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  • #16
    Yep. Both correct. As I said, it as only happened to me once. One of the players even questioned me saying was that correct? The scenario is covered in Section 3 Rule 7(g).

    I have posted this scenario before and there is no rule directly covering it (except perhaps, Section 5 Rule 1 (ii)). Place all the colours on their spots. Remove the blue and place the cue-ball about a balls width from the blue spot in a direct line with the top (black) cushion. Now place all 15 reds in a direct straight line between the cue-ball and top cushion. (takes a while to set up) so that there is no space. If done correctly, you cannot spot the blue. It has to go BELOW (between its spot and the Brown) its own spot. Ideally, Section 3 Rule 7 (g) should also include the blue as well.
    I've not tried to see if all 15 reds can take up all the space between the top cushion and blue spot without the pink and black. I very much doubt it.
    You are only the best on the day you win.

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    • #17
      Originally Posted by DawRef View Post
      Yep. Both correct. As I said, it as only happened to me once. One of the players even questioned me saying was that correct? The scenario is covered in Section 3 Rule 7(g).

      I have posted this scenario before and there is no rule directly covering it (except perhaps, Section 5 Rule 1 (ii)). Place all the colours on their spots. Remove the blue and place the cue-ball about a balls width from the blue spot in a direct line with the top (black) cushion. Now place all 15 reds in a direct straight line between the cue-ball and top cushion. (takes a while to set up) so that there is no space. If done correctly, you cannot spot the blue. It has to go BELOW (between its spot and the Brown) its own spot. Ideally, Section 3 Rule 7 (g) should also include the blue as well.
      I've not tried to see if all 15 reds can take up all the space between the top cushion and blue spot without the pink and black. I very much doubt it.
      It would require 18 balls plus 18½ ball's-widths, so that's 75.2 inches. If we take the distance to be 6 feet, that's 72 inches so it can be done with a little space to spare even if one red has already been potted out of the game.

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      • #18
        It would require 18 balls plus 18½ ball's-widths, so that's 75.2 inches. If we take the distance to be 6 feet, that's 72 inches so it can be done with a little space to spare even if one red has already been potted out of the game.
        Statman? You and I need to get out more!!
        You are only the best on the day you win.

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        • #19
          Originally Posted by DawRef View Post
          Statman? You and I need to get out more!!
          Maybe for a game? We could try to get the balls to achieve this, in an actual game!

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          • #20
            hehe, correct
            Will NEVER happen, but sure, according to the rules, the "exception" for the blue is not covered.

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