During the passage of a bucket coil, a sinusoidal current flow described by the equation

will appear in each drive coil. The angular frequency, of the flow will be 2/period, where the period equals 4m /V with m being the spacing between drive coils and V being the bucket velocity. At any moment, there will be four contiguous drive coils having a current flow within them. Every current will be 30o out of phase with that in each of the adjacent drive coils. Figure 2 illustrates the current flows in neighboring coils during the activation of a particular reference drive coil (coil no. 4). The equations describing the current flows with respect to that reference drive coil current are as follows:

Coil no.1
Coil no.2
Coil no.3 </</TD>
Coil no.4
Coil no.5
Coil no.6
Coil no.7

The force experienced by any two active coils will be given by the product of the instantaneous current in each coil multiplied by the gradient of their mutual inductance (dM/dx) at the distance of their separation:

This force will be attractive between coils with currents of the same sign, and repulsive between coil currents of opposite sign.

Refer again to figure 1 and consider the forces acting on the reference coil no. 4. During its first quarter-cycle (- t - /2), coil no.4 will interact repulsively with coil no. 1, repulsively with coil no. 2, and attractively with coil no. 3. At exactly t = - /2, a weakly felt coil no. 1(a distance 3m away) will turn off and a much closer coil (coil no. 5 only a distance m away) will be activated. Because the force on coil no. 4 due to coil no. 5 is stronger than was the force from coil no. 1, a discontinuity in the gradient of the force acting on coil no. 4 will exist. During the second quarter-cycle (- t 0), coil no. 4 will feel a repulsion from both coils nos. 2 and 3, and an attraction to coil no. 5. Through considerations of symmetry, the forces exerted on coil no. 4 in the third quarter-cycle are the negative of those exerted during the second quarter-cycle. Similarly, forces felt during the fourth quarter-cycle are the negatives of those experienced during the first quartercycle. Over the complete cycle, then, a drive coil receives no net force from its neighboring drive coils.

The program listed below is designed to calculate the reaction force on a drive coil due to all other active drive coils in its neighborhood as a function of the time from which that coil was turned on. It is written to be run on a Hewlett-Packard HP-67/HP-97 calculator.

It is assumed that bucket velocity may be taken as constant during the passage of the bucket through any four successive drive coils (a distance of only a few centimeters). Hence the angular frequency will be constant.

The program is initialized by the input of three pieces of information: key in the dM/dx between drive coils separated by a distance m; ENTER; key in the mat a distance of 2m; ENTER; key in the dM/dx for a distance of 3m (ref.4). Initiate the program by pressing the button labeled [A]. Program execution will begin. Very quickly, the program will pause for a second and the display will show "1.0." During this pause, key in the maximum drive coil current. (this value will default to 1.0 of no entry is made; in this case, all final answers will actually be F/iaib.)

The program will then loop, displaying a zero for a second and then blurring for a second. At any instant when the machine has paused with a zero showing, key in a value of t and the reaction force of the drive coil at that instant will be calculated. The program accepts values of t expressed in degrees rather than radians. The range of values -180o t 180o.

Once a force has been calculated, the answer, (in Newtons) is displayed for 10 sec. the program then branches to the zero/blur input mode, ready to have the next value of t keyed in.

001 *LBLA 034 RCL5 067 x
002 DEG 035 9 068 RCL3
003 ST03 036 0 069 x
004 R 037 - 072 ABS
005 STO2 038 X<0? 071 x
006 R 039 GTOB 072 ABS
007 STO1 040 GSBc 073 RTN
008 CF3 041 STO 076 SIN
009 1 042 GSBb 075 RCL5
010 STO4 043 ST-0 076 SIN
11 PSE 044 GSBa 077 X2
12 PSE 045 ST-0 078 RCL2
13 F3? 046 GTO3 079 x
14 X2 047 *LBLB 080 RCL4
15 ST04 048 GSBc 081 x
16 CF3 049 ENT 082 ABS
17 *LBL1 050 + 083 RTN
18 CLX 051 CHS 084 *LBLc
19 PSE 052 STO 085 RCL5
20 F3? 053 GSBb 086 SIN
21 GTO2 054 ST-0 087 RCL5
22 GTO1 055 *LBL3 088 COS
23 *LBL2 056 RCL0 089 x
24 X<0? 057 F2? 090 RCL1
25 SF2 058 CHS 091 x
26 ABS 059 PRTX 092 RCL4
27 STO5 060 PRTX 093 x
28 1 061 GTO1 094 ABS
29 8 062 *LBLa 095 RTN
30 0 063 RCL5 096 *LBLe
31 - 064 SIN 097 CLX
32 X>0? 065 RCL5 098 1/X
33 GTOe 066 COS 099 RTN
100 R/S
Return to III-2

Table of Contents