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-
I'm no expert, but NO. Remember, when you ride MF it "launches" you over the last seventy or so feet of track. A brake just wouldn't make sense, and will never be added.
Since the Force brakes on the sides with fins in magnet sandwiches, and there are no such sandwiches on the lift, no.
Don't trust hotel managers... only the Lemon Chill guy REALLY knows.
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Jeff
Webmaster/Guide to The Point
Millennium Force laps: 35
Don't trust hotel managers... only the Lemon Chill guy REALLY knows.
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Jeff
Webmaster/Guide to The Point
Millennium Force laps: 35
I swear, if I ever own an amusement park, my PR man is going to do double-duty as the Lemon Chill guy, just to tweak the entire world. =)
'The Lemon Chill guy said...'
'Sure he did.'
'The Lemon Chill guy said...'
'Sure he did.'
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?
Applaud StephenC!!!! lol, i agree with Jeff and Stephen on this. You just have to ask the Lemon Chill guy. He gives out great advice!
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Felt the Force-1
Kicked the Sky-15
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Felt the Force-1
Kicked the Sky-15
Thanks all! :) I thought he might have been wrong considering that I have never noticed it nor felt it.
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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-
I hate Mr. Physics. Jeff open up the rumor section
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Millennium Force Rides:2
Magnum XL-200 Drop:3
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Millennium Force Rides:2
Magnum XL-200 Drop:3
Out of curiousity, does anyone have the coefficient of rolling friction for nylon on steel?
*** This post was edited by Bill on 7/26/2000. ***
*** This post was edited by Bill on 7/26/2000. ***
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.
I knew Dave would chime in on this physics sooner or later
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http://www.msu.edu/~armbrus9/cp.html
http://www.msu.edu/~levydust/mf.html
mf laps = 32
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http://www.msu.edu/~armbrus9/cp.html
http://www.msu.edu/~levydust/mf.html
mf laps = 32
MF might be able to hit 100mph...
...provided the lift sled is going 32.1mph over the top of the hill.
[Yes, the math is accurate. No, I am not going to reproduce it here. Yes, this assumes no friction. No, I did not factor in tailwind. =) ]
*** This post was edited by Bill on 7/26/2000. ***
...provided the lift sled is going 32.1mph over the top of the hill.
[Yes, the math is accurate. No, I am not going to reproduce it here. Yes, this assumes no friction. No, I did not factor in tailwind. =) ]
*** This post was edited by Bill on 7/26/2000. ***
1. I couldn't remember the English value of g.
2. For whatever reason my answers always came out wrong in school whenever I tried to be lazy and stay with English measurements, so I just work in metric out of habit.
I suppose it works without converting. Until I try it, anyway :)
2. For whatever reason my answers always came out wrong in school whenever I tried to be lazy and stay with English measurements, so I just work in metric out of habit.
I suppose it works without converting. Until I try it, anyway :)
I thought MS gave me a headache, now all that numbers is just too much. I gotta go lay down...........
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Visit my RCT CP site.
http://cedarpoint.8k.com
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Visit my RCT CP site.
http://cedarpoint.8k.com
StephenC and Rideman, thank you for finally showing the math that all of us physics nerds have been doing to prove to people the maximum speed of MF. I really appreciate it! :-)
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-The Pants Has Spoken
"Yellow line? What Yellow Line..."
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-The Pants Has Spoken
"Yellow line? What Yellow Line..."
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