. T. i T T

24 SEP 1929

Br AP 9 2e A

A

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 DepartmentalCommittee on Graduate Students

Head of Department

v

STUDY OF THE CARWEN OLSEN TYPE BALANCING MACHINE

&(4

TABLE OF CONTENTS

Page

I. Preliminary Study of Balancing

II. Description and Constants

III.The Dynamics of the Vibrating Bed

IV. Operation and Calibration

V. Balancing Tests

Bibliography

1

17

34

43

70

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. This unbalance is of

two kinds, 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. A single mass would produce a cen-

trifugal force. 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

force. The resultant couple is termed the dynamic un-

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 com-

bined 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, and

an infinitesimal distance apart. The slice will have an

infinitesimal static unbalance and a dynamic unbalance

which is a product of an 'ixiffiitesimal force and an

/

I@2

/ a

B

R

/ I/

/

~ 2

b

.. 3m

Plt

I '1

Pa C

P

Fig. 1.

-4-

infinitesimal length. The dynamic unbalance may there-

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 P2 acting at angles e 1 and E2

respectively with the vertical for a given angular posi-

tion 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 P2b 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) P2 b Sin (9 2-9R)

A C B

Fig. 3.

.-5-

Let the angle between R and the vertical at c be 9R'

Resolving the force system in axial planes con-

taining R and at right angles to it we have the follow-

ing; in the plane containing R, two couples of magnitude

Pia Cos(GR - 91) and P2b Cos(92 - 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

Pia 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

to depend on the arbitrary quantities a and b. Actually

they can be shown to be independent of a and b and inter-

preted in terms of 1 the fixed distance between the two un-

balanced 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

of A, the total moment is equal to

R(a + y) + Pab Cos (9a-9R) - P1 a Cos (OR .

and substituting b = 1-a

R(a + y) + Pol Cos (G2 - GR)- a P1Cos(@R -

+ PSCOs (02 - GRJ

-6-

R is the resultant of Pi and P2 considered as acting in

one plane, hence by reference to Figure 4

RPi

(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 y + P2 1 Cos (Q2 - 9 )= R 17 + IP2 1 COS (92- )

R

This moment would be produced by a single force R

acting at a distance P2 1 Cos (Q2 - QR) to the right ofR

A. 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 + P0 COs (2 - 9 1 ))R

-7-

P2 1 )Ple + P1 Cos (92 -91)

2 2P1> + 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 - R

= P2 1 Sin (92 - GO + a (Pi Sin (R-Q 3.)

Pa Sin (Q2 - QR

From Fig. 4 P, Sin (OR - G) - Pa Sin (Ga - OR) = 0 since

the components of P. and P2 at right angles to their re-

sultant 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)to the right

2 2PI + P2 + 2 PIP2 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 - QR) should evidently

be symmetrical in Pi and Pa. From Fig. 4,

R Sin (Q2 - OR) = Pi Sin (92 - 91)

Hence Pa 1 Sin (92 - =R) P, Pa 1 Sin (92 - 9 1 )

R

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

P, 1 P+P Cos (a-9i)] + P, I {PI+P2 Cos (92-91)

1+ P + 2P 1 P2 Cos(Q2 -@ 1 ) Pi + Pa, + 2 PIPe Cos(Ga-91)

_1 {P1 + P2 + 2 PiP2 Cos (9a - Q 1)Ca = 1, a necessary

Pi + Pa + 2 PiPe Cos (92 - 90)

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-

as previously defined, but is only a component of it. The

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 vibrat-

ing 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 ro-

tating. 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 ef-

fect on the bed is the vibration due to the static unbalance,

since the dynamic unbalance is a pure couple which will pro-

duce 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.

h

) S

/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. This can be read-

ily seen from the diagram. The horizontal components

of the Force S and couple D x 1 would have no moment

about the pivot axis.

D

S

DI-

Fig. 6.

0-

M

-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

the angular position shown. Thie most convenient method

of balancing the shaft is to combine the static unbalance

with the component of the dynamic unbalance in the same

axial plane. It is then possible to balance these two

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 dy-

namic 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.x + Mi so thatS

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 28

where c is the compensating couple. The expression for

the resultant couple is a minimum when Sty = 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 eval-

uation 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 re-

quired 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

M2 = m. and m2 ,d = M2 .

