The Brachistochrone Problem

[notFritzArgelander]

This was a problem proposed by Johann Bernoulli in 1696. The problem was published asking other mathematicians of the day to find the curve of steepest descent with only gravity acting as a force. It was a problem that Galileo tried and failed to solve. Indeed Johann Bernoulli's own solution was flawed.

Newton solved it overnight in 1697.

https://en.wikipedia.org/wiki/Brachistochrone_curve

Since friction and internal degrees of freedom are ignored, it isn't immediately applicable to real world situations. The most efficient curve will depend on additional realistic parameters.

[notFritzArgelander]

PS Typo alert.

The “steepest descent” should read “quickest descent”. The problem is a quickest descent problem which can be solved by using the method of steepest descent. Pardon the momentary dyslexia.

[metastable]

The interesting implication of this problem is a train in an evacuated tunnel covers a distance between 2 points on the ground traveling in a curve underground more efficiently than the straight line on the surface connecting 2 points which is the conventional wisdom.

Interestingly below very roughly 300 meters depth, ordinary water has the same useful energy content as gasoline on a cost basis. When friction and wind resistance is minimized If the train coasts to a depth of -300m, then pushes of a water tank off the back with such force the tank stops on the tracks, the train obtains not only the force of the push but the kinetic energy of the water, which below about 300m is about the same as a same-costing amount of gasoline. After the train moves on geothermal heat can be used to raise the water to the surface, but since the train traveled on a below-ground curve it is more efficient than a geothermal powered train on the surface, because the below ground train uses its own gravitational potential energy as fuel...

33kWh * 20% useable = 6.6kWh
6.6kwh = 23760000 J
Water Gallon = 3.78kg
J = MGH
23760000 J = 3.78 * 9.8 * H
H = J/MG
MG = 3.78 * 9.8 = 37.044
641399.4 m = 23760000/37.044
1 gallons water = $0.0015
1 gallon gas = $3.02
$3.02 / $0.0015 = 2013.33
641399.4 m / 2013.33 = 318.57m

“Elon Musk’s Boring Company appears to have something brewing in between Los Angeles and Las Vegas, and it could have to deal with the company’s plans to link the two cities with a tunnel.

Photos showed a Boring Company machine, along with large tents set up in the middle of the desert in Adelanto, California. The presence of machinery and storage tents hints toward the possibility that the Boring Company could be gearing up for the prospective link between Los Angeles and Las Vegas.”

https://www.teslarati.com/boring-compan ... onnection/

[notFritzArgelander]

The Boring Company scheme as reported is boring because of neglected factors such as risks due to seismic activity.

[Graeme1858]

33kWh * 20% useable = 6.6kWh
6.6kwh = 23760000 J
Water Gallon = 3.78kg
J = MGH
23760000 J = 3.78 * 9.8 * H
H = J/MG
MG = 3.78 * 9.8 = 37.044
641399.4 m = 23760000/37.044
1 gallons water = $0.0015
1 gallon gas = $3.02
$3.02 / $0.0015 = 2013.33
641399.4 m / 2013.33 = 318.57m

That's unbelievable!

$3.02 for a gallon of petrol?

A gallon of petrol here is over a fiver!

Regards

Graeme

[notFritzArgelander]

33kWh * 20% useable = 6.6kWh
6.6kwh = 23760000 J
Water Gallon = 3.78kg
J = MGH
23760000 J = 3.78 * 9.8 * H
H = J/MG
MG = 3.78 * 9.8 = 37.044
641399.4 m = 23760000/37.044
1 gallons water = $0.0015
1 gallon gas = $3.02
$3.02 / $0.0015 = 2013.33
641399.4 m / 2013.33 = 318.57m

That's unbelievable!

$3.02 for a gallon of petrol?

A gallon of petrol here is over a fiver!

Regards

Graeme

Indeed. The US government subsidizes petroleum production. There's a rabbit hole there....

[Lady Fraktor]

Current price locally for 95 octane is 1.152/ liter or 5.24/ imperial gallon (4.546 liters) or 4.36/ american gallon (3.786 liters)

[metastable]

I wrote this little equation to determine how many joules of mechanical energy are required from the vehicle to extract all the kinetic energy from the water tank at the bottom of the ramp.

It’s based on the mass of the vehicle, mass of the tank, a uniform gravity field, the initial velocity, and the length of time it takes an object to free fall in a vacuum to the same depth as the ramp.



At the bottom of the ramp, after applying the mechanical impulse calculated above to the tether, the vehicle ends up with 100% of the kinetic energy the tank had before the impulse, 100% of the energy of the mechanical impulse, and 100% of the kinetic energy the vehicle had before the impulse.

