In this answer to one of my questions about the general physics of "force spreading" of floor protection layers, @oliver pointed out that there is no simple model for this situation. Now I want to understand just another aspect of the problem. Consider rubber gym floor tiles which should in particular protect a (concrete) floor getting damaged by a falling weight. I noticed that most products have a density of $800$ or $900 \, \mathrm{kg}/\mathrm{m}^3$. But products advertised as specially designed for dropping very large weights, often seem to have a much higher density, like $1300$ to $1600 \, \mathrm{kg}/\mathrm{m}^3$. The higher density products also tend to be sold slightly thinner (e.g. $2\,\mathrm{cm}$ instead of $3\,\mathrm{cm}$).
This observation seems to call for the hypothesis that a higher density rubber (say at the same thickness) tile can the concrete floor save from larger weight drops (or larger falling heights).
At first, this seems to be a bit counterintuitive to me since a larger density should lead to a "harder" material which should shorten the "deacceleration distance" and thus increase the maximum force acting on the weight during deacceleration.
If it is true, that the higher density rubber protects the subfloor better, this would imply that the "force spreading" (compare my other question linked above) effect should "overcome" the shorter deacceleration distance, such that the max. pressure on the concrete floor would be lower than in the case of the lower density material.
So is there a simple physical model to explain this qualitatively or semi-quantitatively? Maybe a clever geometric argument considering the more densely packed molecules of the rubber and how they "transmit" the impact force. Beeing more dense leads to a larger "spreading angle" since the molecules are closer together or something like this. But I don't get the details.
