# 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&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

• “The distance the pistons travel in the cylinders (stroke) is related to the power used.” What? That makes no sense. The piston is connected to the crankshaft by a solid rod, so it must always travel the same range during each complete revolution. – Jan Hudec Sep 1 '17 at 4:57
• Yes, I understand you quoted it as you got it. I am saying it does not sound right. And it's not like it was official documentation anyway. And I also understand it is not the actual question—that's why I wrote a comment, not an answer. – Jan Hudec Sep 1 '17 at 5:10
• This is a really good question. Don't feel sorry for its length as it add a lot of precision and make the answers more specific – Manu H Sep 1 '17 at 19:35
• @JanHudec - reading through the linked page, I think there's some confusion. I initially thought the same thing you did - there's no way the piston stroke can change (and a 42" stroke sounds like a marine diesel, not an aviation engine)! I think the 42" refers to manifold pressure. I think that (maybe) increasing that MP to full power (48"?) can cause the piston to hit top/bottom harder? than normal causing the rings to fracture when they hit those ridges. Still not sure that's 100% accurate, but it's the only logical thing I can make of it. (con't... – FreeMan Sep 11 '17 at 17:58
• ... the 42 v 48" MP comes from the DC-3 Throttle's Management section of the link, while the cylinder ridge formation comes from point 1 under the Some differing opinions and interesting information: section. I've gotta admit - it explicitly says the piston stroke varies, and that's just mind-boggling to me... – FreeMan Sep 11 '17 at 18:01

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.

• Thank you @fooot for taking the time to help me with this conundrum. I apologise for my unintentionally (mis)leading you to think both pieces of info came from the same link. I've added the second link now to correct that. But it should still make sense given that both links talk about the same machine, right? Thanks a lot – Electra Sep 12 '17 at 16:05
• Thank you so much for your help. I'd have never figured this out on my own. I didn't think of using two equations to substitute the missing variable and didn't realise it'd be ok to assume constant acceleration, so I'd have never come to that conclusion myself. So 60s doesn't seem that little after all, even at 5000ft altitude, and apparently the weight (according to the second link) doesn't account for that much difference either. So would the wet grass runway (in my example) account for the only reason to use METO for T/O ? thanks a lot – Electra Sep 12 '17 at 17:08
• And how much do you guys reckon the heat would affect that (the first link says "heat gives the 5,000 ft ASL field a density altitude of 8,000 ft") ? Would I calculate the length of the T/O roll as if the altitude of the airport were actually 8,000 ft? Would any of you disagree that the weight makes little difference (second link says so, but I saw a number of links that attributed longer takeoff rolls to the aircraft being fully loaded and thus very heavy)? – Electra Sep 12 '17 at 17:10
• Both heat and a grass runway would reduce the acceleration, it's hard to read the performance document but I didn't see any hints about how much difference those factors would make. – fooot Sep 12 '17 at 17:11
• Yeah, as altitude increases, takeoff distance gets longer, and differences like weight start to be more noticeable. You can see in the takeoff runway required chart, at sea level the lowest two weights differ by 500 feet length, at 6000 feet they differ by 1000. – fooot Sep 12 '17 at 17:27