 # A simple but interesting gravity demo.

#### Astro Pen

##### Write now.
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Now that's what I call counter intuitive.

#### Mike Donoghue

##### Active Member
The reason for this is the wavy ramp is optimized for one of the kinematic equations. Definitely counterintuitive to look at though. Basically, the steep drops increase the vertical acceleration which decreases the time it takes to displace vertically. However, the more waves, the longer the ball takes since it begins to approximate a straight line.

#### Ursa major

##### Bearly Believable
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A practical application of this effect -- well, of the lessons one might learn from it -- is the "counter-intuitive" acceptability of steeper gradients** on high-speed railway/railroad lines than lines used for slower trains (passenger or freight).

** - For obvious reasons, high-speed lines can't use the sharper curves found on slower lines.

• RJM Corbet

#### Eternity_TARO

##### Member
The reason for this is the wavy ramp is optimized for one of the kinematic equations. Definitely counterintuitive to look at though. Basically, the steep drops increase the vertical acceleration which decreases the time it takes to displace vertically. However, the more waves, the longer the ball takes since it begins to approximate a straight line.
Yeah, it's basically the personification of math. Those drops takes advantage of acceleration

#### farntfar

##### I didn't so much fly...,as plummet.
optimized for one of the kinematic equations

You speak of an equation. So is the exact shape of the waves important to match with that equation, or will almost any curve do?

You also speak of one of the equations. Are there other equations, which correspond to other shapes?

#### Wayne Mack

##### Well-Known Member
You speak of an equation. So is the exact shape of the waves important to match with that equation, or will almost any curve do?
I do not believe that the exact shape of the curve is important (by the way, this is an effect of momentum, not gravity). I didn't bother trying to set up any equations, but used a simple visualization to understand the results.

First, simplify the curvy path to a single curve. Second, replace the flat path with the inverse curve. Think of taking a square of wood and using a jigsaw to cut out a deep, concave curve from corner to opposite corner. Flip over the remaining piece to have a convex corner piece.

Draw a pencil line at the midpoint of each curve and compare what happens to the marble in each path. For the falling part, the first half of the concave path and the second half of the convex path, the time to traverse can be considered approximately equal. For the flatter section, however, on the convex curve, the marble starts at zero velocity and only slowly accelerates to the midpoint. For the concave path, the marble is traveling at a fast velocity at the midpoint and carries that velocity through the flatter portion. Thus, the second half of the concave path is traveled more swiftly than the first half of the convex path.

The quick extrapolation to the flat path in the video is to view the flat path as the transition point between convex and concave. If that doesn't quite work, instead, visualize cutting shallower and shallower curves. As the curve gets closer to the flat path, the velocity at the midpoint is reduced and the overall end-to-end time increases.

To restore the multiple curves in the original, simply divide the path creating a series of concave paths racing a series of flat paths. In isolation, each concave path is faster than each flat path. When put together, the concave paths have the advantage of carrying higher velocity into the start of each section.

• farntfar

#### tinkerdan

##### ∞<Q-Satis
This doesn't contain the answer--but it has a clue.

#### hitmouse

##### Well-Known Member
This is interesting. A couple of other thoughts: The ball on the curvy track also appears to come of the track at points, so it has periods of free-fall. The ball on the straight track has to turn to move, since it stays n the the track, so there is the additional factor of angular acceleration as slowing it down.

#### Wayne Mack

##### Well-Known Member
The ball on the curvy track also appears to come of the track at points, so it has periods of free-fall.
I believe leaving the track is actually a detriment. Velocity is being expended travelling horizontally rather than vertically. That little factoid is something I gleaned from watching downhill skiing on the Olympics. Going airborne apparently increases a skier's time.

#### hitmouse

##### Well-Known Member
I believe leaving the track is actually a detriment. Velocity is being expended travelling horizontally rather than vertically. That little factoid is something I gleaned from watching downhill skiing on the Olympics. Going airborne apparently increases a skier's time.
Yes I have heard that. If it is true Then it may be relative to size/surface area/ drag, and so does not necessarily apply to a ball- bearing.
In freefall, the ball is indeed travelling horizontally, but its vertical velocity will also increase at 9.8m/s^2.