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TECHNICAL REPORT ARLCD-TR-81039

COMPUTER SIMULATIONS OF ARTILLERY S&A

MECHANISMS <INVOLUTE GEAR TRAIN AND

PIN PALLET RUNAWAY ESCAPEMENT)

G. G. LOWEN CITY COLLEGE OF NEW YORK

NEW YORK, NEW YORK

F. R.TEPPER ARRADCOM

JULY 1982

~~ ~~ ' , US ARMY ARMAMENT ~~~~C~A~~~E~EVELOPMENT COMMAND '\ ,~w., . . WEAPON SYSTEMS LABORATORY , ' DOVER, NEW JERSEY

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APPROVED FOR PUBLIC RELEASE: DISTRIEUTION UNLIMITED.

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The views, opinions, and/or findings contained in this report are those of the-author(s) and should not be construed as an official Department of the Army position, policy ,·or decision, unless sc designated by other documentation.

Destroy this report when no longer needed. Do not return to the originator.

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BEFORE COMPLETING FORM1. REPORT NUMBER 2. GOVT ACCESSION NO. 3. RECIPIENT'S CATALOG NUMBER

Technical Report ARLCD-TR-81039 - .4934. TITLE (wnd Subtltle) S. TYPE OF REPORT & PERIOD COVERED

COMPUTER SIMULATIONS OF ARTILLERY S&AMECHANISMS (INVOLIUTI: GEAR TRAIN 6. PERFORMING OR. REPORT NUMBERAND PIN PALLET RUNAWAY ESCAPEMENT)

7. AUTHOR(a) S. CONTRACT OR GRANT NUMBER(&)

G. G. Lowen, City College of New YorkF. R. Tepper, ARRADCOM

S. PERFORMING ORGANIZATION NAME AND ADDRESS 10. PROGRAM ELEMENT, PROJECT, TASKAREA & WORK UNIT NUMBERS

ARRADCOM, LCWSLNuclear and Fuze Div (DRDAR-LCN)Dover, NJ 07801II. CONTROLLING OFFICE NAME AND ADDRESS 12. REPORT DATEARRADCOM, TSD July 1982STINFO Div (DRDAR-TSS) '3. NUMBER OF PAGESDover, NJ 07801 20314. MONITORING AGENCY NAME & ADDRESS(If dillerent from Controlling Office) IS. SECURITY CLASS. (of this report)

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17. DISTRIBUTION STATEMENT (of the abstract entered In Block 20, It different from Report)

18. SUPPLEMENTARY NOTES

19. KEY WORDS (Continue on revere. aide If nececemy and Identify by block number)

Involute gears Pin pallet escapementM125AI booster RotorM577 fuze Safing and armingArtillery Timing

20. AGrlACT (Cauihue II e en Idanllyby block nbw& )

Computer simulations of complete safing and arming mechanisms containing pinpallet runaway escapements, two- or three-pass involute gear trains and spindriven rotors were developed. The three motion regimes of the runaway escape-ment, i.e., coupled motion, free motion and impact, were included. In addition,the ability to consider the effects of pallets with arbitrarily located centersof mass has been introduced. Further, the contact forces of the gear trains andthe escapement may be obtained. Simulation runs of actually existing mechanismswere made and good agreement with exverimental data was obtained.

D J 1473 wmorow Or * NOV as IS OssOLETE Unclassified

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CONTENTS

Page

Introduction 1

Description of Computer Programs SANDA3 and SANDA2 I

Program SANDA3 6Program SANDA2 12

Computer Simulations of Example Mechanisms *1Program SANDA3 15Program SANDA2 21

Conclusions 27

References 28

Appendixes

A Dynamics of Rotor Driven S&A Mechanism with a Three-PassInvolute Gear Train and a Pin Pallet Runaway Escapement 29

B Dynamics of Rotor Driven S&A Device with a Two-PassInvolute Gear Train and a Pin Pallet Runaway Escapement 75

C Computer Program SANDA3 109

D Computer Program SANDA2 141

E Friction Forces During Gear Train Motion Reversal 179

P A Computational Rule for the Determination of theSign in the Effective Moment of Inertia IiR 187

Distribution List 195Accession ForNTIS GRA&I

DTIC TABUnannounced 0Justification

Distribution/

Availability Codes

Availi a nd/or__Dist Spocial

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FIGURES

1 Rotor driven S&A device with three pass step-up gear train,configuration no. I

2 Rotor driven S&A device with three pass step-up gear train,configuration no. 2 3

3 Rotor driven S&A device with two pass step-up gear train,configuration no. 1 4

4 Rotor driven S&A device with two pass step-up gear train,configuration no. 2 5

5 Flow chart for simulation of pin pallet runaway escapement 7

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INTRODUCT ION

Computer simulations for complete safing and arming (S&A) mechanisms weredeveloped. Figures 1 and 2 show the essentials of two different configurationsof such mechanisms, consisting of spin driven rotors, three pass involute step-upgear trains, and pin pallet runaway escapements. Figures 3 and 4 illustrate twoconfigurations of similar mechanisms which contain two pass involute step-up geartrains.

The present work draws on ideas developed previously by the authors. Adynamic model of a pin pallet runaway escapement driven by a constant torque wasdeveloped first (ref 1). The ideas of pivot and tooth-to-tooth friction and theresultant etficiencies of multi-mesh gear trains were then formulated (ref 2),and subsequently, the computational technique of describing mesh initializationat arbitrary points was derived (ref 3).

This report reflects many of the ideas originated in references I through3,1 in describing combined mechanism trains which are driven by the variabletorques of centrifugal rotors. Again, the three motion regimes of the runawayescapement, i.e., coupled motion, free motion, and impact, are considered. Inaddition, the ability to include the effect of a pallet which has an arbitrarilylocated center of mass has been introduced. Further, all non-impact contactforces can now be determined for strength considerations.

The complete and detailed derivations of both mathematical models and theirconfiguration variations are given. The associated computer programs are de-scribed in detail, with instructions for running them. These programs were ex-tensively tested, both for the configuration no. 1 data (fig. 1) of the threepass M125A1 booster S&A as well as for the configuration no. 2 data (fig. 4) ofthe two pass M577 SSD mechanism. The M125A1 booster simulation armed in 45 turnsfor average friction conditions at various spin rates. Under similar conditions,the simulation of the M577 551) required 31.7 turns for its arming cycle.

DESCRIPTION OF COMPUTER PROGRAMS SANDA3 AND SANDA2

The essential steps of programs SA.NDA3 and SANDA2, which are listed in ap-pendixes C and D, respectively, follow. This discussion is based on the computerprogram description in reference 1, which deals with a runaway escapement drivenby a constant torque. For the sake of clarity the description in reference 1 isrepeated and, whenever necessary, changes in the programs due to the presence ofthe gear trains and the rotors are identified.

'These mdels can be adapted to spring driven mechanisms with only minor changes.

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Both programs may be used for each of the two possible mechanisms configu-rations. To differentiate between these configurations, a special input para-meter CONFIG must be provided.

The flow chart of the pin pallet runaway escapement part of the programs, isshown in figure 5 which was taken from reference 1. The choice of variable des-ignations was made so that they differ as little as possible from the nomencla-ture used in the various derivations given in the appendixes of the present re-port as well as in appendixes A through F and H of reference I.

Both main programs start with the reading in and writing of the configura-tion type as well as of all relevant physical data. This is followed by thecomputation of gear ratios (app A and B), fuze body angles of the applicableconfiguration, with the help of signum function s6 (app A and B anc ref 2),centrifugal forces (ref 2), earliest and latest possible values of the gear an-gles by way of subroutine ALFA (ref 2), and the initialization of the gear angles(ref 3). In each case, the simulation begins with entrance coupled motion at astarting angle PHID, which represents the angle 0 of the escape wheel which isassociated with the approximate center of the escape wheel tooth. This anglecorresponds to the cumulative escape wheel angle PHITOT of 0 ° .

Program SANDA3

Coupled Motion (Location 1)

To solve the differential equation of coupled motion (eq A-185, ayp A),the main program calls on an available fourth order Runge-Kutta routine. rhemain purpose of the subroutine FCT is to present the second order differentialequation in terms of two first order equations to RKGS. PHI(1) and PHI(2) repre-sent the angle 0 and the angular velocity ; , respectively. The computation ofall parameters of the differential equation takes place in subroutine FCT as wellas in subroutines IN3 and KINEM.

The subroutine KINEM computes current values of g, g, 4i and y, includ-ing the moment arms Al, Bj, C1, and Di (which are primed in the presentreport). The associated expressions for the above values are cited in reference1.

Subroutine IN3 computes various gear mesh parameters and instantaneousmesh contact angles, as well as the signum functions sI to 85. In addition, the

parameters A, to A5 3 , listed in appendix A, are obtained. (Note that the programuses the symbols AA, etc., and that the parameters A 5 4 to A 5 3 are computed insubroutine FCT. This division is dictated by the limited number of argumentspermitted in any one subroutine.) Finally, the gear indexing operation (ref 2)is performed with the help of the angle *.

2 RKGS Routine, IBM System/360 Scientific Subroutine Package, (360A-CM-OX3)

Version III.

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The manner in which the instantaneous rotor angle, C + N4 1 T' of tilecoupled motion differential equation (A-185) must be expresset in subroutine FCTwill now be discussed. Recall that is the initial rotor angle, 4T is the

total angle of rotation of the escape Wheel, and N 4 1 stands for the gear ratiobetween the rotor and the escape wheel. Since the angle *, and consequently theRunge-Kutta variable PHI(I), varies between 1400 and 155* during entrance andbetween 1850 to 2000 during exit coupled motion, the total escape wheel angle

T can only be obtained by continuously adding the increments due to each cycle

of Runge-Kutta computations. Thus

T = TOT + A (1)

where

41'0T total escape wheel angle up to a givenRunge-Kutta cycle. This is representedby PHITOT in the program.

= increment of escape wheel during a given Runge-Kutta cycle.

This increment, A , is calculated as the difference between the latest value ofPil(1) and its previous value, which has been stored as PHIPR. In this manner,equation I becomes

T PHITOT + P1tI(O) - PHIPR (2)

Subroutine FCT also decides on the values of I and IIR as required by

equations A-88 and A-89 as well as equations A-155 and N-156, respectively. Theassociated conditional statements assign the larger values for these combinedmoments of inertia whenever the product of the angular velocity and the angularacceleration is positive, i.e., both quantities have the same sign (app F and

paragraph preceding eq A-148).

The associated subroutine OUTP is responsible for printing out theresults *, $, and * together with the current values of time, g, g, p, $, andPHITOT. Further, all coupled motion contact forces are calculated according to

equations A-192 to A-196, and the maximum values of these forces during one rotorcycle are determined. Finally, this subroutine decides, according to the follow-ing criteria, whether coupled motion is to be continued. These are:

1. The distance g < 0 (ref 1, eq A-12) and because of the natureof the coordinate system g is always negative while the pallet pin makes contactwith the escape wheel tooth, and

2. The contact force P ) 0 (eq A-196)n

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The introduction of the contact force between the escape wheel and the palletallows the use of simpler controls than those given in reference 1.

Thus, when control is returned to the main program, it is either be-cause the pallet pin has left the end of the tooth and there is no further possi-bility of coupled motion, or the pin has disengaged from the inside of thetooth. In either case, free motion results and control is eventually shifted tothe subroutines FCTF and OUTPF through RKGS (location 5). This is done directlyif g < 0. If g > 0, the main program must decide whether the preceding computa-tions have been made for entrance or for exit action and whether the next contactwill occur in enrance or in exit. In the sample mechanism, g = 0 when 0 isapproximately 1550 for entrance action and approximately 2000 for exit action.If g )0 and < 160 0 , all possibility for entrance contact is ended and 0 mustbe incremented by the tooth angle 6 (ref I, figs. 2 and 5), while *must be in-cremented by the angle (2 w - X) in order to get the initial angles for exitmotion. For g ) 0 and 0 > 1600, entrance contact is expected and 0 must bedecremented by the angle 2 6 (ref 1, fig 4, where the new entrance tooth no. 3comes into action). At the same time, the pallet angle P must be decremented by(-2w + A) for the initial angle of entrance motion. This shift from one side of

the pallet to the other must also be observed in connection with the pallet cen-ter of mass angle C (figs. I and 2). During entrance action,

* fi PSICC (3)

in the program, while during exit action

*C X + PSICC (4)

These indexing operations due to pallet position occur both in coupled and infree motion. They have no effect on the continuous computation of the cumulativeescape wheel angle PHITOT.

Free Motion (Location 5)

The differential equations of free motion, as given by equation A-197for the pallet and equation A-203 for the combined escape wheel, gear train androtor, are again solved by the Runge-Kutta routine. In order to obtain the mag-nitudes of the variables 4 and q and their derivatives at identical times, thetwo independent second-order differential equations are transformed into foursimultaneous first-order equations. (While only the two first-order equationsassociated with each of the two variables are actually coupled, the routinetreats all four as if they were coupled and thus produces solutions for identicaltime increments.) These four expressions, which are presented in subroutine FCTFare given in the following form.

DX(l) - X(2) ( $) (5)

9

DX(3) = X(4) (= ,) (6)

1 2DX(2) = A64 [-A 6 5(X(2)) + A6 6X(2) + Ab./

+ A68sin( 1c + N41( T + X(1) - PHIPR))

S (fi) (7)

1 2DX(4) = A [-A 2 ,(X(4)) - A6 1X(4) -Ab2A6 0

+ A6 3sin(y-(= (8)

Subroutine FCTF also computes the parameters A6 0 to A68 and calls on subroutine f

IN3 for the computation of all gear related parameters.3

The associated subroutine OUTPF computes the free motion contact forcesaccording to equations A-209 through A-211, and finds their maxima, after deter-mining the parameters A6 0 to A68. The computation of these latter variables also

requires a call on subroutine IN3. In addition to the above, a continuous countof PHITOT is provided in OUTPF. This angle, time, *, ;, qp, p, and the contactforces are printed. This subroutine also makes the decision of whether or not toremain in free motion. The sensing variables, f and g - = GP (ref I, eq D-7 andD-1i), are used for this purpose.

If f > 0 and g ( 0, free motion is continued without indexing.If f > 0 and g' > 0, free motion is also continued, but since contact is nolonger possible for the component pair for which the previous computations weremade, indexing must take place. This is accomplished in the same manner as incoupled motion.

If f 0, control is returned unconditionally to the main program. Ifit finds that g > 0, indexing takes place and control is given back to the sub-routine FCTF. If gA < 0, contact is about to take place or has just occurred.This program must decide whether this contact represents just a close approachwhich is followed by further free motion, an impending impact, or the beginningof coupled motion. To this end, the quantities Vp and Vs (ref 1, eq F-22 and F-23), are computed for the entrance and exit free motion tests.4 The first three

3 Whenever A60 - IPR - 0, the simulation stops because of division by zero.Should this occur, FCTF prints "IPR EQUALS ZERO - SIMULATION TERMINATED."

4 Under the present circumstances, if * < 1600, only entrance contact can follow;if * > 1600, exit -ontact will occur.

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cases of the entrance free motion test of the main program are illustrated infigure 6. With both angular velocities ( * = PHI(2) - X(2) and - DPSI - X(4))positive, the following three possibilities exist:

1. If I vp > J VS , the contacting surfaces will separate again,free motion will result, and control must be transferred to subroutine RKGS (lo-cation 6), which calls free motion subroutine FCTF.

2. If IV = IVs , the escape wheel will start driving the pal-let in coupled motion, d control must be transferred to subroutine RKGS (loca-tion 1 initiates RKGS), which in this case, calls coupled motion subroutine FCT.

3. If IV I < IV I , impact will occur, and control must be givento subroutine IMPACT (location 1%).

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and ; are positive. Distance f is enlarged.)

