1 x 5
There is a pattern in the table where the number of bounces is going up in increments of 1. The next values in the table should therefore be 1, 6 and 7, representing length, width and bounces respectively.
I’ve also spotted that if I add the length and width together it will give the number of bounces. i.e. L + w = b.
By drawing a 1 x 6 table I can show that this is correct.
1 x 6
I then tried tables with a width of 2. The smallest grid would be 2 x 1 but this has already been drawn as a 1 x 2 table.
2 x 2 2 x 3 2 x 4 2 x 5
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
According to my theory, although the next numbers in the table should be 3, 6 and 9, because there is a common factor of three in the length and width then the number of bounces should be the same as a 1 by 2 table. The next values in the table should therefore be 3, 6, 3. The next diagram shows that this is correct.
3 x 6
How can I define a rule for all sizes of table? It appears that my rule is to add the length and width except where there is a common factor. In these cases I can divide by the common factor after adding.
E.g. In the case of the 3 x 6 table. 3 + 6 = 9 and 9/3 = 3.
In the case of tables where the dimensions have no common factors I can assume that they have a common factor of 1. Therefore my rule becomes:
b = (L + w)/ (common factor of L and w),
where b= the number of bounces, L = the length of the table and w = the width of the table.
I am now going to test this on tables of length 4 but I am going to complete the table according to my rule first.
* Incorrect entry
4 by 4 4 by 5 4 by 6
Clearly the fourth entry in the table is wrong. It should be 4, 4, 2. However, if I divide by 4 instead of 2 as I did, then my calculation would be correct. It may be that I should be dividing by the highest common factor or H.C.F.
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.
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