Dimensions of orbit spaces Let $G$ be a compact Lie group acting effectively on a compact, Hausdorff topological space $X$. I am looking for results of the type 
If $X$ is a ... and the action is ... then $\dim(X/G)\leq \dim(X)-\dim(G)$.
Here $\dim$ denotes the covering dimension.
For example, $X$ is a smooth manifold and $G$ acts smoothly and almost-freely (all isotropy groups finite).
Can anyone point me to a more general theorem of this type?
 A: I'm loathe to answer my own question, but...
Theorem IV.3.8 of Bredon's book Introduction to compact transformation groups (which sadly is on the page google books decided not to include!) seems satisfying.
Let $G$ be a compact Lie group acting locally smoothly on the manifold $M$, such that $M/G$ is connected. If $P$ is a principal orbit (an orbit of maximum dimension) then
$$\dim(M/G)=\dim(M)-\dim(P).$$
A: The following statement is not quite the inequality that you want, but it is for more general spaces than orbifolds. (Note that so-called "good" orbifolds are the quotients of locally linear actions on manifolds with finite stabilizers.)
COROLLARY 1.7.32 of R Palais' 1960 book, Classification of $G$-spaces, p41:

Let $G$ be a compact Lie group. Let X be a separable metric $G$-space.  Then $\dim(X/G) = \sup_{H\leqslant G} \dim(X_H/W_G H) - \dim(G/H)$.  In particular, $\dim(X/G) \leqslant \dim(X)$.

I adapted the statement slightly to be less terse.  Here, $W_G H := N_G H/H$ is the Weyl group of $H$ in $G$, $G_x := \{g \in G ~|~ gx=x\}$ is the isotropy group of $x$, and $X_H := \{x \in X ~|~ G_x = H \}$ is the $H$-stratum.
A: There is a finer approach to the dimension of the quotient when the action of $G$ on $M$ doesn't define a fibration, but has singular orbits. You'll find it here. Of course for the principal orbit you have your formula, but on singular orbits it gives something else. I didn't give a general formula (I should do it however) but I treated the examples $\Delta_n = {\bf R}^n/{\rm SO}(n)$. Topologically, for any $n$ these quotients are homeomorphic to the half line $[0,\infty[$, but not diffeologically, and if we have well (according to your formula) ${\rm dim}_x(\Delta_n) = 1$ for $x \neq 0$, we have however ${\rm dim}_0(\Delta_n) = n$; which proves by the way, that for 2 different quotients, if they are homeomorphic, they are not diffeomorphic. I'm not sure it will help you very much, because it seems that you are interested only in the principal orbits, but your question was too tempting to not underline this construction.
