The coherence bandwidth measures how much a channel is statisticallt flat given a fixed size window through which we watch it.
Imagine a transmitter and a receiver, the transmitted signal being \$x(t)\rightleftharpoons X(f)\$. The receiver ideally gets the same \$x(t)\$ at the antenna, but unfortunately there's the mighty channel \$c(t)\rightleftharpoons C(f)\$. The receiver will then get \$Y(f) =X(f)\ast C(f)\$.
That \$C(f)\$ is ideally 1, i.e. it's perfectly flat for each frequency (please note that \$C(f)\$ is a complex function, i.e. \$C(f) : \mathbb{R} \rightarrow \mathbb{C}\$). That's not what happens in real life. If we measure just \$|C(f)|\$ completaly disregarding the phase, that is fundamental in almost all modulation techniques, we find out that it's everything but flat.
Now just close the math-ish door and let all the engineering folks in. There's already one shouting "Hey, if you look close enough that \$C(f)\$ is flat indeed.". Well, he's right and that's where the coherence bandwidth kicks in: you have the channel transfer function, you look at it close enough, and that becomes flat. The coherence bandwidth tells you how much is close enough: if that's 1kHz, well you've got to magnify your \$x\$ axis to see only a 1kHz portion at a time. If that's 1MHz... Well, you guessed it.
So why is it a statistical parameter? Well, you can't certainly measure all the channels you want to transmit on. Some guys at IEEE one day decided "ok, if you are in a city with tall buildings you should expect a \$B_C\$ of this much, if you are on a flat desert \$B_C\$ would be that much", and so on, and various models were born.
When does the delay spread kick in? Well I see you quite grasp what it is, and as you say \$B_C\$ and \$D\$ are close friend. As you know \$D\$ measures how much delay we should expect between the direct (the most direct) path and the others. That number tells us how long before our signal gets compromised by itself. Well it appears that a nice rule of thumb (read: there's little to no physical meaning associated to the following formula) is \$B_C=\frac{1}{D}\$. And we love to know the \$B_C\$ of a channel because it tells us how much bandwidth our signal can use without using advanced techniques such as an equalizer.
Added after OP request
What is delay spread? Here is what I have in my notes:
$$\Delta\overline{\tau}=\sum_{l=1}^Np_l\tau_l$$
Where:
- N is the number of paths
- \$\tau_l\$ is the delay associated with the l-th path
- \$p_l\$ is the normalised power of the l-th path so that \$\sum p_l=1\$
The delay spread hence express something like the delay after what I expect most of the power is arrived at the receiver. Consider the situation with:
$$N=3\\ \mathbf{p}=[0.7, 0.2, 0.1] \\ \mathbf{\tau} = [1,2,5]ms$$
You get \$\Delta\overline{\tau}=1.6ms\$
While if:
$$N=4\\ \mathbf{p}=[0.3, 0.4, 0.2, 0.1] \\ \mathbf{\tau} = [1,2,4,7]ms$$
You get \$\Delta\overline{\tau}=2.6ms\$
As you can see the delay spread measures when most of the power will arrive, and it's quite useful to appropriately tune the receiving equalizer to get the most out of your signal.
Air lines still exist, with velocity factor very close to 1.0. These are AFAIK mainly used in old-fashioned VSWR measurements. The advantages are that the dielectric constant of air is fairly stable and well-known, and that you can insert a probe (a tiny antenna) into the middle of the transmission line without damaging the dielectric.
ePTFE (aka "Teflon foam") dielectric typically gives a velocity factor of about 0.85. These cables are, in my experience, used because they maintain low loss to fairly high frequencies and their phase delay is quite stable under variations of temperature and flexure, not specifically because of the high phase velocity. I've used them in test and measurement applications, and I imagine they're also used in things like radar and avionics.
I found a reference saying that "foam polystyrene" dielectric gives a velocity factor of 0.91, but I have no experience with such cables, and I don't know what applications they're favored in. In fact I couldn't (with 2 minutes googling) find any vendor actually selling them.
Best Answer
It's the cross sectional area of obstructions in the Fresnel Zone that matters.
Most of the signal power travels in the first FZ. If you obstruct 20 to 40% of that, the remaining 80 to 60% amounts to about 1 to 2dB loss. For 'most' link budgets, that would be acceptable. Rules of thumb is about what 'most' people would find acceptable.
The nice thing about wikipedia is that they tend to give references for claims made in the text. In the text where they mention 40% or 20% obstruction, they refer to this citation [1]
where you can find more detail on it.