T ABLE Il. - Ojfsets of various streamline forms , United States models
Navy B (Goodrich) NavyO NavyF E. P. Parseval P . I Parse val P. II Parse val P. III
-- - - - - - - -.... - --- -
Distance Dlam- Distan ce Dlam- Distance Diam- Distance Diam- Distance Diam- Distance Diam - Distance D!am-from eter from eter from etor from eter from etcr from et.cr from eter nose nose nose nose noso nose nose
--- -- - - - - ---- ·- - - -- -
Pa. L. Pet. D. Pet. L. Pet. D. Pet. L. Pet. D. Pet. L. Pet. D. Pet. L. Pet. D. Pet. L. Pet. D. Pet. L. Pet. D.
2. 36 24. 16 2. 81 32. 47 1. 23 23 . 12 0. ];3 24. 88 1. 25 27. 37 1. 25 27. 27 1. 25 21. 56
4. 73 41. 27 5. 62 55. 06 2. 45 35. 06 2. 59 34. 60 2. 50 37. 92 2. 50 37. 92 2. 50 32. 99
7. 09 55. 14 8. 43 69. 61 3. 68 43. 90 5. 19 48. 44 5. 00 51. 95 5. 00 51. 95 5. 00 47. 79
9. 45 65. 27 11. 24 79. 22 4. 91 50. 61 10. 37 66. 10 10. 00 71. 17 10. 00 71. 17 10. 00 66. 23
11. 81 75. 36 16. 86 91. 17 7. 36 62. 73 15. 56 78. 12 14. 99 83. 38 15. 00 83. 36 15. 00 78. 70
14. 18 81. 94 22. 48 97. 40 9. 81 72. 08 20. 75 86. 66 19. 98 91. 17 20. 00 91. 17 20. 00 88. 05
18. 90 90. 31 28. 11 100. 00 12. 26 78. 57 2 5. 94 92. 73 24. 98 96. 10 25. 00 96. 10 25. ~8 94. 03
23. 63 94; 98 33. 73 100. 00 14. 71 84. 93 31. 12 96. 75 29. 98 98. 96 30. 00 98. 96 30. 0 97. 40
28. 35 98. 09 42. 16 98. 18 19. 62 93. 51 36. 31 99. 40 34. 97 100. 00 35. 00 100. 00 35. 00 99. 22
33. 09 99. 64 50. 59 94. 29 24. 54 98. 05 41. 50 100. 00 39. 96 99. 48 40. 00 99. 48 40. 00 100. 00
~
37. 82 100. 00 59. 02 88. 83 29. 45 99. 61 48. 81 98. 44 44 . 96 98. 18 45. 00 98. 18 45. 00 100. 00
47. 25 98. 44 fr7. 45 81. 56 34. 35 100. 00 56. 12 93. 77 49. 96 94. 81 50. 00 94. 81 50. 00 98. 06
56. 70 93. 06 75. 89 71. 69 39. 27 99. 74 63. 43 86. 23 54. 96 89. 87 55. 00 89. 87 55. 00 95. 86
66. 15 83. 25 84. 32 59. 48 44. 17 98. 96 70. 74 75. 32 59. 96 83. 90 60. 00 83. 90 60. 00 91. 69
70. 88 76. 91 89. 94 48. 57 49. 07 97. 53 78. 05 60. 52 64. 95 76. 36 65. 00 76. 36 65. 00 85. 97
75. 60 69. 38 92. 75 41. 56 53. 98 95. 15 85. 36 44. 16 69. 95 67. 53 70. 00 67. 53 70. 00 78. 96
80. 33 61. 00 95. 56 31. 95 58. 78 62. 34 92. 68 23. 90 74. 94 57. 66 75. 00 57. 66 75. 00 70. 91
85. 05 51. 44 98. 37 18. 96 63. 69 88. 31 100. 00 - 0 79. 94 47. 01 80. 00 47. 01 80. 00 59. 74
89. 78 39. 35 100. 00 .0 68. 69 83. 25 -- - --- - --- - -- 84. 93 35. 84 85. 00 35. 84 85. 00 47. 27
92. 14 31. 94 ------- ------ 73. 60 77. 27 -- ---- - ------ 89. 92 24. 16 90. 00 24. 16 90. 00 23. 25
94. 50 23. 44 ---- -- - ------ 78. 51 70. 26 ------ - ------ 91. 92 12. 21 95. 00 12. 21 95. 00 17. 14
96. 86 14. 00 -- - --- - -- - -- - 83. 41 62. 38 -- --- -- ---- --· 100. 00 . 0 100. 00 .0 100. 00 .0
