What is more important, HP or Torque?

GaryC

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HP, is for top end, torque is how quick you get there.
 

Mike Brunton

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I think you guys are thinking about this the wrong way... torque is just a force. Horsepower is work.

If I can push a 5,000lb box at 10mph across a floor, that requires a given force. If I can push it at 20mph, it required the same amount of force to keep the box moving, but I am getting alot more work done (box is covering more distance).

Neither of them is "more important" for overcoming aerodynamic forces... if you can make good torque at a higher RPM (which means lots of horsepower) then you can take advantage of gearing. A lower/shorter gear is easier to turn (requires less torque), but torque is the twisting force - peak torque is where the most power is hitting the ground.
 

dmora

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Exactly...Your asking the wrong question guys. simply put niether is better than the other, because they are measuring different things. The important part is how quickly you can apply availible torque to produce the HP, and then the gearing to utilize it. F1 cars have absolutely NO torque (small# comparatively) and they they would wail all over a viper.

<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by Mike Brunton:
I think you guys are thinking about this the wrong way... torque is just a force. Horsepower is work.

If I can push a 5,000lb box at 10mph across a floor, that requires a given force. If I can push it at 20mph, it required the same amount of force to keep the box moving, but I am getting alot more work done (box is covering more distance).

Neither of them is "more important" for overcoming aerodynamic forces... if you can make good torque at a higher RPM (which means lots of horsepower) then you can take advantage of gearing. A lower/shorter gear is easier to turn (requires less torque), but torque is the twisting force - peak torque is where the most power is hitting the ground.

<HR></BLOCKQUOTE>
 

GR8_ASP

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I believe the difference in torque and horsepower when used to describe a vehicles tendencies is based on when the torque is developed. I think we all know that 1 hp equal 550 ft-lb/sec. At a given engine speed stating one is all you need. The other is a mathematical derivation of the other. The factor of speed is the difference, both in calculation and in our interpretation of it. Two engines that produce the exact same hosepower may be quite different. For example engine A uses variable valve timing, variable intake manifold tuning and other devices to make its small displacement produce 400 hp. Engine B uses large displacement with minimal tuning effects to produce the same 400 hp. At an engine speed 1/2 of the peak power speed Engine B probably produces 25% or more torque than engine A. The acceleration capability of both right at their peak horsepower points is essentially equal, provided that they are both geared appropriately. However, because engine B has a fairly constant amount of torque available throughout the engine speed range, it provides that same acceleration throughout each gear while Engine A provides a lower level of acceleration initially in each gear, slowly increasing its acceleration rate as the engine speed increases toward its peak torque point, which would be very close to the peak horsepower point.

The most generic comparison would be the engine that produces the most area under the torque curve will provide the best average acceleration, while the peak horsepower engine will produce the highest momentary acceleration and highest vehicle speed (assuming appropriate gearing).

Gets back to the saying "there is no replacement for displacement." Of course displacement alternatives such as turbo or supercharging provide a similar benefit.

Ron
 

SteveBCloud9

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Well, SVS and Hennessy have had their vehicles over 200+. They have proven that the vehicle itself can cut through the air at those speeds, you just need the room and the HP to do it with, I would think. We need a engineer......
 

Serious Eric

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Pull up a chair ... long winded.

Horsepower Horsepower Horsepower ... that's all that matters. It's incorrect to think that your car's peak torque is a measure of it's acceleration capability and that it's peak HP is a measure of it's top end. Maximizing HP is the key to both acceleration and top-end. If you had a continuously variable transmission that you could program to keep your engine at a constant RPM at WOT, would you set it for the torque peak or the Hp peak? Max acceleration AND top-end would be achieved by keeping the motor exactly at the Hp peak. Why? Gearing is the answer. Hp peak occurs at a higher RPM than the torque peak, so running your motor at a higher RPM allows you to take advantage of a higher gear ratio which yields a huge benefit in all-important rear-wheel torque.

The F1 car reference was right on. They make relatively little torque but run their engines at 20,000+ rpms. Why? Are they stupid, do they not care about acceleration? Of course they do ... Motors that turn at those revs are desirable because you can couple them to HUGE gear ratios which also produce HUGE torque multiplication factors at the wheels.

