything else is the same.

^^ EXACTLY! And we already HAVE the axle torque numbers. Bradzq ANSWERED the question.

Vigo wrote: How did bradyzq's post not get any likes before i clicked it? He answered the EXACT question that was being asked!!

Car B is accelerating harder at 6500 than Car A is at 8500RPM. A: 600 5252/8500 = 371 torques @ flywheel * 3.9:1 rear end ratio = 1447 lbft at the wheels B: 6005252/6500 = 485 torques @ flywheel * 3.23:1 rear end ratio = 1566 lbft at the wheels So, Car B is accelerating 8% quicker.

We're looking at the potential acceleration at a given point in time. For this all we need is axle torque [because all other factors are assumed to be the same in both cases]. From there we can easily calculate pounds of thrust - at that moment. ... all we really need to know is axle torque because everything else is the same.

^^ EXACTLY! And we already HAVE the axle torque numbers. Bradzq ANSWERED the question.

Nope. That is 8% more torque. Not acceleration. There is not a 1:1 correlation between torque and acceleration.

Knurled wrote:

Swank Force One wrote: My problem with this is that HP isn't used to figure torque, rather HP is a synthesized number based on torque/rpm. And what i may be going at this incorrectly is where i see that there's two ways to make the same horsepower in this case. They're both making 600hp, just at different times, in different ways, but the end result is the same, is it not?

Simplified to the point where it can't be simplified any further: No matter what the engine's torque or gearing or anything, a given amount of horsepower at a given speed WILL have the same amount of torque at the wheels. If the engine has less torque, to make that horsepower it has to be revving faster, so to be going the same speed it has to have shorter gears, which multiply the torque more. So. Ignore notions of torque as a number. Horsepower is ALL that matters for acceleration.

No wonder I'm confused... this is a direct contradiction to the other most vocal contributor to this thread.

Le sigh. But this sounds like the "gearing doesn't matter" argument again.

Guys, this is quickly turning into a classic internet physics argument, better quit while you still can.

In reply to Swank Force One:

I don't think what Knurled said is disagreeing with me, or with wvumtnbkr for that matter, except perhaps in our interpretations of what you're asking (well, and I never said to ignore torque, but again, that's more down to interpreting the question, not a disagreement about how this works).

I've been focusing on whether power at different rpm levels is still the same amount of power, and the gearing machinations needed to make these equivalent at the wheels.

Note that Knurled says "a given amount of horsepower at a given speed WILL have the same amount of torque at the wheels". That's road speed. It's a good, simple way of putting it, but note that it does require a gearing correction between the two cars being compared so that the horsepower graph aligns on the road speed if they make the same power at different rpm.

If two otherwise identical cars' dyno charts are dead flat at X hp between 50 and 70mph, they will accelerate at the same rate, regardless of the gearing needed to put them in that part of their dyno chart. I'm just rephrasing what Knurled said, which I agree with.

Kenny's probably right, except that I think there's a fair amount of agreeing with different words going on...

Nope. That is 8% more torque. Not acceleration. There is not a 1:1 correlation between torque and acceleration.

You're cherry picking the one little part of that quote that's not true and missing the point.

One setup delivers higher torque to the wheels. Given the same weight, it WILL accelerate harder. THE END.

Ok, I can't help myself.

Torque at the wheels, as a lone value, has NOTHING to do with how hard the car will accelerate. Torque is force vector, it doesn't move jack E36 M3, that force over a distance does. Your car accelerates faster in lower gears because of wind resistance and lower driveline speeds having less losses. If these were ignored, you could go 300+mph with a 5hp go kart motor, or even a hand crank, it would just take all day.

The only thing in this example that effects how fast the car will accelerate is how much power it makes and how often it makes that power. Whichever car has a flatter powerband and better optimized transmission ratios will accelerate faster in the given example.

There is never something for nothing, and you can never solve a problem in which you have more variables than equations. You can not quantify this situation.

Swank Force One wrote: No wonder I'm confused... this is a direct contradiction to the other most vocal contributor to this thread. Le sigh. But this sounds like the "gearing doesn't matter" argument again.

Gearing does matter, just not here. If you want to accelerate faster, gear it up, but at the expense of top speed. If you want more top speed, add gears, lengthen gear splits, or change final drive. There is never something for nothing here, or elsewhere.

I think were most of the differences you'd be thinking of factor in when you start talking powerband and how wide or not wide it could/would be with each engine with regard to shifting, gear ratios, etc, but that's a much more complex problem.

