Hi Esko,

please have a look at the follwoing picture:




You see the two strings, which need not necessarily to be equally long. Assume that the first string is burned from both ends, means from "A" and "B". Then, the left fire spot will run from "A" to "B" and the right fire sport will run from "B" to "A". Even if we don't know how fast the fire spots will travel along their path, we know that the will meet in some point, here called "C", and we know that they will need the same amount of time "t1" to reach point "C".

If we had fired the upper string only from the left side (point "A"), the fire spot, after reaching point "C", would have travelled furtherly to "B" and would have needed the time "t1" from "A" to "C" and the time "t1" from "C" to "B", which, as we know, is 60min in total. From this 2 x t1 must be 60min, or t1 = 30min.

If we burn the two strings as Jan has suggested, means the upper string from both ends and the lower string from the left side (point "D") at the same time, then, when the upper fire spots reached point "C", the lower string burned for 30min, while the fire spot has reached point "F". For the rest of this string the same is valid as for the upper string. So, when burning the string from point "E", when the fire spot has reached point "F", the fire spots will meat at point "G" after "t2", where "t2" = 15min.

Kai