Actually we all ought to forget about flywheel torque completely when it comes to acceleration and top-end. What we should be focusing on is maximizing rear-wheel torque. (I know you just heard me say earlier that maximizing peak Hp yields max accel, but bear with me.) RWT is achieved through a combination of engine torque AND gearing. At any given instant in time, the car is accelerated by a force delivered to the pavement by the tire (Newton's 2nd law F=MA). Force to the pavement is proportional to the axle torque (Force x Moment Arm). RWT is engine torque X total gear ratio (trans and diff). Now let's try an example. Engines typically lose 7-13% of their torque between the torque peak and the Hp peak ... BUT the Hp peak usually occurs at about 30% higher rpm. So if I run my motor at the Hp peak I can gear it down by 30% (a 30% torque multiplication increase to the rear wheels), but I only lose say 10% due to engine torque reduction, for a net 20% more RWT than if I was running at the engine's torque peak.

It's a trade-off between RPM and torque. Keeping a motor's torque alive at a higher RPM is highly desirable when modding the motor. INCREASED torque at higher RPM is cause for a national holiday!

If you're still awake ... hope that helped.
 
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Paolo Castellano

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Originally posted by dmora:
F1 cars have absolutely NO torque (small# comparatively) and they they would wail all over a viper.



Wail all over a Viper? Could it be that the formula one cars weigh around 1200-1400lbs, and have technology superior to that of NASA? Could they wail all over a supra too? Hmm let me see, YES. Can the F1 cars wail all over anything with 4 wheels on a road course? OF COURSE! What is your point here? I dont understand your response. Why dont you try again? Paolo
 
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Paolo Castellano

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Originally posted by Ron Stene:
The most generic comparison would be the engine that produces the most area under the torque curve will provide the best average acceleration, while the peak horsepower engine will produce the highest momentary acceleration and highest vehicle speed (assuming appropriate gearing).

Gets back to the saying "there is no replacement for displacement." Of course displacement alternatives such as turbo or supercharging provide a similar benefit.

Ron

My question more specifically is this: I am curious if a coupe with a 3.73 rear end gear and stock tranny ratios that is in 5th gear at 6,000 RPM @ 164 MPH when shifted into 6th gear will drop the RPM to around 4,000 RPM @162. Now on the dyno graph, @ 4,000 RPM, the car makes 381 RWHP and 500 LBS/ft torque. Normally at this speed, w/a 3.07 gear the car is at 5,000 RPM where the car is putting down 463 RWHP and 486 LBS/ft torque. My question is this, will the car slow down at 162ish @ 4,000 RPM with 3.73 gear? Paolo
 

Serious Eric

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Chill Paolo, I don't think dmora meant any disrespect. Yeah, he's sort of a Supra nerd
smile.gif
but he likes vipers too.


Speaking of nerds, here I go again with another hopefully more succinct example:

Car A: "The GruntMaster" ... 500 ft.lb peak flywheel torque @ 3500 rpm. HP at this rpm is 333. Total gear ratio (trans & diff) = 1:1

Car B: "The Ricer" ... 300 ft.lb peak flywheel torque @ 7000 rpm. HP at this rpm is 400.

Let's say the cars are racing each other from a roll and are both running at their torque peak rpms when they start the race. Since "The Ricer" is at 7000 rpm, he must be in a lower gear (2:1) to be keeping pace with "The GruntMaster" who's engine is at 3500 rpm.

The one that can accelerate the hardest at any given speed will be the one that can put the most RWT to the ground. The ricer (by virtue of it's lower 2:1 gear) can put 600 ft.lb to the rear wheels, while the GruntMaster can only muster 500 ft.lb. The ricer wins ... &lt;I can't believe I said that&gt;
 

GR8_ASP

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Eric,

In your example you are correct. However you missed the reality of the situation. Unlike your example above the vehicles we drive have a set number of gears. Split the total vehicle speed range into 5 or 6 gears (5 for standard Vipers as 6th is not in the picture) to determine how much . The engine needs to provide meaningful torque through the necessary engine speed range for each gear. In computing the optimal shift schedule you optimise the area under the rear wheel torque curve for each gear, changing gears when the next gears wheel torque is equal to the previous gears (the first gear is decreasing while the next is increasing). To do this correctly you must apply the proper gear ratios and system efficiencies (especially for the 1:1 gear as it is significantly more efficient than other gears). Based on this, the wider the gear spacing, the more important the torque range is.

The formula 1 example did not take into account how the gears are selected for each track in such a way to keep the engine in its high wheel torque zone by having very small differences in the gear ratios where the car is through the majority of the track. If you computer modeled the wheel torque example for every possible set of gear ratios for the expected velocity-time plot of the track you would do essentially what the F1 gear selection programs do.

