5th gear review of the GT500
10-26-2006, 03:45 PM
#61
TECH Senior Member
Originally Posted by 25psi
Call it what you want, but it takes a 5.7 liter to make 25 more hp than a 3.5 liter 350z.
The "high tech" VQ 3.5L is LARGER and HEAVIER than the LS1(400+lbs vs 390lbs), and I wouldnt doubt it was more expencive. So which is more "advanced"? Would you not rather take the lighter, smaller, more powerful (and more importantly with more torque at every RPM) LS1?
What about the "omg old tach pooshrod LS7"? It, like the LS1/2/6 is lighter, and smaller than BMW's 5L V10 and Ferrari's 4.3L V8, both of which cant hold a candle to the low and midrange power the LS7 makes either.
Where as most v6 domestics make 200-220. S2000 makes more than a v6 camaro,mustang,etc.... Instead of cramming a large motor into a light vehicle, it would make more sense to utilize a smaller motor and maximize it.
10-26-2006, 03:45 PM
#62
Teching In
Join Date: Apr 2005
Location: Eaton, CO/Lubbock, TX
Posts: 35
Likes: 0
Received 0 Likes
on
0 Posts
you may have a prefence to make tq at a higher rpm... but wouldn't make more sense for drag racing if you had all the grunt from the engine at a lower rpm to get the car moving? you can make all the horse power you want to, but if you cant get the car to move what the hell good is it?
10-26-2006, 03:48 PM
#63
TECH Senior Member
Originally Posted by 25psi
My objective here is not to point out peak numbers, but to illustrate to you that you don't necessarily need a huge motor to gain max hp and torque. Becuase you have a 522 ci motor, doesnt mean you'll make peak torque faster.
10-26-2006, 04:16 PM
#64
TECH Fanatic
Join Date: Dec 2002
Location: htown
Posts: 1,200
Likes: 0
Received 0 Likes
on
0 Posts
Originally Posted by JustaStockHawk
you may have a prefence to make tq at a higher rpm... but wouldn't make more sense for drag racing if you had all the grunt from the engine at a lower rpm to get the car moving? you can make all the horse power you want to, but if you cant get the car to move what the hell good is it?
Theres a misconception that torque accelerates your vehicle, which is not true. The point I'm trying to advocate is that there is no reason to look at the torque. If you want an engine that produces 1000hp a@ 8,500rpm you will need 617ft/lbs of torque. The misunderstanding is that you are thinking that I'm saying you will never need torque at all. This is not the point I was trying to make. I was trying to say that you can throw out your torque numbers, they are unimportant. Now for the car to make X amount of horsepower it will need Y amount of torque at Z rpm. So you will need torque to make horsepower. However the torque, as a number, is useless to brag about. It is useless to say "Oh yeah, well my car makes 400ft/lbs of torque. Now if that same person was to say "My car makes 400ft/lbs of torque at 8000rpm" its a different story. But thats only because that means he's making 609RWHP. Which is definitely a force to be reckoned with.
10-26-2006, 04:24 PM
#65
TECH Fanatic
Join Date: Dec 2002
Location: htown
Posts: 1,200
Likes: 0
Received 0 Likes
on
0 Posts
This is qouted from a member of the mr2oc.com forum
LONG AS HELL BUT VERY INFORMATIVE)
Hate to bring this back up, but a person from the mr2 board took time out to write somethin very interesting regarding torque vs hp. I thought everyone should take a look at this.
These are two identical cars:
http://supraforums.com/forum/showthread.php?t=387509
I am simply copying and pasting a writup I posted on the MR2oc:
__________________________________________________ _________
I have been wanting to do this for a while, but I knew it would take forever and havent had the time. It has taken me about 1.5 weeks to finish with my limited free time.
_________________________
We'll use these two rather extreme examples for this experiment:
First, we have a car that peaks at around 430 ft/lbs at about 4200 RPM, with a peak HP of 395 at about 5700 RPM. 200 ft/lbs is crossed at 2700 RPM, and torque falls to 210 ft/lbs by it's redline of 7300.
Car two has an engine that hits peak torque of 350 ft/lbs at 6600 RPM, with a peak HP of 501HP at 8600 RPM. 200 ft/lbs is reached at 4000, and torque falls to 280 ft/lbs at it's redline of 9100.
Both cars are on 26" tall slicks, with identical, perfectly shifting 93+ LSD transmissions (with a drivetraid loss of, lets say, 16%), ATS Axle cages, 4" skinnies up front, and both cars weigh the same. Both cars weigh 2801.4 pounds, or 87 slugs.
____
Using the E153 gear ratios of-
1. 3.23
2. 1.91
3. 1.25
4. 0.91
5. 0.73
Final gear is 4.28
The following is the process to get the true wheel torque constant.
1st gear: 3.23 x 4.28 = 13.8244 ____ / 1.08333 = 12.26215422 _____ * .85 = 10.423
2nd gear: 1.91 x 4.28 = 8.1748 ____ / 1.08333 = 7.545969463 _____ * .85 = 6.414
3rd gear: 1.25 x 4.28 = 5.35 _______ / 1.08333 = 4.938461554 _____ * .85 = 4.198
4th gear: .91 x 4.28 = 3.89 ________ / 1.08333 = 3.590769232 _____ * .85 = 3.052
___
Fir the first car, using shift point calculations, I have found that the optimal shift points are as follows:
1st > 2nd: 7102 RPM
2nd > 3rd: 6418 RPM
3rd > 4th: 5769 RPM
The optimal shift points for car Two for each gear are past redline, so we will have it shift at its redline of 9100 RPM. This is simply because even at redline where the engine is making only 280 ft/lbs of torque, the wheel torque due to gear multiplication is higher than at any RPM in the next gear higher.