Figure 6. Entrance free motion test

Impact (Location 15)

The subroutine IMPACT uses the current values of the.angular velocitiesad and computes the post impact angular velocities 0.and hf*applyingequations F-20 and F-21, ref I. (The moment of inertia of the escape wheel is

now expressed according to equation A-212, appendix A of the present report,which refers the rotor, as well as the gear train inertias, to the escape wheelshaft. As shown in reference 1, the tangential impact has been neglected and,therefore, E2 m D, and F2 - A1 .)

After control is returned to the main program, it is decided whetherfree or coupled motion follows the impact. This is accomplished by considering

11

the post-impact contact point velocities, Vp and V., in the impact tests, whichare similar to the free motion tests.

If the contact velocities are vectorially equal to each cther, or ifthe absolute value of the difference of their absolute magnitudes is less than2.0 in./sec (5.08 cm/sec), control is transferred to coupled motion. If thesevelocities indicate a subsequent separation, which is more usual, computation istransferred to free motion.

Reversal of Gear Train Motion Due to Impact

If the impact torque on the escape wheel is sufficiently large, themotion of the gear train may be temporarily reversed (i.e., the escape wheelangular velocity ; becomes negative). This causes the friction forces betweenthe gear teeth and at the various gear pivots to be reversed in direction. (Thenormal forces between the gear teeth remain unaffected, and the normal bearingforces are solved for in the usual manner.) This change in the direction of thefriction forces is expressed both for coupled and free motion by letting thecoefficient of friction p of all gear train components become negative (app E).This is accomplished in subroutines IN3 and IN2 by the following use of the sig- [

num function */ :

MU = ABS(MU)* $/ (9)

(The coefficient of friction associated with the escapement interface and thepallet pivot is called pi (read into the program as MUI).1 Any motion reversalat these locations is accounted for by the signum functions s4 and s5 , respec-tively.

Termination of Computations

Computations are terminated whenever the geared motion of the rotorends. This corresponds to * f PHICUTD. (See Program SANDA3, General Parameters,below.) The duration of the subsequent unretarded motion of the rotor is assumedto be negligible.

Program SANDA2

Program SANDA2 is very similar to program SANDA3, and where possible, the

following description will refer to program SANDA3.

Coupled Motion (Location 1)

To solve the differential equation of coupled motion (app B, eq B-113),the main program calls on the fourth order Runge-Kutta routine. Subroutine FCTpresents the second order differential equation in terms of two first order equa-tions to RKGS, which computes PHI(1) and PHI(2) as in SANDA3. The computation ofall necessary parameters takes place in FCT and KINEN. In addition, subroutineIN2 is used. This subroutine, which differs from IN3, computes various gear mesh

12

parameters and contact angles of the two pass train. It further computes the

signum functions sj, s2, s4 , and 95 as well as the parameters A1 to A4 4, listed

in appendix B. The parameters A45 to A50 are computed in subroutine FCT. Final-ly, the gear indexing operation is performed with the help of the angle *.

The instantaneous rotor angle now becomes C+ 431, and is treatedin FCT in the same manner as shown by equations 1 anh 2. Subroutine FCT alsodecides on the values of I., and 11R as required by equations A-88 and A-89 aswell as equations B-87 and B-88. The associated conditional statements assignthe larger values for these combined moments of inertia whenever the product ofthe angular velocity and the angular acceleration is positive, i.e., both quanti-

ties have the same sign. (See discussion following eq 2.)

The associated subroutine OUTP is responsible for printing out the

results *, , and i, together with the current values of time, g, g, 4, i, andPHITOT. Further, all coupled motion contact forces and their maxima are deter-

mined according to equations B-120 to B-123. The coupled motion control criteriagiven in this subroutine are identical to those described in connection withsubroutine OUTP of SANDA3. This is also true of the indexing operations con-nected with the angles *, *, and *C"

Free Motion (Location 5)

The differential equations of free motion, as given in equation B-124for the pallet and equation B-130 for the combined escape wheel, gear train and

rotor, are also solved by the Runge-Kutta routine. The associated four simulta-neous first order differential equations, which are given in subroutine FCTF, aresimilar to those used in SANDA3. (See paragraph before eq 5.).

DX(1) f X(2) : ( ) (10)

DX(3) f X(4) : (= ) (11)

12 -A 2DX(2) [A 56(X(2)) - A 5 7X(2) + A58

-A 5 9sin( ilC + N31(tT + X(1) - PIPR))

(f) (12)

2DX(4) - [-A 2 1(X(4) ) - A 5 2X(4) - A 53A 51

+ A5 sin(y - - ] C (13)

The subroutine FCTF also computes the parameters A5 1 to A5 9 and calls on subrou-tine IN2 for the computation of all gear related parameters.

13

The associated subroutine OUTPF computes the parameters A5 1 to A59 in

order to obtain the free motion contact forces according to equations B-136 and

B-137. In addition, it determines the maxima of these contact forces. Theseoperations also require a call on subroutine IN2. Again, a continuous count ofPHITOT is provided in OUTPF, and time, *, ;, *, p and the contact forces arecaused to be printed together with PHITOT. Whether or not to remain in freemotion is decided by the same free motion tests outlined in SANDA3.

Impact (Location 15)

The treatment of the impact phase of the motion is identical to that

given in connection with SANDA3. (The moment of inertia of the escape wheel isnow expressed according to equation B-138, app B, which refers the rotor as wellas gear and pinion no. 2 to the escape wheel shaft.)

Reversal of Gear Train Motion Due to Impact

Any reversal of the gear train due to impact is treated in the samemanner as that described in SANDA3 (eq 9). The control of the coefficient of

friction, V, now takes place in subroutine IN2. The coefficient of friction

associated with the escapement interface and the pallet pivot is again calledjI and is read into the program as MUl.

Termination of Computations

Computations are terminated whenever the geared motion of the rotor

ends. This corresponds to * = PHICUTD (Program SANDA2, General Parameters,below). The duration of the subsequent unretarded motion of the rotor is assumedto be negligible.

COMPUTER SIMULATIONS OF EXAMPLE MECHANISMS

Computer simulations of modified M125A1 and M577 spin driven S&A devices arediscussed in this section. Both contain appropriately adapted involute gear

trains rather than the actual clock gear trains used.5 Program SANDA3 has been

used for the M125AI mechanism, which has configuration type no. 1. Both theprogram as well as the relevant output are given in appendix C. The simulationof the M577 mechanism, which has configuration no. 2, was accomplished with the

help of SANDA2. This program, as well as associated output, is given in appendixD. Both simulations were run for 30,000 RPM in order to obtain maximum contactforces.

5 All gears and pinions have the original number of teeth and diametral pitches.The individual meshes, with unity contact ratio, are designed to operate

at standard center distaices and have unequal addenda (obtained with programINVOLl of ref 2).

14

The data requirements of both programs, an explanation of the various outputdata, and the manner in which the number of turns to arm results are obtained for agiven spin velocity are presented below.

Program SANDA3

Input Data

The required input data is the first portion of the output of ProgramSANDA3. They represent the mechanism parameters of the M125A1 booster and arelisted below both as computer variables and as symbols.

Escapement Parameters.

A - a = 0.2061 in. (5.234 om) - distance between pivots 0p and Os (fig.)

B = b = 0.192 in. (4.877 mm) = escape wheel radius

C c - 0.0788 in. (2.001 m) - pallet radius from pivot to center of pin(identical for entrance and exit)

R = r = 0.015 in. (0.381 m) - pallet pin radius (identical forentrance and exit)

ALPHA = a - 51* - escape wheel tooth half angle

CONFIG - I - configuration no. I

EREST - C - 0 - coefficient of restitution (high

speed motion pictures of runawayescapement suggest totallyinelastic impacts)

LAMBDA = X - 152.144* - angle between entrance and exit palletradii

DELTA - S - 30* - angle between individual escape wheelteeth

For details of the above nomenclature see reference 1.

Mass Properties of Components.

Ml - ml - 0.1155 x 10- 3 lb-sec 2 /in. (2.022 x 10- 2 kg) - mass of rotor

-2 - m2 - 0.686 x 10- 5 lb-sec 2 /in. (1.201 x 1O- 3 kg) - mass of gear andpinion no. 2

K3 - m3 - 0.282 x 10- 5 lb-sec2 /in. (4.938 x 10- 3 kg) - mass of gearand pinion no. 3

15

M4 = m4 = 0.252 x 10- 5 lb-sec2/in. (4.412 x 10- 4 kg) = mass of escapewheel and pinionno. 4

MP = 0.359 x 10- 5 lb-sec2 /in. (6.286 x 10

- 4 kg) = mass of pallet

II - I = 0.1055 x 10-4 in.-lb-sec2 (0.1189 x 10- 5 kg-m 2 ) ff oi.moent ofinertia of

rotor

12 12 - 0.126 x 10- 6 in.-lb-sec2 (0.142 x 10- 7 kg-m2) moment of

inertia of gear

and pinion no. 2

13 - I = 0.308 x 10- 7 in.-lb-sec (0.348 x 10-8 kg-m2) = moment of inertia

of gear and pinionno. 3

14 - 14 - 0.268 x 10- 7 in.-lb-sec2 (0.303 x 10-8 kg-m2 ) moment of

of escape wheel

and pinion no. 4

IP 1p -0.245 x 10- 7 in.-Ib-sec 2 (0.277 x 10-8 kg-m2 ) = moment of inertia

of pallet

General Parameters.

RCI rCl - 0.106 in. (2.692 mn) - distance from rotor pivot to center

of mass

RCP rcp - 0 - pallet eccentricity

RHOP P p - 0.016 in. (0.406 mm) pallet pivot radius

RPM 30,000 - spin rate

PHICD C - 1440 - rotor angle in starting position

(fig. 1)

PS1CCD _ = 00 = eccentricity angle of pallet

(fig. 1)

PHID - 1450 - escape wheel starting angle

of initial coupled motion

simulation (for choice of

this angle, see ref 1)

16

PHICUTD = 36000 - cumulative escape wheelangle, obtained from productof total rotor rotation

and gear train ratio. Thetotal rotor rotation for theM125AI is 76° while the gear

ratio is 47.25. Thus, PHICUTD- 76 x 47.25 - 36000.

MU = U f 0.15 = coefficient of friction ofgear train (pivots and tooth-to-tooth contacts, includes escape

wheel pivot)

MUI = - 0.15 = coefficient of friction on pallet-escape wheel interface and palletpivot

Gear Parameter.

PSUBD1 = Pdl 44 diametral pitch of mesh no. 1(rotor and pinion no. 2)

PSUBD2 = Pd2 - 65 - diametral pitch of mesh no. 2

(gear no. 2 and pinion no. 3)

PSUBD3 - Pd3 - 77 - diametral pitch of mesh no. 3(gear no. 3 and escape wheelpinion)

NGI W NGI = 42 - number of teeth of rotor (fullgear no. 1)

NG2 M NG2 - 27 - number of teeth of gear no. 2

NG3 = NG3 - 27 = number of teeth of gear no. 3

NP2 - Np2 - 8 - number of teeth of pinion no. 2

NP3 = Np3 - 9 - number of teeth of pinion no. 3

NP4 - Np4 - 9 = number of teeth of pinion no. 4

(escape wheel pinion)

CAPRPl - Rpl - 0.47727 in. (12.1226 mm) - pitch radius of gear no. 1(rotor)

CAPRP2 - Rp2 - 0.20769 in. (5.2753 mm) - pitch radius of gear no. 2

CAPRP3 - Rp3 - 0.17532 in. (4.4531 mm) - pitch radius of gear no. 3

17

RP2 = - 0.09091 in. (2.3091 mm) = pitch radius of pinion no. 2

RP3 W rp3 = 0.06923 in. (1.7584 mm) = pitch radius of pinion no. 3

RP4 = rp4 - 0.05844 in. (1.4844 n) - pitch radius of pinion no. 4(escape wheel pinion)

THETA = 01 = 200 = pressure angle of mesh no. I

THETA2 - 02 = 200 = pressure angle of mesh no. 2

THETA3 - 03 - 200 = pressure angle of mesh no. 3

R1 = I = 0.225 in. (5.715 mm) = distance of rotor pivot fromspin axis

R2 - 6?2 = 0.436 in. (11.074 mm) = distance of pivot of gear andpinion set no. 2 from spin axis

R3 6? = 0.504 in. (12.602 mm) = distance of pivot of gear andpinion set no. 3 from spin axis

R4 = 614 - 0.520 in. (13.208 mm) = distance of pivot of escapewheel from spin axis

R5 6 6i5 = 0.433 in. (10.998 mm) = distance of pivot of palletfrom spin axis

RHOl P1 - 0.062 in. (1.575 mm) = pivot radius of rotor

RHO2 = = 0.025 in. (0.635 mn) = pivot radius of gear and pinionno. 2

RH03 - 03 - 0.018 in. (0.457 mm) = pivot radius of gear and pinionno. 3

RHO4 = P4 - 0.016 in. (0.406 mm) = pivot radius of escape wheel

CAPRB1 - Rbl - 0.44849 in. (11.3916 mm) - base radius of gear no. 1

CAPRB2 = Rb2 - 0.19517 in. (4.9573 mm) = base radius of gear no. 2

CAPRB3 = Rb3 a 0.16475 in. (4.1847 n) = base radius of gear no. 3

RB2 - rb2 - 0.08543 in. (2.1699 mm) - base radius of pinion no. 2

RB3 = rb3 - 0.06506 in. (1.6525 mm) = base radius of pinion no. 3

RB4 rb4 - 0.05492 in. (1.3950 mm) - base radius of escape wheelpinion

CAPROl Rol - 0.48791 in. (12.3929 mm) = outside radius of gear no. 1

18

- o e r u o

CAPR02 = Ro2 = 0.21579 in. (5.4811 mm) = outside radius of gear no. 2

CAPRO3 = Ro3 = 0.18216 in. (4.6269 mm) =outside radius of gear no. 3

R02 = ro2 0.110 in. (2.7940 mm) = outside radius of pinion no. 2

R03 = r.3 = 0.08089 in. (2.0546 mm) = outside radius of pinion no. 3

R04 = ro4 = 0.06828 in. (1.7343 mm) = outside radius of escape wheelpinion

J1 = J = 0 = initialization parameter for meshno. 1. (The value of zero

corresponds to earliest possiblecontact of mesh, ref 3.)

J2 = J2 = 0 - initialization parameter for meshno. 2

J3 = J3 = 0 initialization parameter for meshno. 3

Output Data

Fuze Geometry. The angles BETAID = 13 to BETA4D = 134 and GAMMA21) = Y2 toGAIMA5D = Y5 are printed for checking purposes.

