Interesting Question at Dinner the other night!
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Interesting Question at Dinner the other night!
Whilst engaged in downing some wonderful ethnic cuisine the other night....Chili Cheese Dogs all the way and a glass of American beer....a question arose between mouthfuls of that wonderful delicacy.
While doing turns in your favorite helicopter....and particularly in steep turns....does the helicopter lose lift in the same way that fixed wing airplanes do as a result of bank angle? If so....why...and to what degree? Are there charts or forumla's for calculating that loss?
Help me out here Nick....some free dogs are in the balance here....the Chef suggests that I know not what of I speak. I would like to be able to baffle said Chef with my intricate knowledge of things rotorary when we next dine at his establishment.
While doing turns in your favorite helicopter....and particularly in steep turns....does the helicopter lose lift in the same way that fixed wing airplanes do as a result of bank angle? If so....why...and to what degree? Are there charts or forumla's for calculating that loss?
Help me out here Nick....some free dogs are in the balance here....the Chef suggests that I know not what of I speak. I would like to be able to baffle said Chef with my intricate knowledge of things rotorary when we next dine at his establishment.
Join Date: Apr 2003
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SASless,
The aircraft does not lose lift, it is placed in a maneuver where the lift it has is shared by the need to stay aloft and the need to power the turn.
The extra thrust that the rotor must produce to power the maneuver is calculated the same for any craft, including a car.
For a helo, at 60 degrees bank, the rotor must produce twice the lift it does in level flight, so the power must go up (raise the collective) or the speed must drop (use kinetic energy to power the turn).
The easiest way to calculate the extra power needed is to calculate the thrust increase needed, convert it into equivilent gross weight, and enter the "cruise charts" at that extra weight. The load factor is simply the reciprical of the cosine of the bank angle. At 45 degrees of bank, the aircraft picks up an additional 40% of its weight.
http://www.aerospaceweb.org/question...ce/q0146.shtml
Send your dinner buddies back to basic aero school, partner, and pass the chili dogs!
The aircraft does not lose lift, it is placed in a maneuver where the lift it has is shared by the need to stay aloft and the need to power the turn.
The extra thrust that the rotor must produce to power the maneuver is calculated the same for any craft, including a car.
For a helo, at 60 degrees bank, the rotor must produce twice the lift it does in level flight, so the power must go up (raise the collective) or the speed must drop (use kinetic energy to power the turn).
The easiest way to calculate the extra power needed is to calculate the thrust increase needed, convert it into equivilent gross weight, and enter the "cruise charts" at that extra weight. The load factor is simply the reciprical of the cosine of the bank angle. At 45 degrees of bank, the aircraft picks up an additional 40% of its weight.
http://www.aerospaceweb.org/question...ce/q0146.shtml
Send your dinner buddies back to basic aero school, partner, and pass the chili dogs!
Join Date: May 2005
Location: Missouri, USA
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although my math skills are rusty what Nick says makes sense prima facie. I'm just working on adding my helicopter rating to my fixed wing PPL and asked that question just the other day. Of course, Nicks answer wasn't what my instructor said but I trust Nick slightly more.
As for the dogs and suds, bring 'em on!
As for the dogs and suds, bring 'em on!
