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#1
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On Wednesday, May 7, 2014 1:20:44 PM UTC+12, Bill D wrote:
So, you're saying the pilot will be safer if they don't learn to perform the return to runway maneuver when it's safe to do so? I can assure you that the higher a glider's performance, the safer it is. It's the old, low L/D gliders that can run out of altitude before getting lined up with the runway. I agree with you. I'm shaking my head every time I read this thread. In a modern glass glider (such as the DG1000's I instruct in) with a 40 knot stall speed and being towed at 70 knots you should be able to execute a safe 180º turn with *zero* loss of height. Just slowing down from 70 knots towing speed to 55 knots gains you 80 ft on top of whatever you already had.[1] How much height do you lose in a 45º banked turn at 55 knots? Most modern gliders lose no more than 120 fpm at 45 knots in straight and level flight.. A 45º banked turn gives 1.41 Gs (1/cos(45)), which needs sqrt(1.41) = 1.19 times more speed for the same angle of attack and L/D. 45 knots times 1.19 is 53.6 knots. So 55 knots in a 45º turn has a little more margin above stall than 45 knots in straight and level. The sink rate will be 120 * 1.41 = 170 fpm. Converting to SI and using a=v^2/r, a 45º banked turn at 55 knots (28.3 m/s) has 115.3m radius, or 725m circumference for a full turn. A 180º turn (362m) will take 12.8 seconds. In 12.8 seconds at 170 fpm you'll lose 36 feet. So the height loss in the turn is only about half the height gained from slowing down from towing speed to circling speed! It would actually be better to start turning immediately, but these calculations assume you delay (deliberately or not) and climb straight ahead (no pull-up required) for several seconds before starting the turn. With 18m wingspan in a 45º bank your wingtip is 18m/2*sin(45) = 6.4m or 21 ft below you. So you theoretically could do this from absolutely zero height, with nearly 20 ft to spare. I wouldn't want to try it! But from 100ft? No problem at all. IF you start from normal towing speed and reasonably benign weather conditions. Even if you're releasing from a sick tug that's slowed to 55 knots, you'll be fine from 200 ft. Another post mentioned that glider pilots make mistakes when flying close to the ground because they are not trained to do so and don't do "ground reference" manoeuvres like power plane pilots do. Obviously that person lives in very flat ground, because I can assure them that here we're flying close to ridges and peaks a LOT, from almost the first flight, doing 180º turns at the end of ridge lift runs, or circling low over a peak or head of a gully looking for a thermal. We'd very often be only 100-200 ft or so above the terrain while doing so. [1] handy formula: X knots of kinetic energy is worth (X/5)^2 feet of gravitational potential energy. e.g. 70 knots = (70/5)^2 = 14^2 = 196 ft. 50 knots = (50/5)^2 = 10^2 = 100 ft. Less drag loses of course. You'll never turn speed into quite that much height, and you'll need more height than that to get speed. But the differences are large in a high performance glider at moderate speeds. |
#2
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At 06:55 07 May 2014, Bruce Hoult wrote:
[1] handy formula: X knots of kinetic energy is worth (X/5)^2 feet of gravitational potential energy. e.g. 70 knots = (70/5)^2 = 14^2 = 196 ft. 50 knots = (50/5)^2 = 10^2 = 100 ft. Less drag loses of course. You'll never turn speed into quite that much height, and you'll need more height than that to get speed. But the differences are large in a high performance glider at moderate speeds. ...or you could use my rule of thumb: a change of speed of 10 knots IAS gives you (or costs you) the number of feet in height of the speed you arrive at. For example: 70 knots to 60 knots : plus 60 feet 60 knots to 70 knots: minus 60 feet 70 knots to 50 knots (i.e. 70 to 60, then 60 to 50) : (60+50) = plus 110 feet (N.B. Done without undue delay, and at sea level - more height change at greater altitudes. Height in feet, IAS in knots, works for a change of speed of 10 knots. Results are realistic but approximate.) |
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At 11:49 07 May 2014, James Metcalfe wrote:
60 knots to 70 knots: minus 60 feet sorry - s.b. minus 70 feet |
#4
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On Wednesday, May 7, 2014 11:49:46 PM UTC+12, James Metcalfe wrote:
