wonder what TTD's km/min rate is.....
A day is a drop of water in the ocean of eternity. A week is seven drops.
So 23 and a half seconds per km? Does that sound right?
*** This post was edited by janfrederick 4/8/2003 3:31:22 PM ***
basically....if TTD rose the whole 420' in 2-3 seconds it would be 3.84-2.56 km/min
...a little faster than 1km/min huh?
*** This post was edited by Dale Picolet 4/8/2003 3:37:42 PM ***
UK, home to the King of Inverts, Nemesis, and the soon Heir, Inferno
Hey, do you think they'll have seatbelts on the elevator?
Coaster Art Guy said:
Did someone forget about the deceleration factor?
Not really....my math figures an average rate of climb based on a given time. The velocity at any one point isn't considered...just the rate at which the entire climb is made. In other words....at it's fastest velocity it is above 3.84km/min...and at it's slowest it is below 2.56km/min. 120mph directly corresponds to 3.22 km/min. Which should tell us that there is no way TTD will hit its peak quicker than 2 seconds as that calculates to a rate of 3.84km/min. Let's say TTD averaged 1km/min on it's climb...it would then top out at 7.7 seconds. These times of course do not take into account the launch track parallel to the ground, the height of the track at the base of the tower, nor the physics of the curve to vertical. It simply assumes TTD acting like an elevator.
*** This post was edited by Dale Picolet 4/8/2003 4:23:09 PM ***
Let's assume for a moment that friction and air resistance are arbitrary forces. We are then dealing with an object simply being launched upward. The total distance it must cover is 420 feet, which we have astutely determined is approximately 128 m. We know that it goes at 120 mph, which again, through dimensional analysis...
120 mi 1 hr 5280 ft 1 m
1 hr 3600 s 1 mi 3.281 ft
...comes out to (120x5280) / (3600x3.281) = 633,600 / 11811.6 = 53.64 m/s.
So what we know is that the distance is 128 m, the initial velocity is 53.64 m/s, and g we accept is 9.8 m/s/s. At this point, the problem becomes simple kinematics:
x = vt + (1/2)gt^2 or (1/2)gt^2 + vt - x = 0
In polynomial form, the equation can be very easily solved with the good old TI-83 Graphing Calculator. So first, we substitute in all known values and clean up the equation:
(4.9 m/s/s) t^2 + (53.64 m/s) t - 128 m = 0
Since we know that each unit will inevitably cancel out, and that t is in seconds, units are arbitrary:
4.9 t^2 + 53.64 t - 128 = 0
In polynomial form, this equation can be easily solved using the trusty TI-83 Graphing Calculator. The function's zeroes are at t = -12.96 and t = 2.015. Logic dictates that time cannot possibly be negative, and therefore we see that, in a perfect friction-free situation, a projectile being accelerated upward 420 feet at 120 mph will reach the top of its arc in 2.015 seconds.
Now, we have considered throughout this equation that there is no air resistance and no friction, which is completely false. The truth is, though, that the values are only consequential to real physicists, which we certainly are not. Therefore, let's attribute approximately one to two seconds (which is a gross overestimate to begin with) based on the car's friction with the track, the resistance of air, and the fact that, in a non-vacuum situation, the car's mass affects its acceleration.
Therefore, we can conclude that TTD's train will reach the apex of its 420-foot structure somewhere between 3 and 4 seconds after it begins its vertical ascent.
B for the year, my *ss...eat it, Mrs. Yates! :-P
[Nitro Dave -- 135 Laps] [Track Record: 64 and counting...]
"How can you see into my eyes like open doors leading you down into my core, where I've become so numb? Without a soul, my spirit sleeping somewhere cold, until you find it there and lead it back home."
Krimson n' Kream, Spr 98
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