996 X50 vs. Gallardo..Which is quicker??
Originally posted by Jean
Spot-on. I beat a Gallardo with my heavy (4WD and 6 speed) ex- 964 3.6T and about 460hp. That same 3.6T did the 60-130mph in 10 seconds.
One second is a very substantial difference.
Spot-on. I beat a Gallardo with my heavy (4WD and 6 speed) ex- 964 3.6T and about 460hp. That same 3.6T did the 60-130mph in 10 seconds.
One second is a very substantial difference.
The only problem is when a magazines gets a slow vehicle because it has a problem. This happens often due to a bad clutch or other vehicle problem. Usually the magazine knows this and states that in the article.
On very rare occasions, the magazine gets a faster vehicle. This is very rare, and usually happens when they're testing a European car. Usually, the magazine knows that also and states that in the article.
To help on the last two, I like to compare the times from 2 or 3 magazines to see if something was off. The most recent problems were with the Ford GT and Ferrari 430. Ford bumped the HP from 500 to 550 HP. The earlier tests show much slower times than the more recent tests. Initial Ferrari 430 tests used the European car with launch control. Times for all the other popular sports cars have been very consistant between magazines and on different test dates.
Last edited by Bill S; Nov 2, 2005 at 10:30 AM.
Originally posted by Bill S
Over the years, I've found that the times published in R&T, Motor Trend and C&D magazine give a really good indication of a car's acceleration performance. I've measured or raced many different cars and these magazines were always spot-on with the results I measured. The people that do these tests for these magazines are very experienced and very careful. They have a standard procedure that is proven over many years. 60 to 130 is a great test because it eliminates start-up differences between vehicles and gives a large enough range to eliminate gearing differences. It's also a great way to compare cars because most street and track encounters occur in this range. 1/4-mile times are not good because most people don't know how to launch the car or don't want to launch hard.
The only problem is when a magazines gets a slow vehicle because it has a problem. This happens often due to a bad clutch or other vehicle problem. Usually the magazine knows this and states that in the article.
On very rare occasions, the magazine gets a faster vehicle. This is very rare, and usually happens when they're testing a European car. Usually, the magazine knows that also and states that in the article.
To help on the last two, I like to compare the times from 2 or 3 magazines to see if something was off. The most recent problems was with the Ford GT when Ford bumped the HP from 500 to 550 HP. The earlier tests show much slower times than the more recent tests. Times for all the other popular sports cars have been very consistant between magazines and on different test dates.
Over the years, I've found that the times published in R&T, Motor Trend and C&D magazine give a really good indication of a car's acceleration performance. I've measured or raced many different cars and these magazines were always spot-on with the results I measured. The people that do these tests for these magazines are very experienced and very careful. They have a standard procedure that is proven over many years. 60 to 130 is a great test because it eliminates start-up differences between vehicles and gives a large enough range to eliminate gearing differences. It's also a great way to compare cars because most street and track encounters occur in this range. 1/4-mile times are not good because most people don't know how to launch the car or don't want to launch hard.
The only problem is when a magazines gets a slow vehicle because it has a problem. This happens often due to a bad clutch or other vehicle problem. Usually the magazine knows this and states that in the article.
On very rare occasions, the magazine gets a faster vehicle. This is very rare, and usually happens when they're testing a European car. Usually, the magazine knows that also and states that in the article.
To help on the last two, I like to compare the times from 2 or 3 magazines to see if something was off. The most recent problems was with the Ford GT when Ford bumped the HP from 500 to 550 HP. The earlier tests show much slower times than the more recent tests. Times for all the other popular sports cars have been very consistant between magazines and on different test dates.
Ill have a Gallardo soon and we can find out. Or as I said, I have a couple guys with G cars in South Cali who Im sure would be willing to find out.
And by the way, A couple magazines Sport Auto, and Auto Motor Sport, have the Gallardo hitting 0-100mph in 8.5 sec or better, with 0-125 in 12.8 sec.
