4 ms·

I always had trouble envisioning this. I think part of the reason is the "angle x", which, according to the explanation and video on this page, is vector-valued

I always had trouble envisioning this. I think part of the reason is the "angle x", which, according to the explanation and video on this page, is vector-valued, not scalar. IVT applies to scalar functions, is there an equivalent for a function with a vector domain? The table does not rotate around its own axis, but rather it rotates in "such a way that three legs stay on the surface", i.e. moves around in 3d with a surface-contact constraint, which seems like a motion with more than one degree of freedom to me. Is such a rotation always possible? Is the motion somehow effectively 1D? These questions don't seem to have "obvious" answers to me, and they're only addressed as "assumptions" on this page.

Cerium 2y ago
A three legged table will always sit stably on an uneven surface. Adding a forth table introduces wobble. To remove that wobble, we need to find a location where the fourth leg is exactly the distance between the table top and the ground under the table top. As we rotate the table we know that in some locations a particular leg will be too short (wobble when we press in that location) or become too tall (source of wobble for another leg). As long as the ground is continuous and does not have any cracks to introduce discontinuity, we know that there must be a location that the leg is the exact length. By the intermediate value theorem the length cannot go from too short to too long without a solution in between. I first ran across the Wobbly Table when my wife was studying the Borsuk-Ulam theorem. It is fun since you can effectively solve wobbly tables on many patios.
- ziofill 2y ago
  Maybe I am not understanding something: what if the surface is a plane? The wobble can’t go away in that case right?
  - namibj 2y ago
    I think it assumes a symmetric table on an uneven floor.
  - akubera 2y ago
    In terms of the math: the table legs are assumed to be equal length, and the wobble is caused by variations of the surface. Specifically the feet of the table are in the same plane. So you could rotate your mathematical table until all feet are secure on the plane, then cut the legs to make the top flat again (legs will not be same length, but top and bottom remain planes). As for @Cerium's real-life usage, you have possibility of uneven legs and uneven floor (and discontinuities, like a raised floorboard) so it's obviously not guaranteed, but if the floor is warped and smooth enough, you can try. [EDIT]: Changed wording
  - Cerium 2y ago
    Yes, the math assumes the table is perfect and the floor uneven. In my experience this is frequently true with outdoor patio furniture - think decent glass table on a concrete patio.
- itohihiyt 2y ago
  > A three legged table will always sit stably on an uneven surface. But the table surface might not be level, if my thinking is correct.
  - kqr 2y ago
    That cannot be guaranteed of any table with equal-length legs: imagine the ground as a perfect slope!
    - dark-star 2y ago
      even then the table will stit stably on the slope. The tabletop will not be level with regards to the horizon/gravity but that is not what the theorem requires. It only requires stability (i.e. no "wobbling") Yes this might not be entirely practical if you try it on one of the famous sloped roads in San Francisco but for most other parts of the world (restaurants, cobblestone sidewalks, etc.) it will be good enough from a practical standpoint :)
- lloeki 2y ago
  > It is fun since you can effectively solve wobbly tables on many patios. I use it all the time to secure stepladders into stability before climbing them up.
  - deleted 2y ago
    [deleted]

Cerium 2y ago

A three legged table will always sit stably on an uneven surface. Adding a forth table introduces wobble. To remove that wobble, we need to find a location where the fourth leg is exactly the distance between the table top and the ground under the table top. As we rotate the table we know that in some locations a particular leg will be too short (wobble when we press in that location) or become too tall (source of wobble for another leg). As long as the ground is continuous and does not have any cracks to introduce discontinuity, we know that there must be a location that the leg is the exact length. By the intermediate value theorem the length cannot go from too short to too long without a solution in between. I first ran across the Wobbly Table when my wife was studying the Borsuk-Ulam theorem. It is fun since you can effectively solve wobbly tables on many patios.

ziofill 2y ago
Maybe I am not understanding something: what if the surface is a plane? The wobble can’t go away in that case right?
- namibj 2y ago
  I think it assumes a symmetric table on an uneven floor.
- akubera 2y ago
  In terms of the math: the table legs are assumed to be equal length, and the wobble is caused by variations of the surface. Specifically the feet of the table are in the same plane. So you could rotate your mathematical table until all feet are secure on the plane, then cut the legs to make the top flat again (legs will not be same length, but top and bottom remain planes). As for @Cerium's real-life usage, you have possibility of uneven legs and uneven floor (and discontinuities, like a raised floorboard) so it's obviously not guaranteed, but if the floor is warped and smooth enough, you can try. [EDIT]: Changed wording
- Cerium 2y ago
  Yes, the math assumes the table is perfect and the floor uneven. In my experience this is frequently true with outdoor patio furniture - think decent glass table on a concrete patio.
itohihiyt 2y ago
> A three legged table will always sit stably on an uneven surface. But the table surface might not be level, if my thinking is correct.
- kqr 2y ago
  That cannot be guaranteed of any table with equal-length legs: imagine the ground as a perfect slope!
  - dark-star 2y ago
    even then the table will stit stably on the slope. The tabletop will not be level with regards to the horizon/gravity but that is not what the theorem requires. It only requires stability (i.e. no "wobbling") Yes this might not be entirely practical if you try it on one of the famous sloped roads in San Francisco but for most other parts of the world (restaurants, cobblestone sidewalks, etc.) it will be good enough from a practical standpoint :)
lloeki 2y ago
> It is fun since you can effectively solve wobbly tables on many patios. I use it all the time to secure stepladders into stability before climbing them up.
- deleted 2y ago
  [deleted]