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

In determining the minimum cushion pressure needed to break a given thickness of ice using an air cushion vehicle, Muller (“Ice Breaking with an Air Cushion Vehicle”) in Mathematical Modeling: Classroom Notes in Applied Mathematics, SIAM 1987) they

derived the equation:-

Equation is used to find the minimum pressure to break the ice.

where:-

• p denotes the cushion pressure
• h the thickness of the ice field
• R the size of the aircushion
• σ the tensile strength of the ice
• β is related to the width of the ice wedge

Taking:-

• β = 0.5
• r = 40 feet
• σ = 150 pounds per square inch (psi)
• h = 0.6, 1.2, 1.8, 2.4, 3.0, 3.6and 4.2 feet

Using Newton Raphson method -Python code

#Minimization of the function using Newton Raphsons Method
import matplotlib.pyplot as plt
#fuction
def f(p,h):
	term1=pow(p,3)*(1-pow(beta,2))
	term2=(0.4*h*pow(beta,2)-(sigma*pow(h,2)/pow(r,2)))*pow(p,2)
	term3=(pow(sigma,2)*pow(h,4)/(3*pow(r,4)))*p
	term4=pow((sigma*pow(h,2)/(3*pow(r,2))),3)
	return term1+term2+term3-term4
#function after differentiating
def fprime(p,h):
	Term1=3*pow(p,2)*(1-pow(beta,2))
	Term2=(0.4*h*pow(beta,2)-(sigma*pow(h,2)/pow(r,2)))*2*pow(p,1)
	Term3=(pow(sigma,2)*pow(h,4)/(3*pow(r,4)))
	return Term1+Term2+Term3
#input parameters
beta=0.5
sigma=150
H=[0.6,1.2,1.8,2.4,3.0,3.6]
r=40
#preallocating the memory
p=[]
p_guess=.25
alpha=1
tol=1e-4
iter=1
#while loop minimizes the function in steps
for h in H:
	while (abs(f(p_guess,h))> tol):
		p_guess=p_guess-alpha*(f(p_guess,h)/fprime(p_guess,h))
		iter=iter+1
	p.append(p_guess)
#printing the results	
print(\'Height|\',\'|Pressure\')
for i in range(0,6):
	print(\'|\', H[i],\'|\',\'|\',p[i])
print(iter)
#plots
plt.plot(H,p)
plt.xlabel(\'Height\')
plt.ylabel(\'Pressure\')
plt.title(\'Pressure Vs Height\')
plt.show()

for the thickness of ice -H=[0.6,1.2,1.8,2.4,3.0,3.6] calculation are made

Using relaxation factor of 1 [which is the alpha value].

Results table with h(height) in the first column, and p(pressure) in the second column

For relaxation factor 1 the result is obtained in 13 iterations

PLOT

Using relaxation factor of 0.2

Results table with h(height) in the first column, and p(pressure) in the second column

Thus using relaxation factor 0.2 the result is obtained in 92 iterations

Observations:-

1. Comparing the above two cases alpha=1 & 0.2 we can see the number of itterations differ a lot that is 13 & 92 respectively.
2. In case of relaxation factor of 1 the time of convergence is small the results converges fast.
3. Thus from the second case we can conclude that if the relaxation factor is small the time of convergence for calculation is longer due to greater number of iterations.
4. But if we observe the graph of the two cases there is a lot of variations from which we can conclude that if the relaxation factor is greater then the Newton raphsons method converges to the minimum value with greater accuracy.
5. In the second graph we can see the minimum value of pressure is for height approximately 1.2 feet.
6. The function minimum can be obtained with greater accuracy by taking the value of the relaxation factor less than 1 nearly equal to zero. The iteration in this case increases for smaller value of alpha.