.
e
i T.
T T
24 SEP 1929
A
Br
AP9 2
A
STUDY OF THE CARWEN OLSEN TYPE BALANCING MACHINE
By
Charles M. Perkins
B.A., Cambridge University
1927
Submitted in Partial Fulfillment of the Requirement
for the Degree of
MASTER OF SCIENCE
from the
Massachusetts Institute of Technology
1929
Signature of Author__
Certification by the Department of
Professor in Charge of Research
Chairman of Departmental
Committee on Graduate Students
Head of Department
v
&(4
TABLE OF CONTENTS
Page
I.
Preliminary Study of Balancing
1
II. Description and Constants
17
III.The Dynamics of the Vibrating Bed
34
IV. Operation and Calibration
43
V.
Balancing Tests
70
Bibliography
86
ACKNOWLEDGMENT
I am indebted to ir. I. M. Dow, Course VIA,
for the photographs on pages 84 and 85.
A STUDY OF THE CARVIEN OLSEN TYPE BALANCING MACHINE
Chapter I.
Preliminar
d
of Balanci
The Tinius Olsen machine for dynamic and static
balancing of rotating pieces,
combines the advantages of
rapid operation and very accurate results.
The balancing
mechanism is compact and sufficiently simple in design to
make it unlikely that it will get out of order.
Even the
wear of the parts is unlikely to produce any appreciable
error, unless it were uneven and thus liable to produce
impacts.
To understand the operation and design of the machine
it is first necessary to make some study of the meaning
of unbalance in rotating pieces.
two kinds,
This unbalance is of
static and dynamic, and is due to the piece
not being geometrically symmetrical about its
axis of ro-
tation or to varying density of the material of which the
piece is composed.
The different nature of the two types
of unbalance is best understood by the consideration of a
perfect simple shaft to the periphery of which small
masses are attached.
trifugal force.
A single mass would produce a cen-
The resultant of all such forces is
called the static unbalance, which derives its name from
the fact that theoretically it
could be found on a suf-
ficiently sensitive apparatus without rotating the piece.
-2-
Two equal masses attached to the shaft 1800
apart would produce a centriftgal couple but no resultant
The resultant couple is termed the dynamic un-
force.
balance.
The Olsen machine does not find the total
dynamic unbalance, but merely its component in a plane
at right angles to that of the static unbalance.
The
component in the plane of the static unbalance is combined with the actual static unbalance in finding the
location of the latter.
The location found is not tru-
ly that of the static unbalance, but the point at which
metal must be added or removed in order to compensate
both the static unbalance and the component of the dynamic
unbalance in the same plane.
This will be more fully ex-
plained later.
In an actual test piece the unbalance is not
caused by concentrated masses, but is the resultant of
unbalanced metal distributed over the piece in irregular
fashion.
It will now be shown that the total unbalance,
which is
a combination of all the small amounts of unb&l-
anced metal, can be represented by a force and a couple
in axial planes at right angles to one another.
The design
of the Tinius Olsen machine is based upon this fact.
Consider a thin slice of the test piece cut off
by two planes at right angles to the axis of rotation,
an infinitesimal distance apart.
The slice will have an
infinitesimal static unbalance and a dynamic unbalance
which is a product of an
and
'ixiffiitesimal force and an
/
.. 3m
I@ 2
a
/
B
R
Plt
/
/
'1 ~ 2
I
Pa
C
P
b
Fig. 1.
I
-4The dynamic unbalance may there-
infinitesimal length.
fore be neglected since it is an infinitesimal of higher
order than the static.
Figure 1, p. 3,
shows the static
unbalance of two such thin slices at an axial distance 1,
represented by P;;. and P 2 acting at angles e
1 and E2
respectively with the vertical for a given angular position of the shaft.
A B represents the axis of rotation.
The system of forces is not altered by introducing in
the plane at c opposite forces P? and Pl" parallel and
equal to P, and similar forces P2' and P2" parallel and
equal to P
where c is any arbitrary point on AB.
The
force system is thus equivalent to R the resultant of
Pl" and P2" and two couples Pla and P2 b where a = AC and
b = CB.
R
Pi a Cos (9R-@1)
P2 b Cos
(92-GR)
D
A
C
B
Fig. 2.
P, a Sin (QR-91)
A
C
Fig. 3.
P 2 b Sin (9 2-9R)
B
.5-
Let the angle between R and the vertical at c be 9R'
Resolving the force system in axial planes containing R and at right angles to it we have the following; in the plane containing R, two couples of magnitude
Pia
Cos(GR
-
91) and P2 b Cos(9 2
-
9R), together with R.
The senses of the couples are shown in Figure 2.
In the
plane at right angles vie have the two component couples
Pi a Sin (9R
91) and P2b Sin (9 2 0
R) as in Figure 3, p.4.
These two resolved systems of forces at first
sight appear
Actually
to depend on the arbitrary quantities a and b.
they can be shown to be independent of a and b and interpreted in terms of 1 the fixed distance between the two unbalanced forces being considered.
Referring again to Figure 2, and taking moments
about an axis perpendicular to the plane and passing
through D any arbitrary point at a distance y to the left
R(a + y) + Pab Cos (9a-9R) - P1 a Cos (OR
.
of A, the total moment is equal to
R(a + y) + Pol Cos (G2 - GR)- a P1Cos(@R
+ PSCOs
(02 -
GRJ
-
and substituting b = 1-a
-6-
R is the resultant of Pi and P2 considered as acting in
one plane, hence by reference to Figure 4
R
Pi
(GR-Q1)
(Q2-@R)
Fig. 4.
P1 Cos (0R - 91) + P2 Cos (02 - GR)
R.
Substituting in the last term of the expression for the
total moment, the result is
R (a + y) + P2 1 Cos (Ge - OR) - R a
= R 17
+ IP2 1 COS
(92-
)
)
= R y + P2 1 Cos (Q2 9
R
This moment would be produced by a single force R
acting at a distance
A.
P2 1 Cos (Q2 - QR) to the right of
R
Again referring to Fig. 4, R Cos (Q2 - QR
2+P,
Cos
(Qz -
OX)
and therefore
P2 1 Cos (92 - QR)
R
P 2 1 fP
2
+ P0COs (2
R
-
9 1 ))
-7-
P 2 1 )Ple + P1 Cos (92
2
P1>
-91)
2
+ P2 +
2 Pi Ps Cos (@z - 91)
which is a constant independent of a and b, so that the
force system in this plane is entirely represented by R at
a
fixed distance from
A.
Considering the force system in the plane at right
angles, from Fig. 3. the total couple is
P, a
Sin (QR - @I)
+
P2 b Sin (9a
-
QR)
= P1 a Sin (OR - 91) + P2 (1-a) Sin (92 = P 2 1 Sin (92 -
GO + a (Pi Sin (R-Q 3.)
Pa Sin (Q2
From Fig. 4
R
-
QR
P, Sin (OR - G) - Pa Sin (Ga - OR) = 0 since
the components of P. and P2 at right angles to their resultant R must be equal and opposite.
Thus the total couple
above reduces to Pz 1 Sin (92 - @R) a constant independent
of a and b.
The system of unbalance due to Pi and P2 has thns
been proved to be equivalent to a force R equal in direction
and magnitude to the resultant of P3. and P2 treated as
coplanar forces, and acting at a distance
P2 1 {Pa + Pi Cos (@2- 91)
2
2
to the right
PI + P2 + 2 PIP 2 Cos (92-91)
of A, together with a couple acting in an axial plane at
right angles to that of R, and of magnitude
P2
. Sin (@z - @R)*
-8-.
The force and couple thus obtained can be combined in
a similar manner with the unbalance P. due to any other
thin slice of the test piece and reduced to a force and a
couple.
Successive operations of this type make it evi-
dent that the total unbalance of any piece is equivalent
to a force acting at a specified point and a couple in an
axial plane at right angles to it.
The expression Pa 1 Sin (Q2
From Fig. 4,
be symmetrical in Pi and Pa.
Hence
=
Pi Sin (9 2 - 91)
=R)P,
Pa 1 Sin (92 -
Pa 1 Sin (92
R
-
91
)
R Sin (Q2 - OR)
QR) should evidently
-
The expression for the distance of the line of action of
R from B may be found by interchanging P1 and Pa in the
expression for its
distance from A.
Summing the two ex-
pressions we have
1+
P
+
Cos (a-9i)]
2P 1 P2 Cos(Q2 -@
+ P, I
{PI+P2 Cos (92-91)
Pi + Pa, + 2 PIPe Cos(Ga-91)
1)
_1 {P1 + P 2 + 2 PiP2 Cos
Ca
(9a -
Q 1
Pi + Pa + 2 PiPe Cos (92 - 90)
)
P+P
P, 1
= 1, a necessary
condition which provides a useful check.
The couple in any unbalanced piece which was shovm
to lie in a plane at right angles to that of the resultant
static unbalance is not equal to the total dynamic unbalance
-9-
The
as previously defined, but is only a component of it.
component in the same plane as the
static unbalance merely
shifts the line of action of the latter.
The static un-
balance when shifted in this way will hereafter be called
the 'equivalent
static unbalance'.
In the Carwen Olsen type of balancing machine,
the
principal constituents are a fixed bed plate and a vibrating bed which carries the piece under test and a mechanism
for introducing a couple tending to damp the vibrations of
the bed caused by the unbalance of the test piece when rotating.
There are two sets of pivots, one pair parallel to
the axis of the piece, and one pair at right angles.
The
bed may be supported on either pair at will.
When the longitudinal supports are used, the only effect on the bed is the vibration due to the static unbalance,
since the dynamic unbalance is a pure couple which will produce no moment about these supports.
If the headstock bal-
ancing mechanism is made to introduce a compensating couple
in the correct plane, the vibration is entirely eliminated,
and supposing the amount of the couple in the headstock to
be indicated on the machine,
the static unbalance is given
by dividing the compensating couple by the height of the
centre line of the test piece above the pivot axis.
This
is self evident from the diagram, which is an end view of
the machine.
) S
0M
h
/I
Fig. 5.
S is the static unbalance shown in the position of
its meximum effect, M is the compensating couple and P
the pivot point.
When the transverse pivots are used both the static
unbalance and the dynamic will have a moment effect about
the pivots, unless the static unbalance happens to be
located directly above the pivot axis.
ily seen from the diagram.
This can be read-
The horizontal components
of the Force S and couple D x 1 would have no moment
about the pivot axis.
D
S
I-
D
Fig.
6.
-11.-
However, the vertical components would give rise to a
moment D Sin"ol1 x 1
-
S SinC<2 x a, with the shaft in
Thie most convenient method
the angular position shown.
of balancing the shaft is to combine the static unbalance
with the component of the dynamic unbalance in the same
axial plane.
is then possible to balance these two
It
items by a single weight equal in magnitude to the static
unbalance, and in a certain longitudinal position.
This
leaves only the component of the dynamic unbalance in a
plane at right angles to that of the static unbalance.
