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I have a flagged patio. The flags are odd shapes and sizes and no two are at
the same level. Anywhere I put a four-legged table, it rocks. So I thought
I'd build a three-legged table, but I'd like it to be rectangular.

Here's the question: where do the legs go for maximum stability, assuming they stick straight down underneath?

I think the simplest solution would be for a square top and I suspect that a solution for a rectangle would be proportional to the square. There might be two solutions, one with one leg in a corner and one with it half-way along one edge.

The math is completely beyond me. Is there a genius amongst us?

Nemo

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snipped-for-privacy@nusquam.rete wrote:

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

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

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.

NOT true. Consider the stability of a trestle table. It has only -two- legs, each of which spreads into two widely separated feet at the ground.

It doesn't matter. What counts is where the each foot is in relation to the edge of the table.

WORST CASE: the rotatable legs are perpendicular to the line of the two fixed legs. No worse than the three-legged design.

Best case: the rotatable legs are parallel to the line of the two fixed legs. Equivalent to a regular four-legged table.

Average case. stability is about 70.71% of the four-legged table.

FALSE TO FACT.

Regardless of the shape of the top,***or*** the number of legs, anything -inside-
the polygon described by the footprint of the legs is irrelevant to stability.
Any vertical pressure within that polygon cannot cause the table to tip, short
of induced structural failure, that is.

What matters is that which is__ _outside_ __said polygon -- pressure 'out there'__
_will_ __provide a 'tipping force', with the fulcrum being the nearest segment
of the polygon. Best stability is attained by__ _minimizing_ __the maximum
distance from the__ _edges_ __of the top to the polygon of the legs.

Nonsense.

Consider a table having a top six feet square, and four legs which form a one-foot square exactly centered under the top. The table is "stable" under your definition -- that is, it won't fall over in and of itself -- but it's obviously very susceptible to being tipped by even a modest weight. Probably a single coffee cup at the edge would be sufficient.

Now consider a similar table, with the legs moved outward so that they form a***five***-foot square exactly centered under the top. The center of gravity has
not moved, yet this second table is clearly far more resistant to being tipped
over.

[...]

Sorry, but you're entirely mistaken, because you're considering the wrong problem.

You have completely and entirely correctly described what is necessary to ensure that a table will not fall over__ _on its own_.__

The problem under discussion, however, is how best to make a table that will provide the greatest resistance to tipping over when pressure is applied to the portion of the tabletop that overhangs the base. These are two entirely separate questions.

How much of the edge is outside the base is the***only*** thing that matters.
You're right that an object can't fall over if its center of gravity is over
its base. But you're ignoring the simple and obvious fact that the farther its
edges extend beyond the base, the easier it becomes to tip it far enough that
the center of gravity is no longer over the base.

Consider these two tables.

Table A is a square, 32 inches on a side, with four legs forming a square base 30 inches on a side. Obviously the tabletop overhangs the base by 1 inch all the way around.

Table B is also a square, 32***feet*** on a side, with four legs forming a square
base 30 ***inches*** on a side, and hence the tabletop overhangs the base on each
side by fourteen feet and change.

Same base. COG is in the same place. The distance from the COG to the edge of the base is the same.

According to you, the only thing that matters is the distance from the COG to the base -- which is the same in both cases -- and hence the tables are equally stable. Table B is no easier to tip over than Table A on your planet.

#### Site Timeline

- posted on June 23, 2010, 6:40 am

Here's the question: where do the legs go for maximum stability, assuming they stick straight down underneath?

I think the simplest solution would be for a square top and I suspect that a solution for a rectangle would be proportional to the square. There might be two solutions, one with one leg in a corner and one with it half-way along one edge.

The math is completely beyond me. Is there a genius amongst us?

Nemo

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- posted on June 23, 2010, 6:43 am

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

--

Uno

Uno

- posted on June 23, 2010, 10:31 am

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 1/8" - 1/4" difference in fork/foot lengths.

Sonny

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 1/8" - 1/4" difference in fork/foot lengths.

Sonny

- posted on June 23, 2010, 10:53 am

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.

*or there about: Since the leg rotates, you don't have to be exact (1/8" to 1/4" variance in my first post), hence allowing for adjusting, when rotated, any positioning on uneven surfaces.

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

Sonny

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.

*or there about: Since the leg rotates, you don't have to be exact (1/8" to 1/4" variance in my first post), hence allowing for adjusting, when rotated, any positioning on uneven surfaces.

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

Sonny

- posted on June 23, 2010, 10:20 am

On 23 Jun 2010 06:40:40 GMT, snipped-for-privacy@nusquam.rete wrote:

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.

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.

- posted on June 23, 2010, 8:10 pm

wrote:

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

Nonny

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

Nonny

--

On most days,

it's just not worth

On most days,

it's just not worth

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- posted on June 23, 2010, 1:15 pm

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

- posted on June 23, 2010, 4:07 pm

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.

- posted on June 23, 2010, 4:23 pm

On Jun 23, 12:07 pm, snipped-for-privacy@milmac.com (Doug Miller) wrote:

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

R

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

R

- posted on June 23, 2010, 4:28 pm

On Wed, 23 Jun 2010 09:23:26 -0700 (PDT), RicodJour

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

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

- posted on June 23, 2010, 4:25 pm

snipped-for-privacy@milmac.com (Doug Miller) wrote:

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).

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).

--

Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.

Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.

- posted on June 25, 2010, 12:48 am

NOT true. Consider the stability of a trestle table. It has only -two- legs, each of which spreads into two widely separated feet at the ground.

