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File D 00.11 /29
I AIR SERVICE INFORMATION CIRCULAR
. (AVIATION)
PUBLISHED BY THE CHIEF OF AIR SERVICE, WASHINGTON, D. C.
Vol. V April I • 1923 No. 421
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STANDARD METHOD OF ENGINE
CALCULATIONS
(POWER PLANT SECTION REPORT)
Prepared by H. Caminez and C. W. lseler
Engineering Division, Air Service
McCook Field, Dayton, Ohio
September 28, 1922
WASHINGTON
GOVERNMENT PRINTING OFFICE
1923
NO ·ATING
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CERTIFICATE: By direction of the Secretary of War the matter contained herein is published as administrative
information and is required fr 1 the proper transaction of the public business.
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INDEX.
Object . . ..... .. . . ......... · . .. . . . . .. ..... . . ... . . ...... . . .. ... . ..... . . . . . ... . .... . . . . .... . .............. .
General. .......... . .... .. .... . . ...... ... . . .... ...... ......... .. ... .... . .... . ..... . . ..... .... . . . .. . . .. . .
Kine~atics of engine .. : . . ... . .... : ._. . .. ....... ... ............ . . . ........ . .. ; ..... . .. ... . .... .. ... . . .... _.
PIS ton and connectmg rod pos1t1on _ . _ . . .... . . . .... .. . .... .. ..... . . . . . . . . . ... . . . . . ... . .. ..... . . . .. ... .
Velocity .. . .... .. ..... .... . . .... .. ........ . ..... . ............ : .. ...... .. ............... . ..... .. ... .. .
Acceleration . .... . ... · . ....... . .. ... ........... . . ..... . ... . . . . .. . . . . . .. . . .... . . ... ..... . ..... . . . . . ... .
Inertia and centrifugal forces .. . ..... . ... . .... ... . .. . . .. ... .. .. .. . ..... .. ........... . ... . .. ... .... .. ... .. .
Indicator diagram .... .. ............. . ... . ................ . ..... . .... . ... ... . .. .. . . ... .. ...... . ..... . .. .
Resultant force on piston .. .. . . . .... .. .... . . .. . .... . . . .. . ... .. . ... . .. .. . . .. .. .... .. . ......... . ... . . . .... .
Piston side pressure . . . . . ... . ... .. .. .. .. . . .. ... .. .. . . ... . .... . .... . ..................... . ........ .. ..... .
Torque .. .. ......... . . .. . ... .. ... .... .. .. ... ....... . . . . .. .... . . .. .. ............ . .. · · · · · · · · · · · · · · · · · · · · · · ·
Resultant force on crank pin ... . .. . ... .. . .. . . . . ...... . . . : .. .... . . .. . . . . . .. ... .. . ..... .. .. .. .. . . ... . .... ... .
Resultant force on crankshaft b ~arings .. . .. ......... .... ..... . .. .. . . .. · . . . . . .. .... ... .. .... .. . . ...... .. . .
Bearing analysis . . . . .. .... ........ .. ... . ... ... .. . . .. .. ... .... . ... .. .. . ................. . ...... ..... .... .
Crankpin oilhole location .... . . . .. .............. . ... .. .... . ." . ... .. .... . ... .. . .. . . .... . ... . . . ..... ..... . .
Crankshaft stress analysis .... . ...... . .... .... ..... . ... .. ..... . . ................... .. ..... . . .. .. .. . . ..... .
Connectingrod stress analysis ..... . ... ... . .... . ............ . . .. ... .. ... . ..... .. ... . .. . . ..... . . ........... .
Pistonpin analysis ....... .. . ........... . . .............. . .... . . . .. . ......... .. .... . .... . . . ... . . . .. . .. .. . .
Valve and valvegear analysis . .. . . ... . .. . .. . . ............. .. . . . ....... . . .... . ... .. . . .. . ... .. ... ... . ...... .
Gas velocity . .. . ...... .. ... . .. . . . ...................... '. ............ . . . ....... . .. . ... . . .. .. ... . . .. . . .
Valve spring . . . . .. .. . ........... .. .... . .... ........ . .. . .. . ... . ... .. , . ~ .. ... .. . . . .. ..... . . ... ... ..... .
Cams ...... . ..... . ... .. ... ... . . .. .. .. ... .......... . . .. . .. .. . . ... .. . . .... . . ... . .. . ..... . ............ .
Appendix .. . ... .... . .... .. ......... . .... . . .. . ... . . . . ....... . . .... ... . ...... . .... .. .... ... ..... . . . ... .. . .
Characteristics and dimensions of Liberty 12 engine . .... . . . .. ........ .. ... . . .. . ... ....... . .... .. ..... .
Tables .. . . . . .... .. ... . ..... . .... ...... ... .. .... .. .. .... . . ... . . .. . . . . . .. . .. . . . .... . ..... . . ....... ... .
Standard abbreviations .. . ... ... . ..... . ... . ......... . ............... ... ............... . . . .. . . . . . .... .
Table of symbols .. .. . .... .. . . ... . . . ... .. . . . .. ...... . ... . . .. . . . . .... . .. .... .... .... ... . . ........ . .. . . .
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STANDARD METHOD OF ENGINE CALCULATIONS.
OBJECT OF REPORT.
The object of this. report is to outline and describe the
methods adopted by the Engine Design Branch of the
Engineering Division, Air Service, for determining the
forces and analyzing the stresses existing in the principal
parts of an airplane engine. All reports on engine stress
analysis by the Engine Design Branch will hereafter conform
to the methods described herein without further
special reference to the methods employed.
GENERAL.
An established system of analysis induces a uniformity
in general practice and facilitates checking. It also
makes comparisons between different engines easier and
affords a means of utilizing more conveniently for design
purposes the results obtained in the actual tests of existing
engines. It is, therefore, recommended that airplane
designers and manufacturers use the procedure as outlined
in this repcrt in their stress analysis. The methods employed
are, on the whole, similar to those in general use
and are, therefore, not difficult to fo llow. Some of the
equati.ons are necessarily empirical, since exact stress
determinations are often impossible. The results obtained
by their use afford a means for comparing the actual
stresses in similar parts of engines that are analyzed, and
they are, consequently, of practical value for design
purposes.
KINEMATICS OF THE ENGINE.
The reciprocating motion of the piston is converted into
the rotary motion of the crankshaft through the medium
of the connecting rods. In order to ascertain the principal
forces acting in the engine, the relative position, velocity,
and acceleration of the moving parts must be determined.
The motion of the pistons and the connecting rods is
obtained with relation to the crankshaft, the angular
velocity of which is considered constant because of the
large inertia of the propeller and the uniformity of torque
in multicylinder engines.
The usual arrangement of pistons, connecting rod, and
cunkshaft is shown diagrammatically in Fig. 1, in whichL=
connecting iod length, c. to c.= BD
R=crank radius=OD
IJ=crank angle from top center position.
<f,=augle of connecting rod with centerline of cylinder.
s=piston travel=AB
slightly different. Fig. 2 represents diagrammatically an
articulated connecting rod arrangement, in which
QA and OB=cylinder centerlines.
Lc=length of master connecting rod.
L,=length of link connecting rod .
O=crankshaft center.
D=crankpin center.
E=link pin center (fixed in master rod ).
R=crank radius=OD
M=link radius= DE
a=angle between cylinders= <(A.OB
f3=link radius angle= <)::EDB
/J1 and IJ2=angle. of crank from centerline of link and
master rod cylinders respectively.
</>, and <f,2=angle between connecting rod and centerline
of link and master rod cylinders respectively.
s, and s2=piston travel from top position of link and
master rod cylinders respectively.
,{,=angle between link radius, ED, and link rod
cylinder centerline., OA.
PISTON AND CONNECTING ROD POSITION.
The piston and connecting rod position for various crank
angles can be obtained from a layout drawn carefully to
scale or by mathematical determinations. With the
anangements shown in Figure 1 the piston travel and
connecting rod angle are given by the following equations:
(1) s=RR cos e+L L cos</>
(2) d>=sin  1
(~ sin /J)
By combining equations (1) and (2), the following expression
for piston travel is obtained:
(3) s=R[ lcos o+iJ(!)'  sin2 e]
The per cent of piston travel can be expressed in terms
of crank angle, and the ratio of connecting rod to crank
length (L+R) by the equation
(4) % p. t. = {1 cos o+jiJ({)'sin2 e]
where % p. t.=per cent piston travel.
Equation ( 4) has been solved at every 10° of crank angle
for ratios of connecting rod length to crank length usually
found in airplane engines. The results are given in Table
I in the appendix. Fig. 3 shows the per cent piston travel
vs. crank angle for values of L/R equal to 3.4 and 4.0.
With the link rod arrangement, shown in Fig. 2, the piston
travel and connecting rod angularity can be obtained
from·the equations:
(5) s1=0A0 0A=OA0 (L, cos <1>1+M cos ,t,+R cos B1)
\Vhere two or more cylinders operate on a common
crankpin, an articulated connecting rod construction,
which consists of a master rod and one or more link rods, is
often employed. The piston motion with the master rod
is the same as that obtained by the construction shown in
Fig. 1. In the case of the link rods, which are pivoted at
the big end of the master rod, the resulting motion 1s (6) tf,=<t>2+(f3 a)
(1)
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2
() · i[R . ( )] 7 </>2=sm Le sm 01  a
(8) . _1 [R sin 01  M sin "']
</>1 =Slll L, .
The total piston travel in the link rod cylinders often
differs from that in the master rod cylinder. This must be
taken into consideration when determining the c01μpression
volume.
VELOCITY.
The linear velocity of the crank pin is considered uniform
and is
where ve=crankpin velocity, ft. per sec.
N=r. p. m.
R=crank radius, inches.
With the arrangement shown in Fig. 1, the angular velocity
of the connecting rod and the linear velocity of the
piston are given by the following equations:
(
lO) d<t> R cos o do cos o cos o
dt=L cos</> dt=L COS</> =Ve ..JL2R2 sin2 O Ve
where ~f =angular velocity of connecti.ng rod.
(11) v"=~=ve (sin o+cos O tan </>)=ve (£,.)
where vv=piston velocity
fv=crank' angle factor for piston velocity.
(12) v=Slll o+cos O tan </>=Slll O 1+ / L 2 . 2 f . . [ coso J y (1r) sm O
The crank angle factor for piston velocity, (fv), may be
obtained graphically. Referring to Fig. 1, itis OH divided
by OD. This factor has been determined for values of
L/R generally used in airplane engines and is given in
Table II.
Since </>, the connecting rod angle, is a small angle, but
slight error will be introduced in equations (11) and (12)
by substituting sin </> for tan q,. Making this substitution
and simplifying, the following equations are obtained:
. 1 R . )
(13) vp=Ve (sm o+2 L sm 20
() f
. lR.
14 v=Slil o+2 L Slil 20
1 (Approx. )
(Approx.)
The crank angle factor for piston velocity has been determined
for a ratio of ii,=4 by means of equations (12)
and (14), the results being given in Table IV. Fig. 4,
showing crank angle factor for piston velocity vs. crank
angle, is drawn from this table. The results from both
equations are so nearly alike that no difference can be
shown between the curves drawn. The maximum error of
equation 12 in the example taken is less than .3 per cent.
The expressions for piston velocity for the link rod cylinders
that would be obtained by differentiating equations
(5) to (8) are too involved to be of practical use. The
velocity is therefore obtained graphically by carefully
plotting piston travel vs. crank angle and then determining
the values by means of the tangent of this curve.
ACCELERATION.
The exact expression for piston acceleration for the arrangement
shown in Fig. 1, obtained by differentj.ating
equation (11), is :
dv v 2[ R cos20 J v 2
(15) a= dt ~ If cos o+r cos'<t> sin 0 tan</> = R Xf.
Equation (15) is too complex and is not commonly
· employed. Since the values of angle ct> are always small,
but slight error is introduced in this equation by substit.uting
sin ct> for tan </>, that is,letting cos q, equal unity.
Making this substitution and simplifying, the following
equation for piston acceleration, which is in general use,
is obtained:
v 2( R ) v 2 (16) a=1t cos o+r: cos 20 = R Xfa
where a=linear acceleration of piston.
£.=crank angle factor for piston acceleration.
R cos2 O .
(17) f.=cos o+L ~.,sm O tan ct> cos </>
(Exact.)
R
(18) f. =cos o+ L cos 20 (Approx.)
The crank angle factor for piston acceleration for a
ratio of L/R=4 is given in Table IV as determined by
means of equations (17) and (18). This table shows that
but slight error is introduced by using equation (18).
The crank angle factor for piston acceleration has been
determined by means of equation (18) for values of L/R
generally used in airplane engines and is given in Table
III. Fig. 5 shows crank angle factor for piston acceler:i.tion
vs .. crank angle.
