Is This True?
I was talking to a manager at the Breakers Express hotel about Millennium Force and he told me that on the top of the lift hill there is a break. Is this true? I've looked at many pictures to see if there really was a break, but I could not find it on any of them. If this is true about the break, MF might be able to break 100mph!
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Next CP Trip: 8/10/00
See my Po!nt?
-12E-
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Next CP Trip: 8/10/00
See my Po!nt?
-12E-
There are no brakes on MF other than the final run. And with or without it, you won't break 100mph on MF. Time to summon Mr. Physics (TM):
v(f)^2 = v(i)^2 + 2gd.
Convert everything to metric, of course.
v(i) = 0 (you have no vertical velocity at the top of the hill)
g = 9.8 m/s^2
d ~= 91.5 m (300 ft)
v(f) = your velocity after you've fallen a distance of d, in this case, your speed at the bottom of the drop
Your fastest possible ride, assuming you've refitted MF with frictionless track and wheels and moved it into a vacuum, is a little over 95mph. If you want to claim that the hill gives you an initial velocity, then let v(i) ~= 5.8 m/s (13 mph, the lift hill speed). Now you can get it to 96.5mph. But that's definitely it, unless you have a neat way of bending space-time while you ride. Since friction and freefall air resistance play a small role on the actual ride, trim off a few mph. There's the touted 92mph in all the papers.
But of course, ask the Lemon Chill guy, too. Whatever he says, it's ALWAYS right :) Who said physics was an exact science anyway?
v(f)^2 = v(i)^2 + 2gd.
Convert everything to metric, of course.
v(i) = 0 (you have no vertical velocity at the top of the hill)
g = 9.8 m/s^2
d ~= 91.5 m (300 ft)
v(f) = your velocity after you've fallen a distance of d, in this case, your speed at the bottom of the drop
Your fastest possible ride, assuming you've refitted MF with frictionless track and wheels and moved it into a vacuum, is a little over 95mph. If you want to claim that the hill gives you an initial velocity, then let v(i) ~= 5.8 m/s (13 mph, the lift hill speed). Now you can get it to 96.5mph. But that's definitely it, unless you have a neat way of bending space-time while you ride. Since friction and freefall air resistance play a small role on the actual ride, trim off a few mph. There's the touted 92mph in all the papers.
But of course, ask the Lemon Chill guy, too. Whatever he says, it's ALWAYS right :) Who said physics was an exact science anyway?
StephenC, why mess around with metric? If your input values are in feet and you want your output values in mph, wouldn't it be easier to just do everything in feet and miles?
h = 300 feet; a=32 ft/sec/sec
Vf [mph] = (60/11) * sqr(h [feet])
Vf [mph] = (60/11) * sqr(300 feet)
Vf [mph] = (60/11) * 17.32 = 94.48 mph in a frictionless vacuum with no initial velocity.
That said...
There is no need for a brake at the top of the hill, as, if there were any need to slow or stop the train up there, it could be done using the lift sled, at least until the third or fourth car goes over the top. I don't believe there are any brakes on the lift sled itself, as there are brakes on the cable drum instead.
--Dave Althoff, Jr.
h = 300 feet; a=32 ft/sec/sec
Vf [mph] = (60/11) * sqr(h [feet])
Vf [mph] = (60/11) * sqr(300 feet)
Vf [mph] = (60/11) * 17.32 = 94.48 mph in a frictionless vacuum with no initial velocity.
That said...
There is no need for a brake at the top of the hill, as, if there were any need to slow or stop the train up there, it could be done using the lift sled, at least until the third or fourth car goes over the top. I don't believe there are any brakes on the lift sled itself, as there are brakes on the cable drum instead.
--Dave Althoff, Jr.
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