S

142'll. XII'd

Fig.8.

Wle can replace mi and m2 by a single force R, the vec-

tor sum of the two, its correct angular position being given

by the force diagram Fig. 9.

R -

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 however be

obtained by only two added weights or borings. Consider

the state of unbalance denoted by the diagram Fig.10, where

couples are represented vectorially, with the usual con-

ventions.

St t 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, to-

gether with a couple in the plane of S' of magnitude Slxd.

At the transverse plane containing S" we may add three

weights hit, M2 , m. and at a distance 1 along the shaft

-15-

two further weights m2 and m.. These will cbmpensate the

unbalance entirely if mi = S" = S, m21 = S'.d = S. d, and

m31 = M2 . Solving for mi, m2, m. from these three condi-

tions, we cen find the resultant of mi., M 2 , m. in the trans-

verse 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

R2 's

) m 3in3

(mi+ma) Ri

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 compo-

nent 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 un-

balances 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 to-

gether with the details of its design.

-17-

Chapter 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. The shaft is turned by means of a hand

wheel. 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. Three strips of

*Refer to photographs on page 85.

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 pre-

serve 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

-18-

-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 exam-

ination 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-

the top. Care should be taken in removing the cover since

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, that is, parallel to

the drive shaft of the machine. 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 dial

through a spur gear train. 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 de-

termines the capacity of the machine. If the test piece is

too heavy the critical speed of the machine will lie outside

the lower speed limit and it will be impossible to balance

the piece, since it is essential to run the machine at the

critical speed. 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 follow-

ing manner.

Fig. 12.

The heads of two long bolts were inserted into invert-

ed 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

table of a balance. A dial indicator was mounted in a fix-

ture placed on the table of the balance. The readingsof

this indicator gave the movement of the cross piece relative

to the table of the balance, that is, the amount by which

the spring was compressed. The load was varied by screwing

down the nuts on the two bolts and its amount read on the

scale of the balance. The first reading was taken at a

scale reading of 35 pounds.

Results:

Scale Reading 35 100 175 120 85

Indicator Reading 0 26 53 32.5 19.4

From the graph p. 23a.050" compression = 132. lbs. This

gives the constant for the two side springs. 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 free

from all its pivot points by increasing the compression of

-the two side springs and headstock end spring by screwing

down their adjusting screws. The spring at the tail stock

end was removed. 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 -

A I

-~-7

-r

'1**

**'

[4o

-s-

--7-

---

----

~-

----

}. ~

I i---T EU

EN ___

ZG

E

CO

.. C

HC

O

E YO

R

NO

, 34

* I

I I

tt...-

.

I to

~Alf

tli

Rwaa

tW

&a

-24-

bed given by;

= 2 .2 50 - 1.995 x 132 + 2.250 - 2.058 x 132 + 2.250 - 2.150

.050 .050 .050

.447 .100x 35 = .44x 132 + 00 x 350050 '050 "

= 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

PN

11" ,

Fig.

T P2

11"

14.

P2 x 11 = P1 x 11 + 3F

= 11 (P2 - Pi) = 11 x 132 12.250-1.995-(2.250-2.058)]1250 x .050

11 x 132 x (2.058 - 1.995)

1250 x .050

11 x 132 x .063

1250 x .050

= 1.46"

W

->. < ~

Pa 3IIV

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

P S

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. We have

by taking moments about the axis of rotation, that

W hoC0 + Wi = Pl, where i 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

of the spring, andeC is measured by a dial indicator at a

fixed distance from the axis of rotation. Taking two cor-

responding readings of oc and P and subtracting the result-

ing equations,

W hd + W F = P1

W hoc+ W F = P01

W h (,(-c) = (P - PO) 10

There is only one unknown, namely h, in the above equa-

tion, 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 of the machine,

with its 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 neglect-

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 indi-

cators gave a record of the change in length of the spring,

as may be readily seen by reference to the diagram, Fig.16.

RE=AReading

Reading B=A

(a)

ading Reading+.,0025" =B +.0024"

.0001"

(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

T

025!

-28-

relative to the block, as positive values when taking differ-

ences. 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 seg-

ment, 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. The10

results are tabulated below.