98. 14 8. 97 ---- --- -- -- -- 88. 32 52. 47 -- --- -- -- ---- ---- - ·-- ------ -- -- --- ---- -- - -- --- ------100. 00 . 0 -- ~ ---- ------ 93. 22 40. 52 -- --- -- ------ ------- -- ---- -- ---- - - - -- -- --- --- -- - ---
------ ------ ------- ------ 94. 45 36. 75 ------- ------ ------- ------ ----- -- ------ --- --- -- -- --
------ ------ -- ----- - - - - - - 95. 68 33. 12 - - - - -- - ---- - - ------ - - ---- - -- ----- ------ ------ ------------ --- --- ---- --- ----- - 96. 91 28. 31 -- ---- - --- -- - -- ----- ---- -- - - - ---- ---- -- -- ---- ---- ---- ---- ------ -- -- --- ------ 98. 13 22. 47 ------- ------ ------- --- --- -- ---- - --- --- ---- -- - - - ---
------ ------ ----- -- ------ 99. 36 12. 26 ------- ---------- --- ------ ------- ------ --- --- -- -- --100. 00 .0 . ------ ------ --- --- - -- --- - --- -- - - --- --- ------- --- --- ------- -----.. ------ ------
a. s. T.
Distance Dlam-from eter nose
-
Pet. L. Pet. D.
1. 24 21. 41
2. 51 32. 98
4. 99 47. 83
9. 99 66. 07
14. 98 78. 89
19. 97 88. 07
24. 97 94. 04
29. 96 97. 32
34. 97 99. 11
39. 98 99. 80
44. 99 100. 00
50. 00 98. 75
54. 99 95. 87
59. 97 91. 75
64. 96 86. 24
69. 94 79. 14
74. 93 70. 34
79. 91 59. 76
84. 89 47. 39
89. 87 32. 99
94. 86 10. 83
100. 00 . 0
------------------ --- ------ -- - --- ---
------ -- ---------- ------------ --- ---
--- --- ----- ------ - -- ----
Pony blimp A.A.
Distance Dlam· from eter nose
Pet. L. Pet. D.
2. 09 20. 58
4. 19 33. 49
8. 38 54. 65
12. 57 67. 71
16. 75 77. 50
20. 94 84. 60
25. 13 89. 99
29. 32 94. 18
33. 51 97. 23
37. 70 99. 01
41. 88 100. 00
46. 07 99. 43
50. 26 98. 08
54. 45 95. 88
58. 64 93. 47
62. 83 89. 64
67. 02 84. 81
71. 20 78. 42
75. 40 71. 04
79. 58 63. 52
83. 76 54. 65
87. 96 45. 78
92. 14 35. 49
96. 34 22. 21
100. 00 . 0
- - ---- -- ---------- ----- ----------------- -- ------- ---- - ------
~
.!"d
i t:j
~
z >
~
~
{/l
~
... I-£
... ~
~
TM 1-320
12-14 AIR CORPS
12. Index of form e:fficiency. - In genera l, in design it is desired
to get the greatest volume from the. least surface area as this reduc~
weight and diffusion. Fortunately, good streamlined shapes usually
have high prismatic coefficients, but of course some shapes are r.nore
efficient in this regard than others. In studying relative efficiency
of shapes , both the resistance coefficients and the prismatic coefficients
must be considered . The ratio of the latt er to the former is called
the index of form efficiency, H t· ·
H,=&
13. Dlustrative resistance problem.-a. Proolem .. -Given an
airship whose hull has a length of 200 feet and a major diameter of
43.5 feet 'vith the hull offsets those of the C type airship envelope.
