A fair time ago (perhaps 50 years) I remember reading a question in a physics text book. I may have forgotten some detail, but I think it went
as follows:
You are asked to construct a track which will allow a marble to travel, under the influence of gravity, between two points in a vertical plane. If the start point (A) is at point (0, 1) and the end point (B) is at point
(1, 0) what is the nature of the curve of the track that allows the
journey to be completed in the minimum time? You can ignore friction and
air resistance.
I remember the answer given. Does anyone have any pointers how you may tackle this question?
Solution provided:
N plpybvq...
A fair time ago (perhaps 50 years) I remember reading a question in a
physics text book. I may have forgotten some detail, but I think it went
as follows:
You are asked to construct a track which will allow a marble to travel,
under the influence of gravity, between two points in a vertical plane. If the start point (A) is at point (0, 1) and the end point (B) is at point
(1, 0) what is the nature of the curve of the track that allows the
journey to be completed in the minimum time? You can ignore friction and
air resistance.
I remember the answer given. Does anyone have any pointers how you may
tackle this question?
Solution provided:
N plpybvq...
I think you're remembering The Brachistochrone Problem, which can be solved using "Calculus of
Variations". There are (e.g.) YouTube videos explaining this, and I doubt I could give a better
explanation myself!
The Brachistochrone problem is slightly different from your description - it uses a point particle,
sliding frictionlessly on a track (or wire frame). When you change it to a marble rolling on a
track you introduce an angular rotation component (the rotation of marble) which will distort the
solution so I would expect the solution to be different...
Mike.
A co-worker achieved some of his fame by solving the
"unrestrained brachistochrone" problem. Google will lead you to
the problem statement, along with abstracts of that man's papers on
the topic, though the content may be paywalled.
35 years ago he challenged me with that problem when I stumbled into his >office to say hello; he was impressed when I quickly intuited the solution. >Unfortunately the apogee of my math skill is now well in the past.... :-{
On 04/05/2026 02:00, David Entwistle wrote:
A fair time ago (perhaps 50 years) I remember reading a question in a
physics text book. I may have forgotten some detail, but I think it went
as follows:
You are asked to construct a track which will allow a marble to travel,
under the influence of gravity, between two points in a vertical plane. If >> the start point (A) is at point (0, 1) and the end point (B) is at point
(1, 0) what is the nature of the curve of the track that allows the
journey to be completed in the minimum time? You can ignore friction and
air resistance.
......
I think you're remembering The Brachistochrone Problem, which can be solved using "Calculus of
Variations". There are (e.g.) YouTube videos explaining this, and I doubt I could give a better
explanation myself!
The Brachistochrone problem is slightly different from your description - it uses a point particle,
sliding frictionlessly on a track (or wire frame). When you change it to a marble rolling on a
track you introduce an angular rotation component (the rotation of marble) which will distort the
solution so I would expect the solution to be different...
Can you tell us who the person is? Or, is it classified?
(1) The suggested Googling would have revealed his identity in
less time than it took you to post this.
(2) Is it not rude, at least in many or most contexts, to post
unnecessarily the names of private individuals?
https://en.wikipedia.org/wiki/Brachistochrone_curve
I think you're remembering The Brachistochrone Problem, which can be
solved using "Calculus of Variations". There are (e.g.) YouTube videos explaining this, and I doubt I could give a better explanation myself!
The Brachistochrone problem is slightly different from your description
Mike Terry <news.dead.person.stones@darjeeling.plus.com> posted:
I think you're remembering The Brachistochrone Problem, which can be solved using "Calculus of
Variations". There are (e.g.) YouTube videos explaining this, and I doubt I could give a better
explanation myself!
The Brachistochrone problem is slightly different from your description - it uses a point particle,
sliding frictionlessly on a track (or wire frame). When you change it to a marble rolling on a
track you introduce an angular rotation component (the rotation of marble) which will distort the
solution so I would expect the solution to be different...
Good point.
A co-worker achieved some of his fame by solving the
"unrestrained brachistochrone" problem. Google will lead you to
the problem statement, along with abstracts of that man's papers on
the topic, though the content may be paywalled.
35 years ago he challenged me with that problem when I stumbled into his office to say hello; he was impressed when I quickly intuited the solution. Unfortunately the apogee of my math skill is now well in the past.... :-{
FWIW, he failed when I challenged him back with
"Find the fallacy in this 'proof' of the Four-Color Map Theorem" --
https://fabpedigree.com/james/color4.htm
On 04/05/2026 19:42, James Dow Allen wrote:
Mike Terry <news.dead.person.stones@darjeeling.plus.com> posted:
I think you're remembering The Brachistochrone Problem, which can be solved using "Calculus of
Variations".a There are (e.g.) YouTube videos explaining this, and I doubt I could give a better
explanation myself!
