[bonanova]

It's said that milking stools have exactly three legs so they won't wobble when set on an uneven barn floor.

Not so for tables, since home floors are assumed to be planar.

Which gives rise to an interesting puzzle.

Assume you have a table whose four legs are the same length.

But the floor is warped. Three legs touch the floor; the 4th does not.

Is it always the case that for any mildly warped floor your table can be positioned so that all four legs contact the floor?

Prove your answer.

[Guest]

I don't know!

I can only say from experience that you can, but I cannot prove it...

[Guest]

just chop the legs of the table so that 3 are shorter than 1 of the legs!

[Guest]

yes - if u have enough weight or strength you can warp the table top, alternativly saw the table in half!!

[Guest]

Yes, it is always possible to position a four legged table on a mildly warped floor.

When the floor is normal, the 4 legs of the table rest on the floor as the plane of the floor and the table top are parallel ("ll" used henceforth for parallel).

When the floor is mildly warped and only 3 legs of the table touch the floor, it means that the floor is slightly inclined in one direction and hence not ll to the plane of the table. By rotating the table, there definitely exisit at least 2 positions where the table plane will become ll to the floor and all 4 legs of the table will touch the ground.

For a rectangular table, generally, when you rotate the table by 90 degrees (clockwise or counter clockwise), you should be able to get the table in a position such that two legs of the table are at a higher level than the other two. The table would tilt to an angle such that the plane of the table top is ll to the floor.

Edited June 26, 2009 by DeeGee

[Guest]

It's said that milking stools have exactly three legs so they won't wobble when set on an uneven barn floor.

Not so for tables, since home floors are assumed to be planar.

Which gives rise to an interesting puzzle.

Assume you have a table whose four legs are the same length.

But the floor is warped. Three legs touch the floor; the 4th does not.

Is it always the case that for any mildly warped floor your table can be positioned so that all four legs contact the floor?

Prove your answer.

I think no, it is not always possible.

There is some function f(x,y,z) that defines the floor.

There is a plane F such that at three points defined by the legs f'(x,y,z) = 0.

There is a set of polygons P such that each polygon in P is defined by the set of all points in f'(x,y,z) (all of the points on F where F is tangental to the floor function)

I no longer remember how to do it, But you need find a transform function g(x,y), if it exists such that the corners of the rectangle are rotated into some subset of P.

In layman's terms, there is a plane F that is defined where the legs touch the floor. There is some set of points P which is defined by all the points on F where F touches the floor. There are 4 points on F that define a rectangle. The distances between these points are the same distances that defined the leg points on the table. You need to find some transform function that rotates and/or translates each of these four points simultaneously such that at the end of the transformation, all of the four points are in P. There may or may not be some F where this is possible.

There are also issues with bounding (f(x,y,z) is bounded by 4 functions which define the walls), etc. It seems like a lot of calculus to figure out that you should just shim the fourth leg

[bonanova]

One vote in favor of being able to adjust the table so it does not wobble.

Comments added in the spoiler.

Yes, it is always possible to position a four legged table on a mildly warped floor.

When the floor is normal, the 4 legs of the table rest on the floor as the plane of the floor and the table top are parallel ("ll" used henceforth for parallel).

When the floor is mildly warped and only 3 legs of the table touch the floor, it means that the floor is slightly inclined in one direction and hence not ll to the plane of the table. By rotating the table, there definitely exisit at least 2 positions where the table plane will become ll to the floor and all 4 legs of the table will touch the ground.

An inclined floor can be planar; a warped floor is not.

The table can be ll to an inclined floor but not to a warped floor.

"There definitely exist 2 positions where ..." isn't quite a proof.

For a rectangular table, generally, when you rotate the table by 90 degrees (clockwise or counter clockwise), you should be able to get the table in a position such that two legs of the table are at a higher level than the other two. The table would tilt to an angle such that the plane of the table top is ll to the floor.

"... you should be able to" isn't quite a proof.

[bonanova]

One vote against being able to place the table so it does not wobble.

Comments added in spoiler.

I think no, it is not always possible.

