Producing Howell movements

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PRODUCING HOWELL MOVEMENTS FROM SKELETONS

Taking the 4-table example above we will first of all use the skeleton (a) as a BS. The first step is to balance this base. When doing this, firstly test for balance, then switch one pair of numbers, then test for balance etc., until a balanced case is found. When switching, do not switch the first pair of numbers (8-1), as this is the same as switching all other pairings. The order of switching should follow a pattern so that no one combination will be forgotten.

For simplicity, the recommended pattern is:

1, 2, 1, 3, 1, 2, 1.

For higher numbers of tables just add the next number and repeat the previous pattern:

1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1

So the balanced base is (switching groups 1 and 2):

8-1      2-7      3-6      5-4

The next step in deriving a Howell is to superimpose a skeleton on the base. We will take, for example skeleton (b). The first and second groups of pairs must be switched to be the same as the base, so that now we have:

8-1      4-2      7-3      6-5

Assuming pairs 8-1 are to he seated at table 1, then the tables for the other pairs are found by the calculation:

1 + [N-S(SS)] – [N-S(BS)]

or

1 + [E-W(SS)] - [E-W(BS)]

If the result is negative it is added to (2T-l), which is 7 in the case of 4 tables. As an example take the pairs 4-2 (SS) and 2-7 (BS)

1 + 4(N-S) - 2(N-S) = 3

ie.   4-2 will be at table 3.

7-3 (SS) and 3-6 (BS) gives:

1 + 3 - 6 = -2

As this is negative, add this to 7 giving 5.

The last set is 1 + 6- 5 = 2. Therefore the complete Howell is:

8-1   4-2   7-3   6-5

1   3   5   2

Or in the shorthand method:

8-1   6-5   4-2   R   7-3   2R

As an exercise it is suggested that the reader derive all the 4 table Howell movements via this method.