Coupled motion. For each time increment T of the coupled motion, thefollowing variables are evaluated:

PHI = instantaneous escape wheel angle (deg)

PHIDOT = $ escape wheel angular velocity (rad/sec)

C = g = pallet - escape wheel contact position (in.), ref I

GDOT = g = time rate of change of g used to determine thedirection of relative sliding (in./sec), ref 1

PSID - pallet angle (deg)

PSIDOT = pallet angular velocity (rad/sec)

PHITOT - - cumulative escape wheel angle (deg)

F34 = F34 normal contact force of gear no. 3 on pinion

no. 4 (lb)

F23 = F2 3 - normal contact force of gear no. 2 on pinionno. 3 (ib)

19

F12 F12 = normal contact force of gear no. I on pinionno. 2 (ib)

PN = n normal contact force between escape wheel andpallet (lb), computed according to equation A-195

PNPSI P = normal contact force between escape wheel andpallet (Ib), computed according to equation A-19b

(serves for checking)

DPHI2 = escape wheel acceleration (rad/sec 2), Runge-Kutta output

Free Motion. For each time increment T of the free motion, the following

variables are evaluated:

Pll = = instantaneous escape wheel angle (deg)

PHIDOT = = escape wheel angular velocity (rad/sec)

PSI = pallet angle (deg)

PSIDOT = pallet angular velocity (rad/sec)

PIIITOT = = cumulative escape wheel angle (deg)

FF12 = FF12 = normal contact force of gear no. I on pinionno. 2 for free motion (ib)

FF23 FF23 = normal contact force of gear no. 2 on pinionno. 3 for free motion (lb)

FF34 = FF34 = normal contact force of gear no. 3 on escapewheel pinion for free motion (lb)

Impact. Just preceding the "IMPACT" label, the program prints the valuesVP and VS, which stand for the pre-impact velocity components, normal to the escapewheel tooth, of both the pallet and escape wheel contact points (ref 1). Subsequentto the "IMPACT" label, the following variables are evaluated:

PHI = 4 - instantaneous escape wheel angle (deg), same asbefore impact

PHIDOT = 4 = post-impact escape wheel angular velocity (rad/sec)

PSI = 4 = pallet angle (deg), same as before impact

PSIDOT = -, post-impact pallet angular velocity (rad/sec)

PHITOT = cumulative escape wheel angle (deg), same as

before impact

20

VP = Vp post-impact normal velocity component of pallet(pin) at contact point (ref 1)

VS = Vs post-impact normal velocity component of escapewheel tooth at contact point (ref 1)

In the present program VP is equal to VS since the coefficient of restitution iszero.

Turns-to-Arm and Maximum Contact Forces. The number of turns-to-arm isobtained by the program for that time T3 6 0 0 which corresponds to the escape wheelangle PHICUTD = 36000. Therefore

RPMNumber of turns-to-arm = 60 X T3600

Thus, with T3600 m 0.09068 second (the actual time corresponds to 3600.040) and30,000 RPM

Number of turns-to-arm 30,000 X 0.0906860

= 45.34 turns

The maximum non-impact contact forces for the total cycle, both for coupled and forfree motion, are listed at the end of the output.

Program SANDA2

Input Data

The required input data represent the first portion of the output ofprogram SANDA2. They stand for the following mechanism parameters of the M577 SSD:

Escapement Parameters.

A = a = 0.2097 in. (5.326 mm) = distance between pivots OP

and 0S

B = b = 0.193 in. (4.902 mm) = escape wheel radius

C = c = 0.0788 in. (2.001 mm) = pallet radius from pivot tocenter of pin (identical for

entrance and exit)

R = r = 0.015 in. (0.381 mm) = pallet pin radius (identicalfor entrance and exit)

ALPHA = a - 52* = escape wheel tooth halfangle

21

A&

CONFIG = 2 - configuration no. 2

EREST = E = 0 = coefficient of restitution(high speed motion picturesof runaway escapement

suggest totally inelasticimpacts)

LAMBDA = X = 152.1440 = angle between entrance andexit pallet radii

DELTA = 6 = 30 * = angle between individualescape wheel teeth

Details of the above nomenclature are given in reference 1.

Mass Properties of Components.

Ml = mI = 0.3531 X 10-4 lb-sec2 /in. (6.182 X 10-3 kg) = mass of rotor

M2 = m2 = 0.3226 X 10- 5 lb-sec2 /in. (5.648 X 10- 4 kg) = mass of gear andpinion no. 2

M3 = m3 = 0.2656 X 10- 5 lb-sec 2 /in. (4.650 X 10- 4 kg) = mass of escapewheel and pinionno. 3

MP = mp = 0.2451 X 10- 5 lb-sec2 /in. (4.291 X 10- 4 kg) = mass of pallet

II = I1 = 0.3350 X 10- 5 in.-lb-sec2 (0.3788 X 10-6 kg-m2 ) = moment of inertia

of rotor

12 = 12 - 0.3750 X 10-7 in.-lb-sec2 (0.4240 X 10- 8 kg-m2 ) = moment of inertia of

gear and pinion no. 2

13 = 13 = 0.2490 X 10- 7 in.-lb-sec 2 (0.2815 X 10-8 kg-m2 ) = moment of inertia ofescape wheel andpinion no. 3

IP = Ip = 0.6572 X 10- 7 in.-lb-sec2 (0.7430 X 10- 8 kg-m2 ) = moment of inertia ofpallet

General Parameters.

RU = rCl - 0.115 in. (2.921 m) = distance from rotor pivot to centerof mass

RCP = rCp M 0 - pallet eccentricity

22

RHOP = P = 0.0155 in. (0.394 mm) = pallet pivot radius

RPM 30,000 = spin rate

PHIICD = fIC - -122.230 = eccentricity angle of rotor

in starting position (fig. 4)

PSICCD = *C = 0 . - eccentricity angle of pallet

(fig. 4)

PHID = 150' = escape wheel starting angle ofinitial coupled motion simulation(for choice of this angle, seepp 7 and 14, ref 1)

PHICUTD 16000 - cumulative escape wheel angle,obtained from product of totalrotor rotation and gear train ratio.The total rotor rotation for theM577 is 48.30 while the gear ratiois 33.02. Thus, 48.3 X 33.02 - IbO0.

MU = 0.15 = coefficient of friction of geartrain (pivots and tooth-to-tooth

contacts, includes escape wheelpivot)

MUI = 0.15 f coefficient of friction on pallet -

escape wheel interface and palletpivot

Gear Parameters.

PSUBDI = Pdl = 44 = diametral pitch of mesh no. 1 (rotorand pinion no. 2)

PSUBD2 = P d2 - 63.5 = diametral pitch of mesh no. 2 (gearno. 2 and escape wheel pinion)

NGI - NGI = 41 = number of teeth of rotor (full gear

no. 1)

NG2 = NG2 - 29 = number of teeth of gear no. 2

NP2 = N 6 - number of teeth of pinion no. 2

NP3 - Np3 = 6 = number of teeth of pinion no. 3

(escape wheel pinion)

CAPRPI = Rp -- 0.46591 in. (11.834 mm) - pitch radius of gear no. I (rotor)

23

.... ...... j

CAPRP2 W Rp2 - 0.22835 in. (5.800 mm) - pitch radius of gear no. 2

RP2 = rp2 = 0.06818 in. (1.732 mm) - pitch radius of pinion no. 2

RP3 = rp3 = 0.04724 in. (1.200 mm) - pitch radius of pinion no. 3(escape wheel pinion)

THETA 01 - 200 - pressure angle of mesh no. 1

THETA2 = 02 = 20* - pressure angle of mesh no. 2

RI = R -- 0.225 in. (5.715 mm) - distance of rotor pivot fromspin axis

R2 - 2 0.408 in. (10.200 mm) - distance of pivot of gear andpinion set no. 2 from spin axis

R3 ' 6{3 = 0.368 in. (9.100 mm) - distance of pivot of escape wheel

from spin axis

R4 = 0?4 = 0.361 in. (9.025 mm) - distance of pivot of pallet fromspin axis

RHOl = P1 = 0.0465 in. (1.181 mm) - pivot radius of rotor

RHO2 = P2 = 0.0155 in. (0.394 mm) - pivot radius of gear and pinionno. 2

RHO3 = P3 = 0.0155 in. (0.394 mm) - pivot radius of escape wheel

CAPRBI = Rbl - 0.43781 in. (11.120 mm) = base radius of gear no. I

CAPRB2 = Rb2 - 0.21458 in. (5.450 mm) = base radius of gear no. 2

RB2 = rb2 - 0.06407 in. (1.627 mm) = base radius of pinion no. 2

RB3 = rb3 - 0.04439 in. (1.128 mm) = base radius of escape wheel pinion

CAPROl - Rol - 0.47388 in. (12.037 mm) = outside radius of gear no. 1

CAPRO2 = Ro2 - 0.23387 in. (5.940 mm) - outside radius of gear no. 2

R02 = ro2 - 0.09373 in. (2.381 mm) = outside radius of pinion no. 2

R03 = ro3 = 0.07321 in. (1.860 mm) - outside radius of escape wheel

pinion

Ji = J1 - 0 - Initialization parameter for meshno. 1. (The value of zerocorresponds to earliest possiblecontact of mesh (ref 3).)

24

AW

J2 2 0 - initialization parameter for meshno. 2

Output Data

Fuze Geometry. The angles BETAID - 1 to BETA3D = 83 and GAMKA2D Y2 toGAMMA4D = Y4 are printed for checking purposes.

Coupled Motion. For each time increment, T, of the coupled motion, thefollowing variables are evaluated:

PHI - - instantaneous escape wheel angle (deg)

PHIDOT - = escape wheel angular velocity (rad/sec)

G = g = pallet - escape wheel contact position (in.) (ref 1)

GDOT = g = time rate of change of g used to determine the directionof relative sliding (in./sec) (ref 1)

PSID = = pallet angle (deg)

PSIDOT = 4 = pallet angular velocity (rad/sec)

PHITOT = = cumulative escape wheel angle (deg)

F23 = F2 3 = normal contact force of gear no. 2 on pinion no. 3 (Ib)

F12 = F12 = normal contact force of gear no. I on pinion no. 2 (lb)

PN = Pn = normal contact force between escape wheel and pallet (Ib),computed according to equation B-83

PNPSI = P i= normal contact force between escape wheel and pallet (Ib),n4 computed according to equation B-84, (serves for checking)

DPHI2 = = escape wheel acceleration (rad/sec 2), Runge-Kutta output

Free Motion. For each time increment, T, of the free motion, the follow-ing variables are evaluated:

PHI = - instantaneous escape wheel angle (deg)

PHIDOT - = escape wheel angular velocity (rad/sec)

PSI = - pallet angle (deg)

tt25

_A.

PSIDOT = -, pallet angular velocity (rad/sec)

PHITOT = *T = cumulative escape wheel angle (deg)

FF12 = FF12 = normal contact force of gear no. I on pinion no. 2for free motion (ib)

FF23 = FF23 normal contact force of gear no. 2 on escape wheelpinion for free motion (Ib)

Impact. Just preceding the "IMPACT" label, the program prints the valuesVP and VS, which stand for the pre-impact velocity components, normal to the escape

wheel tooth, of both the pallet and escape wheel contact points (ref I). Subsequent

to the "IMPACT" label, the following variables are evaluated:

PHI = instantaneous escape wheel angle (deg), same as beforeimpact

PHIDOT = post-impact escape wheel angular velocity (rad/sec)

PSI = 4 = pallet angle (deg), same as before impact

PSIDOT = 4 post-impact pallet angular velocity (rad/sec)

PHITOT = O = cumulative escape wheel angle (deg), same as before impact

VP = Vp post-impact normal velocity component of pallet (pin) atcontact point (ref 1)

VS = Vs = post-impact normal velocity component of escape wheel toothat contact point (ref I)

Note that in the present program VP is equal to VS since the coefficient of restitu-tion is zero.

Turns-to-Arm and Maximum Contact Forces. The result of 27.93 turns to armat a spin velocity of 30,000 RPM is obtained by the program in a manner similar tothat shown in Program SANDA3 under paragraph entitled Turns-to-Arm and Maximum

Contact Forces for the M125AI booster mechanism. (Now the escape wheel anglePHICUTD - 16000.)

The maximum non-impact contact forces for the total cycle both for coupledand free motion, are listed at the end of the output.

26

LAi

CONCLUSIONS

While it was not the purpose of the present investigation to undertake para-metric studies of either of the two mechanisms for which the programs were written,both programs were sufficiently tested to confirm that such studies are possible.They may include variations in masses and moments of inertia of all components,variations of gear, escapement and fuze geometries, as well as various frictionconditions.

Test runs for average friction conditions and standard geometries showed thatthe M577 SSD simulation required 27.9 turns to arm regardless of the assumed spinrate. Under similar conditions, the M125AI booster model needed 45.3 turns toarm. Both of these values are well within the military specifications for thesedevices.

The ability to determine the contact forces during coupled and free motionfurnishes the foundation for future work concerning the strength of fuze gearteeth. Such work will also have to include the determination of the appropriatedynamic factors due to the impact of the runaway escapement.

A most interesting phenomenon, which may also serve as the starting point of afuture investigation, occurred following the first exit impact during the initialmotion cycle of the 14577 SSD simulation. The preceding coupled motion was initiatedcloser to the root of the escape wheel tooth than usual, i.e., the starring angle ofthe escape wheel was chosen to be PHID - 1450, when a more usual value during stableoperation is equal to, or larger than 1500. With this entrance starting conditionand the resulting higher than usual pallet angular velocity, the exit impact wassufficiently large to reverse the motion of the escape wheel and the gear train.During this backward motion of the escape wheel, its deceleration (i.e., accelera-tion in the forward direction), due to the influence of the driving rotor, becameapproximately 20 times larger than usual and all contact forces increased substan-tially. This increase in forces subsided as soon as the forward motion of theescape wheel was restored. This phenomenon did not occur during normal operation,since once the escapement has become stable coupled motion starts much nearer to thetip of the escape wheel tooth. The resulting smaller free motion angular velocitiesof the pallet caused less severe impacts and, as a result, prevented motion rever-sals.

27

REFERENCES

1. G. C. Lowen and F. R. Tepper, "Dynamics of the Pin Pallet Runaway Escapement,"Technical Report ARLCD-TR-77062, ARRADCOM, Dover, NJ, June 1978.

2. G. G. Lowen and F. R. Tepper, "Fuze Gear Train Analysis," Technical ReportARLCD-TR-79030, ARRADCOM, Dover, NJ, December 1979.

3. G. G. Lowen and F. R. Tepper, "Fuze Gear Train Efficiency," Technical ReportARLCD-TR-80024, ARRADCOH, Dover, NJ, November 1981.

28

APPENDIX A

DYNAMICS OF ROTOR DRIVEN S&A MECHANISMWITH A THREE-PASS INVOLUTE GEAR TRAINAND A PIN PALLET RUNAWAY ESCAPEMENT

29

DESCRIPTION OF SYSTEMS AND OUTLINE OF DERIVATIONS

The present appendix gives derivations for a complete mathematical model of anS&A mechanism, consisting of a spin driven rotor, a three pass Involute step-up geartrain, and a pin pallet runaway escapement. Figures A-i and A-2 show the two typesof configurations to which this model can be adapted.

This work draws to a considerable extent on previous efforts by the authors,i.e., the dynamics of the pin pallet runaway escapement (ref A-1) and the fuze geartrain analysis (ref A-2). As in reference A-i the following three regimes of themechanism are considered:A-

1

1. Coupled Motion

The escape wheel is in constant contact with one of the pallet pins whileit is driven by the rotor (gear no. 1) through the gear and pinion sets no. 2 and3. The coupled motion differential equation in the escape wheel variable isobtained by combining the solutions to the Newtonian force and moment expressionsfor the individual mechanism components.

2. Free Motion

The pallet and the escape wheel - gear train - rotor systems move inde-pendently of each other in this phase of motion. Thus, there results one differen-tial equation for the pallet in the variable *,and another one for the combinedsystem in the escape wheel variable *

3. Impact

The formulation of the impact regime is taken directly from reference A-I,except that now the moment of inertia of the escape wheel and pinion no. 4 alsocontains the referred mass properties of the rotor and gear and pinion sets no. 2and 3. This impact simulation is based on the classical angular impulse - momentummodel, where a coefficient of restitution is used to account for the energylosses. Since it is assumed that the effect of the impact force between the escapewheel and the pallet is much greater than the effects due to the driving torque ofthe rotor and the various retarding torques due to friction, the latter are disre-garded.

The influence of friction forces is considered bath in the coupled and the freemotion regimes. There is friction at the escape wheel - pallet interface duringcoupled motion, and there is friction between the gear teeth and at all pivotsduring both of these regimes. As in reference 2 the individual pivot friction

A-i For better understanding, consult figures in reference A-I.