I i

HeadstockReading

0

19*5

41.7

65.8

89

112.5

89

67

44*5

20.8

0

12

2nd Indi -cator Reading

0

5.6

8.7

17

23.4

26.1

28.4

22*5

13

4.7

0

2-1/2C= 10 xIi

Correction

0

4.9

10.1

16.4

2262

28.1

22.2

16.8

11.1

5.2

0

LChange'ofLength of

Spring= 12 -C

0

+ 0.7

- 1.4

+ 0.6

+ 1.2

- 2.0

+ 6.2

+ 5.7

+ 1.9

- 0.5

0

Since there should be a straight line connection be-

tween 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:

Ii I2 L = I Ia

0 0

24.6 24.0

50.2 47.7

75.9 72.8

102.4 98.4

75.6 72.5

49.6 47.8

24.0 23.6

0 0

23.4 22.8

50.4 47.9

76.5 73.2

102.5 98.4

Plotting these results and

line, from the graph on p. 30 a.

0

0.6

2.5

3.1

4.0

3.1

1.4

0.4

0

0.6

2.5

3.3

4.1

drawing a mean s traight

.4 = 4I i 100

The distance of the first indicator from the axis of rota-

tion was 11". Referring back to the theory on p. 26.

QC 0 = and the spring constant being

105 lbs. per inch, P - P = 32 x L

r- -- tm Oa

L __ _ _ _77_ _ _ __ _ _

o 04 7 - 1 _ _- 1

0 00

602 -000410

Fig* 17.

z

0

z

0

0

z

-31-

Since Wh (- po) 1 and W = 1250 lbs.

1 = ll1"

Ii 1321250 h x -=--- x L x 11

11 .050

132 121 L-.050 1250 Y,

132 x 121 4.050 x 1250 100

= 10.2"

The statement that corresponding to a small rotation

c< of the bed its centre of gravity moves horizontally hoc,

requires proof.

G? (90*-Q)

/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

compensator from the dynamic pivots. The former of these

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 700 lbs. per inch

Weight of vibrating bed 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 15.06"

Speed Range 250 - 350 r.p.m.

-33-

Moment of Inertia of the Bed

about Static pivot axis

Heaviest permissible test piece

a232,000 lbs.in.

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" 1-1/2", 1-5/8",

1-3/4",. 1-7/8", 2".*

-34-

Chapter 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. 'To find the moment of iner-

tia it is necessary to study the dynamics of the vibrating

bed. 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 po-

sition 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 ver-

tical 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.

It is further assumed that there is a damping effect

B.6 due to the friction at the pivot points, the air resist-

ance, 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 verti-

cally above P the side springs in that position exert a

torque 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 supplied2

by the two springs will be T + 2 a~l.Q.g in the displaced

position.

By taking moments about an axis through P,

(W K+ *h2 + (W) p

= Wg p Sin (9+0) - (T + 2 a li G.g)

+ A Sin (wt+e) - B.G

Rearranging:

(W (K +p ) + mh') 9 W.g.p(Sin 9 Cos + Cos G.Sin /)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

a0 a{W (K2+p')+ mhG9 = (W g p Cos - 2 a l.g.) 9 - B.G + A Sin(ut+E)

or

21W (K +p ) + mhaJ 0 + B.; + (2 al.g.-Wgp Cos ) Q = A Sin (wt+e)

Putting B= XW(K2+pe) + mh

and 2 al 2 g - Wgp Cos

W(K2 +p') + mh2

the equation becomes

9 -r X.; + 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 C Cos (cot +) - D Sin (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 0

d (Y - 4)2 )D. ( X A

D (Y - W2) + C. $JX = 0

o (Y - C02) &)X. - D. e4 .XI = A.tw..2 a

c (Y - 0) X+ D (Y - W)) = 0

D A.ow.X.- a!

C (y - - D (Y - )o x = A (Y -a

oX 2 + D (Y- ) o X = 0

C = -A- --)(Y - top- 18 + 4j'X-

It is immaterial from what instant the time is meas-

ured so that (cot + E) may be written simply as wt.

