# Is the propagation speed of falling dominos scale-independent?

Suppose we scale all the linear space dimensions of falling dominos, like their thickness, width, height, distance (which implies that the volume, which is not a linear dimension, is not scaled by the same factor, so neither is the mass-density) from each other, as well as their mass with the same factor, will the propagation speed of the parallel behind each other placed, falling dominos (without loss of energy to the surroundings), scale with the same factor?

Because the mass density decreases, relatively less potential energy is transferred to the next stone, which therefore reaches the next stone in more time. Which means that the propagation speed diminishes when you scale all the linear dimensions as well as the mass with the same factor (greater than 1; when the factor is less than one the propagation speed increases).

Is my reasoning right?

• If gravity stays the same, then anyone who has ever balanced a stick upright on their palm can tell you the answer. Hint: broomsticks are easier than meter-sticks. – dmckee Feb 5 '17 at 23:11
• @dmckee-I see! A very long stick is much easier to keep upright than a little one. You can see the stick as an inverted pendulum, which you counterforce by your hand movements. So the propagation speed does indeed decrease. – descheleschilder Feb 6 '17 at 0:30
• @dmckee That sounds convincing but boy oh boy would I like to see a complete analysis. – DanielSank Feb 6 '17 at 6:48
• @DanielSank- I think that if we scaled all the linear dimensions with a factor $x$ and the mass (because the volume of the stones increases by $x^3$) by a factor $x^3$, the propagation speed stays the same, so in all other cases it won't. If we also made $g$, the acceleration caused by gravity, $x$ times as large, then also the propagation speed would increase by a factor $x$. – descheleschilder Feb 6 '17 at 8:41
• heavy objects and light objects fall at the same speed. Horizontal acceleration is in proportion to mass, but so is the force from the preceeding domino. Neglecting air resistance the speed will be the same whatever density your dominos. – JMLCarter Feb 6 '17 at 11:00

I think this question is answered in Book domino propagation speed?

There in a comment to the answer by David ben Moshe, Conrad Turner cites a 2008 paper by J M J van Leeuwen. That author found that the speed has the asymptotic form of equation #77 : $$v=\sqrt{gh}Q(h,d,s)$$
where $h$ is height, $d$ is width and $s$ is separation of dominos, and $Q$ is a transcendental function which depends on dimensionless ratios of the parameters $h, d, s$.

As JML Carter notes in the comments above, light and heavy objects fall at the same speed, so the propagation speed is independent of density.

The function $Q$ has scale-invariant arguments, but the factor $\sqrt{gh}$ is not scale-invariant.

• The formula given by David ben Moshe, cited by Efthimiou and Johnson, dates from 1983; it simplifies analysis by assuming the dominoes have zero thickness $d\ll s,h$. Function $G$ is different from the function I have labelled $Q$. Function $Q$ relates to the van Leeuwen article which dates from 2008 and includes domino thickness $d$ in the analysis - that is why the function $Q$ is different from function $G$. For simplicity I have lumped together in $Q$ 3 functions in van Leeuwen's eqn #77. Judging from eqn #15 it seems that function $< \omega >$ is dimensionless. – sammy gerbil Feb 7 '17 at 18:21
• @sammygerbil-So it should be that $Q$ is a function of h, d, s and t (thickness): $G(h, d, s,t)$? – descheleschilder Feb 7 '17 at 18:27