TM 1-320
4o-42 AIR CORPS
be equal to produce motion in a constant turn . From this is derived
the relation-
2a
sin 2 'I!= R(k2-kl)
where 'I!= angle of yaw.
a=di stance from center of volume to center of pressure
of tail surfaces .
R = radius of turning circle.
This relation gives results widely at variance :from the results o:f actual .
tests on full- sized airships, presumably due to the assumption that the·
resultant of the hull forces is zero. Fortunately, the total bending
moment due to a steady angle of turn is only about one-fifth as great
as that due to an equal fixed ang le of pitch where unbalanced weight
and centrifug al force are of equal magnitude.
41. Forces caused by gusts. -a. Very little is known concerning
maximum value of forces caused by gusts. The following statement
very excellently sums up the situation :
1 "The existence of veri table fount ains of upward rushing air whose
sides at times and places are shar ply separa ted from the surrounding
atmosphere must be taken into account in the design of airships. The
most violent of such currents, the tornado, combines vertical velocity
with rotation, but fortunately can be seen from a great distance, and
can and must be avoided. The thunder storm with large fully devel
oped cumulus tops is also conspicuous and avoidable. It would appear
to be :folly to enter such a cloud and subject the ship to the unknown
dangers of wind, rain, hail, and lightning. Barring ·such spectacular
hazard s, there remain convection currents which the ship may run into
at full speed. There is ampl e evidence that upward velocities as high
as 10 feet per second may be encountered. This vertical air velocity u,
combined with the relative horizontal speed v of the airship, will give
-1
the effect of a change of pitch of tan vu.,
b. It rem ains simply for the pilot, as stated in paragraph 38e, to
reduce the speed in bumpy atmosphere, especially if at the same time
the airship is developing large dynamic lift, positive or negative, as
then the stresses are already large.
42. Empirical formulas for maximum aerodynamic bendi n g
moment on hull and for forces on tail surfaces.-a. The follow
ing formula has been developed :for the maximum aerodynamic bend-
1 From "Airship Design" by C. P . Burgess by ·permiss ion of tbe Ronald Press .
AIRSHIP AERODYNAMICS
TM 1-320
ing moment to be expected from such bumpy weather as would be en
countered in mountainou s country:
Mb=0.005 ,W (vol)213L
where Mb= the maximum bending moment in foot-pounds.
L = the length of the airship in feet.
Use of this formula enables the pilot to calculate rapidly maximum
stresses to which his velocity in bumpy air may be subjecting his
airship.
b. Wher e surfaces are designed in approximate accordance with the
formula A = 0.13 ( vol) 213 , the tota l transverse force on either vertical
or horizontal surfaces may be computed quickly by the relati on:
F = 0.026 (vol) 218 p v2•
In above formulas
A=total area of either surface .
F = total force on either surface.
43. Method of calculating shear and bending moment on
hull.-a. The designer and also the pilot in determining shear and
bending moments on the airship must consider both static and dy
namic loads. Both must be computed independently and then added
together algebraically. I t often happens that dynamic loads serve
to reduce stresses due to static loading, but naturally the dangerous
case occurs when stresses are arithmetically additive.
b. The method to be describ ed applies more particularly to rigid
airships, but the principle can be applied to a nonrigid. In the latter
case, the load instead of being distributed throughout the length of
the hull is swung from the envelope by suspension cables which by
their tensions control very larg ely distribution of loading on the
envelope.
c. For calculation of stresses, the hull is considered as a beam loaded
with the weight s acting downward, lift of gas cells acting upward and
aerodynamic forces acting in any longitudinal plane whatsoever. All
loads . are considered as concentrated at the frames rath er than as
uniformly distributed. The calculations may be divided into steps,
as follows:
( 1) Calculation of static load.
(2) Calculation of shear due to static loading.
(3) Calculation of bending moment due to static loading.
( 4) Calculation of load, shear , and bending moments due to aero
dynamic forces.
( 5) Algebraic summation of effects of static and dynamic loading.
TM 1-320
48 AIR CORPS
d. The initial step in the computation is determination of .distribu
tion of weights. This is taken from the detailed weight st~tement,
weights therefrom being distributed to the proper frames. Lift of
the gas in each cell is computed next and distributed a s concentrated
forces on the frames. The static loads on the hull are the differences
between weight and buoyancy at each frame, lift being considered
so
-
.Jo
/. .
tq 000
{Plane conftnl1in? C.G al1d C.8.
90 po 80
/0 000
/()
&olf4ncr <~ncl
1000 Loadi~ ,.,
LBS.
2 000
I""'!•,__---Overall len9rh oF /lir.ship----~
t 1 I Loads'" 1.85. j
2000 2000 •
~----~------~~-----~.r-----~------~.
