Is volume of abstract polytope realisation bounded by length of edges? Suppose we have abstract polytope $F$ of dimension $d$ (that is the greatest rank facet has rank $n$). Such abstract object may have realisations in d-dimensional Euclidean space as polytopes $A_i(F)$, and in some cases such realisation may have properly defined d-dimensional volume $V_i$ (with various values, depending on realisation $i$ shape).
Every such realisation defines realisation of abstract polytope $F_d$ which certainly defines further realisations of all it's facets $F_k$ for all ranks $k$ from maximal one $d$ down to $0$ .
Suppose we restrict our attention to such realisations that polytopes may be bounded within a sphere of finite volume $1.$ So the maximal volume of realisations is no bigger than $1.$ Further assume, that if particular realisation has no defined $d$-dimensional volume (because of missing face perhaps, or being non-orientable particular kind), we would set $V=0$.
We will work with Hasse diagrams of abstract polytope, as described for example here. Important for us would be two facts:

*

*Abstract polytopes with isomorphic Hasse diagrams are isomorphic

*Hasse diagrams are ordered decreasing, from top to bottom by rank of facets.

Suppose we'll assign to the greatest d-facet a number $M_d=\max (V_i)$, equal to the maximum volume of all realisations given abstract polytope $F_d$ (which is clearly finite and no bigger than volume of bounding sphere, that is 1). This is trivial case.
Suppose we don't know maximum volume $M_d$ but instead we have knowledge about such numbers for all N facets of $d-1$ kind, and it is equal to $M_{(d-1)}^s$, where $s \in \{1,2,\ldots, N\}$. Notice that $M_{(d-1)}^s$ are just (d-1) dimensional volumes of all facets.  So we know $N$ numbers instead of one.
Q1: Is it possible to know from this information any bound on $M_d$ (that is is it possible to obtain any information on maximal volume among realisations $A_i(F)$)?
Let's further generalise: suppose we have no knowledge on any facets of rank bigger than n but we know all numbers $M_n^s$ where s numbers all n-dimensional facets of abstract polytope $F$.
Q2: Is it possible to know from this information any bound on $M_d$ (that is is it possible to obtain any information on maximal volume among realisations $A_i(F)$)?
In particular, has the question below known answer?
Q0: Is it possible to know any bounds on maximal volume of realisations of abstract polytope if we know the maximal amount of the length of all the edges (facets of rank $1$) among all its realisations?
 A: When $M_{d-1}^s$ is the facet volume of the $s$-th facet, then it would contribute the largest part to the unknown polytope volume $M_d$ if it would be assumed to be a $(d-1)$-dimensional hyperball volume and thus contribute to $M_d$ as the according cone. That is, the volume of this cone, $V_d^{\operatorname{cone}(s)}$, would provide an upper bound for the respective facet pyramid. (Here using the center of the given $d$-dimensional unit hyperball for the according tip.)
For that aim you'd have to use the volume formula of the general hyperball
$$V_{d-1}^\text{ball}(R)=\frac{\sqrt{\pi^{d-1}}}{\Gamma(\frac{d+1}2)} R^{d-1}$$
and equate that to $M_{d-1}^s$, then solve it for $R=R_s$, resulting in
$$R_s=\frac{\left(\Gamma(\frac{d+1}2)\ M_{d-1}^s\right)^{\frac1{d-1}}}{\sqrt{\pi}}$$
Now calculate the height of that cone according to Pythagoras to
$$H_s=\sqrt{1-R_s^2}=\sqrt{1-\frac{(\Gamma(\frac{d+1}2)\ M_{d-1}^s)^{\frac2{d-1}}}{\pi}}$$
Thus finally the searched for upper bound for the facet pyramid would be the volume of the according cone. That one could be given as
$$V_d^{\operatorname{cone}(s)}=\frac1{d-1}\ M_{d-1}^s\ H_s=\frac{M_{d-1}^s}{d-1}\ \sqrt{1-\frac{(\Gamma(\frac{d+1}2)\ M_{d-1}^s)^{\frac2{d-1}}}{\pi}}$$
Thus you get your searched for bound as
$$M_d \le \sum_s V_d^{\operatorname{cone}(s)}$$
--- rk