( 1) w·hat is total volume of envelope~
(2) "What is hull resistance at 60 miles per hour in standard atmos
phere~
b. Solution .
,(1) Vol =Q.vAL
From Table I, Qv is 0.6562.
7rd2 3.1416 A =4= 4 (43.5)2 = 1,485 square feet.
H ence Vol=0.6562X1,485X200=195,000 cubie feet.
(2) R=0Dp(vol)2/3vl.86
22 60 MP H=60 X15=88 feet per second.
GD from T able 1=0.0136 at 60 M PH
R=0.0136 X0.00237X (195,000)213 X881·86
R=455 pounds.
14. Scale effect.- a. One great reason why so much difficulty is
encountered in determining prior to construction the resistance o£ the
completed hull lies in the fact that the resistance of the model cannot
be multiplied by the ratio of the linear dimensions of the model and
the completed hull to determine the resistance of the latte r. The
discrepa ncy between the calculated resistance and the actual resistance
of the full- sized airship is attributed to scale effect. Often errors
in calculation due to :faulty data or bad theories are so explai ned away
by those responsible :for the mistakes. There are several reasons how
ever, why, even with proper data and theory discrepancies will ·exist
between calculated and actual resistance.
AIRSHIP AERODYNAMICS
TM 1-320
b. The theory of dimensions shows that the coefficients of resistance .
1'L vary directly as -;-· where
v=velocity in feet per second.
L=som e convenient linear dimension of the body such as the
diameter in the case of a cylinder.
v= kinematic viscosity coefficient of the fluid.
a. v, the kinematic viscosity coefficient, is defined as the ratio be
tween the absolute viscosity coefficient and the atmospheric mass
density. Hence-e -
v=~' where v 1s the absolute viscosity coefficient of the air and is a
constant.
vL d. -;' called the Reynolds number after Profes sor Reynolds,
depend s on three variable quantities, p, v, and L : To predict full
scale performance from the model tests, allowance must be made for
the fact tha t the L in the full-sized airship is very different from the
L in the model, and consequently the co-efficient of resistance will be
differ ent.
e. To overcome the effect of this difference a wind tunnel has been
built at Langley Field in which p may be sufficiently increased to
make the product pL for the model equal that of the full- sized small
nonrigid airship, thus eliminating scale effect.
15. Resistance of completely rigged airship. -a. There are·
very little data available showing the relative resistance of the various
parts combining to produce the total resistance of a completely rigged
airship due to the difficulty in obtaining dynamic similar ity between
the model tested and the full-scale airship.
b. Total resistance of airships may be subdivided approximately as
follows for-
(1) Large nonrigids with closed cars: Percent
(a) ~nvelope_____ ____ ____ __ ___ ___ _ _ _ ____ _ ______ ___ _ 45
(b) Surfaces-------- -- ---- -- -- ----- - - ----- - ---- - --- 20
(c) Rigg ing and suspension cables___________________ 15
(d) Cars________________________ ___________________ 15
(e) Accessories --- ------ -- --- --- -- ------------------ 5
(2) Small nonr igids with open cars :
(a) ~nvelope ____ -------- _ ---- ·· _______ -·- ------ _____ . 35
(b) Surface s---- -- -- -- ----- --- --- ---- ---- ------ -- - · 25 (c) Rigging and cables ______________________________ 20
(d) Cars------------------ -- - ----- --- ------- ---- -- _ 15
(e) Accessories---- ------ -- ... _ ------- ------------ 5
TM 1-320
16-16 Am CORPS
(3) Semirigids: Percent
(a) ~vel ope- ------------ ------------- ------ - --- -- ~ 53 (b) Surfaces _______________________________________ 20
(c) Jtiggjng- -- -- -----~-------- ---- -- ---- -------- -- - 7
(d) Cars _______________ ------ ----------- __ - - ------- 13
(e) Accessorie s------------ --- ------- ------------- -- 7
( 4) Large rig ids : (a) IIull ___________________________________________ 60
(b) Surfaces____ ___________________________________ 15
(c) Cars and suspe nsions------ ---- --- ---------- ----- 2q
(d) Miscellane ous rigging and accessorie s____________ 5
16. Deceler ation test.- a. Tests are 1nade frequently on full
sized airships to determine actual risistance of the airshi p at various
speeds . In these tests the airship is brought to a certain velocity
and then the motor s are idled, the velocity being recorded against
time as the airship decelerates.