To prove the statement that the component of the dynamic couple in the static plane together with the static
unbalance is equivalent to a single force, it
is only neces-
sary to take moments about any axis perpendicular to the
plane.
Then if
x is the distance from the static unbalance
line of action and M, represents the dynamic component, we
have a total moment equal to
M, + S.x where S is the mag-
nitude of the static unbalance.
In the diagram A is the
axis of moments.
Mi
A
x
7
Fig. 7.
-12-
Now consider a single force 8' equal and parallel to
S, at a distance y = x + -g from A.
In the two systems
the forces are equal in magnitude and direction, and the
moment in the second case = s
(x +
S
)
=
S.x + Mi so that
the two systems are equivalent.
If A be taken as the pivot point for dynamic balanc.ing, the total maximum moment about the pivot axis will be
the resultant of S'y and.1A2 , the component of the dynamic
unbalance at right angles to the plane of 8'.
If a com-
pensating couple be introduced in the plane of S', the
vibration of the bed will be reduced since the resultant
couple is equal to
(sty
C)2 + M 2 8
where c is the compensating couple.
The expression for
the resultant couple is a minimum when S t y = c.
Therefore
when the vibrations of the bed are reduced to a minimum,
we can find y by reading the value of c since St = S
which is known.
Supposing S and M. to have been corrected, after evaluation by this method, it
is a simple matter to find the
value of MA which is all that is required to complete the
balancing of the piece.
To balance a piece completely only two weights are required in theory.
Referring to Fig.8, where the previous
notation is used for the equivalent static unbalance and the
component dynamic unbalance at right angles to it, consider
three compensating weights added.
These will provide complete balance provided m3 = St = S
and m 2 ,d = M 2
.
M2 = m.
S
142'll.
XII'
d
Fig.8.
Wle can replace mi and m 2 by a single force R, the vector sum of the two, its
correct angular position being given
R
-
by the force diagram Fig. 9.
Fig.9.
In actual practice three or more compensating weights
are generally added or compensating holes bored.
Boring
-14-
is more commonly used than adding metal by welding, and
using a greater number of holes distributes their weakening
effect.
In some cases a piece is corrected by milling or
grinding unbalanced metal instead of drilling.
It will be found in practice that M3 is sometimes
great enough to bring the line of action of St beyond the
end of the shaft.
Complete balance can still
obtained by only two added weights or borings.
however be
Consider
the state of unbalance denoted by the diagram Fig.10, where
couples are represented vectorially, with the usual conventions.
Stt S1 St
Sdd
d
S d
M2
2
Fig. 10.
For St we may substitute S" = S' acting at a point
distant d from St and within the limits of the piece, together with a couple in the plane of S' of magnitude Slxd.
At the transverse plane containing S" we may add three
weights hit, M 2 , m. and at a distance 1 along the shaft
-15-
two further weights m 2 and m..
These will cbmpensate the
unbalance entirely if mi = S" = S, m21 = S'.d = S. d, and
m3 1
=
M2 .
Solving for mi, m2, m. from these three condi-
tions, we cen find the resultant of mi., M 2 , m. in the transverse plane containing S1, and the resultant of mt and m.
in the transverse plane distant 1 from S", thus reducing the
total number of balance weights to two.
The magnitudep
and angular locations of the two weights are given by the
's
R2
)m
3
in3
Ri
(mi+ma)
Fig. 11.
magnitude and direction of Ri and: R 2 in Fig. 11.
The principle of balancing is used commercially in
the manufacture of rotating parts for automobiles, high
speed engines and high speed machinery.
Although the mag-
nitude of the unbalance is frequently not in itself
dan-
gerous, when the pirt is rotating at a speed corresponding
to the natural period of a part or the whole of the engine
or machine, the effects are liable to be very serious.
-16-
Unbalance is liable to cause excessive noise in running at
any speed, while there is a possibility at critical speeds
of serious injury to babbited bearings, and of shaking the
machine loose on its
foundations.
In the case of gener-
ators or motors where pure torque is transmitted,
the
bearings need not be designed to withstand heavy pressures,
so that unbalance of the armature is dangerous.
In a recip-
rocating engine unbalance of the shaft would widen the
range of critical speeds since the first harmonic component of the inertia of the reciprocating parts can always
be eliminated by a system of counterweights; the unbalance
of the shaft would thus cause an additional critical speed,
which is very important in a low speed engine since unbalances of a higher harmonic order may give rise to critical
speeds of such low value as to be unimportant.
The actual operation of the machine and interpretation
of readings obtained in tests will be discussed later together with the details of its
design.
-17Chapter II
Description and Constants*
The Balancing Machine consists of two principal parts,
the base columns and the vibrating bed.
The latter is
supported on the former by means of knife edges and springs.
The knife edges are fixed in the vibrating bed and the bed
is transferred from one pair to the other by changing the
level of the bearing surfaces on the columns.
The base
columns are connected by two side rails of rectangular
cross-section 2-1/2"deep x 1-1/2"1 wide.
These are bolted
to the columns by three bolts at each end,
They carry the
dynamic pivot bearings each of which consists of a hard
steel cylinder sliding in a vertical groove in the rail
and operated by a cam on a transverse shaft passing through
both side rails.
wheel.
The shaft is turned by means of a hand
When the dynamic bearings are raised, they lift
the
bed off the static pivots which are arranged lengthwise.
The columns themselves serve to store the wrenches, bearings,
coupling, etc., of the machine.
Two springs between the
side rails and the vibrating bed supply the restoring couple
in static balancing, while one at each end of the machine
comes into operation in dynamic balancing.
The initial
compression of the springs is controlled by screws.
This
compression should not be so great that the bed does not
rest firmly on either pair of pivots.
*Refer to photographs on page 85.
Three strips of
-18-
1/16"1 metal tie the bed in place so that it cannot move
longitudinally or transversely, or rotate in a horizontal
plane.
The vibrating bed carries the motor, the compensating
mechanism and the bearings for the test piece.
The bear-
ings are three in number and can be moved relative to one
another or may be locked in a fixed relative position and
moved as a single unit.
The bearing standards have a hole
at the side through which a rod passes, and each can be
clamped to the rod by means of a set screw in order to preserve a fixed relative position of the bearings.
Each
standard can be clamped independently to the bed, and is
moved by means of a pinion meshing with a rack on the bed
and operated through a hand wheel.
Half bearings are used,
held in place by dmvels and locked in position by the upper
arch of the standards, which is hinged and is clamped by
means of a screw.
Through the top of the arch a phosphor
bronze pin of 1" diameter passes.
When running a test, the
pin is pressed down on the top of the journal of the test
piece and kept in position by a set screw.
Along the axis
of each pin there is an oil hole.
The compensating mechanism is housed in the headstock
of the machine, with the exception of the static location
compensator which is carried on the drive shaft just outside
the headstock since it must be readily accessible during
-19-
the operation of the machine.
The drive is obtained from
a D.C. shunt wound motor, delivering 1/2 H.P. and absorbing
2-1/4 amperes at 230 volts.
The motor is carried on a bracket bolted to the side
of the bed just below the headstock and drives through an
endless belt.
Provision is made for moving the motor on
the bracket in order to take up any looseness of the belt.
The torque is transmitted from the pulley shaft to the main
spindle by a cone clutch.
At the other end of the spindle
a coupling is attached by clamping with two screws.
The
coupling is split at the further end, and clamped over a
split bushing which fits
on the end of the test piece.
Two
small metal pieces which screw on to the outside of the
bushing prevent it
from sliding out of the coupling.
The
coupling is first assembled on the test piece and then
pushed over the nose of the spindle.
While the latter op-
eration is perforned, care should be taken to rotate the
test piece and advance it
slowly to avoid throwing the cone
clutch out.
The balancing mechanism is readily accessible for examination or repairs by removing the head stock cover.
This
cover constitutes the top and back of the headstock, and to
remove it requires the drawing of four screws and the removal
of the aluminum dial at the top end of the balancing spindle,
when it may be lifted off by the two handles provided at
-20-
Care should be taken in removing the cover since
the top.
it
is heavy and if dropped would be liable to damage the
balancing mechanism.
Holes are drilled in the cover so
that the machine may be lubricated without its removal.
The belt has a heavy metal housing, which is screwed to the
headstock and is provided with a thin sheet cover over the
centre of the pulley so that the clutch may be easily driven
home if
it
should by chance be forced out.
On the balancing spindle is keyed an aluminum dial as
mentioned above.
This is used in locating the angular po-
sition of unbalance either static or dynamic.
Two hand
wheels on the outside of the headstock are geared to two
dials which give readings for angular location and amount
of unbalance.
~The wheel on the left rotates the balancing
spindle independently of the motion of the test piece and
thus alters the phase relation between the compensating
couple and the unbalance.
The wheel on the right moves
one of the weights which give rise to the compensating
couple thus altering the amount of the couple.
A thin rod
rests in a slot on the base column, passes through guides
on the headstock, and is in contact at its upper end with a
dial indicator, screwed to the headstock, which registers
the vibrations of the bed.
The left hand wheel is connected to the dial reading
the angular position of unbalance by a simple train of spur
gears.
The shaft carrying the hand wheel carries at its
-21-
further end a helical gear which mates with a helical gear
on a shaft at right angles to the first,
the drive shaft of the machine.
that is, parallel to
This second shaft carries
a screw thread and as it is turned advances a sliding piece
parallel to the drive shaft.
The sliding piece is forked
about a collar on a long helical gear carried on the drive
shaft, so that the gear is constrained to advance as the
hand wheel is turned.
The gear meshes with a second helical
gear on the vertical compensating spindle so that the latter
rotates as the first gear is advanced.
The long gear is
keyed to the drive shaft and there is unity velocity ratio
between it
and the compensating spindle.
The right hand wheel is also connected to its
through a spur gear train.
dial
On the shaft carrying the wheel
is mounted a helical gear meshing with an equal gear on a
vertical shaft parallel to the balancing spindle.
This
shaft has a screw cut on it and a sliding piece forked about
a collar on the lower balancing weight;
so that turning
the hand wheel moves this weight up and down the balancing
spindle.
This mechanism together with that operated by the
left hand wheel comprises the whole balancing system which
is completely housed in the headstock.
The dimensions of
the various parts will be fully discussed later.
The important constants of the machine are its speed
range, spring constants and certain distances.
The lower
-22-
limit of the speed range is the significant one since it
termines the capacity of the machine.
de-
If the test piece is
too heavy the critical speed of the machine will lie outside
the lower speed limit and it
the piece, since it
critical speed.
will be impossible to balance
is essential to run the machine at the
The highest speed required is the critical
speed when there is no test piece mounted on the bed, and
this lies well below the upper speed limit of the machine.
The upper speed limit is 350 r.p.m. and the lower 250 r.p.m.,
the speed being regulated by a field rheostat.
To find the capacity of the machine, it
is first
neces-
sary to make a study of the dynamics of the vibrating bed.
