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help with movements

#1 User is offline   FrancesHinden 

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Posted 2012-January-01, 09:18

It's a very, very long time since I've directed anything.

I am going to have either 7, 8 or 9 tables or just possibly 10.
I need a pairs movement prepared for all of those in which everybody plays all the boards. It should be a one-winner movement.

The number of boards ideally is 18, but this can be +/- 1 or 2 if really necessary.

It doesn't matter much if it isn't well balanced.

My immediate reaction is that
- with 9 tables I'll play a straight Mitchell
- with 8 tables I can play a hesitation mitchell but 1&8 share, there's a relay between 4 & 5, moving pairs go from EW 8 to NS 8 to EW 1
- with 7 tables I play a double hesitation mitchell with relay between 4 & 5 with two sets of boards on it, moving pairs go from EW 7 to NS 2 to NS 7 to EW 1
- with 10 tables I play a hesitation bowman? (EW move up one table and the boards move as a normal mitchell for tables 1-9. From EW 10 move to NS 9 then EW 1. Table 10 shares with table 8 on the first round, and plays the boards in reverse numerical order, but plays the highest numbered set last).

In all of these I arrow switch one round? Pivot tables don't switch?

The problem with the 8-table movement is that I really will need two sets of boards, else table 1 has to share with the pivot table.
The 10 - table movement looks a bit complicated. Perhaps easier to play 20 boards.

(This is for a teaching day which is why it's so vital that everyone plays all the boards as they are going to be discussed afterwards)
Comments/ better suggestions?
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#2 User is offline   RMB1 

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Posted 2012-January-01, 11:33

The generic solution for we would try these days for 9 rounds "pairs play all the boards in play" would be web mitchells for numbers greater than 9 tables. This would need at least one extra set. For 10 tables this is a bowman.

For 7/8 tables I can see nothing better. For 5/6 tables there are full/three-quarter howells.
Robin

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#3 User is offline   gordontd 

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Posted 2012-January-01, 12:46

View PostRMB1, on 2012-January-01, 11:33, said:

The generic solution for we would try these days for 9 rounds "pairs play all the boards in play" would be web mitchells for numbers greater than 9 tables. This would need at least one extra set. For 10 tables this is a bowman.

Actually, for 11 tables it would be a Bowman. A ten-table Bowman would only have eight boardsets.

A further possibility would be to play 6x3-board rounds, so that board-sharing (if necessary) would be easier. That would give you a Bowman for eight tables.
Gordon Rainsford
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#4 User is offline   FrancesHinden 

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Posted 2012-January-01, 12:54

View Postgordontd, on 2012-January-01, 12:46, said:

Actually, for 11 tables it would be a Bowman. A ten-table Bowman would only have eight boardsets.

A further possibility would be to play 6x3-board rounds, so that board-sharing (if necessary) would be easier. That would give you a Bowman for eight tables.


For 8 tables, do I need to know more than 'it's a Bowman'? Where do I get the movement details from? If I can't get a second set of boards (I"m not sure) then definitely better to share 3-board rounds.

On the 10 table movement, it seems to be a web mitchell according to http://www.ebu.co.uk...rs/?id=8&page=4 ? even number of tables, odd number of board sets. Where do we download these from? (the article just says 'can be downloaded from the ebu website').


p.s. thanks for the very quick replies.
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#5 User is offline   RMB1 

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Posted 2012-January-01, 13:19

View PostFrancesHinden, on 2012-January-01, 12:54, said:

Where do we download these from? (the article just says 'can be downloaded from the ebu website').


Are these links useful?
http://www.ebu.co.uk...les/default.htm
http://www.ebu.co.uk...webmovement.pdf
http://www.ebu.co.uk...webmovement.pdf

The web mitchell movements are included in Jeff Smith's PairsScorer program.
Robin

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#6 User is offline   FrancesHinden 

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Posted 2012-January-01, 15:00

Sorry to sound stupid, but no.

