Is it possible to solve for “t?”

[pbuk]

$$ \begin{align*} x &= r (t - \sin t) \tag 1 \\
y &= r (1 - \cos t) \tag 2 \end{align*} $$

To which we can add $t = [ t_0, 2 \pi - t_0]$ from the context.

 

[pbuk]

But I think $c$ is functionally dependent upon $b$, in order to constrain that trajectory intersects the origin.

Well clearly, [edit: I think actually $c = f(a, b)$] so how does this add to what has already been said ?

 

[pbuk]

I'm not sure that it is a particularly interesting problem anyway - the particle must start with a velocity that is almost vertical (for any significant distance to travel) and what is the easiest way to attain that velocity? Drop it from a height - and then you are back to the classic Brachistrochrone problem.

 

[Devin-M]

the particle must start with a velocity that is almost vertical (for any significant distance to travel)

The optimal path approaches horizontal when the initial velocity is significant, even over a large distance...

 

[pbuk]

The optimal path approaches horizontal when the initial velocity is significant, even over a large distance...

Put the formula =ATAN(C8/(2*PI()*C14-C3))*180/PI() somewhere in the spreadsheet you linked: that is (approximately) the angle to the vertical in degrees. Over 86° with the example data!

 

[Devin-M]

If you put a high enough initial velocity such as 1000 or more m/s as the initial velocity and graph r & the range for t, the path can be almost a straight horizontal line, but not quite.

 

[pbuk]

That's Mach 3.

 

[Devin-M]

The part I find interesting is when the distance is 5000m and the initial velocity 9m/s, the optimal path is more than 10x quicker than the flat route.

 

[pbuk]

The part I find interesting is when the distance is 5000m and the initial velocity 9m/s, the optimal path is more than 10x quicker than the flat route.

Ignoring air resistance and tyre friction at up to 400 mph.

 

[Devin-M]

Right, it would be something more like a magnetically levitated hyperloop carriage in a near vacuum.

 

[pbuk]

Right, it would be something more like a magnetically levitated hyperloop carriage in a near vacuum.

Yes I thought this is where we were going, which is why I said the maths is not very interesting. Either

• The initial velocity is low (just enough to pull out of the station) in which case the solution is not significantly different from zero velocity; or
• The initial horizontal velocity is high and there is little to be gained from a cycloid path which is not significantly different from a straight line.

In either case the practical engineering challenges make the theoretical maths of the truncated cycloid irrelevant.

 

[Devin-M]

True but unless you can actually calculate the brachistochrone w/ initial velocity then it’s impossible to compare the optimal route with the flat route. If the initial velocity is 0 then you simply won’t be leaving the station on the flat route. Also if the brachistochrone starts with 0 velocity, you won’t make it to the end of the route due to friction. But with the Brachistochrone with initial velocity, as long as the friction force * distance is less than the initial kinetic energy, then the brachistochrone between 2 horizontal points is “traversable.”

 

[pbuk]

True but unless you can actually calculate the brachistochrone w/ initial velocity then it’s impossible to compare the optimal route with the flat route. If the initial velocity is 0 then you simply won’t be leaving the station on the flat route. Also if the brachistochrone starts with 0 velocity, you won’t make it to the end of the route due to friction. But with the Brachistochrone with initial velocity, as long as the friction force * distance is less than the initial kinetic energy, then the brachistochrone between 2 horizontal points is “traversable.”

Ah, but friction forces are not constant, they mostly vary in proportion to the square of the velocity and even with low coefficients of friction this means that the solution to the relevant differential equation is no longer a cycloid.

 

[Devin-M]

True again but if friction forces are very low to negligible the answer is close. With null flux coil levitation the friction force decreases significantly with increasing velocity.

 

[pbuk]

We are now way off the topic of General Maths here: to continue you could start a thread in the Engineering forum something like:

I have found that the fastest theoretical path for an underground transport system is a part of a cycloid. What are the engineering concerns that might make this not relevant from a practical point of view?