When the link connecting rod arrangement is used, it
is necessary to obtain the piston acceleration graphically.
A curve of piston velocity vs. crank angle is plotted, and
the values for piston acceleration are determined by means
of the tangent of this curve
Example (Liberty 12):
L 12
R,=3_5= 3.428
. d 1 . 1 (sin 0) q,= connectmg ro ang e=sm
3
.4
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Per cent piston travel:
% p . t.=!(1cos o+3.428 v'3.4282 sin2 o)
Crankpin velocity:
2,,..N R 21rl700 3.5
Ve=50 n=~X 12=5.19 ft. per sec.
Crank angle factor for piston velocity :
f sin O [1+ cos O J v v'3.4282sin2 0
' OH .
fv=oD (see Fig. 1).
Crank angle factor for piston acceleration:
1
f.=cos 0+3.428 cos 2 o
(2)
(4)
(9)
(12)
(18)
The expressions above have been solved for every 15°
of crank angle, and the results obtained are given in
Table V.
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INERTIA AND CENTRIFUGAL FORCES.
The motion of the pistons and connecting rods muse
inertia and centrifugal forces. In a fixedcylinder engine,
the piston motion is linear and the forces set up by' it are
purely inertia forces.
The motion of the connecting rods is more complex.
For an exact solution of the forces produced by the rod, it
is necessary to consider the components into which its
motion may be divided. The motion of the rod may be
analyzed as consisting of a translation of its C. G. with the
linear velocity and acceleration of the piston combined
with an angular velocity and acceleration about the piston
pin. This motion of the rod sets up in each of its elements
three separate forces, consisting of an inertia force due to
its linear acceleration in the direction of the cylinder axis,
a centrifugal force due to its angular velocity about the
piston pin , and ap inertia force due to its angulaf acceleration
about the piston pin. An exact solution of the forces
clue to the connecting rod is veIY complex and unnecessary
. . It is found that these forces are very closely approximated
in the usual airplane engine design by considering
the mass of the connecting rod divided between the piston
pin and the crank pin inversely in proportion to the distance
of the respective pins from the C. G. of the rod.
The portion of the rod at the crank pin produces a cent rifugal
force. The portion at the piston pin produces an
inertia force. The practical. method of obtaining the
weight at the piston pin is to support the rod on knife
edges directly over the center line of the bearings with the
axis of the rod in a horizontal position and the small end
resting on the scales. The weight of the crank pin end
can be obtained in a like manner. The results can be
verified by comparing the sum of the weights of the two
ends with the total weight of the rod.
The centrifugal force acting at the crank pin in the direction
of the crankthrow is
(19) Fc=.0000284 WcR N2
where F c=centrifugal force, lb.
W c=weight of lower encl of conn. rods, lb.
R=crank radius, inches.
N=r.p.m.
The inertia force actingat the piston in the direction of
the cylinder axis is:
(20) Fi==0.0000284 WiRN2f •.
where F 1 =inertia force, lb.
W1=reciprocating weight = weight of piston assembled
plus upper encl of connecting rod,
lb.
f, =crank angle factor for piston acceleration (see
equation (18) and Table III).
Example (Liberty 12):
Centrifugal force at crank pin
F c=0.0000284 W cRN2=0.0000284 X 6.3 X3.5 Xl 7002
(19)
Fc=l811 lb.
Inertia force at piston
F1= 0.0000284 WiRN2f, (20)
F 1=0.0000284X6.2 X3.5Xl7002Xf,=1782Xf,
INDICATOR DIAGRAM.
The gas pressure acting on the piston throughout the
engine cycle is obtained from an indicator diagram. Instruments
have been devised to obtain this diagram directly
from the engine, but for design purposes, it is sufficiently
accurate to determine the diagram analytically.
· Fig. 6 shows a typical indicator diagram. Gas pressure,
lb. per sq. in . gage, is plotted vs. per cent of piston travel.
The gas pressW'e dming compression and expansion is
assumed to vary according to the following law:
(21) P V'>'= constant
where P=absolute gas pressure
V =total cylinder volume
'Y=exponent of expansion and compression curves.
From the above equation, the absolute pressure vs.
cylinder volume is determined. The absolute pressure~
are reduced to gage pressure. The relative total cylinder
volume throughout the cycle is expressed in per cent of
piston travel. The ideal diagram, a b c cl (fig. 6), is then
drawn. The corners of this diagram are rounded off, and
the intake and exhaust lines are added in the manner
shown. The final diagram is thus obtained.
For the construction of the indicator diagram, it is
necessary to know the indicated mean effective pressure,
the compression ratio, and the initial intake pressure. In
new designs, it is necessary to assume the value for the
i .m.e.p., which will be governed by the design feature
contemplated . When the performance of the engine 1s
known , the i. m. e. p. can be obtained by the equation :
(22)
. b.m.e.p.
1.m.e.p.=
12X33000Xb.h.p. 792000b.h.p.
em emXV,><!N = emXV,XN
where i.m.e.p. = indicated mean effective pressure, lb. per
sq . in.
b.m.e.p.=brake mean effective pressure, lb. per
sq . in.
em =mechanical efficiency.
V ,.=total piston displacement of engine,
cu . in .
where D=cylinder diameter, in ches.
L=length ·of stroke, inches.
n=number of cylinders.
The absolute pressure at the beginning and encl of the
expansion curve in the theoretical diagram is obfainecl
from the equations
(24) p _('Y1) (rl)Xi. m. e. P· + p
ct r'>'r f" "
(25) Pc= r'>' Pct
where P,=pressure at beginning of compression, abs.
P c=pressure at beginning of expansion, abs.
Pct =pressure at encl of ,expansion, abs.
r=compressfon ratio.
'Y=exponent of compression and expansion curves.
i.m.e.p.=indicated mean effective pressure.
fct=diagram factor to allow for rounding corners of
theoretical diagram.
The total cylinder volume, expressed in per cent piston
travel, for a given per cent of piston travel x, is x+V0 ,
4
where V0 is the per cent of the total t ravel that the piston
must move to displace a volume equal to the compression
volume. For a compression ratio equal to r, this per cent
of piston travel is
(26) V0= (r~l XlOO) %
The absolute gas pressure for a given per cent of p iston
travel is determined from the equations:
(
10o+v ) ,,
(27) Px1= P. x+v: for the compression stroke .
(
10o+v ) ,, .
(28) Px2= Pd x+ v: for the expansion str.oke.
These pressures (Px1 and Px2) are reduced to lb. per sq.
in . gage, and the curves ab and c d (see Fig. 6) are drawn .
The following are the values used for the constants of the
preceding expressions:
y=exponent of expansion and compression curves=
1.30.
id=diagram factor to allow for rounding of the theoretical
diagram and for the intakeexhaust loop =,
0.90.
P m.x=maximum gas pressure, actual diagram=0.75 Pc·
P .=gas pressure, absolute, at beginning of compression
stroke, for sea level conditions=l3 lb. per sq . in .;
for altitude conditions=90% altitude pres.
Example (Liberty 12):
. 792000 X 421
1. m. e. p.=88.4Xl649Xl700=I34.5 (22)
P.=13, y=l.30, f.=0.90
(1.31) (5 .42  1) 134.5
Pd= 5_421.a_5 .42 X 0.90 +13=68.4 (24)
' T _ 100 _ '0 4.4222.6 (26)
. ( 122.6 ) 1
•
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• Pxi(compress10n)=l3 x+22 _6 lb. per sq. m. absolute.
. (27)
. c .122.6)1.3
• Px2 (expansion) =68,4 x+22 _(J lb. p~r sq. m. absolute.
' (28)
The gas pressures, expressed in lb. per sq. in. abs., are
determined from the above equations for every 10% pisto'ri.
travel and are give:t'I. in Table VI. From this table, the
indicator diagram, Fig. 6, is drawn. A scale showing the
per cent piston travel for various crank angles is also given
on this figure for the c9nnecting rod to crank length ratio
of the Liberty 12, so ihat the gas pressure !It any crank
angle can be determined directlY. from this diagram.
RESULTANT FORCE ON PISTON.
The gas pressure acting on the piston is obtained at
various crank angles from the indicator diagram. The
total gas force is this pressure multiplied by the piston
area. The inertia force at various crank angles is determined
by means of equation (20). The resultant force
acting in the direction of the cylinder axis is obtained at
various crank angles by adding algebraically the inertia
force and the gas force at the crank angles considered.
(29) F,=Fg+F,
where F,,=resultant force along cylinder axis.
Fg=force on piston due to gas pressure.
F 1 =inertia force.
Example (Liberty 12):
The gas pressure throughout the angle is obtained from
the indicator diagram, Fig. 6. The total gas force on the
piston is 19 .63Xgas pressure. The piston acceleration
factor at various crank angles is given in Table V. The
inertia force is (1782)f. (see page 14). The resultant
force along the cylinder axis is the algebraic sum of the
gas and inertia forces. Table VII and Fig. 7 show the gas,
inertia, and resultant forces acting along the cylinder axis
for every 15° crank angle throughout a complete engine ·
cycle.
PISTON SIDE PRESSURE.
The piston side thrust due to the force acting along the
cylinder axis is:
(30) F,=F,Xtan <I>
where F,=piston side thrust.
F,=resultant force along cylinder axis
</>=connecting rod angle.
For the cylinder and connecting rod arrangement shown
in fig. I, the piston side thrust may be expressecl i terms of
crank angle e by the equation
(31) F ,=F;.J(Ls)~ fl ·
R sin2 fl
The piston side thrust is obtained throughout a complete
cycle and is plotted vs. crank angle and vs. piston travel , a~
in fig. 8 and fig. 9. The area of the latter curve is proportional
to the piston friction loss if the coefficient of friction
remain~ constant. From the average h eight of this curve
(fig. 9) the average side thrust dming the power stroke and
dming the complete cycle is obtained. 1
The piston side pressure, lb. per sq. in , is obtained bv
dividing the total piston side thrust by the projected pisto~
bearing area. Since the piston diameter is usually re.lieved
above the lower piston ring, only that portion below
the lower ring is generally considered in determining ,the
effective bearing area.
Where the articulated connecting rod construction is
employed, the forces in the link rod cylinders may produce
piston side thrusts in the master rod cylinder. In a
careful analysis, the effects of these forces should be considered.
Example (Liberty 12):
The piston side thrust was calculated at every 15° of
crank angle and is given in Table VIII. From this table
curves are drawn showing piston side thrust vs. crank
angle (fig. 8) and piston side thrust vs. piston travel (fig. 9).
The average thrusts during the power stroke and during the
complete cycle were obtained from fig. 9 by means of a
planimeter. The effective piston bearing area is given in
the table of dimensions.
u . . b . 870 m.ax. piston earmg pres.=u;. 55= 52 .t> lb . per sq. l.l l .
Aver. pres. during power stroke=1~~:5=43.2 lb. per sq.
in.
Aver. pres. throughout cycle=1~~~5=19.0 lb. per sq. in. •
, I
TORQUE.
The torque caused by the forces acting in the cylinder a
any instant dming the cycle may be determined from the
equation:
(32) T=F.XR Xfv
where T=torque, lb. ft.
F,=resultant force along cyl. axis, lb.
R =crank radius, ft.
f,=crank angle factor for piston velocity. The
value of this factor is given exactly in equation
(12) and approxi1:f1ately in equation (14).
The res ultant torque in a multicylinder engine is the
algebraic _sum of the instahtaneous torques of the individual
cylinders. To obtain this resultant, the angular
relation of the cycles in the various cylinders must be considered.
The torque obtained by\ the above analysis is the indicated
torque, as frictional forces are not considered.
The mean torque can be obtained from a curve of torque
vs. crank angle by means of a planimeter. The ratio of the
maximum to the mean torque of the engine is.determined.
This ratio is used in later calculations.
Example (Liberty 12):
The torque exerted by a single cylinder throughout the
cycle is calculated for every 15° of crank angle from equation
32 and is given in Table VIII. From this table, fig.
10 is drawn, showing the indicated torque of a single cylinder
vs. crank angle. Fig. 11 shows the torque curves for
each bank of six cylinders and the indicated torque of the
twelvecylinder engine.
The ratio of the maximum to mean torque for the engine
(12cyl. 45° Vee) is l.~3. For the single cylinder, tlus
ratio is 8.60.
It should be noted that the character of the torque curve
varies with the engine speed and power output and that it
is also affected by the characteristics of the engine, such a.~
number and arrangement of cylinders, reciprocating
weight, compression ratio, and ratio of bore to stroke.
RESULTANT FORCE ON CRANK PIN.
The resultant force on the crank pin is obtained by
combining graphically the resultant force along the connecting
rod axis with the centrifugal force caused by the
weight of the lower end of the connecting rod . · The
centrifugal force is calculated by means of equation (19).