C Sin ot + D Cos cot = V c2 +D-D Sin (cot + r)

Where =.tanC

Hence the particular integral reduces to

c2+D- Sin (Lot +f)

-t"a+ AaLO2Ka+ -2 02V a Sin (wt +4)

(2t" + 2oK

A Sin (ot + r)2) + 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)

(A -a)a + CO 2X

The value of to which gives the maximum amplitude is

found from the condition that (Y - W2) + wj X must be a

minimum. X and Y are constants and by differentiating

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 = 0

or 2) 2 Y + XS = 0

. .0 W = Y -- 1/2 X2

It 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 Sin (ut + r)- LO3

which apparently would have an infinite amplitude at the

critical speed Y = 1o02

For any given test piece the denominator in the equa-

tion 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. For at the

critical speed o2 = Y - /2 X'

and + (-2) + wax2+ X2-Y . 1/2 X44

If X and Y do not vary greatly with varying sizes of test

piece, / - is nearly constant, and the amplitude at4

the critical speed will be 2 A which is nearly propor-

tional to A which itself is proportional to the unbalance of

the test piece. 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

x 2 = 26.1 and Xz = 68060

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

test piece of known weight it is possible to solve for (K2+p2 )

from the relation

= 2 al g - Wg p Cos / and substituting the

W (Kz+pz) + mh2

result in the equation (a) to find the value of m correspond-

ing 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 weigh-

ing 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 2 a 1 - Wg p Cos_

W (K2+p2) + Ml 2

(262 X 2 2,640 x 121 x 32 x 12 - 1250 x l0.2 x 32 x 12

60 1250 (K2+p2 ) + 237 x (18.5)2

757 = (640,000 - 13,000) 32 x 12

1250 (K2+p2 ) + 81,000

1250 (K2+p2 ) - 627,000 x 32 x 12 - 81,000757

= 318,000 - 81,000

= 237,000 lbs. in2

With the bed running light the critical speed was 307 r.p.m.

so that

307 x 2j7 2 627,000 x 32 x 1260 1250 (k - p2 )

1250(k2 + p2 ) = 627,000 x 32 x 121035

= 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)

680 = 627,000 x 32 x 12

232,000 + 400 m

400M = 627,000 x 32 x 12 - 232,000680

= 354,000 - 232,000

= 122,000

M = 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 exces-

sive 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 Headstock SettingThousands of an inch Inch Ounces

each side of zero

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 dy-

namic 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 read-

ing of the first operation. This gives the distance of the

equivalent static unbalance from the dynamic pivots, meas-

ured 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. For positive location read-

ings leave the pointer in its original position. The fol-

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 slid-

ing pointer on the machine. ThJs. gives the line on which

metal must be removed in order to correct the static unbal-

ance. 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 longi-

tudinal 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 magni-

tude of the remaining component of dynamic unbalance. The

bed is left on the dynamic pivots. 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 head-

stock. 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 locat-

ing 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 un-

balance 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' line into coincidence with

the fixed line on the headstock. 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 vibra-

tions.

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

D

F

m ir t

"

G

c hahi

m B

E

A h K

L

Fig. 20.

am2 r s"'

-49-

test piece are fixed in relation to it and therefore their

planes rotate with it at a uniform velocity. P Q, the bal-

ancing spindle, has a unity gear ratio with A B which repre-

sents the centre line of the test piece. LGHK is a refer-

ence plane while ACDB contains S the static unbalance, and

AEFB contains M the dynamic unbalance. L and K represent

the static pivots. The angle 9 between ACDB and AEFB is

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. The tendency for the bed to

rock about L K is therefore entirely due to S and the centri-

fugal forces of the balance weights m, and m2. 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. The parallel component will

cause no moment. The perpendicular component whose magnitude

is m r i t' Sin (0 +oc ) will cause a moment ml r, 4)g g

Sin (0 +C() x hl. Similarly, M 2 will cause a moment of

-50-

opposite sense equal to mT r 2 Sin-($ -oC ) x h2 .

By adding these three results with proper regard to

signs, we find that the total disturbing couple is

S Sin 4xh -m ri Sin (#4<X)xha - mr _02

g g

Sin (0+0C) x h2

In the machine, m1 ri = m2 re and we may therefore

write mi ri me - m re &) = F and the total moment be-g g

comes:

S.Sin 4xh - 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 CoscB 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-

when the coefficient of Sin ( 6 - ) is a minimum, i.e.,2 2

when (A - B Cosoc) + (B Sinoc) is a minimum.