There is some function f(x,y,z) that defines the floor.

There is a plane F such that at three points defined by the legs f'(x,y,z) = 0.

The legs define four points.

What is f'? Prime usually means derivative with respect to an independent variable. Which?

Does f' = 0 mean that the plane defined by "the three points" is horizontal?

There is a set of polygons P such that each polygon in P is defined by the set of all points in f'(x,y,z) (all of the points on F where F is tangental to the floor function)

This sounds like all the polygons in P are the same polygon.

Namely the polygon whose vertices are the set of all points in f'.

I no longer remember how to do it, But you need find a transform function g(x,y), if it exists such that the corners of the rectangle are rotated into some subset of P.

In layman's terms, there is a plane F that is defined where the legs touch the floor. There is some set of points P which is defined by all the points on F where F touches the floor. There are 4 points on F that define a rectangle. The distances between these points are the same distances that defined the leg points on the table. You need to find some transform function that rotates and/or translates each of these four points simultaneously such that at the end of the transformation, all of the four points are in P. There may or may not be some F where this is possible.

I think I see what you're saying, but it might not be what we need.

If the four legs can be rigidly rotated/translated so they all touch the floor, it's likely that at the four points of contact, the tangential planes will all be different. If we take this approach, we should think of contour lines where the plane F [of the four leg ends] intersects with the function f[x,y,z]. Then we need to find on [at least] one such contour line four points that form a rectangle or square congruent to the four ends of the legs.

We need a proof that such a contour exists [or not.] That's an ambitious approach.

But there is a much simpler proof.

There are also issues with bounding (f(x,y,z) is bounded by 4 functions which define the walls), etc. It seems like a lot of calculus to figure out that you should just shim the fourth leg

[Guest]

One vote against being able to place the table so it does not wobble.

Comments added in spoiler.

In my previous response I probably made too many assumptions:

1) the plane where the three legs currently touched the floor was the plane upon which you wished to target your table (i.e. Any plane that determined the target was required to be parallel to T)

2) The legs need to be stable on this plane (i.e. they all need to be orthogonal to the floor where they contact the floor)...

In addition, when I said the set of polygons, I was assuming that the function that defined the floor may have both peaks and valleys. This means that the set of polygons on F where F is tangental to f' is not necessarily one polygon

If we throw out those assumptions, and assume that the warp is continuous around some point...It's going to depend on where the warp is...and if you are allowed to move the table as well as rotate it as well as the bounds (i.e. walls) of the room

[Guest]

Consider the area of floor to be just smaller than double the table in width and length, so that one leg would always be in each quadrant in any possible position. If all four quadrants were at a different altitude, it would be impossible to position the table so that all four legs touch the floor.

Imagine the rectangle of space available. The top left quadrant would be 2mm higher than the top right, which would be 2mm higher than the bottom left, which would be 2mm higher than the bottom right. The rectangle formed by the legs of the table is larger than an individual quadrant and would be forced to have one leg in each quadrant. It is not possible for all four legs to touch the floor in this example.

[HoustonHokie]

It's said that milking stools have exactly three legs so they won't wobble when set on an uneven barn floor.

Not so for tables, since home floors are assumed to be planar.

Which gives rise to an interesting puzzle.

Assume you have a table whose four legs are the same length.

But the floor is warped. Three legs touch the floor; the 4th does not.

Is it always the case that for any mildly warped floor your table can be positioned so that all four legs contact the floor?

Prove your answer.

so what you're after is a proof of whether or not there always exists 4 coplanar points at set distances from one another on an irregular surface? Further, the points could be anywhere within the surface and at any angle relative to the assumed axis. I'd say it depends upon the degree of irregularity and the size of the surface.

Example: assume the floor is exactly the dimensions of the table legs. I think we'd all agree that for this special case, you can certainly create a mildly warped surface such that 3 legs contact on the same plane and the fourth does not. With no additional room to maneuver the table, a solution with 4 contacting legs is impossible.

The question is, can you improve on this special case by enlarging the floor?

