- #1
Whiteblooded
- 7
- 0
Hi everyone.
I was wondering about the equation of a roller coaster loop. Most people do not realize this, but a roller coaster loop is not a circle.. but rather a 'clothoid loop' shape.. the reason for this being, with a circular loop, the 'G force' varies quite a lot, and is uncomfortable to the rider. The clothoid loop shape gives more of a normal distribution function for G force, where G force is highest at the top of the loop. I'm trying to derive an equation for this shape.
I was originally trying to model a roller coaster car as a point mass, where the velocity is given by:
Etot = \frac{m}{2}v2 + mgh = \frac{m}{2}v02
v2 = v02 - 2gh
Where h is the height (a variable) from the base of the loop, and v0 is the velocity at the base of the loop.
Then one would put this into the centrepital force equatioon to give:
F = \frac{m}{r}v2 = \frac{m}{r} (v02 - 2gh)
Idealy, I'd like to get an equation for r in terms of the angle around the loop. To do this, I need to find h in terms of r and theta. (I think this is where everything went wrong) The value I got for h was:
h = H/2 - rcos(\theta)
Where \theta is the angle from the vertical axis of symmetry, starting at the bottom of the loop, sweeping through to the top. H is the maximum height of the loop. My value for h clearly isn't correct, because it assumes the radius always comes from the centre (H/2) of the loop.
Then when I get this value of h, I'm not sure where to go.. the idea that I have in my head would be to somehow make this force equal to a normal distribution of the form:
A*exp{-Bx2} (Where A and B are some constants/scale factors and x is a variable.. which will be a form of theta).
Then I'd try and equate that with the centrepital force as shown above, and (attempt to) rearrange for r.
Can anyone help me out with this? I've looked all over the web for the solution to this problem. http://physics.gu.se/LISEBERG/eng/loop_pe.html" Has some quite useful things on.. but it seems to skip over a lot of the mathematics and doesn't really explain it very well.
I was wondering about the equation of a roller coaster loop. Most people do not realize this, but a roller coaster loop is not a circle.. but rather a 'clothoid loop' shape.. the reason for this being, with a circular loop, the 'G force' varies quite a lot, and is uncomfortable to the rider. The clothoid loop shape gives more of a normal distribution function for G force, where G force is highest at the top of the loop. I'm trying to derive an equation for this shape.
I was originally trying to model a roller coaster car as a point mass, where the velocity is given by:
Etot = \frac{m}{2}v2 + mgh = \frac{m}{2}v02
v2 = v02 - 2gh
Where h is the height (a variable) from the base of the loop, and v0 is the velocity at the base of the loop.
Then one would put this into the centrepital force equatioon to give:
F = \frac{m}{r}v2 = \frac{m}{r} (v02 - 2gh)
Idealy, I'd like to get an equation for r in terms of the angle around the loop. To do this, I need to find h in terms of r and theta. (I think this is where everything went wrong) The value I got for h was:
h = H/2 - rcos(\theta)
Where \theta is the angle from the vertical axis of symmetry, starting at the bottom of the loop, sweeping through to the top. H is the maximum height of the loop. My value for h clearly isn't correct, because it assumes the radius always comes from the centre (H/2) of the loop.
Then when I get this value of h, I'm not sure where to go.. the idea that I have in my head would be to somehow make this force equal to a normal distribution of the form:
A*exp{-Bx2} (Where A and B are some constants/scale factors and x is a variable.. which will be a form of theta).
Then I'd try and equate that with the centrepital force as shown above, and (attempt to) rearrange for r.
Can anyone help me out with this? I've looked all over the web for the solution to this problem. http://physics.gu.se/LISEBERG/eng/loop_pe.html" Has some quite useful things on.. but it seems to skip over a lot of the mathematics and doesn't really explain it very well.
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