(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 F2 4)

M.R. N x h = m.r.) (hi-h2 ) where M.R is theg 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-h2 )'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

Hmrwl

D r

h -h

mre

M C s 9

C B

/ mE / M Sin

9004

v1

L

G

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

(hi-h2) sinceo is zero. The resultant is therefore:

S Cos $ x 1 + M Cos (+9)- F Cos 0 (h2 -h2 )

It is clear that this expression cannot be zero for all

values of $ unless M is zero or 9 is zero. Let us consider

the effect of varying (hi-h2 ). The moment above may be

written:

(A-B) Cos 0 + M Cos ($+9) where s.1 = A, a con-

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 $ Cos Q

- M Sin $.Sin 9

(A-B+M Cos 9) Cos #

- M Sin @.Sin 5.

This may be written as

(A-B+M Cos 9)2+ (M Sin 9)2 Cos (#+ 8), where 8 is

given by Cot e = A-B+M Cos 9 . The above expression showsM 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 g)2 is a min-

imum. 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,

mrW2 14 R L2 MiRib.2g (hi-h) = x 1 + x Cos 0

Where MiRi is the weight radius of the dynamic couple

and x is its arm. Dividing by 4)28

mr (hi-h2 ) = M R.1 + MiR2. x Cos 9

or 1 + M R3 x Cos 9 _ mr (h-h).2 )M R M R

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 inh

the first operation. Thus for any reading K on the dial,

K = mr (hi-h2 ) or mr (hi-h2 ) = K.h. so that:h

1+ MiRi x Cos 9 = K.h In order to finM R M.R

1 + M.Ri x Cos 9 Which is the position ofM R

the equivalent static unbalance defined in Chapter I,

the reading obtained in the second operation must be mu

plied by h (f or this machine h = 20") and divided by M.

which is the reading obtained in the first operation.

d

lti-

R

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 hand-

wheel, 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. The value of the train is

i8 24 _1

^ = 20 The helical gearsA,, shown in Fig. 29.. have72 120 20

* Page 84.

8-55-

each 12 teeth giving unity velocity ratio between the shafts.

To find the pitch of the screw cut on;S,the following read-

ings were taken. The distance d between the fixed ptPr d d

the back end of the forked piece. Q was measured by a

caliper. 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 = .250" very5

nearly.

One revolution of the pointer on the dial corres-

ponds 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 head stock cover. The hand wheel was then rotated until

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

3.625". If Di is the pitch diameter of the gear on the com-

pensating spindle, since one turn of the hand wheel pro-

duces 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:

p = 2?Ni where Ni is the number of teeth,Li

in this case 15.

p = N = 7rx 15Li fr x 1.59 Sin K3.

D2 and 'c 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

-57-

Dp + D1 = 3.625" so that D2 = 2.035".

Hence = 520, 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 train a total value 18 28 _ . The static112 120 80

dial reads 16 in. ounces per revolution 80 rev olu-3

tions of the hand wheel, so that one inch ounce on the

dial is equivalent to 80 = 5 revolutions of3 x 16 3

the hand wheel. The dynamic dial reads up to 320 oz.in.2

so that 1 oz.in.a is equivalent to 3803 1 revolu-

tion of the hand wheel. 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 separa-

tion 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.

6 inch ounces on dial = 6 x = 10 revolutions on3

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 un-

balance. 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 po-

sition where the unmarked line coincides with the fixed line

on the headstock. Fig.23.,represents a vector diagram of

LCx OD

Y

x

v-0 -o

H

0 -*,T3g

i

-"59-

the tweight radius f of each weight, and the resultant of

the two.

Mr

X R=2mr Sin2

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 m r Sin2 2

and its direction is - above the horizontal line. To bring2

its direction horizontal requires a rotation R of X:. When2

this is done, the resultant weight radius lies in the same

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 elimin-

ate 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 result-

ant weight radius of the static compensator.

2 m r Sin - x 1 = K x 20 where 1 is the dis-2

tance of the centre of the compensator rings from the dy-

namic 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 measure-

ment with a flexible steel scale.

