MF's Ride Speed

I was wondering if anyone had an idea about how fast MF will move through the majority of the circuit. I know it's supposed to hit 92 at the bottom of the hill, but for all we know, the third hill could be one of those slow cresting hills and that'll just kill all the speed. i was just wondering if anyone knew what type of speed averages we're looking at after let's say the third hill or so. I don't know how you'd know, but if you do I was just wondering.

We prob. won't know those facts until they physically test it
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Brian
Greensburg PA
Hometown parks: Kennywood, Cedar Point!
100 days until the real "Millennium

Even if it does burn off some of the speed, it will still probably be going at least 65mph down that hill. Remember, it is only 23 feet shorter than the Magnum's hill.

You can find the average speed by dividing distance over time.

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Jeff
Webmaster/Guide to The Point

I agree if it does slow down after on the third hill it will be going higher than the Mean Streaks first hill and then add on 2 cars

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...We hope you have a great day here at Cedar Point America's Rockin' Roller Coast

Don't forget that it will be starting at 310' The train probably won't lose a lot of speed on the second so you'll probably be around 80-85mph at in the first tunnel. I would expect that the third hill will have a ton of airtime on the top. Same goes with the fourth, naturally. I'm thinking of Shivering Timbers when I say this. The first hill on that is 125' and the second is only 115' or so and there is a load of air all over that thing. It just doesn't lose any speed. Of course there's no turns until the end, but still.

If I said it once, I said it a hundred times:
Is it may yet? :)


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Scott W. Short
sshort@mediaone.net
http://welcome.to/midwestcoastercentral

I wonder if MF will evven reach 92mph?

Why wouldn't it? I wouldn't be suprised if it actually surpassed 92 mph, by one or so. I know this probably isnt a good comparison, but look at Goliath, 85 mph at 255 feet. Although, there is such a thing called "terminal velocity", I believe it will easy reach/surpass 92 mph. Anyways, we shouldn't worry how fast the thing goes, there are many great elements to this ride besides speed...drops, TURNS, and open cars!

Shivering Timbers is cool, I love the Air Time, but lets Face it, Michigan Adventures(excluding the water park, which is practicly everthing there),Is not really that great.

The speed is also going to be complimented by the large amount of time which MF spends close to the ground towards the end of the ride. I keep thinking of a steel "Beast".

Yeah, and it will get even more intense when the trees grow back around those low parts, i can just imagine the blurs of everything whizzing on by at extreme speeds.

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"May the Schwartz be with you..."

yeah, but speed is all relative anyway..it's like the difference between doing 92 MPH in a convertible car, a closed car and a 250HP ski boat. they all FEEL different. the boat feels so much FASTER because you are skimming across the water with spray blowing past you. so MF may actually FEEL faster, even if it isn't 10 years from now, when the trres are all grown in and are whizzing by at close range...

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"I think I scrambled my brain!!"

Ok, anybody that has ever taken physics or algebra should be able to understand this. The law of conservation of energy states that "Energy and not be created or destroyed". This means that the total energy at the top of the hill is the same as at the bottom of the hill. Energy is either kinetic (energy from moving) or potential (energy from height). Kinetic energy is defined as 1/2mv^2 and potential is defined as mgh. No matter what, the kinetic and potential energy at the top of the hill will be equal to what is is at the bottom (actually some is lost to friction but for this, we'll pretend friction doesn't exist). Using the formula 1/2mv^2+mgh=1/2mv^2+mgh you can determine the speed of the trains. M=mass v=velocity g=acceleration due to gravity, 9.8 m/s^2 h=height. The masses cancel so the formula you are left with is 1/2v^2+gh=1/2v^2+gh. For this we will assume that the bottom of the first hill is ground level, meaning it's height is zero. Also remember that the height of the second hill is probably measured from the ground so for this formula it's height is 159 feet. Also, you have to convert the numbers into metric measurements (Well, you don't have to but it makes it easier for me). For this 1 meter per second equals 2.237 mph and one meter equals 3.081 feet. When you plug in the numbers you get something like this: 1/2(92mph or 41.12m/s)^2=1/2v^2+(9.8m/s^2)(159ft or 48.46m). Solve for the v and you get v=27.13m/s or 60.1mph. This means that at the top of the second hill, the trains will be going approx. 61mph plus or minus a few mph. Because this is without friction, I can't figure out the speed at the top of the third hill but someone on URC said it would be going about 31mph and at the bottom it would be going about 73mph. Now I could have come right out and said that but then you wouldn't have learned anything! Isn't physics great! I'll try and get my physics teacher to help me figure out the speeds with friction and when she does I'll post???? ??????????? ??? ???? ???? ??? ?????? ?? ?????????? ?? ????? ? ???

Have you been taking lessons from Rideman?

Apparently not...I would never advocate using metric intermediate measurements when both the input and output measurements are U.S. :)

Incidentally, a quick and dirty formula for top speed assuming zero initial velocity and no resistive forces (friction, etc.) is:

V [mph] = (60/11) * sqr (h [feet])

That is based on the following equations:
a = 32 ft/sec/sec.

v = 32t ft/sec (t = [sec])

d = 16t^2 feet

d/16 = t^2

sqr(d/16) = t = sqr(d)/4 seconds

so... v = 32(sqr(d)/4) feet/sec

Note also that 1 mph = 5280 ft/3600 sec. If you feel like working it all out, you can derive the Vmax = (60/11)sqr(h).

Personally, I don't feel like working it all out right now. :)

--Dave Althoff, Jr.