Let $K$ be a polytope in $\mathbb R^d$, blow it up by a factor $\lambda>0$. For a unit vector $u \in \mathbb S^{d-1}$, $\lambda K$ has 2 support hyperplanes $H_1$ and $H_2$ with corresponding outward normal vectors $u$ and $-u$. I only take into account $u$'s such that $H_1$ and $H_2$ have "good" position, that says they are not parallel to any face of $K$, or equivalently, they contain only one vertex of $\lambda K$.
Consider another parallel hyperplane $H_t$ such that the distance between it and $H_1$ is $t$ (of course $t$ is smaller than the width of $\lambda K$ in direction $u$). So $H_1$ and $H_t$ define a cap of $\lambda K$.
My question is: Is there a chance that we can estimate the volume of that cap, in terms of $u, \lambda$ and $t$?
The easiest case is when $t$ is smaller than the distance from the nearest vertex of $\lambda K$ to $H_1$, then the volume is $\frac{1}{d}at^d$, where $a$ is a constant depends on $u$.
I mean, in general, it is very hard to compute volume of similarly defined caps of $K$. But if we blow up $K$, some factors can be ignored when $\lambda$ is large enough, so I hope there is a bound on the magnitude of the volume of the cap.
