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6.5 seconds of weightlessness?


Gazza
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One thing i have been meaning to post about for a little while is that 6.5 secs of weightlessness claim about the TOT. I guess nobody has really questioned it, but i dont actually think its possibe. Firstly, after watching one of the videos and watching the time from the moment the pod starts falling back down to the moment it reaches the beginning of the launch section (in other words the bottom of the slope to vertical) since this is the time that you are in freefall and hence would be weightless. i got a time of 5 secs, which is below the 6.5 figure stated. Of course the actual time spent weightless would be less than this as on the slope to vertical you would have the G-forces pushing you down into your seat, but just to prove my point i included all of it. Now secondly, if you were to drop an object from a height of 115m (The absolute highest the ride could reach) then: x=ut+1/2at^2 where x=-115, a=-9.8 (value of g) u=0 (since it is starting from rest) and t=? then -115=0t+1/2(-9.8)t^2 -115=-4.9t^2 23.47=t^2 t=4.84 secs Now this is negating friction, but i doubt it would be much for such a heavy pod, but anyway, even if there was friction it wouldnt be able to accelerate towards the ground at g so you wouldnt get true weightlessness anyway. Thoughts?

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Simple answer. You've only calculated the free-fall component, which makes up only half of the weightlessness. Riders experience weightlessness from the moment they hit vertical on the way up because at from this point you are accelerating (negatively) at the same rate as gravity (as there is no upward acceleration from the ride), until such a point Throw a balloon full of water into the sky. It'll become a jelly blob (i.e. weightless) from the moment it leaves your hands until when it returns, not just on the way down. Remember that the height of 115m includes the curve to vertical, during which you are very much being affected by gravity. The height of this curve is roughly 45m, so the length of the vertical track is only 70m. Plug this into the equation and you'll end up with a result of 3.77s x 2 = 7.55s total weightlessness. Factor in track friction, air resistance and the fact that the 45m figure isn't 100% accurate and you'll find the 6.5 seconds.

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  • 1 month later...

It's not maths, it's magic. It does seem kind of strange why you would feel weightlessness going up, and it did confuse me at first (I even considered the possibility that Richard was wrong :o). But alas, he's correct. Weightlessness by definition occurs when a body is accelerating at the same speed as gravity (making the object have no apparent weight, weight being mass by gravity) and this is indeed the case. It's decelerating at the same rate throughout the entire vertical part.

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  • 2 weeks later...

Its a pretty routine calculation though, the sort of thing you can drum out in less than a minute. I guess I was just interested in working out what the go was (which has been done :)) Welcome aboard though, its always good when we get some park employees joining in the conversation. Better get used to people talking about the minutae of theme parks and rides though :P

Edited by Gazza
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