      # A practical use of physics

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### #1 TeleMad

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Posted 16 June 2004 - 05:52 PM

My two sons and I just got back from Disney World/Seaworld. One of the roller coasters we rode was called the Kraken: it was pretty impressive to us. We were wondering how high the first drop was: I said 100 to 150 ft and my oldest son guessed 200 to 250 feet. But nothing we had could answer the question. However, I was given enough info from one sentence in the caption of the photo of us on the ride that we purchased.

"You were hurled over 4,100 feet of track, through seven loops at speeds exceeding 65 mph, while experiencing both negative and positive G-Forces."

When we got back to the motel I grabbed two post-its and using nothing more than standard algebra and two simple physics formulas (v = v0 + at and x = v0t + (1/2)at^2) was able to calculate by hand in just a matter of minutes that the minimum height of the first drop was 160 feet. It was just like solving one of those beginning physics problems from college...guess that stuff can have practical applications after all.

### #2 Tormod

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Posted 16 June 2004 - 06:14 PM

Your guesstimate seems to be closer...according to a non-official guide to Disney World, "Put in simple terms: Kraken sends SeaWorld guests 149 feet high, upside down seven times, underground three times and through the thrill-ride experience of their lives."

<a target=_blank class=ftalternatingbarlinklarge href="http://www.wdisneyw.co.uk/swkraken.html
">http://www.wdisneyw....k/swkraken.html
</a>

If that page is right, the maximum height can't be higher than 149 feet...but perhaps I'm interpreting their text wrong.

Tormod

### #3 TeleMad

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Posted 16 June 2004 - 06:43 PM

I'll take a look into it. But here are the calculations based on the ride's own description ("exceeding 65 mph"). If anyone sees an error, let me know.

The top speed was reached at the bottom of the first drop, for which there was no mechanical assistance (gravity only). The ride's statement said the speeds EXCEEDED 65 mph, which would require a higher drop than for the 65 mph I used in my calculation. In addition, the non-vertical slope of the drop would "dilute" gravity and would INCREASE the minimum height needed to reach even the 65 mph. My calculation does not take these into consideration (assuming they are basically offset by the slight velocity the cars have as they approach the big drop).

1) Convert from mph to ft/sec
60mph = 88ft/sec
65mph = 65mph * [(88ft/sec)/(60mph)] = 95.3 ft/sec

2) calculate time using final velocity
v = v0 + at
v = at
95.3 ft/sec = (32 ft/sec^2) * t
t = (95.3 ft sec^-1) / (32 ft sec^-2)
t = 2.98 sec

3) calculate height from time
x = v0t + (1/2)at^2
x = (1/2)at^2
x = 16 ft sec^-2 * (2.98 sec)^2
x = 16 ft sec^-2 * 8.88 sec^2
x = 159.84 ft
x = 160 ft

PS: Found the problem. Is 8 * 16 = 144? Don't think so. But that's what my scribbling on the second post-it shows. For 8.88 * 16 I have 144 + 1440 + 14400 = 15984 = 159.84. Each term should be a "128", such that 128 + 1280 + 12800 = 14208 = 142 ft. 