How Can I Calculate the Size and Angle of Loops on a Rollercoaster?

[mike_302]

Hi,
I'm supposed to constuct a small rollercoaster using any material i want... I will be using tubing/hosing. A marble is what will be running through it. Now, what i am looking for is a few things that will help me build it to spec for the class. It needs to be at least a 90 second ride. In needs a few loops and turns and stuff. For the most part, I can see how to fix adjust the ride if it doesn't keep the marble on track, but if there are any tricks (simple equations? tips? anything?) that I can apply when building the track so that I can have the ball rolling at a moderate speed, yet not SPEEDING through, that would be great.

On top, a slightly more specific and theoretical question: If I were to make loops (like the type where it goes around and around and around.. on a slight angle to the ground, kind of like a funnel, what would be the size of the circles, and the angle downward of the loops that would keep the ball moving at a decent speed without causing those.. 3 or 4 loops to go by in a flash of 3 -5 seconds. Am I looking at a LARGE circle with a small angle in the loop? a small circle with small angle? Large circle with large angle? etc. Does anyone know what I mean by this ? lol

Thanks in advance for any help!

 

[AFG34]

I am not very advanced with this stuff but some of the stuff i have learned:

-your first drop has to be the biggest so you pick up lots of kinetic energy
-for a circle, Fnet=mv^2/r, where m is mass, r is radius of loop, v is velocity, Fnet is all the forces acting on the marble
-horizontal loops require less energy/speed to overcome; vertical loops perpendicular to the ground require the most kinetic energy; so if you are having trouble with one loop, try tilting it.

90 secs is a lot of time to me, its going to be hard.

 

[mike_302]

I believe it will be hard too. We DO get extra marks for adding some sort of electronic circuit to the whole design and I am thinking of connecting a small DC motor in a battery circuit so that when the marble hits a switch near a certain point at the bottom, it activates the mottor which will be connected to some sort of pulley/conveyor belt unit and bring the marble back to the top to start a second run down (on a different path th the first one... IT will not be a continuous circuit.

Thanks for the reply! IT was exactly wht I was looking for :) If anyone else has any other input, that would be great! Thanks in advance!

 

[Danger]

I believe that it was either Dave or Berkeman who pointed out in another thread that an egg-shaped vertical loop is more efficient than a circular one, so you might want to try that.
90 seconds does sound like a long time, but a little bucket conveyor would certainly help things along. (Some real coasters use boosters part-way through the ride, as well, not just at the start.)
Also, you don't have to start at the highest point if you have a launching device such as a couple of counter-rotating rubber wheels that work like a pitching machine.

 

[mike_302]

I won't be able to go as far as the pitching machine style start, but the egg shape vertical loop is good to know. Of course, it's difficult to undertan in your head because I'm sure that fact was figured out mathemtically. Thanks so far! If anyone else has input, I'll be checking in for a few days.

 

[mike_302]

OK, so I have a 16 mm steel, polished ball bearing, and I'm going to be picking up a bunch of 3/4" rubber tubing/hose very soon for this project. My strategy for this project to "supposedly" get the most marks is to drop the ball from the max height (85 cm) and have it go into as many loops as possible right from the start. My question here is this: How can i tell how many loops I can get the ball to go through, assuming the first drop is vertical, and the bal goes stright form this vertical drop (from 85 cm high) straight into the loops. Also keep in mind that the loop size can vary, but the teacher will multiply that size by 1.5 in his marking scheme if atleast te sides or top of the loop is "open" (half of the tube is cut away), this meaning that it would be to my advantage GREATLY if I can get more loops that conserve as much energy as possibe for the most time, therefor I can AFFORD to hae half the tube on the loop cut away.

Thanks in advance! With the knowledge i have on the subject as it is right now, at the most, I can probably calculate how fast the ball will be going when it reaches a certain point in its vertical drop, but after that, I think it's safe to assume that my work/energy caclulations will have pi or radius in it and that is just WAY beond what I have learned.

 

[Danger]

I'm afraid that I can't help you with any calculations. For myself, I'd just go the trial-and-error route to working out your loops.

 

[mike_302]

yaa, that's what I was thinking at first. I guess I'll just try that unles anyone else has some sort of geniously-simple idea.

 

[AFG34]

check this out: http://www.glenbrook.k12.il.us/gbssci/phys/projects/frig/yepbyrji/coaster.html

 

[Danger]

Wow! Nice one, AFG.

 

[Shekels]

ok a little off the topic of your question but from that same website i got a formula for projection, h = (ax^2)/(2v^2) it says x= distance from center of hill, V=velocity going over the hill, a=acceleration due to gravity(9.8m/s/s), and h= the height from the top of the hill.

What does it mean by h= the height from the top of the hill. ?
does it mean the distance from the top of the hill? because that's what I am thinking.

the picture is attached.