The Brachistochrone problem is slightly different from your description - it uses a point particle,
sliding frictionlessly on a track (or wire frame).a When you change it to a marble rolling on a
track you introduce an angular rotation component (the rotation of marble) which will distort the
solution so I would expect the solution to be different...
Good point.
A co-worker achieved some of his fame by solving the
"unrestrained brachistochrone" problem.a Google will lead you to
the problem statement, along with abstracts of that man's papers on
the topic, though the content may be paywalled.
35 years ago he challenged me with that problem when I stumbled into his
office to say hello; he was impressed when I quickly intuited the solution. >> Unfortunately the apogee of my math skill is now well in the past.... :-{
FWIW, he failed when I challenged him back with
"Find the fallacy in this 'proof' of the Four-Color Map Theorem" --
aaaa https://fabpedigree.com/james/color4.htm
Hmmm,a I like the "proof", because as presented it sounds plausible, and is "nearly" correct.a I
mean, all of the logic is correct apart from one (crucial) step!
I'm not sure if this is a puzzle, but just in case I'll put in some spoiler space...
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......r.
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...l....
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.r......
.s......
..p.....
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.i......
..l.....
...e....
....r...
.....s..
.....p..
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......r.
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[View with fixed font, of course!]
So here is a counter-example for the (fallacious) reasoning presented in your link
aa ---------------1---------------
aa |aaaaaaaaaaaa / \aaaaaaaaaaaa |
aa |aaaaaaaaaaa /aa \aaaaaaaaaaa |
aa |aaaaaaaaaa /aaaa \aaaaaaaaaa |
aa |aaaaaaaaa /aaaaaa \aaaaaaaaa |
aa |aaaaaaaa 2aaaaaaaa 2aaaaaaaa |
aa |aaaaaaa /|aaaaaaaa |\aaaaaaa |
aa |aaaaaa / |aaaaaaaa | \aaaaaa |
aa |aaaaa /a |aaaaaaaa |a \aaaaa |
aa |aaaa /aa |aaaaaaaa |aa \aaaa |
aa |aaa /aaa 3---------4aaa \aaa |
aa |aa /aaaaa \aaaaaa /aaaaa \aa |
aa |a /aaaaaaa \aaaa /aaaaaaa \a |
aa | /aaaaaaaaa \aa /aaaaaaaaa \ |
aa |/aaaaaaaaaaa \ /aaaaaaaaaaa \|
aa 4--------------1--------------3
aaa \aaaaaaaaaaaaaaaaaaaaaaaaaa /
aaaa \aaaaaaaaaaaaaaaaaaaaaaaa /
aaaaa \aaaaaaaaaaaaaaaaaaaaaa /
aaaaaa -----------------------
The numbers above represent the vertices (nodes) of the graph:
a 1 = green
a 2 = red
a 3 = blue
a 4 = yellow
The ascii lines are the edges.a The pentagon (1,2,3,4,2) starting at the top 1 is the main working
pentagon of the (fallacious) proof.a (The temporarily deleted node inside the pentagon is not shown.
aAlso the pentagon is rotated a bit from the proof, but it will make sense if you've followed the
proof.)
The presented proof claims (correctly) that we can change the leftmost 2 to a 4 (and proceeding with
flipping 2<-->4 throughout the associated (2,4)-connected-subgraph).a I.e. specifically in my graph
we can change the leftmost 2 to a 4 and the rightmost 4 to a 2, and the colouring is still valid.
[but the pentagon still has 4 colours!]
Alternatively, /starting from the original graph/ we can perform the similar flipping 2<-->3 of the
rightmost 2 and 3, and the colouring remains valid.a [but of course the pentagon still has 4 colours!]
The presented proof suggests that by doing /both/ of these flips together, the pentagon would
contain only 3 colours [correct], and the new colouring must still be valid [INCORRECT].a As seen in
my graph above, if we apply both flipping operations, the bottom 3 and 4 nodes both become 2 (red)
and are adjacent, so the colouring is broken.
The presented proof chooses to draw its 1-3 and 1-4 "fences" in a way which isolates the regions
they enclose, making the false conclusion sound plausible - it works if the fences are indeed
isolated as the presented proof draws them.a However, the two fences may cross each other at a
1-node, as shown in my example, and then the implied isolation of the to-be-flipped (2,4)- and
(2,3)- subgraphs no longer holds [..so they cannot both be performed together without breaking the
colouring].
In my example graph, the 1-3 and 1-4 fences cross each other at the bottom 1-node.
Mike.
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