BREAKING ICE WITH AIR CUSHIONED VEHICLE - FINDING MINIMUM PRESSURE WITH NEWTON-RAPHSON METHOD

AIM: TO CALCULATE THE MINIMUM PRESSURE OF AN ICE BREAKING AIR CUSHION VEHICLE USING NEWTON-RAPHSON METHOD

THEORY: A hovercraft, also known as an air-cushion vehicle or ACV, is an amphibious craft capable of traveling over land, water, mud, ice, and other surfaces.

Hovercraft use blowers to produce a large volume of air below the hull, or air cushion, that is slightly above atmospheric pressure. The pressure difference between the higher-pressure air below the hull and lower pressure ambient air above it produces lift, which causes the hull to float above the running surface. For stability reasons, the air is typically blown through slots or holes around the outside of a disk- or oval-shaped platform, giving most hovercraft a characteristic rounded-rectangle shape. Typically, this cushion is contained within a flexible "skirt", which allows the vehicle to travel over small obstructions without damage.

INTRODUCTION: NEWTON RAPHSON METHOD –

`(df)/dx=(f(x_(guess) )-f(x_(n)))/(x_(guess)-x_n )`

`x_(guess)-x_n=(f(x_(guess)))/(f'(x_(guess)))`

`x_n=x_(guess)-(f(x_(guess)))/(f'(x_(guess)))`

`x_n=x_(guess)-α{(f(x_(guess)))/(f'(x_(guess)))}`

where,

`α=`relaxation factor

OBJECTIVE:

1. Use the Newton-Raphson method to find out the value of Cushion Pressure for h = 0.6 ft.
2. Find the Optimal Relaxation Factor for this problem with the help of a suitable plot.
3. Tabulate the results of p for h = [0.6, 1.2, 1.8, 2.4,3.0, 3.6, 4.2] assuming a suitable Relaxation Factor. 

GOVERNING EQUATION:

`p^3 (1-β^2 )+(0.4hβ^2-(σh^2)/r^2 ) p^2+(σ^2 h^4)/(3r^4 ) p-((σh^2)/(3r^2 ))^3=0`

where,

`p=` Cushion Pressure

`β=` Related to Width of Ice Wedge

`h=` Thickness of Ice Field

`σ=` Tensile Strength of Ice

`r=` Size of Air Cushion

PYTHON PROGRAM:

import math
import matplotlib.pyplot as plt
import numpy as np

# Inputs
# Size of the Air Cushion (inch)
r = 40

# Tensile Strength of the Ice (pound per square feet)
sigma = 21600

# Related to width of Ice Wedge
beta = 0.5

# Thickness of Ice Field
h = [0.6, 1.2, 1.8, 2.4, 3.0, 3.6, 4.2]

def f(r,h,p,sigma,beta):

	term1 = pow(p,3)*(1-pow(beta,2))
	term2 = 0.4*h*pow(beta,2) - (sigma*(pow(h,2))/pow(r,2))
	term3 = pow(sigma,2)*(pow(h,4))/(pow(r,4))
	term4 = (sigma/3)*(pow(h,2))/(pow(r,2))

	return term1 + (term2)*pow(p,2) + (term3)*(p/3) - pow(term4,3)

def fprime(r,h,p,sigma,beta):

	term5 = 3*pow(p,2)*(1-pow(beta,2))
	term6 = 0.4*h*pow(beta,2) - (sigma*(pow(h,2))/pow(r,2))
	term7 = pow(sigma,2)*(pow(h,4))/(pow(r,4))

	return (term5) + 2*(term6)*p + (term7)/3

"""
Performing Newton Iteration
"""

# Optimal relaaxation Factor
h1 = 0.6

# Array of alpha values for consideration
alpha_val = np.linspace(0.1,1.9,25)

# Empty array of number of iterations
n1 = []

# Creating empty arrays for no. iteration values
for i in alpha_val:
	p_guess = 50

	# Tolerance
	tol = 1e-6

	#Counter variable
	ct = 1

	while(abs(f(r,h1,p_guess,sigma,beta)) > tol):
		p_guess = p_guess - (i*(f(r,h1,p_guess,sigma,beta))/fprime(r,h1,p_guess,sigma,beta))
		ct = ct + 1

	n1.append(ct)
print(n1)

# Plotting Graph for "No. of Iterations vs Relaxation Factor"
plt.figure(1)
plt.plot(alpha_val, n1, color='blue', marker='o', markerfacecolor='yellow', markersize=5)
plt.title('FIGURE 1', fontweight='bold')
plt.xlabel('RELAXATION FACTOR (alpha)')
plt.ylabel('NUMBER OF ITERATIONS')
plt.grid()
plt.show()

# Empty array of cushion pressure (pound per square feet)
pressure = []

# Empty array of cushion pressure (psi)
pressure_2 = []

# Empty array of number of iterations
n = []

# Relaxation factor
alpha = 0.8

# Counter variable
iter = 1

# Creating a cushion pressure array for different values of thickness of ice field
for j in h:
	while(abs(f(r,j,p_guess,sigma,beta)) > tol):
		p_guess = p_guess - alpha*(f(r,j,p_guess,sigma,beta)/fprime(r,j,p_guess,sigma,beta))
		iter = iter + 1

	pressure.append(p_guess)
	pressure_2.append(p_guess/144)
	n.append(iter)

print(iter)
print(p_guess)
print(pressure)
print(pressure_2)
print(n)

# Plotting Graph for "Thickness of ice Field vs Cushion Pressure (pound per square feet)"
plt.figure(2)
plt.plot(h, pressure, color='blue', marker='o', markerfacecolor='yellow', markersize=5)
plt.title('FIGURE 2', fontweight='bold')
plt.xlabel('ICE FIELD THICKNESS')
plt.ylabel('CUSHION PRESSURE')
plt.grid()
plt.show()

# Plotting Graph for "Thickness of ice Field vs Cushion Pressure (psi)"
plt.figure(2)
plt.figure(3)
plt.plot(h, pressure_2, color='blue', marker='o', markerfacecolor='yellow', markersize=5)
plt.title('FIGURE 3', fontweight='bold')
plt.xlabel('ICE FIELD THICKNESS')
plt.ylabel('CUSHION PRESSURE (psi)')
plt.grid()
plt.show()

EXPLANATION OF PROGRAM:

1. Initially, the necessary modules are imported and the inputs are specified.
2. Now, the given equation is broken into terms for easy representation followed by the derivative with respect to Cushion Pressure.
3. Now, a code is written to calculate and plot the optimized value of relaxation factor.
4. An empty array of Cushion Pressure is defined where the values are inserted using the append command. The Cushion Pressure is calculated with respect to the Thickness of Ice Field.
5. The plots of "Cushion Pressure (in pound per square feet and psi)" are plotted against the "Thickness of Ice Field".

RESULTS:

Cushion Pressure (pound per square feet) at Thickness of Ice Field as 0.6 = 4.239 pound per square feet

Cushion Pressure (psi) at Thickness of Ice Field as 0.6 = 0.029 psi

PLOTS:

From the above graph, it can be observed that the optimal relaxation factor correlates to the minimum number of iterations. Therefore, the optimal value of alpha is '1'.

The Cushion Pressure (pound per square feet) varies parabolically with Ice Field Thickness. The Equation mentioned plays a vital role in the result.

The Cushion Pressure (psi) varies parabolically with Ice Field Thickness. The Equation mentioned plays a vital role in the result.

CONCLUSION: The Newton-Raphson proves to be a quick and efficient way of solving complex non-linear equations with a high level of accuracy. It can be employed in solving the equation where the iterations take a large time to converge using a different method.