In the text, a difference equation is obtained from which the energy transfer from the drive to the bucket coil can be found as a function of iD , with the circuit constants n2, Lw and C, and velocity v as parameters. The program in this appendix solves the difference equations, with the help of an approximation for dM/dx. The entire program is designed to run on an HP-25 pocket calculator with 49 program steps. Given a larger calculator or computer, one could generalize the program by using a more exact expression for dM/dx or (more usefully) by including the finite cross sections of the drive and bucket coils.
The simple program consists of four subroutines:
For still higher accuracy, the curve of iD is traced point-by-point by slightly modifying the program: step 47: pause, step 48: pause.
In this program, one or two minor tricks are used to save register space: the constant a, , which equals a3/R, is stored as a sequence of four key strokes in steps 31-34. Register 2 contains a constant whose integer and fractional parts are both used in the dM/dx subroutine. The step interval x appears as a sequence of four key strokes in steps 01-04. (All constants are in MKS units.) In detail, the registers store information as shown in table 5.
As normally used, the program calculates the energy transfer for a half-cycle of oscillation. The total energy transfer for a two-coil bucket, corresponding to the passage of both bucket coils through one drive coil, is then (x)(4n2 iB ) (summation of iDM/dx in register 5). Typically, a step interval x of 1 mm is used, 0.001 in program steps 01-04. Each step interval requires 3.75 sec on an HP-25, and about twice that time on an HP-67, so a half-cycle of oscillation with D = 5 .0 cm requires a little over 1 min (see table 6).
|RO||ao = (x)iB /Lwn2|
|R1||Initial value of position x (in m) at which drive|
coil switch is closed. Zero is allowed , but con-
stant should be chosen so that x update sub-
routine never finishes with zero. R1 stores
|R2||a2 = constant used in dM/dx subroutine. For|
D = 5.0 cm, a2 = -184.000138 (i.e.,
INT = -184, FRAC = - 0.000138). For other
cacalibers D, INT and FRAC are proportional
|R3||a3 = (x) R/vLwn22. Drive winding resistance|
R appears only in this constant
|R4||a4 = (x) vC. Resonant capacity C appears only|
in this constant and is equal to the sum of
the capacities on two adjacent sectprs pf tje actual
|R5||Initialize to zero; stores updated i(dM/dx).|
|R6||Peak voltage to which capicitor is charged before|
drive circuit switches on; R6 stores updated
capacitor voltage Vc.
|R7||Initialize to zero; stores updated current i.|
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