In a chanel of running water, whose breadth is contracted by one or more obstacles; the breadth of the chanel, the mean velocity of the whole stream, and the breadth of the water-way between the obstacles being given; To find the quantity of the fall occasioned by those obstacles.
| Let | b = breadth of the chanel in feet. |
| v = mean velocity of the water in feet per sec. | |
| c = breadth of the water-way between the obstacles. |
| Now 25 : 21∷ c : 21 ⁄ 25 c the water-way contracted. | Principle III. |
| And 21 ⁄ 25 c: b ∷ v : 25b ⁄ 21c v the veloc. per sec. in the water-way between the obstacles. | Princip. V. |
| Also (2a)² : vv ∷ a : vv ⁄ 4a the height fallen to acquire the vel. v. | I. & II. |
| And (2a)² : (25b ⁄21c)² × vv ∷ a: (25b ⁄ 21c)² × vv ⁄4a the height fallen to acquire the vel. 25b ⁄ 21c v. | I. & II. |
| Then (25b ⁄ 21c)² x vv ⁄ 4a - vv ⁄ 4a is the measure of the fall required. | VII. |
Or ((25b⁄21c)² - 1) × vv ⁄ 4a is a rule, by which the fall may be readily computed.
Here a = 16,0899 feet and 4a = 64,3596.
Example I. For London-Bridge.
By the observations made by Mr. Labelye in 1746,
The breadth of the Thames at London-bridge is 926 feet;
The sum of the water-ways at the time of the greatest fall is 236 feet;
The mean velocity of the stream taken at its surface just above bridge is 3⅙ feet per second.