Estimating G Forces on roller coasters

[Joel Kleppinger]

This should be a relatively simple question, but I'm struggling for a frame of reference. What I want to do is estimate the top G at the bottom of the hill within about a half a G or better.
http://www.cedarpoint.com/thrill/millennium/mfheight.htm

My estimation is if we figured the centrifigal force of an object going 90 mph in a circle with a 200 ft. radius, that'd be close enough. (80 degrees down to ~40 degrees up in about 400-500 ft of track) I just don't know what one G would be in terms of speed/radaii. From there, it should be relatively easy to relate it to the question at hand.

Gravity is what, 9.5ft/s^2 or so... or is my memory of physics a bit off (hasn't been since my Junior year of HS)?

I wasn't going to mess with this, but I keep thinking about it... I'm just too curious. My guess is it's going to end up around 6 Gs, but I'm not sure. So I'm really interested in hearing what you all have to estimate.

[SysOpt]

a=(v^2)/r

Gravity = 9.8 m/s^2 = 1 G

Assuming:
v=90mph=40m/s
r=45 meters

a=(40^2)/45
a=35.5 m/s^2

35.5/9.8= 3.6G's

In other words, a 150lb person would feel like they weigh 540lbs http://www.sysopt.com/forum/smile.gif. Better do some strength training http://www.sysopt.com/forum/wink.gif.

And the force on the track at that point in time?

F=ma

Mass of the train is 16,300kg. Mass of people in the train assuming 150lb people (36 per train) is 2,450kg.

F=18750*35.5
F=665,625N (Newtons)

A 3000lb car on earth exerts 13,300N on the ground at rest (only 3,325N on the surface under each of 4 wheels assuming 50/50 weight distribution), so 665kN is a LOT of force.

Scott

[This message has been edited by SysOpt (edited 04-07-2000).]

[Joel Kleppinger]

Duh.. meters, not feet. I really did know that. http://www.sysopt.com/forum/smile.gif

The Gs part is more important than Newtons, at least to riders. Designers might think differently. http://www.sysopt.com/forum/wink.gif

Just wanted to see if we could estimate what percentage would throw up after riding it. http://www.sysopt.com/forum/smile.gif

[seti]

My cats breath smells like cat food.

Ralph (simpsons)

That thing looks awesome! (Webcam (http://www.cedarpoint.com/thrill/millennium/webcamnew.cfm))

[This message has been edited by seti (edited 04-07-2000).]

[Joel Kleppinger]

Just realized something... we assumed in the calculations above that there were no G forces present other than the 90 mph run through a section of the circle (i.e. We assumed no gravity). I think we should add a G to it to account for the act of normal gravity.

So that would put the final result around 4.6 Gs.

[SysOpt]

Yeah that's probably right..

(35.5 m/s^2 + 9.8 m/s^2)/9.8 = 4.6G's

Any Physics buffs want to verify this?

Scott