Post ur *Average* dyno #'s.
09-30-2002, 05:33 PM
#1
Smarter than you
Thread Starter
iTrader: (1)
Join Date: Apr 2001
Location: Third Coast, united states
Posts: 8,240
Likes: 0
Received 2 Likes
on
2 Posts
Post ur *Average* dyno #'s.
I got an idea for a post like this earlier after seeing an article about peak hp #'s vs. real life acceleration.
ill start with my car.
set up- Revhard kit w/ hondata on an LS motor.
My powerband is from 5300-7500 when racing. That is shifting at 7500 and having the RPMs fall down to 5300(LS tranny) and then shifting again at 7500 and so on. SO i picked 6 points(in my racing powerband) on the graph and averaged the hp. I figure this should give a rough estimation of how my car accelerates vs. another 275whp(peak) car.
<U>RPM</U> <U>WHP</U>
5300 237
5700 249
6100 270
6500 274
7100 265
7500 249
so the avg. whp for my powerband is 257whp.
who wants to go next?
[Modified by DIRep972, 5:42 AM 10/1/2002]
ill start with my car.
set up- Revhard kit w/ hondata on an LS motor.
My powerband is from 5300-7500 when racing. That is shifting at 7500 and having the RPMs fall down to 5300(LS tranny) and then shifting again at 7500 and so on. SO i picked 6 points(in my racing powerband) on the graph and averaged the hp. I figure this should give a rough estimation of how my car accelerates vs. another 275whp(peak) car.
<U>RPM</U> <U>WHP</U>
5300 237
5700 249
6100 270
6500 274
7100 265
7500 249
so the avg. whp for my powerband is 257whp.
who wants to go next?
[Modified by DIRep972, 5:42 AM 10/1/2002]
09-30-2002, 05:40 PM
#3
Smarter than you
Thread Starter
iTrader: (1)
Join Date: Apr 2001
Location: Third Coast, united states
Posts: 8,240
Likes: 0
Received 2 Likes
on
2 Posts
Re: Post ur *Average* dyno #'s. (dustin)
its the same thing. just multiply the hp by 5250 and divide it by RPM. either way, it still tells u the same thing.
09-30-2002, 05:43 PM
#4
Honda-Tech Member
Join Date: Jan 2000
Location: Sacramento, CA
Posts: 14,500
Likes: 0
Received 0 Likes
on
0 Posts
Re: Post ur *Average* dyno #'s. (DIRep972)
Hehe, but the "Area under the curve" of a horsepower graph is practically useless. Torque at the wheels is what determines how your car accelerates. I know I can extrapolate the values from your list, but no -- average horsepower and average torque are very different things.
09-30-2002, 05:50 PM
#6
Honda-Tech Member
Join Date: Jan 2000
Location: Sacramento, CA
Posts: 14,500
Likes: 0
Received 0 Likes
on
0 Posts
Re: Post ur *Average* dyno #'s. (DIRep972)
you missed that im not averaging area under the curve. I am avg. the curve itself.
09-30-2002, 06:10 PM
#7
Honda-Tech Member
Join Date: Dec 2001
Location: Filthadelphia Area, PA, USA
Posts: 1,568
Likes: 0
Received 0 Likes
on
0 Posts
Re: Post ur *Average* dyno #'s. (dustin)
you missed that im not averaging area under the curve. I am avg. the curve itself.
Understand that numeric integration across that rev range with X partitions will tell you the area under the curve. Dividing this number by the number of partitions (X) will yield average value.
Understand that numeric integration across that rev range with X partitions will tell you the area under the curve. Dividing this number by the number of partitions (X) will yield average value.
Trending Topics
09-30-2002, 06:16 PM
#8
Smarter than you
Thread Starter
iTrader: (1)
Join Date: Apr 2001
Location: Third Coast, united states
Posts: 8,240
Likes: 0
Received 2 Likes
on
2 Posts
Re: Post ur *Average* dyno #'s. (dustin)
I see what ur saying. When I think of the phrase "area under the curve" I think of "power below turbo" but when you say area under the curve you mean the avg. power between two points.
