For an overlapping layer of long flexible grafted polymers, concentration blobs of width $D\sim {\sigma }^{-1/2}$ contain $g\sim (D/a{)}^{5/3}$ statistical units under the three-dimensional Flory approximation. A chain of $N/g$ blobs has height $h\sim ND/g$, giving the displayed good solvent scaling. Numerical prefactors are not supplied by this argument; extension must remain below the polymer contour length.

In a laterally uniform dense polymer brush with approximately fixed polymer volume fraction $\varphi$, conservation of occupied volume gives $\varphi h=N{v}_0\sigma$, with repeat-unit volume $v_0$. At fixed solvent quality this gives $h\propto N\sigma$, unlike the good-solvent brush height. Isolated collapsed polymer globules and laterally segregated grafted layers need not obey this uniform-layer model.

End-grafted chains are approximately isolated when their lateral spacing ${\sigma }_g^{-1/2}$ exceeds their coil size. When the spacing becomes smaller, excluded-volume crowding favors extension normal to the substrate. The crossover criterion is a scaling estimate, not a sharp phase transition.

A scaling model of overlapping end-grafted polymers assigns each chain a column of lateral width $D\sim s^{-1/2}$ at grafting density $s$. Each concentration blob contains $g\sim (D/a{)}^{1/\nu }$ statistical segments, and stacking $N/g$ blobs gives polymer brush height $H\sim ND/g$. For assumed segment volume $a^3$, the mean volume fraction is $\varphi \sim N{a}^{3}s/H$. It assumes a long many-segment-blob brush and supplies scaling powers rather than a precise chain-end distribution.

In the Alexander-de Gennes brush model, a theta solvent gives $D\sim a{g}^{1/2}$. Stacking $N/g$ blobs then gives the displayed height, with mean polymer volume fraction $\varphi \sim (s{a}^2{)}^{1/2}$ for segment volume $a^3$. Cancellation of the two-segment excluded volume does not remove three-segment repulsion or make the crowded layer freely interpenetrable at no cost.

Conservation of material in a uniform polymer brush gives its mean volume fraction as $N{v}_{0}s/H$, with material volume $v_0$ per counted segment, chain length $N$, grafting density $s$ and height $H$. Identifying $v_0$ with $a^3$ is an extra coarse-graining assumption. The Alexander-de Gennes brush model then gives $\varphi \sim (s{a}^2{)}^{(3\nu -1)/(2\nu )}$, including powers $2/3$ in a good solvent and $1/2$ in a theta solvent.

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 $v_0$.

A line segment along which two faces of a polyhedron meet. A rectangular parallelepiped has twelve edges, four parallel to each of its three side directions.

A polyhedron with two congruent parallel polygonal bases whose corresponding vertices are joined by parallel edges. A right prism has edges between the bases perpendicular to their planes. The optical use of a transparent prism is a separate application of this geometric shape.

The Archimedean solid obtained by truncating an octahedron, with eight hexagonal faces and six square faces. In an fcc nanocrystal the analogous shape can expose both $\{111\}$ and $\{100\}$ facets; the relative facet areas need not equal those of the regular Archimedean solid.

A seven-vertex polyhedron consisting of a five-membered equatorial ring and two axial vertices, each joined to every ring vertex. Removing an equatorial vertex leaves the six-vertex pentagonal-pyramidal skeleton used for some nido clusters.

A ten-vertex cage formed by adding one cap over each square face of a square antiprism. Each axial cap is joined to four vertices in its nearer square; adjacent squares are rotated relative to each other. It is the usual idealized ten-vertex closo cluster skeleton.

The $n$th pentagonal number is $P_n=n(3n-1)/2$ for a positive integer $n$, giving $1,5,12,22,\dots$. A positive integer $x$ is pentagonal exactly when $24x+1$ is a perfect square and $(1+\sqrt{24x+1})/6$ is an integer, by the defining quadratic equation.

The $n$th hexagonal number is $H_n=n(2n-1)$ for a positive integer $n$. Since $H_n=T_{2n-1}$, every hexagonal number is also a triangular number.

The polygonal number $P_{7,n}=n(5n-3)/2$ for a positive integer $n$.

The polygonal number $P_{8,n}=n(3n-2)$ for a positive integer $n$.

The sum of the lengths of its sides of a polygon. For a triangle with side lengths $a,b,c$, the perimeter of a polygon is $a+b+c$.

The semiperimeter is half the perimeter. For a triangle with side lengths $a,b,c$, it is $s=(a+b+c)/2$.

Positive real number lengths $a_1,\dots ,a_k$, with $k\ge 3$, are the sides of a nondegenerate polygon exactly when $2{\max }_i{a}_i<{\sum }_i{a}_i$. Necessity follows from the triangle inequality: the longest side must be shorter than the path formed by the remaining sides.

The kernel of a polygon is the set of interior or boundary points from which every point of the filled simple polygon is visible: the entire joining line segment must stay inside the polygon. A convex polygon is its own kernel of a polygon; a sufficiently bent nonconvex polygon can have an empty kernel of a polygon. For counterclockwise boundary order, intersect the closed half-planes to the left of every directed side. This intersection is exactly the kernel of a polygon, so the kernel of a polygon is always convex. A point in it can serve as a fixed observer seeing the entire room represented by the polygon.