0 like 0 dislike
0 like 0 dislike
[Owner Re-Post] Building a Poker Table

1 Answer

0 like 0 dislike
0 like 0 dislike
Your sketch failed to upload somehow. I'll assume it's a generic round table that essentially consists of a flat top part and a bottom part that touches the ground, and the shaft connecting them is effectively part of the bottom. First of all, better assume that a 150kg man can exert at least 450kg of weight for a few moments when actively trying, like putting his hands on the table and jumping. A poker table is probably going to be heavy enough that the tilt process is too slow for this to matter, the guy will stop exerting force before the table rotates too far - but safety margins are always a good idea.

As a simplification: Things tip over when their center of gravity is outside their support/base horizontally. When saying "distance to edge of base" I mean the minimum distance from the (height 0) center of the table to the convex hull of the points touching the ground (get a string taut around the legs on the floor, see where it's closest to the center, that's it. If you have four legs in a square, use 70% of their maximum horizontal distance to the center). Statically, someone adding their weight to one end of the table is equivalent to shifting the table's center of gravity. Only the horizontal shift matters here. You need to ensure that (distance to edge of top) \* 150kg / ((table weight) + 150kg) < (distance to edge of base), with a bit of margin. Or, after some algebra, maximum force is (table weight) \* (distance to edge of base)/((distance to edge of top) - (distance to edge of base)). E.g. if base is 0.2m and top radius is 1m, max force is (table weight) \* (0.2m)/(1m - 0.2m), or just 1/4th of the table weight.

The reason why it's a good idea to place the most weight at the bottom instead of the top is that if you tilt the table a bit, the center of gravity moves much more horizontally if it's higher to begin with. At the extreme, even the slightest tilt will place it outside the support area - think about an upside down broom. If you approximate the table as being made of two pieces, one base at height 0 and one top at the full table height, the maximum stable tilt angle (without extra weight pushing on it!) is arcsin((table weight) \* (distance to edge of base) / ((top weight) \* (table height))). For small angles you can just approximate that as (table weight) \* (distance to edge of base) / ((top weight) \* (table height)) without the arcsin, giving you the angle in radians with good accuracy.

You can combine the last two criteria to get something like "can take 150kg weight downwards at the edge even if tilted 5 degrees in that direction".
That looks like 0.09 \* (table height) \* (top weight) / (table weight) + (distance to edge of top) \* 150kg / ((table weight) + 150kg) < (distance to edge of base).

Other than that, even if the table is stable when tilted, you don't want to be able to tilt it easily. This time, if you want it to take a 150kg horizontal pushing force without rotating (perhaps slipping on the floor, though), you want to ensure that (table height) \* 150kg < (distance to edge of base) \* (table weight). Or rearranging again, max horizontal force is (table weight) \* (distance to edge of base) / (table height). Also keep in mind that as the force approaches this limit and the table is about to start tilting, it acts like a lever and the force gets concentrated on a single point, and this could snap a table leg if they're the wrong shape. I can't see the sketch so I can't tell if it's a potential problem or not.

Finally, lifting the table up can still tip it over more easily than people pushing down on it, but that's basically impossible to prevent.


Max forces so the table stays where it is (or slips, but doesn't rotate) are (table weight) \* (distance to edge of base)/((distance to edge of top) - (distance to edge of base)) vertically, (table weight) \* (distance to edge of base) / (table height) horizontally. If you have 4 legs touching the ground at distance x from the center, use 0.7x for (distance to edge of base). Subtract 0.09 \* (table height) \* (top weight) / (table weight) from that distance before calculating with it to add a 5 degree tilt safety margin, if you know how much the top part weighs. Worst case would be 0.09 \* (table height). (distance to edge of top) is the top radius.

No related questions found

33.4k questions

135k answers


33.7k users

OhhAskMe is a math solving hub where high school and university students ask and answer loads of math questions, discuss the latest in math, and share their knowledge. It’s 100% free!