I would like to discuss more about how hp vs. torque effect acceleration. I take back my first statement where I said hp and torque tell you the same thing. because they dont.
Hp is more useful in figuring out how fast a car accelerates then torque! what is hp? torque over a given period of time right? torque is only a measure of force and by itself will only tell u how much force is being put on an object, in our case to make it rotate. 200ft/bs of torque on an object 4 times per seconed will result in slower rotation of the object then 200ft/lbs of torque 8 times per seconed. So without factoring in RPM's(rotations per minute) you cannot figure out acceleration. torque by itself is useless.
edited it to make it sound better hehe.
[Modified by DIRep972, 3:35 AM 10/1/2002]
I would like to discuss more about how hp vs. torque effect acceleration. I take back my first statement where I said hp and torque tell you the same thing. because they dont.
Hp is more useful in figuring out how fast a car accelerates then torque! what is hp? torque over a given period of time right? torque is only a measure of force and by itself will only tell u how much force is being put on an object, in our case to make it rotate. 200ft/bs of torque on an object 4 times per seconed will result in slower rotation of the object then 200ft/lbs of torque 8 times per seconed. So without factoring in RPM's(rotations per minute) you cannot figure out acceleration. torque by itself is useless.
edited it to make it sound better hehe.
[Modified by DIRep972, 3:35 AM 10/1/2002]
09-30-2002, 06:16 PM
#9
Smarter than you
Thread Starter
iTrader: (1)
Join Date: Apr 2001
Location: Third Coast, united states
Posts: 8,240
Likes: 0
Received 2 Likes
on
2 Posts
Re: Post ur *Average* dyno #'s. (racerxadam)
gotta love calculus. and i believe it is multiply by rmp and divide by 5252 not the other way around
09-30-2002, 08:35 PM
#14
Honda-Tech Member
Join Date: Jun 2001
Location: Northern, VA
Posts: 7,261
Likes: 0
Received 0 Likes
on
0 Posts
Re: Post ur *Average* dyno #'s. (Stag 2000)
5000 - 230whp
5400- 230whp
5800- 235whp
6200- 235whp
6600- 232whp
6800-225whp
231.16 avg. whp from 5000-6800
B20 STOCK @ 7-8 PSI / S-AFC/440/Vortech 8:1 FMU (My Old Setup)
5400- 230whp
5800- 235whp
6200- 235whp
6600- 232whp
6800-225whp
231.16 avg. whp from 5000-6800
B20 STOCK @ 7-8 PSI / S-AFC/440/Vortech 8:1 FMU (My Old Setup)
09-30-2002, 08:37 PM
#15
New User
Join Date: Dec 2000
Location: Southern, CA, USA
Posts: 1,264
Likes: 0
Received 0 Likes
on
0 Posts
Re: Post ur *Average* dyno #'s.
Dustin,
You're a smart guy, but Chris' explanation makes more sense to me and seems to follow my line of reasoning. Torque is definitely important, but like he said, it alone doesn't tell you anything until examine at what speed of rotation it's at, and by that point you have pure horsepower. Knowing that the car has 250 ftlbs of torque, let's say, is neat but not knowing where it falls on the curve is pretty useless. Say it falls at 2500 rpm, but drops off to 150 ftlbs by 5000. If you didn't know any better, you might think the car would begin accelerating slower as time went on, but in fact it's just the opposite. It's developing 119 hp at the first point and 142 hp at the 2nd mark.
Take the same torque numbers and shift them up by 2000 rpm each, and you now have a car that's making 214 hp and 200 hp. I guarantee the second car will out accelerate the first.
Damn, did I get off track? I think you were making a different point and I'm sure you knew this already, but maybe you can help clarify what you meant in your earlier post.
