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ZC2009

THE SWARM: Speculation & Theories

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I think it would be really cool if at the start there was a loud explosion and it almost launched you up the lift hill and over into a HUGE drop, like your falling and the recover you into lots of twists! Hope they make the ride go underground at some point, they don't have that at Thorpe! Would love to see it! Can't wait for LC12!

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I think it would be really cool if at the start there was a loud explosion and it almost launched you up the lift hill and over into a HUGE drop, like your falling and the recover you into lots of twists! Hope they make the ride go underground at some point, they don't have that at Thorpe! Would love to see it! Can't wait for LC12!

I take it you haven't looked at the plans then? :(PlansIt does not go underground, but does go under parts of the themeing and over water. It does not look like it will launch in any way, as the plans suggest a basic lift hill like Raptor.

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The twitter account seems genuine, and so does all the advertising on park. The Les Coogan Facebook account just doesn't seem real to me though. It's still very interesting however :blush:

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Well I think the marketing for LC12 is going quite well. I have officially been accepted by Les Coogan to be a part of his 'Crack Squad' there are about 17 of us in it I think, possibly something to do with 1st riders when its finished or something, we shall have to see :lol:

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That posters that staff have taken down have the twitter account on them. Thus making the twitter account not fake?

But, if they are taking them down, it means the twitter acount on them is fake? Hence them taking them down?Your probably right, I just thought it was wierd they havn't got it on their official website

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Taking them down, due to end of that marketing time I think.They were there over a month.. if they weren't to do with thorpe, they would've been removed that day or very shortly after... not even thorpe staff are that bad at removing huge things like those posters :lol:

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There's bound to be several stages of the marketing... The posters were just the start of further (and possibly more in your face) schemes...Just wait for the ghosts/priests/burial grounds to appear...

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The LC12.net is legit as the address is shown on the sandwich board of the LC12 actor. (Who is legit, otherwise they'd all have been banned from Thorpe etc. etc.)

I'm also in Crack Squad Alpha, so if you're going to give Echo a like, why not Alpha ^_^ Spread the word, remember, together we can uncover the truth.Facebook page for Crack Squad AlphaThanks :D

Alpha FTW!! :D

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K. So realising how stupidly complicated (and therefore, wrong) my last calculations were, I presented the problem of speed estimation to my Maths/Physics teacher. He showed me this formula, because SUVAT equations are rubbish:gPE = kEWhereby gPE is Gravitational Potential Energy and kE is Kinetic Energy (of the moving train). The Law of the Conservation of Energy states that energy can neither be created nor destroyed. gPE is a stored energy, it is released when an object starts to fall and is converted to kE. Hence the train's gPE at the top of the lift = it's kE at the bottom of the drop. Taking this into account, I suggested that it'd be easier to work out the speed of a train by using the following formula, which is equivalent to the one aforementioned, only it allows us to consider the mass of the train (but in some ways it doesn't, because we don't really need to know the mass of the train) and it also include final velocity (at the bottom of the drop) which ideally is want we wish to know:mgh = 1/2mv2Whereby m is the mass of the train in kilograms, g is the gravitational constant, h is the height from the ground to peak height in metres, and v is the speed of the train at the fastest point on the ride (which we'll consider to be the bottom of the drop, for obvious reasons). Thankfully, having m on both sides of the equation allow us to simply ignore it, because both quantities cancel each other out, so this is the revised formula:gh = 1/2v2We now need to substitute, using the quantities we already know:g = 9.81ms-2 (gravitational constant)h = 38.4m (126ft)Which is about all we need to know really, because, substituting these values into the equation gives us this: 9.81ms-2 x 38.4m = 1/2v2And now, with the only unknowns appearing on just one side of the equation, we can calculate it by: 377J = 1/2v2, because 9.81ms-2 x 38.4m = 377J, we then need to double this quantity to remove the 1/2, and then find it's square root to remove the 2, which will give us the velocity and the bottom of the drop, which is done here:377J x 2 = 754Sq. rt.(754) = 27.5ms-1And 27.5ms-1 in miles per hour is 61.5mph. These calculation do not take into account air speed/velocity, friction or the elasticity of the materials used. Therefore, these calculations are not entirely accurate, but are reflective of a "perfect world" situation.All quantities are drawn to three significant figures.So I say we can expect LC12 to have a top speed of around... possibly 58-59mph. Quite possibly.

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