1. Problem Requirements
-Find the depth of the water when there is an amount of 1000 cubic
feet per second flowing through the channel.
2. Analyze the Problem
-Define width (15), flow of water (1000), N (.014), S (.0015)
-User inputs a depth value so R can be solved
-Input: user guess as to the depth required
-Fundamental process: Manning's equation
Q = (1.486/N)*AR^(2/3)S^(1/2)
R = (depth * width) / (2.0 * depth + width)
-Limitations: depth cannot be negative
3. Design the Algorithm
-After user enters depth, the program must solve for R before it
can solve for Q
-A while loop can be used for when depth is less than 0
ex. while (depth < 0)
{
cout << "Not a valid input.";
}
-An if/else statement can be used for when the percent the user is off
is greater than 1%
ex. if ((flow - target / 10) > 1)
{
cout << "You were off by x percent";
}
else
{
cout << "You were only off by x percent, nice job";
4. Implement the Algorithm
-Program will display the cubic feet per second when the depth is
5.0 feet
-User will be prompted to enter a guess
-If the depth entered would produce a cubic feet per second number
of greater or less than the 1% error given, find the difference
and display how far off the user was
-If the depth entered would produce a cubic feet per second number
equal to 1000, congratulate the user