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The Drag Polar at Maximum cruisespeed of Mach 0.30257at30,000ft.
From Design Data Sheets:
 = 0.0027717 slug/ft3,  = 3.1324010-7 lb.sec/ft2, a = 1116.45 ft/s, then
the airspeed is = 337.8 ft/s
1. Wing contribution:
Assume35% laminar flow on the root and 45% at the tip, Cr = 14.07 ft,
given Ct = 4.11
Rer = 30993843.18
Then 0.02975
Then Cfr = 1.753*10-9
 At tip:
Ret = 9053638.626
Then 0.05523
Then Cft = 5.469*10-9
Cfavg = 3.611*10-9
Exposed wing area:
, SEXPO = 397.0931 ft2
(t/c) = 0.12, λ = 0.25, τ= 1.4, Swet =825.63 ft2
Refu = 2.0045*108
1
Fromdata design sheets:
M Rwf
0.8 0.97
0.78 X
0.7 0.99 , then X = Rwf = 0.986
AR = 7.8, Ʌn = 21.72o
, n =0.25, m = 0.37, λ = 0.25, then Ʌm = 19.8o
at
max
cos Ʌm = 0.94
Fromdata design sheets:
M RLS
0.6 1.15
0.78 x
0.8 1.26
Then at 0.78, RLS = 1.249
L = 1.2
,Then (CDo)w = 1.175*10-8
2). Tail:
Tail; horizontal: (assuming at root45% laminar, at tip 60%)
At root: (Cr = 6.575 ft)
Ret = 14483618.97, Ret =14.48*106
2
0.046, then Cfr = 1.808*10-3
At tip (Ct = 4.11)
Ret = 9.0536*106
0.0661, then Cft = 1.631*10-3
Cfavg = 1.72*10-3
, Sh =120.6 ft2
, from the Previous equation then;
Swet = 250.217 ft2
.
AR = 5.0793, Ʌn =20o
, n = 0, m = 0.37, λ = 0.625, then Ʌm = 16.52o
at
max
Then cos(Ʌm) = 0.95o
M RLs
0.6 1.15
0.78 X
0.8 1.27 then RLs = 1.258, L = 1.2
(CDo
)H =0.001144463
Tail; vertical: (assuming 50% laminar flow at root and 35% at the tip)
At tip: (Ct = 6.49ft)
Ret = 14.29*106
, 0.00397,then Cft = 2.037*10-3
, L =1.2,
3
At root: (Cr =11.73 ft)
Re = 25.839*106
0.0404, Cfr =1.489*10-3
Cavg = 1.763*10-3
, Sexpov = 77.5 ft2
,
(t/c) = 0.12, λ = 0.553, τ= 1.4, then Swetv = 160.84 ft2
AR = 1.6392, Ʌn =40o
, n = 1, m = 0.37, λ = 0.554, then Ʌm = 30.112o
at
max
Cos(Ʌm) =0.86
M RLs
0.6 1.14
0.78 X
0.8 1/24 ,then RLs = 1.23
(CD0)v = 0.000737271
3). Fuselage contribution: (assuming 10% laminar flow)
L = 91.633 ft (correcting value by WebPlotDigitizer),
Refu = 2.15*108, 0.00657,
Then Cffus = 1.47456*10-3,
4
Fuselage wetted Area:
Sfus = (πD2/4), then Sfus = 43.93 ft2
Then Df = 7.38
Assuming (L1 = 12.4 ft, L2 = 68.30 ft, L3 = 10.933 ft):
Then fuselage wetted area = 171.75 + 1810.9 + 155.367 = 2138.017 ft2
dp = 0.95 ft, (CDf)B = 0.00606 ,(CD)b = 0.00005448211
then (CD0)B = 0.00660821
4). the span (Oswald) efficiency factor:
Assuming leading edge radius = 1.505%
then MAC = 11.403
LLER = 11.403*1.505%, then LLER = 0.1716 ft, Re = 3.78*105
ɅLE = 25.42o
P1 = 2.16
5
P2 = 2.4*105, then P2 > 1.3*105
(Assume CLα = 2π)
Then e = 0.993407387

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EASY TUTORIAL OF HOW TO USE CAPCUT BY: FEBLESS HERNANE
 

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  • 1. 0 The Drag Polar at Maximum cruisespeed of Mach 0.30257at30,000ft. From Design Data Sheets:  = 0.0027717 slug/ft3,  = 3.1324010-7 lb.sec/ft2, a = 1116.45 ft/s, then the airspeed is = 337.8 ft/s 1. Wing contribution: Assume35% laminar flow on the root and 45% at the tip, Cr = 14.07 ft, given Ct = 4.11 Rer = 30993843.18 Then 0.02975 Then Cfr = 1.753*10-9  At tip: Ret = 9053638.626 Then 0.05523 Then Cft = 5.469*10-9 Cfavg = 3.611*10-9 Exposed wing area: , SEXPO = 397.0931 ft2 (t/c) = 0.12, λ = 0.25, τ= 1.4, Swet =825.63 ft2 Refu = 2.0045*108
  • 2. 1 Fromdata design sheets: M Rwf 0.8 0.97 0.78 X 0.7 0.99 , then X = Rwf = 0.986 AR = 7.8, Ʌn = 21.72o , n =0.25, m = 0.37, λ = 0.25, then Ʌm = 19.8o at max cos Ʌm = 0.94 Fromdata design sheets: M RLS 0.6 1.15 0.78 x 0.8 1.26 Then at 0.78, RLS = 1.249 L = 1.2 ,Then (CDo)w = 1.175*10-8 2). Tail: Tail; horizontal: (assuming at root45% laminar, at tip 60%) At root: (Cr = 6.575 ft) Ret = 14483618.97, Ret =14.48*106
  • 3. 2 0.046, then Cfr = 1.808*10-3 At tip (Ct = 4.11) Ret = 9.0536*106 0.0661, then Cft = 1.631*10-3 Cfavg = 1.72*10-3 , Sh =120.6 ft2 , from the Previous equation then; Swet = 250.217 ft2 . AR = 5.0793, Ʌn =20o , n = 0, m = 0.37, λ = 0.625, then Ʌm = 16.52o at max Then cos(Ʌm) = 0.95o M RLs 0.6 1.15 0.78 X 0.8 1.27 then RLs = 1.258, L = 1.2 (CDo )H =0.001144463 Tail; vertical: (assuming 50% laminar flow at root and 35% at the tip) At tip: (Ct = 6.49ft) Ret = 14.29*106 , 0.00397,then Cft = 2.037*10-3 , L =1.2,
  • 4. 3 At root: (Cr =11.73 ft) Re = 25.839*106 0.0404, Cfr =1.489*10-3 Cavg = 1.763*10-3 , Sexpov = 77.5 ft2 , (t/c) = 0.12, λ = 0.553, τ= 1.4, then Swetv = 160.84 ft2 AR = 1.6392, Ʌn =40o , n = 1, m = 0.37, λ = 0.554, then Ʌm = 30.112o at max Cos(Ʌm) =0.86 M RLs 0.6 1.14 0.78 X 0.8 1/24 ,then RLs = 1.23 (CD0)v = 0.000737271 3). Fuselage contribution: (assuming 10% laminar flow) L = 91.633 ft (correcting value by WebPlotDigitizer), Refu = 2.15*108, 0.00657, Then Cffus = 1.47456*10-3,
  • 5. 4 Fuselage wetted Area: Sfus = (πD2/4), then Sfus = 43.93 ft2 Then Df = 7.38 Assuming (L1 = 12.4 ft, L2 = 68.30 ft, L3 = 10.933 ft): Then fuselage wetted area = 171.75 + 1810.9 + 155.367 = 2138.017 ft2 dp = 0.95 ft, (CDf)B = 0.00606 ,(CD)b = 0.00005448211 then (CD0)B = 0.00660821 4). the span (Oswald) efficiency factor: Assuming leading edge radius = 1.505% then MAC = 11.403 LLER = 11.403*1.505%, then LLER = 0.1716 ft, Re = 3.78*105 ɅLE = 25.42o P1 = 2.16
  • 6. 5 P2 = 2.4*105, then P2 > 1.3*105 (Assume CLα = 2π) Then e = 0.993407387