And let's add Table C, just like Table A -- but the legs are forty feet long.
And Table D, just like Table A, but the legs are only an inch long.
All four tables have the same lateral distance from the COG to the edge of the
base. That's your phrase, Keith. And according to you, that distance is all
that matters. Thus, according to you, all four are equally stable.
Of course you did. You said it before, and you said it again just now: "the
stability of a table is relative only to its base and COG". Hence two tables
having identical bases, with their centers of gravity in the same place, must
be (according to you) equally stable -- regardless of any other factors such
as the size of the tabletop.
You're correct as far as your definition of the problem. But there's the
ancillary problem of a single fly alighting on the nether edge of a table
whose legs are WAY inside the perimeter.
Come to think on it, I've never seen a dining room table held up by a single
1" pole attached at its center of gravity.
Only if the table surface is unloaded outside the supporting line(s).
It is, as at least two others have already said, also dependent on the
lever arm from the edge to the line of the legs--if that is very long,
it doesn't take much force irrespective of where the legs are wrt the
Simple 1D drawing of table on end or side view (nonproportional font)...
| | |
\ / COG \/
Clearly the amount of force to tip is much less at the "elbows on table"
position owing to the larger distance between that applied force point
and the resistance at leg A attach point as opposed to a butt resting on
the edge closest to leg B.
As noted upthread, the most tipping stability assuming the legs are far
enough apart to avoid the weight of the table alone being a significant
factor is to minimize the maximum normal distance along the lines of
support. That keeps the largest lever arm at the smallest it can be for
any given size of table top.
Let me throw in another factor here. Since this is a real table, the top and
legs will have weight. You really have to factor in the mass of the table on
either side of the tipping line. This means that just finding the shortest
fulcrum doesn't necessarily give the greatest stability. A 1' fulcrum where
the tipping line is parallel to an edge will require a lesser tipping force
than if the line cut across a diagonal.
To make the answer as general as possible, I've been considering a square
slab supported on three points. That way, the weight or length of the legs
don't complicate matters. (But I _still_ haven't found an answer!)
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Then it becomes "part of" the table top. If you want to get pissy, you don't
mention the coefficient of friction between your load and the top (it may fall
of before the table tips).
Obviously I'm simplifying this a lot, for instance ignoring that the table top
is three dimensional and ignoring the mass of the legs. These assumptions
were made above.
You don't think a mass at the end of a lever is going to change the COG?
The distance between my A' and C' defines how much resistance to "Elbows"
there is. Yes, A' to E' is a lever arm and that must be countered by the
table's mass * A'-C'.
I disagree. A little thought experiment here. A table 10' square.
Along one edge are traditional fixed legs, 4' apart. on the opposite
edge is one trestle leg, free to pivot where it meets the table, but
spreading to feet 20' apart. The table will be VERY easy to tip with
pressure 5' away from the trestle leg, as it will just rotate around
the attachment point. Attach that leg rigidly, and it will be VERY
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