31

k-iZC5lmI pA xL..am numh

0

z -,I

4j

LY 00

J2J

z CL

ulO

Lai CL+C

CL )

Z CU)

w~Cho-4

L-.

320

('4

0

0

-IZ

U

C

9L

L.J Cd

C,'-'-0z

4-

ILdw w 0

z-

oCd

tj-

WC

w0

33-

torques are obtained by the algebraic addition of tile two friction moments due tothe x and y components of the normal bearing forces, rather than by the direct useof the resulting normal forces. This conservative approach to friction is necessary

in order to avoid the difficulties which the presence of a square root introducesinto the solutions of the various differential equations.

The following briefly outlines the course of the derivations in the individualsections. While the derivations are specifically for configuration no. I tile re-sults are always applicable to both configurations as long as the appropriate fuzebody angles are used.

GEOMETRY OF FUZE BODY CONFIGURATIONS

The following section contains derivations of expressions for the various fuze

body angles associated with the pivot holes of both configurations of the three passstep-up gear mechanisms. They represent extensions of previous work pertaining toconfiguration no. I given in reference A-2, appendix A. In addition, relationshipsbetween the unit vectors of the body-fixed x-y and x'-y' systems (figures A-I and A-

2) are given in order to make it possible to derive tile escapement relationships inthe primed coordinate system, whose x'-axis is oriented along the escapenent center-

line p-O (ref A-i). Finally, a signum function is introduced so that common

programming expressions can be devised for both configurations.

Fuze Body Configuration No. I

Figure A-3 indicates the following relationships for the angles yi' 5i and

Ri for fuze body configuration no. 1.

Angles y i

From

2 2 2(Rp + I + A?- 26 1 _2cosy 2

the following expression is obtained for y2

-1 1 2 - (R pl+ rp2

Y2 ' cos 26J 22 (A-i)

Similarly, from

= cos - -J2 (A-2)2 d?2 6?3

34

LI I I

lI-

N CC

'4

+ C

I-.c

a.L

in

CD0

CLC

'06

354

Le following is obtained

Y3 =Y2 + yi (A-3)

Lso from

S6? 3 + 63 - (Fp3 + r )p4 (A-4)

2 13 63 4

ie following is obtained

Y4 = Y3 + Y (A-5)

Lth a, the distance between the pallet and the escape wheel pivots, the following

5 obtained

2 2 2a 6 64 + 6'5 - 2 R14 6 sCOSY5 (A-6)

yg = Co 64 2 2 (A-7)

2cos 2 4 65

nd

Y5 = Y4 + Y9 (A-8)

Angles 6.

ince

2 2 2612 = CRpI + rp 2) + 6l1 - 2(R p + rp 2) 631 cos6j

[ Rpu2 2 2

(R + +rp2) + 62 - ?2162 -- cos - I 21( 1- 2 (A-9)

2 R(R pl + r p2)

imilarly,

(R +r)2 + 263 Cs-1 Rp2 --rp3) + 2 --, 3 (A-10)

63 = cos[2 6?2(Rp2 + r p3)

36

and

[(R + 2 2 - 2164 CO co - 1 Rp + re4 + 61 3 - 4 (A1

2 6[3(R p3 + rp4) (A-i)

Further, from

2 2 2635 = a + 6 - 2a 64 cos 6 5 (A-12)

the following is obtained

65 cs I a + 6? 4 -63 5

65 =COS 2a 6J4(A-13)

Angles 0

With equation A-9

61= - 62 (A-14)

Further, with equations A-i and A-10

$2 = Y2 + it - 63 (A-15)

With equations A-3 and A-1I

63 = "Y3 + It - 64 (A-16)

Finally, this furnishes with equations A-5 and A-13:

B = Y4 + 'T - 65 (A-17)

Figure A-3 also shows that the angle y' between the x'-axis and the unit vectorn 5 is given by:

P

Y , t- € (A-18)

37

where

C = N - 5- (A-19)

and therefore

Yp - 65 + Y( (A-20)p

The unit vector n 5 may now be expressed in terms of the primed coordinate system as

follows:

n5 - cos y 1 / + sin yj (A-21)p p

Further, the unit vectors i and j' are given in terms of the x-y system by:

i - cos( - n)i + sin( 04 - W)J

- -COS34 - sin 0 (A-22)

J" - k'x i= -sin( R4 - ii + cos(8 - 70-j

sinO4i - cos 6j (A-23)

Fuze Body Configuration No. 2

Figure A-4 indicates the following relationships between the angles Yi, 6i,

and 6 for fuze body configuration no. 2.

Angles yi

Since these angles are given in the clockwise direction with respect to the

body-fixed x - axis, their values are negative. Thus, with the help of equation A-I:

6 2 + 1 2 - (Rp + rp 2 1- -co 1+ 2 -_Ril r 2 (A-24)

82 i2

38

it

000

a.a C.

41 +0060

CCIZ

390

y' is given by equation A-2, and with that y becomes3 3

Y 3 = Y2- Y3 (A-25)

The negative sign for y' is necessary in order to make y 3 negative.

y is given by equation A-4. This leads to the following expression for Y4

Y4 = Y3- Y4 (A-26)

The angle yg is again taken directly from equation A-7, and with that

y becomes5

Y5 -Y Y5 (A-27)

Angles 6

The angles 62, 63, 64, and 65 may be taken directly from equations A-9, A-10,A-li, and A-13, respectively.

Angles 8i

With equation A-9

$I = 7 + 62 (A-28)

With equations A-10 and A-24

2 = Y2 + iT + 63 (A-29)

where the negative sign of equation A-24 for Y2 is accounted for.

Similarly, with a negative sign of Y3, one obtains with the help of equationsA-Il and A-25:

83 = Y3 + w + 64 (A-30)

40

Finally, with equations A-13 and A-24

4 = Y4 + 1! + 65 (A-31)

According to figure A-4, the angle y' between the positive x'-axis and thedirection of unit vector n 5 is now given by

p

Y= It + E (A-32)p

where

S1- 65- Y5 (A-33)

so that

y- 2i- 65 - Y' (A-34)p

The unit vector n 5, when expressed in the primed coordinate system, is againgiven by equation A-21, i.e.,

n 5 = cosy'i + sinyj (A-35)

Further, the unit vectors iV and J' are given in terms of the x-y system by

i = cos(0 + ir)i + sin(0 4 + W)i

V = -cosR41 - sin015 (A-3b)

J' - k' x i = -sin(fR + W)i + cos(R4 + W)J

J'= sin84 - coSBJ (A-37)

41

Common Computational Expressions for Both Configurations

In order to find common programming expressions for the angles of both config-

urations (with the exception of the angles y' of equations A-20 and A-34), the

following signum function s6 is introduced, whete

s6 f +1 for configuration no. 1 (A-38)

s6 = -1 for configuration no. 2 (A-39)

This leads to the following expressions:

according to equations A-i and A-24

-1 + 6?2 (R _ + r p2Y2 = s6 cos 2 2 (A-40)

according to equation A-2

[,2 2 R r)2S o1 2 + 3 - +

(i C os 2 R R 3 (A-41)2 6?26?3

according to equations A-3 and A-25

Y3 = '2+ s6Y3 (A-42)


3 - + 4 (R P + rp4)SC- rs2 6 4 (A-43)


Y4 Y3 + s6Y'4 (A-44)


- 4 5 a

Y5 cos 2 4 6? " (A-45)

42


Y5 =Y4 + s (A-46)

according to equations A-9, A-10, A-il, and A-13

CR1+ 2 )2 2 21i(RP1+ p )+ V 1 - R

62= cos 2 jl(Rp + r p2 (A-47)

F2 +62 6?2(Rp + r 2) +(3SCos- p2 p3 (A-48)

cos'[ ~~ R ? 2()+dI~663 2 {2(p2 + p3)

(6?3 ( p3 + rt -

2 2 2 "165 -1 a + 64 - R5 (A-50)

2a 6?4


61 w - s662 (A-51)


82 = Y2 + ?F - s 6 6 3 (A-52)


83 m Y3 + i - 8664 (A-53)


04 Y4 + - 8665 (A-54)

43

The signum function s6 will also be useful in conditional statements whichdistinguish between the y s of configuration no. I and 2 as given by equations A-20

and A-34.

Finally, according to equations A-22, A-23, A-36, and A-37, identical expres-sions may be used for both configurations when expressing the unit vectors of theprimed coordinate systems in terms of the unprimed ones, i.e.,

= -cosai - sinQj (A-55)

= sinR4 - cosj (A-5b)

DYNAMICS OF THE PALLET IN COUPLED MOTION

While the following derivation for the dynamics of the pallet in coupled motionis illustrated with figures relating to configuration no. 1, all expressions are

also applicable to configuration no. 2 as long as the appropriate sign for thesignum function s6 (eq A-38 qnq A-39), as well as the correct form of the angle y'(eq A-20 and A-34) are used.

As stated earlier, the dynamic analysis of the pallet is most convenientlyperformed in the primed coordinate system. The coefficient of friction at thepallet-escape wheel interface and at the pallet pivot has the designation 1i.

With the spin wof the fuze body constant, the acceleration of the center ofmass of the pallet is given by figure A-5.

2 2A[cos(+ *c-i( + sin(* + Cj

+ *rcp [-sin( + + cos(i + *c) ] (A-57)

A-2 This simple change in angles is all that is required since the relative

rotation directions of the components are identical for both configurations,and therefore, force and moment expressions retain the same general form.

44

LI1w

0

-~ 02

'4-4 ,

10

0

4j 0

00)d

$40

00

C) 4.

10 )

4 0

-.0 -W)0

X40

91

LA 4,

453

With the above, and according to the free body diagram of figure A-5, substitutioninto Newton's law gives

Pn V IS4P n + F - jIs5F 1'+ Vj+ iIsF n P yp YP' xpi

_M 2 X, 2- m 6'5n5- (w + *) rcP(Cos( P + c ) ' + sill(q + c ) )

P

+ 1 C -sin(!' +pC)i + cost! + !'~~](A-58)

The signum functions s4 and s5 assure the proper directions of the friction forces

on the escape wheel - pallet interface as well as on the pallet shaft, regardless ofthe direction of pallet rotation. Thus,

S = (A-59)

(ref A-i, eq B-I) and

5 (A-60)

The unit vectors n and n are adapted from equations A-I and A-2 (ref 1).t n

The moment equation of the pallet must be written with respect to the acceler-

ated pivot 0 , i.e.,P

Mop = -rOp x mprcp + Hop (A-b1)

where

Mop = sum of all external moments with respect to pivot Op

rop = absolute acceleration of point Op

0p0 time rate of change of angular momentum of the pallet with respect to

point Op.

46

6A:

Since the angular velocity w of the fuze body is constant equation A-61 takesthe fotlowing form:

2 T.-Op = 2(-W 6{5n) x mprcp(cos(4 + q)I' + sin( +t)j + ip (A-62)

Appropriate computations and substitution of all moments, according to figureA-5, furnish the final moment equation (the moment arms A1 , B1 , C,, and Dl of refer-ence A-I are now primed)

2P(Dj + CjUlS) - DP155(F p + Fy) Ip- mprcP 6? Y5 sin(y--pC) (A-63)

where

DI = c cos( - a- ') (A-64)

Cj = -[r + c sin(- a- W) ] (A-65)

As originally discussed in reference A-2, Fp and Fyp represent conservativelyevaluated pivot force components which assure that the pivot frictiou moments areopposed to the rotation at all times. The following illustrates how this goal maybe accomplished.

The pivot friction components Fp and F4 are first obtained from the followingcomponent expressions of equation A-58. Subsequently, they are transformed tobecome F and F , respectively.

xp yp

-P sin(fr-c) - iSPnCOS(fr-a) + F - iiIsy5Fn n xp yp

2 . 2= m [-w dcosy' - (w + ) rcpcos( + c) - _rCpsin( + )] (A-66)

p

and

P CoS(O - a) - 1iS4 Pnsin(4 - a) + F' + 1iJs 5Fpnn yp xp

2 .2M [2 diny w + rcsin() + , + 4r cos(m + C)](A-b7)

47

Simultaneous solution of equations A-66 and A-67 furnishes

2 2 .2F P A I + w A2 2-72-- 4A~k4 A 3 * 'A 4 (A-68)

YP n "

2 2 .2F p P nA5 + W A6 ± 2 w~- ;&7 4 A7 ±* '8 (A-69)

xpW

where

2~I (s4-s5)sin(r-c) - (l+s 4s 5 Jl )cos( ,-ct

A, - __ _ __- (A-70)

I+

= m 635(siny'-tuls~cosy ) + r~p(sin(rI 4') - UsScos(4+,-P,))](A7)

m r [sin(4I) sos1*']A3 = p CP __) - s9 s(* C (A- 72)

21 + il

mr ~c~os4i4~ + v is59in4+c)]A4 (A-7 3)

21 +i

2uJ1( 4-S5)COS(4-i + (1s~s5 0i1 sin4-z)

A5 2 (A-74)

m p [- 6? 5 (cOsy +ijs~siny;) -rCp(cos( -t+u+j~i(+C)A6 = (A-75)

2

A m flr [cos(r+ C) + uls~sin(4+*iC)] (-~

I + Uil

A 8 - 1 + 2 P[i(4 C s5o C (A-77)

48

2The factor w /IwI is introduced in place of w in order to nke sure that the

quantity is positive regardless of direction of spin. (This too, assures that thefriction moments oppose rotation. The driving moment is proportional to w , andtherefore independent of spin direction.) To make the final decision concerning thesigns of equations A-68 and A-69, these forces are substituted into moment equationa-63, and the influence of the direction of rotation on each of the resulting momentcomputations is explored:

2P n[D + C!" 1s4 - Ppiis 5 (A 1 + A)] ± pPlS5w (A 2 + A6 )

2 2S(A+ A7) ± PIdls 5 (A3 + A7 ) ± Pp iss (A + A8 )

2= IP - mprcp 6?5w sin(y-I-C) (A-78)

With s5 positive for positive rotation, and vice versa, while all other para-meters are positive at all times, the following moment components of equation A-78must have negative signs during positive rotation:

-PnppplS5(AI + A5 ) (A-79)

2-P pPS5w (A2 + A6 ) (A-80)

.2-P ils5 (A3 + A7 ) (A-81)

The sign of the term containing is negative and is controlled by the signof p . Therefore, the signum operator s5 is omitted and the term becomes

2

- 2 P 1I W- P (A 3 + A7 ) (A-82)

The choice of sign for the friction moment term in equation A-79, which isproportional to the angular acceleration m , is discussed in detail in appendix F.This work results in the computational rules of equations A-88 and A-89, which dealwith the sign in the effective moment of inertia TPR. (Note that the signum func-tion s5 function has now been omitted.)

49

With these considerations equation A-78 then becomes

22 *.2

PnA1 8 - 1u A 1 9 - 2 - ' A 2 0 - ' A 2 1

2= IPR" mprcp 6?51 [sin(yp - (A-83)

!reA- 3

A 1 8 = DA + C IlS 4 - PPIs 5(A 1 + A 5) (A-84)

A1 9 = p PIs 5(A 2 + A6) (A-85)

A 2 0 = 0PP i(A 3 -+ A 7 ) (A-8b)

A 2 1 = pii s 5(A 3 + A 7) (A-87)

PR = Ip + A 2 2, when ' and p have identical signs (A-86)

IPR = Ip - A22 ,A-4 when and ' have opposite signs (A-89)

A2 2 = OPP 1 (A 4 + A 8) (A-90)

Equation A-83 is now used to find an expression for the contact force Pn. This)ression will later make it possible to establish a single differential equationr the escapement in coupled motion. Thus,

2.2 W 2 2

IpR r+A21tP+ 2 -A20+W- A-mr 5 sin(Yp-r- c) (A-91)A1 8

The parameters An are not sequential at this point. A9 to A1 7 may be foundin equations A-Ill through A-116 and equations A-118 through A-120.

