Rocky table
 

Since Mally has gone off in a huff, I shall put this teaser to you.

A four legged table typically rocks when placed on an unlevel floor.

However, with two simple assumptions, you can rectify the
problem by rotating the table through an angle <90 degrees.

How come? and what are the assumptions?

David

[QUOTE=davieddy;101613]Since Mally has gone off in a huff, I shall put this teaser to you.

A four legged table typically rocks when placed on an unlevel floor.

However, with two simple assumptions, you can rectify the
problem by rotating the table through an angle <90 degrees.

How come? and what are the assumptions?

David[/QUOTE]

I can do better: ONE simple assumption:

The height of the floor at each point is uniformly random. [the interval
does not matter]

Round table, the four legs are inside the table top.

You tilt the table, it rests on three points only and thus does not rock anymore.

Neither of these were the answers I had in mind.
But at least I got some instantaneous response:smile:

[quote=S485122;101615]Round table, the four legs are inside the table top.

You tilt the table, it rests on three points only and thus does not rock anymore.[/quote]

Until you try to cut the steak you're eating.

[quote=R.D. Silverman;101614]I can do better: ONE simple assumption:

The height of the floor at each point is uniformly random. [the interval
does not matter][/quote]

I'm not sure I dined at that restaurant.

[QUOTE=R.D. Silverman;101614]I can do better: ONE simple assumption:

The height of the floor at each point is uniformly random. [the interval
does not matter][/QUOTE]

Is that you, Mr. Silverman?

Uniformly? random? interval? Can you explain what you mean please?

The only thing I can imagine you might be meaning is that the floor say of my office being discribed by the rectangle [0,3]x[0,5], for every point in that space, there is an identically and independently distributed (iid) (uniform) U[0,b] random variable, where b would be the end of the interval you are referring to.
That would be a hell of a floor, and it is not obvious (at least for me) why you should get the table fixed.

Yours, H.

When I first told my good friend (now a professor at Princeton)
of my discovery, he went into deep thoght for a minute
then replied in excitement "Yes. And we can make it rigorous".
(That was about 35 years ago).

He didn't go into the details though.

David

[QUOTE=hhh;101620]Is that you, Mr. Silverman?

Uniformly? random? interval? Can you explain what you mean please?

The only thing I can imagine you might be meaning is that the floor say of my office being discribed by the rectangle [0,3]x[0,5], for every point in that space, there is an identically and independently distributed (iid) (uniform) U[0,b] random variable, where b would be the end of the interval you are referring to.
That would be a hell of a floor, and it is not obvious (at least for me) why you should get the table fixed.

Yours, H.[/QUOTE]

The height of the floor is a uniform random variable taken over some interval
[a,b]. The values of a and b do not matter. Rotate the table using
one leg as a pivot. Any three legs must be co-planar. That there exists
*some* rotation where the 4th point is in the same plane follows from
e.g.: The Ham Sandwich Theorem, or Browder's Fixed Point Theorem

So, for the sake of understanding:

is it like I described, space $R^2$, for any $x\inR^2$ you take a U[0,1] r.v., all iid? This floor would be as I said highly discontinuous.

Or do you mean by "uniformly random" "arbitrary, but continuous". In this case, I can vaguely figure how it should work with what you cited (I think you meant [URL="http://mathworld.wolfram.com/BrouwerFixedPointTheorem.html"]Brouwer's fixed point theorem[/URL]).

[QUOTE=hhh;101631]So, for the sake of understanding:

is it like I described, space $R^2$, for any $x\inR^2$ you take a U[0,1] r.v., all iid? This floor would be as I said highly discontinuous.

Or do you mean by "uniformly random" "arbitrary, but continuous". In this case, I can vaguely figure how it should work with what you cited (I think you meant [URL="http://mathworld.wolfram.com/BrouwerFixedPointTheorem.html"]Brouwer's fixed point theorem[/URL]).[/QUOTE]

Highly discontinuous? Indeed!! It is nowhere continuous.

[QUOTE=R.D. Silverman;101637]Highly discontinuous? Indeed!! It is nowhere continuous.[/QUOTE]

That I know. What I still don't know if of which of the two propositions I gave or what else you were talking. H.

That the floor is continuous (no steps) is one of my assumptions.

[QUOTE=davieddy;101666]That the floor is continuous (no steps) is one of my assumptions.[/QUOTE]

No, it wasn't. Go re-read what you wrote.

[QUOTE=R.D. Silverman;101845]No, it wasn't. Go re-read what you wrote.[/QUOTE]

Bob,

I think that he means that "the floor is continuous" is one of the "two simple assumptions" that he used in his answer.