b. The general theory is that the resistance, or force causing de
celeration, is given by the equation :
R= Mv a, where
ex.= (deceleration in f eet per second) z.
Mv= the virtual mass of Lhe ship .
The virtual mass of an airship is the mass of airshi p and contents
plus the mass of air which is carried along with it. This latter is
computed by the Mun k formula :
AMo=P1, where r is the radius of largest cross section.
c. Observing velocity at end of each second gives the rate of
change of velocity, or deceleration, for each second and by interpola
tion for each air speed . Actually formu las a re employed which in- .
volve calculus and are beyond the scope of this manual.
d. These deceleration .tests are quite valuable as a check against
the resistance formul as developed in this section. They are however
often complica ted by poor instru ments or faulty observation, render
ing it difficult to place a proper value on results so obtained. For
the present more confidence is to be placed on the resistance formulas
and the power requiremen t formulas which will be developed in the
next section.
AIRSHIP AERODYNAMICS
SECI'ION III
POWER REQUIREMENTS
TM 1-320
Paragraph
Power required to overcome airship resistance ____________________________ 17
Result s of various speed trials----- - - ------------- -----'- ---------------- 18
Burgess' formula for horsep ower- --·-- ---- ----- ---- ------ -------- -- - - ---- 19
Speed devel~ped by given horsepower______ ______________________________ 20
Summary ------------------------ -- ------- - - ______________ --·-------- --- - 2l
17. Power required to overcome airship resistance.-a. To
mainta in uniform velocity in flight, resistance of the airship must be
overcome by thrust of the propellers. The work done by the pro
pellers equals the product of the resistance times the distance through
which the airship moves. ·
b. The unit of work in the English system is the foot-pound, or the
quantity of work performed by 1-pound force acting through a dis
tance of 1 foot. Hence work done in propelling the airship in foot
pounds equals resistance in pounds times air distance traveled by
the airship.
a. Power is defined as the rate of doing work, 1 horsepower equaling
550 foot-pound s per second. Therefore the power utilized to over
come hull resistance must equal resistance multiplied by velocity in
feet per second divided by 550.
d. The resistance is given by the equation (see sec. II):
R= CD p (vol)213ut.86
Then the horsepower required to overcome this resistance is given by
the formula:
Cn p (vol)213if·86
H. P.= 550
e. Problem and solution-( 1) Probl em.-What horsepower will be
required to drive an airship of 195,000-cubic-foot capacity at 60 miles
per hour ( 88 feet per second) in atmosphere of standa rd density~
The envelope shape coefficient is 0,0136. The propeller efficiency, E,
is 60 percent. The envelope resistance, F, is 40 percent of the total
resistance of the airship.
(2) Solution .-The horsepower necessary to overcome hull re
sistance is given by-
Cn p (vol)213if·86
H. P = 550
_ (0.0136 X 0.00237 X 3376.4 X 359000)
=71.1 horsepower.