You're a smart guy, but Chris' explanation makes more sense to me and seems to follow my line of reasoning. Torque is definitely important, but like he said, it alone doesn't tell you anything until examine at what speed of rotation it's at, and by that point you have pure horsepower. Knowing that the car has 250 ftlbs of torque, let's say, is neat but not knowing where it falls on the curve is pretty useless. Say it falls at 2500 rpm, but drops off to 150 ftlbs by 5000. If you didn't know any better, you might think the car would begin accelerating slower as time went on, but in fact it's just the opposite. It's developing 119 hp at the first point and 142 hp at the 2nd mark.
Take the same torque numbers and shift them up by 2000 rpm each, and you now have a car that's making 214 hp and 200 hp. I guarantee the second car will out accelerate the first.
Damn, did I get off track? I think you were making a different point and I'm sure you knew this already, but maybe you can help clarify what you meant in your earlier post.
09-30-2002, 08:54 PM
#16
Honda-Tech Member
Join Date: Jan 2000
Location: Sacramento, CA
Posts: 14,500
Likes: 0
Received 0 Likes
on
0 Posts
Re: Post ur *Average* dyno #'s. (degradated)
He's already stated that he is averaging Horsepower in the RPM band where he is racing, so RPM does not matter at all. Within that RPM range, his car will accelerate EXACTLY as his torque curve dictates.
Where you make torque is important because you can use gearing to your advantage -- This is why high redlines allow cars to accelerate faster (if they make torque in high rpms) because they do not lose effective torque to the wheels by shifting to the next gear as early.
However, he already took RPM out of the equation...
Take two vehicles. RPM and horsepower aside... at any given point the vehicle with more torque will have more instantaneous acceleration and will be (inherently) accelerating faster. F=MA.
This is only true if we are measuring effective torque at the wheels... This means we cannot use crank torque numbers corrected for drivetrain losses.... To estimate the outcome of a 2 vehicle race, we must know final drive as well as ratios for each gear. This is why simulators such as cartest require to enter all of this information.
If you would like a more concrete comparison, the easiest way is to attempt to graph effective torque at the wheels for a typical 1/4 mile run for our vehicle. Using the 1st gear ratio and final drive along with the dyno chart, graph ACTUAL torque at the wheels in first gear. "Shift" into second and do the same for all subsequent gears... graphing wheel torque. Do the same for the second vehicle. The car with more area under the curve wins the race. Note that you'll have to do some mathematical magic to discen how far 1/4 mile is, but you shouldn't have too much trouble
This is what a gtech does and what cartest does mathematically.
Dustin
Where you make torque is important because you can use gearing to your advantage -- This is why high redlines allow cars to accelerate faster (if they make torque in high rpms) because they do not lose effective torque to the wheels by shifting to the next gear as early.
However, he already took RPM out of the equation...
Take two vehicles. RPM and horsepower aside... at any given point the vehicle with more torque will have more instantaneous acceleration and will be (inherently) accelerating faster. F=MA.
This is only true if we are measuring effective torque at the wheels... This means we cannot use crank torque numbers corrected for drivetrain losses.... To estimate the outcome of a 2 vehicle race, we must know final drive as well as ratios for each gear. This is why simulators such as cartest require to enter all of this information.
If you would like a more concrete comparison, the easiest way is to attempt to graph effective torque at the wheels for a typical 1/4 mile run for our vehicle. Using the 1st gear ratio and final drive along with the dyno chart, graph ACTUAL torque at the wheels in first gear. "Shift" into second and do the same for all subsequent gears... graphing wheel torque. Do the same for the second vehicle. The car with more area under the curve wins the race. Note that you'll have to do some mathematical magic to discen how far 1/4 mile is, but you shouldn't have too much trouble
This is what a gtech does and what cartest does mathematically.Dustin
09-30-2002, 09:46 PM
#17
Smarter than you
Thread Starter
iTrader: (1)
Join Date: Apr 2001
Location: Third Coast, united states
Posts: 8,240
Likes: 0
Received 2 Likes
on
2 Posts
Re: Post ur *Average* dyno #'s. (dustin)
He's already stated that he is averaging Horsepower in the RPM band where he is racing, so RPM does not matter at all. Within that RPM range, his car will accelerate EXACTLY as his torque curve dictates.
