## Breaking Ice with Air cushion Vehicle - Find minimum pressure with Newton-Raphson method

The Newton Raphson method is an iterative technique that is used to find out the roots of a real valued function. It uses the idea that a continuous and differentiable function can be approximated by a straight-line tangent to it.

In this case Newton Raphson method is used to find out the cushion pressure required to cut a certain thickness of ice using the air cushion vehicle.

An Air cushion vehicle is similar to a Hovercraft which is used to break ice layers on the surface of rivers, lakes and even on the land in some countries .This vehicle uses air pressure to create cracks in the ice surface and then the crack propagates in the direction where there is the least resistance.

To create a crack in the ice surface minimum air pressure is required to be exerted by the vehicle. E.R.Muller presented an equation with few variables to calculate the pressure.This equation can be solved by Newton Raphson method.

Muller’s Equation

p3(1-β2) + (0.4hβ2 – σ h2/r2)p2 + (σ2 h4/3r4)p – (σh2 /3r2)

where,

p = cushion pressure

β = width of the ice-wedge

h = thickness of ice field

σ = tensile strength of ice

r = size of the air cushion

And the newton Raphson equation is given by,

Xn+1 =Xn -f(x)/f’(x)

Where,

Xn+1 = solution of the nth iteration

To increase the accuracy of the obtained solution, a realaxed factor is multiplied to the ratio which gives us,

Xn+1 = Xn – a {f(x)-f’(x)}

PYTHON CODING –

• Python program is shown below which solves the Muller’s equation by the Newton Raphson method.
• All the variables except for the thickness of ice and pressure are assigned certain values.
import matplotlib.pyplot as plt

def f(p,sigma,beta,r,h):
return pow(p,3)*(1-pow(beta,2))+(0.4*H*pow(beta,2)-(sigma*pow(H,2)/pow(r,2)))*pow(p,2)+(pow(sigma,2)*pow(H,4)*p/(3*pow(r,4)))-pow((sigma*pow(H,2)/(3*pow(r,2))),2)

def fprime(p,sigma,beta,r,h):
return 3*pow(p,2)*(1-pow(beta,2))+(0.4*H*pow(beta,2)-(sigma*pow(H,2)/pow(r,2)))*p*2+(pow(sigma,2)*pow(H,4)/(3*pow(r,4)))

alpha=1
sigma =150
beta = 0.5
r = 40
h = [0.6,1.2,1.8,2.4,3.0,3.6,4.2]
tol = 1e-14
iter = 1

#Peforming Newton raphson iteration

p_guess =120
P = []
print(" _______________________")
print("|Thickness |  Pressure |")
print("|__________|___________|")
for i in range (0,len(h)):
H = h[i]
while (abs(f(p_guess,sigma,beta,r,H))>tol):
p_guess = p_guess - alpha*(f(p_guess,sigma,beta,r,H)/fprime(p_guess,sigma,beta,r,H))

iter = iter+1
#print(iter)
P.append(p_guess)
print("|%f  | %f |"%(H,P[i]))
print("|__________|___________|")
"""
print(P[i])
print(f(P[i],sigma,beta,r,H))
print("n")"""
p_guess = 120 

• The below program shows a range of relaxation factors and optimum one is highlighted.
import matplotlib.pyplot as plt
import math

def f(p,sigma,beta,r,h):
return pow(p,3)*(1-pow(beta,2))+(0.4*h*pow(beta,2)-(sigma*pow(h,2)/pow(r,2)))*pow(p,2)+(pow(sigma,2)*pow(h,4)*p/(3*pow(r,4)))-pow((sigma*pow(h,2)/(3*pow(r,2))),2)

def fprime(p,sigma,beta,r,h):
return 3*pow(p,2)*(1-pow(beta,2))+(0.4*h*pow(beta,2)-(sigma*pow(h,2)/pow(r,2)))*p*2+(pow(sigma,2)*pow(h,4)/(3*pow(r,4)))

sigma = 150
beta = 0.5
r = 40
h = 0.6
tol = 1e-2
#Peforming Newton Raphson Iterartion
alpha = []
i = 0.1
j = 0
while(i<5.5):
alpha.append(i)
i = i+0.1
j = j+1
print(alpha[:])
P=[]
Iter =[]

for i in range(0,len(alpha)):
p_guess = 120
ite =1
while(abs(f(p_guess,sigma,beta,r,h))>tol):
p_guess = p_guess-alpha[i]*(f(p_guess,sigma,beta,r,h)/fprime(p_guess,sigma,beta,r,h))
ite = ite+1
Iter.append(ite-1)
P.append(p_guess)
#print(ite)
min_iter = min(Iter)
for k in range(0,len(Iter)):
if(min_iter==Iter[k]):
pos = k
break
print("The optimum relaxation factor is ",alpha[k])
plt.plot(aplha[k],min_iter,'*')
plt.legend(['Optimum Relaxation Factor'])
plt.plot(aplha,Iter)
plt.xlabel('Reduction Factor')
plt.ylabel('Number of iterations')
plt.title('No of Iterations v/s RF')
plt.show()


The above graph shows a decreasing trend in the number of iterations after 0, then the trend increases after a certain value of the relaxing factor. So, there must be minimum value for the number of iterations and its corresponding relaxation factor. This value is considered as the optimum relaxation factor because it reduces the calculation time in a calculating system.

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