Reading 1 = circumferential distancefrom zero mark

0 0

4 1.0156"

9 2.3125"

12 3.148"

15 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 0.= 1 x 180 inr g 3T

degrees.

Reading

00

4

9

12

15

23

24

19.4

44.3

60

77.8

146.3

180

The following is a

Reading

0

4

9

12

15

23

24

tabulation of Sin .9 *.2Sin

20

0.1685

0.377

0.5

0.628

0.957

1.

If the condition

9fille d, Sin

K

2 m r Sin x 1 =2

K x 20 is to be ful-

must be a constant. This is true,, for:

.16854

.3779.512

*62815

.95723

124

.042

.042

= 042

= .042

.042

= .042

-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 202

it is possible to find the value of m r since

Sin 9K 7 = .042.K

m r 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. The centers

of the holes lie on a circle of radius 2 1", the outside16

-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 R Sin 9 = 0

or 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

makes with A B.

Thus m r = M R (2 Sin 0 + 2 Sin 150 + 2 Sin 300 +

+ 2 Sin 750 + Sin 90")

Assuming a density of 0.28 lbs. per cubic inch,

12x (1 x 1 x 0.28 x 16 ounces

R = 211 inches16

m r = E x (1)2 x 1 x 0.28 x 16 x 44_Sin 94 32 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 ob-

tained 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 dy-

namic 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 there-

fore 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 second operation

can be fairly easily determined. If however the component

dynamic unbalance is .large compared with the moment of

the equivalent static unbalance, the minimum vibration con-

dition may be hard to determine accurately. In such a case

it 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 mul-

tiplied 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. If the two results do not

check, it 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 jour-

nals 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 bear-

ings therefore constitutes a test for straightness of the

shaft.

If the vibrations in the various balancing opera-

tions 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 elim-

inating the necessity for a static compensator. It has

been shown that after the second operation the only unbal-

ance 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-

vibrations are entirely eliminated. The readings on the

dials will give sufficient information to enable all unbal-

ance 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 will be constrained to rotate at the

same speed as G. 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 neces-

sary 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 auxil-

iary 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

300

.600

900

120*

150*

18004

2100

24060

225*

1

1-1/10

1-1/10

1

9/10

7/10

5/10

3/10

2/10

1/10

In general it is quicker to advance by 90' at a

time, when the approximate correct position can be estimat-

ed 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-

angular setting. Results:

Headstock Unbalance Amplitude of Vibration

0 4/10

1/8 4/10

1/4 3/10

3/8 1/10

1/2 1/10

7/16 0

Thus the amount of static unbalance is 7/16 in.oz. at 2250

angular setting. 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.

Headstock Unbalance Amplitude of Vibration

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 is

1/2 x20 = 23" towards the headstock.7/16

-72-

This is beyond the limits of the piece and there-

fore 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" diam-

eter. 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. Each weighs about

18 lbs. and the shaft weighs 43 lbs. 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. When no distance is given for one of the bear-

ings, it was not used in that particular test.

Distances of Bearings

Test No.1 No.2 No.3

1 6-1/21" 30" 45"

2 12-1/4" 26" 40-1/2"

3 9-1/4" 22-1/41" 36"

4 6-1/2" 30 " -

5 6-1/2" - 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 LocationPosition Reading

1 1/4 2450 0

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 _ 20 327/32 16 x 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 unbal-

ance without absolute correction. A large number of states

of unbalance can be obtained by altering the relative longi-

tudinal 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

G2 = angular location of counterweight.

Then R = K2 + (23.7)2 + 2 K x 23.7 Cos (9 2 -180 0 -9 1 )

The test results are given below.

Flange No. K 01 92 R

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 accurate-

ly 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

flange from the dynamic pivots.

Test Position Static Location

15"

10"

5"

0

20

Unbalance

6-3/8

6-1/2

6-1/2

6-3/8

6-1/4

Reading

3-5/8

2-1/4

1

- 1/2

6-3/4

of the

Location

11.4"

6.9

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 posi-

tion 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 Static Location Locatio

15"

1511

Unbalance6-1/4

6-1/4

6-1/8

Reading5-1/8

3-5/8

4-1/16

n

16.4"

11.6"

13.3"

*Piece reversed.

