Two questions on making turns and climbing
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Two questions on making turns and climbing
After having placed a question on this forum on the subject of Va (many thanks for the answers), I have two more. It seems that there are more than enough people around who are willing to give good questions a thought (much appreciated!), so I'll throw in two more.
- Firstly: when flying a glider (which I do), how come that during a turn the sinking speed is always more than in 'straight and level' flight? When I learned to fly, which is more than a few years ago, I was taught that 'Lift gets a horizontal component, making the aircraft turn, so the vertical lift-component becomes less and the airplane starts to sink more'. Sounded good, in those days, but now I'm supposed to teach this to newcomers in a while, and I'm beginning to doubt: why can't we just pull the stick enough to create sufficient lift, so we'll maintain the original sink, and not go down faster? What's the limiting factor? Stallspeed? Alpha? Increasing induced drag, reducing speed, so we'll have to pull more on the stick, which means a higher AoA, which again means more Di, which means losing speed again, etc.?
- Secondly: How come an airplane has a lower load factor (lower than one) during a constant, steady climb? That really baffles me: I know that during a climb Lift is lower, because part of it is taken over by Thrust, but still... Me, as a pilot, am taken upwards, against the force of Gravity, so I'd think the loadfactor would even be a little more (Gravity pulling me down, me going the other way). Could it be that I'm experiencing a different loadfactor than my airplane? Would this mean that, when we're going up after take-off in a Boeing, we're all weighing a little less temporarily?
Sorry for the long story, I hope Einstein doesn't have anything to do with this.
Any long or short answers much appreciated.
Eagle
- Firstly: when flying a glider (which I do), how come that during a turn the sinking speed is always more than in 'straight and level' flight? When I learned to fly, which is more than a few years ago, I was taught that 'Lift gets a horizontal component, making the aircraft turn, so the vertical lift-component becomes less and the airplane starts to sink more'. Sounded good, in those days, but now I'm supposed to teach this to newcomers in a while, and I'm beginning to doubt: why can't we just pull the stick enough to create sufficient lift, so we'll maintain the original sink, and not go down faster? What's the limiting factor? Stallspeed? Alpha? Increasing induced drag, reducing speed, so we'll have to pull more on the stick, which means a higher AoA, which again means more Di, which means losing speed again, etc.?
- Secondly: How come an airplane has a lower load factor (lower than one) during a constant, steady climb? That really baffles me: I know that during a climb Lift is lower, because part of it is taken over by Thrust, but still... Me, as a pilot, am taken upwards, against the force of Gravity, so I'd think the loadfactor would even be a little more (Gravity pulling me down, me going the other way). Could it be that I'm experiencing a different loadfactor than my airplane? Would this mean that, when we're going up after take-off in a Boeing, we're all weighing a little less temporarily?
Sorry for the long story, I hope Einstein doesn't have anything to do with this.
Any long or short answers much appreciated.
Eagle
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Hiya Eagle 1,
1) When turning, the glider is subjected to centripetal & centrifugal forces, resulting in what we experience as "G". E.g. in a coordinated turn with 60 deg. bank, the glider is subjected to 2 G. Effectively, if your gliders mass is 400 kg. the wings will now have to create 800 kg. of lift. This is exactly equivalent to flying the glider under 1 G, but with a mass of 800 kg. Now - given they fly the same speed, which of the two is gonna have the higher polar sink? Aaaahh - realisation dawns! That is the reason why the speed for minimum sink increases above the AOM value (which is for s&l) when circling in a thermal.
2) Because the loadfactor is measured perpendicular to the flightpath. Say the flightpath is 90 deg. - i.e. straight up or down. How much load will the wings need to support? None, since in a climb, the kinetic enegy of your glider is converted into potential energy, so the kinetic energy cancels the gravity out, and the wings are unloaded. The same in a dive, only the drag is what counteracts gravity. For the same reason, you'll see that e.g. the K21 will need to be pitched beyond a vertical pitch attitude on the way down to produce a vertical line, since you need a negative AoA to produce zero lift on an assymmetrical profile. The Swift & Fox don't need any such compensations, since they have a (near-)symmetrical profile.
Hope this helps - best regards,
Empty
1) When turning, the glider is subjected to centripetal & centrifugal forces, resulting in what we experience as "G". E.g. in a coordinated turn with 60 deg. bank, the glider is subjected to 2 G. Effectively, if your gliders mass is 400 kg. the wings will now have to create 800 kg. of lift. This is exactly equivalent to flying the glider under 1 G, but with a mass of 800 kg. Now - given they fly the same speed, which of the two is gonna have the higher polar sink? Aaaahh - realisation dawns! That is the reason why the speed for minimum sink increases above the AOM value (which is for s&l) when circling in a thermal.
