A math challenge

Why wouldn't you vary the length of the legs?

As a starting point, I'd suggest making the rectangle not too long (so the ratio of width to length isn't too great - maybe 2:1, max) and not too narrow. Then just place two of the legs at corners of the same long side with the third one in the middle of the opposite side. It'll be "good enough" if not mathematically, precisely optimal. However the advantage of this simple approach is that you'll get your table, rather than be frozen at the design stage trying to extract the last .001% of stability.

You can have 3 legs with 4 feet. Two legs on one end would be attached in a permanent position, as normal. On the other end, one leg with a fork, ie., two feet. This leg is able to be rotated in its "socket" or attachment place.

Either 1) the feet (forks) should be of different lengths and the leg be perfectly verticle or 2) the leg be slightly off verticle and the feet equal lengths.

To stabilize the table on any uneven surface, rotate the forked leg until both feet touch.

This also works if there's one rotating leg on each end, each with forks (or feet), ie., 4 feet total. In sample 1 above, the longer forks, of each leg, should be catacorner from each other.

The levelness of the tabletop won't be significantly off because of a

Sonny

With reference to my post above, for you artistic folks:

The fork/foot lengths can be much greater difference when each fork's/ foot's angle, off the leg, differs also. Each fork's different angle, off the leg, will dictate how long (*or there about) each fork/foot will need to be.

The artistic look of your design can be had when the surface (floor) is level, too, not just for uneven surfaces, LOL.

Sonny

I don't think you have a math problem. The way I see it, if you were to build a square or rectangular table with three straight legs two of the corners would be supported, and the other two would not. If someone leans too heavily, or puts something heavy on the table it may tip. My solution would be to make a rectangular top. Place two straight legs on the narrow end as you normally would. On the other end I would use a " V " shaped leg going from the corners to the center spot on the floor.

Bea

Not so much math as physics. For an object to stand without falling over, its center of gravity must be above its base. The further inside the base the COG is the more stable it will be (you have to tip it far enough to get the COG outside the base).

For your table, assume the mass is in the top (ignore legs). If it's large compared to its thickness, ignore the thickness. The COG is then pretty much the center of the top. The problem then is to place the legs so this point is the furthest from the lines connecting the legs (the "base"). In the case of a square, this is easy (pick two corners and the middle of the opposing side). For a rectangle, I think it becomes clear if the rectangle is exaggerated; two adjacent corners on a long side and the opposite center.

Substituting an inherently less stable form for a slightly wobbly table is a poor trade-off. All you need is to have one leg of the four adjustable.

An easier solution should the three leg choice be preferable is to forgo the square/rectanble table and make it round.

Maybe a matchbook or a rolled up napkin under one leg.

It works at some of the better restaurants I have been in.

Seriously, this group is great, there is no question that can't be asked and get realistic and practical answers.

Larry C

That isn't going to make it any more stable -- it still has only one point of support at that end, and it's still in the same place.

Yes, but it fulfills the new picnic table seismic code requirement for lateral bracing.

Correct. Another "solution" posted here has a similar problem. If you let two legs come together and attach at a single pivot point, they might as well be one leg, as far as stability goes (assuming all the mass is above the pivot point).

Must be a California brand picnic table that's going to be used on a volatile fault line.

Good explanation. Thanks. I particularly like the lines between support points, which are obviously (well, obvious once you suggested them!) directly above the axes of rotation if the table were to tip.

The simpliest solution is to make a small wedge of the wood of your choice and slip it under the leg that appears to be the shortest when the table is standing on the flagstore. It will make it solid. If the table is moved a different leg will appear to be the shortest, so use the wedge on that leg.

Al

alexy wrote: ...

The problem w/ square/rectangular top on three lets still is, of course, that the lever arm for applying a tipping force is longer from the corners at the one-legged end. Minimize the normal (perpendicular) distance form the corners to the lines between the support points; the greatest of those is the highest moment arm and the point most prone to tip. The circular top has no "longest" length in any preferred direction; the maximum is the same in each of symmetric views.

Just ottomh w/o actually doing the geometry, seems to me the disparity is greatest for the square and gradually decreases as the L/W ratio increases for the rectangle. I'd look at the size required/wanted for the purpose and even if don't want round, potentially look at cutting a

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If the flags don't have sharp discontinuities where they meet (i.e. if the surface is uneven but not abrupt) then you can rotate the four-leg table into an orientation where all four legs are on the ground.

Alternately, the table can be made to flex slightly to take up the irregularity; this means your table top cannot be glass or stone... The 'adjustable leg' solution might be simplified with a good McPherson strut.

Make two triangular tables that can be arranged in the size rectangle you want. preserves balance and stability and gives you the shape you want.

basilisk

I agree with this. There's no rule or benefit from the triangle formed by the three legs being equilateral.

Nonny