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#1
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This note is about Vbr (speed for best range) and Vbe (speed for best
endurance) - as well as specific range SR (nm per pound of fuel) and specific endurance SE (hr per pound of fuel) themselves - and how these quantities depend on (fixed pitch propeller) airplane gross weight W and (density) altitude hRho. This was the subject of a recent post about "throttle height." I added my two cents on specific range there, but I think the subject is worth another, and better organized, nickel. I should have included a page or two on this weight-and-altitude effect in Performance of Light Aircraft, but I didn't. These things are important to understand in case you ever find yourself low on fuel and far from home on a dark night. Here are some numbers calculated for a Cessna 172. Units a [W]lbf, [hRho]ft, [Vxx]KCAS, [SR]nm/lb fuel, [SE]hr/lb fuel. 1. W=2400 & hRho=0: Vbr=73, SR=2.40, Vbe=62, SE=0.0357 2. W=2400 & hRho=8000: Vbr=73, SR=2.40, Vbe=62, SE=0.0317 3. W=2000 & hRho=0: Vbr=67, SR=2.88, Vbe=57, SE=0.0470 4. W=2000 & hRho=8000: Vbr=67, SR=2.88, Vbe=57, SE=0.0416. So your altitude doesn't matter for either Vbr or Vbe or SR. For best endurance, however, lower altitude is better by a factor of square root of sigma. Lower weight is best for both range and endurance and lower weight results in lower values for both Vbr and Vbe. The way this was calculated is much too complicated to discuss in ASCII, but the chapter on Cruise and Partial-Throttle Performance in the book mentioned above has all the formulas along with sample calculations. I use none of the usual unrealistic approximations such as constant propeller efficiency eta, constant lift coefficient CL, etc. The partial-throttle bootstrap extension allows one to calculate how much engine torque is needed to fly level at any given airspeed, altitude, and gross weight. Also allows one to find what value of RPM is required to do that. I did assume a constant value of brake specific fuel consumption rate (BSFC), but that is veridical for these low power settings. (For this airplane BSFC takes a step up at around 122 HP, 76% power, from 0.45 to 0.51.) Why is none of this in your POH? The POH cruise table is very poorly designed. It uses the wrong independent variables for entering the tables. (There is a problem with double valuedness of airspeed at low RPM values; that's why the cruise tables don't go down there. In a sense this is the partial throttle version of airplanes' having two speeds for full-throttle level flight, the low one and the high one.) The GAMA-format POH section 5 (Performance) does indeed show some advances over most of its predecessors, but still has quite a ways to go to properly embody brevity, simplicity, and safety. Well, it was a committee thing, and as Robert A. Heinlein's character Lazarus Long said: "A committee is a life form with six or more legs and no brain." Hope this note helps. John -- John T Lowry, PhD Flight Physics 5217 Old Spicewood Springs Rd, #312 Austin, Texas 78731 (512) 231-9391 |
#2
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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 |
#3
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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 |
#4
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![]() "John T Lowry" wrote in message No comments except to say that it's good to have you back, John. -- Jim Fisher |
#5
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if your flying backwards how do you get to the airport to land ?
John T Lowry wrote: Get enough headwind and you'll fly backwards. In which case best range is given by landing immediately! |
#6
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Turn around? (grin)
Dave Jeff wrote: if your flying backwards how do you get to the airport to land ? John T Lowry wrote: Get enough headwind and you'll fly backwards. In which case best range is given by landing immediately! |
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