It will also involve a knowledge of the spring constants.
These constants were obtained by experiment in the following manner.
Fig. 12.
The heads of two long bolts were inserted into inverted V channels so that the shanks were vertical.
A cross
piece through which the bolts passed was in contact with
-23-
the upper end of the spring while the lower rested on the
A dial indicator was mounted in a fix-
table of a balance.
The readingsof
ture placed on the table of the balance.
this indicator gave the movement of the cross piece relative
to the table of the balance, that is,
the spring was compressed.
the amount by which
The load was varied by screwing
down the nuts on the two bolts and its amount read on the
The first reading was taken at a
scale of the balance.
scale reading of 35 pounds.
Results:
Scale Reading
Indicator Reading
35
100
175
120
85
0
26
53
32.5
19.4
From the graph p. 23a.050" compression = 132. lbs.
gives the constant for the two side springs.
This
The constant
found in a similar manner for the spring at the headstock
end is .050" compression = 35 lbs.
The weight of the bed was found by lifting it
from all its
free
pivot points by increasing the compression of
-the two side springs and headstock end spring by screwing
down their adjusting screws.
end was removed.
The spring at the tail
stock
The uncompressed length of each spring is
2.250", which under this load was changed to 1.995" for the
rear side spring, 2.058" for the front side spring, and
2.150" for the headstock end spring.
Using the experimental
values of the spring constants we have the weight of the
-i-
o
'1****'
*
I
-~-7-r
I
~Alftli Rwaa
tW&
-
-
s---7-
[4
A I
I
---- }.~-~I
------
a
EUEN___ZGE
i---T
tt...-.
CO..
CHCO
EYOR
NO,
34
I
to
-24-
bed given by;
=
- 1. 9 9 5 x 132 + 2.250 - 2.058 x 132 + 2.250 - 2.150
.050
.050
.050
2 .2 5 0
.447
.100
.44x
132 + 00
x 35
0050
'050
"
x 35 =
= 1180 + 70
=
1250 lbs.
The distance between the two side springs is 22" and
the end spring lies on the perpendicular bisector of the line
joining them.
Taking moments about the line passing through
the effective point of application of the end spring, we
have
Pa3I
PN
T
IV
~
->.
<
,
11"
Fig.
P2
11"
14.
3F
P 2 x 11 = P 1 x 11 +
= 11 (P2 - Pi) = 11 x 132 12.250-1.995-(2.250-2.058)]
1250 x .050
W
11 x 132 x (2.058 - 1.995)
1250 x .050
11 x 132 x .063
1250 x .050
= 1.46"
This gives one of the coordinates of the centre of
gravity of the vibrating bed.
The coordinate -parallel to the line joining the static
pivot points is not fixed since the bearing standards, which
are of considerable weight, can be moved longitudinally on
the bed.
The principle employed in finding the vertical co-
ordinate is as follows:
G
B
h
S
P
S1
Fig. 15.
Referring to the diagram, fig.14 if the bed B is made
to rotate about the pivot P, G the centre of gravity will
move horizontally by an amount hC * wherec< is the rotation
of B.
This will cause a change of W hor in the moment of
II about the axis of rotation, and therefore a change in the
compression of the spring S since the distance 1 remains
sensibly the same for small values of oC.
The amplitude of
oCis limited to a small amount by the dynamic pivot points
* Proof on p. 31.
-26-
coming into contact with their bearing surfaces.
by taking moments about the axis of rotation,
=
W hoC0 + Wi
where i
Pl,
We have
that
is the hori-
zontal coordinate of G for any fixed arbitrary position of
the bed.
The value of P is found by measuring the length
is measured by a dial indicator at a
of the spring, andeC
fixed distance from the axis of rotation.
Taking two cor-
responding readings of oc and P and subtracting the resulting equations,
+
W F
=
P1
W hoc+
W F
=
P01
W hd
W h (,(-c)
0
=
(P - PO) 1
There is only one unknown, namely h, in the above equation, so that by this means we are able to find its value.
In the actual experiment,
the readings of oC and P were
taken by means of dial indicators.
One of these was mount-
ed on a stand resting on the rear side rail
with its
of the machine,
spindle in contact with the top side of the block
on the vibrating bed, which carries the spring adjusting
screw.
Readings of this indicator can be converted to read-
ings of the rotation of the bed by neglecting the elastic
deformation of the bed and side rail,
due to change of the
system of loads caused by the rotation of the bed. [A
o
second indicator was mounted on a stand on top of the block
1
mentioned above, with its
spindle in contact with the top
-27-
[side of the block on the vibrating bed, which carries the
spring adjusting screw.
Readings of this indicator can be
converted to readings of the rotation of the bed by neglectT
ing the elastic deformation of the bed and side rail,
due to
change of the system of loads caused by the rotation of the
bed.]
A second indicator was mounted on a stand on top of
the block mentioned above, with its spindle in contact with
the top of the adjusting screw.
Its readings gave the move-
ment of the screw relative to the block, which is equal to
the movement of the top of the spring relative to the block
since spring and screw are always in contact.
The differ-
ence between the sets of readings obtained on the two indicators gave a record of the change in length of the spring,
as may be readily seen by reference to the diagram, Fig.16.
Reading
=A
Reading
B
RE ading
Reading
=A +.,0025"
=B +.0024"
025!
.0001"
(a)
(b)
Fig. 16.
Note that the first
indicator was used to read the rise of
the block, while the second read the fall of the screw
-28-
relative to the block, as positive values when taking differences.
Thus to find the change in length between two sets
of readings, find the change in each reading and subtract
the two results.
In the figure the change of reading is
for the first indicator, A + .0025"? - A = .0025"; for
the second, B + .0024"
- B = *0024", and the change of
length = .0025" - .0024" = .0001".
The top of the adjusting screw has a spherical chamfer,
and unless the surface is a perfectly smooth spherical segment, errors will be introduced into the readings of the
second indicator.
Actually the surface is probably somewhat
untrue since the results obtained in this experiment did
not plot into a perfect straight line.
Other methods attempted to find the height of the centre
of gravity were direct measurement of the length of the
spring by calipers, and measurement of the deflection of the
end of a lever arm inserted between the bottom of the screw
and the top of the spring.
In the former case sufficiently
accurate measurements could not be made, while in the latter
a correction had to be applied for the rotation of the lever
arm, and it was uncertain whether the lever kept perfect
plane contact with the bottom of the screw, or rather the
experiment showed fairly definitely that it did not.
In
this case the rotation was read on the indicator at the side
of the headstock.
The distance from the centre of the screw
-29-
to the point on the lever arm where deflections were read
was 2-1/2".
The indicator on the headstock is 10" from the
pivot point so that the correction for the rotation of the
lever arm was 2-1/2 x the headstock indicator reading.
10
results are tabulated below.
Ii
12
2nd Indi
cator Reading
-
Headstock
Reading
2-1/2
C= 10 xIi
Correction
The
L
Change'of
Length of
Spring
=
1 2 -C
0
0
0
19*5
5.6
4.9
41.7
8.7
10.1
-
65.8
17
16.4
+ 0.6
23.4
2262
+ 1.2
26.1
28.1
-
89
28.4
22.2
+ 6.2
67
22*5
16.8
+ 5.7
44*5
13
11.1
+ 1.9
20.8
4.7
5.2
0
0
0
89
112.5
0
+ 0.7
-
1.4
2.0
0.5
0
Since there should be a straight line connection between Ii and the change of length, these results must be
discarded.
In all three experiments the rotation of the bed was
effected by means of the spring adjusting screw.
As the
-30-
screw is screwed down the bed rotates towards the front of
the machine and the spring lengthens slightly since it
will
have to supply a smaller moment.
The results of the first
experiment are tabulated be-
low:
L = I
Ii
I2
0
0
0
24.6
24.0
0.6
50.2
47.7
2.5
75.9
72.8
3.1
102.4
98.4
4.0
75.6
72.5
3.1
49.6
47.8
1.4
24.0
23.6
0.4
0
0
0
23.4
22.8
0.6
50.4
47.9
2.5
76.5
73.2
3.3
102.5
98.4
4.1
Ia
Plotting these results and drawing a mean s traight
line, from the graph on p. 30 a.
The distance of the first
tion was 11".
QC
105
.4 = 4
Ii
100
indicator from the axis of rota-
Referring back to the theory on p. 26.
0
=
lbs. per inch,
and the spring constant being
P - P
=
32 x
L
r-
--
tm
Oa
z
0
L
__
_
_
_77_
_
_
__
_
_
z
0
0
z
o04
7-
0
1
_
1
_-
00
602
-000410
Fig*
17.
-31-
Since
Wh
po) 1
(-
and W = 1250 lbs.
1 = ll1"
Ii
132
1250 h x -=--x L x 11
11
.050
132
-.050
121
1250
132 x 121
.050 x 1250
L
Y,
4
100
= 10.2"
The statement that corresponding to a small rotation
c< of the bed its centre of gravity moves horizontally hoc,
requires proof.
(90*-Q)
G?
/K
G
h
P
Fig. 18.
In the diagram G G' represents the movement of the
centre of gravity and GK is the horizbntal component of
G G'.
-32-
G G' = lod
and G K = G G'
Sin 9
= 1 Sin 9.Oc
= h OC
Two further important constants are the height of
the centre line of the test piece above the static pivots,
and the horizontal distance of the centre of the static
The former of these
compensator from the dynamic pivots.
distances was found by direct measurement to be 20" while
the latter was 15.06".
A suxmary of the constants found is given below:
Side-spring constant
2,640 lbs. per inch
Headstock spring constant
Weight of vibrating bed
700 lbs. per inch
IF250 lbs.
Coordinates of centre of gravity of
vibrating bed
(a) Parallel to dynamic pivots
1.46"
(b) Vertical
10.2"
(c) Longitudinal
Variable
where the origin is taken at the intersection of the static
and dynamic pivot axes.
Height of axis of rotation of test piece
above static pivot axis
20"
Distance from dynamic pivot axis to
centre of static compensator
Speed Range
15.06"
250 - 350 r.p.m.
-33-
Moment of Inertia of the Bed
about Static pivot axis
a
232,000 lbs.in.
Heaviest permissible test piece
300 lbs.
(For the derivation of the last two values above see Chapter
)
III.
Sizes of half bearings
",1-1/8", 1-1/4",s 1-3/8", 1-1/2", 1-5/8",9 1-3/4",.
1-7/8", 2", 2-1/4",
2-1/2".
Sizes of split bushings
7/8",.9 1",2 1-1/8
1-1/4", 1-3/8"
1-3/4",.
1-7/8", 2".*
1-1/2", 1-5/8",
-34Chapter III
The Dynamics of the Vibrating Bed
The capacity of the machine as regards the weight
of the test piece may be calculated, when the moment of
inertia of the bed is known.
is necessary to study the dynamics of the vibrating
tia it
bed.
'To find the moment of iner-
It will be assumed that the disturbing couple has a
simple sinusoidal variation and can be represented by A sin
(wt x e) where t is the only variable.
