Not as fast as a lexus falling from 4000 ft.Originally Posted by Rockefella
One needs also to figure the terminal velocity of the car, but I still doubt it could make it.
Yes, in that scenario a Lexus IS350 can do that
No, Impossible
Not as fast as a lexus falling from 4000 ft.Originally Posted by Rockefella
One needs also to figure the terminal velocity of the car, but I still doubt it could make it.
TOYNBEE IDEA IN KUBRICK 2001 RESURRECT DEAD ON PLANET JUPITER
I forgot how to calculate that, my physics memory is pretty bad right now.Originally Posted by Esperante
Rockefella says:
pat's sister is hawt
David Fiset says:
so is mine
David Fiset says:
do want
anybody realise that we already had a thread on this very commercial?
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Yeah, I know.Originally Posted by cmcpokey
REPOST!
Rockefella says:
pat's sister is hawt
David Fiset says:
so is mine
David Fiset says:
do want
Sure, but that one just posed the scenario, this one has a link for the Ad...
I know, the search feature is our friend
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Thread merging in progress.Originally Posted by PerfAdv
Rockefella says:
pat's sister is hawt
David Fiset says:
so is mine
David Fiset says:
do want
x = ut + 1/2*a*t^2
gives a drop time ~16 s for 4000ft
even including wind resistance, no way could a car on the ground match that
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Sex is what they want. - Frasier
Actually it's much quicker than I thought. There was a recent test in R&T or Motor Trend - don't remember which. They got a 0-60 time of 4.9 seconds. I was shocked.Originally Posted by VtecMini
I'm going to eat breakfast. And then I'm going to change the world.
I just figured it out, it's physically impossible not because of aerodynamics, but because of the friction forces:
F=ma
Friction is what propels the car:
Force of friction: (Coefficient of static friction)*mass*9.8
Theoretically the only way the coefficient of static friction can be > or = 1 is if the two objects are fused together. Even if the car can manage to create the torque required to accelerate quicker than 9.8m/s/s, the tires can't get the grip to do so.
wrong, tires can have coefficient of friction >1. Because tire works on both bonding on molecular level as well as mechanical friction.....Originally Posted by Sweeney921
the thread is irrelavent as the ground vehicle gets a jump anyway, the timing between the start and the drop can be timed...
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Anyone know how to find the time it took for the falling car to reach the ground?
http://www.ultimatecarpage.com/forum...2&postcount=37Originally Posted by sydney427
How can men use sex to get what they want?
Sex is what they want. - Frasier
it coudnt be a standing start, and isnt that the lexus with the electric motor thing, it might be possible but i seriously doubt it was a standing start
Once fanboyism infects you it impares all your judgement.
It's like being drunk, you lack common sense and everyone laughs at you.
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 fallign 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 fatser. 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 Coefficeint
r=atmospheric density
a=area exposed to air(area facing down)
Furthermore, we find terminal velocity by detwermining 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 botom 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 resistence 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(fallign 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 resistence 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 infact 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 Faerrari 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
well cheers. pretty much settles that, ehh?
so what is your affiliation with lexus? seems this had already been thought out before coming here.
oh yeah, welcome to UCP.
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