Regarding the CVT example for maximum acceleration the transmission would indeed be at the maximum horsepower point at all times. However, fuel economy, pass-by noise and general vehicle NVH prohibit that condition for all except WOT conditions. At these other times torque assists the acceleration of the vehicle while the CVT adjusts the effective drive ratio (not instantaneous). This situation is definitely noticeable while driving. Note that the engines available torque must accelerate not only the vehicle but also the driveline. For those who have replaced their flywheels you know that even the torque available with the Viper is resisted quite capably with a heavy flywheel and that in replacing the heavy flywheel with a lighter one the engine acceleration is improved considerably. In neutral, all of the available engine torque is going to accelerate the flywheel, and other rotational components.

This whole discussion would indeed be meaningless if everyone used the horsepower curve (through the full engine speed range) like they use the torque curve. If the full curve is used then it does not matter as both torque and horsepower curves provide the exact same information, only mathematically manipulated. I am sure you are all aware that engine dynamometers do NOT measure horsepower. They measure torque and calculate the horsepower. It is not the maximum horsepower that accelerates the vehicle, it is the rear wheel torque at the instant that you are measuring the acceleration. The horsepower peak occurs only once in each gear.

Ta Ta and Good Night,
Ron
 

Serious Eric

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<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by Paolo Castellano:

My question more specifically is this: I am curious if a coupe with a 3.73 rear end gear and stock tranny ratios that is in 5th gear at 6,000 RPM @ 164 MPH when shifted into 6th gear will drop the RPM to around 4,000 RPM @162. Now on the dyno graph, @ 4,000 RPM, the car makes 381 RWHP and 500 LBS/ft torque. Normally at this speed, w/a 3.07 gear the car is at 5,000 RPM where the car is putting down 463 RWHP and 486 LBS/ft torque. My question is this, will the car slow down at 162ish @ 4,000 RPM with 3.73 gear? Paolo
<HR></BLOCKQUOTE>

Ok, your answer is the 3.07 will give you more RWT and acceleration in 5th at 5000rpm than will the 3.73 shifting to 6th at 4000rpm.

Here's why:
3.07, 5th gear (ratio is .74:1), 486 ft.lb. Let's assume the torque figure is actually flywheel ok?
RWT = 486ft.lb. x .74 x 3.07 = 1104 ft.lb.

3.73, 6th gear (ratio is .50:1), 500 ft.lb.
RWT = 500 x .50 x 3.73 = 932 ft.lb.

Shifting to 6th gear will kill you. That ratio drop from .74:1 down to .50:1 is lousy. And at 5000 rpm, you've still got another grand or so till redline. That's 20 percent more engine revs which should take you from 164mph up to about 197mph (theoretical). Drag will get you a bit before that.

Lose the 3.73 for a top-speed run (unless you re-gear the tranny).
 

GR8_ASP

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An additional note about the "Ricer" example.
&lt;Let's say the cars are racing each other from a roll and are both running at their torque peak rpms when they start the race. Since "The Ricer" is at 7000 rpm, he must be in a lower gear (2:1) to be keeping pace with "The GruntMaster" who's engine is at 3500 rpm.&gt;

In this example if the ricer started at 7000 rpm (torque peak) and had a maximum engine speed of 9000 he would upshift much earlier than the "gruntmaster" which started at 3500 rpm and had until 6000 rpm to shift. The difference between 2000 rpm and 2500 rpm available prior to the next gear is to be divided by the gear ratios, which if actually a 2:1 difference would provide 2 1/2 times as much of the vehicle speed range in the lower gear in the gruntmaster than the ricer. To be equivalent the ricer torque peak would have to be down around 5250 rpm. Not likely. I would hazard a guess that at 5250 rpm the torque of the ricer was 10 - 20% below its peak torque.

Ron
 

Serious Eric

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<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by Ron Stene:
Eric,

In your example you are correct. However you missed the reality of the situation. Unlike your example above the vehicles we drive have a set number of gears. Split the total vehicle speed range into 5 or 6 gears (5 for standard Vipers as 6th is not in the picture) to determine how much .
<HR></BLOCKQUOTE>


No I didn't miss the reality of the situation, I simply chose to ignore the practical implementation details (such as factory gearing), because they cloud the theoretical picture and can easily be manipulated and changed anyway.

All implementation revolves around theoretical first principals ... understand the theory and you can always arrive at a superior implementation. This is why I chose the CVT example. It assumes that any good hot-rodder worth his salt in the real-world can always change gear ratios and ADD gears to suit the characteristics of his engine. I took this to be a given.

