Speed of Ride

I think this ride will be at least 120 mph if the ride is going to be 420 ft up. Because the supports of the launch are to low for you to only go 110 mph and make it up the tower.

I'd bet somewhere in the 80's or 90's. We're not ready for over 100 mph...

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"If we go any faster, she'll blow apart for sure!"

I've always though 117mph sounded like a good number. Yes...let's use that one. I mean...do we really know?

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James Draeger
-Captain Sarcasm

80 miles per hour barely gets Hypersonic over it's 165 foot tower...

420 - 165 = 255 feet

Methinks it will be faster than 80.

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Joshua Wilcox

yeah way faster than 80/90. im sure we're all ready for something over that. if you have ridden MF before you are very ready for something to go a "tad" faster

Duh, yeah I've ridden MF before...several times. But I don't think the GP is ready for over 100 mph. To do that would just feed the trolls in the news media who say coasters kill people.

"2003" will come in at less than 100 mph and less than the infamous 400' mark. Be happy with it.

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"If we go any faster, she'll blow apart for sure!"

Before you eat me alive, the last post was just my GUESS.

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"If we go any faster, she'll blow apart for sure!"

I think the lines for MF speak volumes about what is 'too much' for the public-- MF isn't scaring enough people away in my book;) The Crystal Beach Cyclone attracted gawkers, even if the ridership was low- and all that really matters to CP is gates, not number of riders.

-albert

why would CP want to build a ride that nobody would want to ride??

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Guess who's back....back again ;)

After riding MF over 15 times, I want more height and speed. I think 110 mph, 400 ft would be the perfect ride to step up the coaster records at CP.

Assuming no friction, no air resistance, and a launch at zero feet, in order to go 420 feet up you would need to be going 112.05 mph. How did I get this? A simple conservation of energy problem. When you launch, you have all your energy as kinetic. When you move vertical, it all changes to potential. Keep in mind this launch speed will give you no speed at all on the top, so it will have to be higher to go over the tophat.

KE = PE

1/2mv^2=mgh

m's cancel, h= 420' = 128.016m, g = 9.8m/s^2

(1/2)v^2 = (9.8)(128.016)

v = 50.091 m/s = 112.05 mph

So, in order to even get up to 420 feet, from the ground, you'd have to be going at least 112 mph. Hope this helps!

--James
*** This post was edited by jdoty 10/13/2002 8:13:45 PM ***

Good, so if you started up even just 10 ft above ground, you would need to go about 112 mph.

In theory. Remember, that's assuming no friction and no air resistance :)

--James

Oh Yeah! So it would probably go about 120 mph to get over tophat at a good speed.

That's also assuming it's over 400' tall. I dare to say it will not be. :P

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"If we go any faster, she'll blow apart for sure!"
*** This post was edited by MrScott 10/13/2002 8:28:26 PM ***

Come on you'll....The general public is ready for whatever a park puts out there....Superman was advertised 100mph and over 400 feet high..And yes people thought it was/is a roller coaster....Lines were rediculous.. The lines for S:TE were at times 6 hours long...So, if you don't think the general public is ready for 400 plus feet high and 100 plus mph than you are madly mistaking...

That's right, why wouldn't CP be the first to go over 400 ft. I don't think they would build a ride that is under 400 and is taller than MF. Think about it!
*** This post was edited by broadhurst4 10/13/2002 8:45:24 PM ***

i was wondering what the speed of the ride would be when it hits the bottom of the down side so i used the equation

v^2 = Vo^2 + 2a(Yo-Y)

where v is final velocity, Vo is initial velocity , a is acceleration, Yo is initial height, and Y is final height. To make things easy i figured out what would happen if an object had a free-fall from a height of 420ft (128m) to 0ft, with a velocity of 0m/s at the top.

Therefore:

v^2 = 0^2 + 2(-9.8m/s)(0-128)

= (2)(9.8)(128)

v^2 = 2508 m/s

v = 50 m/s >>>180 kph >>>111.8 mph

i came out with about the same answer as jdoty did so that number is right.

Disclaimer: i would just like to say that i am (by far) not an expert at physics so don't take what i say as fact.

Nice math, is that equation correct? How did you figure out that equation?

physic's teacher..........i have no clue how it all works out i just know that somehow it does...