# Can an empty 747-400 with minimal fuel take off on two engines?

This is a fanciful question. 747s are allowed to take off on 3 engines for maintenance ferries (I did a couple), but a 2 engine takeoff is not a practical or legal consideration. However, my gut level feeling is that it could be done in optimal conditions, but I'm interested in what the more technically oriented people here have to say.

The aircraft for this thought experiment is a 747-400ERF, a freighter with a basic operating weight of 351,088 lbs and a max takeoff weight of 910,000 lbs. Let's say we want to take off and make a 15 minute swing about the countryside around Moses Lake, WA with its 13,500 runway, elevation 1189 feet, on a STP day with no wind. It's flat land, there are no departure or arrival path obstacles.

We want to land with no less than 15,000 lbs of fuel, and the takeoff and flying is going to take around 20,000 lbs. Thus our takeoff weight is going to be around 386,000 lbs, which is roughly 42% of the max takeoff weight.

We have half our engines inoperative, but we are significantly less than half as heavy as the max takeoff weight. To make it easy, let's say that engines 2 and 3 are operating (the inner most engines on each side).

So, what say you?

• Cool thought experiment! I'm assuming you don't need to be able to satisfy usual "V1 cut" performance requirements? (Of course, maybe if you had two of those monster GE-90-115b's...) – Ralph J Jul 2 '15 at 16:20

Hmm... I guess this will come down to how much of the drag is induced drag vs. how much is form drag. The induced drag will obviously be much less, but the form drag will not be much less than flying at the same airspeed at MTOW.

Given the requirement that it must be capable of continuing a takeoff and climbing out at MTOW on 3 engines, though, I'd be very surprised if it weren't capable of doing so at 42% MTOW on 2 engines.

For the most convincing argument in favor of 'yes' I can think of off hand, there's this. GE flew a 747 on one engine. Granted, that engine happened to be a GE90-115b, the world's most powerful commercial jet engine, which is used on the 777-300ER. However, the GE90-115b is rated at 'only' 115,300 lb. of thrust, while a pair of PW 4062s on a 747-400ERF would produce a total of 126,600 lb. GE's aircraft is a modified 747-100. Since Wikipedia lists its operating empty weight at 358,000 lb. and it was hopefully carrying some fuel in addition to that, its weight was likely not much different from the 386,000 lb. in your scenario. Additionally, in GE's case, some of the lift would have to be used to pull against the yawing moment created by flying with only the #2 engine producing thrust.

In light of this, I'd say that the answer is almost certainly, yes, it's possible. Advisable, of course, is a different matter entirely.

GE's Test Bed 747-100 with the GE90-115b mounted in place of its #2 engine. Source: Wikipedia

• That's a phenomenal picture showing just how huge engines have gotten in the last 40-50 years! – FreeMan Jul 2 '15 at 12:42
• @FreeMan Yeah, I've always liked that picture, especially when you compare the GE90 to the JT9D beside it and the humans walking on the ramp. Another one of my favorites is this one with a 777-300ER following a 737-800. The GE90s on the 777 are almost the same diameter as the fuselage on the 737. – reirab Jul 2 '15 at 13:38
• I just noticed that the FlightGlobal blogs (where your image links) are now gone. If you happen to have another source for that image, it would be nice to post it. – FreeMan Dec 10 '18 at 15:38
• @FreeMan Sadly, I don't. IIRC, it was copyrighted, so I never uploaded it to the SE imgur. – reirab Dec 11 '18 at 17:49

I'm not sure if its possible, but I'll try to arrive at a conclusion by using the following mathematical model. Correct me if I'm wrong :

Lets assume all the four engines are operative at 80 p.c. of their full thrust
when they need to take of with MTOW.

So, for MTOW = 910,000 lbs,
if the plane is using PW4062, thrust generated per engine is 63,300 lbf
Net thrust T = 63,300*4*0.8 = 202560 lbf (At 80p.c. of full thrust)

Now, for MTOW, Normal Reaction N of the plane = 29120000 lbf
If, f is the friction force acting on the plane while takeoff, then

T - f = MTOW*(acceleration)
acceleration = (T - f)/MTOW   -(1)

for taking off with two engines and lesser weight, if the plane is able to
achieve the above mentioned acceleration, we can safely assume that it should fly.

so, acceleration in case of using only two engines (at 80 p.c. thrust) and lower weight is:

acceleration = (T' - f')/Weight   -(2)
Now T = 202560lbf
T'= 101280lbf
f = (u)*910,000lbf
f'= (u)*386,000lbf
weight= 386,000lbf
MTOW  = 910,000lbf

from our assumption, lets equate 1 and 2 (and we'll use some constant k to be used later)

so, k(T-f)/MTOW = (T'-f')/weight
k(202560-u*910,000)/910,000 = (101280 - u*386,000)/386,000
202560*k - u*910,000*k = 238768 - u*910,000
k = (238768 - u*910,000)/(202560 - u*910,000)

Now, for u not belonging to the interval (0.222,0.262), k is +ve
and for slip ratio (0,1) , u rarely exceeds 0.5

hence, for k > 0.8, u lies in (0,0.22), which is quite possible.


Hence, acc. to this model, for a coefficient of friction of about 0.23 or below, its possible for an airplane to achieve the velocity required to produce the lift equivalent to the lift required to takeoff with MTOW. If this is possible, then generating the lift required to take off with lesser weight is also possible, theoretically.

p.s. Sorry for some liberal assumptions I made in the model. Please correct me if I was wrong anywhere.