# Physic question: How much height can 180 mph push

Saturday, June 14, 2003 12:08 AM
We know 120 can go up to 420 feet. On a Scream Machines demo I can go about 600 feet with 150 mph. Scream Machines has a 150 mph limit (sad). I saw somewhere that Stan Checkette from S&S wants to do a 180 mph lauch. Now we also know that you don't need to do a very tall ride just throw on some low shallow speed bumps and some fast high speed banked curves before the big hill like on Dodoponda but if you have the money and the zoning clearence to do it you migh as well take full advantage of the oppertunity. So in short how much height is possible with 180 mph. This is assuming a 90 degree incline like on most of the recent launch coasters like TTD and Hypersonic. A think a 700 feet coaster could be possible using that much raw power.
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Saturday, June 14, 2003 1:01 AM
I just worked the calculations, and 180 MPH can theoretically go a lot higher than even that. Assuming a completely frictionless environment, a launch at 180mph directly into a hill should be able to rise to a height of 1082 feet. Of course, there's an enormous amount of wind resistance, and the friction on the wheels would reduce that number considerably, but I'm not sure how much.

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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Saturday, June 14, 2003 4:31 AM
First assume we do make a coaster that fast would it be safe with all g-forces it could posbly push going around a turn?? ive been on hypersonic a few times and even the turn after the hill was intense so i dont think the real question is how high a coaster can go but how fast it can take a turn.
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Saturday, June 14, 2003 4:48 AM
G forces are a function of both speed and turn radius. If you make the radius large enough then you can keep the G forces reasonable. It gets to where it uses up a lot of land though. At 180 mph, a 3 G turn is almost a quarter mile in diameter. This isn't a practical size for most parks. Of course the aerodynamic drag and mechanical friction are such that you won't be going that fast for long.
*** This post was edited by Jim Fisher 6/14/2003 8:49:09 AM ***
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Saturday, June 14, 2003 4:54 AM
exactly my point!!! in my opinion coaster are more exciting when the are compact not with large sweeping turns! so because of the g-forces the ride would be to big to enjoy.

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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Saturday, June 14, 2003 5:59 AM
For reference...

Assuming no resistive forces at all:

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

It logically follows, then, that

Hmax [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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Saturday, June 14, 2003 8:04 AM
Wait a second bighitz, you think Hypersonic's turn is intense? I must have read that wrong. I think the trains crawl through that.

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

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Saturday, June 14, 2003 9:26 AM
interesting... according to rideman's formulas... with 20% resistance... the next barrier (500ft) is achievable with about 140mph. i think both S&S and Intamin currently have the technology to do so....

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

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Saturday, June 14, 2003 11:34 AM
But the question is, where would they build it? Not many parks have the ability to build over 300ft, never-the-less 400, 500, etc. I doubt CP does it cause they have Dragster and the hype for it will be around for at least 5-6 years unless someone else goes bigger in that time. So yes the technology is there, but is there a buyer?

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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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Saturday, June 14, 2003 11:38 AM
Simple approximations yield:
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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Saturday, June 14, 2003 12:30 PM
NOTE: this is using rideman's formulas and assuming 20% resistance.

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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Sunday, June 15, 2003 6:56 AM
As far as the resistance goes, usually it is dependant upon the speed. (1/2*k*v^2) where k is a constant dependant upon the surface area that the wind would strike. so most likely at 180mph, it wouldn't be just 20%.

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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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Sunday, June 15, 2003 7:39 PM
Mechanical friction is a function of the velocity. Aerodynamic drag is a function of the square of the velocity. Of course this means that as you go faster, mechanical drag becomes relatively less important.
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Sunday, June 15, 2003 9:17 PM
just a thought...

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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Sunday, June 15, 2003 10:24 PM
I bet there would be some sort of wind sheild or they will design the trains so that the aerodynamics push the wind away from the faces of the riders. I'm actually a bit surprised as to how much wind is in the riders face on TTD.

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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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Sunday, June 15, 2003 10:46 PM
Rideman's formulas are simply unit friendly versions of basic physics velocity / displacement formulas. Look them up in any physics book.

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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Monday, June 16, 2003 7:28 AM
I built those formulas with a successive integration of "a = 32 ft/sec/sec".

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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