Therefore if you factor out the limitation of gearing as delivered by the factory (which all real racers do), then the principal of keeping the engine turning at the HP peak still stands as the optimal means of achieving max acceleration and top-end.

We have no disagreement about maximizing the integration of rear-wheel torque over usable rpm range. Doing so will show you that you are in effect trying to *center* your rpm range in any given gear around the HP peak. I'm just trying to push home a point that I think some folks miss - that we should focus less on max flywheel torque and think more about gearing, rpm and how it affects RWT. The peak HP concept is simply a way to get this across.
 

Tom Welch

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From years and years..............decades of practical application of weight to power ratios on the dragstrip, I can assure you that torque will win out. Let me use my own viper as an example

I know that my Viper has LESS horsepower than say a few of the Texas NOS cars, but with 950 FT/LBS of torque my car runs very similiar MPH(which by the way is a determination of DRAGSTRIP HORSEPOWER).

Horsepower SELLS.............Torque WINS! Your car came with more torque than horsepower from the factory. Your engine was designed from top to bottom with torque in mind. Torque is most prominent in the lower RPM band, likes long runner intake manifolds with average diameter throttle bodies, etc. To try and reinvent the wheel with the viper engine for high rpm-high horsepower applications is in my opinion as bad as just bolting in a small block chevy!

Tom
Http://btrviper.com
 

Serious Eric

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<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by Ron Stene:

In this example if the ricer started at 7000 rpm (torque peak) and had a maximum engine speed of 9000 he would upshift much earlier than the "gruntmaster" which started at 3500 rpm and had until 6000 rpm to shift.
<HR></BLOCKQUOTE>


You're adding all sorts of hypotheticals (such as where they have to shift), that don't belong in this discussion. My ricer guy's a good mechanic and he has 12 gears, so he can keep his engine right at the HP (or torque) peak no matter what speed. For that matter the gruntmaster has 15 gears ... so what? You see how this escalating "arms race" of gearing goes? As long as the ricer can make decent engine torque at twice the gruntmaster's rpm at any given speed he'll always develop more RWT and will therefore always win.
 

GR8_ASP

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Eric,

Just one comment about the integration of the wheel torque curve. It is far from centered on the horsepower peak. In most engines the torque (call it BMEP if you will) drops off more rapidly after than before the peak power point. This gets into many physical aspects which I will ignore but is definitely a product of the engine designer who desires a torque bias. That is why in all automotive engines the torque peak is at a lower rpm than the power peak.If one was only concerned about power they would be the same.

Regarding the OEM gear selection and that anyone can manipulate them as needed. You must be kidding. Assuming that the owner wants the top gear to produce the top speed possible, and that the vehicle launches from a standstill, the range of possibilities are quite limited. You can add additional gears, with the additional rotating enertia, driveline losses and lost time during shift periods. If an increase in the quantity of gears is not practical (non-CVT in this case) then what I described above still applies. Please don't go into the CVT realm when discussing Vipers. Leave those discussions to fuel economy applications until the technology catches up and can provide reliable 500 lb-ft capacities.

And please, do not mention the "technical aspects" as if they are something you alone understand. Some of us have spent our entire careers dedicated the field of automotive power trains.

Ron
 

Serious Eric

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<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by Tom Welch:
From years and years..............decades of practical application of weight to power ratios on the dragstrip, I can assure you that torque will win out. Let me use my own viper as an example

I know that my Viper has LESS horsepower than say a few of the Texas NOS cars, but with 950 FT/LBS of torque my car runs very similiar MPH(which by the way is a determination of DRAGSTRIP HORSEPOWER).
<HR></BLOCKQUOTE>

Your expertise and experience at getting down the dragstrip faster than most is unquestioned. You must admit however, that doing so involves mastering lots of nasty little practical details such as traction - both at the launch and down the track.

That wasn't really the arena that I was arguing my HP issue in. More in the theoretical realm where "all else is equal" if you know what I mean. Perhaps the Texas NOS cars haven't re-geared properly to take advantage of their HP advantage? I dunno. Perhaps with their higher RPM power they're having to launch at rpms that reduce their off-the-line traction as compared to your own? I'm just saying that with enough gear and traction tuning, HP should be able to beat torque.

No question that mastering gear ratios, traction and finessing the launch counts for a BUNCH in the real world.
 

