Geometry of the Verge and Crown Wheel Escapement by 44Qsr5k

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									                                                      By Alan Emmerson


                                                                                                    APPENDIX C



               GEOMETRY OF THE VERGE AND CROWN WHEEL ESCAPEMENT

                                                                                     Xd
               Datum
                                                                                      d



                                 r          h
                   
                                                                                                R
                                                                                          r

                                                                                        Xr




                                               Direction
                                               of Motion

              xd                    xr


                             Figure C1                                             Figure C2

                                                   Escapement at the Instant of Release.

   The diagrams show, in elevation and plan, an advancing tooth on one side of the crown wheel (nearer to
the reader) about to lift the verge pallet clear of the tooth tip and the corresponding tooth on the other side of
the wheel about to drop onto the other pallet.

   The plane containing the axis of the crown wheel and the axis of the verge is taken as a datum plane.

The angle between the pallets is 
The angular position of the pallet at release is 
The travel or displacement of the axis of symmetry of the pallets from the datum is .
The clearance angle corresponding to the drop is 
The lateral displacement of the tip of the tooth at release is xr
The lateral displacement of the tip of the tooth at drop is xd
The crown wheel has N teeth


   We wish to know the angle of swing at release in terms of the design parameters , h, r and 

                                                      h                 x
   From Figure C1 :                    ,   cos 1 , and   sin  1 r
                                         2             r                  r
                                                   h 
                                     cos 1                                               C1
                                     2              r 2
   There is however a constraint on the free choice of  , h, and r . There is a requirement that the angle of
drop  must be positive as shown.

                                                           
   From Figure C1 :              , (that is               ,)
                       2                                    2
                                                                     xd
                                                 and     tan  1
                                                                     h

                                   xd                              
   Whence            tan  1      and        xd  h tan                                 C2
            2                       h                       2        

   In the symmetrical layout usually adopted, the crown wheel must have an odd number of teeth. Thus the
angular distance between the tip of a tooth and the tip of the tooth closest to diametrically opposite is
½tooth pitch. Consequently, in Figure C2,

                                                                1
                                                         r      pitch   d
                                                                2
                                                                          
                                                         ie r   d 
                                                                          N
                                                                     xr
                                                          r  sin 1
                                                                     R
                                                                     x
   Also from Figure C2,                                   d  sin 1 d
                                                                      R
                                                                  x          x    
                                                           sin 1 r  sin 1 d 
                                                                   R          R N
   Thus for N15 the angles  are small so that :

                                                           xr xd 
                                                               
                                                           R   R N
                                                                                 xd
                                                           ie: x r  R(             )
                                                                          N       R

                                                                 h           
   Substituting from C2 gives                             xr  R  tan      
                                                                N R 2          

                                          
   We may reasonably anticipate that              is small so that:
                                          2
                                                                  h       
                                                          xr  R        
                                                                 N R 2      
        xr                                              R  h       
  But       sin  and thus                   sin             
        r                                               r  N R 2     

                                                        R  h       
  or                                          sin             
                                                        r N r2       

                                              h    R h h
  that is                                                sin 
                                              r    r N r2 r
                                                   R      r
  Whence                                               sin                                       C3
                                                   hN 2     h
                      h                    r          h      R
  But     cos  1     so that           sin cos 1                                              C4
                      r                   2 h          r      hN
                                       r          h R
  Thus the requirement is:              sin cos 1        and   0                               C5
                                      2 h          r hN
   The requirement   0 can be satisfied by pre setting a value for drop. The practicalities of construction
suggest that drop should be expressed as a fraction of the crown wheel tooth pitch. If D is the angular drop
                                                             RD
of the crown wheel, r  RD , for small    , so that  
                                                              r
                                                        2
               So that                        Dk
                                                        N
                                                   RD     R 2
  And thus                                          k
                                                    r     r N
                                               r          h R      R 2 k
  Substituting in C5 gives                      sin cos 1                                         C6
                                              2 h          r hN      r N


  Transforming and collecting terms in C6 gives
                                                    r        h  R    R
                                                   sincos1    2 k                               C7
                                                   2 h        r Nh     r

                                                             h    
  But                                           cos 1                                              C1
                                                             r    2
  There are two approaches to the design of the escapement.

                                                   h r        h  R    R
  Adding C1 and C7 yields            2  cos1      sincos1    2 k 
                                                   r h        r Nh     r
                                                       h
    and this prescribes                     cos  1     
                                       2                r
    Alternatively:

                                                         h r        h  R    R
    Subtracting C1 from C7 yields     0    cos1        sincos1    2 k 
                                                         r h        r Nh     r

                                                   h     r          h      R   R
    ie                                  cos1  sincos1                  2k 
                                                   r     h          r     Nh    r

                                                   h     
    resulting in                        cos  1 
                                                   r     2
    These are the characteristic equations for the escapement. They are accurate only when N 15 and
                                                                       h
        is small . The design criteria for the escapement include      1 or the crown wheel will run free.
2                                                                       r


   Table C1 shows the values of the angle of swing to release for various pallet angles using practicable
values for the other escapement parameters.

 Pallets Included                              Angle of Swing at   Release
      Angle                                       degrees
    degrees                        N=15                                           N=29
                         r                                     r                                   
       70               3.0                40                 1.7                                   29
       80               3.1                36                  18                                   26
       90               3.3                32                 1.9                                   22
      100               3.4                27                 2.0                                   18
      110               3.5                23                 2.2                                   15
For: R=15mm, h=0.75mm and k=0.1
                       Table C1 Verge Escapement , Angle of Swing at Release

								
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