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)? OK, so my explanation is going to be quite long, but please bear with me. My question is in two parts, but first I'll explain what I know.

I was reading the C-47 Pilot's Flight Operating Instructions and the DC-3 Organisation Control Manual, and they both say that takeoff should be performed at full-rated power. 
This link shows a bunch of good reasons why one should always use maximum power for takeoff 

[FAA mandatory for commercial operations; P&amp;W endorse it; 


  'The distance the pistons travel in the cylinders (stroke) is related
  to the power used. The piston rings, over time, wear a ridge into the
  cylinder wall at the top and bottom of the stroke. If 42" is used for
  takeoff, the bottom ridge will develop further up the piston wall than
  in an engine that has seen rated power for Takeoff. If the time comes
  when full rated power is needed (short runway, engine failure, etc.)
  and a power setting greater than 42" is used, the piston is now forced
  past the ridge in the cylinder created by many hours of using de-rated
  power. Forcing the piston rings past this ridge can cause them to
  break'


;carburator has additional cooling feature past 45"; engines last longer],

but also mentions the fact that full-rated should only be used for one minute at a time (which is apparently one of the reasons some people choose to take off with METO power instead). 

One minute sounds like a very short time to me (and to the person defending T/O with METO power), but I'm not a pilot, so I do not know how long a  normal T/O roll with a DC-3 at maximum takeoff power  would take. 

The DC-3 performance information sheet (found here) mentions that (at sea level and normal runway conditions) a 1700 ft T/O roll is enough to reach 100 mph (which I am assuming is the minimum takeoff speed ), but it says nothing about how long  the DC-3 would take to reach that. 

They also mention that at 5000 ft altitude it'd take a 2300 ft T/O roll to attain that same speed. But that still doesn't tell me how long that would take because I don't know the acceleration rate the DC-3 has. I've checked the charts, but I admit I couldn't conclude anything from them (doc is semi-readable, plus, it might have been way over my head anyway). I'm sorry if this question is stupid, maybe I need to brush off my physics , but I really am not managing to figure this one out on my own ... 

I am aware there are a number of variables that come into play here, but I'd be happy if anyone could give me an approximation , or even an example of what conditions need to exist for the DC-3 to take only a minute to do the T/O roll (if one wanted to use full-rated power).
And, more importantly: Would a wet grass runway at an almost 6000 ft altitude airport, plus having the aircraft at maximum takeoff weight (or close to it) justify choosing to take off with METO instead of full-rated power? I mean, such a T/O roll would definitely take over a minute, right?

Thanks a lot
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.