speed...
- More
- 1 of 2
- 2
is it hard to keep your eyes open and keep your arms up? do you get any good air time on the coaster and if you do were is the best place?
All three rides I had no problem keeping my hands up the entire day. It will be hard to hold on all the time during media day. :) Personally I thought that the best airtime was in the front seat, especially on the third hill.
-------------
daniel j. haverlock
'00 Magnum Count: 010
'00 M. Force Count:003
spiritofthepoint.com
-------------
daniel j. haverlock
'00 Magnum Count: 010
'00 M. Force Count:003
spiritofthepoint.com
Ugh... it goes 92, just like they said it would.
-------------
Jeff
Webmaster/Guide to The Point
Park visits May 6, 7, 10, 11, 13, 14, 16, etc...
-------------
Jeff
Webmaster/Guide to The Point
Park visits May 6, 7, 10, 11, 13, 14, 16, etc...
Actually it was clocked between 102-107 mph......keeping my arms up didn't happen just for the fact that i was a wee bit scared but you wait until i get another chance to ride it....though I would wait till employee nite when i get another chance to ride it with only a 20 minute wait :)
-------------
Wagons Supervisor 2000
-------------
Wagons Supervisor 2000
107!!!!! I find that hard to believe!!
-------------
http://www.msu.edu/~armbrus9/cp.html
Jeffrey Spartan and Proud New PAPPA
MSU National Champs 2000
-------------
http://www.msu.edu/~armbrus9/cp.html
Jeffrey Spartan and Proud New PAPPA
MSU National Champs 2000
People! Are you kidding me? 107 mph? You dont spend 25 mil on a ride and calculate the speed using the latest technology in computer design and then have it go 15 mph faster than you thought it would! I'm goin with Jeff on this one.
-------------
Millennium Force: As fast as a male horsefly!
-------------
Millennium Force: As fast as a male horsefly!
The ride was never clocked that high. Didn't happen. In fact, someone already went in to the physics of this in another thread. It's as true as the construction worker who lost a hand on the tunnel.
-------------
Jeff
Webmaster/Guide to The Point
Park visits May 6, 7, 10, 11, 13, 14, 16, etc...
-------------
Jeff
Webmaster/Guide to The Point
Park visits May 6, 7, 10, 11, 13, 14, 16, etc...
It is possible to go faster, but not much at all.
-------------
Millennium Force will be the next to sink!
-------------
Millennium Force will be the next to sink!
Sorry for not seeing the other thread Jeff. Just made it back to a computer today. I'm just telling ya all what everyone was talking about all weekend after we got off work! I seriuosly don't think it would make a difference if it went faster or not. I mean come on people 10 more mph is not that much more. Ever go 65 in a 55...it just happens. Maybe one of MF operators here could clear this up for us.
-------------
Wagons Supervisor 2000
-------------
Wagons Supervisor 2000
I heard it from several ride ops, but there was a lot of gossip about "fast" things that weekend. ;) I asked a silver tag who told me point blank it's going the speed it should.
-------------
Jeff
Webmaster/Guide to The Point
Park visits May 6, 7, 10, 11, 13, 14, 16, etc...
-------------
Jeff
Webmaster/Guide to The Point
Park visits May 6, 7, 10, 11, 13, 14, 16, etc...
As said in a previous post, It's impossible to go over a hundred.
-------------
Visit my RCT CP site.
http://cedarpoint.8k.com
-------------
Visit my RCT CP site.
http://cedarpoint.8k.com
Maybe Intamin stuck in some LIM's on the first drop. :)
-------------
...and enjoy the rest of your day at Cedar Point, The Am-aaaa-zement park!
-------------
...and enjoy the rest of your day at Cedar Point, The Am-aaaa-zement park!
If the laws of physics are correct, and the last time I checked they were, the fastest it can drop at 32 meters a second.
Rideman, care to chime in?
Rideman, care to chime in?
Objects falling near the earth's surface tend to accelerate at a rate of about 32 ft/sec/sec. With this bit of information, you can take indefinite integrals and get:
a = 32 ft/sec/sec.
v = 32t + v0 feet/second
d = 16t^2 + v0t + d0 feet
If we assume that d0=0 (we're falling the whole distance) and v0=0 (we're falling from a dead stop) we can reduce the equations as follows:
a = 32 ft/sec/sec
v = 32t ft/sec
d = 16t^2 feet
If we know the distance (310 feet, for instance) we can solve for time:
d = 16t^2 feet
d/16 = t^2
sqr(d)/4 = t
And we can substitute that expression for t in the velocity equation:
v = 32t feet/second
v = 32 * (sqr(d)/4) feet/second
We can do one more conversion because 1 mph is equal to 5280 feet per 3600 seconds:
v = (3600/5280) * 32 * (sqr(d)/4) miles/hour
If you reduce all the fractions you end up with:
V [mph] = 60/11 * sqr (d [feet])
For a 310 foot drop, this works out to 96.037 mph.
