Height/Speed input

Reverse freefall at 6Flags Magic Mountain(415' tall). Oh, my Browns....

We can all speculate or do the math on how much speed it takes to reach 400 feet. However without knowing if the top of the tophat will be invertedat all or not. theres no way to tell how fast it needs to go. You do have to figure that Cedar Point and Intamin will make sure the train is still traveling at a relatively good speed just in case seeing that in 400 feet of rise alot of factors such as friction due to the temperature of the track, wind, and other things. you dont want a train falling 400+ feet back into a station. so it obviously will need to go faster than it takes to get there. Also if the tophat is inverted it will need a tremendous amount of speed in order to make riders feel comfortable while inverted. I doubt that Cedar point would cause people to be discouraged from riding because of their fears if it should hang at a slow speed while inverted. They would not cut down on the rider base. I Also feel that the L:TH:B theory is not likely either. Cedar Point is not Six Flags and will make sure this ride has some dramatic element to it other than a 400 ft tophat. even if its a bunny hill or two. I will be going to CP on Saturday and will take my Digital Camera with me so i will be back with a link to some new pictures sometime next week.

"I feel the need......The need, for speed!"-Top Gun

S: TE = Superman: The Escape @ Six Flags Magic Mountain

Majin and tservo - To really put the exclamation point on this concept; if I were to shoot a gun and I knew nothing would ever be in the path of the bullet while pointing it exactly parallel to the ground and at the same time at the same height I dropped a ball straight down, the bullet would hit the ground exactly at the same time as the ball. :)

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cyberdman

SteveHockey4 said:
However without knowing if the top of the tophat will be invertedat all or not. theres no way to tell how fast it needs to go.

Why? Did someone change the way gravity works when the train is upside down?

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Jeff
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Jeff - Webmaster/Admin - CoasterBuzz.com
"There's nowhere to run, nowhere to hide, when it's all in your mind. You gotta let go." - Ghetto, Supreme Beings of Leisure

I Said:Also if the tophat is inverted it will need a tremendous amount of speed in order to make riders feel comfortable while inverted. I doubt that Cedar point would cause people to be discouraged from riding because of their fears if it should hang at a slow speed while inverted. They would not cut down on the rider base.

Sorry i dont mean to be mean , but i just wanted to clear this up. I didnt say that it would need more speed to make it over the top.

"I feel the need......The need, for speed!"-Top Gun

The funny thing is that Mr. Freeze is slow paced in the inverted top hat. If parks put too much rider confort issues in place all the coasters in the world would be a bore.

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Chris Knight
Why does CP only have one Bird of Prey and a whole bunch of Sea Gulls?

From the looks of the giant crane today (Tuesday 10-08 1 pm), the halfway point on the crane seems to be a little higher than the tower is now. The tower is estimated at about 175 feet up, so the halfway mark of the crane must be around 200 feet up. The top half of the crane doesn't seem longer than the bottom half, so at the most the crane could only be around 375 feet tall...any other thoughts on this observation??? Jump all over me...

Hi folks, I'm a new user here - just some thoughts on the height/speed question. Please excuse the metric figures, I will convert them back to imperial.

Simple Newtonian laws of motion show some interesting results.

To work out how high an object will travel when launched straight up, we need to solve the equation

s = (u * t) + (0.5 * a * t * t)

where

u is the initial velocity

a is acceleration due to gravity

t is the elapsed time

s is the height the object will travel

a is easy: acceleration due to gravity is near enough a = -10m/s/s.

Assuming this rumoured initial launch speed of 110mph is accurate, we know u as well. 110mph is near enough 183km/h = 183000 meters per hour = 3050 meters per minute = 50.8m/s.

Lets just call this 50m/s to make the math easy.

We know that the intial launch speed is 50m/s, and that acceleration due to gravity would be -10m/s/s.

So the length of time that is required to accelerate from 50m/s (at the launch) to 0m/s (the speed at the highest point which the train could possibly get to) is 50/10 which is 5. So we know that t = 5.

We can now calculate the theoretical height which can be achieved from a 110mph launch

s = (u * t) + (0.5 * a * t * t)

s = (50 * 5) + (0.5 * -10 * 5 * 5)

s = ( 250) + ( -125)

s = 250 - 125

s = 125m

125m is near enough 410ft. (actually, if we used 50.8m/s for the launch speed, we get a height of 129m, which is about 423ft)

This calculation ignores the effects of friction from the rails, and drag induced by a 110mph wind going past the train. And it also assumes that the train is launched straight up the hill - there will be considerably losses in velocity as the train goes through the curve from horizontal to vertical.

Taking these effects into account, I would guess that realistically to make the top-hat, the probable height difference between the launch station and the top-hat would be around 350ft or so, not much more.

Alternatively, the launch speed would have to be higher than 110mph....which would be pretty cool!