The resultant force along the connecting rod axis (Fd) is
(33) F  F,
d  cos cf,
where F,=resultant force along cyl. axis
ct,=connecting rod angle.
In the above equation, a plus ( + ) force denotes a force
producing compression in the connecting rod, a negative
(  ) force one producing tension.
When the cylinder and connecting rod arrangement
shown in Fig. 1 is used , Fd can be expressed in terms of
the crank angle (0) by the equation:
(34) Fct= :•
,v1  ( 1, ) sin zo
Figure 12 shows the method employed for determining
graphically the resultant force on the crank pin. In thiR
/ 39757~23~2
5
figure, OD represents the crank throw, and AD and BD
represent two connecting rods operating on a common
crank pin. The centrifugal force, F0 , is laid off to a suitable
scale along the center line of the crank throw OD
from the point 0. The force, F01 , represents a positive
force acting along the connecting rod AD and is. drawn
parallel to AD. Fct2 represents the force acting along BD
and is_ drawn parallel to this rod. This force has been
taken as negative and is, therefore, shown acting upward
toward the piston pin. The resultant force, F,, is now
drawn from O to the end of F<lz· This resultant represents
to the scale chosen the magnitude of the force acting on
the crank pin at a particular cral)k i£ngle. The direction
of this resultant force with respect to the engine axis
is given by the angle a, with respect to the crank throw
by the angle. fl, and with respect to the connecting rod
AD by the angle 'Y.
The resultant force is obtained at various crank angles
throughout the cycle. Figs. 13 and 14 show polar diagrams
of the resultant forces acting on the crank pin of the
Liberty 12 engine at 1,700 r. p. m. In Fig. 13 the direction
of the forces is taken with respect to the engine axis,
and in Fig. 14 with respect to the crank throw. Fig 15
sh~ws the wagnitude and direction of the forces acting
on the connecting rod.
RESULTANT FORCE ON CRANKSHAFT BEARINGS.
The load distribution on the crankshaft bearings depends
upon the rigidity of the crank shaft aud crank case, the
alignment of the bearings, and the clearances between the
journal and the bearings. Since these factors can not be
predetermined, it is impossible to make an exact analysis
of the load distribution between the various crankshaft
bearings where more than two main bearings'are employed.
The following empirical method is employed for
computing the forces acting on the main bearings of a
crank shaft where there is a main bearing at each side of
the crank pin. The forces acting on the crankshaft
bearings are obtained by considering the force at the
crank pin together with the centrifugal force resulting
from the weight of the crank pin and crank cheeks to be
equally divided between the two crankshaft bearings at
each side of the crank pin. Examples of polar diagrams
of the forces on the main bearings of crank shafts of this
type are shown in Figs. 16, 17, and 18. These diagrams
show the forces on the end, center, and intermediate
bearings of the Liberty 12 engine. The end bearings are
loaded on only one side by half the force on the crank pin
combined with half the centrifugal force due to the crank
pin and crank cheeks of the crank throw. This same
loading is applied .__on both sides of the center and the
intermediate bearings. But the forces acting on both
sides of the center bearings are 300° out of phase, while
in the case of the intermediate bearings these forces are
240° out of phase.
In crank shafts where more than one crank pin ,bearing
or crank throw is located between each pair of main
crankshaft bearings, the load distribution on the main
bearings is obtained by treating each s~ction of the crank
shaft between two main bearings as a uniform beam
rigidly supported at the center line of the main bearing
bolts. The reactions on the main bearings A and B due
to a force F, on a crank pin located between them is
6
b2 (3a+b) a2 (a+3b)
(35) R,=Fr (a+b)3 and Rb=Fr (a+b)3
where R, and Rb=bearing reactions at bearings A and B,
respectively.
Fr=resultant force on crank pin.
a and b=distance between center line of crank
pin and center line of main bearings
A and B, respectively.
The resultant force on a crankshaft bearing is the vector
sum of the reactions due to the crank pin loads. In
determining the resultant force, due consideration must
be taken of the direction to the engine axis f each separate
reaction. Both the relative positions of the crank pins
and their cyclic relation must also be considered .
BEAR1NG ANALYSIS.
The maximum and mean bearing pressures are determined
by dividing the maximum and mean resultant
forces acting on the bearing by the projected area of the
bearing. In determining the projected bearing area, only
the straight portion of the bearing length is considered as
effective. The resultant pressures are expressed in lb.
per sq. in.
The rubbing factor on the bearing is the product of the
rubbing velocity of the jou'rnal and the mean bearing
pressure. The rubbing velocity in ft. per sec . is determined
by the equation,
(36)
D,r.p.m.
v,=,r f2 X 60
where v,=rubbing velocity of bearing, ft. per sec.
D=bearing diameter, inches .
The rubbing factor is expressed in lb. per sq. in. Xft.
per sec.
The magnitude of the bearing pressures and rubbing
factors that can be successfully sustained depends largely
upon the design of the bearing. The factors of greatest
consequence are the rigidity of the bearing, the provision
for proper lubrication, the relation of bearing length to
diameter, and the maintenance of proper bearing clearances.
Example (Liberty 12):
Crank pin bearing:
Bearing area
Rubbing velocity
Max. force
Mean force
=2.19X2.375=5.2 sq. in.
,rX2.375Xl700_
17 6
f
12X60  · t. per
= 5380 lb.
=3900 lb.
sec. (36)
Max. bean.n g pressure=5T38I0 =l035 lb . per sq. m. .
Me an b ear.m g pressure= 3T90I0= 75 O lb . per sq. m. .
Rubbing factor
Main bearings:
Bearing area
Rubbing velocity
=750Xl7 .6=13200.
= l.625X2.625=4.27 sq. in.
=,r X2.625Xl700=19 45 f
12X60 · t. per
· sec. (36)
End crankshaft bearing:
Max. force =3610 lb .
Mean force =2660 lb ..
. 3610
Max. beanng pressure= 4.27=845 lb. per sq . in.
Me an b eari. ng pressure= 2660 l . 4_27=623 b. per sq. m.
Rubbing factor
Center crankshaft bearing:
Max. force
Mean force
= 19 .4;5 X 623=12100.
=6750 lb.
=4975 lb .
Ma x. b ean.n g pressure= 6750 1 8 lb · 4_27= 5 0 . per sq . rn .
M ean b ean.n g pressu' re= 4975 lb . 4_27 =1165 . per sq. m.
Rubbing factor =19.45Xll65=22650.
[n termediate crankshaft bearing:
Max. force
Mean force
=4900 lb.
=3075 lb.
I
M ax. b eari. ng pressure= 49oo 11 lb . 4_27= 50 . per sq. ill.
M ean b ean.n g pressure= 3075 lb · . 4_27=720 . per sq. ill.
Rubbing factor =19.45X720=14000.
CRANK PIN OIL HOLE LOCATION.
In engines having oil supplied to the crank pin bearings
through the crankshaft, the location of the oil hole in the
crank pin has a marked effect upon the operation and
durability of the bearing. The following method for
crank pin oil hole location in which consideration is given
to the magnitude, direction, and duration of the forces
acting on each element of crank pin area is employed.
The bearing pressure is considered as evenly distributed
over an arc of 180° on the crank pin. The magnitude of
the force and its direction with respect to the crank throw
is obtained at equal intervals throughout a complete cycle
from a polar diag~am of resultant forces on the crank pin.
These forces are plotted as a series of half rings having
their radial thicknesses proportional to the magnitude of
, the force and their midpoints falling on I} line through
the center of the crank pin in the direction cf the application
of the force considered. The summation of these
rings produces an area which is termed the comparative
.wear on the crank pin. The best location for the oil hole
is that point on the crank pin where the radial thickness
of this resulting area is at a minimum. · The comparative
wear diagram for determining the best location of the oil
hole in the Liberty 12 engine is shown in Fig. 19.
CRANK SHAFT STRESS ANALYSIS.
It is impossible to determine the exact nature and magnitude
of all the stresses in a crank shaft, since the deflections
in the engine members and the tortional vibration
in the crank shaft produce complicated strains which can
not be calculated. The magnitude of these stresses is, of
course, largely dependent upon the rigidity of the crank
shaft and the crank case and upon the engine speed.
Fatigue action caused by the reversal of stresses becomes
serious where there are sudden changes in section or where
keyways, holes, or sharp corners occur. At these sections
the stress distribution is unequal, causing localized
stresses of double or more the normal stress. It is essential
that the crankshaft material possess a high elastic limit,
adequate ductility, and a high impact factor (Izod or
Charpe) to withstand the type of loading imposed upon it
in an airplane engine.
The following method of crankshaft analysis, although
empirical and giving only approximate results, is considered
most useful, since it gives comparative values for
the stresses in the various sections of a welldesigned shaft
and is therefore adequate for design purposes. The allowable
stresses as found by this analysis are limited to values
dictated by experience with the type of crank shaft
considered.
Since the usual airplane engine crank shaft has a main
bearing on either side of every crank pin, this will be the
first case discussed.
The crank throw nearest the propeller transmits the
maximum. instantaneous torque of the engine. There is
also an axial force acting on the crank pin which has no
additional power component. It is assumed that the
maximum instantaneous torque is transmitted through the
crank throw at the same time that the maximum axial
force parallel to the crank throw acts upon the crank pin.
The character of the various strains in the journals, crank
pin, and crank cheeks as a result of these forces is shown
in Fig. 20, in which the deflections are exaggerated. The
crank throw is treated as a beam with rigid supports in
the plane of the main bearing bolts, the assumption of
rigid supports being justifiable in the usual case since the
end bearing is generally long or an outboard bearing is
provided.
The maximum instantaneous torque transmitted by the
crank shaJt is:
(37)
T 12X33000Xb.h.p.XK 63000X b.h.p.X K
m hN . N
where Tm= maximum instantaneous torque, lb. in.
b . h. p.=brake horse power
N=r. p. m.
K= ratio of maximum to mean torque, obtained
from torque diagram.
The maximum axial component of force acting in the
direction of the crank throw is obtained from the polar
diagram of resultant forces acting on the crank pin, showing
the direction of the force with respect to the crank
throw, as in Fig. 14.
In the following analysis, reference will be made to Fig.
20, in which:
(38)
F n=maximum force parallel to crankthrow.
F 1=tangential force obtained from the maximum
instantaneous torque.
FTm
t r
Fb=resultant of F, and F n·
7
(39) Fh=.JF/+F n'
a=distance from center of crank pin to center of
bearing bolts.
r=crank radius.
c=distance from center line of crank pi11 to tip of
journal.
e=distance from center of crank pin to center of
crank cheek.
f=onehalf crank pin length.
STRESS IN' JO URN AL CASE I.
Considering the crankthrow as a uniform beam with
rigid end supports and loaded at the center, the bending
moment in the journal, Mb, due to the force, Fb, is:
The equivalent bending moment, Me, resulting from
Mb and the maximum tortional moment in the crankshaft,
Tm, is:
(41) M0=} Mb+h/Mb2+Tm2~
The journal usually has a hollow round section.
Let D=outer diameter of section.
(42)
d=inner diameter of section.
1r D4d4
Rectangular section modulus, Z=
32
 D
(43) Polar section modulus, Zp=2Z.
(44) Area of section, A= f (D2d2) .
'fhe maximum fiber stress in ihe section is:
(45) Sb=~•.
The total shear stress is:
STRESS IN CRA.NKPINCASE I.
Considering the crankthrow as a uniform beam with
rigid end supports, the bending moment at the center
section of the crankpin, A A, due to the load F"' is:
(47) M1,=} F"Xa.
Considering the crankpin as a guided cantilever loaded
at the end by a force F,, the bending moment at the end
section, GC, is:
(48) M,,=F,Xf.
The equivalent bending moment, M., resulting from the
larger value of · the bending moment, Mb, as determined
from equations 47 and 48, together with the maximum
tortional moment in the crankshaft, Tm, is:
M.=t M,,+t.JMb2+Tm2
•
The maximum fiber stress in the section is:
S,,=~·
The total shear stress is:
S Tm+ Fn
,=zp 2A·
\

(41)
( 45)
(46)
STRESS IN CRANKCHEEKSCASE I.
The stress in the crankcheek is determined for section
EE, Fig. 20.
In transmitting the maximum torque of the crankshaft,
a twisting moment is set up in the crankcheek, which is:
(49) M,=F,Xe.
A bending moment is also
1
set up about the HH axis
and is:
Mb=F,Xc.
In a rigidly supported crankshaft of the type discussed,
·the bending moment in the crankcheek due to the force
F" ·is considered negligib le. This force produces only
direct compression in the crankcheek.
The usual crankcheek crosssection is rectangular.
Let b=width of crankcheek.
t=thickness of crankcheek.
the polar section modulus is
(51) Zp=kbt2
•
where k is a constant depending upon the ratio of b to t.