___
Fowl smoke billows from behind the cars as they smoothly reverse back to the starting line. With tires now hot, they begin to rev their engines. Car one, with the aid of 2-step technology, builds boost at 5000 RPMs, while car two builds boost at 6000. The pair of identical robotic drivers, programmed to drive, shift, and react exactly alike, slowly release their clutches as the last yellow dims.
We have seen cars on drag radials and slicks lift the fronts from a dig. If these MR2s weigh 2801.4 pounds, and have a wheelbase of 94.5 inches, with a weight bias of 42/58, that would take 1176.588 pounds of force to lift the front wheels up. With the fulcrum of the car being at the rear wheels, it would take 9265.6305 pounds of force at the contact patch. Speaking that neither of these cars is capable of producing that much torque, we will say that these tires, in this case, are unspinnable (although they technically are spinable, it is not relevant to the case.
As they launch, both cars slip the clutch at their launch RPMs, with, say, approxomately of 300 pounds of torque making it through the clutch, at which time the clutches are fully released and full torque is applied. This means that up until 4200 RPM, both cars have a linear force at the contact patch of 3126.9 pounds, and accelerate at a rate of 35.94 ft/s^2. 4200 RPM in first with these tires is 23.5 MPH, or 34.467 ft/s. Between 0 and 23.5 MPH, the cars both cover a distance of 16.527 feet, in a time of .96 seconds.
At this point, car One's clutch is fully released, with it's full 430 ft/lbs being forced through the drivetrain. Car Two still has it's same 300 ft/lbs at the wheels due to the still slipping clutch. while car one remains at 3126.9 pounds with the same acceleration of 35.94 ft/s^2. At 5000 RPM, it's clutch is also fully released. 5000 RPM in these cars is 28 MPH, or 41.067 ft/s. Between 23.5 and 28 MPH, car Two covers 6.94 feet in .184 seconds. At the same time, car One, with an average force of 4325.545 pounds, has accelerated to 29.74 MPH, and covered a distance of 7.18 feet.
Po' Kid
View Public Profile
Send a private message to Po' Kid
Visit Po' Kid's homepage!
Find More Posts by Po' Kid
Add Po' Kid to Your Buddy List
Today, 02:53 AM #2
Po' Kid
Nautical Blue MR2T
Join Date: Jul 2004
Location: Central Oklahoma
Posts: 1,336
Trader Rating: (0)
T: 1.144 seconds.
Car One has covered a distance of 23.707 feet and reached a speed of 29.74 MPH at 5300 RPM in 1st gear.
Car Two has covered a distance of 23.467 feet and reached a speed of 28 MPH at 5000 RPM in 1st gear.
Between 5300 RPM and car One's optimal 1st > 2nd shift point of 7102 RPM car One has an average crank torque of 315 ft/lbs, or 3283.25 pounds of force at the tires. This means an average acceleration of 37.74 ft/s^2. Between these speeds of 29.7MPH (43.56 ft/s) and 39.7 MPH (58.23 ft/s), car One covers a distance of 19.784 feet, in .389 seconds. In the same time, car Two, with an average crank torque of 330 ft/lbs, or 3439.59 pounds of force at the tires, and an acceleration of 39.54 ft/s^2, reaches a speed of 38.49 MPH, and covers a distance of 18.967 feet.
T: 1.533 seconds.
Car One has covered a distance of 43.491 feet and reached a speed of 39.7 MPH at 7102 RPM in 1st gear.
Car Two has covered a distance of 42.434 feet and reached a speed of 38.49 MPH at 6860 RPM in 1st gear.
Car two shifts to 2nd gear. We can say that it takes .25 seconds to make this shift. At 39.7 MPH, or 58.23 ft/s, it travels an additional 14.5575 feet. In this .25 seconds, car Two has the same average acceleration of 39.54 ft/s^2. Car Two's speed increases to 45.2 MPH (66.305 ft/s), and it covers a distance of 15.383 feet.
T: 1.783 seconds.
Car One has covered a distance of 58.049 feet and maintains the same speed of 39.7 MPH, but at 4200 in 2nd.
Car Two has covered a distance of 57.817 feet and has reached a speed of 45.2 MPH in 1st at 8080 RPM in 1st.
Car two, between 8080 RPM and 9100 RPM averages 300 ft/lbs at the crank. This translates to 3126.9 pounds at the contact patch, and and acceleration of 35.94 ft/s^2. Between these speeds of 45.2 and 50.9 MPH (66.293 and 74.653 ft/s), it covers a distance of 16.393 feet in .233 seconds. At the same time, car One is averaging about 425 ft/lbs at the crank, which translates to 2725.95 pounds of force at the tires, or an acceleration of 31.33 ft/s. In the same .233 seconds, car One reaches a speed of 44.7 MPH (65.531 ft/s), and covers a distance of 14.419 feet.
T: 2.016 seconds.
Car One has covered a distance of 72.468 feet and reached a speed of 44.7 MPH at 4830 RPM in 2nd.
Car Two has covered a distance of 74.210 feet and reached a speed of 50.9 MPH at 9100 RPM in 2nd.
