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.