How Can Steel Dragon 2000 be faster Than Millenniu
Great, now we have people saying that the steeper the drop the better. Superman: ROS at SFA doesn't have a steeper drop and IMO its a better ride than MF. Steepness doesn't make a great ride, the whole ride makes a great ride.
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Parks for 2000: CP, Great Escape, Great Adventure, SFA, Islands of Adventure, Kennywood
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Parks for 2000: CP, Great Escape, Great Adventure, SFA, Islands of Adventure, Kennywood
It doesn't matter how heavy the trains are. You could drop a car and a penny off the empire state building and they would hit at the same time.
Todd said:
"It doesn't matter how heavy the trains are. You could drop a car and a penny off the empire state building and they would hit at the same time."
Not true...
Under conditions where there would be no wind, the car and penny would hit at the same time. However, since the weight of a car is enough to overcome the friction produced by wind resistance, in almost any real-life situation, the car would always hit first. The wind would interfere with the gravitational force pulling the penny more than it would affect the car, so the car would hit first.
A great majority of physics laws only work under impossible conditions. Therefore engineers must always take things like friction into account when they make their calculations.
The weight of the car is a non-factor. If anything, the fact that it has more surface area, therefore causing more wind resistance and friction, the penny would likely hit the ground first.
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Jeff
Webmaster/Admin - CoasterBuzz.com
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Jeff
Webmaster/Admin - CoasterBuzz.com
Jeff is right in saying that the "weight" of the car is a non-factor. HOWEVER, the MASS of the car is a HUGE factor. You all are neglecting one of Newtons laws, The Law of Inertia (i.e. an object at rest tends to stay at rest, an oject in motion stays in motion....). Anyway, the force of wind resistance might be greater in magnitude on the larger car, but it will have little effect on slowing the car down. The wind force on the penny, though smaller in magnitude, will have a better chance of slowing the penny down. Thus, the car will hit first.
lata,
jeremy
--Who is suprised that with all these enthusiasts, no one seems to have tried the "Penny Trick" on Intamin 1st Gen Freefalls
lata,
jeremy
--Who is suprised that with all these enthusiasts, no one seems to have tried the "Penny Trick" on Intamin 1st Gen Freefalls
If a drop is long enough an object will eventually reach a terminal velocity. This speed is a function of the mass/drag ratio. A penny weighs about 3 grams while a car weighs 1,000,000-2,000,000 grams. That's about 500,000 time as much mass. The frontal area of a tumbling penny is about .5 sqin. The frontal area of a tumbling car is 5000 to 10000 sqin. Thats only about 15,000 times as much area, so we can expect the car to have a higher terminal velocity. (Without getting into drag coefficients, it's reasonable to assume that they aren't different enough for a penny and a car to change the result.)
in a vacuum all objects fall at a rate of 9.8 meters/sec Squared. That means every second of freefall you gain 9.8m. EX: 1st sec. 9.8meters of freefall, 2nd sec. 9.8+9.8m of freefall, 3rd sec. 9.8+9.8+9.8m a sec. and so on. Steepness has nothing to do with it. you still will get to the same speed. but there is a maximum velosity of around 120mph.
coaster_fanatic said:
"in a vacuum all objects fall at a rate of 9.8 meters/sec Squared. That means every second of freefall you gain 9.8m. EX: 1st sec. 9.8meters of freefall, 2nd sec. 9.8+9.8m of freefall, 3rd sec. 9.8+9.8+9.8m a sec. and so on. Steepness has nothing to do with it. you still will get to the same speed. but there is a maximum velosity of around 120mph."
This is true in free fall. If you throw an object to side as it falls in free fall then the net acceleration down ward will be the same. However if it is rolling down a ramp(track) then there will be friction applied. There will be less friction on a steeper slope because the object is applying less force to the ramp and more force to acceleration. The ramp then through Newton's third of motion provides the force that pushes the object to the side.
The things you learned in your high school physics class work under ideal conditions. In real world situations things start to get a little more interesting. There are a great many factors that influence terminal velocity in the real world. They include air resistance, wind, and friction.
CrystalKat said:
The speed of the lift will affect the ride because it is starting out from that initial velocity, rather than zero.
There is an effect, although not nearly as significant as many would suppose. An acceleration is a change in velocity *per unit time*. Objects which begin their descent at a faster initial velocity will traverse the distance of interest (here: the height of an initial drop on MF) in less time. As a result, the integrated acceleration during their descent is smaller. This has quite an equalizing effect when one considers the objects' velocities at the end of their descents.
Using simple finite differences, and using g=10 m/s, you can verify that for a frictionless free-fall of 90 m (about the height of MF):
initial v=0 m/s:
time to descend~4.2 s
final v~42 m/s
initial v=5 m/s:
time to descend~3.8 s
final v~43 m/s
initial v=10 m/s:
time to descend~3.4 s
final v~44 m/s
So, projecting an object downward at 10 m/s (far faster than MF's initial downward velocity) in a frictionless atmosphere gets the object through the 90 m depth more rapidly, but doesn't add much at all to the final velocity.
In a viscous atmosphere, drag on a falling object increases as the object's velocity increases (this is how an object asymptotically approaches its terminal fallspeed). Accordingly, an object which is introduced into its descent with a non-zero downward velocity actually experiences an even smaller net downward acceleration than an object which which begins its descent at rest. This effect further serves to minimize the differences between objects whose initial fallspeeds vary.
Although it isn't intuitive, the high speed lift of MF probably doesn't add much at all to its peak velocity at the bottom of the hill.
i'm going to say mf goes faster than they say it does. i got the chance to ride in the front seat and it sure does feel like it goes faster than 92 or 94 or whatever it is. but anyways who cares how fast it goes it is still fast and thrilling. that i what we ride coasters for it the thrill right?
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front car on mf. hope you have the chance to do it.
oh yeah and the car and penny thing. the penny wouldn't hit the ground because the wind blowing up the side of the building would blow the penny back up.
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front car on mf. hope you have the chance to do it. *** This post was edited by MFfan on 3/24/2001. ***
The trains are heavier, and SD2k is built at a higher elevation which means the air is thinner.
Since the air is thinner it means less resistance.
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Get wrapped in the coils of Viper at SFGAm. *** This post was edited by geicu on 3/24/2001. ***
Since the air is thinner it means less resistance.
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Get wrapped in the coils of Viper at SFGAm. *** This post was edited by geicu on 3/24/2001. ***
The weight of the trains has no effect on the speed during the drop. Two objects will fall at the same rate regardless of mass, with the exception of air resistance. If anything those bulky trains on SD2K will cause more resistance than on MF.
Well to put it simply the formula is:
ax³*yt³=ng³*¾*yr²*90°*234*318+310*yx³(876859)-2313³=Speed of drop
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TITAN IS GONNA ROCK TEXAS
ax³*yt³=ng³*¾*yr²*90°*234*318+310*yx³(876859)-2313³=Speed of drop
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TITAN IS GONNA ROCK TEXAS
force312 said:
"So what can we conclude from this thread?"
That we all have much too much time on our hands?
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Don't touch the watch.
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