At 06:55 07 May 2014, Bruce Hoult wrote: [1] handy formula: X knots of kinetic energy is worth (X/5)^2 feet of gravitational potential energy. e.g. 70 knots = (70/5)^2 = 14^2 = 196 ft. 50 knots = (50/5)^2 = 10^2 = 100 ft. Less drag loses of course. You'll never turn speed into quite that much height, and you'll need more height than that to get speed. But the differences are large in a high performance glider at moderate speeds. ..or you could use my rule of thumb: a change of speed of 10 knots IAS gives you (or costs you) the number of feet in height of the speed you arrive at. For example: 70 knots to 60 knots : plus 60 feet 60 knots to 70 knots: minus 60 feet 70 knots to 50 knots (i.e. 70 to 60, then 60 to 50) : (60+50) = plus 110 feet It's nearly the same formula. If you made it "every 12.5 knots IAS" instead of 10 then it'd be nearly exact. The derivative of my formula (X/5)^2 i.e. X^2/25 is X/12.5. So gaining or losing X feet at X knots takes roughly 12.5 knots of speed change. Doing it using the speed you're changing *to* is clever, as it makes some allowance for drag losses. However your height estimates are systematically biased 20% too large. If you said 70 to 50 is (60+50)-20% = 110-22 = a gain of 88 ft then that would be good. And 50 to 70 is (60+70)-20% = 130-26 = a loss of 104 ft, which is pretty good too. If you're doing something like "how high can I zoom to after a 140 knot low pass and still have 60 knots for the circuit?" then I'm not convinced that adding up eight numbers is easier than calculating two squares and subtracting them :-) By me: 28^2 - 12^2 = ~800 - ~150 = 650 ft (784-144 = 640 if you do it exact) By you: 130+120+110+100+90+80+70+60 = 760 ft By you with my -20% correction: 760 - 152 = 608 (remember folks, that's 140 knots at the END of the low pass, not the start!) |
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At 23:50 07 May 2014, Bruce Hoult wrote:
However your height estimates are systematically biased 20% too large. Well, about 8% for a pull-up, and 17% for a dive - at the high-speed end of the scale. (Better and worse, respectively, at the low-speed end.) Whereas your rule is about 10% low throughout. By me: 28^2 - 12^2 = ~800 - ~150 = 650 ft (784-144 = 640 if you do it exact) By you: 130+120+110+100+90+80+70+60 = 760 ft By you with my -20% correction: 760 - 152 = 608 And the truth (from v^2 = u^2 + 2gh ... whence we both started, I'm sure) is 708.66 ft. (I knew you'd like the ".66" ;o) ) Spreadsheet he http://tinyurl.com/mzokpyk for those amused by such things. But of course these are both rules of thumb, intended more to give better understanding than for exact calculation. I have more often referred to mine in discussing recovery of lost speed low on finals (or after a low winch-launch failure) than for the 120kt beat-up case! |
#6
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On Friday, May 9, 2014 4:44:38 AM UTC+12, James Metcalfe wrote:
At 23:50 07 May 2014, Bruce Hoult wrote: (Better and worse, respectively, at the low-speed end.) Whereas your rule is about 10% low throughout. Yes, it's deliberately biased a little to the low side from the true (as you note) /4.7516 to make it more of a "guaranteed to pull up that far". Also useful for "I'd like to be down on that ridge line, but I don't want to overspeed". Of course in that case the height difference is going to be pure guess anyway. I've contemplated using /4 for dives where you want to make sure you have some minimum speed, but I think that overdoes it. /4.5 would be better but not easy to calculate in your head. |
#7
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On 5/7/2014 2:55 AM, Bruce Hoult wrote:
On Wednesday, May 7, 2014 1:20:44 PM UTC+12, Bill D wrote: I can assure you that the higher a glider's performance, the safer it is. It's the old, low L/D gliders that can run out of altitude before getting lined up with the runway. I agree with you. I'm shaking my head every time I read this thread. In a modern glass glider (such as the DG1000's I instruct in) with a 40 knot stall speed and being towed at 70 knots you should be able to execute a safe 180º turn with*zero* loss of height. The comparison isn't quite as simple as just looking at L/D. Turn radius also has a lot to do with your chances of making it back to the field, and turn radius is proportional to the SQUARE of airspeed. Compare your example (40 knot stall) with a (horrors) 2-33. The highest stall listed for a 2-33 is around 30 knots. If you do the math, you will find that your DG1000 has nearly double the turn radius of the slower glider. Vaughn |
#8
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....now redo your calculations while flying through 8 knots of sink.
Mike |
#9
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On Wednesday, May 7, 2014 6:28:41 AM UTC-7, Mike the Strike wrote:
...now redo your calculations while flying through 8 knots of sink. Mike ...and calculate the height loss while making a 180-degree turn. Taking the 12.8 seconds just mentioned at 800 feet per minute gives you a height loss of 170 feet from the airmass movement alone, plus whatever you add for the glider itself. Safety margins in severe sink disappear frighteningly quickly. Mike |
#10
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On Wednesday, May 7, 2014 9:35:46 AM UTC-4, Mike the Strike wrote:
On Wednesday, May 7, 2014 6:28:41 AM UTC-7, Mike the Strike wrote: ...now redo your calculations while flying through 8 knots of sink. Mike ..and calculate the height loss while making a 180-degree turn. Taking the 12.8 seconds just mentioned at 800 feet per minute gives you a height loss of 170 feet from the airmass movement alone, plus whatever you add for the glider itself. Safety margins in severe sink disappear frighteningly quickly. Mike Now throw in wind shear and tailwind component when failure to anticipate leads to turning in the wrong direction. Double AARRGGHH!! UH |
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