And btw, I dont claim the Gallardo to be the worlds fastest car, but they run pretty decent. my only experience running a Gallardo was in my stock Murcielago, racing my firends 04 6 Speed G. His car is modified, with Kreisigg exhaust (no cats), Gruppe M Intake, and reprogrammed Ecu. Going from 30- to approx 120-130 my Murci beat him by maybe 1 1/2 cars.
Last edited by allanlambo; Nov 2, 2005 at 10:35 AM.
Originally posted by allanlambo
So lets see, 1st you quote that tha tyour CS eats Gallardos for lunch, then you, yourself show magazine times showing the Gallardo will slap a 360CS around like no tommorow, and now you find magazines a good indication of what a car will do?
So lets see, 1st you quote that tha tyour CS eats Gallardos for lunch, then you, yourself show magazine times showing the Gallardo will slap a 360CS around like no tommorow, and now you find magazines a good indication of what a car will do?
I only said the two cars are exactly the same in the 1/4-mile which is what I've found when I run with Gallardos. I've never raced a Gallardo above 120 or so, so I don't know what happens there. However, with a few second different to 150, no doubt the Gallardo will be ahead with identical drivers. However, there's absolutely no way of knowing how far ahead without trying.
Last edited by Bill S; Nov 2, 2005 at 11:08 AM.
Originally posted by Bill S
Allan, it's funny how you change what a person writes. You would really screw-up those "telephone" games kids play!
I only said the two cars are exactly the same in the 1/4-mile which is what I've found when I run with Gallardos. I've never raced a Gallardo above 120 or so, so I don't know what happens there. However, with a few second different to 150, no doubt the Gallardo will be ahead with identical drivers. However, there's absolutely know way of knowing how far ahead without trying.
Allan, it's funny how you change what a person writes. You would really screw-up those "telephone" games kids play!
I only said the two cars are exactly the same in the 1/4-mile which is what I've found when I run with Gallardos. I've never raced a Gallardo above 120 or so, so I don't know what happens there. However, with a few second different to 150, no doubt the Gallardo will be ahead with identical drivers. However, there's absolutely know way of knowing how far ahead without trying.
Originally posted by SCvet00
If a car is ~14' (Z06 is 15' exactly, and a Viper is about 14') long, that is
220/14 = ~16 cars. *** whoopin..
If a car is ~14' (Z06 is 15' exactly, and a Viper is about 14') long, that is
220/14 = ~16 cars. *** whoopin..
The distance a vehicle travels is:
X = Xo + Vo t + .5 A t²
Xo = initial position
Vo = initial velocity
A = vehicle acceleration
t = time
Now, let's assume both vehicles accelerate fairly evenly from the end of the quarter mile (where they are even) to 150 MPH. That assumption gives the most distance between the vehicles. Then:
A (Gallardo) = ((150-118)/(21.4-12.4)) * (5280/3600) = 5.21 ft/sec²
A (CS) = ((150-115)/(23.9-12.4)) * (5280/3600) = 4.46 ft/sec²
So, assuming the CS and Gallardo race for another 10 seconds after the quarter mile:
X (Gallardo) = Xo + 118*(5280/3600)*10 + .5*5.21*10² = 1991 feet
X (CS) = Xo + 115*(5280/3600)*10 + .5*4.46*10² = 1910 feet
So, the Gallardo will be 81 feet ahead. Assuming about 15 feet for a car length, this is about 5 cars. In other words, there is only 4 car-lengths between the Gallardo and CS after racing for 10 seconds beyond the quarter mile (where the Gallardo is going over 150 MPH).
In reality, there actually could be a lot less distance. For example, if the Gallardo and CS are even at 130 MPH, the difference is smaller. However, these equations are a worst case. There will not be more than 4 cars lengths between the Gallardo and the CS.
So, I guess MEGA BUSLENGTHS is correct if you mean 4 VW buses.
Last edited by Bill S; Nov 2, 2005 at 09:18 PM.
