Total drag questions
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Disgusted of Tunbridge
Disgusted of Tunbridge
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ft, did you not replicate with
with what I wrote in the previous post:
?
We can produce all sorts of fancy formulae (which really won't mean anything to anybody without a diagram!), but the answer will be- AoA will be in proportion to Cos climb angle. A 15 degree climb will have .97 of level flight lift- you know it, I know it- I feel it in me bones!
Rainboe,
todays nitpicker comment: In a vertical climb most aircraft (which have cambered wings) will in fact have a zero-lift AoA which is negative rather than zero, giving you a few additional degrees nose down relative to vertical to add to those caused by the angle of incidence.
todays nitpicker comment: In a vertical climb most aircraft (which have cambered wings) will in fact have a zero-lift AoA which is negative rather than zero, giving you a few additional degrees nose down relative to vertical to add to those caused by the angle of incidence.
Originally Posted by Rainboe
In a vertical climb, AoA for all intents and purposes would be zero,meaning the fuselage would have negative pitch (ie be pointing at about 86 degrees pitch whilst plane is climbing vertically)
In a vertical climb, AoA for all intents and purposes would be zero,meaning the fuselage would have negative pitch (ie be pointing at about 86 degrees pitch whilst plane is climbing vertically)
We can produce all sorts of fancy formulae (which really won't mean anything to anybody without a diagram!), but the answer will be- AoA will be in proportion to Cos climb angle. A 15 degree climb will have .97 of level flight lift- you know it, I know it- I feel it in me bones!
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Rainboe, yes, I was replying to you. AoA will typically be a few degrees negative in a vertical climb rather than zero, due to camber. Fuselage pitch relative to the airflow will be another few degrees more negative due to angle of incidence.
Of course this means I goofed earlier. AoA in a climb won't be given by
AoA_climb = AoA_level*L_climb/L_level
at all, even though the relationship is essentially linear.
(AoA_climb-AoA_zero_lift) = (AoA_level - AoA_zero_lift)*L_climb/L_level
should hold more or less true though!
As for diagrams, you have them earlier in this thread. In addition, this is rather obvious to a lot of people without diagrams so you're making a rather broad statement there. After all, this is only the basic T/W/D/L diagram.
Of course this means I goofed earlier. AoA in a climb won't be given by
AoA_climb = AoA_level*L_climb/L_level
at all, even though the relationship is essentially linear.
(AoA_climb-AoA_zero_lift) = (AoA_level - AoA_zero_lift)*L_climb/L_level
should hold more or less true though!
As for diagrams, you have them earlier in this thread. In addition, this is rather obvious to a lot of people without diagrams so you're making a rather broad statement there. After all, this is only the basic T/W/D/L diagram.
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It seems we've now dealt with the original question, and we've all agreed lift and AoA decrease in a steady climb, even if only very little for small climb angles in a 'conventional' aircraft.
So we're now dealing with Rainboe's pet situation, a steady vertical climb.
Admittedly, at first sight we are just talking T=W+D+L, but since none of those vectors point in exactly the right direction, another diagram may be called for?
CJ
So we're now dealing with Rainboe's pet situation, a steady vertical climb.
Admittedly, at first sight we are just talking T=W+D+L, but since none of those vectors point in exactly the right direction, another diagram may be called for?
CJ
