How????

Don't worry this isn't just another post complaining about Steel Dragon over MF, but a question.

How does Steel Dragon go faster than MF???.If MF has a 300 foot drop at 80 degrees. 80 degrees, nearly straight down. But then Steel Dragon is only a few feet higher, but the drop is Magnum's. I wonder if Steel Dragon's speed is right.

Angle is irrelevant. Fall 300 feet at 10deg, and 300 feet at 90deg, and you'll still be going about 95 mph at the bottom. It just depends on how fast you want to get to that top speed.

For further information, visit your local library and consult your friendly neighborhood physics book.

Yeah but MF gets pulled over the top almost three times faster then Steel Dragon.

RideMan?? help!!!

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ride early, ride often!!!

i was looking at the first drop over at thrillride.....the pullout is GIGANTIC! it looks like the only steep track is 5 sections long (rct jargon here). it doesnt really matter, we got the very best rollercoaster ever made right in our home state.
-Mike
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3 Force rides:
3-2, 5-1, 9-2 *** This post was edited by punk on 8/8/2000. ***

In a perfect world, angle is irrelevant. This not being a perfect world, the total length of track as well as magnitude of resistance [drag and friction] factor in. How else could MF valley on a 180ft hill after a 300ft drop?

However, top lift speed becomes less and less relevant to bottom speed as drop height increases. It's still important at the tops of hills, however.

Ok, remember how some people said that the only way MF hit 92 mph was because of the lift. Well SD2000 hit 95mph with just a few more feet. My point is how does a train going about 16mph at the top of the hill not go faster than a train only going about 5mph.

Now explain that one.

Simple, actually.

Horizontal Kinetic Energy = [.5]m*v^2, with m being mass and v being velocity [simplified to the scalar 'speed']

Mass can be neglected, as it's constant.

So... 13mph = 5.8m/s.
KE at top of lift = 16.82 units

KE values can be added, so we add in the drop factor, which -- neglecting friction, etc -- is mgy. [Mass, Gravity {9.8m/s/s} and height in meters]

Drop KE = [neglecting mass] 9.8 * 91.44 = 896.1 units

Add the chain and you get 912.93 units. Back to horizontal gives 42.73m/s or 95.58mph max.

SD's is [assuming 5mph chain]
KE top = 2.5
KE drop = 9.8*94.488 = 925.98
Total = 928.48

This gives a max of 96.40mph

So both running at maximum speed, SD is about 0.8mph faster.

Must agree with Punk that pullout is ridiculous. Mf all the way baby!!!

Rideman won't read this thread because the topic is devoid of any meaning.

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Jeff
Webmaster/Guide to The Point
Millennium Force laps: 35

You may be forced over the top of MF at 13 mph, but your velocity in just the vertical vector (which is the only velocity that factors into the gravity/speed equation) is negligible. It barely affects the top speed at all.

it is true that the inital speed barely affects the top speed, but not for the reason you state...
the actual reason is that the initial speed results in a drop that takes less time, which results in less time for the car to gain speed (i won't go into detail, since it's all been discussed in other threads)

think of it this way... a coaster car given an initial speed of 13 mph and dropped 300 feet straight down will be going 95.3657 mph, with all of it being in the vertical direction (and assuming no friction of course)...

a coaster car rolling down a 45 degree slope and given the same 13 mph inital speed will reach the same 95.3657 mph after dropping 300 vertical feet... the difference is that it will have a speed in the horizontal vector of 67.4337 mph and a speed in the vertical vector of 67.4337 mph (speed / square root of 2)

Assuming there is friction, I would assume that the car rolling down the 45 degree angle would be going a little slower due to it having to cross more track....