Sidders

Yes, and it's that attention to detail on the word that makes me wonder whether it is in fact a legitimate logo and therefore the ride's official name. I hope not because surely, there's better names out there?That was not an invitation to start guessing at names.

I'm very intrigued by the looks of this. At first there were theories that this would be something linked to the Mayans but it seems that theory is dead and buried now and instead we've got some kind of man-made massacre bringing about the disintegration of "the world as we know it" . I get a Twenty-Eight Days Later meets The Crazies theme... which I'm not too disappointed about it must be said. Interesting to note the emphasis of "WAR is coming" in such heavily-stylised font. There's already some commenters asking whether that's the name of the ride. Odds are it's not as I doubt the name would be released this soon. And also, the date in the video is August 1st, which is one week away, and we've known from the start that this is just a day when something big is meant to happen with regards to the ride; it's not exactly going to be ready for the public by then.

I appear to have killed this thread. HURRAH.

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.

Dayum! Necks will be broken at 1:10! That kind of jolt can't be safe at that speed?And that coaster sounds awful!

You know we're joking, right? T'internet fails at sarcasm.

Judging by the "flaps" and the angle the supports are designed to be bolted into place beneath the track, I'd say yes.

No. I got a D in my AS January exam. -BIG SAD FACE.

There's no doubt about it: there's definitely no way of calculating the exact speed of the train because, as you point out, the train will not be accelerating uniformly as in, moving a certain distance (in metres) over each period of time (seconds). So the only way we can estimate the speed is to assume that it is accelerating uniformly about a fixed origin (also doubtable). There's no need for assumption in free-fall because all objects in free-fall undergo gravitational acceleration, which is equal to 9.81ms-2, which therefore means that mass of the object is negligible and has no effect on the final velocity whatsoever. Remember, speed and velocity are not affect by the mass of an object, and the only thing that sepearate speed and velocity is that the latter has a given direction, which was then put into a = vω / r (not a suvat equation) to accommodate the concept of circular motion about a fixed origin (again, if we assume the train remains a perfect 19.3m from the origin at all points during the drop).Remember the famous "If I dropped a cannon ball and a strawberry from x height, which would hit the ground first?" scenario? Well, they both hit the ground at the exact same time, as both undergo the same acceleration. The common misconception is that the cannon ball would land first, and whilst this is not the case, the cannon ball would reach terminal velocity first, where terminal velocity is when the decelarative force on the object is equal to the accelerative force (uniform acceleration) and this is directly due to the larger mass (which means that terminal velocity is easier to reach as it is a lower quantity on larger objects) but the two objects still hit the ground at the same time.The suvat equations, as with most mechanics equations, are based on the assumption that wind speed/velocity, friction and other extraneous forces are negligible. The only equations in which the initial velocity = 0 are the suvat ones where linear motion (free-fall) is calculated. Using the concept of free-fall, the moment you drop a body of mass from x height, it starts it's descent with "zero initial velocity" and will then continue to accelerate downwards due to the force dissipated by gravity. And because in the first equation I was calculating what the velocity would be if the train was in free-fall, then substituting it into a circular motion equation a = vω / r, we then get the speed over a chord of 60.6m, about an origin 19.3m from the train..Naturally, there is some momentum from the train being booted off the lift hill, but that's horizontal motion, not vertical (free-fall). Horizontal motion does not affect vertical motion, and vice versa, hence why when you drop a package from a plane, it will continue to travel at a constant horizontal velocity equal to the velocity of the plane until it hits the ground. Horizontal motion is calculated by s =vt and is a much easier concept that vertical motion as gravity is not taken into account, but it does mean the speed is a constant, and that, if there were no extraneous forces, an object provided with the a force will continue to travel in a perfectly straight line, and never stop. Hence why, in space, if you dissipate a force on an object, that object will travel in a straight line "to infinity" as there is no air resistance or friction stopping it.I hope this helps

My pleasure, Josh :)Obviously, there are some omitted data that couldn't possibly be calculated like wind speed/velocity and friction at any given moment, so this would be in a perfect world where extraneous forces were negligible. I'd say it'll definitely be within the 54-55mph range, with more of a possibility of that estimate increasing than decreasing when we find the true value. :PAnd Adamm, you cannot use a free-fall calculator because, believe it or not, the ride is not going to free-fall directly to the Earth in a perfectly linear motion.

It's not faster than that. If you have a problem with Physics then I don't quite know what to say...You could write to Thorpe in your neatest hand-writing and ask if they could strap a rocket to the back of the train to make it go faster down the drop and make it a world's first?