The first link gives 16 and 19 table movements (plus some articles I wrote!).
The second is for 14-26 tables, the third for 15-25 tables.

Am I being thick?
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#7 User is offline   RMB1 

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Posted 2012-January-01, 16:04

View PostFrancesHinden, on 2012-January-01, 15:00, said:

The first link gives 16 and 19 table movements (plus some articles I wrote!).
The second is for 14-26 tables, the third for 15-25 tables.


Replace 13 by 9 (or 12 by 8) and then
the web mitchell odd movement will do 9 rounds at 11-17 tables
the web mitchell even movement will do 9 rounds at 10-18 tables
the web mitchell skip movement will do 8 rounds at 10-16? tables

Replacing 12 by 6 then
the web mitchell skip movement will do 6 rounds at 8-12 tables

Jeff Smith's program has the 13x2, 12x2, 9x3, 8x3 movements.
But it will do 9x2 and 8x2. It doesn't have 6/7 round web movements.
Robin

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#8 User is offline   pgrice 

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Posted 2012-January-01, 16:59

View PostFrancesHinden, on 2012-January-01, 09:18, said:

It's a very, very long time since I've directed anything.

I am going to have either 7, 8 or 9 tables or just possibly 10.
I need a pairs movement prepared for all of those in which everybody plays all the boards. It should be a one-winner movement.

The number of boards ideally is 18, but this can be +/- 1 or 2 if really necessary.

It doesn't matter much if it isn't well balanced.

My immediate reaction is that
- with 9 tables I'll play a straight Mitchell
- with 8 tables I can play a hesitation mitchell but 1&8 share, there's a relay between 4 & 5, moving pairs go from EW 8 to NS 8 to EW 1
- with 7 tables I play a double hesitation mitchell with relay between 4 & 5 with two sets of boards on it, moving pairs go from EW 7 to NS 2 to NS 7 to EW 1
- with 10 tables I play a hesitation bowman? (EW move up one table and the boards move as a normal mitchell for tables 1-9. From EW 10 move to NS 9 then EW 1. Table 10 shares with table 8 on the first round, and plays the boards in reverse numerical order, but plays the highest numbered set last).

In all of these I arrow switch one round? Pivot tables don't switch?

The problem with the 8-table movement is that I really will need two sets of boards, else table 1 has to share with the pivot table.
The 10 - table movement looks a bit complicated. Perhaps easier to play 20 boards.

(This is for a teaching day which is why it's so vital that everyone plays all the boards as they are going to be discussed afterwards)
Comments/ better suggestions?


All the above seem reasonable solutions to the balance/duration/complexity problem you have. Movements for playing 9-rounds at 7, 8, 9 and 10 tables are in Manning and you should be able to print table cards from your scoring program. The 10 table Hesitation Bowman works best with a phantom pair (on table 10) to avoid the need for Table 10 to share with other tables in turn. Of course if you have 2 board sets this problem goes away. Manning doesn't always just arrow-switch one (the last) round, but if balance is not a primary objective this will work fine.

Peter
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#9 User is offline   gordontd 

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Posted 2012-January-01, 20:09

View PostFrancesHinden, on 2012-January-01, 12:54, said:

For 8 tables, do I need to know more than 'it's a Bowman'? Where do I get the movement details from? If I can't get a second set of boards (I"m not sure) then definitely better to share 3-board rounds.

On the 10 table movement, it seems to be a web mitchell according to http://www.ebu.co.uk...rs/?id=8&page=4 ? even number of tables, odd number of board sets. Where do we download these from? (the article just says 'can be downloaded from the ebu website').


p.s. thanks for the very quick replies.

For the eight table Bowman, put out boardsets 1-6 on tables 1-6. Table 7 will share with table 1 throughout. Table 8 will share with tables 6,4,2,6,4,2 so that they play boardsets in reverse order. There will be a skip after three rounds, but otherwise EW pairs just move up one table. Arrow-switch the last round.

