Diamond Enthusiast

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Let's see: 15 reds at one point each =15 + All the colours (7+6+5+4+3+2 ) =27 makes 42 If we consider a foul on black by the opponent, before our player starts, as part of our player's score then we have 7 points in hand to add making 49. The only way I can see of this happening would be if all the balls were lined up, in the way of a trick shot.There should be enough space, given that the table is twelve feet long and there would be the whole long diagonal to a corner pocket.If not there would have to be two lines, one for each of two pockets, and two cannons of the cue ball, one on each line, off the first onto the second. You'd need to have no bag on each receiving pocket too because the bag would only accept a few balls before it was full. If it's legal then this must be technically possible, mustn't it ? There may be some argument about whether the laws allow such a fluke to be treated as one break on one shot, there then being no respotting of the colours between each one going in, but that's a question of the laws.It could be that the law sets a maximum of seven on a shot (one black) or eight (red and black or vice versa) and states that the other colours must be respotted in such a case as this. 
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| Posts: 8340 | Location: Newmarket, UK/ Antibes, S.France | Registered: 07-14-02 |    |
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Diamond Enthusiast


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Ny interpretation of what could happen ...bear with us Somehow all the red balls on a break off shot all get pocketed and all colours remain on the table (Score 15 points) Fred? on a full sized table the pockets empty into rails underneath...capacity 8 to 10 balls? Moving on if you then cleared the Colours with a black as the nominated ball the total break will be 49 or 44 with a yellow (2pts) Question. Impossible to make an aggregate frame score below 44?  (no fouls) My record "reds pocketed" is 3 off the "break off" on an uneven toy snooker set
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| Posts: 13326 | Location: 6 miles west of Wigan UK | Registered: 06-05-02 |    |
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