RCT Bob wrote:I'm quite sure that those curves are without friction.

Indeed they are. You can incorporate friction in the equation, but like

mathworld says "the problem can also be solved analytically, although the solution is significantly messier". (though I'd say the parametrization in x and y is still quite nice, except for the part where you probably have to vary both θ and k to optimize for time)

speising wrote:If A and B are on the same height, that's right. But that is not the situation we have here (or in the animation on the wikipedia page)

Watch this:

https://youtu.be/skvnj67YGmw?t=16m43s

I used the same height as the simplest counter-example to bondsbw. That doesn't mean it's the only counter-example.

What you see in that video is, once again, a slope of approximately -2/π, so the brachistochrone curve is monotonically decreasing and ends horizontally. It's a pity no one on youtube has an experiment showing that the fastest trajectory on a shallow slope goes below the end point.

And since I don't have a wood/acryllic workshop, we'll have to do some math. Assuming a gravity of g=1 and a slope of 1:20*

Let's have a look at the green line. It would take 10 units of time to traverse it (because the velocity at height B is

1/2mV²=mgh => 1/2V²=1*1/2 => V=1 and the horizontal distance is 10). And we can agree that the green path is faster than any path from A to B that doesn't go below B. (It's shorter than the shortest distance between A and B and it's at maximum velocity all over the 'curve'.)

Now take the red path: it takes

y=2=1/2gK² => K²=2/(1/2)/g => K=2 to get down,

x=10=vt=(g*2)L => L=10/2/g L=5 to get across and

M=K-Z, 1/2=1/2gZ² => Z²=1 => M=K-Z=2-1 =>** M=1 to get back up. So the red path takes a total of t=K+L+M=8, which is definitely less than the green path.

And as

this page explains, having tight corners takes longer than a smooth curve, so you can improve on t=8 by rounding the corners of the red path a bit (and thereby making them physically feasible, in case you don't accept corners in a counter-example).

Since the paths that stay above B are slower than the green line and there exists a path faster than the green line, the fastest path (=brachistochrone curve) must go below the end point for this shallow slope.

*because it makes both the calculations and the counter-example path trivial. Changing gravity only means you have to go deeper/less deep on the red path. Making the slope even shallower works in favour of the red path (try it out with a horizontal distance of 20). And for steeper slopes, at some point, you need paths that approximate the cycloid, rather than some horizontal and vertical lines to provide a counter-example.

**It takes 2s to get from 0m to 2m and 1s to get from 0m to 0.5m, so it must take 1s to get from 2m to 0.5m. (arbitrarily using seconds and metres as the units of time and length here)