Poincaré inequality on a dilated ball

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I know we have the following Poincaré inequality :

Theorem : Let $\Omega \in \mathbb{R}^d$ an open regular set. There exists a constant $C(\Omega,d)$ such as, for all $u \in H^1_0(B_0)$ :

$$||u||_{L^2(\Omega)} \leq C(\Omega,d) ||\nabla u||_{L^2(\Omega)^d}.$$

The constant $C(\Omega,d)$ is only linked to the geometry $\Omega$ and $d$.

My question

Let $B_0=B(0,1)$ a ball in $\mathbb{R}^d$ of radius $1$ and $B_r=B(0,r)$ a ball of radius $r$. I am looking for a result that links $C(B_0,d)$ and $C(B_r,d)$, more precisely, I am wondering if we could have an equality between those two Poincaré constants.

Have you ever heard of such a result ? Any help or references is welcome !

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Basically all you need is the multivariable chain rule for dilations.

For any $u \in H^{1}_{0}(B_{r})$ you can consider $v(x)=u(x/r)$ instead - note that $v \in H^{1}_{0}(B_{1})$.

Now check how the $L^2$-norm of the function and its gradient changes if we compute it for $v$ instead of $u$.

The constant will blow up as $r \rightarrow \infty$, since the Poincaré Inequality does not hold on $\mathbb{R}^d$.

For $d=1$ a counterexample is a sequence of piecewise linear functions $u_{n}$ such that $u_{n}(x)=1$ on the interval $[-n,n]$ and $u_{n}(x)=0$ for all $|x|>n+1$.

$||\nabla{u_{n}}||^2_{L^2} = 2$ for all $n \in \mathbb{N}$ but obviously $||u_{n}||^2_{L^2} \rightarrow \infty$ as $n \rightarrow \infty$.