The values for k are ·given in Fig. 21, deten;nined from
equations obtained from Burr's " Elasticity and Resistance
of Materials of Engineering," Appendix I.
The rectangular section modulus about the HH axis is:
(52) Zb= b:t.
The rectangular section modulus about the EE axis is:
(53) Z\=bt
The direct tensile stress in the crankcheek is:
(•4) S =M"+F".
a ' Zb 2A
The shear stress is:
(55) S,=~i,.
p
The maximum equivalent stress resulting from S, and
S, is:
(56) s.=t s,+v't s,2+s;.
~xample (Liberty 12):
B. h. p.=421 at 1,700 r.p.m.
Ratio of maximum to mean torque, K=l.23
Tm=63,00ox/.J t 0Xl.23=19,200 lb. in.
F n=max. axial force (fig. 14)=5,200 lb.
F,= max. tangential force.
F =!_9,2oo =5 490 lb
' 3.5 ' .
Fb=v'5,4902+5,2002=7,560 lb.
a=3.25 in.
c=2.188 in.
e=l.75 in.
f=l.25 in.
;
(37)
(38)
(39)
8
s;rRESS IN JOURNAL.
Mb=t 7,560X3.25=6,140 in. lb .
M0=t 6,140+iv'6,1402 +19,2002 =13,200 in . lb .
D=2.625 in. d = l.375 in.
(40)
(41)
Z= l.642 in.3 ZP=3.284 in.3 A=3.927 in.2
Sb=maximum fiber stress.
(42 44)
13,200
Sb= 1.642 =8,000 lb. per sq. in. (45)
S,=total shear stress.
S 19,200+ 5,200 6 500 lb _ .
,= 3.284 2X3.927= ' · per sq: m. ( 46)
STRESS IN CRANK PIN.
Mb (center section)=t 5,200X3.25=4,230 in. lb. (47)
Mb (end section)=5,490Xl.25=6,860 in. lb. (48)
M.=t 6,860+tv'6,8602+19,2002=13,600 in. lb . (41)
D=2.375 d=l.25 .
Z= l.215 in.3 Zp=2.43 in.3 A=3.203 in.2 (4244)
Sb=max. fiber stress.
Sb=\\
6
i°i=ll,200 lb . per sq. in. (45)
S, = total shear stress.
S 19.200+ 5,200 8 00 lb .
· ,= 2.43 2X3.203= ,7 · per sq. m.
STRESS IN CRAKK CHEEK.
M,=5490 Xl.75=9600 in. lb.
Mb=5490X2.188=12000 in. lb.
F 0 =5200 lb.
b=3.47; t=l.0; 'b/t=3.47; k=.275.
Zp=.275X3.47Xl2=.954 in. 3
Z _<!.:_472 9 01 . 3
b  6  . lll.
A=3.47.
S,=tensile stress.
12000 5200 .
S,= 2_01 + 2X3.47=6730 lb. per sq . rn.
S,=shear stress.
9600 .
S,= _954 =10000 lb. per sq. m.
S0=maximum equivalent stress.
(46)
(49)
(50)
(51)
(52)
(54)
(55)
6730 .
S0 = 2 +v't X67302+100002=14000 lb. per sq. rn .(56)
CRANKSHAFT STRESS ANALYSISCASE 2.
In this analysis, the type of crankshaft which has two
or more crank throws or crank pins supported between
each pair of main crankshaft bearings is considered.
9
This type includes the two and three bearing fourthrow
crankshafts and the three and four bearing sixthrow
crankshafts.
Each portion of the crankshaft located between two
main bearings is separately considered . It is assumed
that the portion of the shaft investigated transmits the
maximum instantaneous torque of the engine at the time
that the stresses in it due to the axial forces acting on the
crank pins are at a maximum. The maximum instantaneous
torque is determined by means of equation (37).
The axial forces acting on the crank pin are obtained from
a polar diagram of the resultant forces, as shown in Fig. 14.
STRESS IN JOURNALCASE 2.
The bending moments in the journal due to the axial
forces acting on the crank pins of that portion of the
crankshaft under consideration are separately determined.
In the section of the journal under analysis, the number
of bending moments due to axial forces that must be de .
termined are equal to the number of crank pins between
each pair of main crankshaft bearings. Denoting the
!:)ending moments due to the axial forces F n', F /', F / 11
,
acting on crank pins 1, 2, 3, by the symbols Mb', Mb11
,
Mb111
, the value of each bending moment at the journal
can be determined by the equation
ab2
(57) Mb=F "(a+b)2
where a=the distance from the center of the crank pin on
which F n is actin,$ to the center of the journal
in which the bending moment is determined.
b=the distance from the center of the crank pin to
the center of the other journal.
It is considered that a bending moment M, is also produced
in the journal by a tangential force F, due to the
maximum instantaneous torque Tm (see equation 38).
This force is assumed to act on the crank pin adjacent to
the journal and to produce a bending moment at right
angles to the crank throw whose magnitude is:
(58) M F c(dc)2
, t d2
where c=the distance between the center of the journal
and the center of the adjacent crank pin.
d=distance between the journal centers.
The bending moments due to the axial forces on the
crank pins and the bending moment due to the tangential
force are combined vectorially. The resulting bending moment
Mc is thereby obtained. In combining the bending
mo.ments, it is necessary to consider the angular relation
at the crank pins together with the phase relation of the
forces acting upon them. It is generally necessary to
determine the value of Mc throughout a cycle in order to
obtain its maximum value.
The equivalent bending moment acting on the journal
is obtained . from Mc and the maximum instantaneous
torque, Tm.
The maximum fiber stress is Sb=~·
...
T F
The shear stress is S.= t+2.A
where F,=resultant of bearing reactions as determined by
equation (35) .
STRESS IN CRANK PINCASE 2.
The bending moment, Mb, at the center of a crank pin
located between main bearings A and B, due to the force
F n acting on the crank pin, is computed by means of the
equation
(59)
where a=distartce from center of crank pin to center of
bearing A
b=distance from center of crank pin to.center of
bearing B
The bending moment Mb' at the center of this same
crank pin due to a force Fu', acting on another crank pin
located between the same main bearings and at a distance
s from the first, is
(60) M '=F , b (as)
b n a+b
(61) M '=F 'a (bs)
b n a+b
Equation (60) should be used when the force acts
between the crank pin, where the bending moment is
determined, and main bearing A . Eq~ation (61) should
be used when the force acts between _the crank pin and
bearing B.
The vector sum of the bending moments M1, and J\'li,1 is
the combined bending moment Mc. In determining
Mc, the angular relation of the crank pins, together with
the cyclic relation of the forces acting on the crank pins,
must be considered. It is generally necessary to determine
the value of Mc throughout a cycle in order to obtain
its maximum value. ,
The equivalent bending moment resulting from Mc
and the maximum torsional moment Tm is:
(41)
The maximum fiber stress is
(45)
The total shear. stress is
/
(46)
STRESS IN E ND CRANKCHEEKCASE 2.
In transmitting the maximum torque of the crank shaft,
a twisting moment is set up in the crank cheek, wh,ich is
Mt=FtXe (49)
where Ft=Tm (38)
r
e=distance from center of crank pin to center of
crank cheek .
10
The maximum bending moment in the crank cheek
due to· the tangential force, F., about an axis parallel to
the crankshaft (HH axis, Fig. 20) is
(50)
where c=distance from center line of crank pin to top of
journal.
The maximum_ axial force on the crank pin, F n, produces
a bending moment in the crank cheek about an
axis perpendicular to the crank shaft (EE axis, Fig. 20)
which is
(62) M/=F '!Xe
The tensile stress is
(63) S  Mb+Mb' +F n
tzb Zb' 2A
The shear stress is
I
(55)
The values of Zp, Zb, and Zb' are given in e1uations (51),
(52), and (53).
The maximum equivalent stress resulting from S, and
st is
s.=, s,+Jt s.2+s,2. (56)
STRESS IN INTERMEDIATE CRANKCHEEKCASE 2.
The torsional moment in the crankcheek due to the
tangential force, Ft, is
Mt=FtXe. (49)
The tangential force, Ft, also sets up a bending moment
about an axis parallel to the crankshaft (HH axis, Fig. 20),
which is
where h = distance between the centers of the two crankpins
on either side of the crankcheek.
ll=angle that a line joining the crankpin centers
makes with a line through a crankpin and
journal center.
The bending moment, due to the axial force on the
crankpin, F n, about an axis perpendicular to the crankshaft
(EE axis, Fig. 20) is
CONNECTING ROD STRESS ANALYSIS.
STRESS IN CONNECTING ROD SHANK.
The stress in the <;onnecting rod shank due to direct
compression is calculated by means of Rankine's column
formula, and is
(65) S0= 1+0.000§26 ~ F, or i+0.000526 ~:( F,,
where F,=resultant force along connecting rod axis.
A=area of shan~ cross section.
1, and ly=moment of inertia of area about x x and yy
axes, respectively (see Fig. 22).
L= length of connecting rod c. to c.
L,= length of shank as in Fig. 22.
0.000526=constant for steel column.
The following relation, which gives the characteristics
of the shank cross section for equal compressive strength
in both planes, is obtained from the preceding equations.
(66) I =(~)2 !.e.
Y L 4
WHIPPING EFFECT ON CONNECTING ROD .
Considering the rod as•a uniform section extending from
the piston pin center to the crl!nk pin center, the whipping
force that results, due to the transverse acceleration of the
ro_d, is approximately
(67) F =~:",~ c~N) 2 ~ sin 0=0.00000402 ALN2R sin o
w 2g 30 12 '
where Fw= whipping force, lb.
A=area of shank cross section, sq. in.
L= length of connecting rod, c. to c., inches.
N=r. p. m.
R =crank radius, inches.
O=crank angle.
w=weight of material, lb. per cu . in.=0.283.
g=acceleration due to gravity=32.2.
The whipping force is considered as a dis tributed load
varying uniformly from zero at the pistonpin to a maximum
value at the crankpin end of the rod. Such a loading
is assumed to produce a reaction of 1/3 the total load
at the piston pin and 2/3 at the crank pin. The reactions
are small and are neglected in determining bearing loads.
The bending moment produced by the whipping force in a
section of the connecting rod · shank at a distance x from
the piston pin is
(62) (68) Mw=F:t (1 f,)
The tensile stress is
S _Mh+M\ Fn
' tzb Z\ +2A·
The sh ear stress is
(63)
(55)
The maximum equivalent stress resulting from Ss and
S, is
s.=t st+Jt si2+s.2. (56)
The maximum bending moment, M' w, occ urs at the
section where x=0.577 L, and is
(69) M',,=0.1283 L F w
The stress set up in the shank cross section due to the
whipping action is determined by the equation
(70)
where Sb=fiber stress due to whip.
Z=section modulus of shank area about axis
parallel to piston pin .
....
{
/
/
11
BENDING STRESS IN MASTER ROD.
In an articulated connecting rod design, the action of the
link rods sets up bending stresses in the master rod . The
line of action of the force at each link pin does not pass
through the crankpin center at all crank positions, with
the result that turning moments are caused about the
crank pin and a bending action is produced in the master
rod shank. Since the intersection of the line of action
of the resultant force at each link pin with the center line
of the master connecting rod falls within the big end of the
master rod, the bending moment that results at any cro~s
section of the shank is proportional to the distance of the
section from the pistonpin center. The resulting stress
set up in any cross section is
(71)
where Sb=fiiber stress due to bending action.
, M,=resultant turning moment about crank pin due
to resultant forces acting on link pins.
L=length of master connecting rod, c. to c.
X=distance of crosssection from pistonpin center.
Z=section modulus of are:i, about axis parallel to
piston pin.
Example (Liberty 12):
At slow speed, the ip.ertia forces are negligible and the
force along the connecting rod axis is due to gas p~essure.
The force due to the maximum gas pressure as obtained
from the indicator card is
F 450Xl9.63=8860 lb.
r COS 4° (33)
Shank cross section (see Fig. 22, section AA).
B=0.938, H=l.375, C=0.125, D=0.094, E=l.125
A=area of section=0.34 sq. in.
l=moment of inertia
lx=0.103
ly=0.0172
L= 12.0
0.000526 L2
I,
0.000526 L/
4I,
0.735
0.672
(65)
(65)
Sc=F, ( 0.~4+ 0.735 )=8860 X3.676=32,570 lb. per
sq. in. (65)
The variation of stress in the connecting rod shank
during normal running conditions is shown in Table IX
and Fig. 23. The total stress in the shank is the sum of
the stress due to the force acting along the rod and that
due to the whipping force. It is assumed that the maximum
stresses resulting from these forces occur at the same
cross section.
The force along the connecting rod axis is
F F,.