Where you make torque is important because you can use gearing to your advantage -- This is why high redlines allow cars to accelerate faster (if they make torque in high rpms) because they do not lose effective torque to the wheels by shifting to the next gear as early.
However, he already took RPM out of the equation...
Take two vehicles. RPM and horsepower aside... at any given point the vehicle with more torque will have more instantaneous acceleration and will be (inherently) accelerating faster. F=MA.
Where you make torque is important because you can use gearing to your advantage -- This is why high redlines allow cars to accelerate faster (if they make torque in high rpms) because they do not lose effective torque to the wheels by shifting to the next gear as early.
However, he already took RPM out of the equation...
Take two vehicles. RPM and horsepower aside... at any given point the vehicle with more torque will have more instantaneous acceleration and will be (inherently) accelerating faster. F=MA.
this one is an LS turbo @ 7psi of boost. (sorry its big, friend scanned it that way and im too lazy to fix it.) http://visualxposure.net/7lbs.jpg (made it a link just for u b20c5 turbo)
and this one is a GSR turbo @ 7psi of boost.
(dont let the uneven torque and hp axis's confuse u when looking at this graph. torque is on the right)
http://us.f1.yahoofs.com/users/e081b...__hr_dynoi.jpg ok i dont know what happened to the graph, all of a sudden yahoo says we dont have access. I guess u will just have to trust me on the tq values.
now compare the powerbands of the two cars. 5-7k for the LS and 6-8k for the GSR. put RPM aside and I will just pick 7 points of tq. for each car in the RPM band each uses for racing.
LS GSR
180 180
188 181
185 180
185 179
182 178
175 176
165 172
avg. tq for LS is 180ft/lbs. avg. tq. for GSR is 178ft/lbs.
the amount of torque each car is making is very similar and from what u stated above, these cars should be accelerating at very much the same pace. I have a feeling the GSR is gonna win the race not because it doesnt have to shift as early, but because it is putting out more torque per seconed then the LS. and tq per seconed, is hp(in general terms). also, lets assume both cars are running the same transmission so that will be a constant between the two set ups.
[Modified by DIRep972, 4:11 PM 10/1/2002]
09-30-2002, 10:21 PM
#18
Smarter than you
Thread Starter
iTrader: (1)
Join Date: Apr 2001
Location: Third Coast, united states
Posts: 8,240
Likes: 0
Received 2 Likes
on
2 Posts
Re: Post ur *Average* dyno #'s. (dustin)
If you would like a more concrete comparison, the easiest way is to attempt to graph effective torque at the wheels for a typical 1/4 mile run for our vehicle. Using the 1st gear ratio and final drive along with the dyno chart, graph ACTUAL torque at the wheels in first gear. "Shift" into second and do the same for all subsequent gears... graphing wheel torque. Do the same for the second vehicle. The car with more area under the curve wins the race. Note that you'll have to do some mathematical magic to discen how far 1/4 mile is, but you shouldn't have too much trouble This is what a gtech does and what cartest does mathematically.
Dustin
This is what a gtech does and what cartest does mathematically.Dustin
09-30-2002, 11:07 PM
#19
Join Date: Sep 2002
Location: Where Geos Go Fast, 95355
Posts: 1,969
Likes: 0
Received 0 Likes
on
0 Posts
Re: Post ur *Average* dyno #'s. (DIRep972)
GEO Storm GSI W/ Impulse RS Engine, Stock rods/crank Arias custom pistons portwork, stock valvetrain Stock rods T3/T04e Turbo 18 PSI Boost (low boost) and rough tuning (Haltech), AFR's were in the 10-11:1 ratios and had misfires ~6k and valve float problems at above 8k.
New valve springs have been fitted, MSD DIS4HO, and more dyno time has solved the problems Power at 18 psi is near 390 hp and Power at 28 psi is unkown due we can't get traction on rollers! power is estimated to be 500+ to the ground.