T.M 1-320
17-18 Am CORPS
Since hull resistance is but 40 percent of tota l, the horsepower to over
come tota l resistance
Since propeller efficiency is 60 _percent-
Total horsepower required= (71.1)(o.40~o. 60)
=296 horsepower .
f. As illustrated in the problem in d above, the following is a con
venient formula for the horsepower required when the percentage of
resistance due to the hull and the propeller efficiency are known.
g. A commoner method of determining the horsepower requirements
is to determine a shape coefficient by wind tunnel test of the completely
rigg ed model. In this case the body in question is not as perfe ct
a streamlined shape as the hull itself so the resistance varies more
nearl y as the square of the velocity. Then the horsepower required
becomess--
0 D p(vol)213v8
H. P.= 550E
where 0' D is the shape coefficient of the model.
( 1) Problem.- What horsepower wil~ be required to drive an air
ship of 195,000-cubic-foot capacity at 60 miles per hour (88 feet per
second) when the atmosphe ric density is standa rd, the coefficient of
resistance 0' D of the completely rigged ship is 0.0165, and the propeller
installation efficiency is 60 percend
(2) Sol1JJtion.
H P = 0' D p(vol)213v8
. . 550 E
0.0165 X0.00237 X 195000213 888
~----~~5s=o~x~o~.=6o~-----
= 275 horsepower, nppro:-dmately.
18. Results of various speed trials. -a. The following data
were obtained by progressive speed trials made on the United States
Navy C class nonrigid airship of 180,000-cubic-foot capacity:
•
AIRSHIP AERODYNAMICS
R- pounds
V in
foot- R . P M. B. H. P . E ' sec-
onds • Total Hull
66. 6 1, 100 109 60 540 334
73. 3 1,200 143 60 643 394
80. 1 1, 300 183 60 754 457
87. 7 1,400 231 60 875 517
Ap-
pend-
ages
TM 1-320
C'D
0.02 0
. 019
. 019
. 018
The value 0' D is the corrected coefficient of resistance , but its ac
cura cy is somewhat uncerta in, also the proportions o£ hull resistance
app ear high . The value o£ 0' D obtained from the win d t unnel test
was ·o.027. The proportional value of the appendages or parasiw
resistance was computed from the wind tunnel data.
b. The follo w·ing data were obtained from deceleration tests of
Germa n r igid airshi ps:
Num- Maxi- Pro-
ber of mum por-
Name Cub ic feet D L B. H. P. tional C' D veloc-en- effi-gmes ity • cten cy
Foot-. Feet Feet seconds
J-'Z 10 706,000 45. 9 460 3 62. 4 450 67 0. 107
L 33 2,1 40,00 0 78. 3 645 6 92. 5 1, 440 49 . 039
L 36 2,140,000 78. 3 645 6 92. 5 1, 440 62 . 045
L 43 2,140,000 78. 3 645 5 88. 9 1, 200 56 . 047
L 44 2,140 , 000 78. 3 645 5 94. 0 1, 200 56 . 031
L 46 2, 140,000 78. 3 645 5 95. 5 1, 200 58 . 031
L 57 2, 640,000 78. 3 745 5 94. 8 1, 200 69 . 034
L59 2, 640,000 78. 3 745 5 94. 6 1,200 66 . 038
L 70 2,400,000 78. 3 694 7 113. 5 2,000 65 . 031
19. Burgess formula for horsepower. -a. A very h andy for
mula for determining the horsepowe r r equired to drive an airship
of any given volume and speed is furni shed by the Natio nal Advisory
Committee for Ae1·onauti cs Report No. 194, as follows:
v3p_(vol)2' 3
H. P.= Op
~8
TM 1-320
19-20 AIR CORPS
where Op is a constant which can be take n from the compilation
below:
N O"nrigid airships .
50,000 to 200,000 cubic feet ------------- - - ----- -------- Op=20,000
200,000 to 300,000 cub ic feet_ ________ .. . -·---- ------ -- Op=21, 000
300,000 to 400,0J :> cuui<: feeL--- -----·-·--- -------- ------ Op= 22,000
R igid airsh ips
1,000,000 to 2,000,000 cubi c feeL _______________________ Op=3 0,000
2,000,000 to 3,000,000 cubic feeL- -------- ---------- ---- Cp=32,000
3,000,000 to 4,000,000 cubic feeL _______________________ Cp=33, 000
4,000,000 to 6,000,()()() cubic f eet_ _______________________ Cp= 34,000
6,000,000 to 10,000,000 cubic f eeL ____________ -- - ----- Cp=35,000
b. Solving the problem given in p aragraph 17g (1) by the Burgess
formula gives-
•
P ( vol )213 va
H. P.= Cp
- (0.00237) (195,000 )11' 3 (88)8
20 ,000
= 273 horsepow er.
