I Is it possible to solve for “t?”

[Devin-M]

2 comments...

By other work referenced, we know that the solution is part of a cycloid defined by

I believe that equations 1 & 2 specify a cycloid that proceeds above the x-axis from the origin, but the needed path is an upside-down or inverted cycloid that proceeds below the x-axis from the origin.

we know the distance X between A and B and that xA=xB so

I believe you meant yA=yB instead of xA=xB

 

[Devin-M]

If I solve for t and r to 5 decimal places in 9.8m/s^2 gravity where the horizontal distance between the start and endpoint is 1 meter and the initial velocity is 1m/s I get the following parametric graph:

Parametric Graph: https://www.desmos.com/calculator/0k2dfjrnbw

Excel Calculator: https://u.pcloud.link/publink/show?code=kZq8VHXZIECHjF10TLJpQhMs7l1WEROKqDt7

...& travel time of 0.60214 seconds.

 

[Devin-M]

and we know the distance X between A and B and that xA=xB so
(4)2xA+X=2πr⟹X=2(πr−xA)

We now have two unknowns, t0 (the value of the parameter t at X)

There may be an error in equation 4... I think the equation is true for a complete cycloid when you know the horizontal distance between the cusps (except 2xA should simply be xA), but isn’t true when the cycloid is truncated by a certain distance in the y-axis when you know the horizontal distance between the truncated cusps.

I believe xA + X = 2pi * r only when the cycloid hasn’t been trimmed in the y-axis by specifying a range for t other than 0<=t<=2pi

 

[pbuk]

I believe that equations 1 & 2 specify a cycloid that proceeds above the x-axis from the origin, but the needed path is an upside-down or inverted cycloid that proceeds below the x-axis from the origin.

You can flip the sign of equation (2) if you want and work with negative values of $y$ (and g) everywhere, but the maths is the same if we keep equation (2) as it is and measure $y$ downwards (in other words flip your drawing upside down).

I believe you meant yA=yB instead of xA=xB

I had the right picture in my head and the conclusion is right but of course $x_A \ne x_B$. The workings should have been $$ x_B = x_A + X = 2 \pi r - x_A \implies X = 2 (\pi r - x_A) \tag 4 $$

 

[pbuk]

There may be an error in equation 4... I think the equation is true for a complete cycloid when you know the horizontal distance between the cusps (except 2xA should simply be xA), but isn’t true when the cycloid is truncated by a certain distance in the y-axis when you know the horizontal distance between the truncated cusps.

I believe xA + X = 2pi * r only when the cycloid hasn’t been trimmed in the y-axis by specifying a range for t other than 0<=t<=2pi

View attachment 273021

No, for a complete cycloid the horizontal distance is simply $2 \pi r$.

Setting $x = x_A, t = t_0$ in equation (1) and substituting it into equation (4) you get
$$ X = 2 (\pi r - r (t_0 - \sin t_0)) \implies r = \frac{X}{2(\pi - t_0 + \sin t_0)} $$ which is exactly what you have in cell C14: =(C3/(2*(PI()-C162+SIN(C162)))).

 

[etotheipi]

Apologies if I missed this, but how did we prove that the optimal path is a cycloid? When I tried to solve the variational problem, I ended up with a differential equation$$y'^2 y''(v_0^2 - 2gy) = (1+y'^2)((v_0^2 - 2gy)y'' - g)$$that I don't know how to solve. Although I can try the cycloid ansatz and see if it works

 

[pbuk]

Apologies if I missed this, but how did we prove that the optimal path is a cycloid?

That was my first question too, but if you think about it then it must be so because all we are doing by giving the particle an initial velocity is considering its position at some point along the solution to a different Brachistochrone problem with a higher stationary origin. The potential energy difference at that origin must equal the new kinetic energy.

Also the OP gave a reference:

I got that one from pg 116:
http://classicalmechanics.lib.rochester.edu/pdf/vpcm.pdf

Edit: this does of course assume familiarity with the classic Brachistochrone problem and it's cycloid solution as set out by Bernoulli and by Newton.

 

[etotheipi]

Thanks, interesting problem! As a sanity check, I let $v_0 = 0$ in my differential equation and then trialled$$(x,y)= \left( a(\theta - \sin{\theta}), a(1-\cos{\theta}) \right), \quad y' = \frac{\sin{\theta}}{1-\cos{\theta}},\quad y'' = \frac{-1}{a(1-\cos^2{\theta})}$$ and that satisfies the equation. For non-zero $v_0$, I think it will be solved by something of the form $(x,y)= \left( a(\theta - \sin{\theta}) - b, a(1-\cos{\theta}) - c \right)$

 

[pbuk]

For non-zero $v_0$, I think it will be solved by something of the form $(x,y)= \left( a(\theta - \sin{\theta}) - b, a(1-\cos{\theta}) - c \right)$

You have three unknowns there (a, b, c) but only 2 constraints (initial velocity and distance). So the correct equations are (1) and (2) in my post a few up.

 

[etotheipi]

You have three unknowns there (a, b, c) but only 2 constraints (initial velocity and distance). So the correct equations are (1) and (2) in my post a few up.

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

 

[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.