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Discussion Starter · #1 ·
Say a 3000lb Ferrari that has

Say a 3000lb Ferrari that has 360 hp and 275 lbs torque vs the same vehicle similar weight and different engine with 300hp and 330lbs torque.

Which would be faster on the top end and in acceleration say on a road circuit?

What would be the advantages and disadvantages of each?
 
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Discussion Starter · #2 ·
torque is what wins races. Hor

torque is what wins races. Horsepower is what sells cars. The real question is the width of the power band. A driver must keep the engine rpm's in the sweet spot of the power curve. But, in road racing, things happen and its torque and a wide power band that will pull you through. On a high speed oval, where the speed doesn't vary much, that high end horsepower engine will be faster. but, on most road racing course's that we have raced on, I would take the engine package that has the wider torque band. Match it up with the proper 6 gears and it would be easier to drive faster than an engine that makes more power. but in a much smaller higher rpm range. These days you have to match the engine cams, fuel maps, and timing curves to the track you will be running.
 
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Discussion Starter · #3 ·
I agree with Bob:

TQ wins r


I agree with Bob:

TQ wins races (and makes for better street performance) while HP sells cars and bragging rights. TQ also has the property that when accompanied by a wide power band it lets the driver avoid shifts by "riding" the TQ curve. For a manually shifted car, every shift you don't have to perform saves 0.5 seconds of full power acceleration.

Turn 2 at TWS is a prime example: Entry speed is slow enough that one could choose 3rd gear and then shift into 4th right at trackout (F355 street tires). However, it is faster to simply use 4th gear throughout the whole turn and down the next straight since A) you aren't accelerating at full power in most of turn 2, B)you don't end up running out of revs at the critical trackout point when the rear end might not be exactly aligned with the direction (arc) of travel (if you know what I mean) and C) then having to shift up to 4th for 2 seconds and back to 3rd for turn 3.

Oval tracks where speed is (essentially) constant and the last 1/16 mile of a 1/4 mile drag race are won by HP.

Given the same engine and two cam timing arrangements; I am willing to sacrifice 10-20 HP at the top end for 10-20 more lb-ft in the midrange. The package is simply easier to drive.

However, given the same top end power (say 350 HP versus 350 HP) the car with the bigger TQ curve will win. A prime example would be a C5 Vette versus a F355. Both cars weigh in at 3200 pounds, the Vette has 350 TQ/350 HP while the Ferrari has 265 TQ and 380 HP. The Vette powerband is simply bigger than the Ferrari powerband and wins the vast majority of the drag races.
 
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Discussion Starter · #4 ·
Horsepower is your Torque mult

Horsepower is your Torque multiplied by your engine RPMs.

If you bump up your torque, then you have higher available horsepower at each RPM level.

On the other hand, you can bump up your horsepower by simply bringing your engine at a higher RPM (i.e. raising your redline)...but, you aren't getting that extra horsepower except for those brief moments when you are at that high RPM level.

The extra torque is available all of the time. The extra horsepower, however, is only available if and when you get the engine back up to that peak RPM level.
 
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Discussion Starter · #5 ·
Horsepower was first used by J

Horsepower was first used by James Watt during a business venture where his steam engines substituted horses. It was defined that a horse can lift 33,000 pounds with a speed of 1 foot per minute: 33,000 ft?lbf?min-1. This is roughly equivalent to lifting 147,000 Newtons (the weight of a 15,000 kg mass) at a speed of 0.005 metre per second.
 
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Discussion Starter · #6 ·
The Irony of all of this is th

The Irony of all of this is that James Watt was using water vapor to pump water out of mines.
 
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Discussion Starter · #7 ·
Interesting info, Thanks

So


Interesting info, Thanks

So your torque curve is where maximum power is at for fastest acceleration, when it begins to fall considerably you should shift vs going up all the way to redline in addition you keep the revs in the highest torque output range?

Is the max torque dynamic? Meaning in different gears it will be at different rpms?
 
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Discussion Starter · #8 ·
first question.. YES You &#34

first question.. YES You "may" waste time reving the engine to its red line as the torque drops off, the engine will gain its rev's more slowly.

