1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What shape ramp for maximum marble jump?

  1. Oct 16, 2012 #1

    I'm in the process of building a marble run out of brass wire using 15mm steel ball bearings. For ease of visualisation, it will be similar in principle to this:

    I want to include a "jump" where the marble travels up a ramp at speed and is launched in the air and eventually lands on another piece of track/catching element.

    I'm envisaging a an approx 40cm near vertical track drop with a curve at the bottom that launches the marble up at around 45 degrees. My question is - what size/shape curve do I need to maximise the size of the jump the marble makes?

    If the curve is very small - the marble will bounce. If the curve is very long, the marble will lose too much momentum on its way up the curve and fail to jump at all.

    Should the curve be of just a circular shape? Or have some sort of exponential change in angle?

    I did a couple of mockups:

    My rusty phsyics knowledge gives me the hunch that I need a ramp that minimises friction - is that right?

    Can anyone point me in the right direction? thanks
  2. jcsd
  3. Oct 16, 2012 #2
    Last edited: Oct 16, 2012
  4. Oct 17, 2012 #3


    User Avatar
    2017 Award

    Staff: Mentor

    I would expect that the shape does not matter much, as long as the curve radius is significantly larger than the marble diameter (factor 3-5+ or something like that).
  5. Oct 18, 2012 #4
    Thank-you for the reply :)

    I read some about Brachistochrone cruves and from what I can make out, it seems that they would get the marble from one point to another fastest but I don't think this would help with the actual size of the marble jump.

    If the start of the ramp is point A and the launch point of the ramp is point B, my concern is not how to get from A->B fastest, but rather how to maximise speed at point B.

    It might be that I just end up building a few variations and going with whatever happens to work best, but since I've started digging in to it, I'm kind of curious what the optimal ramp shape would actually be.

    On the subject of marbles rolling, would that suggest that the entry to the ramp shouldn't be too steep so that the marble does have time to roll and build momentum. If the entry ramp was vertical, the marble would drop and only start rolling when hitting the curve.

  6. Oct 18, 2012 #5
    Good point Jeffage, fastest total time doesn't mean fastest at point B. (Note, you don't just want fastest at B but also an incline of 45 degrees at B if you ignore air friction).

    However, I am pretty sure that the Brachistochrone is also the fastest at point B. But I don't know really.
  7. Oct 18, 2012 #6


    User Avatar
    Homework Helper

    Another issue is that the marble is rolling, so some of the gravitational potential energy goes into angular energy of the marble, which will reduce it's linear speed somewhat.
  8. Oct 18, 2012 #7
    We'd probably want to ignore the ball physics for simplicity and think of it as a point.
    Another problem is that the point would be going faster at angles lower than 45 degrees, so the optimal jump length might be a little less than 45 degrees.
    Help, does anyone know the answer?
  9. Oct 18, 2012 #8


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    If the marble is assumed rolling at take-off, its rotation rate will be entirely determined by its linear speed (for a given marble size). So assuming a uniform solid sphere, the fraction of KE that will go into rotation is also fixed. This means that we can concentrate on maximising linear speed and optimising the angle, and not worry about rotation too much (but see later).
    Ensure that the last part of the ramp makes contact contact with the lowest part of the ball. If there's a groove in the ramp (to ensure the ball stays on it), and the groove is a bit too deep, the ball will be rolling on the edges of the ramp and so rotating faster. That will sap linear speed.
    Losses aside, the take-off speed, then, should entirely depend on the vertical drop. It will not be affected directly by the take-off angle, so 45 degrees is a good place to start.
    That leaves us with the issue of minimising losses.
    If any slipping occurs, there will be energy loss. This may happen if the first part of the slope is so steep that the traction is insufficient to make rotational acceleration keep up with linear acceleration.
    There will be losses due to rolling resistance. This encourages keeping the track short, but I'm unsure how rolling resistance varies according to the normal force at each point of track. If the curve is tight, this force will be high and may cost more.
    There will be losses due to air resistance. This again suggests keeping the track short, particularly in the lower sections where the ball is moving quickly.
    Note that keeping the track short while not making the curve too tight suggests lowering the take-off angle a little.
  10. Oct 19, 2012 #9


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Based on my earlier post, I would suggest a track that consists of a straight ramp followed by, roughly, an arc of a circle. In principle, the bottom of the curve should be a little flatter since the ball is travelling faster there and because g is directly added to the normal force there, but if the height of the curved section is small compared to that of the ramp it shouldn't matter much.
    The ramp should be inclined as steeply as possible without leading to skidding. The tangent to the arc should, of course, start off matching the ramp and finish at 45 degrees. But how tight the curve should be I cannot say. It will depend on details of the materials and drag.
  11. Oct 19, 2012 #10
    If there is no friction and no slipping then the take-off velocity will depend only on the gravitational potential energy difference between the start point and the take-off point, no?

    However, the 45 degree take-off angle is optimised for longest possible horizontal travel distance to the same vertical position as at take-off. So you really want to make sure that your take-off point is as low as possible, because then you can convert maximum GPE into kinetic energy AND then catch the ball at the same height.

    It makes no sense to say the angle must be 45 degrees if your launch point is 10cm higher on the table than your catch point. If you superimpose two parabola trajectories with the same initial speed but different initial angles, and the take-off point is at y=0, then the 45 degree parabola will intersect the y=0 line a second time at maximum x value. However, it will NOT necessarily intersect the y = -1 line at maximum x value... a slightly smaller angle will.

    Hence if your take-off position is 5cm above the table, because you used a very large ramp, the 45 degree condition is not optimal. You could get more distance by (a) using a smaller take-off angle and positioning your catcher at a lower height, or (b) by using a smaller ramp, to attain more kinetic energy, and positioning your catcher level with the take-off (using 45 degrees).

    Conclusion- there is definitely some maths to be done!

    This image might help explain what I mean- the smaller take-off angle attains greater horizontal distance if the particle falls below take-off height.
  12. Oct 19, 2012 #11


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That's exactly what I meant when I said it "should entirely depend on the vertical drop". Sorry if that wasn't clear.
    There's no benefit in making the take-off point lower if the catch point remains the same height. In fact, you'd be worse off because it would increase losses. I'm assuming the vertical difference between the start of descent and the catch point is a given.
    You're quite right, thanks for picking that up. I was making the unwarranted assumption that the take-off point and catch point were at the same height. There are standard equations for the more general case, which back in school was called "range on an inclined plane". See e.g. "Maximum range" at http://cnx.org/content/m14614/latest/
  13. Oct 20, 2012 #12


    User Avatar
    2017 Award

    Staff: Mentor

    If the vertical difference between starting point and catching point is fixed and we neglect losses, the optimal launching height is "as low as possible" with a very steep angle - this is easy to see with a time-reversed setup.
    With losses, launch height ≈ catch height looks like a reasonable setup.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook