Cubman
Member
Hi Folks, Best regards to All - it's time for my biennial question:ermm:
Same book but different chapter. In Chapter 5, on page 76, there is table 5.5; in the last two columns for E and G points on the envelope Mr Hiscocks has given for
H. tail load, P,lb. n=1 the values -137 and -22 which I presume are n=-1.
When I crunch through the calculation for example n= -1, q= 23.3, Aft CG= -0.25 and column G, the equilibrium equation comes out as:
-839 -(-0.25 * -1440) + ( 2 * 74) - (P * 15) = 0 whence P= -70 not -22
Am I correct in making the second term minus 0.25 because it is aft, and minus 1440 for the lift as n=-1? I have worked through for n=-2.4 with no problem, in that my answers agree with the table.
All the Best
Cub
Same book but different chapter. In Chapter 5, on page 76, there is table 5.5; in the last two columns for E and G points on the envelope Mr Hiscocks has given for
H. tail load, P,lb. n=1 the values -137 and -22 which I presume are n=-1.
When I crunch through the calculation for example n= -1, q= 23.3, Aft CG= -0.25 and column G, the equilibrium equation comes out as:
-839 -(-0.25 * -1440) + ( 2 * 74) - (P * 15) = 0 whence P= -70 not -22
Am I correct in making the second term minus 0.25 because it is aft, and minus 1440 for the lift as n=-1? I have worked through for n=-2.4 with no problem, in that my answers agree with the table.
All the Best
Cub