"However, with two simple assumptions, you can rectify the
problem by rotating the table through an angle <90 degrees.
How come? and what are the assumptions?"

I suspect that "The feet of the table are coplanar" is sufficient, but not necessary, to serve as the other assumption.

This was my other assumption (feet at corners of a square).
It is most definitely necessary though.

David

[QUOTE=Wacky;101856]I suspect that "The feet of the table are coplanar" is sufficient, but not necessary, to serve as the other assumption.[/QUOTE]

Actually, this assumption is not needed under my solution.

I can also replace "height is a u.r.v." with "height is a continuous function",
but the function will be nowhere differentiable. The height will be a
lacunary function. All that is required is that within a circle of radius epsilon
of each point, the height of the floor varies sufficiently. "sufficiently"
depends on how far the 4 feet of the table depart from a plane. If the
feet are coplanar, then I believe that the height varying by k * epsilon
for any k > 1 in the neighborhood of the point on which the 4th leg will
sit is sufficient to let one apply the Ham Sandwich Theorem.

However, while continuous, the floor will not be *smooth*

[quote=davieddy;101858]This was my other assumption (feet at corners of a square).
It is most definitely necessary though.

David[/quote]
Unless you place a minimum limit on the "unlevelness" of the floor.
Wacky, did you get necessary and sufficient the wrong way round?

[QUOTE=davieddy;101877]Unless you place a minimum limit on the "unlevelness" of the floor.
Wacky, did you get necessary and sufficient the wrong way round?[/QUOTE]

No, I don't think so. I believe that the requirement is that the floor be more uneven than the feet of the table.

If my assertion is correct, then coplanar feet is sufficient. However, if would not be necessary.

Uneven floor !
 

:sad:

All that 'higher maths' when all one has to do is place a wedge under the leg that's off the floor. Plain common sense!

Mally :coffee:

To be fair, I think it is only RDS who has introduced the higher maths.
I suppose your solution (if they provided a sufficiency of paper napkins)
would be at least as practical as rotating the table a bit.
I was in a restaurant whe I first had this insight.
David

Napkins !
 

[QUOTE=davieddy;101897]To be fair, I think it is only RDS who has introduced the higher maths.
I suppose your solution (if they provided a sufficiency of paper napkins)
would be at least as practical as rotating the table a bit.
I was in a restaurant whe I first had this insight.
David[/QUOTE]

:rolleyes:

Well David paper napkins if they got slightly wet the legs would sink and the rocking would start again. Surely the restaurant could provide damask napkins that would do the trick.

Hey Davy you are keeping me away from the Sri Lanka-India cricket match which is on live right now. Its a crucial match in the series and right now India is sweating to keep the score at a reasonable total which we could catch up with and beat.

RGDS,

Mally :coffee:

[QUOTE=Wacky;101856]I suspect that "The feet of the table are coplanar" is sufficient, but not necessary, to serve as the other assumption.[/QUOTE]

Actually, I will now retract this as being "sufficient". The area between the feet also has to be sufficiently concave to preclude any points other than the feet from touching the floor.

But I suspect that that is an assumption that David failed to consider necessary.

Although he didn't state it, I think that he expects the "four legged table" to be of a conventional configuration which includes a reasonable degree of symmetry and "feet" that extend sufficiently below their supports to provide the necessary concavity.

Isn't it "a bitch" to describe the problem without ANY loopholes for the lawyers and mathematicians? :)

(Note to xilman: REAL WORLD is implied.)

[QUOTE=R.D. Silverman;101869]However, while continuous, the floor will not be *smooth*[/QUOTE]

Bob,

For what factor base?

:)


OTOH, we all know that FLOOR is a step function.


(As in "you step on the floor"?)

Forget this thread- Let's revert to "four on the floor".

Richard

[quote=mfgoode;101903]:rolleyes:

Hey Davy you are keeping me away from the Sri Lanka-India cricket match which is on live right now. Its a crucial match in the series and right now India is sweating to keep the score at a reasonable total which we could catch up with and beat.

RGDS,

Mally :coffee:[/quote]

Oh dear. Your guys took one hell of a beating.

Time to state my argument.
Rotating the table through 90 degrees replaces one of the
diagonal pairs of feet with the other, so the rocking is on the
other pair of feet.
Somewhere in between, the rocking changes from one pair
of feet to the other. At this point, all four feet are grounded.

I have been applying this for 35 years with 100% success!
(With the exception of Wacky's concavity condition
or coplanarity failing).

David