Post script: As a drag racer, assuming traction, the most horsepower over time wins. In a road racing scenario there are many, many other things in play, as you guys tend to slow down for some reason, close the throttle, hit it again, and repeat many times per lap. There are a ton of transient effects that are present here that make things even more complicated. Add things like boost threshold into your thinking and you're head will explode. I suggest just making a metric @#)($*TON of power and you'll be good no matter what.

In reply to Vigo:

See Kenny's post directly below yours.

It is not that simple.

If it was, a steam engine would win all drag races.

I think if you want to accelerate faster you should just add more boost.

Paul_VR6 wrote:

Swank Force One wrote: No wonder I'm confused... this is a direct contradiction to the other most vocal contributor to this thread. Le sigh. But this sounds like the "gearing doesn't matter" argument again.

Gearing does matter, just not here. If you want to accelerate faster, gear it up, but at the expense of top speed. If you want more top speed, add gears, lengthen gear splits, or change final drive. There is never something for nothing here, or elsewhere. I think were most of the differences you'd be thinking of factor in when you start talking powerband and how wide or not wide it could/would be with each engine with regard to shifting, gear ratios, etc, but that's a much more complex problem. Post script: As a drag racer, assuming traction, the most horsepower over time wins. In a road racing scenario there are many, many other things in play, as you guys tend to slow down for some reason, close the throttle, hit it again, and repeat many times per lap. There are a ton of transient effects that are present here that make things even more complicated. Add things like boost threshold into your thinking and you're head will explode. I suggest just making a metric @#)($*TON of power and you'll be good no matter what.

I think i get it...

The gearing doesn't matter BECAUSE i've geared both cars to reach the same exact top speed.

What i'm still confused over is the concept that some people seem to be throwing around that it's all about torque, and not horsepower.

Which would mean that somehow, this car is less powerful:

Than this:

I can't tell which one is more powerful.

You need to calculate the area under the graphed lines for horsepower. You need to do this over the same RPM range.

Just looking at a few point, the bottom graph is making about 315 hp at 3600 rpm. The top is only making 120 hp at the same 3600 rpm.

At 4400 rpm, the bottom vehicle is making about 340 HP. The top car is making 180 hp at 4400 rpm.

However, once you get to 6000 rpm, the top vehicle is making 400+ hp while the bottom car is making only 250 hp.

Whichever vehicle has more area under the curve makes more power. The peak HP number is for bragging rights only and has little real impact in teh real world.

Rob R.

Top vehicle has a good 2000rpms where it's making more power than the second one does at peak. So during an acceleration run, at no point would that car make less power than the 2nd one.

Which would tell me the 1st car is more powerful where it matters.

Am i doing that right?

Ok, I can't help myself. Torque at the wheels, as a lone value, has NOTHING to do with how hard the car will accelerate

Missing the point. Torque is not a lone value in this situation. The other values are known and/or assumed to be constant!

Look, if Ben wants to know which one is faster @ a given road speed, the answer is both/neither because the required gearing would counteract any differences in engine torque at the 600hp rpm and it will just be hp to weight and it will be equal. To hit the same road speed the 6500 rpm motor would need a ~2.98 ratio and when you multiply 485 by ~2.98 you get the same wheel torque at the same road speed as 371 X 3.9. Same torque, same speed... same HP. Same hp, same weight, same acceleration.

But given the info that he gave us, the answer is that car B is delivering more torque to the wheels at a lower road speed and can accelerate harder at that speed than the other one is at its higher speed.

I am gonna be the ass here.

Faster as a noun infers top speed.

B given traction can be obtained as the same as A.

B is Quicker and Faster

Flight Service wrote: I am gonna be the ass here. Faster as a noun infers top speed. B given traction can be obtained as the same as A. B is Quicker and Faster

We don't know the coefficient of drag, or the transmission ratios, thus we cannot know that either.

We know the transmission ratios (posted earlier this page) and the drag will be the same, since they're the same car.

<-- did not read whole thread.

Given the stipulations in the first post, the short answer is "they will both put the same power down to the ground"

The longer answer is:

Car B will have less driveline loss. The larger the gear difference, the larger the driveline loss. It will be minimal, but it will be different.

Car B will also be faster @ the peak HP point, because at peak HP it will be traveling a lower speed (6500/3.20) vs (8500/3.9) and thus have lower aerodynamic drag.

SFO: I have a spreadsheet I built for calculating what you really want to know here... what's your email?