It is reasonable to expect that the formula will give a higher number than reality, because the formula represents an ideal case, where there are no resistive forces (friction, wind resistance, etc.). But it will get you into the ballpark for most rides. It also fails to take into consideration the initial velocity which, on Millennium Force, is 22 ft/sec. I didn't manage to work that one out because when v0 is non-zero, the equations get complicated:
a = 32 ft/sec/sec
v = 32t + v0 ft/sec
d = 16t^2 + v0t feet
This means that the time required is now a quadratic polynomial and I always hated solving those. I got as far as...
310 = 16t^2 + 22t
155 = 8t^2 + 11t
155 = t(8t+11)
If anybody wants to solve that one for t, then you can plug it into...
v = 32t + 22 ft/sec
then multiply by 3600/5280 to get the ideal maximum speed that Millennium Force can possibly attain in an ideal world.
--Dave Althoff, Jr.
a = 32 ft/sec/sec.
v = 32t + v0 feet/second
d = 16t^2 + v0t + d0 feet
If we assume that d0=0 (we're falling the whole distance) and v0=0 (we're falling from a dead stop) we can reduce the equations as follows:
a = 32 ft/sec/sec
v = 32t ft/sec
d = 16t^2 feet
If we know the distance (310 feet, for instance) we can solve for time:
d = 16t^2 feet
d/16 = t^2
sqr(d)/4 = t
And we can substitute that expression for t in the velocity equation:
v = 32t feet/second
v = 32 * (sqr(d)/4) feet/second
We can do one more conversion because 1 mph is equal to 5280 feet per 3600 seconds:
v = (3600/5280) * 32 * (sqr(d)/4) miles/hour
If you reduce all the fractions you end up with:
V [mph] = 60/11 * sqr (d [feet])
For a 310 foot drop, this works out to 96.037 mph.
It is reasonable to expect that the formula will give a higher number than reality, because the formula represents an ideal case, where there are no resistive forces (friction, wind resistance, etc.). But it will get you into the ballpark for most rides. It also fails to take into consideration the initial velocity which, on Millennium Force, is 22 ft/sec. I didn't manage to work that one out because when v0 is non-zero, the equations get complicated:
a = 32 ft/sec/sec
v = 32t + v0 ft/sec
d = 16t^2 + v0t feet
This means that the time required is now a quadratic polynomial and I always hated solving those. I got as far as...
310 = 16t^2 + 22t
155 = 8t^2 + 11t
155 = t(8t+11)
If anybody wants to solve that one for t, then you can plug it into...
v = 32t + 22 ft/sec
then multiply by 3600/5280 to get the ideal maximum speed that Millennium Force can possibly attain in an ideal world.
--Dave Althoff, Jr.
It does the hill in 22s, right?
300 vertical feet at 45 degrees = 486ft = 148.15m
in 22s = 6.73m/s
(ignore mass, as it cancels)
Ep0 = energy = gh = 9.8*91.46
Ek0 = .5v^2 = (1/2) * 6.73^2 = 22.67
Ebottom = 1/2v^2
.5v^2 = 918.978
v^2 = 1837.96
v = 42.87m/s = 95.897mph
300 vertical feet at 45 degrees = 486ft = 148.15m
in 22s = 6.73m/s
(ignore mass, as it cancels)
Ep0 = energy = gh = 9.8*91.46
Ek0 = .5v^2 = (1/2) * 6.73^2 = 22.67
Ebottom = 1/2v^2
.5v^2 = 918.978
v^2 = 1837.96
v = 42.87m/s = 95.897mph
Thank you, Jmstuckman, for finishing that. I just double checked your calcs and they appear correct.
Bill, I'm curious: When all of the information you need to solve the problem is given in US measurements, and the result you want is also in US measurements, why convert to metrics, a conversion which does nothing for you but introduce rounding error? Just curious...I mean, even if you only knew "g" to be 9.8 m/sec^2, you could have just converted that to 32 ft/sec^2??
--Dave Althoff, Jr. *** This post was edited by RideMan on 5/10/2000. ***
Bill, I'm curious: When all of the information you need to solve the problem is given in US measurements, and the result you want is also in US measurements, why convert to metrics, a conversion which does nothing for you but introduce rounding error? Just curious...I mean, even if you only knew "g" to be 9.8 m/sec^2, you could have just converted that to 32 ft/sec^2??
--Dave Althoff, Jr. *** This post was edited by RideMan on 5/10/2000. ***
- More
- 1 of 2
- 2
You must be logged in to post