For a ten-table web mitchell you really do need two sets of boards. Put boardsets 1-5 on tables 1-5, then
boardset 4 on table 6
boardset 3 on table 7
boardset 2 on table 8
boardset 1 on table 9
boardset 9 on table 10

One set of boards feeds in at table 5 and comes out after table 1; the other set of boards feeds in at table 10 and comes out after table 6. Note that the top half of the movement plays boards in reverse order, so boards are fed in to table 10 as boardset 9,8,7,6...

EW pairs move up one table throughout the entire movement, and all tables arrow-switch the final round. Boards move down one table each round, but remain within their own half of the movement - ie boards from table 6 go to a relay and then to table 10; boards from table 1 go to a relay and then to table 5.

Will you know in advance how many pairs you have?
Gordon Rainsford
London UK
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#10 User is offline   FrancesHinden 

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Posted 2012-January-02, 04:21

View Postgordontd, on 2012-January-01, 20:09, said:



Will you know in advance how many pairs you have?


Yes in theory a day or so before, but in practice it's always possible for someone not to turn up, or (less likely) for someone to turn up not having booked. And I'm busy every day between tomorrow and the day we play so I want to have things sorted for all contingencies now.

p.s. hope you are enjoying Mexico.
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#11 User is offline   Xiaolongnu 

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Posted 2012-January-02, 08:04

A Howell movement or some subset of it sounds good. My club practises it, never had any problems with it. You get to play all boards.
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#12 User is offline   gordontd 

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Posted 2012-January-02, 08:47

View PostXiaolongnu, on 2012-January-02, 08:04, said:

A Howell movement or some subset of it sounds good. My club practises it, never had any problems with it. You get to play all boards.

Howell movements are ideal when you have one fewer round than the number of pairs, so for nine rounds (18 boards) a full Howell would only be suitable for five tables. There are also 3/4 (or reduced) Howells when you don't meet all the other pairs, and that might be a good solution for six tables playing 9 rounds. However, with larger numbers of tables, as Frances is expecting, a Howell is not really ideal.
Gordon Rainsford
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#13 User is offline   axman 

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Posted 2012-January-02, 10:58

View PostFrancesHinden, on 2012-January-01, 09:18, said:

It's a very, very long time since I've directed anything.

I am going to have either 7, 8 or 9 tables or just possibly 10.
I need a pairs movement prepared for all of those in which everybody plays all the boards. It should be a one-winner movement.

The number of boards ideally is 18, but this can be +/- 1 or 2 if really necessary.

It doesn't matter much if it isn't well balanced.

My immediate reaction is that
- with 9 tables I'll play a straight Mitchell
- with 8 tables I can play a hesitation mitchell but 1&8 share, there's a relay between 4 & 5, moving pairs go from EW 8 to NS 8 to EW 1
- with 7 tables I play a double hesitation mitchell with relay between 4 & 5 with two sets of boards on it, moving pairs go from EW 7 to NS 2 to NS 7 to EW 1
- with 10 tables I play a hesitation bowman? (EW move up one table and the boards move as a normal mitchell for tables 1-9. From EW 10 move to NS 9 then EW 1. Table 10 shares with table 8 on the first round, and plays the boards in reverse numerical order, but plays the highest numbered set last).

In all of these I arrow switch one round? Pivot tables don't switch?

The problem with the 8-table movement is that I really will need two sets of boards, else table 1 has to share with the pivot table.
The 10 - table movement looks a bit complicated. Perhaps easier to play 20 boards.

(This is for a teaching day which is why it's so vital that everyone plays all the boards as they are going to be discussed afterwards)
Comments/ better suggestions?



You have given the impression you intend to use teaching deals and perhaps that scoring [if any] is merely incidental. Perhaps a barometer would deal with the challenges. Basically, you can run the movement for as long or as short as suits your purposes and whoever is not a half table will play all the boards. As for the ideal of 18 boards boards 19-36 can be duplicated as 1-18 if a duplicating machine is used needing only N/4 sets of [36] boards when the boards are shared
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