At this point, car Two shifts to 2nd, taking another .25 seconds to complete the shift. In these .25 seconds, it travels a distance of 18.663 feet and maintains its speed of 50.9 MPH. Car One, in these .25 seconds, with an average force of 2405.25 pounds at the contact patch, accelerates at a rate of 27.647 ft/s^2, to a speed of 49.4 MPH, and covers a distance of 17.221 feet.
T: 2.266 seconds.
Car One has covered a distance of 89.689 feet and reached a speed of 49.4 MPH at 5220 RPM in 2nd.
Car Two has covered a distance of 92.873 feet and maintained its speed of 50.9 MPH at 5380 RPM in 2nd.
Car One accelerates to 6418 RPM, its optimal shift point into 3rd gear. Between 49.4 MPH (65.531 ft/s) and 60.7 MPH (89.027 ft/s), it has a linear force of 2212.83 pounds, making its acceleration 25.435 ft/s^2. It covers 71.388 feet, and does so in .924 seconds. In those same .924 seconds, car Two has an average linear force of 2116.62 pounds, making its acceleration 24.329 ft/s^2. It reaches 66.2 MPH, and covers a distance of 79.365 feet.
T: 3.190 seconds.
Car One has covered a distance of 161.077 feet and has reached a speed of 60.7 MPH (89.027 ft/s) at 6418 RPM in 2nd.
Car Two has covered a distance of 172.238 feet and has reached a speed of 66.2 MPH (97.133 ft/s) at 7000 RPM in 2nd.
Car One shifts from 2nd to 3rd, and again takes .25 seconds to do this. At a speed of 60.7 MPH (89.027 ft/s), it travels a distance of 22.257 feet. In this time, car Two has an average linear force of 2116.62 pounds and 24.329 ft/s^2. Its speed increases to 70.4 MPH, and covers a distance of 25.044 feet.
T: 3.440 seconds
Car One has covered a distance of 183.334 feet, and has maintained its speed of 60.7 MPH at 4200 RPM in 3rd.
Car Two has covered a distance of 197.282 feet, and has reached a speed of 70.4 MPH (103.215 ft/s), at 7440 RPM in 2nd.
Car Two accelerates from 70.4 MPH to 86.1 MPH (126.28 ft/s) at its 9100 RPM redline. It has an average linear force of 1956.27 pounds, giving it an average acceleration of 22.486 ft/s^2. It covers a distance of 117.702 feet, in a time of 1.026 seconds. In that same time frame, car One, with an average linear force of 1763.16 pounds, giving it an average acceleration of 20.266 ft/s^2, reaches a speed of 74.9 MPH (109.815 ft/s), and travels and distance of 101.982 feet.
T: 4.466 seconds.
Car One has covered a total distance of 285.316 feet, and has reached a speed of 74.9 MPH at 5180 RPM in 3rd.
Car Two has covered a total distance of 314.984 feet, and has reached a speed of 86.1 MPH at 9100 RPM in 2nd.
LONG AS HELL BUT VERY INFORMATIVE)Hate to bring this back up, but a person from the mr2 board took time out to write somethin very interesting regarding torque vs hp. I thought everyone should take a look at this.
These are two identical cars:
http://supraforums.com/forum/showthread.php?t=387509
I am simply copying and pasting a writup I posted on the MR2oc:
__________________________________________________ _________
I have been wanting to do this for a while, but I knew it would take forever and havent had the time. It has taken me about 1.5 weeks to finish with my limited free time.
_________________________
We'll use these two rather extreme examples for this experiment:
First, we have a car that peaks at around 430 ft/lbs at about 4200 RPM, with a peak HP of 395 at about 5700 RPM. 200 ft/lbs is crossed at 2700 RPM, and torque falls to 210 ft/lbs by it's redline of 7300.
Car two has an engine that hits peak torque of 350 ft/lbs at 6600 RPM, with a peak HP of 501HP at 8600 RPM. 200 ft/lbs is reached at 4000, and torque falls to 280 ft/lbs at it's redline of 9100.
Both cars are on 26" tall slicks, with identical, perfectly shifting 93+ LSD transmissions (with a drivetraid loss of, lets say, 16%), ATS Axle cages, 4" skinnies up front, and both cars weigh the same. Both cars weigh 2801.4 pounds, or 87 slugs.
____
Using the E153 gear ratios of-
1. 3.23
2. 1.91
3. 1.25
4. 0.91
5. 0.73
Final gear is 4.28
The following is the process to get the true wheel torque constant.
1st gear: 3.23 x 4.28 = 13.8244 ____ / 1.08333 = 12.26215422 _____ * .85 = 10.423
2nd gear: 1.91 x 4.28 = 8.1748 ____ / 1.08333 = 7.545969463 _____ * .85 = 6.414
3rd gear: 1.25 x 4.28 = 5.35 _______ / 1.08333 = 4.938461554 _____ * .85 = 4.198
4th gear: .91 x 4.28 = 3.89 ________ / 1.08333 = 3.590769232 _____ * .85 = 3.052
___
Fir the first car, using shift point calculations, I have found that the optimal shift points are as follows:
1st > 2nd: 7102 RPM
2nd > 3rd: 6418 RPM
3rd > 4th: 5769 RPM
The optimal shift points for car Two for each gear are past redline, so we will have it shift at its redline of 9100 RPM. This is simply because even at redline where the engine is making only 280 ft/lbs of torque, the wheel torque due to gear multiplication is higher than at any RPM in the next gear higher.
