Dominoes, when placed upright, remain that way. Sometimes, even if you tip them a little bit, they will go back to their upright position.

However, if you tip them too far, they will fall over.

After trying this with many different sized/shaped dominoes and some textbooks, I've noticed that this angle of "maximum tippage" varies for each one, depending on its dimensions.

Taller dominoes seem to have a lower maximum tippage angle. Dominoes with wider bases seem to have a higher maximum tippage angle.

What are other factors are involved?

Is there a way to compute this maximum tippage angle, given the height of the domino, and the width of its base, and any other factors that might be involved? If so, what is this relationship (mathematically)?

(Dominoes start acting weirdly when their weight/density is unevenly distributed, so for this question, assume that dominoes are of constant density)


2 Answers 2


To make it fall you need a torque. This torque is provided by the weight force acting on the center of mass of the object and by the offset between the center of mass and the edge of the object.

Imagine your domino standing upright then tilt it. You are moving the center of mass. When the center of mass (blue) is on the right of the edge (red) then you have a torque, represented by the triangle.

alt text

The torque is $\tau = m g D$ so to make it fall you need $D$ greater than zero.

If the domino (of uniform constant density) as base of width L the center of mass is located at L/2. For an height H, the center of mass is located at height H/2. That his other sketch:

alt text

Taking this one in the limit case where $D=0$ you obtain the last sketch.

Solving the trigonometry you obtain $\alpha = Atan \frac{L/2}{H/2}$.

The angle of the domino with respect to the "table" is $90-\alpha$ degrees.

  • $\begingroup$ I'm...not exactly sure what your diagram is trying to illustrate, sorry. What is the 90 degree angle representing? Where is $\alpha$, and where does it come from? $\endgroup$
    – Justin L.
    Commented Nov 5, 2010 at 0:45
  • 1
    $\begingroup$ Am I wrong in thinking that $Atan \frac{L/2}{H/2}$ is the same as $Atan \frac{L}{H}$ ? $\endgroup$
    – Justin L.
    Commented Nov 5, 2010 at 22:12

The direction that the domino falls is determined by the location its center of mass. It falls to the left or right depending on whether the center of mass is to the left or right of the bottom-most edge. If the center of mass is precisely above the edge, then it balances on that edge in an unstable equilibrium.


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