Discharge Corrected | Python Fiddle

#!/user/bin/python

import math
from math import degrees

#inputs
print ("---- Step 0: Input Values.")

width = float(raw_input("Width in meters? >>"))
print (width)

depth = float(raw_input("Depth in meters? >>"))
print (depth)

slopeDeg = float(raw_input("Slope in degrees? >>"))
slopeRad = math.radians(slopeDeg)
print (slopeDeg)
print ("Slope in radians >> %s" %slopeRad)

Qkleinhans = float(raw_input("Kleinhans Discharge (Q) in m^3/s? >>"))
print (Qkleinhans)

#1st we find the hydraulic radius: 
print ("----Step 1/5: Finding the hydraulic radius of the wetted perimeter.")

hydraulicradius_value = width * depth / (2 * depth + width)
print ("R (hydraulic radius) = [width]x[depth]/[[2x[depth]+[width]]")
print ("R (hydraulic radius) = %s m" %hydraulicradius_value)
    
#2nd we find the mean value of the Strickler coefficient:
print ("----Step 2/5: Finding the K sub m, or mean value of Strickler coefficient for the wetted perimeter.")

Km_beforeQ = depth * (hydraulicradius_value ** 0.667) * (slopeRad ** 0.5)
print ("Km = Q / [depth]x[R^0.667]x[S^0.5]")
print ("Km = Q / %s" %Km_beforeQ)

#compute the value of Qcorrect
Qcorr_value = 18

Km = Qcorr_value/Km_beforeQ
print ("Km = %s" %Km)
print ("Q Corrected = [[Qkleinhans in m^2/s]x[width in m]x[110^1.5 in m^0.33/s]]/[[width in m]x[110^1.5 in m^0.33/s]+2x[depth in m]x[km in m^0.33/s] = m^2/s")
print ("Q Corrected = %s" %Qcorr_value)

#3rd we find the Strickler coefficient without wall drag
print ("----Step 3/5: Finding the K sub s, or Strickler coefficient without wall drag.")
Ks = (((width ** 0.667) * (Km) * 110) / ((((width * (110 ** 1.5)) + 2 * depth * ((110 ** 1.5) - ((Km) ** 1.5))) ** 0.667))
print ("Ks = %s" %Ks)

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