___
Fowl smoke billows from behind the cars as they smoothly reverse back to the starting line. With tires now hot, they begin to rev their engines. Car one, with the aid of 2-step technology, builds boost at 5000 RPMs, while car two builds boost at 6000. The pair of identical robotic drivers, programmed to drive, shift, and react exactly alike, slowly release their clutches as the last yellow dims.
We have seen cars on drag radials and slicks lift the fronts from a dig. If these MR2s weigh 2801.4 pounds, and have a wheelbase of 94.5 inches, with a weight bias of 42/58, that would take 1176.588 pounds of force to lift the front wheels up. With the fulcrum of the car being at the rear wheels, it would take 9265.6305 pounds of force at the contact patch. Speaking that neither of these cars is capable of producing that much torque, we will say that these tires, in this case, are unspinnable (although they technically are spinable, it is not relevant to the case.
As they launch, both cars slip the clutch at their launch RPMs, with, say, approxomately of 300 pounds of torque making it through the clutch, at which time the clutches are fully released and full torque is applied. This means that up until 4200 RPM, both cars have a linear force at the contact patch of 3126.9 pounds, and accelerate at a rate of 35.94 ft/s^2. 4200 RPM in first with these tires is 23.5 MPH, or 34.467 ft/s. Between 0 and 23.5 MPH, the cars both cover a distance of 16.527 feet, in a time of .96 seconds.
At this point, car One's clutch is fully released, with it's full 430 ft/lbs being forced through the drivetrain. Car Two still has it's same 300 ft/lbs at the wheels due to the still slipping clutch. while car one remains at 3126.9 pounds with the same acceleration of 35.94 ft/s^2. At 5000 RPM, it's clutch is also fully released. 5000 RPM in these cars is 28 MPH, or 41.067 ft/s. Between 23.5 and 28 MPH, car Two covers 6.94 feet in .184 seconds. At the same time, car One, with an average force of 4325.545 pounds, has accelerated to 29.74 MPH, and covered a distance of 7.18 feet.
Po' Kid
View Public Profile
Send a private message to Po' Kid
Visit Po' Kid's homepage!
Find More Posts by Po' Kid
Add Po' Kid to Your Buddy List
Today, 02:53 AM #2
Po' Kid
Nautical Blue MR2T
Join Date: Jul 2004
Location: Central Oklahoma
Posts: 1,336
Trader Rating: (0)
T: 1.144 seconds.
Car One has covered a distance of 23.707 feet and reached a speed of 29.74 MPH at 5300 RPM in 1st gear.
Car Two has covered a distance of 23.467 feet and reached a speed of 28 MPH at 5000 RPM in 1st gear.
Between 5300 RPM and car One's optimal 1st > 2nd shift point of 7102 RPM car One has an average crank torque of 315 ft/lbs, or 3283.25 pounds of force at the tires. This means an average acceleration of 37.74 ft/s^2. Between these speeds of 29.7MPH (43.56 ft/s) and 39.7 MPH (58.23 ft/s), car One covers a distance of 19.784 feet, in .389 seconds. In the same time, car Two, with an average crank torque of 330 ft/lbs, or 3439.59 pounds of force at the tires, and an acceleration of 39.54 ft/s^2, reaches a speed of 38.49 MPH, and covers a distance of 18.967 feet.
T: 1.533 seconds.
Car One has covered a distance of 43.491 feet and reached a speed of 39.7 MPH at 7102 RPM in 1st gear.
Car Two has covered a distance of 42.434 feet and reached a speed of 38.49 MPH at 6860 RPM in 1st gear.
Car two shifts to 2nd gear. We can say that it takes .25 seconds to make this shift. At 39.7 MPH, or 58.23 ft/s, it travels an additional 14.5575 feet. In this .25 seconds, car Two has the same average acceleration of 39.54 ft/s^2. Car Two's speed increases to 45.2 MPH (66.305 ft/s), and it covers a distance of 15.383 feet.
T: 1.783 seconds.
Car One has covered a distance of 58.049 feet and maintains the same speed of 39.7 MPH, but at 4200 in 2nd.
Car Two has covered a distance of 57.817 feet and has reached a speed of 45.2 MPH in 1st at 8080 RPM in 1st.
Car two, between 8080 RPM and 9100 RPM averages 300 ft/lbs at the crank. This translates to 3126.9 pounds at the contact patch, and and acceleration of 35.94 ft/s^2. Between these speeds of 45.2 and 50.9 MPH (66.293 and 74.653 ft/s), it covers a distance of 16.393 feet in .233 seconds. At the same time, car One is averaging about 425 ft/lbs at the crank, which translates to 2725.95 pounds of force at the tires, or an acceleration of 31.33 ft/s. In the same .233 seconds, car One reaches a speed of 44.7 MPH (65.531 ft/s), and covers a distance of 14.419 feet.
T: 2.016 seconds.
Car One has covered a distance of 72.468 feet and reached a speed of 44.7 MPH at 4830 RPM in 2nd.
Car Two has covered a distance of 74.210 feet and reached a speed of 50.9 MPH at 9100 RPM in 2nd.
At this point, car Two shifts to 2nd, taking another .25 seconds to complete the shift. In these .25 seconds, it travels a distance of 18.663 feet and maintains its speed of 50.9 MPH. Car One, in these .25 seconds, with an average force of 2405.25 pounds at the contact patch, accelerates at a rate of 27.647 ft/s^2, to a speed of 49.4 MPH, and covers a distance of 17.221 feet.
