Under which conditions is full rated power instead of METO power required for takeoff (DC-3/C-47 with Pratt&Whitney 1830 twin wasp engines)?
Let's consider your case of a 2300 ft (700 m) takeoff roll. Your link lists the procedure as reducing power upon reaching 90 kt (46 m/s). To make things easier we can assume constant acceleration. That won't be true in real life but it will give us an approximation.

Two basic equations for acceleration, relating distance $s$, acceleration $a$, time $t$, and velocity $v$:

$s=\frac{1}{2}at^2$

$a=\frac{\Delta v}{t}$

Substituting our known values in, we get:

$700\ \mathrm m=\frac{1}{2}at^2$

$a=\frac{46\ \mathrm{m/s}}{t}$

Soving for $t$, we get about 30 seconds (and $a=1.5\ \mathrm{m/s^2}$). Even accounting for acceleration reducing as speed increases, that suggests there's plenty of room to reach 90 knots in less than one minute.

If we look at what takeoff roll would result if we need 60 seconds to reach 90 knots, we get 1380 m, or 4500 ft.

To verify this you can also look up videos of DC-3 takeoffs. Like this one, which seems to take about 30 seconds:


                
            

I agree with the comments, piston travel changing significantly with power setting does not make sense.