# Investigate, for different size pool tables, the number of contacts, (including the start, rebounds and finish) when a pool ball is projected from one corner and bounces off the sides of the table until it can enter a pocket.

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Introduction

Rhys Griffiths 11B

February 2003

Mathematics Coursework

Intermediate Tier

Four Pocket Pool

I intend to investigate, for different size pool tables, the number of contacts, (including the start, rebounds and finish) when a pool ball is projected from one corner and bounces off the sides of the table until it can enter a pocket.

The ball is projected at an angle of 45o degrees to the side of the table and rebounds at the same angle.

To carry out my investigation I intend to follow the following steps

- Try a simple case
- Use some helpful diagrams
- Organise in order from the simplest case and increase in small steps
- Put my results in tables
- Spot a pattern and test it
- Find a rule and test it
- Explain why it works

For a 4 x 15 pool table, the ball has contact with the table 19 times.

The simplest case is a 1 x 1 table. I will then increase this to a 1 x 2 then 1 x 3 etc.

1 x 1 1 x 2 1 x 3 1 x 4

1 x 5

Length (L) | Width (w) | Bounces (b) |

1 | 1 | 2 |

1 | 2 | 3 |

1 | 3 | 4 |

1 | 4 | 5 |

1 | 5 | 6 |

Middle

Length (L) | Width (w) | Bounces (b) |

2 | 1 | 3 |

2 | 2 | 2 |

2 | 3 | 5 |

2 | 4 | 3 |

2 | 5 | 7 |

The same patterns and rules do not apply in the same way as for the tables of width 1. However, there appears to be an increasing pattern of 3, 4, 5, 6, 7, etc. even though 4 and 6 are missing. So the next row of the table should be 2, 6, 8. The next diagram shows that this is also not correct. The next number of bounces is 4.

2 x 6

This however does fit the pattern in that where the length and width have a common factor the number of bounces is the same as tables where the length and width have the same ratio. e.g. A 2 x 6 and a 1 x 3 both have 4 bounces.

I decided to look at tables with a width of 3 to see if this pattern re-occurred.

The smallest table of width 3 which I have not already drawn is 3 x 3.

3 by 3 3 by 4 3 by 5

Length (L) | Width (w) | Bounces (b) |

3 | 1 | 4 |

3 | 2 | 5 |

3 | 3 | 2 |

3 | 4 | 7 |

3 | 5 | 8 |

According

Conclusion

To test this I’ll try a table where the length and width have more than one common factor. The table is 6 by 12. The number of bounces should therefore be (6 +12)/6 = 3. The diagram below confirms this.

6 x 12

I’ve discovered that for rectangular tables, where a ball is projected from a corner at 45o, the number of bounces (including the starting and finishing points) can be calculated from the rule:

b = (L + w)/ H.C.F.

where b= the number of bounces, L = the length of the table, w = the width of the table and H.C.F. is the highest common factor of L and w.

e.g. for a 10 x 80 table b = (10 x 80)/ 10

= 80.

Why does this work? If you enlarge a table, for example a 1 by 3 table by a factor of 4, then the new table is 4 by 12. The ratio 1:3 = 4:12. The pattern of the lines will be exactly the same on both tables. They will be just longer (or shorter). The number of bounces will not increase as there are still only four sides.

Y:\svn\trunk\engine\docs\working\acumen3\58533.doc

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