Care must be taken that I p - A2 2 does not become negative. If this occurs,1PR must be set equal to zero. See footnote A-8 where the possible value oflPR in the free motion equation of the pallet is discussed.

50

As in reference A-I the variables p and are now expressed in termsof $ and * the escape wheel variables (ref A-1, eq A-15 through A-18 and B-6)

.2 (-2p= u V (A-92)

and

Z $ (Z is not defined in ref A-1) (A-93)

where

1 [Q + Asin( - a)] (A-94)c Cos I'l

Equation A-91 then becomes

2A [ + (A 2 1 Z2 + IpRV); + L A2 0 Z

n Aj 1 R~ I,

2 2+ w Ai 9 - m r pd 5? sin(y-4- C) (A-95)

DYNAMICS OF ESCAPE WHEEL IN COUPLED ROTION(ESCAPE WHEEL INCORPORATES PINION NO. 4)

While figure A-6 shows a free body diagram of the escape wheel and pinion no. 4of configuration no. 1, the following derivations hold for both configurations aslong as the appropriate signum function s 6 is chosen. The pivot forces F.4 and Fy 4

as well as forces F3 4 and T4 are now defined in the general x-y system. This makes

it necessary to give the unit vectors nt and n in terms of the x-y system.

From equations A-1 and A-2, reference A-1 (and using x'-y' coordinates)

nt a cos(o - a) i + sin(¢ - a) j' (A-96)

n n -sin( - a) 17 + cos( - ) j (A-97)

51

0K,

oo**

A.0CZu

0

CDu

CL

fo

ca-

a- ~-52

Equations A-55 and A-56, which hold regardless of configurations, are used forthe following transformed expressions:

nt = -cos(- + 64) i -sin(- a + 4) j (A-98)

and

n = sin(- a + 84) i - cos(- a + 4) j (A-99)

The expressions for the unit vectors H34 and RN34 , as used in the analysis ofpinion no. 4 in reference A-?, section A-la, are of further interest

n34 - sin(8 3 + 03) i - cos(0 3 + 83) j (A-100)

and

nN34 = cos (83 + 03) 1 + sin(8 3 + 03) j (A-lO)

Using the concept of the D'Alembert force

2T4 = m4 IR4 W (cosy 4 i + sinY 4 j), (A-102)

the force equation of the escape wheel assembly becomes for counterclockwise rota-

tionA-5 (ref A-2, eq A-35)

-P n + pis 4Pnt + F 3 4A3 + Ps3F 34n.34 + T4 + Fx4i + ljFy4l

+ IFx4J - Fy 4 J = 0 (A-103)

Note that the coefficient of friction i is now used for all pivots and gear tooth

contacts A-6 of the remainder of the mechanism train.

A-5 See appendix F for description of motion reversal, i.e., clockwise escape

wheel rotation. This may occur after severe impacts.

A-6 The signum functions sj, s2 , and s3 are defined in reference A-2 in

connection with the tooth contact friction of the various meshes.

53

The moment equation of the escape wheel for counterclockwise rotation is writ-ten with the help of figure A-6

-P n(Al + BP 1 s 4 ) - PPp4(F + F y4) + rb 4 F 34 U s 3 (d 3 - a 3 ) F 34 =10 (A-104)

where

A, - b coso + g (A-105)

B, = b sina (A-10b)

The pivot forces F and F4 are treated again in the manner discussed ear-

lier. They are obtained lrom thX component expressions of equation A-103, i.e.,

-P sin(fr-c$+ 4 ) - s 4j 11PnCoS(4-cdF8 4 ) + F 3 4sin(0 3+0 3 )n

+ s3F 34cos(+0) + T 4cosy4 + Fx4 + Fy 4 = 0 (A-107)

and

P cos(O-c0+4 ) - s41Pnsin(-c-r 4 ) - F 3 4 cos(O3+8 3 )

+ ts 3F 34sin( 01" 3) + T siny 4 - Fy 4 + IFx 4 ' 0 (A-108)

Simultaneous solution of equations A-107 and A-108 gives

F . P A 9 + F 3 4 A 1 0 + T4 All (A-109)y4 n

Fx4 ' Pn A 1 2+ F 3 4 A 1 3 + T4 A 1 4 (A-110)

54

IL_

_

whe re

-(pls 4 j)sin(a +84) + (l+IJPs4)cos(4-0'8 4 )A9 = (A-Ill)

21+ P

2-(l-s3)sin(f3+03) - (I+P s3)cos( t+0 3)

A 1 0 = 2 (A-I12)

1+ i

siny 4 - UcOSY 4

All = 2 (A-I 13)

I+

(l+IJlVB 4)sin(r-cdI-fR 4) + (s4Ul-I)cos( -crFOO)

A 1 2 - (A-114)2

1 +1

2-(I+uj s 3)sin( 3+ 3) + u'(l-s 3)cos(l 3+0 3)

A 1 3 (A-I 15)

1+ P

-cosy 4 - sinY 4A 1 4 = (A-116)

1+ P

Substitution of the pivot force component expressions, with the indicated

signs, into the moment equation A-104 will furnish the correct pivot friction mo-ments. The resulting expression is then solved for the contact force P.

-14 * + F 3 4 A 1 5 - T4 A 1 6P = (A-117)n A1 7

where

A1 rb 4 - J [s 3(d 3 - a 3) + P4(A 1 0 + A 1 3)] (A-118)

A 1 6 - pp 4(A II + A14 ) (A-I)

A 1 7 A' + jiIs 4B+ uP4(A9 + A 12) (A-120)

55

COMBINED COUPLED MOTION DIFFERENTIAL EQUATIONFOR ESCAPE WHEEL AND PALLET

(APPLICABLE TO BOTH CONFIGURATIONS WITH APPROPRIATE s6)

To obtain a combined coupled motion differential equation for the pallet and

escape wheel in terms of the escape wheel angle 4, equations A-95 and A-117, both

for the contact force Pn are set equal to each other. This furnishes

22 2

2.

A15AIF34 - A 16A18T4 - A 17W A 1 9 - mprCp 5sin( yp- - iC) (A-12 1)

To complete the system differential equation, it is necessary to find an ex-

pression for the contact force F34 in the above. This will be accomplished by

combining the appropriate differential equations for gear and pinion no. 2 and 3 andfor the rotor.

DYNAMICS OF ROTOR (GEAR NO. 1)(APPLICABLE TO BOTH CONFIGURATIONS WITH PROPER s6 )

A free body diagram of the rotor for configuration no. I is shown in figure A-7. The acceleration ACR of its center of mass, with constant spin velocity w , is

given by

- 2 2 r-1ACR rffi1CjA -R w dii- (u*4)) [ C cos(4)IC +4))i + sin(4)1C+b)j

+ 4)rC 1 I--sin(lC+il)I + cos(4IC+0)J ] (A-122)

Since the motion of the rotor must be expressed in terms of the escape wheelvariables, the following transformations are now introduced:

- (A-I123)

- N 4 1 4 (A-124)

whereNP4 NP3 Np2 (A-i2)

N41 nN G3N G2N GA15G3 G2 NGI

56

co

Is-

IIc

57-

The rotor angle IC + * , is expressed as follows:

IC + tj - IC + N4 1 *T (A-12b)

where T represents the total rotation of the escape wheel from the inception of

motion. (The section on Geometry of Fuze Body Configuration describes the manner in

which *T is obtained as a function of the instantaneous angle *.)

Equation A-122 then becomes

2 2A CR ji (* N 4 1) r C [cos( IC+N41tT)i + sin(lc+NI41 T)j]

+ N 4 1 0 rCl [-sin(OICIN41T)i + cos( lC+N41T)j] (A-127)

Newton's force equation is now written for clockwise rotation of the rotor witil

the help of equation A-127 and the free body diagram of figure A-7:A-7

-F 1 2n 12 - 1F - FX, + pFY1i + FFI + jiF = mjACR (A-128)

where

n12= sin(81 + 6 1)i - cos(8 1 + e1)j (A-129)

nN12 =cos(S1 + 0 1)i + sin(8 1 + 6 1 )j (A-130)

(ref A-2, eq A-78 and A-79)

Further, in figure A-7

F 2 1 -F 1 2n 12 (A-131)

and

F f2l fi - 1 F 1 2 nNl 2 (A-132)

(ref A-2, eq A-103 and A-104)

A-7 Description of motion reversal, appendix F.

58

The moment equation must be written in the manner of equation A-61, with re-spect to the accelerated point 01. The pivot reactions F x and F are treated sothat the associated friction moments retard the clockwise rotatitn of tlhe rotor.This leads to

RbIF12k - islaIFjI + lp(F X+F yl)k

6?i ) x m rcI [cos(OiC+N41T)i + sin(,ic+N41T)j

+ I IN4 1#c (A-133)

This becomes

RBIF 12 - W la IF 12+ jp I(F X + F y)

2 "

= m 1w2 6 Iir csin(¢ic + N4 1 T) + 1 1N410 (A-134)

The forces F I and F are obtained from the simultaneous solution of thefollowing component expressions of equation A-128:

-Fl 2sin( 1 +0 1 ) - F, SlFl2os(8+0 1) - FX1 + pFy I

= mI - W2 I - rC1Cos( 1c+N41T) - N41rClsin( iC+N 4 1h)] (A-135)

F 1 2cos(Bj+O1 ) - islF 1 2sin( $1+0) + F 1 -+ x

.2ml [-(c+N4 1 ¢) rClsin(O 1C+N41T) + N410rCjcos( 1C+N40,r) ] (A-13b)

Simultaneous solution of the above gives

2

2 2w * 2FYI - F 1 2 2 3 * wA 2 4*-- 4 1; 2 5 * (N41 ;) A 2 5 * N4 1OA 2 b (A-1I7)

and~2 2

2 AWA 2 9 N4 1A 3 0 (A-138)

59

whe re

2u(l+s )sin( R1+61) - (1-u s 1)cos( 01+0 1)

A2 3 (A-1 39)

+ ~ +

A24 (A-140)

2 +

A 2 S(A-14 41)

1+ pi

A26 41)=i(,+4T (A-142)

and

(1-t sl)sinU6 1+01) + uls1cs +1A 2 ;= (A- 143)

1+

ml[ 6?1 + r~l(cos(r , 0+N4l4h) - join( ic+N 4 10T))A2 8 = - -2 (A-144)

A2 9 = m1r~j [cos( jlc+N1lT - win( Ic+N41T) (A- 145)2

Aj m~rC, [sin( Ic+N4 14r) + Wros( 1 0 +N4 14h)](-1

I1+ pi

60

Equations A-137 and A-138 are now substituted into the moment equation A-134 asfollows:

2RbIF 12 - ts 1a 1F 12 + UPI[F 1 2(A 2 3+A 2 7) * w (A 24+A 2 8 )

2.2

* 2-- N14 14(A 25+A2 9) -k (N41,) (A 2 5+A 2 9) * N4 14(A 2 6+A3 0 ) ]

2 "

= ml r d{irsin(Cic+N4 1T) + IIN 41 I (A-147)

In order to have the pivot friction moments positive for clockwise rotation,the following terms must be positive:

W P1 F 1 2 (A 2 3 + A2 7)

2PP PI (A 2 4 + A2 8)

*20 P1 (N 1 ) (A 2 5 + A2 9)

The term2

2 j p, "w N4 1 (A 251-A2 9 )

must also be positive for clockwise rotation of the rotor. This can only be accom-plished by making the term negative since N4 1 is negative. To avoid the use of ji asa signum function, as required for motion reversal according to appendix E, itsvalue is made absolute. Possible motion reversal is then only affected by the sign

of ; , the angular velocity of the escape wheel, which is normally positive. Thenthe resulting term has the following form:

2

-2 Itilp' N4 1 ; (A25 + A2 9)

In order to determine the sign of the pivot friction moment which is propor-tional to the angular acceleration * of the escape wheel in equation A-147, theideas presented in appendix F are used. To be able to do this let the value of the

61

coefficient of friction of this term become absolute, so that it ceases to serve asa directional signum function in the sense of appendix E. Further let the expres-sion be changed, for the time being, so that it becomes a function of the rotor

acceleration j . With the above and in the sense of equation F-2, appendix F, the

absolute value of this friction moment MAA may be expressed as

MAA = I 'P1 (A 26 + A 30) 1

From this point on one may use the reasoning of appendix F directly, keeping in

mind that I wIPI(A26 + A 30) and I are now used instead of A 2 2 and i, respectively.

As for the four cases in appendix F, the effective moment of inertia 1I, takestwo forms:

IR =I + I IP1 (A26 + A 30 ), when I and I have the same sign,

and

IIR = I- lpIP1 (A 26 + A30), when ;I and Il have opposite signs

Equation A-147 gives all relevant expressions in terms of the escape wheel var-

iables and ; . Since they both are proportional to j and j by the identicalgear ratio N4 1 (or in the case of the two pass train by way of N3 1), one may readily

extend the above computational rule to the escape wheel variables. This is re-flected by equations A-155 and A-156.

The above considerations give the moment equation A-147 the following form:

2F 1 2 [Rbl - vs 1a I + op(A 23+A 2 -) ] + w pIju(A 24+A 28 )

2 W 2

-Ii*Ii I N (A(A254-A2 29)I 41(A25+A29) + 41 (A 2 5 A2 9 )

2 "

= mrCl 6?li sin( Ic+N 4 1T) + (1 1 + vjpl(A 26 +A 3 0 ))N41 (A-148)

L 62

Finally the above expression is solved for F,2

2 "

-A 3 2+N4 1 A 3 3, - A 3 4(N 4 1$) + A35sin( 1 c+N41T) + IIRN4101F1 2 A 3 1 (A-149)

where

A 3 1 = RbL - s ia I + lip (A 2 3 + A 2 7) (A-150)

2A 3 2 = w pj1(A 2 4 + A2 8 ) (A-151)

2

A 3 3 = 2 Wk- P, (A 2 5 + A2 9) (A-152)

A 34 = uop 1 (A 2 5 + A 2 9) (A-153)

2

A 3 5 nm 6 1rCL W (A-154)

IR =1 1 + IHlpi (A 2 6 + A 3 0 ), when 4 and 4 have the same signs (A-155)

IR =I1 - liuP1 (A 2 6 + A3 0 ), when $ and 4 have opposite signs (A-156)

DYNAMICS OF GEAR AND PINION SET NO. 3(APPLICABLE TO BOTH CONFIGURATIONS WITH PROPER sb )

The moment equation of gear and pinion no. 3 may be taken from reference A-2 ifits right hand side is set equal to 13,3 rather than to zero. (The dynamic forceequation was already satisfied by the inclusion of the inertia force T3 .) rhus,with the help of equation A-65, reference A-2, and letting

03 N4 3 ( (A-157)

63

N 43 N 4 (A-I 58)N G3

following expression is obtained

3 F 3 4 - Ps 3 a 3 F 3 4 - rb 3 F23 + Is 2 (d 2 - a 2 )F 2 3

+ t0p3 [FZ 3(A3 6 + A 3 9 ) + T 3 (A 3 7 + A4 0 ) + F34(A38 + A4 1)

I 3N 4 3 (A-159)

re

2(I + UI s 2 )cos(6 2 - 62) + P(s 2 - I)sin(B2 - 62)

(A-160)b+ I 21 + i

sinY3 + tcosy31 =-~ (A-lb 1)

2

1 + P

2(1 + 0 s 2)sin(B6 2 02) + i(I - s 2)cos(032 02)

(A-163)2

1+ P

2(1 - U s 3 )sin(B3 + 03) + U(1 + s 3 )coS( 3 - 03)

= - (A-163)

9 2

1+ U

64

If equation A-159 is solved for F23 , the following is obtained:

F 3 4 A 4 2 + T 3 A43 - 13 N4 3 .$

F 2 3 =A44 (A-166)

where

A 4 2 = Rb 3 - "[s 3 a 3 - P3(A38+ A 4 1 )] (A-17)

A 44 = 0p3 (A3/ + A 4 0) (A-168)

A 4 4 = rb 3 - V[s 2 (d 2 - a 2 ) + pj(A36 + A 3 9)] (A-169)

DYNAMICS OF GEAR AND PINION SET NO. 2(APPLICATION TO BOTH CONFIGURATIONS WITH PROPER s6 )

The moment equation of gear and pinion set no. 2 may also be taken from refer-ence A-2 with the provision that its right hand side is set equal to 12$, ratherthan equal to zero. (Again, the force equation was already satisfied by the inertiaforce T2 .)

