Maximal number of vectors with (pairwise) negative scalar product

Consider $\mathbb{R}^n$ equipped with the standard scalar product. Let $f(n)$ denote the maximal cardinality of a set of vectors in $\mathbb{R}^n$ with a pairwise negative scalar product.

What is $f(n)$?

Some thoughts: WLOG one can assume, that all vectors have length 1. This turns this problem in a space packaging problem in $S^{n-1}$: How many discs of radius $\pi/4$ can be placed inside $S^{n-1}$, so that the do not touch each other. If one allows touching, than the $2n$ vectors $\pm e_i$ (where $e_i$ denotes the standard basis) seem to be quite optimal.

The vertices of the regular $n$-simplex have pairwise negative scalar products, so we know, that $f(n)\ge n+1$.

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You are actually meaning their inner product. –  Xiaochuan Jan 12 '11 at 13:57
It depends whether you allow zero scalar product. $e_1 \cdot e_2=0$ so is not negative. –  Ross Millikan Jan 12 '11 at 14:02