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I forgot to mention, in that more recent post, that I was only considering
the case of no headwind, no tailwind. In the earlier post I did specify that. A sample result is shown for a 20-knot headwind (for a C172 at 6000 ft) on pages 335-336 of Performance of Light Aircraft. It's essentially a trial-and-error calculation. The 20-knot headwind raises Vbr=73 KCAS to Vbrhw=78 KCAS and depressed the maximum specific range about 20%. Specific range, in general, is (V-Vhw)/cP, where V is airspeed, Vhw is headwind speed, c is brake specific fuel consumption rate, and P is engine power. So a simple formula. But the optimal SOLUTION to that simple formula, especially in the partial-throttle bootstrap approach, is far from simple. That's because the formulas for engine torque and for RPM required for level partial-throttle flight are somewhat complicated. Numerical solution, to optimize, is easiest. Everything Roger said about headwind effects is correct. Get enough headwind and you'll fly backwards. In which case best range is given by landing immediately! John -- John T Lowry 5217 Old Spicewood Springs Rd, #312 Austin, Texas 78731 (512) 231-9391 "Roger Long" om wrote in message ... What about wind? If Vbr is equal to the headwind component, range will be zero. With one hour fuel. Vbr + 1 will give you a range of a mile. The faster you go, the more range. At some speed though, increase in fuel consumption offsets the gain. Could you give us a simple and usable formula for the Skyhawk in your example? -- Roger Long |
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