Stupid and unrealistic "Which is faster?" question

In reply to bradyzq:

car b is making 8% more torque at its peak torque. This doesnt not tell you anything about acceleration.

Your units are Lb*Ft. No "time" measurement in there at all.

Now... i DID say it was stupid and unrealistic... making it smart and realistic requires data that is not available.

I guess in a simpler form:

Which car puts down more power to the tire at the point of peak HP? I suppose that's the question i'm asking.

Swank Force One wrote:

bradyzq wrote: 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 lb*ft at the wheels B: 600*5252/6500 = 485 torques @ flywheel * 3.23:1 rear end ratio = 1566 lb*ft at the wheels So, Car B is accelerating 8% quicker.

Hrm. That's interesting. So i guess my followup question is: despite making the same peak horsepower, these cars are not making the same power at peak?

What do you mean by power? Hp? They are not making the same torque at peak (by your stipulation).

wvumtnbkr wrote:

Swank Force One wrote:

bradyzq wrote: 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 lb*ft at the wheels B: 600*5252/6500 = 485 torques @ flywheel * 3.23:1 rear end ratio = 1566 lb*ft at the wheels So, Car B is accelerating 8% quicker.

Hrm. That's interesting. So i guess my followup question is: despite making the same peak horsepower, these cars are not making the same power at peak?

What do you mean by power? Hp? They are not making the same torque at peak (by your stipulation).

I can't really answer that, i don't understand the question.

Car A makes 600hp/371ftlbs @ 8500rpms at the crank, 3.9 rear end.
Car B makes 600hp/485ftlbs @ 6500rpms at the crank, 3.23 rear end.

Which one is putting more power/force/whatever you want to call it at the point outlined above to the tire?

I think Brad did some math that made sense to me, i just don't know why the torque number was used instead of horsepower. I'd look at the two examples and say that they're making the same power, just in different ways. They're both over 5252rpms, they're both making 600hp.

Swank Force One wrote: Now... i DID say it was stupid and unrealistic... making it smart and realistic requires data that is not available. I guess in a simpler form: Which car puts down more power to the tire at the point of peak HP? I suppose that's the question i'm asking.

Well, if you know the torque numbers at the HP peak, then it is as simple as multiplying the torque figure at that peak HP number times the rear end ratio. (assuming same trans). This will give you the torque being applied to the ground at your HP peak. Or in other words, a meaningless number.

I am not sure exactly what you are after.

You keep saying "power". Power has a time unit with it. Therefore, this snapshot idea does not work.

Rob R.

In reply to Swank Force One:

485tq is going to be epic trying to put down through a miata...... I definitely choose option B on the simple grounds of being more of a deathtrap.

Swank Force One wrote:

Kenny_McCormic wrote: Insufficient data, faster how? Traction? Gearing? The 3.23 car might go to 60, or down the 1/4 faster, having one less shift. It might even have a higher top speed if the 3.9 car runs out of gears before aero.

None of those were the question, though. I just want to know which accelerates faster at the point of peak HP. Nothing else matters.

Your instantaneous acceleration(i.e. the slope of the line tangent to the acceleration curve), known in physics by the extremely technical term "jerk", at peak HP, will be higher on the 3.9 geared car.

Horsepower does not exist in a static condition, its a product of torque and RPM. Go back to what you learned in physics.

bradyzq wrote: 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 lb*ft at the wheels B: 600*5252/6500 = 485 torques @ flywheel * 3.23:1 rear end ratio = 1566 lb*ft at the wheels So, Car B is accelerating 8% quicker.

Going back to this example:

Why is it not:

A: 600hp * 3.9:1 rear end ratio = 2340 "magical number" to the wheel
B: 600hp * 3.23 rear end ratio = 1938 "magical number" to the wheel

That's what i'm not understanding. Using the math based on torque, and i become confused why an F1 car out-accelerates a Cummins equipped Dodge dually. (Gross over-simplification)

In reply to Swank Force One:

Car B makes more torque at the HP peak.

However, that says nothing about acceleration.

In reply to Swank Force One:

Because torque is instantaneous. Horsepower has a time dependancy.

A steam engine can make 4000 ft lbs of torque ..... With 7 hp. Seriously.

They spin WAY too slow to make any hp. If they could spin to 5252 rpm, the numbers for HP would be quite high. They can't. This is why they can haz no horspowerz.

An F1 car weighs 1/3 as much as a dually. The gearing on a dually is E36 M3 and the engine doesn't accelerate as fast.

Still going with Car B.

Car B is blue.

Swank Force One wrote: Car "A" 2100lbs 600hp @ 8500rpms 3.9 rear end Carb "B" 2100lbs 600hp @ 6500rpms 3.23 rear end Both cars have identical transmissions. Consider this a simplified HP vs. TQ discussion, with a twist. I am ONLY interested in which car offers faster acceleration at the "peak" HP numbers given in the above examples. I say Car A is faster at 8500rpms than Car B is at 6500rpms. Sure, at 6500rpms, Car B is making 485ftlbs while Car A is only making 371ftlbs, but over 5252, isn't it the HP that matters? Discuss and learn me.

It's not "over 5252", it is all the time.

