Isnt construction meant to start this month?.... Anyone seen anything on lc12's island???

Vertical construction on big projects like this usually starts around August/ September time, what you get in the run up to that is ground works and initial infrastructure for the construction area.

I wonder when we will see footers............

Hang on, I thought European B&M coasters were manufactured at B&M's plant in Switzerland! Confused!!

You are partly correct Javs, yes some of the coaster parts are made in Switzerland (trains etc.), and I think they are designed there too, but the track is made in a B&M production facility in Ohio which as far as I know, only do the track.

Well, I learned something new there!!! I really thought that they only built the U.S facility to cut costs in shipping huge amounts of steel out from Europe! I find it incredible that they construct the steel across the pond and ship it over when they could produce it in Europe.Exciting to see the track for sure!!!!

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?

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?

Thanks for taking the time to do all that Mr Hicks!

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.

its gotta be faster than hat just use a freefall calculator and I'm NOT A TROLL

As our dear Mr Hicks has just ably demonstrated, there are just a few more factors to be considered here than acceleration in freefall...Hicks, I hereby applaud you! That was excellent :PEdit: You posted while I was writing, you ninja

Merged the "How fast will LC12 be" thingy topic with this one as there is no need for a separate topic, and it helps keep the place tidy

right................ anyway does anyone know where the supports are made or is that also in Ohio?

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.Insert all the maths calculationsWhich is about 55.7mph.Lovely. Can you stop trolling now?

You are really smart and that is actually a plausible speed because:LC12 - 126ft Nemesis Inferno - 95ft Nemesis's speed (which I'll assume) is it's top speed which would be the drop is 47.8 mph so LC12's being 8 mph faster at 21ft higher so 55.7 mph is very plausible. Look up facts Adamm before posting it'd called rcdb.com and looking at what I've just said it will be around that speed, if not them please PM me and I'll apologise if you last that long.

its gotta be faster than hat just use a freefall calculator and I'm NOT A TROLL

You are a troll Troll One who purposely and deliberately (that purpose usually being self-amusement) starts an argument in a manner which attacks others on a forum without in any way listening to the arguments proposed by his or her peers. He will spark of such an argument via the use of ad hominem attacks (I.e. 'you're nothing but a fanboy' is a popular phrase) with no substance or relevence to back them up as well as straw man arguments, which he uses to simply avoid addressing the essence of the issue.Yes my dear you fit that description.Now onto other discussions that troll's have not started. Has anyone thought that Les Coogan might be like a scientist that's been fired for his 'crackpot' theories such as the one's that we've heard about etc and LC12 will be the proof or the events happening?

Now onto other discussions that troll's have not started. Has anyone thought that Les Coogan might be like a scientist that's been fired for his 'crackpot' theories such as the one's that we've heard about etc and LC12 will be the proof or the events happening?

No. I doubt there is going to be a story. This is just the marketing, when the ride comes out, it will be forgotten. My question now Will this make up for it's own numbers? What's its throughput? And is it a step towards beating Alton ?

A coaster going at speed in the 50s, is one fast coaster.Considering raptor goes around the speed of air, and thats 45mph. Very interesting.Throughput wise, I'd imagine it'd be in the low thousand. Hopefully on par with towers throughputs, to the like of air.

There are more factors to consider than just using a suvat equation...Gotta take the weight of the train, air resistance and a few other bits and bobs as well when we're working out the actual speed... Constant acceleration is a decent enough assumption, but very rarely do rides do that...Conservation of Energy and using Potential and Kinetic Energy equations could well give us another answer...

Sorry to try and pick holes, but just something I thought about... :)Why have you used the initial velocity as 0? Surely there'd be 'some' velocity to start off with due to the lift hill? Unless I'm missing something obvious..

Sorry to try and pick holes, but just something I thought about... :)Why have you used the initial velocity as 0? Surely there'd be 'some' velocity to start off with due to the lift hill? Unless I'm missing something obvious..

This is arguable actually... It's a definite assumption, and is quite a regular occurance in such equations mainly as it helps simplify the equation and usually won't cause toooooo much of an effective difference...Using both an actual value of velocity for the lift hill and assuming it is zero are valid methods...

A coaster going at speed in the 50s, is one fast coaster.Considering raptor goes around the speed of air, and thats 45mph. Very interesting.Throughput wise, I'd imagine it'd be in the low thousand. Hopefully on par with towers throughputs, to the like of air.

Really? RCDB say otherwise .

Length: 2526' 3" Height: 108' 3" Inversions: 3 Speed: 55.9 mph Elements: CorkscrewZero-G RollIn-Line Twist

Although I'm not trying to argue with physics, I think it's to be faster than 55mph.

^Peter said LIKE air... Which is (according to rcdb) 46.6mph...Besides, it'll hit it's top speed for what? A handful of seconds? Like every other ride in the world? Does it matter that much?

There are more factors to consider than just using a suvat equation...Gotta take the weight of the train, air resistance and a few other bits and bobs as well when we're working out the actual speed... Constant acceleration is a decent enough assumption, but very rarely do rides do that...Conservation of Energy and using Potential and Kinetic Energy equations could well give us another answer...

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.

Sorry to try and pick holes, but just something I thought about... :PWhy have you used the initial velocity as 0? Surely there'd be 'some' velocity to start off with due to the lift hill? Unless I'm missing something obvious..

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

Was the mechanics textbook you ate tasty?

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

Oh UVAST <3