1

2

3

4

5

*6

*7

8

F I: I

I ~1~~ IIII

It I

I -r--'--~- 2112I. -

_____________ _____________ t

I I

______ ______ I TT I -

K- I I ~--F---~ I-"- ~H-~"m-I I I

I F I - -

'T -~

__________ - r~r~r

I I I

I II I

I

_____

II 1___

77 72 ~ Liii '-~ -

- -,--.-----.------------ I Ii

F I t--. FI L - I I

--------------------------------------------------------------------------t ~-~------r--f----*--~- -~ -

____ -- -~ ~-------~-- F

-~ - --

-71

_____ _____ _____ _____ _____ _____ _____ _____ --1 ----t -~ ___ ___ ___ ___ ___ ___ ___ F ___

__________ __________

~~1*- II I F

F F II -

-F- 0o - - -- - F

-lot' I

fUOfNf OIETZQ~N Cu.. C1IICAeO N~W YORK Nc. ~46 A

IUF~ FITeNC. HCG E OKNl 4

of FtAA

Fig, 26.*

-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 un-

balance of flange No. 2. Results are tabulated below:

Test Position Static Location LocationUnbalance Reading

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"

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 prob-

able distance of the unbalance from the machined face is

2.4 + 3.92 or 3.1". The next series of tests gave a con-

siderably lower value, and the average of the two is approx-

imately the correct value.

*Piece reversed.

.....

...

4 4

-It

4I

03

L

4_

__

_ 44

'-'aa

t

--

K .1

-79-

Test Position Static Location LocationUnbalance Reading

*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

1.2 + 2.6 = 1.9" from the machined face.2

The results of these tests were so unsatisfactory

that the accuracy of the machine was checked in the follow-

ing 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 x 20 = 17.4". The7-3/4

actual position was 19.1" and therefore the machine read in-

accurately. The initial compression of the end springs

was then reduced and a further series of tests run on flange

No. 2.

* Piece Reversed.

L I4

L

I

04__

-_ ___

- I A--

~*----*

I"j

- ----

CD

~ i

7K

-..

.....

to~1

~~-_

__

~ -~

II _ _ _ t

____

IT

ZE

C.

CICj4eO

,-NE

YOKN

t.34

-80-

Test Position Static Location LocationUnbalance Reading

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 Location LocationUnbalance Reading

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".

The average value for location measured towards the

hub from the machined face is 0.2 + 0.3 - 0.3 = zero, very3.

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 un-

balance 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 unbal-

ance 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 Location Componentunbalance Dynamic Unbalance

No. 1 6-1/4 1.6" 0

No. 2 4 2.5" 0

No. 3 8-5/8 0 0

The unbalance of flanges No. 1 and No. 3 were re-

duced 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 - x 1 X x r4 Cx x

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

14 M RSd 2( r

d = 3/4" r = 1 - 3 / 8 "

' = 4 ozs. per cu. in.

YR = 2-1/4 in. ozs.

. . 1 = 0.931.

For flange No. 3

r =4 M R

, -d2d = 3/4 " 1 = 3/4"1

MR = 4-5/8 in.ozs.

. .r= 3.511

In the previous tests no account was taken of th4

angular location of the flange relative to the shaft. Si

the unbalance of the shaft is 1/4 in.oz. an error was int

duced 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 Static Location Loca

Flange Position Reading Unbalance Reading

1 0 3-3/4 4 - 5/16 -1

*1 Q 3-3/4 4 + 3/8 + 1

2 0 3-7/8 4-1/8 -1/2 -2

3 0 3-5/8 3-7/8 1/4 1

*3 0 4-3/8 4-1/8 -3/8 -

ace

ro-

tion

.611

.9"

.4"1

.3 t

211

* Piece Reversed.

-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 Location

1 4 1.75"

2 4-1/8 2.4"?

3 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. read-

ing on the static dial, provided the machine is properly

adjusted.

The Inside of the Headstock

Fig. 29.

-85-

Side View of Machine End View of Machine

Headstock

Balance Weights Separated Balance Weights at Zero

Headstock with Cover Removed

Fig. 30.

-86-

BIBLIOGRAPHY

1. Mechanics of Machinery, Ham and Crane

2. Vibration Problems in Engineering, Timoshenko

3. Aeronautics, Wilson.