With the help of equation A-92, reference A-2, and letting

02 = N 4 2 0 (A-170)

where

N4 NpN42 N 4NP (A-171I)

NG3 NG2

the following is obtained:

-Rb 2 F23 + ps 2a 2F 2 3 + rb 2 F 12 - (d -a )F12

-P2 [F 12(A 4 5+A4 8) + T 2(A 4 6+A 4 9) + F 2 3(A 4 7+A 5 0) ] = I 2N4 2$ (A-172)

65

whe re2

(I - s 1)sin( + 01) + (1 + p s 1)Cos( + e1)

A4 5 = (A-173)

I+

A 4 6 - i 2 1 (A-174)

2

)J(1 + s 2 )sin(R2 - 02) + (I - j s 2)cos(0 2 - 02)A 4 7 = (A-I75)

I1+ ii

21( - s )cos( 1 + O1) - (I + tj s )sin( O + 0 I)

A 4 8 =2 (A-17)

1+ i

isinY2 + cosY2

A 4 9 -- (A-177)

21+

2(I - i s 2 )sin( 2 - 02) - 1(I + s 2 )cos(6 2 - 02)

A5 0 2 (A-178)

1 +i

The following is found if equation A-172 is solved for F12 :

F 2 3 A 5 1 + T 2 A 5 2 + 12 N4 2F 1 - A53 (A-179)

whe re

A5 , = Rb2 - P [s 2a 2 - P2(A4 7 + A5 0 )] (A-180)

A 5 2 = PjP2 (A 4 6 + A4 9) (A-181)

A 5 3 r -b2 [ s (d I - a 1) + p 2(A 4 5 + A4 8) ] (A-182)

66

DYNAMICS OF COMBINED SYSTEM IN COUPLED MOTION

(APPLICABLE TO BOTH CONFIGURATIONS WITH PROPER s6 )

In order to obtain a single differential equation for the total system in

coupled motion, in terms of the escape wheel variable $, an appropriate expressionfor the contact force FA4 must be substituted into equation A-121. To this end,equations A-149 and A-i, both expressions for the force F,2 and originating from

the expressions for the rotor and gear and pinion set no. 2, are first set equal toeach other and then are solved for the force F2 3 :

F 23 = A 151 -A32A53 - A 3 1A 5 2T 2 + A35As3sin($IC+N415T)

2 .2+ A 3 3A5 3N 4 1 - A 3 4A5 3N41 4

+ (As31IRN41 - A 311 2N 4 2)O] (A-183)

The just determined expression for F2 3 is set equal to equation A-16b for F23 ,

which originates from the dynamics of gear and pinion set no. 3. This results in:

F A -A 3 ,A 4 4 A 5 3 - A 3 1A 4 4 A5 2T 2 - A 3 1A4 3A 5 1T 3F =A31A42A51

+ A3sA44A53in( IC+N 4 14h) + A 3 3A44A 5 3 N4 1 ;

. 2

- A 3 4A 4 4A5 3(N 4 1 0)

+ (A44A53I1RN41 - A 3 1A4 41l 4 2 + A 3 1A 5 11 3N 4 3)o ] (A-184)

67

Finally, the above is substituted into equation A-121. The following system

differential equation for coupled motion is obtained:

A 5 4 + A 550 + A 5 6 ; = A5 7 + A58sin(* - *-*C ) + A59sin(tic+N41) (A-185)

where

A 1 0A1 BA4 4A 5 3A 5 4 14A 18+ 'PR A1 -/ - A 3 lA 4 A1N4111R

AIS~es,,AISAld

N4 212 A4 2 N4 31 3 (A-186)+ 4A51A2

2 A 1 5A I A 3 4A 4 4A 5 3 2A 5 5 IpR A 17V + AIA 2IZ + -A31A42A51 N4 (A-187)

2 A 15A jA 3 3A 4 4A 5 3

As5 = 2-w A 17A2OZ- N4 1 (A-188)

A5 7 = -A 1 6A 1 T4 A 4 A4 A5

A 15A 1 A 3 2A44A 53 2A_ AIB 32 4 -A 17A I,} qu {A-189)

A 3 1A4 2A5 1

2A 5 b = AI-1w mprcp 6?5 (A-19U)

A1 IA 1 A3 A44A 5 3A 5 9 = A3 1A4 As5 (A-191)

68j

CONTACT FORCE EXPRESSIONS FOR COUPLED MOTION(APPLICABLE TO BOTH CONFIGURATIONS WITH PROPER s6)

The following force expressions for various contact forces are useful forstrength calculations, and are applicable to both configurations.

According to equation A-184

F 3 4 [ -A 3 2A 4 4A 5 3 - A 3 1A 44A 52T 2 - A 3 1A 4 3A 5 1T 3A 3 1A4 2A 5 1

+ A 3 5A 4 4A53sin( Ic+N 4 1 T ) + A 3 3A 44 A5 3N 4 1;

.2- A 3 4A4 4A 5 3(N 4 10) + (A 44 A531 1 RN41

- A 3 A 4 4'LN 42 + A 3 1AsiljN4 3) ] (A-192)


F 34A4 2 + T 3A43- I3N430F 2 3 - A44 (A-193)


F2 3A5 1 + TA 52 + 1 2N 42$F 1 2 A (A-194)

A5 3

The contact force P between escape wheel and pallet, may either be expressed

in terms of the escape weel variable * according to equation A-117, or in terms of

the pallet variable * according to equation A-91. Thus,

69

-I4 + F 3 4A1 5 - T4A 1 6P n (A-195)

or

2.2 2

S- [ Ip + A 2 1 p + 2- A 2 0 + W A 1 9n* A 1 8 PR

2

M s 6 (A-196)

DIFFERENTIAL EQUATIONS FOR FREE MOTION REGIME(APPLICABLE TO BOTH CONFIGURATIONS WITH PROPER s6 )

Both the differential equations for the free motion of the pallet and thesystem composed of the escape wheel, gear and pinion no. 3, gear and pinion no. 2,and the rotor may be obtained from certain coupled motion expressions.

Free Motion of Pallet

Let Pn = 0 in equation A-91. After some rearrangement, the followingdifferential equation results:

.2A60* + A21 + A 6 1 = - A 6 2 + A 6 3sin(y' - k- C) tA-197)

where

A6 0 = IPR A (A-198)

A-8 For free motion, IPR cannot be zero since it would make the value of in

the Runge-Kutta solution indefinite (footnote A-4).

70 I1

2

A G 1 • 2 w -- Azo (A-199) lwl

2 Ali2 • w A l:J (A-200)

2 61 A 63 = w mprCP 5 (A-201)

Free Motion of the Escape Wheel, Gear Train, Rotor System

For the pr~dent case start by letting Pn • 0 in tha escape wheel expression A-117, i.e.,

(A-202)

Equation A-18LI, which furnishes an expression for force F34 , is than substituted i.nto equation A-202. This provides the desired second free motion differential e<l.uation

(A-203)

(A-204)

(A-205)

(A-20b)

71

A 5A 4 3 A 15A 4 4A 2 A 1 5A 3 2A4 4A 5 3AA7T 2 - A31 A (A-207)

A6 7 -A 16 T- A4 2 T3 -A - T21 _AA42As1

A 1 5 A 3 5A 4 4 A 5 3A 6 8 - A 3 1 A 4 2 A 5 1 (A-208)

CONTACT FORCE EXPRESSIONS FOR FREE MOTION

(APPLICABLE TO BOTH CONFIGURATIONS WITH PROPER s6)

The contact force FF34 may be obtained from equation A-117 by setting Pn equalto zero, so that

'44 + T 4A 1 6FF34 A15 (A-209)

The expression for the contact force FF23 has the same form as equation A-193since it originates from the dynamics of gear and pinion set no. 3

F F34 A42 + TA 4 3 - 1 3N4 3 AFF2 4 (A-210)F23 A4 4

Similarly, the form of FFI 2 is like that of equation A-194

FFL2 F 2 3 51TA522N42 (A-211)

CHANGES IN IMPACT EXPRESSIONS(APPLICABLE TO BOTH CONFIGURATIONS WITH PROPER s6 )

While generally the impact description of reference A-I remains unchanged, thetotal moment of inertia ISTOT of the escape wheel is now increased, since it mustalso account for the presence of the gear train and the rotor. Therefore,

72

2 2 2 (A-212)ISTOT = 14 + 13N4 3 + 12N4 2 + 11N4 1

where

14 escape wheel - pinion no. 4 moment of inertia

13 gear and pinion set no. 3 moment of inertia

12 = gear and pinion set no. 2 moment of inertia

1 = rotor moment of inertia

See equations A-125, A-158, and A-171 for the transmission ratios.

73

REFERENCES

A-I. G. G. Lowen and F. R. Tepper, "Dynamics of the Pin Pallet RunawayEscapement," Technical Report ARLCD-TR-77062, ARRADCOM, Dover, NJ, June1978.

A-2. G. C. Lowen and F. R. Tepper, "Fuze Gear Train Analysis," Technical ReportARLCD-TR-79030, ARRADCOM, Dover, NJ, December 1979.

74

APPENDIX B

DYNAMICS OF ROTOR DRIVEN S&A DEVICE WITH A TWO-PASSINVOLUTE GEAR TRAIN AND A PIN PALLET RUNAWAY ESCAPEMENT

75

DESCRIPTION OF SYSTEMS AND OUTLINE oF DERIVATIONS

Figures B-I and B-2 show the two configurations of the S&A mechanism discussed

in this appendix. These devices differ from those described in appendix A because

they contain only two step-up gear meshes. Since the escape wheel still is assumed

to have counterclockwise rotation, the motion of the rotor must also be counter-

clockwise, and the gear and pinion set no. 2 must turn in a clockwise direction.

This requires certain changes in the rotor geometry as well as in the direction of

the lines of action of the gear contact forces F1 2 and F2 3 . Naturally, all quanti-

ties previously associated with gear and pinion set no. 3 are now absent.

The general approach to the modeling of these devices is the same as that used

in appendix A. While the derivations are specifically for configuration no. 1, the

results are applicable to both configurations as long as the appropriate fuze body

angles are used.

GEOMETRY OF FUZE BODY CONFIGURATIONS

Fuze Body Configuration No. I

Figure B-3 indicates the following relationships for the angles yi,

and 8. for configuration no. 1:

According to equation A-I (app A)

-1 + 6 2 2] + 1Y2 = cos 1 2 (R1 2+ r 2 (B-1)

Further, according to equation A-2 (app A)

1 [ 62 2 21

= cos- 2 + 3 - (Rp2 + r p3 ) (B-2)y = cos 2 R2 6?3 -2

Finally, from equation A-3 (app A)

Y3 =2 + Y3 (B-3)

77

'41

L4..

44

0 4.

4J41

C Z

r-

I-J0

CCCL)

SY to

-j C-

-A V) ad

L<0

79C

'C

0 41

+9

6-4

'1

-C

800

The angle y must be newly derived. Thus,

3 + ?4- a 2

S 2 COS2 (-4

Then, as in equation A-5 (app A)

Y4=Y + Y (B-5)

Angles 6i

The angles 62 and 63 can be taken directly from equations A-9 and A-10(app A), respectively, i.e.,

62 = Cos-I p + p2) 1 f2t2 g I(R p + rp 2 )

and

(R +r ) + 6?2-

I p2 (2p + rp)63 = Cos 2 ,(Rp3 -r - (B-7)

p2 p3 J

The angle 64 must take the pivot-to-pivot distance a of the escapement into

account. Therefore,

[2 + 2 2]

64 = cos- 1 a + (3 - )?42a 6? 3 ( - )

Angles 6

The angles R, and 62 may also be taken directly from equations A-14 and A-

15 (app A), respectively, i.e.,

61 = 'T - 6 (B-9)

and

a2 = Y2 + I - 63 (B-10)

81

The angle 03 is found with the help of angle 64 of equation B-8. Then, similar to

equation A-16 (app A)

y3 Y 3 + 7r - 64 (B-li)

Figure _H-3 shows the angle y' between the positive X- axis and the unitvector n 4 It is given by

y- = - E(B-I2)p

where e is obtained with the help of equations B-4 and B-8

and therefore

y = 4+ (-14)p

The unit vector n4 is expressed in terms of the primed coordinate system as follows:

n4 = cosy17 + sinyj (-15)p p

Further, the unit vectors i and j', when expressed in the x-y system, become

i" = cos(0 3 - w)i + sin(a 3 - )j

or

- -cos 3i - sina'i (B-16)

and

J' = k' x C sin 3 - cog13 j (B-17)

82

Fuze Body Configuration No. 2

Figure B-4 is used to define the angles yi , and a of configuration no. 2.

Angles yi

Since these angles are defined in the clockwise direction with respect tothe body fixed X- axis, their value must be negative, as is the case for the threepass configuration no. 2 of appendix A.

According to equation A-24 (app A)

I 2 + [ 2 2 - (R + r p2

Y2 -cos 2 2 (B-I)

Angle yj is given by equation B-2, and with that y 3 becomes

Y3 = Y2 - Y3 (B-19)

The negative sign for y5 is necessary in order to make y 3 negative.

The angle y4 is given by equation B-4. It allows the following expression for

angle Y4

Y4 = Y3 - Y4 (B-20)

Again, y' must have a negative sign to make Y4 negative.

Angles 8i

The angles 62, 63, and 64 can be taken directly from equations B-6, B-7,and B-8, respectively.

Angles Ri

As in equation A-28 (app A)

71 + 62 (B-21)

Further, according to equation A-29 (app A)

B2 = Y2 + ' + 63 (B-22)

83

01CL V)

CLC

cl0

~0

CL CL

84

Finally, according to equation A-30 (app A)

03 = '3 + iT + 64 (B-23)

Figure B-4 shows the angle y' between the positive x'- axis and the unitvector n 4 . This angle is givenpy

Y' = + (B-24)p

where

E= - - (B-25)

Therefore,

y' = 2ir - 64- y (B-26)

p

The unit vector n 4 can again be expressed in terms of the primed coordinate systemas

n 4 = cosy'i" + siny'j (B-27)p p

In r manner parallel to that used for equations A-36 and A-37 (app A), the unitsvectors i" and j" in the X-Y system are expressed as follows:

- -cos 3 i - sinBj (B-28)

and

j sin6 3i - cos 63 (B-29)

Common Computational Expressions for Both Configurations

In order to find common programming expressions for all angles of both configu-rations, with the exception of the angles y" of equations B-14 and B-26, the follow-

ing signum function s6 is again introduced p

s +1 for configuration no. 1 (B-30)

85

s6 • -1 for configuration no. 2

ThiA lends to the following expressions:

according to equations B-1 and B-18

tR-.:2:----(_R.~:.P=-1 _+_r P.c:.:2:....)_ 2 2] 2tRliR2

according to 12quation B-2


nccording to e~uation B-4

-1 ':'4 =: cos


nc<":ordinr, to equations B-6 through B-8, respectively

-1 02 .,. COB

(B-31) ·

(H-32)

(B-33)

(B-34)

(B-35)

(B-36)

(U-37)

(R + r ) + 62 6? 2

(S3= Cos - 1 p2 p3 z R 2 r 3) (B-38)

2 2 p2 p3

[2 +6?2 _ 6j2

= COS [ 2a 6 3


1= i - s662 (B-40)


R2 = Y2 + 7T - s66 3 (-

according to equations B-I and B-23

93 = Y2 + 1T - s6 4 (8-42)

The signum function s 6 will also be useful in conditional statements which distin-

guish between the y*Ws of configuration no. I and 2, as given by equations B-14 and

B-26, respectively.p

The expressions for the unit vectors i" and j' are identical for both configu-

rations according to equations B-16 and B-17 as well as equations B-28 and B-29.

