Brachistochrone Problem w/ Initial Velocity

[Devin-M]

Hello,

There is a physics problem called the Brachistochrone problem which I know has been solved for 0 initial velocity (assumes 0 friction and only gravity) and I know the answer is a cycloid. My question is: is there is an existing formula for finding the portion of a cycloid which is the quickest route between 2 horizontal points when there is an initial velocity.

I haven’t been able to find a source for one. I believe I’ve found the correct geometric solution for trimming the cusps off the cycloid by a vertical height which gives that initial velocity, but I don’t necessarily want to present the formula here if there’s an existing formula.

For example, I calculate the quickest route between 2 horizontal points separated horizontally by 1 meter, in 9.8m/s^2 gravity with 1m/s initial velocity is 0.602... seconds along a portion of a cycloid with a certain cycloid generating radius, and a certain initial and final theta.

Thanks

 

[sysprog]

This paper discusses variants of the problem in which "the initial angle of inclination is given (Section (5)), or that in which the initial speed (kinetic energy) is specified (Section (6))": https://mate.uprh.edu/~urmaa/reports/cbrach.pdf

The following paper is available; however, I'm not sure of the copyright status, so per PF policy, I shan't post a link to the pdf:

Some remarks on the brachistochrone problem (May, 2011) Marian Muresan

Abstract
This paper introduces some results on the brachistochrone problem with Coulomb friction and nonzero initial velocity. The following boundary conditions are considered hereafter: fixed end points and fixed initial and final velocities.​

Also:
https://dspace.mit.edu/bitstream/handle/1721.1/32840/61357053-MIT.pdf
http://sites.millersville.edu/rumble/StudentProjects/Gemmer/ThesisFinal.pdf

 

[Devin-M]

Since posting here I found this:
https://aapt.scitation.org/doi/abs/10.1119/1.11441?journalCode=ajp

Can anyone verify my calculated time of 0.602...s (optimal) for 1m horizontal distance with 1m/s initial velocity?

 

[wrobel]

https://www.physicsforums.com/insights/general-brachistochrone-problem/

 

[PeroK]

https://www.physicsforums.com/insights/general-brachistochrone-problem/

Can you beat $0.602s$?

 

[epenguin]

This is a moment when I can't spend too much time thinking about this hazy just about grasped knowledge of mine, but sometimes little slogans or mottoes are usefu. I recall 'The brachistochrone is also the tautochrone'. In other words 'the time taken by an object sliding without friction in uniform gravity to its lowest point is independent of its starting point on the curve.'

See https://en.wikipedia.org/wiki/Tautochrone_curve which has a nice animation.

And example of this is probably already known to you from elementary physics: the period of a simple pendulum is independent of the amplitude. That is usually established using a linear approximation (valid for small angles of initial displacement) of the pendulum dynamics but is actually wider in validity.

 

[Devin-M]

“If the mass starts with a non-zero initial velocity, then the starting point is not at the cusp of the cycloid, but down a distance such that the kinetic energy equals the potential energy difference from the cusp.” (pg 116)

http://classicalmechanics.lib.rochester.edu/pdf/vpcm.pdf

So the problem becomes finding the cycloid generating radius, initial and final theta and calculating the travel time, for which I get 0.602...s for 1m horizontal with 1 m/s initial velocity:

https://www.desmos.com/calculator/ekbzcrkbtm

I used:

(A is -0.05102 m is the depth that accelerates objects to 1m/s in 9.8m/s^2 gravity and B is the horizontal distance or 1 meter)

A=-(B/2(pi-t+sin(t)))(1-cos(t))

-0.05102=-(B/2(pi-t+sin(t)))(1-cos(t))

therefore:

t=0.8117451055281... rad

&

r=B/(2(pi-t+sin(t)))

r=0.1636480760880076136181... =B/(2(pi-t+sin(t)))

My understanding is:

A=-(B/2(pi-t+sin(t)))(1-cos(t))

^Can’t actually be arranged for t and so t & r are irrational

 

[Devin-M]

Since I have been told the equation can’t be rearranged to solve for t, I used a series of logic steps to find it, starting with an initial guess of 1 radian. The algorithm finds the initial depth and compares it with the desired initial depth, then determines whether an increase or decrease is called for, and by what increment (increment divided by 10 any time a change in the increase or decrease determination occurs), to arrive at a change in guess, this new change is tested and the revisions continue until t is found to a satisfactory accuracy...

 

[Devin-M]

To calculate the time I used:

dtime=sqrt(r/g)*dtheta

dtime = sqrt(0.163.../9.8)* 0.811...

0.104... seconds = sqrt(0.163.../9.8)* 0.811... (untruncated cusp to initial theta)

time_whole_cycloid=2*pi*sqrt(r/g)

0.811... =2*pi*sqrt(0.163.../9.8) (untruncated whole cycloid)

0.602...=0.811...-0.104...-0.104... (untruncated cycloid minus the truncated cusps time)

0.602...s

Does it look correct?

 

[Devin-M]

Assuming it’s true, then by extension when friction is negligible, a vehicle with an initial 9 m/s can cover 5000 meters horizontal more than 10x quicker on the optimal curved route than the straight/flat route, without using any additional energy. (Flat time ~10.1x longer than optimal time).

 

[Devin-M]

A=-(B/2(pi-t+sin(t)))(1-cos(t))

Can anyone please explain why or why not this equation can or can’t be rearranged to solve for t?

 

[A.T.]

A=-(B/2(pi-t+sin(t)))(1-cos(t))

Can anyone please explain why or why not this equation can or can’t be rearranged to solve for t?

Have you tried this?
https://www.symbolab.com/

 

[Devin-M]

Have you tried this?
https://www.symbolab.com/

Yes, it will rearrange for B but not t.

The best answer I’ve gotten so far (on a trigonometry help forum) was:

“I'm afraid it can't be done algebraically. There is a linear term in t and a sin(t). You can't undo that.”

 

[Devin-M]

Assuming I’ve been doing the calculations correctly, it appears that the initial velocity has to increase at the square root of the horizontal distance increase factor to keep the ratio between the flat time and the optimal time constant. In other words if one quadruples the horizontal distance, the initial velocity needs to double to keep the same ratio between flat and optimal time. If one doubles the initial velocity and quadruples the horizontal distance, the optimal travel time doubles.