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 4inch bore rather
than a 3inch bore. This results in the relation so that . When the quantity of water is
doubled, the time doubles to hours. The 4inch bore comes from Nicholas
Pike’s A New and Complete System of Arithmetick, p. 174. Perhaps Macdonald copied the
statement of the problem incorrectly.