Well, to answer the troll's question in, which is one of many other pointless topics currently spamming up the Quick Questions forum: the only way we could tell the speed, in miles per hour, was if we knew the peak height in metres (38.6m), the circular acceleration (or centripetal force), the angle of descent (no information as of yet but read on and I'll calculate it for you), and the period, in seconds, taken for the train to move over the chord (chord = section of a circumference, e.g. chord = drop section of the track). It's not impossible to calculate, but it's rather challenging, is all I'm saying. And it's also a common misconception that you'll need to know the weight/mass of the train, which has no effect whatsoever if the object is uniformly accelerating.Here's the Physics Speak:From ground level to peak height, the highest point of LC12 is 126ft, or 38.6m, so if we consider how fast the train would fall if it were simply dropped from that height, we can then calculate the acceleration about the origin, and then ultimately, the velocity about this origin. To calculate the velocity of the train if it were to be dropped and left to undergo uniform gravitational acceleration (which is a constant at 9.81ms-2), we use v2 = u2 + 2as, whereby:v = v (unknown quantity)u = 0ms-1 (initial velocity)a = 9.81ms-2 (uniform gravitational acceleration)s = 38.6m (displacement)So let's substitute:v2 = u2 + 2as IS THE SAME AS v2 = 0ms-1 + (2 x 9.81ms-2 x 38.6m) IS THE SAME AS v2 = 757.3. So, if this equals v2, then v is the square root of 757.3, yes? Which means v = 27.5ms-1.And knowing this, we can now calculate the circular acceleration of the train as it moves about the origin of the "circle" - which is always 19.3m from the train at any given time in the cycle (because r = d / 2 = 19.3m) - using this formula: a = vω / r, whereby:a = a (unknown quantity)v = 27.5ms-1 (velocity)ω = 3.3 (constant)r = 19.3m (radius)Now let's substitute again:a = vω / r IS THE SAME AS a = 91.8 / 19.3m IS THE SAME AS a = 4.7ms-2. So if we consider a = 4.7ms-2 to be the train's uniform acceleration about the origin of the "circle". Let's re-use the first formula to find the velocity over the chord, whilst using the uniform velocity as this new-found quantity.But first, we need to consider the drop section of the track to be exactly half of a perfect circle for the purposes of the following. From ground level to peak height, the highest point of LC12 is 38.6m, as we've already discussed, right? So this would be the diameter of the "circle". Well, using the equation Pi x d (where d is diameter in metres), we can calculate the circumference of this virtual circle. So logically, half of the calculated circumference would be the length of the track that makes up the drop (the chord of the circle), would it not? So:Pi x d IS THE SAME AS 3.14 x 38.6, which equals 121.2m. Divide by 2 to get the length the chord and we get 60.6m.So, the overall displacement of the train as it travels from A (peak) to B (trough) is 60.6m. And here's the equation again: v2 = u2 + 2as, whereby:v = v (unknown quantity)u = 0ms-1 (initial velocity)a = 4.7ms-2 (uniform acceleration of train)s = 60.6m (displacement)ONE MORE TIME WITH FEELING v2 = u2 + 2as IS THE SAME AS v2 = 0 + (2 x 4.7ms-2 x 60.6m) IS THE SAME AS v2 = 576.1. And if we square root it, we get 24.0ms-1.Which is about 55.7mph.Lovely. Can you stop trolling now?

This honestly looks like someone with a few missing brain cells painted on the word "the" twice, then painted over it and left it.Loving the viral tweets and the mini-site. YES THORPE WELL DONE YOU CAN DO PROPER PROMO. *applauds*.

This comment clearly states that you intended to scrap Storm Surge the year it was built, 2009/2010. Or do you mean you wanted to scrap the plans for it to be built in the first place, before any construction? For the record, I love that idea.

Yes, yes, I'm sure Thorpe would remove a ride they built earlier the very same year.

Let's just hope they do something more innovative than what they did with the un-inspired layout that turned out to be Colossus.I'm for an Intamin Mega-Lite or B&M 4D prototype similar to the concept at Gardaland, considering the chance of a Woodie is well and truly out the window and in the next street by now.

Welcome magic mill! To Thorpe Park Mania... we're quite nice people to be honest.And I think I speak on behalf of the majority of the forum that we agree with you said about Thorpe needing Air-Time. A piddly little hill on Colossus and SAW's 'Air-Time' hill are all we have. I won't even include Stealth, you get more Air-Time on a see-saw.

Granted, but you'll always teleport to your school.I wish my ass wasn't numb right now (plenty of opportunies for double entendres here...)

Oh ma dayyyys, this is a tough one....I pick Goldfrapp :DKe$ha or 3OH!3? (oooh difficult )

Roar, let's add some man-ness to proceedings, shall we? :L:LProfessor Green OR Lily Allen

Only across Europe, New Zealand and Australia, so it wasn't an international single as it never got a UK, Irish, American or Asian release. Just a regional release to selected countries.Besides, cover art isn't proof! Just look at 'Alejandro' and 'Dance In The Dark', they're covers were released back in 2009 following promotion on iTunes to get people to buy the album. But of course, then 'Alejandro' got officially released internationally earlier this month.For those of you who haven't seem the AMAZING 'Dance In The Dark' cover... here it is, in all it's GaGa glory:*splat*

FREE TO LOVE YES YOU'RE FREE TO LOVE AGAIN BOOM BOOM BOOM FREE TO LOVE YES YOU'RE FREE TO LOVE AGAIN AUTO-TUNE BOOM BOOM BOOM :DOn a more serious note, I second that notion, Mark.