Breguet Range of a Jet Aircraft

How far can a jet fly on the fuel it burns? In steady level flight the thrust equals the drag and the lift equals the weight, so the fuel flow is the thrust specific fuel consumption times the weight divided by $L/D$. As fuel burns the aircraft gets lighter and the flow falls with it, and integrating that loss of weight over the flight gives the Breguet range equation. It is a derived result, not a fit, and the open MIT course notes by Spakovszky (Unified Engineering, Thermodynamics and Propulsion, section 13.3) derive the same expression.

$V$ is the true airspeed, held steady, and $L/D$ is the lift-to-drag ratio in cruise. $c'$ is the fuel weight burned per unit of thrust per unit of time, $g\,c_T$, written as a number per hour: 0.57 pound of fuel per pound of thrust per hour is 0.57 per hour in either system (16.1 g/(kN s) in the metric form). $m_0$ is the mass at the start of the cruise and $m_1$ the mass at the end, so the fuel burned is $m_0 - m_1$. The logarithm of the mass ratio is there because the fuel flow shrinks as the weight does. The factor $V(L/D)/c'$ is the natural scale of the range: a weight ratio of $e$, burning 63.2 % of the weight, gives exactly this, 24 694.7 km for the inputs below.

Take $V = 230$ m/s (828 km/h), $c' = 0.57$ per hour, $L/D = 17$, and a mass falling from 250 t to 200 t, so a fifth of the weight is burned (a ratio of 1.25). The range is 5 510.47 km. A step-by-step integration of the weight loss at one-second steps gives 5 510.446 km, which confirms the closed form. Solving backwards from 5 510.47 km recovers 828 km/h for the speed, 0.57 per hour for the consumption and 17 for the lift-to-drag ratio.

It is the range of the cruise only, in still air. Climb, descent and reserves all cut into it, and so does the variation of the specific consumption with the weight, which the equation treats as steady; a real flight plan also carries a margin for the wind. The speed is the true airspeed, so a ground distance needs the wind triangle.