Finding a stable placement of an irregular (non-convex) shape

note: this answer assumes your shape is a proper polygon.

For our purposes, we'll define an equilibrium position as one where the Center of Mass is directly above a point that is between the leftmost and rightmost ground-contact points of the object (assuming the ground is a flat surface perpendicular to the force of gravity). This will work in all cases, for all shapes.

Note that, this is actually the physical definition of rotational equilibrium, as a consequence of Newtonian Rotational Kinematics.

For a proper polygon, if we eliminate cases where they stand on a sole vertex, this definition is equivalent to a stable position.

So, if you have a straight downward gravity, first find the left-most and right-most parts of it that are touching the ground.

Then, calculate your Center of Mass. For a polygon with known vertices and uniform density, this problem is reduced to finding the Centroid (relevant section).

Afterwards, drop a line from your CoM; if the intersection of the CoM and the ground is between those two x values, it's at equilibrium.

If your leftmost point and rightmost point match (ie, in a round object), this will still hold; just remember to be careful with your floating point comparisms.

Note that this can also be used to measure "how stable" an object is -- this measure is the maximum y-distance the Center of Mass can move before it is no longer within the range of the two contact points.

EDIT: nifty diagram made hastily

So, how can you use this to find all the ways it can sit on a table? See:

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EDIT

The programmable approach

Instead of the computationally expensive task of rotating the shape, try this instead.

Your shape's representation in your program should probably have a list of all vertices.

Find the vertices of your shape's convex hull (basically, your shape, but with all concave vertices -- vertices that are "pushed in" -- eliminated).

Then Iterate through each of pair of adjacent vertices on your convex hull (ie, if I had vertices A, B, C, D, I'd iterate through AB, BC, CD, DA)

Do this test:

1. Draw a line A through the two vertices being tested
2. Draw a line perpendicular to A, going through CoM C.
3. Find the intersection of the two lines (simple algebra)
4. If the intersection's y value is in between the y value of the two vertices, it stable. If the y values are all equal, compare the x values.

That should do the trick.

Here is an example of the test being running on one pair of vertices:

If your shape is not represented by its vertices in your data structure, then you should try to convert them. If it's something like a circle or an ellipse, you may use heuristics to guess the answer (a circle has infinite equilibrium positions; an ellipse 4, albeit only two "stable" points). If it's a curved wobbly irregular shape, you're going to have to supply your data structure for me be able to help in a program-related way, instead of just providing case-by-case heuristics.

First find its center of mass (CM). A stable position is one in which the CM will be higher if you make a slight rotation. Now look at the hull, the smallest convex region that encloses the shape:

_{(source: walkytalky.net)}

If the hull is a polygon, then a stable position is one in which the shape is resting on one of the sides, and the CM is directly over that side (not necessarily over the midpoint of the side, just somewhere over it.

If the hull has curves (that is, if the shape has curves which touch the hull), they must be give special treatment. The shape will be stable when resting on a curved edge iff the CM is directly above the lowest point of the curve, and the radius of the curve at that point is greater than the height of the CM.

Examples:

1. A rectangle. The hull is simply the rectangle, and the CM is at the center. The shape is stable on each of the four sides.
2. A rectangle with the sides hollowed, but the corners still intact. The hull is still the original rectangle, and the CM is close to where it used to be. All four sides of the hull are still stable (that is, you can still rest the shape on any two corners).
3. A circle. The CM is in the center, the hull is the circle. There are no stable positions, since the radius of the curve is always equal to the height of the CM. Give it a slight touch, and it will roll.
4. An ellipse. The CM is at the center, the hull is the shape. Now there are two stable positions.
5. A semicircle. The CM is somewhere on the axis of symmetry, the hull is the shape. Two stable positions.
6. A narrow semicircular crescent. The hull is a semicircle, the CM is outside the shape (but inside the hull). Two stable positions.

_{(source: walkytalky.net)}

(The ellipse position marked with an X is unstable, because the curvature is smaller than the distance to the centre of mass.)