New valve springs have been fitted, MSD DIS4HO, and more dyno time has solved the problems
Power at 18 psi is near 390 hp and Power at 28 psi is unkown due we can't get traction on rollers! power is estimated to be 500+ to the ground.
09-30-2002, 11:14 PM
#20
New User
Join Date: Nov 2000
Location: Tampa, FL, USA
Posts: 3,467
Likes: 0
Received 0 Likes
on
0 Posts
Re: Post ur *Average* dyno #'s. (dustin)
Hehe, but the "Area under the curve" of a horsepower graph is practically useless. Torque at the wheels is what determines how your car accelerates. I know I can extrapolate the values from your list, but no -- average horsepower and average torque are very different things.
10-01-2002, 12:34 AM
#21
New User
Join Date: Feb 2001
Location: SLC, Ut
Posts: 548
Likes: 0
Received 0 Likes
on
0 Posts
Re: Post ur *Average* dyno #'s. (dustin)
Dustin: You're a mart futher smucker and I would side with you, but:
I absolutely agree that the vehicle with more torque will have more instantaneous acceleration, as long as both vehicles have the same mass and both vehicles have the same overall gear ratio. (newtonsApple == newtonsApple)
The problem with your argument is that the "Force" variable in your equation is measured in Newtons, Kilowatts, Horsepower, ETC..., which includes time and distance and can easily be manipulated into one equation: (HP = RPM * torque / 5252). From that we can easily see that: torque * RPM / 5252 = mass * acceleration. This of course proves you to be half right. Torque is very important, but so is RPM. HP (read work done) is nothing without either one.
Fortunately most of the Honda vs. Chevrolet (or Ford, or Dodge, or ...) battles that I have fought stand to prove that half the torque at twice the RPM still equals victory for the lighter car. So the straight line race still goes to man with the most HP per lb (given identical drivers, traction, and mass...).
Vapor: Torque (ability to do work) ultimately leads to acceleration. HP is a measurement of torque over time (work done).
I would ultimately choose the car that makes the most torque at the highest RPM and has the best gear ratio(s) to use it.
Gear ratio is extremely important, but I do not have time to discuss that right now. G'night.
EDIT: DIYRep.. If you EVER post a graphic that big again, I will personally drive down to "The Lone Star State" and bitch slap you into the "Twin Star State" myself. [mumbling] "Too lazy to fix it..."... my ***.... mumble, something, mumble
[Modified by B20C5 Turbo, 2:58 AM 10/1/2002]
Take two vehicles. RPM and horsepower aside... at any given point the vehicle with more torque will have more instantaneous acceleration and will be (inherently) accelerating faster. F=MA.
The problem with your argument is that the "Force" variable in your equation is measured in Newtons, Kilowatts, Horsepower, ETC..., which includes time and distance and can easily be manipulated into one equation: (HP = RPM * torque / 5252). From that we can easily see that: torque * RPM / 5252 = mass * acceleration. This of course proves you to be half right. Torque is very important, but so is RPM. HP (read work done) is nothing without either one.
Fortunately most of the Honda vs. Chevrolet (or Ford, or Dodge, or ...) battles that I have fought stand to prove that half the torque at twice the RPM still equals victory for the lighter car. So the straight line race still goes to man with the most HP per lb (given identical drivers, traction, and mass...).
Vapor: Torque (ability to do work) ultimately leads to acceleration. HP is a measurement of torque over time (work done).
I would ultimately choose the car that makes the most torque at the highest RPM and has the best gear ratio(s) to use it.
Gear ratio is extremely important, but I do not have time to discuss that right now. G'night.
