Movement for 17 tables
Would this be a suitable movement.
Play two 9-table (8.5) 3/4 howells with 1 pair from each movement meeting at table 17 (or the equivalent of any table where there is a stationary NS). If so, can we keep the movements the same (other than one pair would play as NS on the table instead of the allocated EW
It is just that this would cope with a 24-board session rather than a pure 27 one.
Comments
It would work, but you might have trouble persuading 68 players that none of them will be stationary. Otherwise you could just do a curtailed 13-round Web, or a 16-table skip Web with a full rover table.
A straight Mitchell just playing 13 rounds would work but you would need 34 boards.
Alan
I think we can take it as given that the original poster is aware of that and wants to avoid that outcome.
Actually I think he wants to just play 12 rounds, which would be even worse.
Would an 9 table 3/4 howell be totally non-stationary? I would have thought 12 pairs would be moving which leaves 5 stationary in each half?
Our room can (just) cope with 17 tables in two rows of 8 and one at the end - playing this movement would allow our pre-dealt boards to be shared accross tables and save the expense of getting in two sets when we may not need them. (The regional main club charges a nominal fee for each set. But £5.00 a set for 50 weeks comes out at £250.00.)
What about two sections and 8 3-board rounds?
If you want to share one set of boards, play the combined mitchell (two 8.5 table mitchells, merged at table 9)
You are quite right - you would be able to have ten stationary pairs. However, you would still need two sets of boards unless you were prepared to have sharing in two-board rounds.