T: 2.266 seconds.
Car One has covered a distance of 89.689 feet and reached a speed of 49.4 MPH at 5220 RPM in 2nd.
Car Two has covered a distance of 92.873 feet and maintained its speed of 50.9 MPH at 5380 RPM in 2nd.
Car One accelerates to 6418 RPM, its optimal shift point into 3rd gear. Between 49.4 MPH (65.531 ft/s) and 60.7 MPH (89.027 ft/s), it has a linear force of 2212.83 pounds, making its acceleration 25.435 ft/s^2. It covers 71.388 feet, and does so in .924 seconds. In those same .924 seconds, car Two has an average linear force of 2116.62 pounds, making its acceleration 24.329 ft/s^2. It reaches 66.2 MPH, and covers a distance of 79.365 feet.
T: 3.190 seconds.
Car One has covered a distance of 161.077 feet and has reached a speed of 60.7 MPH (89.027 ft/s) at 6418 RPM in 2nd.
Car Two has covered a distance of 172.238 feet and has reached a speed of 66.2 MPH (97.133 ft/s) at 7000 RPM in 2nd.
Car One shifts from 2nd to 3rd, and again takes .25 seconds to do this. At a speed of 60.7 MPH (89.027 ft/s), it travels a distance of 22.257 feet. In this time, car Two has an average linear force of 2116.62 pounds and 24.329 ft/s^2. Its speed increases to 70.4 MPH, and covers a distance of 25.044 feet.
T: 3.440 seconds
Car One has covered a distance of 183.334 feet, and has maintained its speed of 60.7 MPH at 4200 RPM in 3rd.
Car Two has covered a distance of 197.282 feet, and has reached a speed of 70.4 MPH (103.215 ft/s), at 7440 RPM in 2nd.
Car Two accelerates from 70.4 MPH to 86.1 MPH (126.28 ft/s) at its 9100 RPM redline. It has an average linear force of 1956.27 pounds, giving it an average acceleration of 22.486 ft/s^2. It covers a distance of 117.702 feet, in a time of 1.026 seconds. In that same time frame, car One, with an average linear force of 1763.16 pounds, giving it an average acceleration of 20.266 ft/s^2, reaches a speed of 74.9 MPH (109.815 ft/s), and travels and distance of 101.982 feet.
T: 4.466 seconds.
Car One has covered a total distance of 285.316 feet, and has reached a speed of 74.9 MPH at 5180 RPM in 3rd.
Car Two has covered a total distance of 314.984 feet, and has reached a speed of 86.1 MPH at 9100 RPM in 2nd.
10-26-2006, 04:25 PM
#66
TECH Fanatic
Join Date: Dec 2002
Location: htown
Posts: 1,200
Likes: 0
Received 0 Likes
on
0 Posts
Car Two shifts from 2nd to 3rd, and, of course, takes .25 seconds. At a speed of 86.1 MPH, or 126.28 ft/s, it travels a distance of 31.57 feet in that time. In that time, car One, with an average linear force of 1616.23 pounds, giving it an average acceleration of 18.577 ft/s^2, reaches a speed of 78 MPH (114.459 ft/s), and covers a distance of 28.034 feet.
T: 4.716 seconds.
Car One has covered a total distance of 313.350 feet, and has reached a speed of 78 MPH at 5393 RPM in 3rd.
Car Two has covered a total distance of 346.554 feet, and maintains its speed of 86.1 MPH, but at 5955 RPM in 3rd.
As car One approaches its 3>4 shift RPM of 5769 RPM in 3rd (83.4 MPH or 134.219 ft/s), it has an average linear force of 1532.23 pounds, giving it an average acceleration of 17.612 ft/s^2. It covers a distance of 139.504 feet in a time of 1.122 seconds. In this time, car two, with an average linear force of 1426.32 pounds, giving it an average acceleration of 16.406 ft/s^2, reaches a speed of 144.688 ft/s or 98.651 MPH, and covers a distance of 152.013 feet.
T: 5.838 seconds.
Car One has covered a total distance of 452.854 feet and has reached a speed of 83.4 MPH at 5769 RPM in 3rd.
Car Two has covered a total distance of 498.567 feet and has reached a speed of 98.7 MPH at 6830 RPM in 3rd.
Car One shifts to 4th and takes .25 seconds to do so. At a speed of 83.4 MPH, or 134.219 ft/s, it travels 33.555 feet in this time. In the same .25 seconds, car Two, with the same average acceleration of 16.406 ft/s^2, increases its speed from 98.651 MPH to 101.4477272727, or 148.790 ft/s, and covers 36.689 feet.
T: 6.088 seconds.
Car One has covered a total distance of 486.409 feet and maintains its speed of 83.4 MPH, but at 4200 RPM in 4th.
Car Two has covered a total distance of 535.256 feet and has reached a speed of 101.5 MPH at 7015 RPM in 3rd.
Between 7015 RPM and 9100 RPM in 3rd, car Two would reach 131.6 MPH (193.013 ft/s) in 3rd. With an average linear force of 1322.37 pounds, giving it an average acceleration of 15.200 ft/s^2, it would cover a distance of 497.209 feet, in 2.909 seconds. In the same 2.909 seconds, car One, with an average linear force of 1144.500 pounds, giving it an average acceleration of 13.155 ft/s^2, would reach a speed of 172.487 ft/s (117.605 MPH), covering a distance of 446.110 feet.
