joecar

I'll try the quiz...

1) TQ at flywheel at power peak rpms:
F1: FWTQ = 750 * 5252/18000 = 219 ftlb
NC: FWTQ = 750 * 5252/9000 = 438 ftlb

2) F1 is geared 2:1 relative to NC, so assuming 1:1 gearing for NC:
F1: RWTQ = 2 * 219 = 438 ftlb
NC: RWTQ = 1 * 438 = 438 ftlb

3) Both cars have same driveline losses, so both cars have the same RWHP.

From what I answered, the rear tires would not know the difference if the engines were cross-transplanted.

Old SStroker

You get the cigar!

At any given rpm, more torque = more hp. I'm not sure everyone buys into torque @ the tires making it accelerate, however. Fortunately for us, vehicles and engines play by Mother Nature's rules, not ours.

Big_D

work over time

racer7088

The F1 engine has more frictional losses with double the rpm and also double the amount of valves and cams which are also doing double the rpm. Friction is mega important in Nextel Cup whereas friction is god in F1. Everything in physics is trying to kill the F1 engine which sees basically double the g forces that a cup engine sees.

1FSTZ71

iTrader: (11)

Geez what's that have to do anything. I'm working overtime right now

Ed Wright

iTrader: (1)

Yep, "you sure this dyno is correct??" Never failed to see one dissapointed.
I tell them "I was tuning and racing these cars when they were new, they just weren't as fast as this new stuff". They don't remember it that way. It took some work to put a new 409" into the 13s, I know because I was there doing it.

eLTwerker

Joecar- how can FWTQ and RWTQ be the same in the thought experiement(for each individual car). There would have to be some losses from the FW to the RWs. Right?

joecar

You are correct... since the quiz says both drivelines have the same losses, I chose to ignore ignore them to make the comparison easier and the calculations easier.

Sigforty

Here is a good explaination of what is happening in cars and gearing as well.

http://vettenet.org/torquehp.html

eLTwerker

Son of a bitch...I knew there was something in there I missed.

MrEracer

iTrader: (3)

Quote by Wnts2go100
"umm..? are you sure? the mower weighs what? 500 lbs? less most likely... thats nasty power:weight. theres more to something like that than just how much power each can pump out. imagine if the corvette had almost 200hp for every 500 lbs.. itd be a 8-7 second ride".

Yes, I'm sure the Vette will beat the mower even if the mower has a 20,000 rpm rev limit. This is because the mower would have a top speed of only 15.47 mph. As OldStroker said, the mower might beat the Vette to it's top speed, but this is due to power to weight reasons (as you mentioned), not torque at the wheels... The key words in that last statement are HORSEPOWER to WEIGHT...

Having said that, you might get the mower to beat the Vette by REDUCING torque to the wheels by giving it a transmission that would keep the mower at its maximum HORSEPOWER (191 hp at 20,000 rpm) all the way down the quarter mile. This would take advantage of the better power to weight ratio held by the mower. Don't forget that in all the example's above, the mower only has 50 lb-ft of torque at the flywheel. This is a classic example of why HORSEPOWER is the star and Torque is a supporting cast member...

Also, these examples should demonstrate that manipulating torque with gear ratios does not help. However, maximizing the available HORSEPOWER with gearing is huge.

Lessons to be learned from this discussion:
1. Torque at the rear wheels means nothing without the horsepower and proper gearing to back it up.
2. Power to weight with the proper gearing is what wins races.

Want to win races: Raise your ENGINE torque curve (I told you how to do this in one of my earlier posts) which increases ENGINE horsepower. Reduce weight and optimize your gearing, traction, suspension, and driver skills, and forget about torque to the wheels.

As a side note to all the horsepower denegraters who say that HP is not real and is just a calculated number, Torque is also just a calculated number. It is derived from a force applied at a distance and calculated as; Tq = force X Distance...

I hope all these examples have not veered too for from the original post...
Shirl

Old SStroker

Sometimes you confuse me, Shirl.

You are technically correct about Torque being "calculated" on many engine dynos. It's the FORCE that is measured at the end of the torque arm on the absorber. The length if that arm is some fixed distance from the center of rotation so Force times that fixed distance = Torque. The Force is very real.

Sometimes torque is measured more directly by instrumenting a shaft attached to the engine's output. Strain gages would be one way to do this. Any shaft is a torsion bar and can be strain gaged and calibrated in Lb-ft (or N-m) per degree of twist. That might be the easiest way to measure the output of huge engines like in a supertanker. Come to think of it you could also instrument half shafts or axle shafts and measure rear wheel torque. Wonder if anyone has ever done that? If you also measured the shaft rpm you could calculate hp at the wheel from the measured torque.

Increasing torque at any rpm raises the hp at that rpm, as you said. Torque at the drive wheels at a given vehicle speed is, of course, horsepower at the wheels.

