Math Puzzle - The Nelson Column

Martha was visiting her grandmother, who lived in the country. At the back of the farmhouse was a very large porch, and in the front of that a garden in which grew all kinds of flowers. One afternoon, when everyone else was taking a nap, ... Read more of What The Flowers Told Martha at Military Training.ca

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The Nelson Column

(MISCELLANEOUS PUZZLES)

During a Nelson celebration I was standing in Trafalgar Square with a friend of puzzling proclivities. He had for some time been gazing at the column in an abstracted way, and seemed quite unconscious of the casual remarks that I addressed to him.

"What are you dreaming about?" I said at last.

"Two feet——" he murmured.

"Somebody's Trilbys?" I inquired.

"Five times round——"

"Two feet, five times round! What on earth are you saying?"

"Wait a minute," he said, beginning to figure something out on the back of an envelope. I now detected that he was in the throes of producing a new problem of some sort, for I well knew his methods of working at these things.

"Here you are!" he suddenly exclaimed. "That's it! A very interesting little puzzle. The height of the shaft of the Nelson column being 200 feet and its circumference 16 feet 8 inches, it is wreathed in a spiral garland which passes round it exactly five times. What is the length of the garland? It looks rather difficult, but is really remarkably easy."

He was right. The puzzle is quite easy if properly attacked. Of course the height and circumference are not correct, but chosen for the purposes of the puzzle. The artist has also intentionally drawn the cylindrical shaft of the column of equal circumference throughout. If it were tapering, the puzzle would be less easy.

Answer:

If you take a sheet of paper and mark it with a diagonal line, as in Figure A, you will find that when you roll it into cylindrical form, with the line outside, it will appear as in Figure B.

It will be seen that the spiral (in one complete turn) is merely the hypotenuse of a right-angled triangle, of which the length and width of the paper are the other two sides. In the puzzle given, the lengths of the two sides of the triangle are 40 ft. (one-fifth of 200 ft.) and 16 ft. 8 in. Therefore the hypotenuse is 43 ft. 4 in. The length of the garland is therefore five times as long—216 ft. 8 in. A curious feature of the puzzle is the fact that with the dimensions given the result is exactly the sum of the height and the circumference.