(For a justifica-
tion of this assumption see Chapter IV, p.50.).
Further,
the motion of the bed when resting on the static pivots
will be studied without reference to the motion when on the
dynamic pivots, since with the bed in the former position
the machine's critical speed is lower than in the latter
position;
so that with the heaviest permissible piece fcr
the static position the critical speed will coincide with
the lower speed limit of the motor, and for the dynamic position it will be above this lower limit.
Fig. 19 p.Ma is a diagrammatic representation of the
bed when it
has rotated through an angle Q from its equil-
ibrium position.
S. and So represent the reactions due to
the side springs, R that due to the static pivots, and G is
the centre of gravity of the bed.
P is the pivot axis and
PG = p.
The equation of motion can be obtained directly by
equating the applied system of forces to the accelerational
-35-
system by taking moments about the instantaneous axis of
rotation P.
The more orthodox method is to equate the two
systems by taking moments about G and to equate the two vertical systems of forces.
This would involve two equations
from which R would have to be eliminated, while the former
method gives the result diectly.
Let W = weight of bed
m =
" test piece
K = radius of gyration of bed about G
h = distance from static pivots to centre of
gravity of test piece.
is further assumed that there is a damping effect
It
B.6 due to the friction at the pivot points, the air resistance, and the hysteresis loss in the springs.
This is prob-
ably not true but no serious errors will result since the
damping is very slight, although it must be present.
Since G in the equilibrium position is not vertically above
torque
P
the side springs in that position exert a
T = Wg p Sin 9.
In the displaced position the right
spring will have an increased compression
a 1 9
where a is
the spring constant while the left spring will have its
compression decreased by
a 1 9;
so that the torque supplied
2
by the two springs will be
position.
T + 2 a~l.Q.g in the displaced
By taking moments about an axis through P,
(W K+
2
*h + (W) p
-
= Wg p Sin (9+0)
(T + 2 a li G.g)
+ A Sin (wt+e) - B.G
Rearranging:
+ Cos G.Sin /)
W.g.p(Sin 9 Cos
(W (K +p ) + mh') 9
20
+ A Sin (ct+e) - T -2
a
.g -B.
= W.g p Cos 0 x 9 + W g p Sin
+ A Sin (w~t+S) - T - 2 a 12
g.g
- B.4
since 9 is always small.
T = Wg p. Sin g, so that the equation reduces to
{W
a0
(K2+p')+ mhG9 = (W g p Cos
-
a
2 a l.g.) 9
-
B.G + A Sin(ut+E)
or
1W
2
(K +p ) + mhaJ 0 + B.; + (2 al.g.-Wgp Cos
Putting
B=
) Q = A Sin (wt+e)
X
W(K2+pe) + mh
2 al 2 g - Wgp Cos
W(K2 +p') + mh2
and
the equation becomes
9
-rX.; + Y.g = A Sin (ojt+6)
The Oompleme.tgry Function is given by:
9 = e
(P Cos
Y -
t + Q Sin /
x
t)
-37-
To find the particular integral investigate
G = C Sin (cot +6 ) + D Cos (cot +c)
=Q CCos (cot +) - DSin (w
t + 5)
#
'
=(h-f.
C Sin (cot + E
-
D Cos (capt +-e
To satisfy the equation we must have
- 6C -
.D.
X + C.Y =A
- 6'D + h.C. X + D.Y
d (Y - 4)2 )D.
0
X
(
A
= 0
D (Y - W2) + C. $JX
o (Y - C02) &)X.
D. e
4 .XI = A.tw..
a
X+ D (Y - W)) = 0
2
c (Y - 0)
-
A.ow.X.
D
a!
C
(y -
-
D (Y
oX 2 + D (Y-
)
)o
-
x = A (Y
-a
= 0
oX
)
C = -A-
--
(Y - top- 18 + 4j'X-
It is immaterial from what instant the time is measured so that (cot + E) may be written simply as wt.
C Sin ot + D Cos cot =
V c2 +D-D
Sin (cot + r)
=.tan
Where
C
Hence the particular integral reduces to
c2+D-
Sin (Lot +f)
-t"a+ AaLO 2 Ka
+ -2 02Va
2oK
+
(2t"
A
2)
Sin (wt
+4)
Sin (ot + r)
+ to X
-38-
The complementary function has an exponential factor
with a negative index so that it tends to disappear as the
time increases.
Thus the steady motion of the bed is rep-
resented by the particular integral,
Q =
A
Sin (tt+r)
+ CO 2X
-a)a
(A
The value of to which gives the maximum amplitude is
found from the condition that (Y - W2)
+ wj X
must be a
X and Y are constants and by differentiating
minimum.
with respect to w and placing the result equal to zero, we
find this value of 6).
The result obtained is defined as
the critical speed, and is that at which the amplitude of
vibrations becomes a maximum, that is at which the effect of
unbalance is most pronounced.
Solving for W
2 (Y - W') (-2-to) + 2 to X
or
2)
2 Y + XS
. 0.
It
= 0
= 0
W
= Y -- 1/2
X2
is shown on p. 70 of Wilsonts 'Aeronautics" that X
may be neglected if the free motion damps to one half its
original amplitude in one complete vibration.
An experi-
ment on the machine showed that it required between 30 and
35 complete vibrations to perform this damping, which means
that X is very small ard may be neglected in calculating
the critical speed.
It is however, necessary to include X
-39-.
in the theoretical discussion in order to explain why the
amplitude of vibration does not tend to become infinite at
the critical speed, since if
X be omitted the equation of
motion becomes
Q
A
2
LO3
-
Sin (ut + r)
which apparently would have an infinite amplitude at the
critical speed Y = 1o0 2
For any given test piece the denominator in the equation of motion is a constant apart from variations of 60.
If
the disturbing agency is merely due to the unbalance of the
test piece and no compensating couple is introduced, the
amplitude of vibration at the critical speed is directly
proportional to the amount.of unbalance.
critical speed
(-2) +wax2+X2-Y . 1/2 X4
o2 =
Y
and
-
For at the
/2 X'
+
4
If X and Y do not vary greatly with varying sizes of test
piece,
/
-
is nearly constant, and the amplitude at
4
the critical speed will be
2 A
which is nearly proportional to A which itself
the test piece.
is proportional to the unbalance of
This is the principle upon which the
Gisholt balancing machine is designed.
Returning to the Olsen, machine, for a given test piece
the amplitude at the critical speed will be
, an
V 4
-40-
expression with an absolutely constant denominator,
so that
the amplitude will vary directly with A, a result used in
Chapter IV.
In order to find the capacity of the machine it
is
necessary to investigate Y when the critical speed coincides
with the lower speed limit pf the machine.
The lower limit
is 250 r.p.m. and the corresponding value of rh is
60
x 2
=
and Xz = 680
26.1
2
( = 680 = y =
2 a.)l g
-
W.g.p. Cos
......
(a)
W(K2+pz) + m h2
If the critical speed be found corresponding to any given
is possible to solve for (K2 +p2
)
test piece of known weight it
from the relation
=
2 al g - Wg p Cos /
and substituting the
mh 2
W (Kz+pz) +
result in the equation (a) to find the value of m corresponding to the lowest obtainable speed.
The actual experiments made, consisted of measuring the
critical speed of the machine when running light, and with
a test piece consisting of a shaft and two fly wheels weighing in all 237 lbs.
By reference to Chapter II,
a = 2,640
1 = l"
7 = 1250
p005
/
= 10.2"
With the test piece mentioned above m was 237 lbs. and
h was 18.5",, and the critical speed 262 r.p.m.
Substitut-
ing in the e quation
2 a 1
2
- Wg p Cos_
W (K2+p2) + Ml
2,640
(262 X 2
60
757
=
2
x 121 x 32 x 12 - 1250 xl0.2 x 32 x 12
1250 (K2 +p2 ) + 237 x (18.5)2
(640,000 - 13,000) 32 x 12
1250 (K2+p2 ) + 81,000
1250 (K2+p2 ) - 627,000 x 32 x 12 - 81,000
757
= 318,000 - 81,000
= 237,000 lbs. in2
With the bed running light the critical speed was 307 r.p.m.
so that
2
627,000 x 32 x 12
1250 (k - p 2
)
307 x 2j7
60
1250(k 2 + p 2 ) = 627,000 x 32 x 12
1035
= 232,000 lbs. in. 2
These two results check within 2 1/2%, and the second is
more accurate since in the experimant with the. test piece
mounted in the machine, the inside diameter
of the fly
wheels was larger than the diameter of the shaft, and
18.5" can be considered only an approximation for the
height of the centre of gravity.
In general the height
of the centre of gravity of a test piece is 20".
-42-
Returning to equation (a)
12
680 = 627,000 x 32 x
232,000 + 400 m
400M = 627,000 x 32 x 12 - 232,000
680
M
=
354,000 - 232,000
=
122,000
=
305 lbs.
The maximum permissible weight for a test piece is
therefore about 300 lbs.
In setting up the equation of motion no account was
taken of the gyroscopic couples due to the balance weights.
The axis of rotation of the balance weights is vertical,
the axis of precession is the pivot axis, and therefore the
gyroscopic couple lies always in a plane parallel to the
pivot axis.
Hence its only effect will be to alter the
reactions on the pivots without affecting the vibratory
motion of the bed.
-43-
ChapterIV
Operation and Calibration
To operate the machine in any specific test it is
first necessary to run it up to its critical speed, at
which point the effect of the unbalance will produce excessive vibration.
The machine is thus more sensitive than
at any other speed.
The first
operation is to find the
magnitude of the static unbalance, which is done with the bed
resting on the longitudinal or static pivots.
Wait till the
amplitude of the vibrations steadies down, and then read its
amount on the dial indicator, which should be set to read
zero when the bed is at rest.
It is useful to have a second
indicator in a fixture resting on the rear side rail of the
machine, with its
spindle in contact with the block on the
side of the bed which carries the spring adjusting screw.
This indicator can be watched while finding the critical
speed since the field rheostat controlling the speed of the
motor is situated on the wall in the rear of the machine.
Having read the amount of the amplitude of vibration
at the critical speed, set an unbalance in the headstock by
turning the right hand wheel.
Set a reading on the static
scale (the one which reads up to 16 inch ounces, and does not
rotate) according to the table-below.
Note that this table
is only intended to give an approximate setting,
to give
044-
some idea of the amount of unbalance and thus save time.
Amplitude at Critical Speed
Thousands of an inch
each side of zero
Headstock Setting
Inch Ounces
12
1
16
2
19
3
22
4
26
6
28
8
30 or over
10 or over,
Amplitudes of more than 60 thousandths, i.e., 30
each side of zero, are difficult to read and vary for any
given unbalance unless the exact critical speed can be
found, which is difficult at such an excessive vibration.
Introducing the headstock unbalance may cause the
vibrations to increase or decrease, according to the phase
relation between the headstock and the test piece unbalances.
