How come the new Xs only go 130?
#16
Wrong!! The terminal velocity of a car would be about 52 mph. This is assuming it's falling flat. If the car was falling pointing down, the terminal velocity would be something more like 90mph.
#17
just an out of box opinion, could it be the incoming wind speed was also strong while you were driving at a high speed?
it's pretty easy for any car to go from 0 - 100 with little horse power, but to go from 100 - 200, you need an engine of a super car cuz the wind resistance doubles for every 10 or 20 increase of speed. so, you if were hitting around 10 to 20 below the max speed, you could be running against the wind.
it's pretty easy for any car to go from 0 - 100 with little horse power, but to go from 100 - 200, you need an engine of a super car cuz the wind resistance doubles for every 10 or 20 increase of speed. so, you if were hitting around 10 to 20 below the max speed, you could be running against the wind.
Last edited by badassflip; 12-30-2008 at 09:27 AM.
#18
#19
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#20
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Then I found the vid above.
Then I found this on the www.ultimatecarforum.com:
Lexus IS Commercial
--------------------------------------------------------------------------------
Actually, since the car gets a head start we can infer that the car at most is traveling its maximum speed(2006 Lexus IS350 is 142mph). Therefore the car is at most traveling 208 ft/sec. It would take the car on the ground 19.231 seconds to travel the 4000 feet(1219.2 meters).
When we look at the falling car, we can disregard air friction and say that the falling car accelerates at 9.81 m/s^2 or 32.19 ft/s^2. From the equation: x=1/2at^2 we can conclude that the car will take 15.765 seconds to travel the 1219.2 meters(4000 feet). However, air friction will give the car a terminal velocity. Without the terminal velocity, we can figure that the car was traveling 507.475 ft/sec or 346.006 mph. It is obvious that the falling car will be faster. But lets not overlook terminal velocity.
Terminal velocity can be calculated using the equation:
V = sqrt ( (2 * W) / (Cd * r * A)
v=terminal velocity
W=weight in Newtons
Cd=Drag Coefficient
r=atmospheric density
a=area exposed to air(area facing down)
Furthermore, we find terminal velocity by determining our variables. The weight of the 2006 Lexus IS350 is 3,435 lbs. The atmospheric density is a main determining factor, and since the density is decreased by higher temperatures, for the fact of covering margins, lets say the experiment was done at 80 degrees F. This would make the atmospheric density 1.184. Again, for the sake of covering margins, lets say the car fell with its bottom facing downward, its surface area of exposure would be equal to the factory dimensions of 180 inches by 70.9 inches, or 88.625 square feet. Lastly, the drag coefficient must be calculated. First, take into account that the drag coefficient of a falling cube is 0.8. this is roughly the shape of the falling car. Also, consider the drag resistance of the front of a Lexus IS350(the drag coefficient if the car was falling head first) is very close to 0.3. Once again, for the sake of covering margins, lets use a coefficient of 0.6. After calculating, you will come up with a table looking like this:
altitude terminal velocity
3000 ft 243.795 ft/s
2000 ft 240.198 ft/s
1000 ft 236.679 ft/s
0 ft 233.235 ft/s
Since we will not reach terminal velocity until an altitude of 3000 feet(falling for 1000 feet), we can disregard the 4000 ft mark. We can then take the average terminal velocity from 3000 ft to 0 ft, about 238.515 ft/s. To cover margins, lets use 240 ft/s.
Therefore, we can use the equation velocity=acceleration*time. We find that we reach terminal velocity(240 ft/s) using the acceleration of gravity of 32.19 ft/s^2 after 7.456 seconds. We then use the equation distance=1/2*acceleration*time^2 and find that the car had traveled 894.688 feet when it reached terminal velocity.
Using this information, since the car had traveled 894.688 feet in 7.456 seconds, and since the car is now traveling 240 ft/sec for the remainder of its trip, we conclude that the car will take an additional 12.939 seconds to travel the remaining 3105.312 feet.(dist/veloc=time, (4000-894.688)/240=12.939)
Through addition we find that the car will take about 20.395 seconds to hit the ground.
Therefore, we see that(with our margins covered for variations) when we take air resistance into account and factor in a terminal velocity, a Lexus IS350 on the ground takes 19.231 seconds to travel 4000 feet. And as the commercial demonstrates, a Lexus IS350 takes about 20.395 seconds to reach the ground and travel its distance. Illustrating how the Lexus on the ground can, in fact, travel 4000 feet quicker than gravity can accelerate the same car 4000 feet to the ground.
These statistics(weight, surface area, Cd) are based on the 2006 Lexus IS350 manual transmission. Also, I over estimated many of my values, such as the temperature and the Cd of the Lexus. If they were to conduct the experiment at lets say 30 degrees F, the time the car would take to reach the ground would increase by about 1 second. Also, lets say the Cd of the car was in fact close to one of a cube, the terminal velocity of the car would be an average of about 205 ft/sec. This would actually increase the time of the falling car by about 2.301 seconds from 20.395 seconds to 22.696 seconds.
As you can see, the demonstration done in the commercial is completely plausible as long as the Lexus on the ground starts from near its maximum speed.
However, if the Lexus did not start from near 140 mph, it would never be able to compete with gravity. Just to compare this car with the Ferrari 333 Sp. This car can accelerate from 0-60 in 3.8 seconds. The Lexus IS350, 5.6 seconds. Just for a fact, the Ferrari 333 SP is a modified race car, not stock. Based on this, and the fact that the Ferrari 333 SP can accelerate at an average of about 20 ft/sec over a quarter mile and gravity, 32.19 ft/sec demonstrates that the Ferrari 333 cannot conquer the feat of starting from a stop, let alone the Lexus IS350.