'
9 cos cf,
(33)
The compressive stress due to compression in the rod is
(65)
The tensile stress due to tension in the rod is
F
S,=f=2.94 F,
The whipping force is
Fw=0.00000402X0.34Xl2Xl7002X3.5Xsin e (67)
Fw=l66 sine
The fiber stress due to the maximum bending moment
caused by the whip is
Sb=(0.1283Xl2Xl66 sin O) ~:~~~
Sb=l710 sine
The total stress in the shank is
S/=Sc+S., (compressive).
St' =S,+S" (tensile).
(69 and 70)
From Table VII and Fig. 24, it is seen that the stress due .
to the whipping force is at all times quite small, and that
it is only 0.2 per cent of the total stress when the latter is ·
at its maximum. As a rule, it is unnecessary to consider
the whipping force in an engine of the usual design.
STRESS IN CONNECTING ROD FORK.
Straddle and link connecting rods are sometimes constructed
with forked ends as in Fig. 24. Considering the
fork to be' free from end wise constrainment, the following
~q uation is derived to determine the stress at any crosssection
of the fork.
F [ sin/3 y . J (72). S= f x_+r: (Cxy, sm /3)
where S=maximum stress.
F,=force along conn. rod.
A=area of crosssection considered.
C=distance of center of fork bearing to centerline of
rod.
/3=angle of section to centerline of conn. rod (see
Fig. 24).
x=distance from inner edge of section to centerline
of connecting rod.
y,=distance from C. G. of section to inner edge.
y2=distance of C. G. of section to outer edge.
l x=moment of inertia of section about gravity axis
XX.
Example (Liberty 12):
Section of forked rod ( weakest section).
Referring to Fig. 24. · 1
•
X= 0.20, /3=60°, C=0.89
H=l.375, E=0.125, K=l.125, D=0.56, G=0.13
A=0.286
y,=0.171
y,=0.389
lx=0.00717
12
Max. force along rod, due to gas pressure= 8860
8860 [0.866 0.389 J S =2
0
_286+ 0_00717 (0.89  0.20  0.171X0.866) (72)
S=4430 (3.03+29.4) = 144,000 lb. per sq. in.
NoTE. The ends of the fork in this rod are constrained
from endwise deflection by the bearing shell. The
greater portion of the bending stresses are thereby eliminated
from the fork and are taken up by the sh ell. In the
derivation of equation (72), no consideration is given to
this supporting action, so that t he stresses in the fork in
this case are much lower than those determined above in
the case where the fork is a good fit on the bearing shell.
STRESS IN BEARING CAP.
In the following analysis the bearing cap is considered
as a curved beam of uniform croAs section, hingesupported
at the bolt centers, and carrying a uniformly distributed
load. The support that the cap would receive from the
journal if sufficient flexure would occur in the cap is
neglected, since such action is undesirable. In making
the above assumptions, it can be shown that the maximum
stress in the bearing cap is
(73) S=F co.0;30 + o:.)
where S=maximum stress in bearing cap
F = load on bearing cap
C=distance between bolt centers
A= area of cap cross section
Z=section modulus of area
The maximum load on the bearing cap is due to centrifugal
and inertia loads. It occurs when the piston is at
top center and while the engine is running at its highest
speed and there is no gas pressure acting on the piston.
In determining the centrifugal load on the cap, the weight
of the bearing cap and bearing cap bushing is subtracted
from the total centrifugal weight at the crank pin. With
the articulated connecting rod design, the components of
the forces in the link rod cylinders in the direction of the
master rod axis are added to the inertia and centrifugal
forces due to the master rod and its piston.
Example (Liberty 12):
Stress in bearing cap of forked rod.
At weakest section of cap, A=O. 37, Z=O. 021,
0=3. 375
Inertia force (maximum) at 1700 r.p.m.=2300 lb.
Centrifugal weight at . crank pin minus weight of
bearing cap and bearing cap bushing=4.4  l.6=
2.8 lb.
F0= 0. 0000284X2.8X 3.5Xl7002=810 lb.
F=2300+810=3110 lb.
(19)
S=3110 (0.023 X3.375+ _ 1_ )=15 700 lb. , per
0.021 2X0.37 '
sq. in. (73)
The stress in the cap for the plain end rod , calculated in
the same manner, is 27,000 ib. per sq. in.
STRESS IN BEARING CAP BOLTS .
In aqdition to the stress due to the working load, considerable
stress is induced in the bolts in screwing up the
units. The magnitude of this stress cannot be predetermined
and ~ay vary from 20,000 to 80,000 lb. per sq. in.
Whether the total stress in the l;>olt is the sum of the stress
due to screwing up and the stress due to t.hll working load,
or whether it is only the greater of these stresses, depends
upon the relative amount of deflection that occurs in the
bolt and in its supporting boss when under load. In the
usual design, screwing up the nut is accompanied by an
elongation of the b olt ·without any appreciable deflection
or compression of the bolt support. In this case, the maximum
stress in the bolt can only reach the value of either
the stress due to screwing up or the stress due to the working
load.
In order that the load on the bearing cap does not cause
the bearing cap to separate from the rod and thereby allow
a leakage of oil from the bearing, it is necessary that the
stress in the bearing cap bolts due to the working load be
less than the stress due to screwing up the bolts. This
condition limits the allowable stress in the bolts due to
the working load.
The maximum load on the bearing caps is also the maximum
working load on the bearing cap bolts. The resulting
stress due to this load is
(74) S= _E_
nA
where S=tensile stress in bolts.
F=load on bearing cap.
n = number of bearing cap bolts.
A=minimum crosssectional area of bolt, usually at
root of threads.
Example (Liberty 12):
Forked rod bolts
4i\in. bolts S. A. E. thread. (A=0.052 sq . in.)
P=3110 lb.
3110 .
S=4x0.052=14,900 lb. per sq. m. (74)
PISTONPIN END.
When the piston pin is supported by a bearing in the
connecting rod, the rod end must have sufficient stiffness so
that it will not pinch the pin when under load. In a uniform
ring under a concentrated load, the decrease in diameter
of the ring perpendicular to the load is
(75) ~=2 FR3 (!__!_) El 1r 4
where F=force on ring
R = mean radius of ring.
E=modulus of elasticity of material.
!=moment of inertia of ring cross section.
\
,..
13
From the preceding equation the following expression
can be obtained for the relative deflection per inch of diameter
in the piston pin end
FD2
(76) Ko=EI
where Ko=constant times deflection per inch of diameter
D=mean diameter of pistonpin end.
The maximum force on the upper half of the pistonpin
end of the connecting rod, which is used in the above equation,
is the maximum inertia force of the piston a.ssembly.
Example (Liberty 12):
Weight of piston assembly=4.9 lb.
Fi (max,)=0.0000284 X 4.9X3.5Xl7002Xl.292=1820
Th. ~~ ..
Maximum bearing pressure in piston (B1=2Xl.055):
8840
P l. 25X2Xl.055 3350 lb. per sq. in. {77)
Maximum tensile stress:
32 1.25 8840
St=; (1. 25._0_81254)  8 
(2X,U82.0)=29300
lb. per sq. in. (78)
Maximum shear stress:
Ss. 8840
2 X i(l.252 0.81252
)
6240 lb. per sq. in. (79)
VALVE AND VALVE GEAR ANALYSIS.
GAS VELOCITY.
l=0.000279, D=l.56, E=3Xl07
K . 1820Xl.562
53
The comparison of valve sizes in different engines is
' usually made on the basis of gas velocities. The methods
0=3x101 x.000219= · (76) for calculating gas velocities are necessarily empirical.
PISTONPIN ANALYSI~
An investigation is made of the stresses in the piston pin
and the bearing pressure in both the connecting rod and '
piston bearings. In determining stresses, the pin is con ,
sidered as a beam supported at the center of the piston
bearings and loaded uniformly at the upper connecting
rod bearing.
The bearing pressure is
I F
(77) P=DB
The maximum tensile stress is
)
S F(2LB) ~~ _ D_ F(2LB)
(78 t 8 z 1r D4d4 8
The shear stress is
(79)
where
F=force on piston.
D=outer diameter of pin.
d=inner diameter of pin.
L=distance between centers of bearings in piston.
B=length of bearing in connecting rod.
B1=total length of bearings in piston.
Example (Liberty 12):
F (maximum) =450 Xl9.63 =8840 lb.
D=l.25, d=.8125, L=3.18
Of the various methods in use, the three methods outlined
below have been found most useful, and it is recommended
that in engine analysis the gas velocity be determined by
all three methods. ·
1. Gas velocity through valve port. This method
considers the mean velocity in the valve port on a basis
of filling or exhausting the cylinder during 180° of
crankshaft rotation.
(80)
where v 1=gas velocity through valve port, ft. per sec.
D=cylinder diameter, inches.
s=piston stroke, inches.
N=r. p. m. ·of engine.
d=diameter of valve port (nominal valve diameter
or bore of mouth of port), inches.
n=number of intake or exhaust valves.
2. Gas velocity through yalve annulus on basis of maximum
lift. This method considers the mean velocity
through a cylindrical annulus of a diameter equal to the
valve port diameter and a height equal to the maximum
valve lift, this annulus being taken as constant throughout
the period of filling or exhausting. The time of filling
or exhausting is taken as 180° of the crankshaft rotation.
,r
4D2s 2N D2sN
(3l) V2=121rdhn 60 1440 dhn
where v2=gas velocity through annulus on basis of maximum
lift, ft. per sec.
h =total valve lift, inches.
3 Gas velocity through valve annulus on basis of mean
Maximum bearing pressure in connecting rod (B=2.0): li'ft.· This method considers the mean velocity through
8840 .
P=1.25 x 2= 3540 lb. per sq. m.
39757233
a cylindrical annulus of a diameter equal to the valve
(77) port diameter and a height equal to the mean valve lift
14
on a ha.sis of filling or exhausting the cylinder during an
interval equal to the total time of valve opening. The
mean lift of the valve can be obtained from a valve lift
diagram.
(82)
where v3 = gas velocity through valve annulus on ha.sis
of mean valve lift, inches.
q 0 =period of valve opening expressed in degrees
of crank shaft rotation.
hm=mean valve lift.
In determining the ga.s velocitieR by methods 2 and 3
the valve annulus is considered a.s a cylindrical surface,
which is the case for a flat seated valve only. Where the
valve seat is tapered, the actual minimum annulus area
is a truncated conical surface, in which ca.se the actual
annulus opening is less and the .gas velocity is greater than
that considered by the methods given. When the valve
lift is not great, the reiation of the conical annulus to the
cylindrical annulus considered in equations (81) and (82)
is given by the equation
(83) A1= A2 (cos a+lcos2 a sin a)
where A1=area of conical annulus.
A2= area of cylindrical annulus.
a=angle of valve seat.
For a 30° valveseat equa,tion (83) becomes
(84) A1= A2 ( 0.866+ 0.375~)
For a 45° valveseat equation (83) becomes
(85) A1=A2 ( 0.707+0.353~)
It is not thought desirable to consider the angle of the
valve seat in determining the ga.s velocities through the
valve annulus. If the ga.s velocity is figured by equations
(81) and (82), a slightly higher ga.s velocity is allo}V,
able when the valve seat is 30° than when it is 45°, for a
comparison of equations (84) and (85) shows that the restriction
in the annulus due to the angle of the valve seat
is less for the 30° valve seat .
Example (Liberty 12):
Intake. Exhaust.
Valves per cy!inder n ..... .. __ ...... _ ..... ____ .. .
Nominal valve diameterd .......... _ . . .... . .. _ .. ~:1;:mk ::~':'..%'.:'::~~:::::::: :: : : : : : ::: : : : : : : :
Total period of openingq ..... . . . .. . . ... . . .. _. _ . _
1
2. 5
.43.S
. 245
215°
1
2. 5
.368
. 230
236°
Cylinder boreD=5. Stroke  s=7. r. ,ll· m .N=
1700.
Inlet valve:
v, 52x 1x 1100
360 x 2_52Xl =132.2 ft. per sec. (80)
V2
52x1x1100
190 ft. per sec. 1440X2.5X0.435Xl (81)
Va
52x1x1100
8X215X2.5X0.245Xl 282. 5 ft. per sec. (82).
Taking into consideration the 30° angle of valve seat
V'.~V2A2=190 1
 A, (o.866+ 0.375X0.435)
2.5
204 ft . per sec. (81 and 84)
V'a= v3!~=282.5X ( o.866+ o.3\~0.245)
312 ft. per sec. (82 and 84)
Exhaust valve :
V1 =132.2 ft. per sec.
V2=224.5 ft. per sec.
V3=274 ft. per sec.
Taking into consideration the 30° angle of valve seatV'
2=244 ft. per sec.
V'3=304 ft. per sec .
VALVE SPRING.
