What is the tower of fields a number must be in in order to be constructed by MARKED RULER and compass? I know that for unmarked ruler and compass a number (distance) is constructible from $\mathbb Q$ iff it lies in a finite tower of field extensions $\mathbb Q=K_{0}...K_{n}$, where $[K_{i} :K_{i+1}] =2$ for all $i$. And an angle $x$ is constructible iff $\cos x$ meets the same conditions. 
My question is, for marked ruler and compass, a number is constructible iff what? Must we merely impose the condition that $[K_{i} :K_{i+1}] = 2$ or $3$? Can you give an explanation and proof please?
Edit: To be more specific, the ruler is marked with only points that may be constructed with compass and unmarked straightedge, and points that may be constructed with ruler and compass. This may also be relevant: we only allow verging between lines, not lines and curves or curves and curves.
 A: The important question is: What kind of markings does your ruler have. Suppose, that you have the perfect ruler, which has a mark for each number in $ \Bbb{R} $. In this case you can construct each point in the plane. Because your ruler has each real number as a marking you can construct the real line - just lie your ruler through $ 0 $ and $ 1 $ than you can make a point for each real number. Therefore $ \Bbb{R} $ is constructible. Now we have $ [\Bbb{C} : \Bbb{R}] = 2 $. Now the rule you have pointed out does work for any extension $ K \subset \Bbb{C} $ of $ \Bbb{Q}$: 
A point $ x $ of $ \Bbb{C} $ is constructible out of $ K $, iff $ x $ lies in an iterated quadratic extension of $ K $
Thus also $  \Bbb{C} $ is constructible. Which just means, that you can construct every point in the plane.
Remark: You don't really have to worry about marking off uncountably many points of your ruler. Because if you want to construct an explicit point it is easy to see, that you only have to use finitely many marked off points.
A: It depends on how it is marked, but a single notch does not not really help, so the least number of notches that's interesting is two. Already Archimedes proved that one can do trisection with that. If we only use it to do trisection than $[K_{i} :K_{i+1}] = 2, 3$ is right. This is proved by Gleason in Angle Trisection, the Heptagon, and the Triskaidecagon. This exactly covers what ancient Greeks called "solid constructions", i.e. those doable by using conic sections. As a result, the regular heptagon becomes constructible (although Archimedes's construction of it is based on a different kind of neusis). But the twice-notched straightedge is more powerful. Here is Theorem 5.1 from Baragar's Constructions Using a Compass and Twice-Notched Straightedge:

Suppose $α\in\mathbb{C}$ is constructible using a compass and twice-notched straightedge. Then $α$ belongs to a field $K$ that lies in a tower of fields 
  $\mathbb{Q}=K_0⊂K_1⊂K_2⊂···⊂K_n=K$ for which the index $[K_j:K_{j−1}]$ at each step is $2,3,5$, or $6$. In particular, if $N=[K:\mathbb{Q}]$, then the only primes dividing $N$ are $2,3$, and $5$.

This has consequences for constructing regular polygons as well, not even twice-notched straightedge (and compass) can construct the regular $23$-gon. As of Baragar's writing in 2002, it was not known if the regular $11$-gon can be so constructed, but in 2014 this construction was found by Benjamin and Snyder.  For a more general discussion of neusis see Lyle Ramshaw's answer on Math Overflow.
