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- Physic question: How much height can 180 mph push

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For comparison, the theoretical top height of TTD with a 120 mph launch is 479 feet, and as you know, it doesn't go too much over 420. That's a 15% reduction in height. Lets assume even greater resistance on the new ride: so assuming a 19% reduction, you end up with about 875 feet, which would be a good estimate in my opinion.

I do not know the terminal velocity of a coaster train, but it is quite possible that you might go 180mph up, but only 150 mph down...but I think I remember someone calculating it, and the heavy, aerodynamic coaster trains won't reach terminal velocity until about 250 mph or so...but I'm not sure.

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Jman

Webmaster: Gravibulb Coasters - Home of Professional Quality Coaster Photos

http://balder.prohosting.com/gravbulb/coasters/

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*** This post was edited by Jim Fisher 6/14/2003 8:49:09 AM ***

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Now if it went straight up a hill and flipped lie ttd then it would be exciting and fesible but having it comes straight down would be to intense for my liking

*** This post was edited by bighitz 6/14/2003 8:55:54 AM ***

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Assuming no resistive forces at all:

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

It logically follows, then, that

H*max* [feet] = ((11/60) * v [mph])^2

Those formulas handle the conversion for you, as they factor in both the 5280 feet/mile and the 3600 seconds/hour conversions, and are based on an acceleration of 32 ft/sec/sec.

--Dave Althoff, Jr.

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If at first don't succeed, find out if the loser gets anything.

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"It's like something out of that twilighty show about that zone"

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It's my time to Fly! - URGE - Titan A. E.

Is it yours?

Superman: Ultimate Flight

SFGam 2003

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100ft = 50mph

200ft = 75mph

300ft = 95mph

400ft = 115mph

500ft = 140mph

600ft = 160mph

700ft = 180mph

Of course, these aren't perfectly accurate, but I could imagine them being similar to what you are looking for.

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i put the formula into excel and ran all heights from 100 to 1000, then graphed. what you notice is its not a straight line... the additional speed required for each successive 100ft drops. from 100ft to 200ft the difference in speeds is 25mph. from 200 to 300 its 19... then 16... 14... 13... 12.. but the amount that the additional speed drops also drops. like a quadratic equation...

so as coasters get higher... speed becomes more efficient... but the rate at which it becomes more efficient slowly decreases

very weird if its true... but its based all on rideman's formulas which i dont know how he derived.

so theoretically in the case of a 10,000ft coaster vs. an 11,000ft coaster... the latter would travel only 30mph faster (610 and 640mph respectively). but those speed would likely not be achievable due to terminal velocity and all that.

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"It's like something out of that twilighty show about that zone"

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It's my time to Fly! - URGE - Titan A. E.

Is it yours?

Superman: Ultimate Flight

SFGam 2003

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On TTD, people's faces are pushed back, and people are having their contacts rget pushed back into their eyes. This is all at 120mph. If we got something at 150-180mph, could it potentially be dangerous to the rider? Especially the rider in the front seat

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It's my time to Fly! - URGE - Titan A. E.

Is it yours?

Superman: Ultimate Flight

SFGam 2003

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Oh, and Enfynet, your estimates get quite off on your higher velocities...see my original work for 180MPH on the second post in this thread, it's much closer to 900 feet than 700.

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Jman

Webmaster: Gravibulb Coasters - Home of Professional Quality Coaster Photos

http://balder.prohosting.com/gravbulb/coasters/

*** This post was edited by Jman 6/16/2003 2:50:20 AM ***

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a = 32 ft/sec/sec

integrate to get velocity...

v = 32t + v0 ft/sec <- but assume v0 = 0

v = 32t + 0 ft/sec

integrate to get distance, remembering that v0=0...

d = 16t^2 + v0t + d0 <- but assume d0 = 0

d = 16t^2 + 0t + 0 feet

d = 16t^2 feet

Now we still have a problem, in that we don't know time. But algebraically we can get the time in terms of distance...

d/16 = t^2

sqr(d)/4 = t [seconds]

So in the velocity equation, we can substitute sqr(d [feet])/4 for time...

v[ft/sec] = 32((sqr(d)/4)) = 8(sqr(d [feet]))

Except that given an input value in feet, that gives us feet per second. Next we need to multiply by 3600 to get feet per hour...

v[feet/hour] = 28800 * sqr(d)

...then divide by 5280 to get miles per hour

v[miles/hour] = (28800/5280) * sqr(d)

It just so happens that 28800/5280 = 60/11.

Hence the reduced, unit-friendly formula is:

v[mph] = (60/11) * sqr(d[feet]).

And yes, the relationship between speed and height is quadratic in nature. Doubling the speed requires quadrupling the height if you are falling, but doubling speed will quadruple height if you are launching.

--Dave Althoff, Jr.

Not really a math geek, really!

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