If two drag cars were virtually identical, with identical torque & hp curves (another thought experiment, but with Pro Stock it might not be too far off), why would one car run more gear and run the engine another 500 rpm or so down the back side of the hp curve at least in top gear (thru the traps)? Dumb or smart idea? Why?

Jon

dynocar

Quote, "If two drag cars were virtually identical, with identical torque & hp curves (another thought experiment, but with Pro Stock it might not be too far off), why would one car run more gear and run the engine another 500 rpm or so down the back side of the hp curve at least in top gear (thru the traps)? Dumb or smart idea? Why?"

Answer, good idea, so you average more HP in each gear in that when your RPM drops in the next gear, it doesn't fall so far off of the HP curve. (More area under the curve).

Should add this to lessons learned which I don't think has been mentioned yet. At any given MPH, selecting a gear that puts your engine's RPM as close to the peak HP RPM will provide you with the most axle torque due to gearing's torque multiplication.

MrEracer

iTrader: (3)

Jon,
I'm not sure what is confusing about what I said. Torque is calculated from two real world measurable parameters, Force and radius. Torque is the calculated product of these items. When you use a torque wrench or strain gauge, the tool does the math for you. The scale on the torque wrench takes the radius of the tool into account and is calibrated to 'read' as Torque. Same with the strain guage which uses equations and calibrations to 'read' the torque. Torque is real and so is horsepower. Just because they are both 'calculated' does not make them any less real... Which was the point of my comment...

With regard to your quote:
"Increasing torque at any rpm raises the hp at that rpm, as you said. Torque at the drive wheels at a given vehicle speed is, of course, horsepower at the wheels".

I must take exception to this... If you mean that increasing ENGINE torque at any RPM raises ENGINE horsepower at that RPM, then I agree that this is true. If you mean that increasing torque at the wheels (using a gear ratio change) also raises horsepower at the rear wheels, then your statement is not true. Horsepower through any gear ratio remains unchanged (minus parisitic losses) while torque can be multiplied to any level at the expense of RPM. Per my mower example; torque was multiplied from 50 to 5000 using a 100:1 gear ratio. The engines horsepower remained the same, 20 HP at the engine and 20 HP at the wheels... This is why the mower lost to the Vette.
Shirl

Old SStroker

Yes, I meant engine torque. Horsepower never gets multiplied by gearing. Only torque gets multiplied.

joecar

Gearing multiplies torque and divides angular velocity, so power remains the same across gearing (other than losses).

Old SStroker

OK, but what if you threw a bit more gear at the combination so that you went well past the power peak approaching the finish line and your torque at the wheels was less than your "identical" car in the other lane at the line. Could you still out accelerate him (out ET him) for the full 1/4 mile? Remember you had more gear (torque) multiplicaton all the time in each gear (or just in top gear to simplify things). Yes, you used a different rpm range, but not necessarily more average power. Hmmmm.

I suggest that in some cases this might be a viable way to be quicker without necessarily making more power than the competition. Let's say the engines were rpm limited by something like valvetrain. If you could figure how to spin it higher than your rivals without even without changing the shape of the torque/hp curves, you might have the advantage. It goes to "more torque at the wheels more of the time" gettin' her done.

Could it be that the quicker ProStock cars might have tried this, and may still be doing so? They aren't making more BMEP or more power, they just might be making a few more r's. At the 10K+ revs they seem to be turning, the next few hundred revs are very costly. AFAIK, NHRA hasn't yet put a gear rule in ProStock.

We're not talking a big difference here when PS races are won by a few thousandths of a second. A millesecond at the finish line is about 3-1/2 inches. It still amazes me how close the PS folks are with all the variables they juggle.

This isn't an original thought...there are few of them, and I have absolutely no contact with any PS team, but it seems possible to me, and a few others I've talked with who seem to know more than I.

Truly there is an interesting relationship between torque and power and how it is applied. Different ways of explaining things fortunately doesn't change how things work. Give the engine/car what it wants, not what you think it wants. Figuring out what it wants is the challenge.

P Mack

iTrader: (6)

I remember a thread almost like this one a few years ago. One of the admins of this site was arguing that with gearing held constant, a car will accelerate harder the instant it's at its power peak than it will at its torque peak. Haha, good times.

Adrenaline_Z

Nice thread guys. No matter how many times this question is asked, there's
always a new twist on how to explain it. Very cool.

PMack: Have you ever stumbled across those guys that swear up and down
that Torque peak is the best shift point? Then you ask why they shift
at 6000-6500 RPM instead of ~4000 RPM during a race? It's fun to have them explain
their view point.

Adrenaline_Z

I see what you're getting at now. Pass up some upper track edge and trap
speed to get the E.T.

That's a clutch tech's worst nightmare I would think. A little more gear...adjust the clutch...
and hope she hooks through the 1/4 mile with that extra ratio.

Interesting strategy for sure.