By comparison, a wheeled vehicle applying the same mechanical impulse to the ground obtains 100% of the impulse, but no energy from the water tank and no gravitational potential energy from descending the ramp, so travels more slowly for the same impulse.

[metastable]

Suppose there are 2 points A & B that are 1 meter apart at the same elevation and I have 2 frictionless test masses acted on only by gravity both with initial velocity 1 m/s.

There is both a path of shortest distance (straight line) and a path of shortest duration (curve). The shortest distance path (straight line) takes exactly 1 second to traverse.

Is there a formula for calculating the travel time of the test mass which travels the shortest duration curve (sliding w/ no friction acted on by gravity)?

[notFritzArgelander]

For a frictionless bead on a wire the link in the OP and references therein have the formulas.

[metastable]

It says "If the body is given an initial velocity at A [...] then the curve that minimizes time will differ from the tautochrone curve."

I want to compare 1m/s straight line time for 1 meter vs initial 1m/s on optimal curve duration.

[notFritzArgelander]

It says "If the body is given an initial velocity at A [...] then the curve that minimizes time will differ from the tautochrone curve."

I want to compare 1m/s straight line time for 1 meter vs initial 1m/s on optimal curve duration.

Then use the supplied formulas.

[metastable]

found it here-- pg 116
http://classicalmechanics.lib.rochester ... f/vpcm.pdf

"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 d such that the kinetic energy equals the potential energy difference from the cusp."

[notFritzArgelander]

found it here-- pg 116
http://classicalmechanics.lib.rochester ... f/vpcm.pdf

"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 d such that the kinetic energy equals the potential energy difference from the cusp."

That is also obvious from the OP and conservation of energy, courtesy of Noether's Theorem.

[metastable]

I think I'm descending into madness trying to solve this. If anyone can help it would be most appreciated.

I'm trying to find the length from A to B where curve AEB is a cycloid.

Line GH is exactly 1 meter, and line DF is 0.05102 meters which I believe is the change in height that accelerates things to 1m/s.

As a bonus I'd also how long it takes to get from G to H assuming the bead enters at point G with 1m/s.



I'm also struggling with a different issue. Assuming the curve is a ramp and the bead initially travels with 1m/s horizontally before entering the curve, how does it instantly change vectors to follow the curve if only acted on by gravity?

[notFritzArgelander]

I'm only going to answer selected bits......

I think I'm descending into madness trying to solve this. If anyone can help it would be most appreciated.

Help is available in upper division (junior or senior year) courses in the analytical dynamics of point particles and rigid bodies. Multivariate calculus is the only prerequisite.

It is against forum rules or preferences to do solutions of possible homework problems. What you are asking for is close to a common homework problem.

I'm also struggling with a different issue. Assuming the curve is a ramp and the bead initially travels with 1m/s horizontally before entering the curve, how does it instantly change vectors to follow the curve if only acted on by gravity?

It's not "only acted on by gravity" the bead also experiences mechanical contact forces due to the constraint provided by sliding on the wire. The Lagrangian and Hamiltonian reformulations of Newtonian mechanics (taught in the above mentioned course) enable and justify the clean and uncluttered handling of forces of constraint which are extremely difficult to deal with in a view that is restricted to Newton's formulation.

[metastable]

this is as close as I could get from trial & error... but its approximate so not exact. a bead that takes 1 sec to travel from 0 to 1 meter horizontally travels in (yet to be determined) shortest possible time on the orange slope from 0 to 1. a bead lifted to the cusp should cross 0 traveling with initial velocity of 1m/s.

[notFritzArgelander]

this is as close as I could get from trial & error... but its approximate so not exact. a bead that takes 1 sec to travel from 0 to 1 meter horizontally travels in (yet to be determined) shortest possible time on the orange slope from 0 to 1. a bead lifted to the cusp should cross 0 traveling with initial velocity of 1m/s.

.....

This doesn't look right at first glance.

Trial and error is not reliable. One should solve the equations of motion.

[metastable]

[notFritzArgelander]

Fancy flashing graphics are meaningless. It is not clear that you are interpreting the terms in the equation correctly. Without showing your work it is impossible to say it's correct. It's on you to establish that you have not abused the formulae from the link you cited in post 13.

For instance there is a related problem, the tautochrone, which might help refine one's thinking.

https://en.wikipedia.org/wiki/Tautochrone_curve

But without defining the assumptions that go into your assumed equation, no one should believe it.