Turn the left hand wheel, which alters this phase relation,
until the vibrations become minimum in the estimation of
the operator.
Then make fine adjustments with both hand
wheels until the vibrations are reduced to zero or as near it
as possible.
The reading on the static dial now gives the
amount of the static unbalance in inch ounces, and is the
product of the weight of unbalanced metal and the distance of
its centre of gravity from the axis of rotation of the test
-45-
piece.
This product will in future be called the 'weight
radius I of the unbalance.
Having found the magnitude of the static unbalance,
the position of the equivalent static unbalance, as defined
in Chapter I, is found by the next operation.
Stop the
machine and suspend the bed on the dynamic pivots.
Run the
machine up to its critical speed which is always slightly
higher than the critical speed when the bed is supported on
the static pivots.
Leaving the left hand wheel in the same
position as in the first operation, turn the right hand
Wheel until the vibrations damp down to a minimum.
In this
condition the moment of the static unbalance about the dynamic pivots plus the component of the dynamic unbalance in
the plane of the static unbalance, have been compensated by
the unbalance in the.headstock.
Multiply the reading on the
static dial by 20 and divide the result by the static reading of the first
operation.
This gives the distance of the
equivalent static unbalance from the dynamic pivots, measured towards the tail
stock.
If the vibrations increase as unbalance is applied on
the headstock, turn the left hand wheel until the pointer
denoting angular position has moved through 1800.
Then
apply unbalance in the headstock until the vibrations are
reduced to a minimum.
Perform the same arithmetical calcu-
lation as in the former case;
the result will give the
position of the equivalent static unbalance measured from
the dynamic pivots towards the beadstock of the machine.
-46-
If the last operation is necessary,
that is the
location reading is negative, set the angle reading pointer
back to its original position.
ings leave the pointer in its
For positive location readThe fol-
original position.
lowing operations apply to the case where the reading is
positive;
if it is negative, use A scale where R scale is
mentioned, and R scale where A scale is mentioned.
Set the aluminum dial on the vertical shaft in the
headstock, so that the line on it marked 'Static Front'
coincides with the fixed line on the headstock.
Draw a
line along the front of the piece with the aid of the sliding pointer on the machine.
ThJs. gives the line on which
metal must be removed in order to correct the static unbalance.
If'
it is desired to add metal instead of removing,
make the line on the aluminum dial, which is 1804 from the
'Static Frontt line, coincide with the fixed line on the
headstock, and mark the piece as before.
This second line
will in future be called the 'unmarkedt line.
On the line
dravm along the test piece in either case, mark the longitudinal location of the static unbalance.
This is readily
done by using the scale marked on the vibrating bed of the
machine.
The third and last experiment is to find the magnitude of the remaining component of dynamic unbalance.
bed is left on the dynamic pivots.
The
Before starting the
machine bring the 'unmarked' line into coincidence with the
fixed line on the headstock and set the static tompensator
-47-
which is located on the main shaft to the left of the headstock.
The compensator has two scales each of which is
divided into two parts, one marked R and the other A.
If
metal is to be removed to correct for static unbalance, use
the R scales, if added use the A scales.
Set the left hand
scale to the reading of the static dial obtained in locating the static unbalance;
then set the right hand scale
against the fixed pointer on the headstock so that it gives
the same reading.
This setting of the static compensator
entirely eliminates the effect of the equivalent static unbalance when the bed is supported on the dynamic pivots.
Run the machine up to the critical speed and turn the left
hand wheel until the angular reading is increased by 90'.
Apply unbalance in the headstock until the vibrations are
entirely eliminated.
Read the moving or dynamic dial which
gives the unbalance as a product of the weight radius and
arm of the dynamic unbalance mhich is in a plane at right
angles to that of the static unbalance.
If the vibrations
cannot be eliminated by the method described above, rotate
the left hand wheel until the pointer on the dial has moved
through 1800, then apply unbalance in the headstock until
the vibrations are eliminated.
If this is also unsuccessfiil
some error has been committed and the whole experiment must
be repeated.
The plane of the unbalance thus found can be
marked on the piece as in the case of the static unbalance,
-.48-
by bringing the 'Static Front'
the fixed line on the headstock.
line into coincidence with
Whenever the aluminum dial
is to be brought into one of the positions used in locating
planes of unbalance,
it
must be done by turning the main
spindle of the machine without altering the setting of the
lef-t hand wheel.
This is most easily effected by pulling
on the driving belt.
Points to notice in operating the
machine are:
a.
Oil the machine and half bearings before starting.
b.
Adjust the rheostat occasionally during balancing
so as to be sure that the critical speed is
maintained.
c.
Be sure that all compensating mechanisms are at
zero positions before starting.
d.
Form a rough estimate of the unbalance before
correcting a piece and see that the readings
obtained are possible.
e.
Do not touch the vibrating bed while operating
as this is liable to eliminate small vibrations.
The principles underlying the various operations can
be simply explained by means of diagrams.
The first dia-
gram Fig.2o0, pA8a, represents the system of forces tending
to rock the bed in static balancing.
The unbalances of the
-48a-
P
H
m ir t
"
D
m2 r s"'
F
G
c
hi
m
B
E
A
h
K
L
Fig. 20.
ha
a
-49-
test piece are fixed in relation to it
planes rotate with it
and therefore their
at a uniform velocity.
P
Q,
the bal-
ancing spindle, has a unity gear ratio with A B which repreLGHK is a refer-
sents the centre line of the test piece.
ence plane while ACDB contains S the static unbalance, and
AEFB contains M the dynamic unbalance.
L and K represent
The angle 9 between ACDB and AEFB is
the static pivots.
constant, while 0 increases uniformly with the time and may
be written as Wt.
d is the phase difference between the
balancing spindle and the test piece.
It
is evident that the
dynamic couple M can cause no moment about L K since it lies
always in a parallel plane.
rock about L K is therefore
The tendency for the bed to
entirely due to S and the centri-
fugal forces of the balance weights m, and m 2 .
P Q rocks
with the bed and the masses ma and ma always lie in planes
perpendicular to PQ.
Resolving S into components in the
plane AGHD and perpendicular to it, it is evident that only
-the perpendicular component can c aus e a moment about. L K.
The amount of this component is S Sin 0 and the moment is S
sin 0 x h, the distance between A B and L K.
The moment of
the centrifugal force of mi is found by resolving parallel
to L K and perpendicular to it.
cause no moment.
is m
r i t'
g
Sin (0 +C()
The parallel component will
The perpendicular component whose magnitude
Sin (0 +oc ) will cause a moment ml r, 4)
g
x hl.
Similarly, M 2 will cause a moment of
-50-
r2
Sin-($
-oC
) x h2
.
opposite sense equal to mT
By adding these three results with proper regard to
signs, we find that the total disturbing couple is
Sin (#4<X)xha
ri
S Sin 4xh -m
-
g
Sin (0+0C)
mr
_02
g
x h2
In the machine, m1 ri = m 2 re and we may therefore
me - m re
g
write mi ri
g
comes:
S.Sin
4xh
&)
= F and the total moment be-
- F Sin (O+OC)
x (hi-ha).
This expression can never be zero for all values of
# unless c is zero,
that is, the static unbalance and the
compensating couple are in phase.
To bring them in phase
the left hand wheel is turned and the correct position is
found when the vibrations become a minimum.
This can be
proved mathematically as follows.
Let S x h = A
and F (hi-ha) = B
where A and B
are constants.
A sin
- B Sin (+)
A Sin- B Sin
CosO- B Cos/Si
(A-B CosoC ) Sin # - B Sina-.Cos 0
(A-B Cosoc) 2+(B Sino) (Sin ($-E)
where Cot S = A - B Cosc
B Sin Oc
The resultant disturbing moment about the pivot axis
can therefore be written in a simple sinusoidal form by
introducing the lag angle 8.
The moment becomes a minimum
-51-
6 - ) is a minimum, i.e.,
when the coefficient of Sin (
2
when (A - B Cosoc)
2
is a minimum.
+ (B Sinoc)
(A - B CosOc)2
+
(B SinQC)
= A2+B2 - 2 A.B CosoC
which is a minimum when Cosoc= 1
i~e.,oC = 0.
Thus the in-phase condition is denoted by'minimum
vibrations of the bed f or any setting of the right hand
wheel.
There will evidently be no vibration when Qc is zero
and S x h = F (hi-h2 ).
Writing more explicit expressions for S and F
2
4)
M.R. N
x h = m.r.) (hi-h2 ) where M.R is the
g
g
Weight radius' of the static unbalance and m.r = miri=msra.
Then M R x h = m r (h1 -ha).
In this equation m.r is a con-
stant and h is a constant, so that (h1 -h 2 )'0 M.R in the
balanced condition.
The dial on the right hand wheel has
been rotated through an angle proportional to (hi-h2 ) so
that it
can be calibrated to read M.R directly.
In the second operation the bed rests on the dynamic
pivots X Y.
The dynamic unbalance M may be represented by
its components 14 Cos Q in the plane of S and
'.M
Sin 9 in an
axial plane at right angles to it.
We shall now show that by reducing the vibrations to
a minimum we have a measure of the longitudinal position of
S, if the phase setting is allowed to remain the same as in
the first operation.
The moment about X Y produced by S
-51 a-
P
H
mrwl
D
r
h -h
mre
M C s 9
G
C
B
/
E
/
m
M Sin
9004
v1
L
Fig. 21.
-52-.
and M is s x 1 x Cos 0 + M Cos (0 + 9),
and that produced
by the balance weights is by referring to fig.21.-F Cos 0
$
S Cos
is zero.
The resultant is therefore:
x 1 + M Cos (+9)-
F Cos 0 (h2 -h 2
)
(hi-h2) sinceo
It is clear that this expression cannot be zero for all
values of
$
unless M is zero or 9 is zero.
the effect of varying (hi-h2 ).
Let us consider
The moment above may be
written:
where s.1 = A, a con-
(A-B) Cos 0 + M Cos ($+9)
stant and B = F (hi-hz) which can be varied at will.
(A-B) Cos 0 + M Cos (0+9)=(A-B) Cos O+M Cos
M Sin $.Sin 9
(A-B+M Cos 9) Cos
-
Cos Q
#
-
$
M Sin @.Sin 5.
This may be written as
(A-B+M Cos 9)2+ (M Sin 9)2
given by Cot
e
=
Cos (#+ 8), where 8 is
The above expression shows
A-B+M Cos 9 .
M Sin 9
that the disturbing moment is of a simple sinusoidal nature
involving the angle of lead g.
Minimum vibrations obtained
by varying B mean that (A-B+M CosQ)2+ (M Sin
imum.
g)2
is a min-
The value of B at which this occurs is found by
differentiating with respect to B, giving as a result
2(A-B+M Cos 9) x (-1) = 0
or B
=
A+M Cos Q.