- posted on June 25, 2010, 1:53 am

snipped-for-privacy@host122.r-bonomi.com (Robert Bonomi) wrote:

But the two legs don't swivel at the top to allow them to conform to irregular surfaces. I.e., the contact points are always the same distance from the table above them, just as they would be if the contact points were at the end of separate legs.

But the two legs don't swivel at the top to allow them to conform to irregular surfaces. I.e., the contact points are always the same distance from the table above them, just as they would be if the contact points were at the end of separate legs.

--

Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.

Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.

- posted on June 26, 2010, 2:33 am

It doesn't matter. What counts is where the each foot is in relation to the edge of the table.

WORST CASE: the rotatable legs are perpendicular to the line of the two fixed legs. No worse than the three-legged design.

Best case: the rotatable legs are parallel to the line of the two fixed legs. Equivalent to a regular four-legged table.

Average case. stability is about 70.71% of the four-legged table.

- posted on June 26, 2010, 2:52 am

On Fri, 25 Jun 2010 21:33:10 -0500, snipped-for-privacy@host122.r-bonomi.com (Robert
Bonomi) wrote:

Actually, what matters is where each foot is in relation to the***center*** of
the table.

Actually, what matters is where each foot is in relation to the

- posted on June 26, 2010, 3:22 am

FALSE TO FACT.

Regardless of the shape of the top,

What matters is that which is

- posted on June 26, 2010, 5:07 am

On Fri, 25 Jun 2010 22:22:00 -0500, snipped-for-privacy@host122.r-bonomi.com (Robert
Bonomi) wrote:

Wrong.

Right. But it's the CENTER OF GRAVITY that must be above the base, which for a normalish table will be at the center of the table. Not that for a really weird table the CoG doesn't even have to be on the surface. The entire table can be outside the legs (think donut) and it can still be stable. The CoG***must*** be within the base for it to be stable.

Wrong. What matters is the***center*** of the table (assuming a "normal" shaped
table) must not be outside the base. It doesn't matter how much of the edge
it outside, as long as the center of gravity is inside.

The edges don't matter at all.***ALL*** of the mass acts as if it is located at
the center (of gravity).

Wrong.

Right. But it's the CENTER OF GRAVITY that must be above the base, which for a normalish table will be at the center of the table. Not that for a really weird table the CoG doesn't even have to be on the surface. The entire table can be outside the legs (think donut) and it can still be stable. The CoG

Wrong. What matters is the

The edges don't matter at all.

- posted on June 26, 2010, 4:29 pm

Nonsense.

Consider a table having a top six feet square, and four legs which form a one-foot square exactly centered under the top. The table is "stable" under your definition -- that is, it won't fall over in and of itself -- but it's obviously very susceptible to being tipped by even a modest weight. Probably a single coffee cup at the edge would be sufficient.

Now consider a similar table, with the legs moved outward so that they form a

[...]

Sorry, but you're entirely mistaken, because you're considering the wrong problem.

You have completely and entirely correctly described what is necessary to ensure that a table will not fall over

The problem under discussion, however, is how best to make a table that will provide the greatest resistance to tipping over when pressure is applied to the portion of the tabletop that overhangs the base. These are two entirely separate questions.

How much of the edge is outside the base is the

- posted on June 27, 2010, 4:42 am

On Sat, 26 Jun 2010 16:29:04 GMT, snipped-for-privacy@milmac.com (Doug Miller) wrote:

Nonsense yourself. Physics doesn't lie.

It is stable. However, the measure of stability, which I've stated***many***
times here, is the distance from the ***CENTER*** (of gravity) to the edges of the
base (defined as lines between the feet, whether these lines or the center are
in fact within the table, or not.

Do learn to read. The edges of the table have nothing to do with stability. The stability is defined only by the center of gravity and the base. Moving your legs changed the base.***Obviously*** it's more stable. The ***fact*** is that
the stability is defined by the COG and the base.

I don't know what universe you're in. That's the way physics works in this.

Nonsense.

...and that is measured by the distance between the point on the "floor" under the COG and the base line(s),***ONLY***.

Wrong. What matters is the distance form the COG to the base. Period!

It's not a matter of falling over by itself at all. The force it requires tip the table over is determined by the distance from the COG to the base edge as the moment arm and the mass. The edge of the table top has***nothing*** to do
with it.

Nonsense yourself. Physics doesn't lie.

It is stable. However, the measure of stability, which I've stated

Do learn to read. The edges of the table have nothing to do with stability. The stability is defined only by the center of gravity and the base. Moving your legs changed the base.

I don't know what universe you're in. That's the way physics works in this.

Nonsense.

...and that is measured by the distance between the point on the "floor" under the COG and the base line(s),

Wrong. What matters is the distance form the COG to the base. Period!

It's not a matter of falling over by itself at all. The force it requires tip the table over is determined by the distance from the COG to the base edge as the moment arm and the mass. The edge of the table top has

- posted on June 27, 2010, 12:41 pm

Consider these two tables.

Table A is a square, 32 inches on a side, with four legs forming a square base 30 inches on a side. Obviously the tabletop overhangs the base by 1 inch all the way around.

Table B is also a square, 32

Same base. COG is in the same place. The distance from the COG to the edge of the base is the same.

According to you, the only thing that matters is the distance from the COG to the base -- which is the same in both cases -- and hence the tables are equally stable. Table B is no easier to tip over than Table A on your planet.

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