20. Speed developed by g iven horsepower. -a. By transpos
ing t he horsepower for mula s the following formul as are obtained for
the speed developed by a given horsepower :
2.sn /H. P. X550XE X F (H. P.X550 XEX F\
'
v=-y CvXPX (vol)2ta = CvXP X (vol)2/3- ) from paragraph 17}.
3 /H. P. X 550X E
= -y 0 , X P X ( vol )2'3 from paragraph 17 g.
3fH.P. X 01)f h 9 = V P ( vol) 213 rom p aragrap 1 a.
b. Problem and. solution.-( 1) Problem.-An airship of 195,000-
cubic-foot capacity has a power install ation of two motors developing
150 horsepower each, or a tot al of 300 horsepower. The atmospheric
density is standard. What speed should be obtained at full powed
(2) S olution.- Using Burgess' formula.
3 /H. P. X 01)
v=-y p(vol)21a
3 I 300 X 20000
- v o.oo237 x (195ooo)2' 3
=90.8 feet p er second = 6L9 miles per hour.
..
AIRSHIP AERODYNAMICS
c. Problem UJrU1 solution.
TM 1-320
2Q-21
(1) Problem.-An airship of 195,000-cubic-foot capacity is to be
equipped with two .engines developing a total of 300 horsepower.
What speed can be expected using the following data~
(a) Standard atmospheric density.
(b) Shape coefficient, 0 n is 0.0136.
(e) Propeller efficiency, E, is 60 percent.
(d) Envelope resistance is 40 percent of total resistance of com
pletely rigged airship.
(2) Solution.-Using Prandtl coefficient.
( 300X550X0.60X0.40 ) 0·86
V= 0.0136X0.00237X3,376.4
=88.4 feet per second=6 0.3 miles per hour.
d. Experience has shown the lower figure, as determined by Prandtl
coefficients, to be more generally correct than the higher figure as
determined by the Burgess formula. .
21. Summary.- a. From study of the formulas it appears that
the speed of an airship is proportional to the cube root of the horse
power, or vice versa the horsepower varies directly as the cube of
the speed. Since power plant weights vary directly as the horse
power, the weight of the power plant varies also as the cube of the
speed. A point is readily reached therefore beyond which it is not
economical to increase the speed due to the excessive weight"s involved.
b. In still air the higher the speed the less economical the fuel
consumption and the shorte r the radius of action. This is not true
when the airship is traveling against adverse winds. The study of
just which air speed is the most economical will not be discussed in
this manual as it properly belongs to the subject of navigation.
8EariON IV
STABILITY
Paragraph
Variation of pressur-e distri bution on airship bull ________________________ 22
Specific stability and center of gravity of airshiP------- ------ - ------- - ---- 23
Center of buoyanCY- ------------------- - - --- ----- - --- --- ---- - ---------- 24
Description of major axis of airshiP - ----- - ----- - ------------- --- - --- --- 25
Types of stability----------- ---- - ---- ---- - - ---------·--- ------ - ------ 26
Forces and moments acting on airshiP---------- - ---------- ----------- --- 27
Damping moment------- ------ ---- ----- --- ----- - ------------------ ---- 28
Longitudinal stabi I i ty - ------ --- --- --- --- ------ ---- - ------ --- ------- --- -- 29
Directional stabilitY---------------------- - ----------- - - -------------- 30
Lateral stability--------------------- - - - - -- - - --- ------------------- ---- 31
S1JmJuarr-----·-----·--------- ---- ..... . ····- ---------------- ... -·fl"'---·-·- ~