Originally posted by Bill S
Um, not exactly. Here's some useful equations for you bench racers:
The distance a vehicle travels is:
X = Xo + VoT + .5 A T T
Xo = initial position
Vo = initial velocity
A = vehicle acceleration
T = time
Now, let's assume both vehicles accelerate fairly evenly from the end of the quarter mile (where they are even) to 150 MPH. That assumption gives the most distance between the vehicles. Then:
A (Gallardo) = ((150-118)/(21.4-12.4)) * (5280/3600) = 5.21 ft/T T
A (CS) = ((150-115)/(23.9-12.4)) * (5280/3600) = 4.46 ft/T T
So, assuming the CS and Gallardo race for another 10 seconds after the quarter mile:
X (Gallardo) = Xo + 118*(5280/3600)*10 + .5*5.21*10*10 = 1991 feet
X (CS) = Xo + 115*(5280/3600)*10 + .5*4.46*10*10 = 1910 feet
So, the Gallardo will be 81 feet ahead. Assuming about 15 feet for a car length, this is about 5 cars. In other words, there is only 4 car-lengths between the Gallardo and CS after racing for 10 seconds beyond the quarter mile (where the Gallardo is going over 150 MPH).
In reality, there actually could be a a lot less distance. For example, if the Gallardo and CS are even at 130 MPH, the difference is smaller. However, these equations are a worst case. There will not be more than 4 cars lengths between the Gallardo and the CS.
So, I guess MEGA BUSLENGTHS is correct if you mean 4 VW buses.
Um, not exactly. Here's some useful equations for you bench racers:
The distance a vehicle travels is:
X = Xo + VoT + .5 A T T
Xo = initial position
Vo = initial velocity
A = vehicle acceleration
T = time
Now, let's assume both vehicles accelerate fairly evenly from the end of the quarter mile (where they are even) to 150 MPH. That assumption gives the most distance between the vehicles. Then:
A (Gallardo) = ((150-118)/(21.4-12.4)) * (5280/3600) = 5.21 ft/T T
A (CS) = ((150-115)/(23.9-12.4)) * (5280/3600) = 4.46 ft/T T
So, assuming the CS and Gallardo race for another 10 seconds after the quarter mile:
X (Gallardo) = Xo + 118*(5280/3600)*10 + .5*5.21*10*10 = 1991 feet
X (CS) = Xo + 115*(5280/3600)*10 + .5*4.46*10*10 = 1910 feet
So, the Gallardo will be 81 feet ahead. Assuming about 15 feet for a car length, this is about 5 cars. In other words, there is only 4 car-lengths between the Gallardo and CS after racing for 10 seconds beyond the quarter mile (where the Gallardo is going over 150 MPH).
In reality, there actually could be a a lot less distance. For example, if the Gallardo and CS are even at 130 MPH, the difference is smaller. However, these equations are a worst case. There will not be more than 4 cars lengths between the Gallardo and the CS.
So, I guess MEGA BUSLENGTHS is correct if you mean 4 VW buses.
Originally posted by allanlambo
You better go and re-do your math.
You better go and re-do your math.
It's OK with me if you'd rather use that math. Whichever works best for you.
Originally posted by Bill S
Allan, the math is correct. Saying the Gallardo will be 220 feet ahead assumes the CS stopped dead and then the Gallardo continued going for another 2 seconds at 150 MPH.
It's OK with me if you'd rather use that math. Whichever works best for you.
Allan, the math is correct. Saying the Gallardo will be 220 feet ahead assumes the CS stopped dead and then the Gallardo continued going for another 2 seconds at 150 MPH.
It's OK with me if you'd rather use that math. Whichever works best for you.
