F4F-4 #1 – 2425, S, 12.0, 6.0, LB, LVL, 0
Mil(4), Tr: none, 6H/0V
MC= 0.0, V= 0.0, OC= 0.0
3H 3/3 HTL to SSE into 2125
1H 1/2 BTL to 2026
1H 2/2 BTL, 1/2 RL to SE in 1926
1H 2/2RL to IL in 1827/1927
(0c+4p+0da+0dv)-(3BT+2HT+0mv+0os) = 4 – 5 = -1 cr
F4F-4 #1 – 1827/1927, SE, 12.0, 6.0, IL, INV, -1
F4F-4 #2 – 3027, SSW, 12.0, 6.0, LB, LVL, 0
P: E(4), Tr: none, FP:6, 6H , 0V
MC= 0.0, V= 0.0, OC= 0.0: Net( 0.0)
3H 3/3 HTL=>S
3H 3/3 HTL=>SSE, fall into 2426
(0c+4p+0da+0dv)-((0 -4)t+0mv+0os)=4 – 4 = 0
F4F-4 #2 – 2426, SSE, 12.0, 6.0, LB, LVL, 0
1a = F4F #1 (Miquel) - 1827/1927, SE, 12.0, 6.0, IL, -1.0, INV
1b = F4F #2 (Bing) - 2426, SSE, 12.0, 6.0, LB, 0.0, LVL
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1c: Zero#2, 2853, W, 12.0, 6.0, LV, 0, LVL
Em(5), HT2 pull to STC, 4H2V [1/2H..]
MC=0 V=+0.6 OC=0
3H: RL 3/3 to LB
1H2V: RL 3/3 to ILB, HT 3/3 face SW
5p-(4tr+6cl+(2ht-2cr*))= -0.5spd
Zero#2, 2831, SW, 12.6, 5.5, IL, 0, ISTC
*cr = Transition Turn Decel Credit
Power = Em (5), Pull BT-2 to STC, 2H+2V+2H
(MC= 0.0, V=0.6, OC= 0.0 )= +0.6 removed OC as Zero #1 cannot lose altitude via OC when it goes into a steep climb.
2H removed initial roll left as it is not needed
V as above
VH RL (2/3), BTL (2/2) to E
H RL(3/3) to IL, BTL(1/2)
(5p+0c+0da+0dv) - (0c+6BTL+6tr-3decelcredit+0cd+ 6cl +0m+0mv+0os)= 5- 15 = -10
A6M2 #1 (Colugo) 1730, E, 12.6, 5.0 , IL, 0.0 , ISTC, 1BT
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