Serious Eric

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<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by Ron Stene:
Eric,

Regarding the OEM gear selection and that anyone can manipulate them as needed. You must be kidding. Assuming that the owner wants the top gear to produce the top speed possible, and that the vehicle launches from a standstill, the range of possibilities are quite limited. You can add additional gears, with the additional rotating enertia, driveline losses and lost time during shift periods. If an increase in the quantity of gears is not practical (non-CVT in this case) then what I described above still applies. Please don't go into the CVT realm when discussing Vipers. Leave those discussions to fuel economy applications until the technology catches up and can provide reliable 500 lb-ft capacities.

And please, do not mention the "technical aspects" as if they are something you alone understand. Some of us have spent our entire careers dedicated the field of automotive power trains.

Ron
<HR></BLOCKQUOTE>


Did I step on your toes? When did I mention "technical aspects" as if I alone understood them? I THOUGHT we were having a good old HP vs torque theory debate here and you get pissy? What's up with that?

One doesn't have to be a transmission design expert like yourself to know how the physics of HP vs torque works. Furthermore I don't CARE whether a CVT is practical or not or whether 12 gears or 15 gears is practical or not. That doesn't have anything to do with my point. NOTHING. Sheesh. Horse dead. Done.
 

Tom Welch

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Eric,

Yes, there are lots of varibles in drag racing, as in any form of motorsports. like the fella said about the IRL cars, he just failed to mention that it takes them a lap or two to get going.

It is all about application. All of this usefull information may send Paolo on his way to bonneville, I don't know. For sure though, 6th gear is a no-no.

Best of luck to you all,

Tom
 

YellowSnake

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Looking at top speed, horsepower wins. Making more torque at high rpm's means you can use a stiffer gear for any given car speed, and thus have more effective torque at the drive wheels.
 

dmora

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Because the initial question was about torque. Everyone knows that big displacement motors make gobs of torque. A supra does not have a big displacement motor compared to a vette or viper or Mack truck. And besides... this is ViperClub.org... not supra forums.com . I said viper because you know what your car (viper) will do. If this were supraforums, i would have said supra. Dont think i was picking on the viper directly.

<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by Paolo Castellano:
Originally posted by dmora:
F1 cars have absolutely NO torque (small# comparatively) and they they would wail all over a viper.



Wail all over a Viper? Could it be that the formula one cars weigh around 1200-1400lbs, and have technology superior to that of NASA? Could they wail all over a supra too? Hmm let me see, YES. Can the F1 cars wail all over anything with 4 wheels on a road course? OF COURSE! What is your point here? I dont understand your response. Why dont you try again? Paolo
<HR></BLOCKQUOTE>
 

dmora

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I dont know if this has been answered clearly and easily yet by anyone because frankly, theres too much reading going on.
smile.gif
The simple answer is when you shift to 6th on stock gearing, whatever it maybe, your shifting into the cruising gear. From what i have gathered and been told of vipers is that this gear is just used for putting around on the highway.

To contrast to this, we'll use my supra for example... My supra doesnt have an over drive. 5th gear is 1:1 and still pulling hard at around 150mph, i shift to 6th, and it keeps pulling. the ratio is still close so i can accelerate to 200ish mph.
On LOW boost, my car will hit 180mph. On HIGH boost, a supra will do over 200mph on stock gearing. This is why torque is relevant only to the rate of acceleration and gearing.


<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by Paolo Castellano:
Originally posted by Ron Stene:
The most generic comparison would be the engine that produces the most area under the torque curve will provide the best average acceleration, while the peak horsepower engine will produce the highest momentary acceleration and highest vehicle speed (assuming appropriate gearing).

Gets back to the saying "there is no replacement for displacement." Of course displacement alternatives such as turbo or supercharging provide a similar benefit.

Ron

My question more specifically is this: I am curious if a coupe with a 3.73 rear end gear and stock tranny ratios that is in 5th gear at 6,000 RPM @ 164 MPH when shifted into 6th gear will drop the RPM to around 4,000 RPM @162. Now on the dyno graph, @ 4,000 RPM, the car makes 381 RWHP and 500 LBS/ft torque. Normally at this speed, w/a 3.07 gear the car is at 5,000 RPM where the car is putting down 463 RWHP and 486 LBS/ft torque. My question is this, will the car slow down at 162ish @ 4,000 RPM with 3.73 gear? Paolo
<HR></BLOCKQUOTE>
 

Sniper

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So how would a car match up at the strip that has 405 hp / 498 torque to a car that has 435 hp / 473 torque?
 

dmora

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We dont have enough information....ontop of the issue that theres too many other variables that would be at issue at a drag strip, like launch/ and coef -drag.