Second question: Engine torque will be the same value regardless of the gear. But, torque at the driven wheels will be mupliplied by each gear ratio. So, if your engine has 300 ft/lbs torque, and a first gear and final gear ratio combined, of 7:1 you would have 2100 ft/lb at the wheels.(minus transmission losses) In racing we always use the lowest ratio's (high numerical) possible. That will give you the highest rate of acceleration in each gear and just enought final gear overall to reach the highest top speed the car can do at the end of the longest straight.
 
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Discussion Starter · #9 ·
To add to what Mitch said in a

To add to what Mitch said in a few posts up. Shifting eats up major time in lapping a road race course. At Sears Point (Infineon) we shift 17 times per lap. Typical half hour sprint race equals 18 or 19 laps. That is 323 shifts in a 30 minute race. Half of those shifts are up. so that is about 160 times a half second or one minute and 20 seconds that the car is not accellerating, it really just rolling.
 
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Discussion Starter · #10 ·
I computed up a few examples t

I computed up a few examples to show where to shift. First I take the TQ curve of an F355 and perfore TQ multiplication through the gear box , differential, and tire diameter to get a thrust graph. Since each gear ratio changes the relationship between RPM and velocity, I chart the Multiplied TQ curve by the common element :: velocity.



A few simple equations later we can see the HP versus Speed relationship.



Notice the actual information content is identical, it is just presented differently. Note, also, that one comes to the same conclusion:

A) you shift when the thrust in the next higher gear is greater than the thrust at higher RPMs in the current gear

and

B) you shift when yo have more power in the higher gear

Are both true and duals of each other.
 
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Discussion Starter · #11 ·
In the above graphs, I include

In the above graphs, I included the second set of data to illuminate what the car feels like when aerodynamic losses are present rather than the idealized world. For this case, I set the aerodynamic loss to the cubic with respect to power and then set the scale such that it consumes all engine power at the factory specified max speed. This data is represented in dashed lines.

While 1st gear is practically unhindered by aero losses, aero-drag increases rapidly, and should be feelable even in 2nd gear.

While 4th gear seems peppy in the 90-120 range, 5th gear feels flat as represented in the rather horizontal dashed brown line.

6th gear is lacidazical.

Notice: however; that peak acceleration in all three of these gears (4th, 5th, 6th) occur at practically the same velocity (shown by the arrows).
 
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Discussion Starter · #13 ·
Mitch,
Very interesting graph


Mitch,
Very interesting graphs! If you have time, please explain more on how you came up with the thrust. You wrote "multiplication through the gear box , differential, and tire diameter,". I assumed that you transfered the engine torque from gear to gear (knowing the radius of the gears) and then divided the final number by the tire radius to get the forward force at edge of the tire. But this is just a guess....
--tony
 
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Discussion Starter · #14 ·
Tony:

Correct::

Thrust[R


Tony:

Correct::

Thrust[RPM] = engine_TQ[RPM]*overall*Tranny_Ratio[gear]/Wheel_rolling_Radius

Where overall = Drop_ratio*crown_ratio*differential_ratio {= 4.33}

Since engine_TQ[RMP] is in lb-ft and Wheel_circumfrence is in ft, we end up with thrust in lb.

Then::

Thrust[speed] = Thrust[RPM]*Wheel_circumfrence/(overall*Tranny_ratio[gear])

Thrust is the amount of force delivered to the contact patches on the rear tires (in lb). Dividing the Thrust by the weight of the car (in lb) yeilds the acceleration in Gs; If (and only if) the tires do not loose traction.

By the way; I set up the aerodynamic resistance such that the entire engine HP is consumed at 183 MPH (factory top speed), and that the consumption is cubic in HP or quadradic in TQ. Actually, you can use similar logic as above to show that aero is cubic in HP if it is quadradic in TQ and vice versa. This is not strictly correct as there is a linear rolling resistance (in TQ), but close enough for this presentation {Ah, the magic of eXcel...}
 
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Discussion Starter · #15 ·
agree with all I read, except

agree with all I read, except that transmission losses do not effect torque, only HP
 
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Discussion Starter · #16 ·
Nope:

Since HP = TQ * R


Nope:

Since HP = TQ * RPMs / 5252

Rear_wheel_HP = Rear_wheel_TQ * RPMs / 5252

So if HP goes missing at one place, exactly the comparative amount of TQ had to go missing at the same time. So there is no way to loose one without suffereing a loss in the other. {Hint RPMs remain constant and so does 5252}
 
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