Using a compass and straightedge, what is the shortest way to divide a line segment into $n$ equal parts?

Let starting segment is $AB$. As one can see from first link, starting condition is "segment-on-the-line". Anyway one can add $1$ line at the start to get this starting condition.

Consider odd $n$.

Let coordinates of starting points are: $A(-1,0)$, $B(0,0)$.

If point $C$ has coordinates $C(n,0)$, then (see Figure 1) coordinates of other points are: $D\left(\dfrac{1}{2n},\dfrac{\sqrt{4n^2-1}}{2n}\right)$; $\qquad$ $E(1,0)$; $\qquad$ $P\left(-1+\dfrac{1}{n},0\right)$.

Figure 1:

Consider even $n$.

Let coordinates of starting points are: $A(0,0)$, $B(1,0)$.

If point $C$ has coordinates $C\left(\dfrac{n}{2},0\right)$, then (see Figure 2) coordinates of other points are: $D\left(\dfrac{1}{n},\dfrac{\sqrt{n^2-1}}{n}\right)$; $\qquad$ $E\left(\dfrac{1}{n},-\dfrac{\sqrt{n^2-1}}{n}\right)$; $\qquad$ $P\left(\dfrac{1}{n},0\right)$.

Figure 2:

Main idea: For given $m$ to draw point $C(m,0)$ as fast as possible.

As I checked (up to $m=2048$), it is possible to draw point $C(m,0)$, applying

$$ 1+\lfloor \log_2 m \rfloor $$

steps, where $\lfloor \cdot \rfloor$ is floor rounding function.

According to described construction, upper bound of (total) steps is $$ 3+\lfloor \log_2 n \rfloor, \mbox{ if } n \mbox{ is odd} ; $$ $$ 2+\lfloor \log_2 n \rfloor, \mbox{ if } n \mbox{ is even}. $$

Upper bounds of steps (starting with $n=1$) are: $$ 0; ~~ 3, 4; ~~ 4, 5, 4, 5; ~~ 5,6,5,6,5,6,5,6; ~~ 6,7,6,7, ... $$

Examples:

$n=11$: build point $C(11,0)$ and follow figure 1 (total $6$ steps).

$n=12$: build point $C(6,0)$ and follow figure 2 (total $5$ steps).