-53-
More explicitly,
mrW 2
g
(hi-h) =
14 R
L2
x 1 +
MiRib.2
x Cos 0
Where MiRi is the weight radius of the dynamic couple
and x is its arm.
Dividing by
4)2
8
mr (hi-h2 ) = M R.1 + MiR2. x Cos 9
1 + M R3 x Cos 9
M R
_
mr (h-h).
2
M R
)
or
The static dial
gives a reading proportional to mr (hi-h2 ).
Actually it
read:s mr (hi-h2 ) in order that M R may be read directly in
h
the first operation. Thus for any reading K on the dial,
K = mr (hi-h2 ) or
h
mr (hi-h2 ) = K.h. so that:
1+ MiRi x Cos 9 = K.h
M R
M.R
In order to fin d
1 + M.Ri
Which is the position of
x Cos 9
MR
the equivalent static unbalance defined in Chapter I,
the reading obtained in the second operation must be mu ltiplied by h (f or this machine h = 20") and divided by M. R
which is the reading obtained in the first
operation.
The static compensator is set so as to introduce a
moment about X Y (Fig.21) in the same plane as S and M cos
0-- and equal to s.1 + M (os 0, but of opposite sense.
When
this has been done the only moment about X Y that can exist,
due to unbalance is that aris ing from M Sin Q which lies in
-54-
a plane at right angles to the plane of S and M Cos
@.
In
order to balance M Sin Q we shall therefore require to
change the phase angle of the compensating spindle by 900
advanced or retarded according to the sign of M Sin 9.
The unbalance M Sin 9 and the compensating couple will then
be in phase, so that by changing the amount of the latter,
vibrations can be entirely eliminated.
Reading the dynamic
dial gives the component of dynamic unbalance at right angles
to the static as a product of weight radius and couple arm.
Since a perfectly general state of unbalance was taken to
illustrate the theory of the machine,
it
is evident that the
three operations described will give sufficient information
to enable any piece to be corrected for unbalance.
The study of the machine is incomplete without some
discussion of its calibration and the mechanisms included
in the headstock.
There are three main subjects, namely,
the phase changing mechanism, the mechanism for moving the
balance weights apart, and the static compensator.
The phase changing mechanism was described in
Chapter II.
The gear train connecting with the dial consists
of an 180tooth pinion mounted on the shaft carrying the handwheel, and meshing with a 72-tooth wheel compounded with a
24-tooth wheel which meshes with a 120-tooth wheel on the
spindle carrying the pointer.
i8
72
*
24
^
120
Page 84.
_1
= 20
20
The value of the train is
The helical gearsA,,
shown in Fig. 29..
have
8-55-
each 12 teeth giving unity velocity ratio between the shafts.
To find the pitch of the screw cut on;S,the following readThe distance d between the fixed ptPr
ings were taken.
the back end of the forked piece. Q
caliper.
dd
was measured by a
The reading of the dial is given by 9.
9
d (inches)
900
1.195"
1800
2.448"
2700
3.699"
2.448 - 1.195 = 1.253
3.699 - 2.448 = 1.251.
This gives an average of 1.252" advance per 900 change of
reading;
90* change of reading = 1/4 x 20 = 5 revolutions
of hand-wheel.
Therefore the pitch = 1*252 =
5
.250" very
nearly.
One revolution of the pointer on the dial corresponds to one revolution of the vertical compensating shaft.
Knowing this and the distance between the centre lines of
the drive shaft and compensating spindle, it is possible to
find the helix angles and the pitch diameters of the gear C
and the one on the compensating spindle which meshes with it.
The centre distance was found by tying one end of
a fine piece of string to the pulley rim and the other to
the end bearing pin.
By rotating the pulley the string
could be lined on the centres of the other two bearing pins,
and was then directly above the centre line of the drive shaft.
-56-
The left hand wheel was rotated until the 'static front'
line on the aluminum dial coincided with the fixed line on
The hand wheel was then rotated until
the head stock cover.
the dial reading was changed by 900.
The centre distance
could then be found directly by scaling the distance between
the string and the ?static front' line on the aluminum dial,
the result being
1
3= 1.8125".
16
The sum pf the pitch diameters of the two gears is therefore
If Di is the pitch diameter of the gear on the com-
3.625".
pensating spindle, since one turn of the hand wheel produces 0.25" advance of the gear on the drive shaft and 1/20
revolution of the compensating spindle,
irD,
=
0.25"
20
or D, =
= 1.59".
Taking Ori as the helix angle of this gear and Li
as the length of its normal helix, the pitch is given by:
where Ni is the number of teeth,
p = 2?Ni
Li
in this case 15.
p
D2 and
'c
=
=
N
Li
7rx 15
fr x 1.59 Sin
K3.
being the corresponding diameter and helix angle
of the sliding gear, we have
=
tan ci (since the speed ratio is unity)
Di
'oK
=90*
-
ohi
-57Dp + D 1 = 3.625"
so that D2 = 2.035".
= 520,
Hence
o
=
380, and p = 11.98.
Allowing for a slight experimental error this means that the
gears must be 12 pitch with helix angles and pitch diameters
as specified above.
The mechanism used to separate the balance weights
is of somewhat simpler form.
The train of spur gears con-
necting the hand wheel and dial is of the type similar to
that used for the phase changing dial, but the number of
teeth on the respective wheels is 18, 112, 28, 120, giving
.
The staticthe train a total value
18
28
_
112
120
80
dial reads 16 in. ounces per revolution
80 rev olu3
tions of the hand wheel,
so that one inch ounce on the
dial is equivalent to
the hand wheel.
80
=
5
revolutions of
3 x 16
3
The dynamic dial reads up to 320 oz.in.2
so that 1 oz.in.a is equivalent to 3803
tion of the hand wheel.
1
revolu-
The two helical gears connecting
the hand wheel spindle with the vertical shaft on which the
screw is cut, have each 16 teeth so that one turn of the
hand wheel produces one turn of the vertical shaft.
To find
the pitch of the screw two readings were taken of the separation of the weights, measured by a caliper.
The separation
was measured between the lower face of the upper weight and
the upper face of the lower.
At 1 inch ounce reading on
the static dial the separation was
.289 inches,
and at
7 inch ounces reading it was 2.790 inches.
Thus 6 inch oun-
ces is equivalent to 2.5 inches separation of the weights
very nearly.
= 10 revolutions on
6 inch ounces on dial = 6 x
3
handwheel
= 10 revolutions of screw shaft, so that the
pitch of the screw is given by 2*5 or a quarter of an inch.
10
The actual value of the weight radius of the balance weights
multiplied by the distance between them was shown (in the
earlier part of this chapter) to be the static dial reading
multiplied by 20.
Calling the weight radius of each m r
inch ounces,
m r x 2.5
6 x 20
i.e., m r = 48 ounce inches.
The static compensator consists of two concentric
cylinders each bored so that the two will have an equal unbalance.
Th6<cylinaers aro rhown in Fig, 22,p,59a,
The ring y is a simple annular fitting over the
ring X which is flanged at the left end, the flange carrying
a zero mark.
Before the rings are rotated relative to each
other, their centrifugal forces will be 1804 apart and since
they are equal there will be no resultant.
Now suppose the
ring Y to be rotated through 9 relative to X.
The aluminum
dial on the balancing spindle is supposed to be in the position where the unmarked line coincides with the fixed line
on the headstock.
Fig.23.,represents a vector diagram of
0-
*,T3g
Y
-0
LCx
OD
v
-o
i
H
x
-"59-
the
tweight radius f of each weight, and the resultant of
the two.
Mr
X
R=2mr Sin
2
9/2
Y
mr
Fig. 23.
The rotation 9 is supposed to be such as to bring
some value on the scale marked R onY, opposite the zero
mark on X.
From the diagram the resultant weight radius
is
2 m r Cos (90
-
-)
2
= 2 m r Sin
2
To bring
- above the horizontal line.
2
its direction horizontal requires a rotation R of X:. When
2
this is done, the resultant weight radius lies in the same
and its direction is
axial plane as the equivalent static unbalance St
and both
are directed horizontally towards the back of the machine.
Since S1 and the static compensator lie on opposite sides
of the dynamic pivots, by a suitable calibration of the
-60-
compensator scales, a setting can be made which will eliminate the moment effect of St about the dynamic pivots.
In order to bring the resultant weight radius of
the compensator into the same plane as St by setting the
right hand R scale against a fixed point directly in front
of the centre line of the axis of rotation, it is evident
that a point on this scale must subtend half the angle that
the corresponding point on the left hand R scale subtends.
The reading of the second balancing operation must
be multiplied by 20 to find the maximum moment about the
dynamic pivots of ti', and if this is balanced by the resultant weight radius of the static compensator.
2 m r Sin
-
2
x
1 = K x 20
where 1 is the dis-
tance of the centre of the compensator rings from the dynamic pivots and K is the reading obtained in finding the
position of St.
In this machine 1 = 15.06 by direct measure-
ment.
The following results were obtained for the
cir-
cumferential calibration of the left hand scale by measurement with a flexible steel scale.
Reading
0
1 = circumferential distance
from zero mark
0
4
1.0156"
9
12
15
2.3125"
3.148"
4.0625"
23
7.6875"
24
9.4375"
-61-
The outside diameter of Y is 6" so that the values
of 9 corresponding are given by 9 = 1
r
degrees.
g
0.= 1 x 180
3T
in
Reading
0
0
4
19.4
9
12
44.3
60
15
77.8
23
146.3
24
180
*
The following is a tabulation of Sin .9
.2
Sin
Reading
0
2
0
4
0.1685
9
12
0.377
0.5
15
0.628
23
0.957
24
1.
x 1 = K x 20 is to be fulthe condition 2 m r Sin
2
9
fille d, Sin
This is true,, for:
must be a constant.
K
.1685
.042
4
.377
.042
9
.5
= 042
12
*628
= .042
15
.957
.042
23
1
=
.042
24
If
-62-
Hence the compensator scales must be calibrated
according to an inverse sine law.
From the relation
2 m r Sin a x 15.06 = K x 20
2
it is possible to find the value of m r since
Sin 9
K 7 =
K
m r
.042.
20
= 15.8 inch ounces.
2 x 15.06 x .042
This value was checked by taking the compensator
apart and calculating the weight radius from the measured
dimensions of the outer ring.
R
A
0B
Fig. 24.
There are 13 equal holes 15/32" diameter spaced
150 apart and the depth of the ring is 1 inch.
of the holes lie on a circle of radius 2 1",
16
The centers
the outside
-63-
radius of the ring being 3".
The.weight radius required is
equal to the weight of the ring multiplied by the distance
of its centre of gravity from the centre of the circle 0.
Taking moments about A B we have the condition that m r, the
weight radius, minus the moment of the metal required to
fill
the holes, must equal
zero since an unbored ring
would balance about A B.
m r -M
or
R Sin 9 = 0
mr =j7MR Sin 9 = MR jSin 9
where M is the weight of metal required to fill
one hole
and R is the radius of the circle on which its centre lies.