EDIT: DIYRep.. If you EVER post a graphic that big again, I will personally drive down to "The Lone Star State" and bitch slap you into the "Twin Star State" myself. [mumbling] "Too lazy to fix it..."... my ***.... mumble, something, mumble
[Modified by B20C5 Turbo, 2:58 AM 10/1/2002]
10-01-2002, 07:33 AM
#22
Honda-Tech Member
Join Date: Jan 2000
Location: Sacramento, CA
Posts: 14,500
Likes: 0
Received 0 Likes
on
0 Posts
Re: Post ur *Average* dyno #'s. (B20C5 Turbo)
Dustin: You're a mart futher smucker and I would side with you, but: Take two vehicles. RPM and horsepower aside... at any given point the vehicle with more torque will have more instantaneous acceleration and will be (inherently) accelerating faster. F=MA. I absolutely agree that the vehicle with more torque will have more instantaneous acceleration, as long as both vehicles have the same mass and both vehicles have the same overall gear ratio. (newtonsApple == newtonsApple)
The problem with your argument is that the "Force" variable in your equation is measured in Newtons, Kilowatts, Horsepower, ETC..., which includes time and distance and can easily be manipulated into one equation: (HP = RPM * torque / 5252). From that we can easily see that: torque * RPM / 5252 = mass * acceleration. This of course proves you to be half right. Torque is very important, but so is RPM. HP (read work done) is nothing without either one.
Fortunately most of the Honda vs. Chevrolet (or Ford, or Dodge, or ...) battles that I have fought stand to prove that half the torque at twice the RPM still equals victory for the lighter car. So the straight line race still goes to man with the most HP per lb (given identical drivers, traction, and mass...).
Vapor: Torque (ability to do work) ultimately leads to acceleration. HP is a measurement of torque over time (work done).
I would ultimately choose the car that makes the most torque at the highest RPM and has the best gear ratio(s) to use it.
Gear ratio is extremely important, but I do not have time to discuss that right now. G'night.
EDIT: DIYRep.. If you EVER post a graphic that big again, I will personally drive down to "The Lone Star State" and bitch slap you into the "Twin Star State" myself. [mumbling] "Too lazy to fix it..."... my ***.... mumble, something, mumble
The problem with your argument is that the "Force" variable in your equation is measured in Newtons, Kilowatts, Horsepower, ETC..., which includes time and distance and can easily be manipulated into one equation: (HP = RPM * torque / 5252). From that we can easily see that: torque * RPM / 5252 = mass * acceleration. This of course proves you to be half right. Torque is very important, but so is RPM. HP (read work done) is nothing without either one.
Fortunately most of the Honda vs. Chevrolet (or Ford, or Dodge, or ...) battles that I have fought stand to prove that half the torque at twice the RPM still equals victory for the lighter car. So the straight line race still goes to man with the most HP per lb (given identical drivers, traction, and mass...).
Vapor: Torque (ability to do work) ultimately leads to acceleration. HP is a measurement of torque over time (work done).
I would ultimately choose the car that makes the most torque at the highest RPM and has the best gear ratio(s) to use it.
Gear ratio is extremely important, but I do not have time to discuss that right now. G'night.
EDIT: DIYRep.. If you EVER post a graphic that big again, I will personally drive down to "The Lone Star State" and bitch slap you into the "Twin Star State" myself. [mumbling] "Too lazy to fix it..."... my ***.... mumble, something, mumble
10-01-2002, 08:02 AM
#23
Honda-Tech Member
Join Date: Jan 2000
Location: Sacramento, CA
Posts: 14,500
Likes: 0
Received 0 Likes
on
0 Posts
Re: Post ur *Average* dyno #'s. (dustin)
Let me see if I can clarify...
Lets take two cars. Integra LS. Same Weight. Same final drive. Same gearing. Same amount of available traction. Understand that this is entirely hypothetical.
Assume that a shift at 7000 drops the car to 5000rpm in every gear.
Car 1
RPM torque horsepower
5000 300 285
5500 250 260
6000 200 228
6500 150 185
7000 100 133
Avg 200 218
Car 2
RPM torque horsepower
5000 100 95
5500 150 157
6000 200 228
6500 250 309
7000 300 400
Avg 200 238
Looks like car #2 has 20 average horsepower more through the RPM band, correct?
