Suppose $X,Z$ und $T$ are instantaneous observers and we know the velocity of $X$ with respect to $Z$ and the velocity of $Z$ with respect to $T$. What is the velocity of $X$ with respect to $T$?
$X=\a(Z+uE)$ and $Z=\b(T+vF)$ where $E\perp Z$ and $F\perp T$ are unit vectors. The energy and the momentum of $X$ with respect to $T$ are given by
\begin{eqnarray*}
\g
&\colon=&-\la X,T\ra
=-\a\la Z+uE,T\ra
=-\a\la\b(T+vF)+uE,T\ra\\
&=&-\a\b\la T,T\ra-\a\la vF+uE,T\ra
=\a\b-\a u\la E,T\ra
\end{eqnarray*}
and
\begin{eqnarray*}
P
&\colon=&X+\la X,T\ra T
=X-\g T
=\a(Z+uE)-\g T\\
&=&\a(\b(T+vF)+uE)-\g T
=(\a\b-\g)T+\a\b vF+\a uE~.
\end{eqnarray*}
This looks strange at a first glance, for $P$ doesn't seem to be orthogonal to $T$. However a straightforward computation shows that it is in fact orthogonal to $T$. Therefore the velocity $V$ of $X$ with respect to $T$ is given by
$$
V=(\tfrac{\a\b}{\g}-1)T+\tfrac{\a\b v}{\g}F+\tfrac{\a u}{\g}uE~.
$$
Obviously, that's not an important formula; keep in mind that computing velocities is straightforward but somehow annoying!