 Dense-brush packing limit

A dense uniform polymer brush with $\varphi \simeq 1$ must have height $H\simeq N{v}_{0}s$ by polymer-brush volume fraction. The contour bound $H\le Na$ therefore requires $s\le a/v_0$. If $v_0=a^3$, the limit is $s{a}^2\le 1$; beyond it the putative incompressible branch demands chains longer than their polymer contour length. This is a failure of that microscopic packing model. A Kuhn length does not alone determine $v_0$.