Thus,

iV - cos6 3i -sin 3j (8-43)

and

j = sin$ - Cos 3J (8-44)

87

DYNAMICS OF PALLET IN COUPLED MOTION(APPLICABLE TO BOTH CONFIGURATIONS WITH APPROPRIATE s6 )

Equation A-95, appendix A, which gives the contact force Pn between pallet andescape wheel, may be readily adapted to the present two pass mechanism. It is onlyrequired that, depending on configuration, the angle y' is computed according toeither equation B-14 or equation B-26. Further, the distance from the spin axis tothe pallet pivot is now given by 6 4 rather than by 6?5" Thus equation A-95 be-comes

2RU Z2 .2 W 2

= +A PRUO+ (A2 IpV) + 2 -- A20Z + w A 1

mprc p R% W sin(ypq.-.C ) (B-45)

The parameters A1 to A8, A18 to A 2 1 and I may essentially be taken directly fromPRequations A-70 through A-77 and equations A-84 through A-90, appendix A. In equa-tions A-71 and A-75 it is necessary to substitute 614 for 635 o

DYNARICS OF ESCAPE WHEEL IN COUPLED MOTION(APPLICABLE TO BOTH CONFIGURATIONS WITH THE APPROPRIATE s6)

The analysis of the escape wheel can be taken over from appendix A, once theappropriate signum function s 6 has been chosen for the given configuration and tuefollowing changes in nomenclature have been made:

4 becomes 3

03 becomes 02

m4 becomes m3

6 4 becomes ? 3

14 becomes 13

Y4 becomes Y3

S3 becomes s2

88

d 3 becomes d2

a 3 becomes a 2

n 34 becomes n 2 3

F becomes F 2 3

Fx4 becomes F2 3

F becomes F 3

F 4 becomes F 3

r b4 becomes rb3

T 4 becomes T3

Note that the meanings of s4 and s5 remain unchanged (eq A-59 and A-60).

Similar to equation A-117, the contact force equation, as derived for the

escape wheel, then has the following form:

-10 + F 2 3AI5 - T 3A 1 6p = -(B-46)n A 1 7

where now

A15= rb3 - [s 2(dZ - a2 ) + p 3(A1 0 + A 1 3)i (1-47)

A 1 6 = lip 3(A 1 1 + A 1 4 ) (B-48)

89

Al 7 - Al I + +pP 3 (A 9 + A 1 2 ) (B-49)

and similarly

-(JIS4 -P )sin( -a-If3) + (1+1Pi s4 )cos( -c- 3 )A9 = (B-50)

21+ l

2-(l-s C)sin(032+ 2 ) (1+P s2)Cos(02+02 )A10 =2(B-51)

I ++j

Sin s)3 - Pcosys3)A . .. ...... (B-52)

22

1 ++

( l+P 1s )sin(-a"+ 3) + (si 1-P )cos(W-a+03)A12 (B-53)

S+

2-4j~ s2 )sin(62+)2) + IJ(I-S2 )COS(02+()2)

13 = 1(B-54)

I + 2

-cosy 3 - Wsiny 3A1 4 . .. (B-55)

2I +

90

COMBINED COUPLED MOTION DIFFERENTIAL EQUATION FOR ESCAPE WHEEL AND PALLET

(APPLICABLE TO BOTH CONFIGURATIONS WITH APPROPRIATE SO

Equations B-45 and B-46, are now set equal to each other in order to obtain acombined differential equation for the escapement. This leads to

[IA1d+ IpRAI7U ;+ [IpRA.7V + A1 7 A2 1 Z

2

+ 2 w Ai 7A2 0Z

= A1IBAF2 3 - A1 6 Aj 8 T3- A, 7w [A,, - mprCpd?4Sin(y;tjMPC)1 (B-56)

Again, it is necessiry now to determine an expression for the contact force F2 3 .

DYNAMICS OF ROTOR (Gear No. 1)

(APPLICABLE TO BOTH CONFIGURATIONS WITH APPROPRIATE s6 )

Figure B-5 shows a free body diagram of the rotor of configuration no. 1, whichmoves in a counterclockwise direction. The acceleration of its center of mass isgiven by equation A-122, appendix A.

Since the motion of the rotor must be expressed again in terms of the escapewheel variables, use must be made of

and

j = N 3 1 (B-58)

where

N x N-P3 P2 (B-59)N 3 1 xG

91

0D4-

iz0

oo

-A

I 0

The rotor angle t C + j is expressed in terms of the total escape wheel

rotation OT

sIc + 01 = ¢I¢ + N 3 1 T (8-60)

(For details, see program description.)

Newton's force equation may now be written in the form of equation A-128. Thetooth forces

f fi -F 1 2n, 2 (B-61)

and

Ff 2 1 =s1F12nN1 2 (B-62)

have the appropriate directions to account for friction.

Thus

-F 12n 12 + 'a 1F I2nN12 + Fxli - uF yl + Fylj + pFxli

2 . 2__

mI - 2 11 - (w + N3 1 ) r [cos( IC+N3%)! + sin( lC+N3h)j]

+ N3I0rCI [-sin(sic+N314T)l + cos(5 1 C+N310T)j]} (6-63)

where

n12= -sin( - 81)1 + cos(Ol - e1)j (B-64)

nN12 -cos(l - e1)i - sin(O8 - el)j (B-65)

93

The moment equation must be written in the manner of ejuation A-61 with respectto the accelerated point 01. The pivot reactions IF and F are written such thatthe associated friction moments retard the countercybckwiseyrotation of the rotor.This leads to

-F12Rbl + IsIaIF 12 - UP1(FX1 + F yl)

2= m1w RirClsin( iC+N31h) + IN 3 1 (B-66)

The forces F and F are obtained after the simultaneous solution of thefollowing componen

1 expreskons of equation B-63 for F and F

X1 yl

F2 sin(BI-0) + slBF 1 2cos(8 1-6 1 ) + F -

12 X1 .2m1 j-u2 6 Oi - (o*N 3 1 ) rclcos(ic +N311) - N31rClsin( lc+N3l4) (B-67)

and

-Fl 2cos(BI-6 1 ) - usiFl 2sin(a1-61) + Fy I + pF

r -(s*N3J)rC2rlsin(.iC+N3 I4T) + N310Ccos( I c+N 3 T) (B-68)

Simultaneous solution of the above furnishes

2 W 2

F F 1 2A2 3 * w A 2 4 -k 2 -I N314 25 * (N 3 ) A 2 5 * N3 14A 2 6 (B-69)

22 2 2

F F 1 2A 2 7 * (A 2 8 2 w N314A29 * (N3 J) A 2 9 * N3 10 3 0 (B-70)yl IWI

94

where

2

(I- t s )sin( 01-01) p( 1~+9 )cos~a r-61)A23

1+ I

mlin [611 +4 rCj(cos( lC+N31hi) +I jzin( ,d+N3l4h))]2

1+

inirCl [cos(ojd+N31k) + jsin(,ic+N 3l4)]A2 5 2 (8-73)

21+

2

ujl+s 1 )sin( 01-01) + (1-13s)o( 161A2 7 =(B-75)

2++

+j~ po~j+3h -nr~ [izo( iCn31 t-sN30,,Nl4)]A2 9 2 (5-77)

+ ii

A3 - m~rC, [m~in(fl6+N3 llr) + Cos( $I cN3 l#T)j

1+ p

95

Equations B-69 and B-70 are now substituted into the moment equation B-66

2-F12Rbl + Is la 1 F 1 2 - vIl [F 1 2(A 2 3+A 2 7) * w (A 24+A 2 8 )

2

2 w N3 ji(A2 5*A2 9) (N 31 ) (A25+A29) * N3 1;(A 2 6+A3 0 )]

2mI 6 ?IrClsin(.iC+N3 1h) + IIN310 (B-79)

In order to have the pivot friction moment negative, the following terms on theleft hand side of equation B-79 must be negative:

ap1FIz (A 2 3 + A2 7)

2i1Pw (A 2 4 + A 2 8)

221ipi I N(N3 1 (A25 + A29)

(With N 3 1$ positive, j must be absolute. See discussion following equation A-147).) The term 'ppN 3 1 I (A 2 6 + A3 0) is treated in the same manner as was shown inconnection with equation A-149, appendix A. (See the discussion in appendix F.)

96:!I.

The above considerations give the moment equation the following form:

2-F 1 2 [Rbl - m ja1 + lip(AZ'IA2 7)] - w ppj(A24 +A 2 8)

22ip1Iw N 3 1 2

(A 2 5+A 2 9 ) - IPl (N 3 1 ) (A 2 5 + A 2 9 )

2"

mira lrClsin( lC+N314T) + (I 1 *tip 1(A 2 6*A 3 0 ))N 3 10 (B-80)

Finally, the above expression is solved for F 1 2 , i.e.,

2 "

-A 32 -A 3 3N 3 1$ - A3 4(N 3 iJ) - A3 5sin(Oic+N3 0%) = IlRN310F 1 2 f A31

where

A 3 1 Rbl - jts ia 1 I up(A 2 3 + A 2 7) (B-82)

2A 3 2 = ij Pw (A 2 4 + A 2 8) (B-83)

2

A3 3 - 21J0pip- (A 5 + A2 9 ) (B-84)

A 34 - lipI(A 2 5 + A2 9) (B-85)

A35 2 m I rClW (B-86)

97

IIR ' Il + Iilp l(A26 + A3 0), when * and * have the same signs (B-87)

I1R I uIpl (A2 6 + A3 0), when * and * have opposite signs (B-88)

DYNAKICS OF GEAR AND PINION SET NO. 2(APPLICABLE TO BOTH CONFIGURATIONS WITH APPROPRIATE s6)

The free body diagram of gear and pinion set no. 2 of configuration no. I isshown in figure B-6. Its force equation is written as follows with the D'Alembertforce

2T 2 m2 6?2 w (B-89)

Thus

-F2I 2 3 - s2IiF2JN23 + F 1 ;p 1 2 -s IF 12N12 Fx2' + Fy 2

+ Fy2J - Fx2J + T 2 (cosY2i + siny 2j) - 0 (B-90)

where

n23- sin(0 2 + 02)1 - cos(8 2 + 0 2)J (B-91)

nN2 3 C 0os(12 + e 2 )i + sin( 0 2 + e 2 )J (8-92)

98

C.3 r

N4j

co

4)

(4-a. 0

-4

C4U

ci 4)m 0

'-4

k -

C4)

acc

99

The unit vectors n 12 and N1 2 were given by equations B-64 and B-65. Hence

F3= -F2 3 n2 3 (B-93)

f32= -s2F23nN23 (B-94)

To write the moment equation, let

;2 -N32 (B-95)

where

N32 (8-96)NG2

Then

F 2 3Rb2 - s 2F 23a - F 12rb2 + sF (d-a 1) + 2(F+Fy2)

- 1 2N 32t (B-97)

The bearing forces F and F are obtained from the simultaneous solution ofx2

the following component equations Y1 the force expression B-90:

-F 2 3sin(B2+02) - F 2 3 W.e2cos(O~f 2 ) - F12sin(0 1 -0 1 )

+ 1i8F12cos(8 1- 1) + Fx2 + I 'y2 + T 2PosY2 - 0 (B-98)

and

100

F 2 3cos(0 2+82 ) - us 2F 2 3sin(S 2+8 2) + F, 2cos(I-O1 )

+ is 1F Lstn( 8 1-0 1 ) + F 2 - Sx2 + T 2siny 2 0 (B-99)

This leads to

Fx2 ffi F 2 3 A 3 6 + F 1 2 A 3 7 + T A 3 8 (B-IOU)

F = F 2 3 A 3 9 + F 1 2 A 4 0 + T 2 A 4 1 (B-IU1)y 2 2

(all signs are left positive in order to furnish positive friction moments in equa-

tion B-97) where

2(1-p s2)sin(62+82) + p(l+s 2)cos(B2+82)

A 36 2 (B-02)

1+ tp2(l+ij Sl)sin(61 -61 ) + pCl-s l)COS(8 1-61)

3=i(B-f03)

1+ p

A38 .... (B-104)2

1J(l+s 2)sin( BZF02) -1 at S2)COS( 021+02)

1+ p

2

A4- l-S)sin(1-0l) - (I+ S )Cos(8)-8 2 ) 1A9 = (8-105)

2A 4 0 = 2 (B-107)

++ p

- 1(-107)

1+ p

101

I

Substitution of equations B-100 and B-101 into the moment equation B-97 and

subsequent solution for the contact force F2 3 gives

F 12 A 42 - T 2 A 4 3 + 1 2 N 32 °

F 23 = A4 (B-108)

where

A 4 2 - rb2- i [s i(d I- a 1) + p2(A 37 + A 40)] (B-109)

A4 3 = 1P2(A38 + A 41) (B-10)

A 44 = Rb 2 - 1 [s 2a 2 - p 2(A 3 6 + A 3 9) ] (B-I>

DYNAMICS OF COMBINED SYSTEM IN COUPLED MOTION(APPLICABLE TO BOTH CONFIGURATIONS WITH APPROPRIATE s6 )

In order to obtain a single differential equation for the total system incoupled motion, in terms of the escape wheel angle *, an appropriate expression for

the contact force F2 3, which also contains the contribution of the rotor, must besubstituted into equation B-56.

Thus, first substitute equation B-81 for F12 into equation B-108, which is theabove expression for F2 3 :

1 -A4 2A 32 A 4 2A 3 3 A 42A 34 2 .2F2 3 A- A3 1 T 2A 4 3 - A 31 N3 1 - A3

F2 AA44311

35 A 42 1A---i- sin( ,C+N314r) + (120 32 - IILRN31)J (B-112)

102

Now equation B-12 is substituted into equation B-56, and the final differentialt equation of coupled motion results

A40 + A460 + A47

= A4 8 - A4 9sin( ,c+N 3 1i) + A50sin(p--4'C) (B-113)

where

A1 5A1 8 A 42

A 4 5 = 1 3A 1 8 + IPRA7U - A44 (1 2 N 3 2 - - IlN 31) (B-114)IPA ~ A 4 A 3 1 i

2 A1 5 A1 8A 3 4A 4 2 2A4 6 = IPRA17V + AIA 2IU + A3 1A4 4 N 3 1 (E-115)

A47- 2 2 A 15A 1&k 33A 42N3(B16A4 7 = 2 - - A 17A20 + A31A 3 -llb)

A1 5A 1 8A32A4 2 A i5A 18A 4 3 2

A4 8 = - - - A4 - T2 - A1 6A 18T3 - w A1 7A1 9 (B-117)A4 A31A 44 A4

A 1 5A 1 3 5A 4 2

A4 9 - -__A31A44 (B-118)

2

A 5 0 = mp rCP 6?4A1 7,w (B-119)

103

CONTACT FORCE EQUATIONS FOR COUPLED MOTION

(APPLICABLE TO BOTH CONFIGURATIONS WITH APPROPRIATE s )

The contact force F23 is given by equation B-112

A 3 2 A 4 2 A 3 3A 4 2 A 4 2 A 34 2 .2F23 A31 A3 1 A3 I A31 N3

A35A2 A42

A3 1 sin( iC+N31h) + (I 32 - A31 IIRN31)] (B-120)

1The contact force F1 2 is found with the help of equation B-108

F 2 3 A 44 + T 2 A 4 3 - 1 2 N 3 2

A4 2 (B-I21)

The contact force Pn is given by equation B-46

-13 * + F 2 3 A 1 5 - T 3 A 1 6p - (B-122)n A1 7

If this force is desired in terms of the pallet variable I, equation A-91 must be

used with the understanding that it now has the following form:

2.2 2 2

+ A 2 1 * + 2 - A 2 0 + wA 19 - mprcp 04W sin(yp-r-) C )= W A8aC (B-123)n A1 8

where 64 is used, together with y; as defined earlier in this appendix. As stated

previously, all other parameters may be taken from appendix A.

