In the Beginning was Information

p = 0.006/h) to produce propulsion energy and heat. During the first hourthe amount of fuel x 1 it thus requiresx 1 = G 0 x p = 200 x 0.006 = 1.2 g fat. (1)At the beginning of the second hour it weighs G 0 - x 1 = 200 - 1.2 = 198.8 g,so that it consumes somewhat less energy during the second hour:x 2 = (G 0 - x 1 ) x p = G 1 x p = 198.8 x 0.006 = 1.193 g (2)x 3 = (G 0 - x 1 - x 2 ) x p = G 2 x p = 197.6 x 0.006 = 1.186 g (3)For the 88th hour the fuel consumption is down tox 88 = (G 0 - x 1 - x 2 - x 3 - … - x 87 ) x p = G 87 x p (4)because of its reduced weight. We can now calculate its weight at the endof the journey, after subtracting the hourly weight reduction:1st hour: G 1 = G 0 - x 1 = G 0 - G 0 x p = G 0 (1-p) (5)2nd hour: G 2 = G 1 - x 2 = G 1 - G 1 x p = G 1 (1-p) = G 0 (1-p) 2 (6)3rd hour: G 3 = G 2 - x 3 = G 2 - G 2 x p = G 2 (1-p) = G 0 (1-p) 3 (7)…zth hour: G z = G z-1 - x z = G z-1 - G z-1 x p = G z-1 (1-p) = G 0 (1-p) z (8)…88th hour: G 88 = G 87 -x 88 = G 87 - G 87 x p = G 87 (1-p) = G 0 (1-p) 88 (9)The hourly weights G 0 , G 1 , … G 88 form a geometric sequence with commonratio q = 1 -p < 1. This computation is somewhat oversimplified 29 ,but, by substitution in (9) we find the final weight after 88 hours to be:G 88 = 200 x (1 - 0.006) 88 = 117.8 g (10)The total fuel consumption is given byG 0 - G 88 = 200 - 117.8 = 82.2 g (11)which is appreciably greater than the 70 g of fat the bird started out with!The bird’s weight cannot go below 130 g (see Figure 45). In spite of flyingat the optimum speed for minimal fuel consumption, the bird would29 A more exact calculation involves the differential equation dG/dt = - G(t) x p withG(t=0) = G 0 instead of the hourly steps considered above. The solution of this equationis a continuous function G(t) = G 0 x exp(-px t) where p = 0.006/h, which differsonly in a few insignificant decimal places from the result given in equation (10).242