# How did shortening the 707 increase its range?

According to the Wikipedia entry for the 707-120:

The 707-138 was a −120 with a fuselage 10 ft (3.0 m) shorter than the others, with 5 ft (1.5 m) (three frames) removed ahead and behind the wing, giving increased range. Maximum takeoff weight was the same 247,000 lb (112,000 kg) as the standard version

With the same MTOM between the -120 and the -138, what made the difference in the range? Was it simply less parasitic drag from the shorter fuselage?

I would presume that the fuel capacity (stored in the wings - center section tanks didn't come about until the -320) would have been the same.

I see notes on the -120B which switched to the P&W JT3D-1 engines which were more fuel efficient. That would account for greater range, but that doesn't apply to the -138 (at least not at launch).

• Could it be as simple as less fuselage mass means you can fill up the tanks more if the (unchanged) MTOM rather than tank volume is the limiting factor? – Sanchises Jan 31 '19 at 16:11
• – ymb1 Jan 31 '19 at 16:27

If the fuse was shortened that much (reducing both the weight of the airframe and its cargo-carrying capacity) and the maximum takeoff weight stayed the same, the most natural explanation is that the plane's allowable fuel load at takeoff or the tank's capacity was increased.

• Notice the robustly overdesigned wings and tail of the 707 readily allowed shortening of the fuselage. An interesting study would be shortening vs thinning (reduction of cross section) of fuselage. – Robert DiGiovanni Aug 4 '19 at 11:45

MTOW is Empty Weight + Payload + Fuel Weight, and is limited by aircraft geometry factors. For a shortened aircraft, empty weight will be lower, which means either more payload or more fuel weight allowed. Some aircraft are payload limited at their extreme range.

Range is a function of MTOW and empty weight, according to the Breguet Equation for a jet: $$R=\frac{V}{c_T} \cdot \frac{L}{D} \cdot ln\frac{W_i}{W_i-W_f}$$

with $${c_T}$$ = specific fuel consumption, $$W_i$$ = initial weight, $$W_f$$ = fuel weight.

For a given fuel fraction and engine type, the primary operational parameter in the equation is L/D: the drag polar. Shortening the fuselage will indeed lower the $$C_{D0}$$ as OP states. Torenbeek section 5.3.1 gives some statistical data for high subsonic jets of the B707 era:

• Typical $$C_{D0}$$ = 0.014 - 0.020
• $$C_D$$ (including induced) = 0.02 - 0.12

At the same speed, engine type, MTOW (and L), fuel weight: a reduction of D of 10% results in an increase in range of about 10%.