The usual type of valve spring is the helical spring
made from round steel wire. Volute springs made from
steel strip, laminated cantilever springs, and various other
spring types are also occa.sionally employed. An analysis
of the valve spring sh~uld include the force exerted by the
spring at various valve lifts and th·e stress in the spring
material at zero and maximum lift. In some cases, the
natural period of spring vibration should also be determined.
The following are the formula, used in the analysis of the
usual type of cylindrical helical valve spring made from
round steel wire.
15d4G 8 PD3n
(86) P= 8 D"n or 15 =cFG
Where P =force along spring axis, lb.
15=total deflection o.f spring ( deflection at zero
valve lift plus valve lift), inches.
d=diameter of wire, inches.
D=mean diameter of coil (at center of wire),
inches.
G=transverse modulus of elasticity of material (for
steel, G=l2.5Xl06) .
8Pll PD
(87) S= 1rd" =2.546 cl"
where S=shear stress, lb . per sq. in .
Example (Liberty 12):
Calculations for inlet out8i,]P valve spring:
Diameter of wire, d
Mean diameter of coil, D
Number of effective coils, n
0.120? in.
1. 45, in.
=12
Length of spring, valve closed = 2. 25 rn.
Force along spring, valve closed=23. 5 lb.
Total valve lift
Deflection of spring at zero valve lift :
8X23. 5Xl.4533Xl2
15 0.1205'Xl 2.5Xl06 2·626 in.
. 435 in .
(81i)
I
•
/
15
Deflection of spring at full valve lift:
01 =2.626+0.435=3.061 in.
Force along spring at full valve lift:
P,=PXt=23. 5X;:~~!= 27.4 lb.
Stress in spring material, valve closed:
of the valve is away from the cam when the cam follower
is in contact with the tf p radius of the cam, and is •
(90) __ !: (21rN) 2 _:Q ·c k2 cos 2/3+sin• /3]
a L 1 60 X 12 X cos /3 + (k2 sin2 /3)'< .
where a=acceleration of valve, ft. per sec.2
N =r. p . m. of cam shaft. (
L/L, =ratio of valve lift to cam follower lift.
S 8X23 .5Xl.453 lb . D= distance from cam center to center for the tip
= ... x.12~=49,800 . per sq. m. (87) radius, inch es (see Fig. 25).
Stress in spring material, valve open:
/3= angle of cam rotation from position of full lift .
k (r+ R,)
D
s, 8X27.4X l.453 1r X .120a3 58,000 lb. per sq . in. (87) r=radius of cam roller.
Valve spring .
Inlet. Exhaust.
Outer. Inner. Outer. Inner.
 1· 
Diameter of wire . . ................... 0.1205
Mean diameter of coil. ............. , . l. 45:l
Number of effective coils ...... .. ..... 12.
Length of spring, valve closed........ 2. 25
Force along spring, valve closed ....... 2:i. 5
Tota l valve lift.......... . ....... .. ... . 435
Stress in wire, valve closed.. . . . . . . . . . 49800
0. 1205
1. 016
12.
2.25
26.5
. 435
39200
55900
0. 1483
. 1.453
LO.
2. 25
45.
. 36R
51400
61600
0.1205
I. 016
12.
2.25
26.5
. 36R
39200
Stress in wire, valve open .. .... . . . . . ·I 58000 51400
'~' "'
CAMS.
The shape of t he tip of the cam must be such that at th"
maximum engine speed the force due to the valve acceleration
away from the cam iR less {han the pressure exerted
by the spring.
(88) P> W•a
g
where,P = force exerted by valve spring at a given lift, lh
W0 =total equivalent reciprocating weight at valve,
lb.
R1=tip radius of cam.
Where a flat cam follower is used instead of a roller, the
radius R, and the ratio k are infinite. For this case equa tion
(90) becomes
(91) a= _ _!: ( 2 ... N)
2
D cos /3
L, 60 12
. The cam analysis should contain curves o:f valve lift,
valve velocity, and valve acceleration. The mean valve
lift is obtained from the valve lift curve. The valve
velocity at opening and seating can be obtained from the
velocity curve. The acceleration curve is necessary to
determine the forces acting on the valve and on the cam
shaft.
The formulae for determining mathematically the lift,
velocity, and acceleration of the cam follower are given
below for the usual geometric cam shapes. These formulae
hold true only where the motion of the cam follower is
along a straight line that passes through the cam center.
The lift, velocity, and acceleration of the valve is L /L, '
times that of the cam follower.
Tangentional cam with roller tappet (Fig. 25b):
Roller in contact with tangential face :
(92) s=(R+ r)X(sec a  1)
g=acceleration due to gravity=32.2
a=acceleration of valve at lift corresponding
force P, ft . per sec. 2
to (93) v=(2;:) x R1t r xtan a Xsec a
The total equivalent reciprocating weight lat the valve,
w., includes the weight of the valve and all movine fittings
attached to same, plus ! the weight of the valve spring,
plus the equivalent weight at the valve of the other parts
of the valve gear that reciprocate with the valve, such as
rocker lever, push rod , and valve tappet. The weight at
the valve, equivalent to a weight in a part of the valve gear
which reciprocates in unison with the valve, can be
obtained from the equation
(89) W'.=W' (~Y
where W'. =equivalent weight at valve corresponding to
weight W' .
L' = lift of weight W' corresponding to valve lift L.
The acceleration of the valve away from the cam occurs
when the cam follower is operating on the valve tip. I ts
value depends upon the valve mechanism employed and
the engine speed. In the usual case, shown in Fig. 25 a,
where the path of the cam follower is along a straight line
passing through the cam center, line PO, the acceleration
(94)
(95)
a=(~ ... N)2 XR + r X2cos2
"'.
60 12 cos3 a
Roller in contact with tip radius (Fig. 25 a):
s=.J(r+R,)2(D sin /3)2+D cos /3 (r+ R)
(96) v=(2;:) R [ sin /3 ( 1+ .Jkzc~:!2 13) J
(97) =  (21rN)
2 _Q [ "f k
2
co~+sin4 /3]
a 60 12 cos ,.., (k2 sin2 /3)i
where s=lift of cam roller, inches.
v=velocity of cam roller, ft. per sec.
a=acceleration of cam roller, ft. per sec .2
R=radius of cam base circle, inches.
r=radius of cam roller, inches.
R, = tip radius of cam.
D=distance from cam cen ter to center for
radius, inches.
k_(r+R,)
  DN
=r.p. m. of cam shaft.
the tip
a =angle of cam rotation from position of zero lift.
/3=angle of cam rotation from position of full lift.
16
Convex flank cam with roller tappet (Fig. 25c):
Roller in contact with convex flank:
(98) s=.f(R2+r)2(D1 sin a)2  D1 cos a(R+r)
(99) v= (21r N) D, sill__« [1
60 12
(100) a=(2;~)
2
~2 [cos a
+ Di{(R2+r)2 (12 cos2 a)D2
1 sin4 a}]
{.f(R2+r)2 (D, sin a)2}i
where R2=flank radius of cam, inches.
D1=distance from cam center to center for the
flank radius, inches.
Convex flank cam with flatfaced follower (mushroom
cam) (Fig. 25d):
Follower in contact with convex flank:
(101) s=(R2R) X (1cos a)
(102) v=(2;~)xsi;2 °' x(R2 R)
(103) a=(
2;~)2 x ci2 °' x (R2R)
Follower in contact with tip radius:
(104) s=D cos i3+R, R
(105) (21rN) D sin /3
V  60 12
(106) _ (21rN)2 D cos /3
a 60 12
Example (Liberty 12):
The valve lift, velocity, and acceleration curves, together
with the shape of the ~am and the spring pressure
curve, are given in Figs. 26 and 27 for the inlet and exhaust
valves, respectively. The valve lift, velocity, and
acceleration curves were determined by means of the
formulae given above for a tangential cam, in which it
was assumed that the path of the roller lies along a straight
line passing through the cam center. These curves are,
therefore, slightly in error, since in the example the path
of the roller is along an arc. The rocker ratio employed
is such that the valve lift is 1.49 times the cam follower
lift. The velocity and acceleration curves ar shown for
the normal engine speed of 1,700 r . p.m. The force scale
employed in the diagrams is such that the spring force
curve also shows the acceleration of the moving valve
parts that the spring is capable of counteracting.
Inlet valve, lift, velocity, and acceleration:
R=0.5525
k=l.22
r= 0.4062
N=850
R,=0.2869
D=0.5673
Rollerin contact with tangential face:
s= l.49X0.9587X(sec a1)
v=l.49X88.9X0.0799Xtan a seca
\ 2 cos2 a
a= l.49X7900X0.0799X cos3;:;
(92)
(93)
, (94)
Roller rn contact with tip radius:
s= l.49 (.f0.480.322 sin2/3+0.5673 cos /30.9587) (95)
v=l.49{88.9X0.0473x[sin /3 (1+, ~ ~  )J} (96)
v l.49  srn2 /3
a= l.49{7900X0.0473X [ cos /3
+ 1.49 cos 2/3+sin4 /3]}
(1.49sin2 /3)i
(97)
Spring force. Valve Valve
closed. open.
Outer spring._. __ . ___ . __ .•. _______ ._______________ 235 lb.
Inner spring ______ _ .. __ . . . . . . . . _, __ . ... ____ __ ______ 26.5 lb.
Total.. .. _, ______________________ . _ _ _ _ _ _ _ _ 5Q_Q lb.
27.41b.
37.8 lb.
65.21b.
Equivalent reciprocating weight of valve mechanism:
Valve, spring collar, and key ______________ 0.70 lb.
tinner and outer springs ____ ___ . _________ .17 lb. 
Rocker assembly. _ .. ___ . _. _. _. _______ : _ .15 lb.
Total__ ________________  .. ____ .... 1.02 lb.
Exhaust valve, lift, velocity, and acceleration:
R=0.5525.
k= l.49.
r= .4062.
N= .850.
R1= .3236.
D= .4889.
Roller in contact with tangential face:
s=l.49X0.9587X(sec a1).
v=l.49X88.9X.0799Xtan a sec a.
2cos2a
a= l.49X7900X .0799 X cos3a
Roller in contact with tip radius:
(92)
(93)
(94)
s=l.49(.f0.5330.239 sin2 /3+0.4889 cos /3  0.9587) (95)
v=l'.49{88.9X.0407 X[sin /3(1+ .f cos~ )]} (96)
2.22sm2 B
J [ 2.22 cos 2 i3+sin• /31}
a= 1.49 t7900X.0407X cos 13+ (2.22 _sin2 /3) t J (97)
Spring force. Valve Valve
closed. open.
    
Outer spring _____ __ ... _. _ ... ___________ .. _ _ _ _ _ 45.0 lb. 54.0 lb.
Inner spring____ _ __ __ __ _ __ __ _ _ __ __ __ ___ _ __ _ _ _ _ 26.5 lb. 36.l lb_
Total.. ___                 7L5lb, 1~
Equivalent reciprocating weight of valve mechanism:
Valve, spring collar, and key .. _._ .. _ ...... 0.70 lb.
! inner and outer springs ___ .............. .21 lb.
Rocker assembly .. _._. __ ...... :......... .15 lb.
l Total__ ____________________________ 1.06 lb.
•
APPENDIX.
CHARACTERISTICS AND DIMENSIONS LIBERTY
12 A. ENGINE.
Number of cylinders ............... . . 12°.
Arrangement of cylinders ............. 2 banksat45° Vee.
Method of numbering cylinders.
{
o o o o o oR.
Propeller end. . 0 0 0 0 0 0 L.
 654321.
Firing order .. .. ................... . ..... .
6R, 5L, 2R, 3L, 4R, 6L, lR, 2L,
Bore ................ ....... . ... . . .. .... .. .
lL,
5R, 4L, 3R.
5 in.
Stroke ..... .... .............. . . . . . .. . . ... 7 in.
Piston area . ... .. ... ....... .. ........... .. . 19.63 sq. m.
Total piston displacement. ................ 1,649 cu. in.
Brake horsepowernormal. . . . . . . . . . . . . . . . 421.
R. P . M.Normal.. . .. ..... : ............. 1,700.
Compression ratio ......................... 5.42:1.
Mechanical efficiency.... ... .... . . . . . . . . . . . 88.4 % .
Connecting rod length, center to center ..... 12 in.
Connecting rod to cranklength ratio ........ 3.428:1.
Reciprocating and centrifugal weights:
Piston; complete with rings and pin. . . 4.9 lb.
Upper end of connecting rod . . . . . . . . . . 1.3 Jb.
Lower end of forked connecting rod .... 4.4 lb.
Lower end of plain connecting rod ..... 1.9 lb.
Total reciprocating weight per cylinder ..... 6.2 lb.
Total centrifugal weight per crank pin ..... 6.3 lb.
Crankpin bearing:
Effective length.............. . . . ...... 2!,, in.