T: 8.997 seconds.
Car One has covered a total distance of 932.519 feet and has reached a speed of 117.6 MPH at 5920 RPM in 4th.
Car Two has covered a total distance of 1032.465 feet and has reached a speed of 131.6 MPH at 9100 RPM in 3rd.
Car Two shifts from 3rd to 4th and takes .25 seconds to do so. At a speed of 193.013 ft/s, it travels 48.253 feet. In the same time, car One, with an average acceleration of 11.401 ft/s^2 reaches a speed of 175.337 ft/s (119.548 MPH), and covers a distance of 43.474 feet.
T: 9.247 seconds.
Car One has covered a total distance of 975.993 feet and has reached a speed of 119.5 MPH at 6015 RPM in 4th.
Car Two has covered a total distance of 1080.718 feet and has maintained its speed of 131.6 MPH, but is now at 6625 RPM in 4th.
Car Two must travel another 239.282 feet to finish the race, while car One must travel another 344.007 feet to finish. With an average acceleration of 11.577 ft/s^2, car two reaches a speed of 206.868 ft/s - 141.046 MPH - in a time of 1.197 seconds.
T: 10.444 seconds.
Car Two has finished the race at a speed of 141.046 MPH at 7105 RPM in 4th.
Car One, with an average acceleration of 11.050 ft/s^2, in the additional 344.007 feet required, reaches a speed of 195.821 ft/s (133.514 MPH) in a time of 1.854 seconds.
T: 11.101 seconds.
Car One has finished the race at a speed of 133.514 MPH at 6720 RPM in 4th.
T: 4.716 seconds.
Car One has covered a total distance of 313.350 feet, and has reached a speed of 78 MPH at 5393 RPM in 3rd.
Car Two has covered a total distance of 346.554 feet, and maintains its speed of 86.1 MPH, but at 5955 RPM in 3rd.
As car One approaches its 3>4 shift RPM of 5769 RPM in 3rd (83.4 MPH or 134.219 ft/s), it has an average linear force of 1532.23 pounds, giving it an average acceleration of 17.612 ft/s^2. It covers a distance of 139.504 feet in a time of 1.122 seconds. In this time, car two, with an average linear force of 1426.32 pounds, giving it an average acceleration of 16.406 ft/s^2, reaches a speed of 144.688 ft/s or 98.651 MPH, and covers a distance of 152.013 feet.
T: 5.838 seconds.
Car One has covered a total distance of 452.854 feet and has reached a speed of 83.4 MPH at 5769 RPM in 3rd.
Car Two has covered a total distance of 498.567 feet and has reached a speed of 98.7 MPH at 6830 RPM in 3rd.
Car One shifts to 4th and takes .25 seconds to do so. At a speed of 83.4 MPH, or 134.219 ft/s, it travels 33.555 feet in this time. In the same .25 seconds, car Two, with the same average acceleration of 16.406 ft/s^2, increases its speed from 98.651 MPH to 101.4477272727, or 148.790 ft/s, and covers 36.689 feet.
T: 6.088 seconds.
Car One has covered a total distance of 486.409 feet and maintains its speed of 83.4 MPH, but at 4200 RPM in 4th.
Car Two has covered a total distance of 535.256 feet and has reached a speed of 101.5 MPH at 7015 RPM in 3rd.
Between 7015 RPM and 9100 RPM in 3rd, car Two would reach 131.6 MPH (193.013 ft/s) in 3rd. With an average linear force of 1322.37 pounds, giving it an average acceleration of 15.200 ft/s^2, it would cover a distance of 497.209 feet, in 2.909 seconds. In the same 2.909 seconds, car One, with an average linear force of 1144.500 pounds, giving it an average acceleration of 13.155 ft/s^2, would reach a speed of 172.487 ft/s (117.605 MPH), covering a distance of 446.110 feet.
T: 8.997 seconds.
Car One has covered a total distance of 932.519 feet and has reached a speed of 117.6 MPH at 5920 RPM in 4th.
Car Two has covered a total distance of 1032.465 feet and has reached a speed of 131.6 MPH at 9100 RPM in 3rd.
Car Two shifts from 3rd to 4th and takes .25 seconds to do so. At a speed of 193.013 ft/s, it travels 48.253 feet. In the same time, car One, with an average acceleration of 11.401 ft/s^2 reaches a speed of 175.337 ft/s (119.548 MPH), and covers a distance of 43.474 feet.
T: 9.247 seconds.
Car One has covered a total distance of 975.993 feet and has reached a speed of 119.5 MPH at 6015 RPM in 4th.
Car Two has covered a total distance of 1080.718 feet and has maintained its speed of 131.6 MPH, but is now at 6625 RPM in 4th.
Car Two must travel another 239.282 feet to finish the race, while car One must travel another 344.007 feet to finish. With an average acceleration of 11.577 ft/s^2, car two reaches a speed of 206.868 ft/s - 141.046 MPH - in a time of 1.197 seconds.
T: 10.444 seconds.
Car Two has finished the race at a speed of 141.046 MPH at 7105 RPM in 4th.
Car One, with an average acceleration of 11.050 ft/s^2, in the additional 344.007 feet required, reaches a speed of 195.821 ft/s (133.514 MPH) in a time of 1.854 seconds.