I'd say yes, you can improve the likelihood of having all four table legs contact the floor, but you can't guarantee it. It's disproven in the one special case I mentioned, therefore it cannot happen in all cases.

[bonanova]

Spoiler for seeking an answer:

so what you're after is a proof of whether or not there always exists 4 coplanar points at set distances from one another on an irregular surface? Further, the points could be anywhere within the surface and at any angle relative to the assumed axis. I'd say it depends upon the degree of irregularity and the size of the surface.

Example: assume the floor is exactly the dimensions of the table legs. I think we'd all agree that for this special case, you can certainly create a mildly warped surface such that 3 legs contact on the same plane and the fourth does not. With no additional room to maneuver the table, a solution with 4 contacting legs is impossible.

The question is, can you improve on this special case by enlarging the floor?

I'd say yes, you can improve the likelihood of having all four table legs contact the floor, but you can't guarantee it. It's disproven in the one special case I mentioned, therefore it cannot happen in all cases.

OK.

The question is about topology, not boundaries.

Think of a normal table in a normal room.

But to make it somewhat rigorous ... assume that

1. The floor is infinite in extent.
2. The surface is continuous and differentiable.

[Guest]

My vote is that it should be possible for tables of 4 legs:

For a rectangular table

A----B

|    |

D----C

A plane is uniquely defined by 3 points, and so, assuming that the floor is not planar (where the solution is trivial), we can always position any three of the legs on the floor such that the fourth in the air. We are trying to show that there is a plane which intersects with the floor in four places.

We arbitrarily choose a position such that three legs (A, B and C) touch the floor and one (D) is in the air. Unless the floor is so warped that it comes into contact with the bottom of the table, and because the floor is continuous, there is also a position to which we can smoothly move, in which either A or C is in the air.

However, in order for either to leave the floor, D must be in contact with the floor, because otherwise the flat plane on which the table rests is not well defined, ie it is only resting on two legs! I would conclude that therefore there must a point such that A,B,C and D are all in contact with the floor.

Not mathematical enough for my liking, but I think it concludes that there is a solution....

Obviously, the reason why a three legged chair is better, is that it is easier to position it such that the stool is flat. Though all four legs are on the floor, the table is very unlikely in a horizontal position. In addition, the legs are certainly not necessarily square to the part of floor on which they stand.

[bonanova]

My vote is that it should be possible for tables of 4 legs:

For a rectangular table

A----B
| |
D----C

A plane is uniquely defined by 3 points, and so, assuming that the floor is not planar (where the solution is trivial), we can always position any three of the legs on the floor such that the fourth in the air. We are trying to show that there is a plane which intersects with the floor in four places.

We arbitrarily choose a position such that three legs (A, B and C) touch the floor and one (D) is in the air. Unless the floor is so warped that it comes into contact with the bottom of the table, and because the floor is continuous, there is also a position to which we can smoothly move, in which either A or C is in the air.

However, in order for either to leave the floor, D must be in contact with the floor, because otherwise the flat plane on which the table rests is not well defined, ie it is only resting on two legs! I would conclude that therefore there must a point such that A,B,C and D are all in contact with the floor.

Not mathematical enough for my liking, but I think it concludes that there is a solution....

Obviously, the reason why a three legged chair is better, is that it is easier to position it such that the stool is flat. Though all four legs are on the floor, the table is very unlikely in a horizontal position. In addition, the legs are certainly not necessarily square to the part of floor on which they stand.

Yes. That does it.

Of legs A, B, C and D, the first three touch while D does not.

If the table is rotated 90^o clockwise [assuming it's square] around an axis perpendicular to the plane of the table top,

we reach the position where either A or C [but not both] can touch the floor.

Let us instead rotate the table clockwise in a way that keeps legs A, B and C in contact with the floor.

We cannot continue this rotation for a complete 90^o because we have seen that after 90^o A and C cannot both touch the floor.

This can only be true because at some point D comes into contact with the floor, and after which either A or C breaks contact.

At this point, all four legs contact the floor.

The proof extends to rectangular tables by simply shortening or lengthening [say] the y dimension to make the table square.