<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by Matt Kovac:
So how would a car match up at the strip that has 405 hp / 498 torque to a car that has 435 hp / 473 torque?

<HR></BLOCKQUOTE>
 

dmora

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<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by Matt Kovac:
Two Vipers.

<HR></BLOCKQUOTE>

assuming everything is equal? Driver, launch, weight, i would wager on the higher hp car. But theres really only 1 way to find out.
smile.gif
 

GR8_ASP

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Regarding the chassis dynomometer it is measuring the torque applied to it. That is if the engine is producing 500 lb-ft, you have it in 4th gear (1:1) and have a 3.07 final drive ratio the chassis dyno will measure 1535 ft-lb of torque applied to the dyno drum. It then mathematically translates it back into engine parameters by knowing what the rotational speed of the engine is in comparison to the drum speed and outputs all the information in terms of engine speed. If you were not able to translate it into engine speed parameters the output would be drum torque and drum speed. The horsepower through the run would be identical to the engine results but the torque would be off by the gearing multipliers, and of course the x-axis would be dyno drum speed.

Regarding the earlier question &lt;My question more specifically is this: I am curious if a coupe with a 3.73 rear end gear and stock
tranny ratios that is in 5th gear at 6,000 RPM @ 164 MPH when shifted into 6th gear will drop the RPM to around 4,000 RPM @162. Now on the dyno graph, @ 4,000 RPM, the car makes 381 RWHP and 500 LBS/ft torque. Normally at this speed, w/a 3.07 gear the car is at 5,000 RPM where the car is putting down 463 RWHP and 486 LBS/ft torque. My question is this, will the car slow down at
162ish @ 4,000 RPM with 3.73 gear? Paolo&gt;

If I understand the conditions correctly you are applying 932.5 lb-ft of torque to the rear wheels at 162 mph in 6th gear with a 3.73 ratio. While in 5th at a comparable speed and a 3.07 ratio you are applying 1104.1 lb-ft. If all 1100 lb-ft are required to maintain that speed the 3.73 gear example would be slowing down as it would not be applying adequate force.

Simpy put the acceleration potential can be derived by using F=ma where the F is the rear wheel torque divided by the rolling radius of the rear tire and m is the mass of the car. To complete the picture you could apply rolling resistance and velocity related wind resistance factors. If you also had a rear wheel torque curve for each gear, based on chassis dynamometer measurment, you would be able to compute most of the aspects reagrding your cars performance. But that would not be nearly as fun as evaluating it yourself. Vehicle performance predictive programs have become quite sophisticated and can apply that information and more (engine inertia, driveline inertia, axle and wheel inertia, rolling radius at various speed conditions taking into account applied load and centrifugal force, frontal area and drag coefficients based on vehicle speed and environmental conditions, etc.). Way to much for my fragile mind but just additional computing for a program.
 
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Paolo Castellano

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Originally posted by Eric Schieve:
Ok, your answer is the 3.07 will give you more RWT and acceleration in 5th at 5000rpm than will the 3.73 shifting to 6th at 4000rpm.

Here's why:
3.07, 5th gear (ratio is .74:1), 486 ft.lb. Let's assume the torque figure is actually flywheel ok?
RWT = 486ft.lb. x .74 x 3.07 = 1104 ft.lb.

3.73, 6th gear (ratio is .50:1), 500 ft.lb.
RWT = 500 x .50 x 3.73 = 932 ft.lb.

Shifting to 6th gear will kill you. That ratio drop from .74:1 down to .50:1 is lousy. And at 5000 rpm, you've still got another grand or so till redline. That's 20 percent more engine revs which should take you from 164mph up to about 197mph (theoretical). Drag will get you a bit before that.

Lose the 3.73 for a top-speed run (unless you re-gear the tranny).

Eric, this is exactly what I was looking for. I have to say I have had my car to 190 MPH when it was stock( about 3 years ago)
I looked at the dyno sheets for my stock #'s @ 5,000 RPM and I had 405 RWHP and 425 lbs/ft RWT. If you plug in the torque # from the stock configuration, you get 425 RWT x .74 x 3.07= 965.5 ft.lb. Now if you take my headers/exhaust/K&N's/no cats # of 500 ft.lb. x .5 x 3.73 = 932.5 ft.lb. which is close, no? Paolo
 

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