Diameter..... . . . . . . . . . . . . . . . . . . . . . . . . 2j in.
Bearingarea .......................... 5.2 sq. in.
Crankshaft bearing (center, intermediate,
and end): •
Effective length .................. .I .•.. li in.
Diameter ............................. 2i in.
Bearing area ............... ... . . ..... . 4.27 sq. in.
Crankshaft:
Diameter of journal... . . . . . . . . . . . . . . . . 2i in.
Bore through journal.. ..... . .......... li in.
Diameter of crank pin ............... .. 2j in.
Bore through crank pin ................ 11 in.
Width of crank cheek (at top of journal) . 3H in.
Thickness of crank cheek ............... 1 in.
Length of crank pin .... •. . . . . . . . . . . . . . . 2! in .
Distance between journal centers . .... . 61 in.
Piston:
Total length of skirt ................... 5.08 in.
Effective bearing length . .. . . .. .. . . .... 3.31 in.
Effective bearing area . ....... ..... ..... 16.55 sq. in .
TABLE I.
Per cent piston travel.
Crank
angle.
,    ,   
0 360
10 350
20 340
30 330
40 320
50 310
60 300
70 290
80 280
90 270
100 260
110 250
120 240
130 230
140 220
150 210
160 200
170 190
180 180
_o~o .
1. 0
4.0
8.8
15.2
22.8
31. 4
40.5
49.6
58.6
67.0
74. 7
81.4
87.1
91. 8
95.4
98.0
99.5
100.0
0.0
1. 0
3. 9
8. 7
15. 0
22.5
31. 0
40.0
49.1
58.0
66.5
74.2
81.0
86.8
91.6
95.3
97.9
99. 5
100.0
0. 0
1. 0
3. 8
8.6
14. 8
 22.3
30.6
39.6
48. 7
57. 5
66.0
73.8
80.6
86.5
91.4
9957.. 28 1
99.4
100.0
o. 0
1. 0
3.8
8.5
14.6
22.1
30.3
39.2
48.2
57.1
65.6
73.4
80.3
86.3
91. 2
95.1
97.8
99.4
100.0
(17)
c. 0
.9
3. 7
8.4
14.4
21. 8
30.0
38.8
47.8
56. 7
65.2
73. 0
80.0
86.0
91. 0
95.0
97. 7
99.3
100.0
o.o
. 9
3. 7
8.3
14.3
21.6
29. 7
38.5
47.5
56.4
64.8
72. 7
79. 7
85.8
90.9
94.9
97. 7
99.3
100.0
o.o
.9
3. 7
8.2
14.2
21. 4
29. 5
38.2
47.2
56.1
64.5
72.4
79. 5
85.6
90.8
94.8
97. 7
99.3
100.0
0.0
.9
3.6
8.1
14.1
21. 2
29. 3
37.9
46.9
55.8
64.2
72.1
79. 3
85.5
90. 7
94. 7
97. 7
99.3
100.0
18
TABLE II.
Crank angle factor for piston velocity.
Crank
i=3.8 I i=4.0
angle. L L L L L L
R=3.o R=3.2 R=3.4 R=3.6 R=4.2 R=4.4
       
0 0
0 360 o.o 0. 0 o.o 0. 0 o. o o.o 0.0 0.0
10 350 .231 . 227 .224 .221 . 218 .216 . 214 .212
20 340 .450 . 443 . 437 .432 .427 .423 .419 . 415
30 330 . 646 .637 .629 .622 .615 .609 . 604 . 599
40 320 . 811 .199 . 790 . 782 . 775 • 768 • 762 . 756
50 , 310 . 936 . 924 . 915 .906 . 898 . 891 .885 . 880
60 300 1. 017 1.007 .998 .990 .983 . 977 • 971 .966
70 290 1. 053 1. 045 • 1. 038 1. 032 1. 027 1. 022 l. 018 I. 015
80 280 1. 045 1. 041 1. 037 1. 034 1. 031 I. 029 1. 027 1.025
90 270 1.000 1. 000 1. 000 1.000 1. 000 1.000 1. 000 1. 000
100 260 . 925 .929 .9:J2 .935 . 938 . 941 . 943 . 945
110 250 . 827 .835 . 841 .847 . 852 .857 . 861 . 865
120 240 . 715 . 725 . 734 . 742 • 749 . 755 . 761 . 766
130 230 . 596 :~ .617 . 626 . 634 .641 . 647 .652
140 220 .475 . 495 .503 . 511 .518 .524 . 530
150 210 .354 . 363 .371 . 378 . 385 . 391 . 396 .401
160 200 . 234 . 241 . 247 . 252 . 257 . 261 .265 . 269
170 190 . 117 .120 .123 .126 .129 .131 .133 . 135
180 180 0.0 . 0.0 0.0 0.0 o. 0 0.0 0.0 0.0
TABLE III.
Crank angle factor for piston acceleration.
Crank angle.
I L L L L L L L L
n: = 3.0 R = 3. 2 n:=3.4 n=3.6 R=3.8 R = 4.o R=4.2 R=4. 4
             
0 0
0 360 1.333 1. 313 1.294 1. 278 1.263 1. 250 1. 238 1.227
10 350 1. 298 1. 279 1. 261 1.246 1. 232 1. 220 1. 209 1.198
20 340 1.195 1.179 1.165 1. 153 1.142 1.131 1.122 1. 114
30 330• 1. 033 1.022 1. 013 1.005 .998 . 991 .985 .980
40 320 .824 . 820 . 817 . 814 .812 . 809 . 807 .806
50 310 . 585 .589 . 592 . 595 . 597 .600 .602 .604
60 300 . 333 . 344 .353 .361 . 368 . 375 . 381
I
. 386
70 290 . 087 . 103 . 117 .129 .140 . 151 . 160 . 168
80 . 280  . 139  . 120
I
 .103  .087 .073 . 061  .050  .040
90 270  . 333  .313  . 294 . 278  .263  .250 .238  .227
100 260  . 486  . 467  .150 . 435  .421  . 409 .397  .387
llO 250 . 597  .581  .567  . 555  .544 . 534 .524 I  . 516
120 240  .667  . 6.56  .647  . 639  .632  .625 .619  .614
130 230 . 701  . 697 . 694 . 691  .688 . 686  . 684  .682
140 220 . 708  .712 ·. 715  . 718  . 720  723  . 725  . 727
150 210  .699 . 710 . 71!J ·. 727  . 734 . 741 . 747 . 753
mo 200  . 684 . 700  .711 . 727  . 738  . 749  . 757  . 765
170 190 . 672 . 691  . 708  . 724  . 738 . 750  . 761  . 771
180 180 .666 . 688  . 706  . 722  .737  . 750 . 762 . 773
TABLE IV.
Factor for piston I Factor for piston
velocity, fv i = 4 acceleration, f, i = 4
Crank
angle 8.
Equation Equation Equation Equation
(10) (12) (17) (18)
(exact). (approx.). (exact) . (approx.).
   
0
0 0. 0 0.0 1. 2500 1.2500
10 . 2164 . 2164 1. 2202 1. 2197
20 .4226 . 4224 l. 1336 1.1312
30 . 6091 .6083 .9950 . 9910
40 . 7676 . 7659 . 8139 .8094
50 . 8914 .8891 .6026 . 5994
60 .9768 • 9742 . 3752 . 3750
70 1. 022·3 1. 0200 .1468 . 1505
80 1. 0289 1.0276  . 0682 .0613
90 1. 0000 1. 0000 =.2583 . 2500
100 . 9407 . 9420 . 4155 . 4085
110 .8571 • 8594 . 5372 . 5335
120 • 7552 • 7578  . 6248 .6250
130 . 6406 . 64i9  .6830 . 6862
140 • 5180 . 5197 . 7182 . 7226
150 .3909 . 3917  . 7370  . 7410
160 . 2614 . 2616  . 7458 . 7482
170 . 1308 .1308 . 7494  . 7499
180 , 0.0 0.0  . 7500 . 7500
,
/
19
TABLE V.Piston motionLiberty 12 enginr.
Per I
cent
II Crank Per cent p · I Piston
6 an~ie, </> travel. factor, fv. t10n
I
angle, Con~. rod piston ve/~~7fy acc~lera
% p. t. factor, r •.
,  . 0  
0
15
30
45
60
75
90
105
120
135
150
165
180
0 0
4 20
8· 23
11 54
14 38
16 22
16 58
16 22
14 38
11 .54
8 23
4 20
0 0
0.00
2.20
8.53
18. 33
30.57
44. 00
57.46
69.89
80. 57
89.04
95.13
98.80
100. 00
TABLE VI.
0.000
.332
.628
. 856
.997
I. 042
1.000
. 890
.715
.558
. 372
. 186
. 000
•
Gas pressure lb.
per sq. in. (ab
I. 292
I. 229
1. 012
. 707
. 354
.006
.292
' .511
 .646
. 707
 . 720
. 713
.708
Gas pressure lb.
per sq. in. (gage).
( 122.6 )'"' solnte) .
piston x+vc travel x+22.6
'Expan
 
x . Expan Com Comsion.
pression. sion. pression .
   
0 22. 6 9.000 615.6 117. 0 600. 9 102.3
10 32. 6 5. 594 382. 7 72. 7 368.0 58. 0
20 42. 6 3. 956 270.5 51. 4 255.8 36. 7
30 52.6 3. 003 205. 3 39.1 190. 6 24. 4
40 62.6 2. 398 164. 0 31. 2 149. 3 16. 5
50 72. 6
1. 978 1 135. 2 25. 7 120. 5 11. 0
60 82. 6 1. 671 114.3 21. 7 99.6 7. 0
70 92. 6 1. 440
98. 51 18. 7 83. 8 4. 0
80 102. 6 1. 260 I 86.2 16.4 71. 5 1. 7
90 112. 6 1.117 76. 5 1~.5 61.8  .2
100 122. 6 I. 000 68.4 13.0 53. 7 1.7
'
T ABLE VII.
Gas A 1 Inertia Resultant
Crank pressure, Gas force, fagf~r· force, I force
angle. lbs. per F •· f,. ' F along cyl.
sq . m. ,. [ axis, 1<'..
~ i~ u,g ti =J
1:.Jt6
8
0
I !J:z 45 269 5, 280 . 707 ·~ 4, 020
60 187 3,670 . 354 631 3, 040
75 138 2,710 . 006 11 2, 700
90 105 2, 060  . 292 52J 2, 580
105 85 1, 668 . 511 910 2,580
120 71 1,393 . 646 1, 150 2, 54)
135 60 1, 177  . 707 1,260 2,440
150 40 785 .720 1,283 2,070
165 23 I' 451 . 713 1, 270 1, 720
180 12 236  . 708 1, 261 1,500
195 4 79  . 713 1,270 1 350
210 2 39 . 720 l , 283 1; 320
225 2 39  . 707 1,260 1,300
240 2 39 . 646 1, 150 1,190
255 2 39  . 511 910 950
270 2 39 . 292 520 560
285 2 39 . 006  11 30
300 2 39 . 354 631  590
315 2 39 . 707  1. 260 1, 220
330 2 39 1.012 1,804 1.760
345 2 39 1. 229  2, 190  2, 150
360 2 39 1. 292  2, 300  2, 260
375 0 0 1. 229 2, 190  2,)90
390 1.7 33 1.012  l,804 1,840
405  1.7  33 .707'  1,260  1,290
420 l. 7 33 . 354 631 660
435  1. 7  33  . or6  11 40
450  1. 7  33 . 292 520 490
465  1. 7 33 . 511 910 880
480 1.7 33  .646 1,150 1,120
495 1. 7 33 . 707 1,260 1,230
510  l. 7 33 . 729 1, 283 1 250
525  l. 7  33 . 713 1,270 1; 240
54J 1.7  33 .708 1 1,261 1,230
555  1  20 . 713 1,270 1,250
570 0 0 . 720 1,283 1,280
585 1 20 . 707 1,260 1, 280
600 3 59 . 646 1,150 1,210
615 5 98  . 511 910 1,010
630 9 177 . 292 520 700
645 14 272 . 006 11 260
660 24 471 . 354 631 160
690 66 1, 295 1. 012 1, 804 510 I
675 40 785 .707 1. 260 I <180
705 139 2, 728 I. 229  2, 190 540
· .,,,
'
"' ' \ 20 ·
TABLE VIII.
Crank Per cent Resultant Tangent Pi5ton Piston Single·
angle, piston force of conn. side velocity cylinder
0. travel, along cyl. rod angle, thrust, factor, torque.
% p.t. axis, Fa. tl!.n <t,. F,. fv. ft.lb.T.
I .
