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That the growth is asymptotically linear is clear. But if you keep the width $2R$ of the strip (over which the centers are chosen uniformly) fixed, then the growth speed $c_R$ does depend on is not strictly proportional to $R$. R^{-1}$. This is clear when thinking to the small width case: if$R<1$, then there is only one branch in the tree because two consecutive disks always touch. If$d$is the distance between their centers, then they will arrange with a height gap equal to$\sqrt{4-d^2}$. Because the distribution of$d$has density $$\frac{2R-d}{2R^2},$$ we find that the growth average speed $$\frac1n\sum_1^nd_j$$ tends to the expectation of$\sqrt{4-d^2}$: $$c_R=\int_0^{2R}\sqrt{4-x^2}\frac{2R-x}{2R^2}dx.$$ Calculus gives $$c_R=\frac2R\sin^{-1}R+2\sqrt{1-R^2}+\frac4{3R^2}\left((1-R^2)^{3/2}-1\right).$$ As expected,$c_R\rightarrow2$as$R\rightarrow0$. On the other hand,$c_1=\pi-\frac43$. Presumably, the OP is interested with$\gamma=\lim_{R\rightarrow+\infty}Rc_R$. 2 deleted 1 characters in body That the growth is asymptotically linear. But if you keep the width$2R$of the strip (over which the centers are chosen uniformly) fixed, then the growth speed$c_R$does depend on$R$. This is clear when thinking to the small width case: if$R<1$, then there is only one branch in the tree because two consecutive disks always touch. If$d$is the distance between their centers, then they will arrange with a height gap equal to$\sqrt{4-d^2}$. Because the distribution of$d$has density $$\frac{2R-d}{2R^2},$$ we find that the growth average speed $$\frac1n\sum_1^nd_j$$ tends to the expectation of$\sqrt{4-d^2}$: $$c_R=\int_0^{2R}\sqrt{4-x^2}\frac{2R-x}{2R^2}dx.$$ Calculus gives $$c_R=\frac2R\sin^{-1}R+2\sqrt{1-R^2}+\frac4{3R^2}\left((1-R^2)^{3/2}-1\right).$$ As expected,$c_R\rightarrow2$as$R\rightarrow0$. On the other hand,$c_1=\pi-\frac43$. Presumably, the OP is interested with$c_\infty=\lim_{R\rightarrow+\infty}c_R$. \gamma=\lim_{R\rightarrow+\infty}Rc_R$.
That the growth is asymptotically linear. But if you keep the width $2R$ of the strip (over which the centers are chosen uniformly) fixed, then the growth speed $c_R$ does depend on $R$. This is clear when thinking to the small width case: if $R<1$, then there is only one branch in the tree because two consecutive disks always touch. If $d$ is the distance between their centers, then they will arrange with a height gap equal to $\sqrt{4-d^2}$. Because the distribution of $d$ has density $$\frac{2R-d}{2R^2},$$ we find that the growth average speed $$\frac1n\sum_1^nd_j$$ tends to the expectation of $\sqrt{4-d^2}$: $$c_R=\int_0^{2R}\sqrt{4-x^2}\frac{2R-x}{2R^2}dx.$$ Calculus gives $$c_R=\frac2R\sin^{-1}R+2\sqrt{1-R^2}+\frac4{3R^2}\left((1-R^2)^{3/2}-1\right).$$ As expected, $c_R\rightarrow2$ as $R\rightarrow0$. On the other hand, $c_1=\pi-\frac43$.
Presumably, the OP is interested with $c_\infty=\lim_{R\rightarrow+\infty}c_R$.