 

[Devin-M]

https://www.mdpi.com/2073-8994/11/10/1239/htm

 

[etotheipi]

Oooo, well the problem would be even more interesting with friction! I'm not sure how best to solve that! I imagine, you consider a trajectory $\left(x(t), y(x(t))\right)$, and then the frictional force at any position is going to be$$\boldsymbol{F} = -\frac{\mu v^2}{\rho} \boldsymbol{t} = -\mu v^2 \frac{(1+y'^2)^{\frac{3}{2}}}{y''} \boldsymbol{t}$$Let $E = \frac{1}{2} (\dot{x}^2 + \dot{y}^2) + gy$, then$$\dot{E} = \dot{x} \ddot{x} + \dot{y}(\ddot{y} + g) = -\mu (\dot{x}^2 + \dot{y}^2)^\frac{3}{2} \left[\frac{(1+y'^2)^{\frac{3}{2}}}{y''} \right]_{x=x(t)}$$I'm not sure how we can re-cast such a differential equation in $t$ into one without any time dependence that we could then solve with variational methods?

 

[Devin-M]

https://mathcurve.com/courbes2d.gb/brachistochrone/brachistochrone.shtml

w/ initial velocity:

w/ friction:

 

[etotheipi]

Well, it would seem I screwed up a little bit! ☺ My assumption was that, in the equation defining the friction$$F = \mu R_n = \mu m \left[g\cos{\theta} + \frac{v^2}{\rho} \right]$$that the $g\cos{\theta}$ could be neglected, whilst in fact the article has done the exact opposite and said the $v^2 / \rho$ can be neglected! The other key part that I didn't see was$$\frac{dv}{dt} = v \frac{dv}{ds} = \frac{1}{2} \frac{d}{ds} (v^2)$$Still end up with a horrible equation, but surprisingly nice solution to the variation problem!

 

[pbuk]

Oooo, well the problem would be even more interesting with friction! I'm not sure how best to solve that! I imagine, you consider a trajectory $\left(x(t), y(x(t))\right)$, and then the frictional force at any position is going to be$$\boldsymbol{F} = -\frac{\mu v^2}{\rho} \boldsymbol{t} = -\mu v^2 \frac{(1+y'^2)^{\frac{3}{2}}}{y''} \boldsymbol{t}$$Let $E = \frac{1}{2} (\dot{x}^2 + \dot{y}^2) + gy$, then$$\dot{E} = \dot{x} \ddot{x} + \dot{y}(\ddot{y} + g) = -\mu (\dot{x}^2 + \dot{y}^2)^\frac{3}{2} \left[\frac{(1+y'^2)^{\frac{3}{2}}}{y''} \right]_{x=x(t)}$$I'm not sure how we can re-cast such a differential equation in $t$ into one without any time dependence that we could then solve with variational methods?

I don't think you can. In any case the mathematical solution is useless, this is an engineering problem and yopu need to non-neglect all the neglected variables.

The resulting problem is only solvable (but trivially solvable) by computational methods.

 

[etotheipi]

I don't think you can. In any case the mathematical solution is useless, this is an engineering problem and yopu need to non-neglect all the neglected variables.

The resulting problem is only solvable (but trivially solvable) by computational methods.

There is nice approximate form of solution starting on bottom of 332 here:
https://www.jstor.org/stable/2974953?seq=5#metadata_info_tab_contents

Also, I think this is not an engineering problem, but quite a purely mathematical problem!

 

[pbuk]

There's tons of open sourced info about this in Musk's hyperloop project, no need to pursue it further here.

 

[fresh_42]

There's tons of open sourced info about this in Musk's hyperloop project, no need to pursue it further here.

... where still the mathematical formula should be discussed, and not the physical system which led to it.

If someone wants to open a new thread about Hyperloop, please feel free to do so. Contact a mentor if you want some of the late posts here to be moved into the new thread in order to get it started.
This one will be closed.