I 0 o.o 3,590 . o.o 0 o.o 0
15 2.2 6,640 .0757 502 .332 643
30 8.5 5,810 . 1474 857 .628 1,064
45 18. 3 4, 020 . 2107 847 . 856 1,004
60 30.6 3,040 . 2612 794 . 997 884
75 44.0 2,700 . 2937 793 1. 042 820
90 57.5 2,580 . 3051 787 1.000 752
105 69.9 2,580 .2937 758 .890 670
120 80. 6 2,540 . 2612 664 . 735 544
135 89.0 2,440 . 2107 514 .5.';8 397
150 95.1 2,070 . 1474 305 .372 225
16.5 98.8 1,720 . 0757 130 . 186 93
180 100.0 1,500 .o 0 .0 0
195 98.8 1,350  . 0757 102 .:...186 73
210 95.1 1,320  .1474 195 .372 143
225 89.0 1,300  . 2107 274 .558 212
240 80.6 1,190 . 2612 311 . 735 255
255 69.9 950  . 2937 279 .890 247
270 57. 5 560  ·.3051 171 l. 000 163
285 44. 0 30 . 2937  9 1. 042 9
300 30.6 590  2612 154 .997 172
315 18.3 1,220 . 2107 257  . 856 305
330 8.5 1, 760 . 1474 260 .628 322
345 2.2  2,150 . 0757 1m .332 208
361) 0.0 2,260 .o 0 .0 0
375 2. 2 2, 190 . 0757  166 .332 2i2
390 8. 5 1,840 .1474  271 .628 337
405 18.3 1,290 .2107 272 . 856 322
420 30. 6 660 .2612  172 .997 192
435 14.0 40 . 2937 12 1. 042 12
450 57.5 490 .3051 150 1.000 143
465 69.9 880 . 2937 258 .890 228
480 80.6 1,120 .2612 293 . 735 240
495 89. 0 1,230 . 2107 25!1 . 558 200
510 95.1 1,250 .1474 184 .372 136
525 '98:s 1,240 .0757 94 .186 67
540 100.0 1,230 .0 0 .o 0
555 98.8 1,250  .0757 95  . 186 68
570 95.1 1,280 .1474 189 . 372 139
585 89.0 1, 2'lO .2107 270 .558 208
600 80.6 1,210  . 2612 316 . 735 259
615 69.9 1,010 .2937 297  . 890 262
630 57.5 700 .3051 214  1.000 204
645 44.0 260 . 2937 76 1. 042 79
660 30.6 160 . 2612 42 .997 47
675 18. 3  480  .2107 101 .856 120
I
690 8.5 510 . 1474 75 .628 9~
705 2.2 540  .0757 ·  41 .332 52
'.!1
TABLE IX.
I Result
Crank Force Conn. Force Stress in Stress angle alongcyl. rod angle along conn. rod Sine. due to Ia 1nn tc sotrnens.s 0. axis fa. cos ef,. conn. rod due to axis, Fd. Fc1. whip. I rod I shank.
 
0
0 3590 1.000 3590 13200 o. 0000 0 13200
15 6640 . 9971 6660 24500 .2588 440 24940
30 5810 . 9893 5880 21600 .. sooo 850 22450
4.5 4020 . 9785 4110 15100 . 7071 1210 16310
60 3040 .9675 3140 11550 . 8660 1480 13030
75 2700 . 9595 2810 10320 . 9659 1650 11970
I
90 2580 . 9565 2700 9930 1. 0000 1710 11640
105 2580 . 9595 2690 9890 . 9659 1650 11540
120 2540 . 9675 2630 9670 .8660 1480 11150
135 2440 • 9785 2490 9160 . 7071 1210 10370
150 2070 . 9893 2090 7680 . 5000 850 8530
165 1720 . 9971 1730 6360 .2588 440 6800
180 1500 1. 0000 1500 5520 .0000 0 5520
195 1350 . 9971 1350 4970 . 2588 440 5410
210 1320 . 9893 1330 4890 . 5000 850 5740
225 1300 . 9785 1330 4890  . 7071 1210 6100
240 I 1190 . 9675 1230 4530  . 8660 1480 6010
255 950 • 9595 990 3640 . 9659 1650 5290
270 560 • 9565 590 2170  1. 0000 1710 3880
285 30 • 9595 30 110  . 9659 1650 1760
300 590 .9675 610 1790 . 8660 1480 3270
315 1220 . 9785 1250 3680  . 7071 1210 4890
330 1760 . 9893 1780 5240 .5000 850 6090
345 2150 . 9971 2160 6350 . 2588 440 6790
360 2260 1. 0000 2260  6650 .0000 o 6650
375 2190 . 9971 2200  6470 . 2588 440 6910
I
390 1840 . 9893 1860 5470 . 5000 850 6320
405 1290 . 9785 1320 3880 . 7071 1210 5090
420  660 . 9675 680 2000 . 8660 1480 3480
435 40 . 9595 40  120 . 9659 1650  1770
450 490 . 9565 510 1880 1.0000 1710 3590
465 880 . 9595 920 3380 . 9559 1650 5030
480 1120 .9675 1160 4270 .8660 1480 5750
495 1230 .9785 1260 4630 . 7071 1210 5840
510 1250 . 9893 1260 4630 . 5000 850 5480
525 1240 • 9971 1240 4560 .2588 440 5000
540 1230 1. 0000 1230 4520 .0000 0 4520
555 1250 . 9971 1250 4600 . 2588 440 5040
570 1280 • 9893 1290 4740 . 5000 850 5590
585 1280 . 9785 1310 4820  . 7071 1210 60.30
600 1210 . 9675 1250 4600  . 8660 1480 6080
615 1010 . 9595 1050 3860  . 9659 1650 5510
630 700 • 9565 730 2680  1.0000 1710 4390
645 260 . 9595 270 990 . 9659 1650 2640
660  160 . 9675  170
I
500 . 8660 1480  1980
675 480 . 9785 490  1440  . 7071 1210 26.50
690 510 . 9893 520 1530  .5000 850 2380
705 540 .9971 540 1990 . 2588 440 2430
I
/
STANDARD ABBREVIATIONS.
abs. ________ _______ ._. ___ . ___ ___ . __ ._. ___ _ . . __ .. absolute in .. ______ _ ... _ . . __ . ___ .. __ . __ ._ . ____ . __ . _._. ___ . _ .inch
b. h. p ..... ___ . _ ... ... . . _. _. _ .. __ . __ . _brake horse power i. h . p. _ .... ___ ........ ___ .... _____ .indicated horsepower
b. m. e. p __________________ brake mean effective pressure i. m. e. P _. __ ___ ____ __ .indicated mean effective pressure
r;"i••••••··••••••·••••••••• r;;1§ 1 ~~L. ) ••.•.•.•. ) ..................... ~;] cm. _. __ . _____ . ______ _______ . __ ____ _________ _ centimeter mm. ___ . __ _ .. . _ . ________ ___ . __ ___ . __ . __ . ___ . millimeter
cu .. _ .. _._ .... _ . . .. ........ _ ..... _ ... _  ..   .. _ .. cubic
c. to c . . . . .. . . ..... _ .. . ___ . _. ____ . ______ .center to center
deg ... ____ . ..... . . . .. .. . . ... . .... . . . ............. degree
eff ..... ___ . . .. _ . .. . ............... _ ........... efficiency
F ......... . ....... _ . . .... _ .... . ........ . . . . . Fahrenheit
fig ..... ____ ._ . ... _ ... .. .. .. .......... . .. .. .. _ ... _ .figure
ft. ... . . .. ... ......... _ . ... . ... _ ... _ . ... __ ... _ ... . .. . foot
' ~ g ...  . .. .. .. ...... . . . . . . .. . .. · . . . . . ..... .. . . ..... gram
gal. ..... ___ .. __ . . . . _ .. . . . _ . ........ .. . . _ ...... . .. gallon
hr . . ... ____ .. .... . . . ... ................  . . ... _ ...  .hour
h. P .............. .. .. ....  ..  . _ .... ... horsepower
m. e. p. _ .... .. .... .. ... .. ... __ . _ mean effective pressure
no ...... . .. . .. __ ... _._ ·. .. . .. _ . . .... .. ......... . number
oz. __ ................ . ..... . ... .. .. ....... . .. .... . ounce
pt ... .. .. . .. . ... . .. __ . . . _. _ .. __ . . ..... . ..... . __ . . . _ .pint
qt .. .......... . ............ ... .. ... .. ... ... .. . . ... quart
r. p. m .... . .... .. .. , . . . ... . ... . .. revolutions per minute
sec . .. . . ............ .. ... .. .. .... . . .. . ..... ... ... second
sq. __ . . _._ . . . . . . . . . . . .. . . .. . . .... .. ..... . ........ square
sp. gr ..... .... ... . _ . .. .. .. . .. . ..... . .... specific gravity
tens. str. __ ... . . . . . . __ . __ . .. . .. __ . ___ . ___ t en sile strength
vol. _ . . ____ .... _ . ____ .. _ .... __ . . ... _ .. .. ........ volume
NoTE.Abbreviations apply to both singular and plural.
(22)
/
I

_;
TABLE OF SYMBOLS.
A area, sq. in.
a linear acceleration, ft. per sec. 2
Eb
breadth.
('.,constant.
c coefficient.
Ddiameter.
cldifferential.
F..Modulus of elasticity (Young's ).
e efficiency.
e. air standard efficiency.
eh hydraulic efficiency .
emmechanical efliciency .
e, thermal efficiency.
evvolumetric efficiency.
F force, lb.
F.resultant force along cylinder axis.
Fbresultant of F, and F,,.
F0 centrifugal force.
Fct force along connecting rod a xis.
Fgforce due to gas pressure.
F, inertia force.
F ., force parallel to crank throw.
F ; resultant force.
F, piston side thrust.
F, tangential force.
Fw force due to whip.
f factor.
f,, acceleration factor.
fctdiagram factor.
£_. velocity factor.
G torsional modulus of elasticity.
g acceleration due to gravity.
H head.
h height.
I  moment of inertia, rectangular.
i  '
Ji
KIP
polar moment of inertia.
Ixmoment of inertia about x x axis.
I ymoment of inertia about y y axis.
k!.,
length.
1
M ___:moment.
Mb moment due to bending.
M0equivalent bending moment.
M,moment due to torque.
Mwbending moment due to whip.
m mass.
• Nrevolutions per minute.
n number.
0  reference center.
Pp
unit pressure, lb. per sq. in.
p, pressure at beginning of compression.
Pb pressure at end of compression.
p0 pressure at bfginning of expansion,
Pctpressure at end of expansion,
Qquantity.
q
Rradius (crank radius) .
r compression ratio.
S stress, lb. per sq. in.
Sbstress due to bending.
S0stress due to compression.
S0 equivalent stress.
Sqstress due to torque.
S, combined stress.
S,stress due to shear.
S, stress due to tension.
s piston travel.
T torque.
Tm maximum torque.
t time.
t 0  temperature.
Ut
0 , temperature, absolute.
t 0
0 temperature, centigrade.
t 0 itemperature, Fahrenheit.
uV
volume.
V 0 compression volume.
Yd volume displaced by piston per stro:,e.
V, total piston displacement of engine.
vvelocity.
v 0 crank pin velocity.
VP piston velocity.
v, rubbing velocity.
W weight.
XxYW
0 centrifugal weight.
W, inertial weight.
ydistance of extreme fiber from neu'tral axis.
Z section modulus, rectangular.
Zμ section modulus, polar.
(alpha) a  constant angle.
(beta) ,:'I constant angle.
(gamma) y coefficient of compression or expansion.
(Delta) <1.
(delta) Ii deflection.
(epsilon) <
( eta) 17 coefficient of vi, cosity.
(theta) 0 variable angle (crank angle) .
(Lambda)>.
(mu) μ coefficient of friction.
(pi) 1r constant=3.1416.
(rho) p variable radius.
(Sigma) ~summation.
(Phi) </,variable angle.
(Psi) ,t, variahle angle.
(omega) wangular velocity.
(23)
24
FIG. l.
FIG. 3.
. l •'JG. 2.
·>+++++++_..........
..,ceCclYT P/6TO/Y T/2RVe~l++++++HI
V6. Ci:'19/YK RIYGLE
25
FIG. 4.
FIG. 5.
CRANK A/'/4Ll: r//CTO!e
FOR PlfjTOl'I Vei.OC !TY
CR""1/'IK A/'14U: FAC m.e.. ·
F0/2. P15roH /ICCELER!9T!d/Y
26
FIG. 6.
FIG. 7.
_ RC!'.'5 ,4Crt!Ytf "91.()/Ytf tH++ttt1,...,...,.,...,rtrr11
r'l/lYOcl? AX~
27

f'l!!JTOI"/ 5/CJE THR(/.5,
 V.5. C,{¥1/YK /IIYl7L.E
'
FIG. 8.
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revolutions per minute.
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