Later....
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Steve

Found the Point in 2002 and love it!
*** This post was edited by Panman 10/8/2002 6:45:01 PM ***

You forgot the fact that the launch is inclined and will probably end at about a height or 50-60 feet off the ground.

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June 28th: LocoBazooka Tour (Sevendust headlining)
July 11th: Korn, Puddle of Mudd, and Deadsy

It actually doesn't (significantly) matter what angle the launch is - at 110mph the difference between the end of the launch and the top-hat can't exceed 410ft. On re-reading my previous post I realise I didn't make this clear.

Remember, these calculations assume there are no losses in velocity due to friction etc. An inclined launch will result in a slightly higher ride - the difference in attainable height between a flat launch and an inclined launch will be due to whatever losses in velocity are incurred as the train changes angle from flat to the same angle as the inclined launch.

As an aside, it's also worth noting that these figures apply to the center of gravity of the train. In theory, each car in the train could go higher than the 410ft height provided that the c.o.g of the the train as a whole passes below 410ft - much like a high-jumper can clear a 6ft bar, although his/her c.o.g actually passes well under the bar. If the train were long enough, and the top-hat shaped correctly, the height of the ride could be significantly extended without the train having to go faster to clear the top-hat.

If the end of the launch is 50-60 ft above ground, then clearing 400ft or more at the top of the ride is easy. But the absolute limit ( for the c.o.g. of the train) will be 460 to 470 ft.

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Steve

Found the Point in 2002 and love it!

The fact that the ascent will start 50-60 feet above ground is key. The incline of the launch doesn't matter, since it is the velocity at the end of the launch which is used for the initial velocity.

Also, the train will need some speed to get over the crest of the top hat, so the final velocity shouldn't quite be 0. I just use the equation y = (V^2 - Vo^2) / 2g, where y is the height, V is the final velocity, Vo is the initial velocity, and g is the acceleration do to the force of gravity (9.8 m/s/s). Taking the 50-60 feet of initial height and extra speed needed at the top, for a 420 foot top hat, I calculated around a 115 mph launch, give or take a few mph.

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AP Physics is awesome.
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*** This post was edited by Majin Heero 10/8/2002 9:47:44 PM ***
*** This post was edited by Majin Heero 10/8/2002 9:51:18 PM ***

Panman said:

As an aside, it's also worth noting that these figures apply to the center of gravity of the train. In theory, each car in the train could go higher than the 410ft height provided that the c.o.g of the the train as a whole passes below 410ft

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Steve

I've been wanting to say that for weeks, but you said it quite eloquently. It's nice to know that more than one 'phijucks' geek reads the forum.

As far as the 'too slow for an inverted tophat theory goes' frankly, I doubt the public cares how fast a train travels through a tophat, loop, or anything else. There are plenty of restraint systems that hang riders upside down safely with no problem. While a coaster can obviously keep the Gs positive if the designers want it that way, there isn't the *need* anymore. Check out screamscape's last update if you need any proof of that!!

-albert

The limiting factor here is the curve they have already constructed, as you physics geeks already know. I don't think Cedar Point will risk going much higher than 4g's. Having said that...

Assuming a maximum of 4.5g's on the bottom radius, a bottom radius of 150 ft (I think this is a bit large), a top hat radius of 30ft, a maximum of -0.5g's over the top hat, along with a 3% total energy loss due to frictional effects I figure the following:

max speed: 100 mph

max drop height: 320ft

total height (from the ground): 350ft

Of course it's nothing but a guess and it means little but isn't physics fun?!

(On a side note, a 400ft drop with the same conditions would require a bottom radius of 186 ft. Is this a possibility?)

*** This post was edited by mdeng 10/9/2002 8:41:19 PM ***

I'm confused...;)

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That wasn't my intention, sorry. :)

The radius isn't circular, at least I dont think so, mdeng. I really don't have a concept of what the curvature of the pullup (or out) is at the base, but the numbers seem about right, if a smidge low. I'd be surprised if the highest point wasn't higher than a certain other ride in California-- but I would say the pure physics don't quite justify being sure of that yet-- just a hunch -- but a good scientist doesn't just look at numbers;)

-albert

Yeah I realize the radius isn't circular, it was just an approximation because I don't have the equation yet for the radius of curvature, lol. If Jeff would just get a perfect side view of the track and some dimenstions to go with it then maybe....

I know everyone (including me) wants/thinks Intamin is going build higher than Superman but I just can't see that yet. Patience is a great virtue, I wish I had it.

Well, to find the radius of a pullout for something coming from vertical, take the starting height and subtract the ending height.

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If you meet me and forget me, you've lost nothing. If you meet Jesus and forget Him, you've lost everything.

It isn't a quarter circle. I'm pretty sure the curve is more complex than that-- more gradual at the bottom and tighter as it approaches vertical and the train is going slower.

-albert