1000 /000 /000 1000
I 1+1000
- 1000
I ...__ ____ __,,_,QOQ
fjft1. IN
LB. Meter.s
F tcu•n: :.!7.-Loads, shear , a1111 bending moments ca used by static loading.
positive and loads negative . When the airship is in stat ic equilibrium,
the algebraic sum of the loads must equal zero. Figur e 27 illustrates
the computa tion of loads at each frame of an airship 50 meters long,
having four fram es spaced 10 meters apart. The method shown is
applied to the large st airships.
e. Commencing at either end of the airship, the shear at any frame
equals the algebraic sum of loads up to that frame. This system
gives a constant shear between frames, changing at each frame QY
AmSHIP AERODYNAMICS
TM 1-320
the amount of load at that frame. The shear in figure 27 was com
puted in this manner.
(1) For instance, the load at station 0 is -1,000 pounds. Then
the shear between stations 0 and 10 equals -1,000 pounds. At station
10 the load is + 2,000 pounds. Hence the shear between stations 10
and 20 is -1,000 pounds + 2,000 pounds, or + 1,000 pounds.
(2) For an airship in static equilibrium, when centers of buoyancy
and gravity are vertically disposed, areas under the shear curve must
add algebraically to zero. This should be checked before proceeding
to computat ion of bending moments.
f. For calculation of bending moments, all loads between ends of
the airship and any frame are considered as supported by cantilever
action from that frame. In the case illustrated by figure 27 starting
at station 0, the bending moment for-
(1) Station 0= 0.
(2) Station 10= - 1,000X 10= -10,000 meter-pounds.
(3) Station 20= ( -l,OOOX20) + (2,000X 10) =0.
g. An easier method of computing bE ding moments is to sum up
the areas under the shear curve. Thus in figure 27, for station 20,
the bending moment= 10,000- 10,000= 0. For an airship in static
equilibrium, when the center of gravity is vertically below the center
of buoyancy, the bend-ing moment c_urve returns to zero at both ends
of the airship, since the summations of positive and negative areas
under the shear curve are numerically equal.
h. Table IV, extracted from "Airship Design," by C. P. Burgess,
of the Bureau of Aeronauti cs, United States Navy, shows loads, shear,
and bending moments on the ZR- 1, computed in accordance with
the method described therein.
i. In computing aerodynamic loads, shear, and bending moments, a
method. somewhat similar to that described above is employed.
(1) Upturning dynamic forces on the hull are computed , using the
Munk formula. This formula is omitted here as it involves mathe
matical computation beyond the scope of this manual. The forces
so determined are distributed to the frames as concentrat ed loads.
(2) Excess static weight or buoyancy is then distributed to the
frames in proportion to the cross-sectional area at the frames, unless
known eccentric loading shows this distribution to be greatly in error.
(3) Dynamic force on surfaces is then distributed to proper frames.
This force, as shown in paragraph 39c, is given by the relation -
F= (Vol) v2[a(kz- k1) sin 28
TM 1-320
43 AIR CORPS
TABL E IV.-Loads, shear, and bending moments in U. S. S. ZR-1
when the gross lift is 136,631,. pounds
[This table reproduced from Airship Design, by C. P. Burgess, by permission of
the Ronald Press]
Station Gross Fixed Dispos- Total Load Shear
Bending
able moment meters lift weight weight weight m.
Pounds Pounds Pounds Pounds Pounds Pounds Pounds o _________ 307 2, 618 0 2, 618 - 2, 311 - 2, 311 10 _________ 1,453 1,877 0 1, 877 - 424 -23, 110 - 2, 735 20 _________ 2,812 1, 902 0 1, 902 910 - 50,460 . - 1, 825 30 ___ ______ 4, 496 1, 991 2, 276 4, 267 229 -68,710 - 1, 596 40 _________ 5, 789 2,328 2,200 4, 528 1, 261 -84, 670 - 335 50 _________ 7, 128 2,389 5, 182 7, 571 - 443 - 88, 020 - 778 60 _________ 8, 218 5,858 1, 512 7, 370 848 -95,800 70 70 _________ 8, 985 2, 708 2,378 5, 086 3, 899 - 95, 100 3, 969 80 _________ 9, 402 3, 091 5, 656 8, 747 655 -55, 410 4, 624 90--------- 9,510 9,483 6, 100 15,583 - 6, 073 - 9, 170 - 1, 449 100 _________ 9,540 3, 224 6, 055 9, 279 261 -24 660 - 1, 188 ' uo __ . ______ 9, 584 3,069 5, 704 8, 773 811 -36, 540 -377 120 __ _______ 9, 560 8, 183 5, 016 13, 199 - 3, 639 -40, 310 - 4, 016 130 ___ ______ 9, 536 3, 096 1, 790 4,886 4, 650 -80,470 634 140 ___ ______ 9,417 3, 064 5, .562 8, 626 791 -74, 130 1,425 150 ___ ______ 9, 003 2, 712 5, 406 8, 118 885 - 59, 880 2, 310 }6() _________ 8, 169 8,057 2, 259 10, 316 - ~. 147 -36, 780 160 170 _________ 6, 778 3, 076 2, 653 . 5, 729 1, 049 -35, 150 1, 212 180 ___ __ ____ 4, 467 3, 212 1, 227 4, 439 28 -23,030 1, 240 188- ---- ---- 2, 222 1, 520 0 1, 520 702 - 13, 110 1, 942 194.75 __ ____ 258 1, 100 1, 100 2, 200 - 1, 942 0
136, 634 74, 558 62. 076 136, 634 0000
. I I
( 4) The load on each fram e, shearing forces, and bending moments
are then computed and tabulated as explained in d, e, f, g, and h
above. A table so prepar ed, extracted from Airship Design, is given
below.
