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Don't forget the Caproni A-21J/A-21SJ. I believe that was the first Jet
sailplane. They addressed the 'melting the tail" problem by splitting the exhaust into two vectors, each having a slight angle away from the tail. It was moderately heavy at the time (early 70's I think), ~1400 lbs, w/ 220 lbs of thrust, but climbed well, particularly at altitude. Only a few (+/-3) still registered. Mike Martin Hellman wrote: Jim Culp wrote in message ... How many pounds thrust might be needed to keep a std class glider with std pilot aboard such as a Std Libelle or Ls4 or Discus a) continuing in level flight at 60-70 knots b) climb at rate of 3oo fpm at 60-70 knots CH gave formulas that work in a later post in this thread, but there's a simpler (at least to me) way to calculate these numbers. a) L/D is the ratio of Lift (which equals glider weight) to Drag. So, if the ship weighs 705 lb and L/D=30 (as in CH's example), then the drag is 705/30 = 23.5 lb. You have to use the L/D at the speed you are operating, so drag will increase at higher speeds, where the L/D drops off. b) To get the additional, "climb thrust" needed for a given climb rate, take the true air speed (60 kts), divide by the climb rate (3 kts) to get 20. Then divide the weight of the glider (705 lbs) by 20 to get the additional thrust needed for this climb rate (35.25 lb). Add this additional, climb thrust to the level drag to get total required thrust, 58.75 lb. Note that if the desired climb rate equaled the true air speed, you would need to add 100% of the glider's weight to the level drag. Which makes sense. A thrust to weight ratio of a little more than 1 is needed for vertical flight. It's interesting that nautical units are better than even metric (km/hr and m/sec) for calculations like this, at least within 1%. That's because, within 1% (OK, more like 1.01%) a nautical mile is 6000 feet, a number which is evenly divisible by the number of minutes in an hour. That's why 300 fpm = 3 kts (within 1%). Hope someone else finds this simpler too. Martin |
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