T: 11.101 seconds.
Car One has finished the race at a speed of 133.514 MPH at 6720 RPM in 4th.
10-26-2006, 04:49 PM
#67
Banned
iTrader: (32)
Join Date: Mar 2005
Location: Fountain Inn, SC
Posts: 1,157
Likes: 0
Received 0 Likes
on
0 Posts
Another thread that go stolen by a ricer to prove a point that makes no sense Please start your own thread for stupid (bigger isnt always better) arguments in another server for ricers !! sounds like the little man syndrome to me . Back to the topic at hand i have a hard time believing half the crap that any of these magizines they all have there own opinions which usually end up wrong !
Please start your own thread for stupid (bigger isnt always better) arguments in another server for ricers !! sounds like the little man syndrome to me . Back to the topic at hand i have a hard time believing half the crap that any of these magizines they all have there own opinions which usually end up wrong !
10-26-2006, 04:56 PM
#68
TECH Fanatic
Join Date: Dec 2002
Location: htown
Posts: 1,200
Likes: 0
Received 0 Likes
on
0 Posts
What have I posted up that has anything to do with magazine racing? I've pointed out, calculation,data to prove all my responses. All you have done is taken the easy way out, instead of "disproving" anything that I have said. Prove me wong!
10-26-2006, 04:57 PM
#69
TECH Senior Member
Originally Posted by 25psi
Theres a misconception that torque accelerates your vehicle, which is not true. The point I'm trying to advocate is that there is no reason to look at the torque. If you want an engine that produces 1000hp a@ 8,500rpm you will need 617ft/lbs of torque. The misunderstanding is that you are thinking that I'm saying you will never need torque at all. This is not the point I was trying to make. I was trying to say that you can throw out your torque numbers, they are unimportant. Now for the car to make X amount of horsepower it will need Y amount of torque at Z rpm. So you will need torque to make horsepower. However the torque, as a number, is useless to brag about. It is useless to say "Oh yeah, well my car makes 400ft/lbs of torque. Now if that same person was to say "My car makes 400ft/lbs of torque at 8000rpm" its a different story. But thats only because that means he's making 609RWHP. Which is definitely a force to be reckoned with.
What your saying about torque is true though, BUT you can throw the hp numbers out the window and be fine as well.
For example, your car DOES accelerate according to the torque curve, if your making 300ft.lbs at 2000RPMs, and 300ft.lbs at 4000RPMs, in the same gear you will NOT pull any harder, although hp will be doubled.
10-26-2006, 05:06 PM
#70
Teching In
Join Date: Jun 2006
Location: Auburn/Phenix City Alabama
Posts: 39
Likes: 0
Received 0 Likes
on
0 Posts
Originally Posted by 25psi
The bigger your motor doesnt necessarily represent more hp and torque. Becuase you have a 7.0 liter V8 motor, doesnt mean you'll make more torque per/lb vs a 3.5 liter V6.
You are thinking becuase you have a v8, that you will crank out more torque. Not true! Let me ask you this. Which do you think will make more power. A 3.5 liter V6 or a 3.0 liter V8?
69 Mach. It's not that simple to calculate which you think is faster being all things equal. If you had a 5.0 liter 4cyl vs a 5.0 liter V8 in the same car, I'll put my money on the 4cyl. For one, you will have a higher revving motor. Two, less rotation mass. Three, lighter internal parts. Four, a more even distribution to place the motor in the car. Dont think just becuase you have a V8 you make more torque, that a theiry my friend. Stop looking at Hondas as your benchmark for this discussion.
My 350z made 412rwhp and 478ftlb of torque at 8psi. So whats your point?
You are thinking becuase you have a v8, that you will crank out more torque. Not true! Let me ask you this. Which do you think will make more power. A 3.5 liter V6 or a 3.0 liter V8?
69 Mach. It's not that simple to calculate which you think is faster being all things equal. If you had a 5.0 liter 4cyl vs a 5.0 liter V8 in the same car, I'll put my money on the 4cyl. For one, you will have a higher revving motor. Two, less rotation mass. Three, lighter internal parts. Four, a more even distribution to place the motor in the car. Dont think just becuase you have a V8 you make more torque, that a theiry my friend. Stop looking at Hondas as your benchmark for this discussion.
My 350z made 412rwhp and 478ftlb of torque at 8psi. So whats your point?
Last edited by 69mach; 10-26-2006 at 05:12 PM.
10-26-2006, 05:15 PM
#71
TECH Fanatic
Join Date: Dec 2002
Location: htown
Posts: 1,200
Likes: 0
Received 0 Likes
on
0 Posts
What racing are you talking about? The only thing that a 4cyl or 6cyl hasnt done is run in the 5's. Tell me why did it take an LS1 so long to touch 6's, when a supra did it with half the displacement and 2 less cylinders.
Is it more practical to use a V8 for a truck, sure. But can it be done with a 4cyl, why not. The reason for using a larger displacement engine, is to make torque down low, not necessarily making more torque than a smaller motor.
Is it more practical to use a V8 for a truck, sure. But can it be done with a 4cyl, why not. The reason for using a larger displacement engine, is to make torque down low, not necessarily making more torque than a smaller motor.