AIRSH I P AERODYNAMICS
TM 1-320
TABLE V .-Aerodynamic forces, shear, and bending moments in U. S.
S. ZR- 1 at 85 foot/seconds and 5° 42' pitch
Turning Unbal- I I Bending ' anced I
I moment Station meters force s ' L Load Shear stati c I ' .
on hull I
weights 1
pounds
-
1 Pounds , Pounds o _________________ -820 -57 - 877 0 ~---- ---- . --------10 _____ ____________ - 1,032 - 179 0 2, 3oo 1 l. 089 - 877 -8, 770 20 __ __ _______ ______ -1, 200 - 334 I 2, 300 ! 766 212 -6,650
ao _________________ -1, 228 -500 3. 754 2.026 978 3, 130 40 _____ ______ __ ____ - 1, 180 - 665
4. 937 30 092 3, 004 33, 170 0 ' 50 _____ _______ __ ___ -985 ..:.... 816
2, 300 499 6. 096 I 94, 130 60 _____ ____________ - 755 - 933 - 1. 688 I 6, 595 160, 080 -------- I 70 _________ ____ ____ - 494 - 1, 020 ---- ---- - 1, 514 4, 907 209, 1~0
80 _______ __ ___ _____ - 151 - 1,067 -------- -1. 218 3. 393 243,08 0
90- ---- ----- ------ - - 55 - 1, 077 --- --- -- - 1. 132 2, 175 264,830 100 _____ ____________ 0 - 1, 077 -------- -1.077 1, 043 275, 260 110 _____ ____________ 0 - 1,077 ----- --- - 1, 077 - 34 274, 920 120 _________________ 0 - 1, 077 ---- ---- -1.0 77 - 1. 111 263,810 130 ____________ _____ 41 - 1, 077 -------- -1. 036 - 2, 188 241, 930 140 _____ ___ ___ ______ 151 - 1, 067 ------- - - 916 - 3, 224 209, 0690 150 _____ __ ________ __ 494 -1, 030 ---- -- -- -536 - 4, 140 168, 290
-L60 _______________ __ 851 - 933 ---- ---- - 82 - 4, 676 121, 530
170- - - - - -- - - - - - - - - . - 1,346 - 783 -- ------ 563 - 4, 758 73, 950 180 _________ ________ 1,891 -563 ----- --- 1, 328 - 4, 195 32, 000 188 ____________ _____ 1, 780 - 250 --------, 1, 530 - 2, 867 9,02 0
194.75 __ , - - -0 -- ------ 1. 346 - 9 . 1. 337 -1. 337 0 --------,
-15, 591 15. 591 000
j. To determine total shear or bending m<?ment at any frame, it is
necessary to add results obtained from static loading to those computed
from aerodynamic forces.
TM 1-320
43-44 AIR CORPS
( 1) To obtain total shearing force between stations 30 and , 40:
Pounds ·
Shear due to aerodynamic forces from table V = 3, 004
Shear due to static loading from table IV =-1, 596
Total shearing force = 1, 408
(2) To obtain total shearing force between station s 80 to 90:
Shear due to stat ic loading
Shear due to aero dynamic forces
Total shearing force
(3) To obtain total bending moment at station 130:
P ounds
- 4 624 ' 2,175
6,799
Meter pounds
Bendi ng moment due to static loading from table IV = - 80, 470
Bending moment due to aerodynam ic forces from table V = 241, 930
Tot al bending moment = · 161, 460
44. Conclusion.- While static means of sustentation and control
are available to lighter than air aircraft, the intelligent pilot should
constantly bear in rriind the effects of aerodynamic for ces on his air
ship. He must understand the relation of velocity to resistance, power
requir ements, and fuel consumption. He must be cognizant of the
characteristics of his prop ellers and be able to make utmost use of
variable pitch should his propell ers be·capable of adjustment in that
regard. He should comprehend the theory of airship stability and
be alert to augment that stabi lity by use of his controls. It is essen-·
tial that he at all times appre ciate effect of dynamic forces on his flight
path in regard to both direction and altitude, and be able to assist his
static control by dynamic means whenever necessary. F inally, he
· must be aware of the stresses to which his airship is being subjected
and, knowing maximum performance for which his aircraft was de
signed, so vary the velocity as to preclude possibility of exces.S
structural stresses.
[A. G. 062 .11 (9- 11-40 ) .]
BY ORDER OF THE SECRETARY OF wAR:
OFFICIAL:
E. S. ADAMS,
Major General,
The Adjutant General.
DISTRIB OTION :
G. C. MARSHALL ,
Ohief of Staff.
D 1 ( 3) ; B 1 ( 2) ; IR 1 ( 5) -; IBn 1 ( 10).
II.!. GOVERNMENT PRINTING OFFICE: 1$41
For sale by the Superintendent of Documents, . W.ashington, D. C. · · - .. . - Price, 15 cents