what determines the size and shape of a loop of the roller coaster?

Answer 1

Uhm, perhaps things like inertia and momentum?

Obviously if the loop is too small the car will stress on the track/frames too much. the faster a cart is going as it rounds the loop the more stress (G's) it will exert on the "loop". (i am also sure that there are limitations to the G's that fun parks/coaster builders are willing to inflict on the passengers while keeping them within a safe limit)

If the loop is too big (and the kart isn't going fast enough) then the kart wont make the loop (unless assisted in some other way?) and will either slide back down the way it came, or get stuck half way and either fall or just dangle there.

I don't build coasters but i think those are the sorts of things that dictate the size of the loop in a coaster.

I am sure someone more fluent in physics will come along with the formulas for such things.

Answer 2

i possibly think this can be an answer:
look,
The frictionless circular roller coaster loop with negligible train length is a popular textbook problem. The speed is then obtained directly from the conservation of energy, i.e. mv2/2=mgh. At any given part of the frictionless roller coaster, the centripetal acceleration is thus given by ac= v2/r = 2gh/r where h is the distance from the highest point of the roller coasters and r is the local radius of curvature.
Assume that you pass the top of a loop with a speed v0, obtained e.g. by starting from rest at an height h0=v02/2g above the top of the circular loop. If the loop has a radius, r, the centripetal acceleration at the top will be a0=2g h0/r. The centripetal accelerations at the side and at the bottom are immediately obtained from the value at the top as (a0 + 2g) and (a0 +4g), respectively.

The limiting case of weightlessness (0g) at the top, where no force is needed between train and track, nor between riders and train, occurs when h0=r/2, so that the centripetal acceleration is given by a0=g (and the centripetal force is thus provided exactly by the gravitational force from the earth). The centripetal acceleration at the sides and bottom will be 3g and 5g, respectively. (What is the total acceleration of the rider for these cases?) The corresponding "g-forces" are 3g and 6g.

Trains moving slowly across the top would fall of the track, were it not for the extra sets of wheels on the other side of the track. Similarly, the riders would depend on the restraints to remain in the train. However, even if the train were nearly at rest at the top, with riders hanging upside down, experiencing -1g, the riders would still be exposed to 5g at the bottom (and 2g at the side) if the loop were circular.

Children's roller coasters may be limited to 2g, family rides often reach 3g, sometimes more, whereas many of today's large roller coasters exceed 4g. Depending on the individual's "g-tolerance", the oxygen supply to the head may stop completely at 5 to 6 g, resulting in unconsciousnes if extended in time. Although higher g-forces can be sustained with special anti-g-suits, e.g. for pilots, 6g for any extended period of time would not be acceptable for the general public

However, the disadvantages of circular loops are not limited to the maximum g-force at the bottom: Entering the circular loop from a horizontal track would imply an instant onset of the maximum g-force (as would a direct transition a circular path with smaller radius of curvature). An immediate transition from one radius of curvature to another would give a continuous, smooth track, but with discontinuous second derivatives. Clearly, a function with continuous higher derivatives would be preferrable.It is obvious that different approaches have been used to achieve the desired transition from a smaller radius of curvature at the top to a larger radius at the bottom.

Just a small piece of help,
Thanks

Answer 3

Safety is the number one concern of a coaster builder, and the loop size is often related to the speed of the cars as it approaches that part of the run.

Answer 4

The engineer who designs it .