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The trick here is to note that in trimmed level flight the load factor is one but the stick force is zero. So the first 1g is free in terms of stick force.
Limiting load factor for a large transport aircraft in the clean condition is 2.5g. So to go from 1g straight and level to 2.5g we require an increase of 1.5g. Each 1g reuqires a 280N stick force, so an increase of 1.5 g from 1g to 2.5g requires a 420 N stick force. The only other limit load factors that appear in the JAR ATPL exams sare: 2.0g for large transport aircraft with flaps down. 4.4g for a light utility category aircraft. |
Ok, that's clear now !
Thanks and Good night. |
Incorrect algebra!
I know this topic/question is 4-5 years old, but in coming across it today, as an engineer, I tried to work it, and kept coming out with a different answer than what was explained by those that replied to the original poster. The algebra does not work out, and the replies, while getting the published known answer, got the actual problem (_as posed_) wrong.
The original poster's problem stated that 2Vs gives a load factor of 2. If you solve for Lf2 with v1=2Vs, Lf1=2, and v2=1.3Vs, taking weight, surface area, air density, and drag coefficients to be all constant for the purposes of this problem (they don't change with respect to the difference in velocities): then, load factor is just a function of velocity (Lf1=f(v1^2)): So we compare the ratios: Lf1/Lf2 = v1^2 / v2^2. so, substituting known values: 2/Lf2 = (2Vs)^2 / (1.3Vs)^2. Vs cancels, and solve for Lf2 = 0.845. [In fact, at 2Vs, the load factor should be 4, not 2 as posed!! If you work it with putting Lf1=4 above, then you get the correct answer of Lf2=1.69 for 1.3Vs] Stated better: Any aeroplane flying at a multiplicand of Vs and at the angle of attack to attain the maximum lift will experience a load factor equal to the square of the multiplicand as a g force, e.g 1.3Vs creates 1.69g. from Principles of Flight for Pilots by Peter Swatton Principles of Flight for Pilots - Peter J. Swatton - Google Books |
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