Originally posted by Bill S
Um, not exactly. Here's some useful equations for you bench racers:
The distance a vehicle travels is:
X = Xo + VoT + .5 A T T
Xo = initial position
Vo = initial velocity
A = vehicle acceleration
T = time
Now, let's assume both vehicles accelerate fairly evenly from the end of the quarter mile (where they are even) to 150 MPH. That assumption gives the most distance between the vehicles. Then:
A (Gallardo) = ((150-118)/(21.4-12.4)) * (5280/3600) = 5.21 ft/T T
A (CS) = ((150-115)/(23.9-12.4)) * (5280/3600) = 4.46 ft/T T
So, assuming the CS and Gallardo race for another 10 seconds after the quarter mile:
X (Gallardo) = Xo + 118*(5280/3600)*10 + .5*5.21*10*10 = 1991 feet
X (CS) = Xo + 115*(5280/3600)*10 + .5*4.46*10*10 = 1910 feet
So, the Gallardo will be 81 feet ahead. Assuming about 15 feet for a car length, this is about 5 cars. In other words, there is only 4 car-lengths between the Gallardo and CS after racing for 10 seconds beyond the quarter mile (where the Gallardo is going over 150 MPH).
In reality, there actually could be a a lot less distance. For example, if the Gallardo and CS are even at 130 MPH, the difference is smaller. However, these equations are a worst case. There will not be more than 4 cars lengths between the Gallardo and the CS.
So, I guess MEGA BUSLENGTHS is correct if you mean 4 VW buses.
Um, not exactly. Here's some useful equations for you bench racers:
The distance a vehicle travels is:
X = Xo + VoT + .5 A T T
Xo = initial position
Vo = initial velocity
A = vehicle acceleration
T = time
Now, let's assume both vehicles accelerate fairly evenly from the end of the quarter mile (where they are even) to 150 MPH. That assumption gives the most distance between the vehicles. Then:
A (Gallardo) = ((150-118)/(21.4-12.4)) * (5280/3600) = 5.21 ft/T T
A (CS) = ((150-115)/(23.9-12.4)) * (5280/3600) = 4.46 ft/T T
So, assuming the CS and Gallardo race for another 10 seconds after the quarter mile:
X (Gallardo) = Xo + 118*(5280/3600)*10 + .5*5.21*10*10 = 1991 feet
X (CS) = Xo + 115*(5280/3600)*10 + .5*4.46*10*10 = 1910 feet
So, the Gallardo will be 81 feet ahead. Assuming about 15 feet for a car length, this is about 5 cars. In other words, there is only 4 car-lengths between the Gallardo and CS after racing for 10 seconds beyond the quarter mile (where the Gallardo is going over 150 MPH).
In reality, there actually could be a a lot less distance. For example, if the Gallardo and CS are even at 130 MPH, the difference is smaller. However, these equations are a worst case. There will not be more than 4 cars lengths between the Gallardo and the CS.
So, I guess MEGA BUSLENGTHS is correct if you mean 4 VW buses.

Craig
Last edited by Craig; Nov 2, 2005 at 04:52 PM.
Originally posted by Bill S
Allan, the math is correct. Saying the Gallardo will be 220 feet ahead assumes the CS stopped dead and then the Gallardo continued going for another 2 seconds at 150 MPH.
It's OK with me if you'd rather use that math. Whichever works best for you.
Allan, the math is correct. Saying the Gallardo will be 220 feet ahead assumes the CS stopped dead and then the Gallardo continued going for another 2 seconds at 150 MPH.
It's OK with me if you'd rather use that math. Whichever works best for you.
Honestly, im terrible at math, but do yourself a favor, imagine yourself standing at the end of a 150mph run way. Now imagine one car going buy at 150mph and you instantly click your stop watch and wait to 2 seconds until the next car passes. If you think only 4 car lengths will seperate the 2, you are delerious.
Also factor into your math, that when when the Gallardo its 150mph marker, the 360 is going no where near that same speed.
Talking with some drag racers, they said the general rule was figure just in the 1/4 every tenth of a second= 2 car lengths.
Also factor into your math, that when when the Gallardo its 150mph marker, the 360 is going no where near that same speed.
Talking with some drag racers, they said the general rule was figure just in the 1/4 every tenth of a second= 2 car lengths.