2) Because the loadfactor is measured perpendicular to the flightpath. Say the flightpath is 90 deg. - i.e. straight up or down. How much load will the wings need to support? None, since in a climb, the kinetic enegy of your glider is converted into potential energy, so the kinetic energy cancels the gravity out, and the wings are unloaded. The same in a dive, only the drag is what counteracts gravity. For the same reason, you'll see that e.g. the K21 will need to be pitched beyond a vertical pitch attitude on the way down to produce a vertical line, since you need a negative AoA to produce zero lift on an assymmetrical profile. The Swift & Fox don't need any such compensations, since they have a (near-)symmetrical profile.
Hope this helps - best regards,
Empty
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Hi Empty,
Thanks very much for the trouble you took answering my questions, but... there's still something that doesn't convince me (might be my limited ability to understand these things):
- isn't it so that , for instance in a turn with a 60 degree bank angle, indeed the loadfactor is higher (which I fully understand), but that it is also compensated by the fact that Lift is really increased to two times its original value (by pulling yoke or stick)? In that case I'd say that there would be no extra sink; indeed, of you would [I]not[I] increase Lift, then you'd be sinking faster! I mean: during a turn in a glider the resultant lift-vector is increased to twice the original Lift, which compensates the resultant weight-vector; you'd keep your original sinkspeed, but how come we're always going down faster?
- If the loadfactor decreases with increasing pitch-angle (climb/dive) - I get that one: lower lift, more drag/thrust - how come that I, as a pilot, don't feel it? It seems that my body is still experiencing a 1-G situation during a steady climb, as far as I'm aware.
If anyone has anything to add, thanks very much!
Eagle
Thanks very much for the trouble you took answering my questions, but... there's still something that doesn't convince me (might be my limited ability to understand these things):
- isn't it so that , for instance in a turn with a 60 degree bank angle, indeed the loadfactor is higher (which I fully understand), but that it is also compensated by the fact that Lift is really increased to two times its original value (by pulling yoke or stick)? In that case I'd say that there would be no extra sink; indeed, of you would [I]not[I] increase Lift, then you'd be sinking faster! I mean: during a turn in a glider the resultant lift-vector is increased to twice the original Lift, which compensates the resultant weight-vector; you'd keep your original sinkspeed, but how come we're always going down faster?
- If the loadfactor decreases with increasing pitch-angle (climb/dive) - I get that one: lower lift, more drag/thrust - how come that I, as a pilot, don't feel it? It seems that my body is still experiencing a 1-G situation during a steady climb, as far as I'm aware.
If anyone has anything to add, thanks very much!
Eagle
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Eagle1, re your first question to do with the turn, you said you understand the lift vector must be doubled in magnitude in a 60 degree bank to support the weight of the a/c. As the lift is increased so is induced drag, which increases rather sharply as more and more bank goes on. As induced, and so total, drag increases, then we either a) open up the throttle to compensate, or in your case, b) point a little more weight down the flight path to balance the increased drag. The option of thrust in a powered a/c allows us to stay level if we want, in a glider the only way to point more weight along the flight path is to point it down slightly, and voila, increased sink!!
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Load Factor
Hi Eagle,
Empty Cruise is correct, but perhaps I can explain it in a slightly different way.
You are right that gravity continues to act on your body in the same way at all times in flight. However, the definition of load factor is Lift divided by Weight, and it is in the trickery of the definition that we end up with load factor being less than one in a steady climb.
In a steady climb, the aircraft is in equilibrium - total "up" forces exactly equal to total "down" forces. "Up" forces are partly thrust (say 10 units) and partly lift (say 90 units). Assume that "down" forces are all due to weight (100 units), and calculate your new load factor.
Weight does not change - just lift. So your body is quite correct in feeling the same weight as always.
Cheers,
O8
If the loadfactor decreases with increasing pitch-angle (climb/dive) - I get that one: lower lift, more drag/thrust - how come that I, as a pilot, don't feel it? It seems that my body is still experiencing a 1-G situation during a steady climb, as far as I'm aware.
You are right that gravity continues to act on your body in the same way at all times in flight. However, the definition of load factor is Lift divided by Weight, and it is in the trickery of the definition that we end up with load factor being less than one in a steady climb.
In a steady climb, the aircraft is in equilibrium - total "up" forces exactly equal to total "down" forces. "Up" forces are partly thrust (say 10 units) and partly lift (say 90 units). Assume that "down" forces are all due to weight (100 units), and calculate your new load factor.
Weight does not change - just lift. So your body is quite correct in feeling the same weight as always.
Cheers,
O8
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Many thanks to you all, you've made things definitily clearer to me. Exams coming up the day after tomorrow, so I'll keep it short: back to the books!
Eagle
Eagle
