Problem 4

If a leaden pipe or other conductor will fill a cistern in a given time it is required to find the diameter of another pipe which will fill the same cistern in any required time.

Comment: The problem is an exercise in geometrical proportions and square roots.

Rule

The squares of the diameters as inversely as the times.

Examples

1.  If a leaden pipe of an inch in diameter will fill a cistern in 3 hours I want to find the diameter of another pipe which will fill the same cistern in 1 hour.

Solution: . Let be the desired number (the square of the desired diameter). Since the time varies inversely with the diameter, then we have the geometrical proportion or or , which yields . The desired diameter is to three decimal places.

Comment: Macdonald did not pay close attention to what he was copying. He wrote the solution in the form without any intermediate steps. The statement as given appears odd. If he meant instead of (which he probably did not) the statement is incorrect. He was probably just hurrying to write down the answer, which is not quite correct as he has given it.

2.  If a pipe whose diameter is inches will fill a cistern in 5 hours at what time will a pipe whose diameter is inches fill the same cistern.

Solution: and . Let be the desired time. Then or hours. Macdonald expresses his answer in minutes so that the answer in minutes is .

Comment: (1) This problem has nothing to do with finding a square root. (2) Macdonald gives minutes as his answer because in his longhand division he incorrectly converts the hours to minutes at one point. Here is what he did.

He first found to get . Then he converted to minutes by multiplying by for which he correctly obtains . Then he divides into . Macdonald’s longhand division looks like this:

 55.6 1225 67500 6125 6250 6125 125 ×60 7500 7350 150

Where is his error? Can you explain why he made this error?

3.  If a pipe 6 inches bore will take 4 hours in running a certain quantity of water in what time will 3 pipes each 3 inches bore be in discharging double the quantity?

Solution: for the one pipe and for one of the 3 pipes which yields . Without doubling, let be the quantity of water. Therefore so that . Since the quantity of water must be doubled then the solution is hours or hours and minutes.

Comment: (1) Once again, this problem has nothing to do with finding a square root. (2) Macdonald answer is hours. Where he erred was in the size of the set of 3 pipes. In his calculation he gives the 3 pipes a 4-inch bore rather than a 3-inch bore. This results in the relation so that . When the quantity of water is doubled, the time doubles to hours. The 4-inch bore comes from Nicholas Pike’s A New and Complete System of Arithmetick, p. 174. Perhaps Macdonald copied the statement of the problem incorrectly.