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![]() "Kevin Neave" k wrote in message So ((U*U) - (V*V)) / (2*a) = s For the light glider V=19m/s, U=50m/s, a=7.31m/s/s ((50 * 50) - (19 * 19)) / (2 * 7.31) = 146.31 metres. So Our height gained = 1/sqrt(2) * 146.31 = 103.46m For the heavy glider V=22m/s, U=50m/s, a=7.23m/s/s ((50 * 50) - (22 * 22)) / (2 * 7.23) = 139.24 metres !! Height gained = 98.46 metres !! ------------------------------------------------------------------ --------------------------- OK, So there's some assumptions in the above, but I think all of them were made in favour of the heavy glider. But I say once again, for a pull up from 100kts with 100kgs of ballast, 'It's too close to call'... Over to you Todd :-)) Kevin. I don't see anywhere in your in your math the higher initial rate of sink for the lighter ship. If you look at polars for ballasted vs empty, for any given speed, the rate of sink is higher unloaded. The assumption that both gliders are at the same height when they reach the the 0 fps up point of the pullup looks flawed in my mind. You must wait longer in the unloaded glider to establish a climb. In fact I think the softer pullup, the greater the difference in starting heights of the newtonian decayed climb. Scott. |
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