10-26-2006, 05:19 PM
#72
Originally Posted by 69mach
Why make a V8 engine for a truck when you can make a 4cyl truck with just as much torque??? Also in almost every single proffesional racing sport what do you see....usually a big displacement V8. There is obviously something wrong with your theory on torque. Instead of us having to prove you wrong, why dont you prove us wrong with your theory??? Their is plenty of video evidence on this website to back up our theory. Just my opinion though...
Oh, by the way, You guys should listen to me, I made into this month's GMHTP mag.
Last edited by BLOODTA; 10-26-2006 at 05:27 PM.
10-26-2006, 05:26 PM
#73
TECH Fanatic
Join Date: Dec 2002
Location: htown
Posts: 1,200
Likes: 0
Received 0 Likes
on
0 Posts
Originally Posted by BLOODTA
Dude, at least we are on the same page when it comes to good ol' American displacement, Torque is the key.
Your brainwashed! Ok, for arguement sake. Take a civic with 700hp and 400ftlb of torque vs a Viper with 500hp and 700ftlb of torque, who would win? The difference is, the civic is a 9 sec car, vs a low 11 sec car.
10-26-2006, 05:30 PM
#74
Originally Posted by 25psi
Your brainwashed! Ok, for arguement sake. Take a civic with 700hp and 400ftlb of torque vs a Viper with 500hp and 700ftlb of torque, who would win? The difference is, the civic is a 9 sec car, vs a low 11 sec car.
Imagine the sound that a 700 hp Civic's exhaust would make.. Sounds like Megatron letting one rip.
10-26-2006, 06:01 PM
#75
Banned
iTrader: (3)
Join Date: Mar 2006
Location: Upstate NY
Posts: 1,292
Likes: 0
Received 0 Likes
on
0 Posts
Originally Posted by 25psi
Take a civic with 700hp and 400ftlb of torque vs a Viper with 500hp and 700ftlb of torque, who would win? The difference is, the civic is a 9 sec car, vs a low 11 sec car.
Not to mention, 700 hp on a V-10 is just getting started.
With 700 hp, the viper would be close to 800 tq.
You also asked me earlier what my car peaks at, I think stock its 325 @ 4500, but I push 290+ from 2400 RPMS on.
Its there constantly....
Fact of the matter is, youre pro-import.
If you want a smaller engine to keep up, it needs better tuning, TONS of boost, lightened bodywork, etc... and always more money.
Either way in today's world, you can now use the same technology on newer engines, like the Cobra motor that can push 600+hp Stock and be reliable, or the new small blocks that reach 500+hp out of the box, and have the new technology and design to back it up.
more cubes = more power.
You want to see youre import dreams die?
Check out some real engines.
For how much some of you import guys spend, you could buy a house.
V-8's are not only more bang for your buck, thier easier to make fast, SOUND WAY BETTER, and have effortless bottem end torque, like my LT1.
Thier is no lag, and they will always have the fastest track times.
The only reason select 4 and 6 bangers keep up is solely on technology, boost, and tuning.
want to see some real race engines? **** putting 30K into a civic to make it 800hp... how about this? Check out some of these.
http://www.shafiroff.com/pro_engines.asp
10-26-2006, 06:43 PM
#78
Teching In
Join Date: Sep 2005
Location: florida
Posts: 36
Likes: 0
Received 0 Likes
on
0 Posts
megatron letting one rip... priceless...
ok 25psi... you got issues lol... you cant just make up equations and put numbers in there to sound smart.. you dont need x abmount of horsepower to get x amount of torque... proof of that is everyone post up your dyno numbers, i bet there is two people with the same horsepower but different torque, or vice versa.
its all about combination, dont you know that. i could make 1000hp but get 500 to the ground, or i could make 500 horsepower and get it all to the ground.. does that mean the cars are the same speed? not necessarily, it takes more than just an engine and how much power that engine makes to make a car go fast. if you dont belive me race a stock 350z vs a vette, and watch that 6cyl get its a$$ spanked all day.
o but your right.. id take the 5.0L 4 cylinder, so i can rev that P.O.S to 12k rpms and cook steak on it
still cant get over megatron.. that made my day.
ok 25psi... you got issues lol... you cant just make up equations and put numbers in there to sound smart.. you dont need x abmount of horsepower to get x amount of torque... proof of that is everyone post up your dyno numbers, i bet there is two people with the same horsepower but different torque, or vice versa.
its all about combination, dont you know that. i could make 1000hp but get 500 to the ground, or i could make 500 horsepower and get it all to the ground.. does that mean the cars are the same speed? not necessarily, it takes more than just an engine and how much power that engine makes to make a car go fast. if you dont belive me race a stock 350z vs a vette, and watch that 6cyl get its a$$ spanked all day.
o but your right.. id take the 5.0L 4 cylinder, so i can rev that P.O.S to 12k rpms and cook steak on it
still cant get over megatron.. that made my day.
Last edited by grandsport96; 10-26-2006 at 08:43 PM.
10-26-2006, 06:52 PM
#79
TECH Senior Member
Originally Posted by 25psi
Your brainwashed! Ok, for arguement sake. Take a civic with 700hp and 400ftlb of torque vs a Viper with 500hp and 700ftlb of torque, who would win? The difference is, the civic is a 9 sec car, vs a low 11 sec car.
10-26-2006, 06:54 PM
#80
TECH Senior Member
Originally Posted by 25psi
What racing are you talking about? The only thing that a 4cyl or 6cyl hasnt done is run in the 5's. Tell me why did it take an LS1 so long to touch 6's, when a supra did it with half the displacement and 2 less cylinders.