Look at it this way. 371 times 3.9 is 1446 ft-lb at the axle. 485 times 3.23 is 1566 ft-lb at the tire. At the proscribed engine speed, engine B will accelerate harder. I'm too lazy to do the math, but I'd also bet that it's going slower as well.

All HP is, is a convenient way to figure out torque vs. speed. A high winding engine gets more torque by having more gear multiplication.

Let me try it another way...

You would need more points on the graph to be able to tell anything that deals with a time dependancy.

Both Horsepower AND acceleration require that time dependency. Therefore, I can't tell you which one is accelerating faster.

You have already stated the HP is 600hp at those rpms. It is still going to be 600 hp no matter what you do with the gearing.

914Driver wrote: Still going with Car B. Car B is blue.

I'm picking whichever one is white (I'm a closet racist).

Swank Force, you need to play around with this: http://vlsicad.ucsd.edu/~sharma/Potpourri/perf_est.html

Okay, I'll math it. 8500 / 3.9 = 2179 axle RPM. 6500 / 3.23 = 2012 axle RPM.

So the 6500rpm engine will be accelerating harder because it is effectively geared shorter than the other engine. I makes peak HP at a lower road speed. It would have to have 2.98 gears to have the same 2179 axle RPM at peak HP.

And if you did that, it would be making 1445 ft-lb at the axle, same as the other engine.

In reply to Knurled:

Disagree.

So a steam engine making 2000ft*lbs at 200 rpm with a rear end ratio of 4.00 would out accelerate both?

I'm thinking that the rear-end numbers were picked out of thin air. By making the question slightly more stupid and unrealistic (in terms of unavailable rearend gears), we may be able to make it easier to understand.

Let's take your 600hp @ 6500 rpm, run it through a 4:1 rearend, and make it 600hp @ 1625 wheel rpm. That's effectively a speed.

Now let's take your 600hp @ 8500 rpm, and run it through a ratio that results in the same 1625 wheel rpm, which would be very very close to 5.23:1

Now we have two cars, we're taking a snapshot at whatever mph 1625 wheel rpm is. Each is putting out 600hp.

hp = (tq * rpm)/5252
so
tq = (hp * 5252)/rpm

8500 rpm car is putting down ((600 * 5252)/8500) * 5.23 ft-lb of torque, or about 1938.9 ft-lb of torque at the axle.

6500 rpm car is putting down ((600 * 5252)/6500) * 4 ft-lb of torque, or about 1939.2 ft-lb of torque at the axle. I put it to you that the .4 ft-lb difference (EDIT: .3, hahaha) is down to ignoring digits on my calculator watch.

I give up...

You can't do it.

You need more than 1 data point to discuss something happening over time. Hp and acceleration are based on time.

Rob R.

wvumtnbkr wrote: In reply to Knurled: Disagree. So a steam engine making 2000ft*lbs at 200 rpm with a rear end ratio of 4.00 would out accelerate both?

For a brief, shining moment, yes. But the vehicle speed would be extremely low.

First gear accelerates harder than fifth gear, yes?

wvumtnbkr wrote: I give up... You can't do it. You need more than 1 data point to discuss something happening over time. Hp and acceleration are based on time.

We're looking at the potential acceleration at a given point in time. For this all we need is axle torque (not engine torque - AXLE torque - clue) and tire diameter and vehicle weight. From there we can easily calculate pounds of thrust - at that moment.

Because we're comparing two different scenarios, all we really need to know is axle torque because everything else is the same.

wvumtnbkr wrote: You need more than 1 data point to discuss something happening over time. Hp and acceleration are based on time.

It's an academic exercise, so not very real world. But fundamentally, velocity is the slope of the position graph, and acceleration is the slope of the velocity graph. The rate of change at that point...

While power is expressed as work over time, you can still take a snapshot. If you ignore everything going on, the result may not reflect the real world.

In ideal terms F=MA; We have F, so if we're holding M constant, we can observe whether A changes as we fiddle with definitions of F.

pounds of thrust (torque) does not tell you acceleration. Torque (multiplied or not) doesn't tell you speed, direction, anything useful unless you are trying to lift or drag a specific weight and you need to know if you can do it.

Acceleration units are = length * time ^2 example m/s^2

The only thing you can get from a snap shot is (torque) length * force. No time.

Can't do it says the dimensional analysis.

In reply to ransom:

What is your F? There is an F in torque. It is stuck with a linear measurement as well (which cant be ignored).

Swank Force One wrote: Which car puts down more power to the tire at the point of peak HP? I suppose that's the question i'm asking.

There is plenty of opportunity to run around in circles, but the answer is in your question, and it really is this simple: If each of them is making 600 peak horsepower, and we're comparing peak horsepower, they are equal. That's ignoring all the practical auto racing questions, which I think you were trying to leave out in favor of understanding torque and power.

It's just another leverage question: You can double your force by giving yourself twice the lever arm, but it doesn't come for free; you have to move twice as far. It's the same here. 600hp is 600hp. If you need 600hp at 5252 rpm, you'll need a twisting force of 600 ft-lb at 5252 rpm. If you have an engine that makes 300 ft-lb at 10504rpm, you're all set, just gear it 2:1. Have one that makes 1200 ft-lb at 2626? Just gear it 1:2.