104

DIFFERENTIAL EQUATIONS FOR FREE MO'TION

(APPLICABLE TO BOTH CONFIGURATIONS WITH APPROPRIATE SO

Free Motion of Pallet

Yet P = 0 in equation B-123. After some rearrangement, the free motion

differential equation for the pallet becomes

.2A5j + AZI1 + A5 2 = -A5 3 + A54Sin(y;-r.-PC) (i-124)

where

A 5 , = iPR (i-125)

2

A 5 2 = 2 - w A 2 0 (B-12b)

2A 5 3 = A 19 (B-127)

2A 5 4 W m p rCp 64 (B-128)

(Again, the angle y; is defined for both configurations by eq B-14 and B-2b.)

Free Motion of the Escape Wheel, Gear Train, Rotor System

By setting Pn - 0 in equation B-46, the differential equation of the systemwithout the pallet is obtained. This leads to

13 * = A 1 5 F 2 3 - T 3 A 16 (B-I29)

Equation B-112 is then substituted into the above for F23. The resulting

differential equation is given by

105

.2A5 5 # + A 56 * + A 5 7# , A5 8 - Assin( iC + N31h) (B-130)

where

A 1 5 A 15A4 2A 5 5 = - N3 21 2 + A 3 1A4 4 N3l1lR (B-131)

A 1 5A 34A4 2 2A 56 A44 N 3 1 (B-132)

A 1A 3A42

A5 7 N3 A3 1A. N 1 (B-133)

A57 A3 2 AL 4 2A 15A4 3 A 15A 3244,2

A 5 8 -A 16T 3 -A T 2 (B-134)44 A 3 1A4 4

AlA 35 A4 2

A 5 9 A31A44 (B-135)

CONTACT FORCE EQUATIONS FOR FREE MOTION

(APPLICABLE TO BOTH CONFIGURATIONS WITH APPROPRIATE 56 )

Equation B-129 may be solved for the free motion contact force FF23, once * isknown

FFZ3 = 13 * + T 3 A 16 (B-136)F3 A 1 5

Equation B-108, which was derived from the force and moment equations of gear

and pinion set no. 2, may be modified to obtain the free motion contact force FF12

FF23 A44 + T 2 A 4 3 - 12 N3 2 (B-137)FF12 A4 2

106

It must be understood that, both in equations B-136 and B-137, the angular acceler-ation of free motion *, as obtained from the solution of equation B-130, is used.

CHANGES IN IMPACT EXPRESSIONS(APPLICABLE TO BOTH CONFIGURATIONS)

Both the rotor and gear and pinion set no. 2 must be referred to the escapewheel shaft. Thus, similar to equation A-212, appendix A

2 21 3 3+ 1N32+II31(B-138)

STOT 1 3 +I 3

See equations B-59 and B-96 for the above transmission ratios.

107

APPEtIDIX C

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177

AP~PENDIX E

FRICTION FORCES DURING GEARTRAIN MOTION REVERSAL

179

k-MED1ldO PAGN BLANK-NO0T 71 I&

Since all friction forces of the escape wheel to rotor gear train reversedirection during a possible motion reversal following an impact, the introductionof a negative coefficient of friction, i.e., a -i, will satisfy for computational

purposes the associated force and moment equilibrium expressions. (This methodneed not be resorted to for the pallet-escape wheel interface, since there motion

reversal is accounted for by the signum function s4.)

The above will be proven to hold for both the tooth contact and the pivot

friction.

Tooth Contact Friction

The direction of the friction force of the pinion on a gear of a mesh is the

same as that of the relative velocity of the contact point on the pinion with

respect to the contact point on the gear (ref E-1). This relative velocity is

obtained by way of the vector equation

VC = Cp/C + VC

where (fig. E-1)

V = absolute velocity of contact point Cp of pinion tooth, with

p direction normal to line connecting pivot point 0 and contact

point Cp (contact points Cp and C are coincident5

VC absolute velocity of contact point C of gear tooth with

g direction normal to the line connecting the pivot point 0 gand the contact point Cg

VC /C relative velocity between contact points C and C . The

p g direction of this velocity is normal to the line of action

T -Tp.

Velocity triangles according to equation E-1 for contact of a simple mesh

above and below the pitch point for two cases is shown in figure E-1. In figur,

E-1(a), the gear is the driver. In figure E-li(b), where the motion is rever ,cd,

the pinion becomes the driver. Figures E-l(a) and E-lI(b) differ with respect to

the direction of the relative motion, and with that, the direction of the contact

friction force for comparable contact. This difference in the direction of the

friction force ;say be expressed by reversing the sign of the friction force. For

computer purposes, this is best accomplished by making p positive when the gear

is the driver and making it negative when the pinion becomes the driver. The

normal force between the gear and the pinion, which has the direction of the line

of action, is not dependent on the driver.

1 8 1 I -A = 1 P A G J L --N O T ] a L

.,r.

to

0U

4) •

.4)_ ___ p . w -

ob I> -

bo 4)0s- E-4p

u 0

S ,r4 t-

04,)\ -S.--/\ ;'\ -) o°°

> 0

Ao bo 0 r

0

4Jq-4(I4

0 U U

,p..

1820

S0 .

540 ti -4 -

p. - 0

toE-) t. C) J

-40. V.0-0

0

E-4C -

I>)

040

1>0U~

5..

1823

Pivot Friction

The free body diagram of the combined gear and pinion set no. 2 in clockwiserotation is shown in figure E-2. The associated equilibrium equations B-59 and

B-60, which are used here as a general example, may be rewritten in the followingmanner:

F + W y = A 1 + W 2 COS2 - IF 12 CO S 1 (E-2)

ad x yand

-F x + F = A2 + a F2 3SIN2 - jF 1 2SINI (E-3)

where A, and A2 stand for the normal tooth contact and pivot forces which do not

contain the coefficients of friction and do not change direction on motion rever-

sal.* The trigonometric functions refer to the associated angles j - G 1 and

R2 + 02. The signum functions sI and s2 are omitted for the present discussionsince their influence is not relevant in this context. They change the signs of

all friction terms on the right sides of equations E-2 and E-3, depending on

whether contact is made before or after the pitch point.

Assume now that the direction of the gear set is changed to be counterclock-

wise. This causes the tooth contact friction forces (as shown earlier) to be

reversed in sign. Furthermore, the two pivot friction forces are also reversedin sign. The above force equilibrium equations now have changed signs in all

friction terms, i.e.,

Fx - ijFy= A 1 - F 2 3COS2 + 0 1 2COS1 kE-4)

and

I + F = A 2 - tiF 2 3SIN2 + jiF 12SIN1 (E-5)x y

Clearly, this effect of the change of the direction of all friction forces can be

expressed by making the coefficient of friction negative for computational pur-

poseS.

Equation A-2, reference E-1 shows that if the assumptions of the

directions of the normal pivot forces prove wrong, the associatedfriction forces still furnish friction moments which have the

appropriate direction of rotation.

183

Ic

4.)

0

C4 co

ot

C)0:Li

CCo

1844

This idea also holds true in general for the moment equation and is iLLus-trated with the help of the moment equation B-58 for gear and pinion set no. 2,which for clockwise rotation is of the following general form:

B - F 2 3a 2 + pF 12(d i-a j) + UP2(Fx 2 + Fy 2 ) = 1 2 N 3 24 (E-b)

where B is similar to A1 and A2 in equations E-2 and E-3 by representing momentterms which do not contain the coefficient of friction, and wthose sign does notchange on motion reversal. In the same vein, the signum functions sI and s.2 areomitted.

If the direction of rotation is counterclockwise, the above expression be-comes

B F23a 2 - F 1 2 (d 1-a 1) - pp2(Fx2 + Fy 2 ) = IZN3 2 (E-7)

Thus, the change in the direction of the friction moments can also be ex-pressed by making the coefficient of friction negative for computational pur-poses.

Finally, the results of these new force and moment equations for the variouscomponents may be obtained by changing the signs of all coefficients of frictionin the results of the original expressions which were derived for forward motionof the train.

I

REFERENCE

E-1. G. G. Lowen and F. R. Tepper, "Fuze Gear Train Analysis," Technical ReportARLCD-TR-79030, US Army ARRADCOM, Dover, N~J, Decem-,ber 1979.

186

APPENDIX F

A COMPUTATIONAL RULE FOR THE DETERMINATION OF THE SIGNIN THE EFFECTIVE MOMENT OF INERTIA IiR

187

I 87

Whenever the center of mass of a fuze train component, such as a pallet or a

rotor, does not coincide with its pivot axis, there will be a pivot frictionmoment which is proportional to the angular acceleration of the component. Justlike any other friction moment, its direction must be opposite to the angularvelocity of such a rotating part. The following shows how the appropriate choiceof the sign in the expression for the effective moment of inertia IiR* of thecomponent assures that this friction moment will always retard eotion. Whilethis discussion holds in general, it makes specific use of the pallet momentequation A-78, appendix A. This expression becomes in condensed form

Ms M, I (F-1)S A A p

where

MS = the sum of all moments on the pallet, with the exception of thefriction moment MAA which is proportional to the angular

acceleration p of the pallet. The moment MS is sufficientlylarger than the moment MAA so that it will at all times decidethe direction of the angular acceleration 'i

MAA ilI P p(A4+A8) = A 2 2 1P

i.e., A2 2 is always positive. Note also that the signum functioiiS5 has been omitted.

Figure F-i, a through d represents free body diagrams of a simplified pal-let, or rotor, with various directions of , MS, and MAA.

In figure F-la the moment MS is assumed to be CCW, i.e., positive. Theangular velocity $ is also assumed to be positive and therefore the frictionmoment must be negative. Equation F-I becomes

MS - MAA = I p (F-3)

* The letter i in the subscript stands for the component number or letter.

189

Ms us

NAA NM

iS U AA Ip* M1+ AA p

a b

MS

-MS

\01

nAA -AA

+I

C d

Figure F-i. Free body diagram of simplified pallet or rotor

190

In general, the sign of p is never predetermined in a differential equation ofthis type, but it is decided by the sign of the resultant moment. Ln this spe-

cific work it must be remembered that the direction of ; is decided by MS aloneand that the friction moment MAA must act opposite to the direction of the atyu-lar velocity p.

If equation F-2 is substituted into the above expression with apositive , it becomes

S = (l + A2 2) (F-4)

Let equations F-3 and F-4 now be used to find equivalent expressions for theangular acceleration p. Thus, from equation F-3

H - NAA

p

and from equation F-4

S (V-tu)I p + A 2 2

In the first forma the absolute value of the angular acceleration is reduced, fromthe value it would attain due to the moment MS alone, by the presence of "AA. i1the second form, which must be used in the computations, y is reduced iai abholut'magnitude by the equivalent presence of a larger effective moment of inertia PR

I lp + A2 2 . (The subscript PR is used for the pallet.)

Now consider the condition of figure F-lb where MS is alSo posiLive,

bu is negative, and with that MAA must be positive. This leads to the follow-ing moment equation:

M S + MAA= I p 4 7

or, with equation F-2 and a positive q)

MS = (i - A2 2 ) (F-8)

The above expressions are again used to find two equivalent forms for q.From equation F-7 one obtains

MS + HAAS A (F-9)

pp

191

while equation F-8 gives

• S!J) = -( F-10) I

I - A2 2p

[n the first form the absolute value of 4 is increased by the added presence ofMAA over that due to MS alone. The second form shows this increasa in , by wayif the reduced effective moment of inertia IpR = Ip - A2 2 .

In figures F-Ic and F-Id MS is shown to act in a clockwise, i.e., negativedirection, and with that 4 is also negative. Consider the condition of figure F-Ic first. The moment equation must take the negative into account by way of apositive MAA. Thus, from the free body diagram

- S +M, = t (F-Il)S AA p

To keep all signs correct on substitution of equation F-2 and the negative , tileabove expression must take the form

- M- A 2 2 = I p)

or

- M = -00 + A 2 2 ) (F-13)p

Te absolute value of 4 is then given by

'M (I 4I + A22

Again, this value is reduced in absolute magnitude by the itcreaso of the etlee-tive moment of inertia to IpR = Ip + A2 2 . This is the equivalent to a decreasein the absolute value of the r'. sultant moment.

igure. F-Id shows the pivot frictioii moment MAA iin the same dire'Lion ts MSSirl)(( n P is positive. The mlu. n,,t equation becomes

- I -) AA p ( I u

tpon snhst~itiition of eqVal , Io -2 togelther with the Ie;itwLi' , one liuds

1q2

-M + A22 I-'> = I - p (-It)p

or

-MS = - 22) (F-17)

The absolute value of i now becomes

MS

I - A 2 2 (F-18)

It is increased by the decrease in the effective moment of inertia to is, =

I - A2 2 .

These four cases allow the following computational rule for the determina-tion of the sign of the parameter A2 2 in the effective moment of inertia.

From the equations F-6 and F-14, together with the associated figures s-Iaand F-ic, respectively

IPR = I + A 2 2, when i and ' have identical signs (F-19)

Further, from equations F-10 and F-18, together with the associated figures F-lb

and F-id, respectively

I = I - A 2 2 , when ' and ' have opposite signs (F-2u)

193

DISTRIBUTION LIST

CommanderU.S. Army Armament Research and

Development CommandVIN: DR)AR-LCN, Ii. Grundler

G. Demitrack

L. HorowitzA. Nash

R. BrennanB. Frey

S. Israels

C. JanowH. Rand

R. Shaffer

G. Taylor

F. Tepper (15)

L. WisseDRDAR-TSS (5)

DRDAR-GCL

Dover, NJ 07801

Administrator

Defense Techiical Information Center

ATTN: Acces:;ious Division (12)

Cameron Stat onAlexandria, VA 22314

DirectorU.S. Army Materiel Systems

Analysis Activity

ATTN: DRXSY-MP

Aberdeen Proving Ground, MD 21005

Commander/Director

Chemical Systems LaboratoryU.S. Army \rmament Research and

Development Coimand

ATTN: DRDAR-CLJ-LDRDtXR-CLB-PA

APG, Edgewood Area, MD 21010

DirectorBallistics Research Laboratory

U.S. Army Armament Research and

Development Command

ATTN: DRDAR-TSB-S

Aberdeen Proving Ground, MD 21i105

hLk.l- -A &A, .

ChiefBenet Weapons Laboratory, LCWSLU.S. Army Armament Research and

Development CommandATTN: DRDAR-LCB-TLWatervliet, NY 12189

CommanderU.S. Army Armament MaterielReadiness Command

ATTN: DRSAR-LEP-LRock Island, IL 61299

DirectorU.S. Army TRADOC Systems

Analysis ActivityATTN: ATAA-SLWhite Sands Missile Range, NM 88002

CommanderHarry Diamond LaboratoriesATTN: Library

DRXDO-DAB, D. OvermanWashington, DC 20418

City College of the City Universityof New York

Mechanical Engineering DepartmentATTN: Dr. G. Lowen (10)137 St. and Convent AvenueNew York, NY 10031

196

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