The Lexus IS Started near its maximum speed, and was therefore able to beat gravity, as the commercial demonstrated.
Any questions or comments, feel free to email me at lexus@alexnwolf.com
Thank You,
Alex Wolf
--------------------------------------------------------------------------------
Actually, since the car gets a head start we can infer that the car at most is traveling its maximum speed(2006 Lexus IS350 is 142mph). Therefore the car is at most traveling 208 ft/sec. It would take the car on the ground 19.231 seconds to travel the 4000 feet(1219.2 meters).
When we look at the falling car, we can disregard air friction and say that the falling car accelerates at 9.81 m/s^2 or 32.19 ft/s^2. From the equation: x=1/2at^2 we can conclude that the car will take 15.765 seconds to travel the 1219.2 meters(4000 feet). However, air friction will give the car a terminal velocity. Without the terminal velocity, we can figure that the car was traveling 507.475 ft/sec or 346.006 mph. It is obvious that the falling car will be faster. But lets not overlook terminal velocity.
Terminal velocity can be calculated using the equation:
V = sqrt ( (2 * W) / (Cd * r * A)
v=terminal velocity
W=weight in Newtons
Cd=Drag Coefficient
r=atmospheric density
a=area exposed to air(area facing down)
Furthermore, we find terminal velocity by determining our variables. The weight of the 2006 Lexus IS350 is 3,435 lbs. The atmospheric density is a main determining factor, and since the density is decreased by higher temperatures, for the fact of covering margins, lets say the experiment was done at 80 degrees F. This would make the atmospheric density 1.184. Again, for the sake of covering margins, lets say the car fell with its bottom facing downward, its surface area of exposure would be equal to the factory dimensions of 180 inches by 70.9 inches, or 88.625 square feet. Lastly, the drag coefficient must be calculated. First, take into account that the drag coefficient of a falling cube is 0.8. this is roughly the shape of the falling car. Also, consider the drag resistance of the front of a Lexus IS350(the drag coefficient if the car was falling head first) is very close to 0.3. Once again, for the sake of covering margins, lets use a coefficient of 0.6. After calculating, you will come up with a table looking like this:
altitude terminal velocity
3000 ft 243.795 ft/s
2000 ft 240.198 ft/s
1000 ft 236.679 ft/s
0 ft 233.235 ft/s
Since we will not reach terminal velocity until an altitude of 3000 feet(falling for 1000 feet), we can disregard the 4000 ft mark. We can then take the average terminal velocity from 3000 ft to 0 ft, about 238.515 ft/s. To cover margins, lets use 240 ft/s.
Therefore, we can use the equation velocity=acceleration*time. We find that we reach terminal velocity(240 ft/s) using the acceleration of gravity of 32.19 ft/s^2 after 7.456 seconds. We then use the equation distance=1/2*acceleration*time^2 and find that the car had traveled 894.688 feet when it reached terminal velocity.
Using this information, since the car had traveled 894.688 feet in 7.456 seconds, and since the car is now traveling 240 ft/sec for the remainder of its trip, we conclude that the car will take an additional 12.939 seconds to travel the remaining 3105.312 feet.(dist/veloc=time, (4000-894.688)/240=12.939)
Through addition we find that the car will take about 20.395 seconds to hit the ground.
Therefore, we see that(with our margins covered for variations) when we take air resistance into account and factor in a terminal velocity, a Lexus IS350 on the ground takes 19.231 seconds to travel 4000 feet. And as the commercial demonstrates, a Lexus IS350 takes about 20.395 seconds to reach the ground and travel its distance. Illustrating how the Lexus on the ground can, in fact, travel 4000 feet quicker than gravity can accelerate the same car 4000 feet to the ground.
These statistics(weight, surface area, Cd) are based on the 2006 Lexus IS350 manual transmission. Also, I over estimated many of my values, such as the temperature and the Cd of the Lexus. If they were to conduct the experiment at lets say 30 degrees F, the time the car would take to reach the ground would increase by about 1 second. Also, lets say the Cd of the car was in fact close to one of a cube, the terminal velocity of the car would be an average of about 205 ft/sec. This would actually increase the time of the falling car by about 2.301 seconds from 20.395 seconds to 22.696 seconds.
As you can see, the demonstration done in the commercial is completely plausible as long as the Lexus on the ground starts from near its maximum speed.
However, if the Lexus did not start from near 140 mph, it would never be able to compete with gravity. Just to compare this car with the Ferrari 333 Sp. This car can accelerate from 0-60 in 3.8 seconds. The Lexus IS350, 5.6 seconds. Just for a fact, the Ferrari 333 SP is a modified race car, not stock. Based on this, and the fact that the Ferrari 333 SP can accelerate at an average of about 20 ft/sec over a quarter mile and gravity, 32.19 ft/sec demonstrates that the Ferrari 333 cannot conquer the feat of starting from a stop, let alone the Lexus IS350.
The Lexus IS Started near its maximum speed, and was therefore able to beat gravity, as the commercial demonstrated.
Any questions or comments, feel free to email me at lexus@alexnwolf.com
Thank You,
Alex Wolf
I would say that the IS350 and the G have about the same airodynamic drag falling top up.
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