Which car will accelerate faster? At the end of EVERY gear, these imaginary cars will be exactly equal -- perfectly synchronized. They will have accelerated exactly as their torque curves show. At 5000 rpm, car 1 will be ahead, but by 7000 rpm car 2 will have caught up. This is because average torque is equal.
Dustin
Lets take two cars. Integra LS. Same Weight. Same final drive. Same gearing. Same amount of available traction. Understand that this is entirely hypothetical.
Assume that a shift at 7000 drops the car to 5000rpm in every gear.
Car 1
RPM torque horsepower
5000 300 285
5500 250 260
6000 200 228
6500 150 185
7000 100 133
Avg 200 218
Car 2
RPM torque horsepower
5000 100 95
5500 150 157
6000 200 228
6500 250 309
7000 300 400
Avg 200 238
Looks like car #2 has 20 average horsepower more through the RPM band, correct?
Which car will accelerate faster? At the end of EVERY gear, these imaginary cars will be exactly equal -- perfectly synchronized. They will have accelerated exactly as their torque curves show. At 5000 rpm, car 1 will be ahead, but by 7000 rpm car 2 will have caught up. This is because average torque is equal.
Dustin
10-01-2002, 09:26 AM
#24
Smarter than you
Thread Starter
iTrader: (1)
Join Date: Apr 2001
Location: Third Coast, united states
Posts: 8,240
Likes: 0
Received 2 Likes
on
2 Posts
Re: Post ur *Average* dyno #'s. (dustin)
Let me see if I can clarify...
Lets take two cars. Integra LS. Same Weight. Same final drive. Same gearing. Same amount of available traction. Understand that this is entirely hypothetical.
Assume that a shift at 7000 drops the car to 5000rpm in every gear.
Car 1
RPM torque horsepower
5000 300 285
5500 250 260
6000 200 228
6500 150 185
7000 100 133
Avg 200 218
Car 2
RPM torque horsepower
5000 100 95
5500 150 157
6000 200 228
6500 250 309
7000 300 400
Avg 200 238
Looks like car #2 has 20 average horsepower more through the RPM band, correct?
Which car will accelerate faster? At the end of EVERY gear, these imaginary cars will be exactly equal -- perfectly synchronized. They will have accelerated exactly as their torque curves show. At 5000 rpm, car 1 will be ahead, but by 7000 rpm car 2 will have caught up. This is because average torque is equal.
Dustin
Lets take two cars. Integra LS. Same Weight. Same final drive. Same gearing. Same amount of available traction. Understand that this is entirely hypothetical.
Assume that a shift at 7000 drops the car to 5000rpm in every gear.
Car 1
RPM torque horsepower
5000 300 285
5500 250 260
6000 200 228
6500 150 185
7000 100 133
Avg 200 218
Car 2
RPM torque horsepower
5000 100 95
5500 150 157
6000 200 228
6500 250 309
7000 300 400
Avg 200 238
Looks like car #2 has 20 average horsepower more through the RPM band, correct?
Which car will accelerate faster? At the end of EVERY gear, these imaginary cars will be exactly equal -- perfectly synchronized. They will have accelerated exactly as their torque curves show. At 5000 rpm, car 1 will be ahead, but by 7000 rpm car 2 will have caught up. This is because average torque is equal.
Dustin
10-01-2002, 10:01 AM
#25
Honda-Tech Member
Join Date: Jan 2000
Location: Northern, CA, USA
Posts: 1,119
Likes: 0
Received 0 Likes
on
0 Posts
Re: Post ur *Average* dyno #'s. (DIRep972)
I dont believe so dustin. We are all thinking along the same lines, which is, that torque in the right place is what makes a car accelerate, but I think you are overlooking the fact that hp is just a representation of how many times you can apply torque over a given time period. In your hypothetical situation, I honestly think the car with more hp would win. why? because the two motors are applying the same amount of torque to the crank/drivetrain/wheels. but in the given time of the race, motor 2 will have applied the same amount of torque on the crank more times then motor 1. causing the drivetrain to rotate and rotating the wheels more times then car one, thus it will be at a higher mph.
Damn I havent posted here in a while.