9 is the angle that the radius to the centre of the holes
Thus m r = M R (2 Sin 0 + 2 Sin 150 + 2 Sin 300
+
makes with A B.
+ 2 Sin 750 + Sin 90")
Assuming a density of 0.28 lbs. per cubic inch,
(1
12x
x 1 x 0.28 x 16 ounces
R = 211 inches
16
m r
=
E x (1)2 x 1 x 0.28 x 16
4 32
x
44_Sin 9
16
Sin 9 = 0 + .5176 + 1.0000 + 1.4142 + 1.7320 + 1.9318 + 1
= 7.5956
Hence
mr=
15.7 inch ounces.
The slight discrepancy with 15.8 the figure previously obtained is probably due to the fact that the density is not
quite 0.28 lbs. per cubic inch.
-64-
For the right hand R scale the circumferential
distance of each scale mark from the zero mark is exactly
half the distance of the corresponding mark on the left hand
scale for the reason previously explained.
Exactly similar
to the two scales marked R, are two scales marked A plotted
in the opposite direction from the zero mark.
These are to
be used when it is desired to add metal instead of removing
it
to correct static unbalance.
For if the machine is set
for adding metal, and the 'unmarked line' is toward the
front of the machine,
the compensator will have 'to supply
a centrifugal force towards the front in order to overcome
the moment of the equivalent static unbalance about the dynamic pivots.
Setting on the R scales gives a centrifugal
force towards the rear, so that the A scales must be used.
Also if in a 'remove metal' test, the reading of the
second operation is negative, the equivalent static unbalance
is on the headstock side of the dynamic pivots, and to balance
it
the compensator will have to supply a centrifugal force
towards the front of the machine, and the setting must therefore be made on the A scales.
Similarly if in an 'add metal'
test the reading of the second operation is negative, the
compensator setting must be made on the R scales.
It will generally be found that the component of
dynamic unbalance obtained by the third operation is small
compared with the moment of the equivalent static unbalance
-65-
and the minimum vibration condition of the
can be fairly easily determined.
second operation
If however the component
dynamic unbalance is .large compared with the moment of
the equivalent static unbalance, the minimum vibration condition may be hard to determine accurately.
it
In such a case
is best to alter the angular setting by 90' and reduce
the vibrations to a minimum, which should be more clearly
defined.
Stop the machine and set the static compensator
to the reading obtained.
Restore the angular setting to its
original position and eliminate the residual vibration of the
bed.
Vith this cycle of operations, the second reading
gives the value of the component dynamic unbalance when multiplied by 20, While the third reading when multiplied by 20
and divided by the reading of the first
operation gives the
location of the equivalent static unbalance.
It
is generally advisable when finding the location
of the equivalent static unbalance to reverse the piece if
an accurate result is desired.
check, it
If the two results do not
is probably owing to the fact that the bed is not
vibrating aboit a fixed axis due to movement of the pivot
points on the bearing surfaces.
When this is the case the
first operation will give readings which involve not only
the magnitude of the static unbalance but also that of the
dyhamic unbalance, and hence an untrue value will be found
for the static unbalance.
This value is used in calculating
-66-
the location of the equivalent static unbalance from the
reading of the second operation and will therefore lead to
erroneous results.
To counteract this effect the initial
compression of the end springs should be reduced so that the
pivots will bear more firmly on the bearing surfaces and thds
eliminate any tendency to slip.
Shafts undergoing test should be made with journals of very accurately cylindrical form.
Any tendency to
an oval shape of journal will cause impacts which will make
the bed vibrate and preclude any idea of accurate balancing.
A shaft that is curved should have the bearings on the
machine in the same relative positions during the test as
the bearings it is to run in when in actual operation.
By
moving the shaft longitudinally relative to the bearings
different states of unbalance can be obtained.
Balancing a
shaft in two different positions with respect to the bearings therefore constitutes a test for straightness of the
shaft.
If the vibrations in the various balancing operations cannot be brought down to zero within close limits,
the bearing pins are probably not holding the test piece
tightly enough in the half bearings.
This is liable to
happen when the half bearings are of larger diameter than
the journals of the test piece.
To remedy the defect force
the bearing pins down harder on to the journals.
It should
-67-
be possible to bring the amplitude of vibration down to less
than half a thousandth of an inch, that is, a quarter of a
thousandth each side of the zero on the indicator.
The design of the machine could be modified to give
more accurate results and quicker operation, besides eliminating the necessity for a static compensator.
It has
been shown that after the second operation the only unbalance that is not compensated by the setting of the balance
weights is the component of the dynamic unbalance in a plane
at right angles to that of the static unbalance.
This could
be balanced by a shaft carrying balance weights exerting a
couple leading or lagging the principal balancing shaft by
900 and driven through spur gears connecting the two shafts.
The shafts must rotate at the same speed.
A third hand
wheel would be required to adjust the auxiliary compensating
couple, and a third dial to read its amount.
The method of
operation would be as follows:
Find the amount and angular position of the static
unbalance using the principal balancing spindle by the usual
method.
Stop the machine and support the bed on the dynamic
pivots.
Run at the critical speed and without altering the
angular setting reduce the vibrations to an approximate
minimum by means of the principal balancing couple.
Operate
the hand wheel of the auxiliary compensating spindle until
the vibrations are almost zero.
Make final adjustments with
the hand wheels of both balancing spindles until the
-68-
The readings on the
vibrations are entirely eliminated.
dials will give sufficient information to enable all unbalance of the test piece to be corrected.
The auxiliary bal-
ancing spindle must be capable of being indexed 1800 to
compensate for component dynamic unbalance of negative
sense.
This can be effected without stopping the machine
by means of the mechanism sketched in fig. 25., page 6$.
The pin P can be lifted clear of the slot A in the
gear G by depressing the lever L which lifts the whole shaft
relative to G.
The gear G meshes with an equal gear on the
principal balancing spindle.
When P is clear of the slot G
continues to rotate at uniform speed while the shaft will
decelerate due to friction until P comes into contact with
the block C when it
same speed as G.
will be constrained to rotate at the
The lever L can then be released and the
shaft will have been indexed 1800.
Before starting the
machine P must always be brought back to A if it was necessary to index to B in the previous test.
The advantages and disadvantages of the modified
design are listed below.
Advantages.
(a) Elimination of static compensator,
demanding
that the machine be stopped only once in a
complete test.
(b) Saving of time
(c) More accurate results
so-68a-m
A
Fig. 25.
-69-
Disadvantages.
(a) A more expensive machine
(b) Greater intelligence required in
interpreting readings.
The objection (b) is not of great importance, since
the interpretation of the reading on the third dial involves
only the principle that two minuses make a plus.
That is
to say, if the location reading is positive and the auxiliary spindle does not need to be indexed, the sense of the
component dynamic unbalance is positive; if indexing is
necessary,
the sense of the unbalance is negative.
When
the location reading is negative and the auxiliary spindle
is not indexed the sense of the component dynamic unbalance
is negative;
if the shaft is indexed, it is positive.
-70-
Chapter V.
Balancing Tests
A number of varied tests were run to investigate
the behavior of the machine in actual operation.
The first
of these was on a Ford four throw crank shaft with 1-1/4"
line bearings.
An initial unbalance of 1/2 in. oz. wasset
in the headstock and readings of the amplitude of vibration
taken at intervals of 300 for the left hand wheel setting.
The results obtained are tabulated below.
Angular reading
Amplitude of Vibration
0
1
300
1-1/10
.600
1-1/10
900
1
120*
9/10
150*
7/10
18004
5/10
2100
3/10
24060
2/10
225*
1/10
In general it is quicker to advance by 90' at a
time, when the approximate correct position can be estimated and final accurate adjustments made.
The headstock unbalance was then removed and the
hand wheel reset at intervals of 1/8 inch oz. at 2250
-71-
Results:
angular setting.
Amplitude of Vibration
Headstock Unbalance
0
4/10
1/8
4/10
1/4
3/10
3/8
1/10
1/2
1/10
0
7/16
Thus the amount of static unbalance is 7/16 in.oz. at
angular setting.
2250
Its position was found by a similar tabu-
lation, starting with a headstock unbalance of 2 oz. in.
The reading could not be obtained with the same setting for
angular position so this was changed by 1800.
Amplitude of Vibration
Headstock Unbalance
2
4.6
1
1.3
1/2
0.6
1/4
0.4
0
0.3
-1/4
0.2
-1/2
0.1
-3/4
0.2
The location reading is therefore -1/2 so that the position
of the equivalent static unbalance
1/2 x20
7/16
=
is
23" towards the headstock.
-72-
This is beyond the limits of the piece and therefore indicates a considerable component of dynamic unbalance
in the plane of the static.
No appreciable reading could
be obtained for the component at right angles.
The piece
was not corrected.
The second test was on a 6-throw automobile crank
shaft.
This was an old piece with very poorly finished
bearings, which were untrue enough to cause severe impacts
and mask the effects of unbalance on the vibration of the bed.
No useful results could be obtained.
The next test was an
attempt to balance a pressed steel pulley.
The shaft used
was an unfinished piece of two inch cold rolled stock.
By
finding the unbalance of the shaft alone, and the shaft with
the pulley mounted on it, it should be possible to find the
unbalance of the pulley by subtraction of vectors.
In ad-
dition it is necessary to measure the untrueness of the
centre line of the shaft at the point where the pulley is
attached, and the weight of the pulley, since there will be
an unbalance due to the fact that the centre of gravity of
the pulley is describing a circle.
The centrifugal force
due to this can be calculated and subtracted vectorially
from the former result to give the actual unbalance of the
pulley.
In the experiment no reliable results could be ob-
tained owing to the roughness of the shaft which caused
impacts sufficient to make it impossible to repeat results.
-73-
The remaining tests were carried out on a specially
designed test piece intended for demonstration purposes.
The piece consists of a finish ground 1-7/8" shaft, 48"
long with the last two inches turned down to 1-1/4" diameter.
On the shaft may be mounted any or all of three sim-
ilar cast iron flanges, the hubs of which are provided with
clamps so that they may be fixed to the shaft in any given
angular position.
The outside diameter of each flange is
10" and one face of each is finished.
18 lbs. and the shaft weighs 43 lbs.
Each weighs about
By using three unbal-
anced masses of this nature, a great variety of states of
unbalance can be obtained.
The first series of tests was run on the shaft alone.
The bearings on the machine were placed in different relative
positions for each test, the distance of the centres of the
bearings from the right end of the shaft being given in the
table below.
was not used in that particular test.
Distances of Bearings
Test
No.1
No.2
1
6-1/21"
30"
2
12-1/4"
26"
3
9-1/4"
4
6-1/2"
30 "
5
6-1/2"
-
22-1/41"
No.3
45"
40-1/2"
36"
-
ings, it
When no distance is given for one of the bear-
45"
-74-
The distance of the right end of the shaft from the
dynamic pivots was 11".
The unbalance is very slight which
accounts for the variation in the angular readings;
unless
the unbalance is large the shaft can be apparently balanced
for a considerable range of angular readings.
However, if
the unbalance is as great as 4 or 5 inch ozs. the angular
position can be found within 1 or 20.
The results of the
tests are tabulated below.
Tests
Static Unbalance
Angular
Location
Position
Reading
0
1
1/4
2450
2
7/32
2400
1/16
3
3/16
2500
1/16
4
1/4
2500
1/16
5
3/16
2600
1/32.
Taking an average of the above readings, the static
unbalance is 7/32 in.oz. at 2500 with a location reading of
1/16.
To find the location from this reading, multiply by
20 and divide by 7/32 which will give its distance from the
dynamic pivots measured towards the tailstock of the machine.
1/16 x 20
7/32
_
20
16
x
32
7
=
5.7"
The distance from the right end of the shaft is
therefore 11 + 5.7 = 16.7".
In no case was there any appre-
ciable reading for the component of dynamic unbalance in a
plane at right angles to that of the static.
-75-
Originally the flanges were intended to be corrected
entirely for unbalance, and unbalanced masses of any desired
amount attached by means of bolts through the half inch
holes drilled in the flanges.
Since the major part of the
unbalance is due to the hubs, it would be possible to correct
the static unbalance by removing metal from the flange,
but this would automatically introduce a dynamic unbalance
which it
would be impossible to correct.
The three pieces
were therefore reduced to. the same state of static unbalance without absolute correction.
A large number of states
of unbalance can be obtained by altering the relative longitudinal and angular locations.of the three flanges on the
shaft.
The static unbalance of each piece was found to be
greater than the capacity of the machine, namely 16 in.ozs.
A counter weight in the form of a bolt carrying three nuts
was inserted into the 1/2" hole in the flange, in order to
reduce the static unbalance to a measureable amount.
The
weight of the bolt and nuts was 7.30 ozs. while the centre
of the hole was at a radius of 3-1/4" so that the weight
radius of the counterweight was 23.7 in.ozs.
The actual
unbalance of each piece can be found by subtracting this
weight radius vectorially from the unbalance found in the
test.
To perform this operation, change the angular posi-
tion of the weight radius by 180' and add vectorially to the
results of each test.
-76-
Let R equal actual unbalance of any one piece
K = test result
9, = angular location of K
Then R =
= angular location of counterweight.
K2 + (23.7)2 + 2 K x 23.7 Cos (9 2 -180 0 -9
1
)
G2
The test results are given below.
Flange No.
K
R
92
01
1
9-1/8
780
2150
32.5
2
8-1/4
550
2130
31.5
3
5-1/4
3500
3170
19.6
These values of R show that the unbalance in every
case is too great to be corrected by boring the hubs, where
it is located.
Permanent counter weights in the form of
steel studs turned down to
1/ 2 "
diameter and threaded at the
end were made and bolted to the flanges.
The weight radius
in each case was approximately 25 inch ounces.
Further tests were then run to determine accurately the unbalance of each piece.
Flange No. 1 gave contra-
dictory results in the location readings but appreciably
constant readings for the amount of static unbalance.
A
number of readings were taken with the piece mounted on the
shaft so that its hub pointed towards the headstock.
A
brake test was being run on apparatus alongside the machine,
causing considerable vibration vihich partly qccounts for the
errors in the first group of results.
The column headed
-77-
'position' gives the distance of the finished face of the
flange from the dynamic pivots.
Location
Reading
Location
Position
Static
Unbalance
1
15"
6-3/8
3-5/8
11.4"
2
10"
6-1/2
2-1/4
6.9
3
5"
6-1/2
4
0
6-3/8
- 1/2
5
20
6-1/4
6-3/4
Test
1
3.1"
-1.6"
21.6"
In test No. 5 the piece was reversed, i.e., the
hub was towards the tailstock;
also the readings of this
test were taken after the completion of the brake test
mentioned before.
In the graph on p.77a location is plotted against
position.
This should result in a straight line at 450.
From the graph there is an error proportional to the position since the straight line is not inclined at 450.
This
error will be smallest at the position 0, giving the most
probable distance of the equivalent static unbalance from
the machined face of the flange as 1.6".
The result is
borne out by the readings of test No. 5 and further tests
made on the following day.
Test
Position
*6
15"
*7
8
*Piece reversed.
1511
Location
Reading
5-1/8
Locatio n
6-1/4
3-5/8
11.6"
6-1/8
4-1/16
13.3"
Static
Unbalance
6-1/4
16.4"
-r--'--~-
2112
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-78-
The average of the three results above gives the
location as 1.6" from the machined face.
In no case was
there any appreciable reading for the component dynamic
unbalance.
This was also true for the other two flanges.
The next series of tests was run to find the unbalance of flange No. 2.
Results are tabulated below:
Test
Position
Static
Unbalance
1
15"
4-5/16
2-1/2
11.6"
2
10"
4-3/8
1-3/8
6.3"
3
20"
4-1/4
3-1/2
16.5"
*4
20"
4-1/4
4-1/4
20"
*5
10"
4-1/4
2-3/8
11.2"
Location
Reading
Location
Lines are plotted for location against position
in the graph on p. 78a.
The two lines make intercepts
on the 'location' axis of + 2.4 and - 3.9, so that the probable distance of the unbalance from the machined face is
2.4 + 3.9
2
or 3.1".
The next series of tests gave a con-
siderably lower value, and the average of the two is approximately the correct value.
*Piece reversed.
4_
__
_
- -
K
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'-'aa
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03
.....
...
4
44
4
4I
t
L
-79-
Position
Test
Static
Unbalance
Location
Reading
Location
*6
15"
4-1/4
3-1/4
15.31
*7
10"
4-1/4
2-1/4
10.6"
*8
2011
4-1/4
4-1/4
20"
9
20"
4-1/4
3-3/8
15911
10
15"
4-1/4
2-3/8
11.2"
11
10"
4-1/4
1-3/8
6.5"
From the graph, p. 79a, the probable location is
= 1.9" from the machined face.
1.2 + 2.6
2
The results of these tests were so unsatisfactory
that the accuracy of the machine was checked in the following manner.
A bolt was lashed to the shaft, the location
of the unbalance found and compared with the position of
the bolt.
In this case there could be no appreciable dynam-
ic unbalance, so that location and position should be the
same.
The static unbalance was 7-3/4 and the location read-
ing 6-3/4, giving as the location
6-3/4
7-3/4
x
20 = 17.4".
The
actual position was 19.1" and therefore the machine read inaccurately.
The initial
compression of the end springs
was then reduced and a further series of tests run on flange
No. 2.
* Piece Reversed.
LI
I4
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-_
-80-
Test
Static
Unbalance
Position
Location
Reading
Location
12
10"
3-7/8
1-1/2
7.7"
13
15.2"
3-3/4
2-3/8
12.7"
14
20"
3-7/8
3-3/8
17.4"
*15
20"
3-7/8
4-3/8
22.6"
*16
15"
3-3/4
3-1/4
17.3"
*17
10"
3-3/4
2-3/8
12.6"
The average result for location measured from the
machined face is
2.3 + 2.3 + 2.6 + 2.6 + 2.3 +
2.6 = 2.5",
6
which is approximately the average of the results of the two
previous series of tests.
Flange No. 3 caused no difficulty, and since the
first three tests on it checked within reasonable limits, no
further readings were taken.
Test
Position
Static
Unbalance
1
10"
8-5/8
4-1/4
9.8"?
2
20"
8-5/8
8-1/2
19.7"
*3
20"
8-5/8
8-1/2
19.7".
Location
Reading
Location
The average value for location measured towards the
hub from the machined face is 0.2 + 0.3 - 0.3 = zero, very
3.
nearly. This is due to the fact that the flange is considerably
*Piece Reversed.
-81-.
thicker on one side than the other, so that the static unbalance is largely due to the flange;
the angular position
of the unbalance is approximately 1800 from the unbalanced
portion of the hub, and hence the effect of the hub unbalance is to move the line of action of the 'equivalent static
unbalance' towards the machined face.
The results for the three flanges are tabulated
below.
Locations are measured towards the hub from the
machined face.
Flange
Static
unbalance
No. 1
6-1/4
Location
4
No. 2
No. 3
8-5/8
Component
Dynamic Unbalance
1.6"
0
2.5"
0
0
0
The unbalance of flanges No. 1 and No. 3 were reduced to approximately 4 inch ounces.
Let M R = amount of unbalance to be corrected
r = radius at which correction is applied
d = diameter of drill
C =density of the metal
1 = depth of hole
Then
M R
4- x
1 X Cxxx r
M R and
are known, r and d may be conveniently chosen,
so that it
is possible to solve for 1, in correcting flange
No. 1.
Flange No. 3 has its unbalance located in the flange
which is 3/4" thick so that 1 is known and the value of r
may be solved.
-82-
For flange No. 1
d = 3/4"
r =
1 - 3/ 8
"
M R
Sd 2(
14
r
' = 4 ozs. per cu. in.
YR = 2-1/4 in. ozs.
.
. 1 = 0.931.
For flange No. 3
M R
, -d2
=4
1 = 3/4"1
d = 3/4
"
r
MR = 4-5/8 in.ozs.
.r=
.
3.511
In the previous tests no account was taken of th4
Si ace
angular location of the flange relative to the shaft.
the unbalance of the shaft is 1/4 in.oz. an error was int roduced in the results.
This unbalance of the shaft was
added to the unbalance of the flanges in the final tests
on the corrected pieces, since in each case the unbalance
of the shaft was made to oppose directly the unbalance of
the flange.
Static
Reading
Static
Unbalance
Location
Reading
Loca tion
0
3-3/4
4
- 5/16
-1 .611
Q
3-3/4
4
+ 3/8
+ 1 .9"
2
0
3-7/8
4-1/8
-1/2
-2
3
0
3-5/8
3-7/8
0
4-3/8
4-1/8
1
*1
*3
Position
* Piece Reversed.
-3/8
.4"1
1 .3 t
1/4
-
Flange
211
-83-
In the second test on No. 3, the unbalances of
shaft and flange acted in the same direction so that 1/4
was subtracted from the static reading.
Tabulation of Final Results
Flange
Static Unbalance
1
4
4-1/8
2
4
3
Location
1.75"
2.4"?
-1.7"?
There was not sufficient time to run further tests
on the shaft when carrying two or all of the f langes.
The results
show that the machine may be expected
to give results which can be repeated within close limits.
The error should in no case exceed about 1/8 in.oz. reading on the static dial, provided the machine is properly
adjusted.
The Inside of the Headstock
Fig. 29.
-85-
End View of Machine
Side View of Machine
Headstock
Balance Weights at Zero
Balance Weights Separated
Headstock with Cover Removed
Fig. 30.
-86-
BIBLIOGRAPHY
1.
Mechanics of Machinery, Ham and Crane
2.
Vibration Problems in Engineering, Timoshenko
3.
Aeronautics, Wilson.