OBJECTIVE :

• determining the minimum cushion pressure needed to break a given thickness of ice using an air-cushion vehicle, Solve by using Newton Raphson method and Write a Python code
  
  
• Use the relaxation factor of 1 and explain your results
• How will you choose the optimum relaxation factor for this case? and what is the optimum relaxation factor for this case? Show with the help of a graph, assume h = 0.6

• Explain your results with respect to the appropriate relaxation factor

  Also, create a small report summarizing your observation

WHATS IS NEWTON RAPSON METHOD?

In numerical analysis, Newton's method, also known as the Newton–Raphson method, named after Isaac Newton and Joseph Raphson, is a root-finding algorithm which produces successively better approximations to the roots (or zeroes) of a real-valued function

Newton's method in optimization

PROCEDURE :

In Newton Rapson, method is used to minimize the problem and get the optimize value of an equation

we are solving equations for a slope which is nonlinear, nonlinear is a curve which is

y = x^2 it is important for finding the drag coefficient values at various parameters

as to find the cd at various parameters we need a function to solve there comes CFD

and we can find the drag coefficient with the help of a nonlinear solver.

it not only calculate drag coefficient but also minimizes it

newton Rapson method is an iterative technique and is objective is to find the roots

fig.1 minimum value to be check by nuton Rapson method

guess a value of a minimum of x and calculate by the newton Rapson method it gives values near to the range or it troubleshoots the values as it is an optimization process and it can troubleshoot the values so we need to take absolute values of f(guess)

PROGRAM FOR PLOTTING HEIGHT WRT TO PRESSURE

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



B =0.5 #beta
r =40
sigma = 21600
height =[0.6,1.2,1.8,2.4,3.0,3.6,4.2]

def f(p,h):
    a1 = pow(p,3)*(1-pow(B,2))
    a2 =(0.4*h*pow(B,2)-(sigma*pow(h,2)/pow(r,2)))*p*p
    a3 = (pow(sigma,2)*pow(h,4)*p)/(3*pow(r,4))
    a4 =pow((sigma*pow(h,2))/(3*pow(r,2)),3)
    eq =a1+a2+a3-a4
    return eq

   


def fprime(p,h):
	b1=3*pow(p,2)*(1-B*B)
	b2=2*p*((0.4*h*pow(B,2))-(sigma*pow(h,2)/(pow(r,2))))
	b3=(pow(sigma,2)*pow(h,4))/(3*pow(r,4))
	eq_r =b1+b2+b3
	return eq_r



x_guess = 80
alpha = 1
tol =1e-5
iter =1
it =[]
p =[]

iter = 1
for h in height :

	while(abs(f(x_guess,h))>tol):
			x_guess = x_guess -alpha*(f(x_guess,h)/fprime(x_guess,h))
			iter=iter+1
	p.append(x_guess)
for i in range(0,7):
	   if i ==0:
	    print("height:p")
	    print(height[i],"|",p[i])
plt.plot(height,p)
plt.xlabel('height(feets)')
plt.ylabel('pressure(psi)')
plt.show()​

p= to calculate

h =0.6,1.2,1.8,2.4,3.0,3.6,4.2

r = 40

σ =21600 (144*150)

β = 0.5

make a function of the formula given

calculate the values of p with h

take the derivative of the formula and again calculate the values of p with h

take a guess value of x

find the x optimized value as x guess by running the formula in for loop

print and plot height vs pressure change

fig.2 plot of height vs pressure

PROGRAM FOR PLOTTING alpha WRT TO Iteration

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



B =0.5 #beta
r =40
sigma = 21600
h =[0.6]

def f(p,h):
    a1 = pow(p,3)*(1-pow(B,2))
    a2 =(0.4*h*pow(B,2)-(sigma*pow(h,2)/pow(r,2)))*p*p
    a3 = (pow(sigma,2)*pow(h,4)*p)/(3*pow(r,4))
    a4 =pow((sigma*pow(h,2))/(3*pow(r,2)),3)
    eq =a1+a2+a3-a4
    return eq

   


def fprime(p,h):
	b1=3*pow(p,2)*(1-B*B)
	b2=2*p*((0.4*h*pow(B,2))-(sigma*pow(h,2)/(pow(r,2))))
	b3=(pow(sigma,2)*pow(h,4))/(3*pow(r,4))
	eq_r =b1+b2+b3
	return eq_r



x_guess = 10
alpha = np.linspace(0.01,1.99,100)
tol =1e-5
iteration =[]

p =[]

h= 0.6
iter = 1
for a in alpha :
	x_guess=50
	print(a)
	iter=1
	while(abs(f(x_guess,h))>tol):
			x_guess = x_guess-a*((f(x_guess,h)/fprime(x_guess,h)))
			iter=iter+1
	p.append(x_guess)
	iteration.append(iter)
for i in range(0,99):
	   if i ==0:
	    print("alpha:iteration")
	   print(alpha[i],"|",iteration[i])
plt.plot(alpha,iteration)

plt.xlabel('alpha')
plt.ylabel('iteration')
plt.show()

We have to keep all the parameters same except height we have to change as 0.6

and the range of alpha is defined as 0.1 to 1.99 in 100 values to plot alpha vs iteration

so that we get an optimized value of alpha at which iteration.

fig.3 alpha vs iteration

we got the optimum value at 13 iterations ie. 1.130000000000001

A report summarizing your observation :

we have an observation that as the guess is likely near we can get the result

the change the alpha values change the number of iteration

if we increase the iteration the as tolerance is change

and as the alpha is changing the iteration change but up to some extent so need to fix an absolute value or it will be overwritten and troubleshoot and give u a negative results

CONCLUSION :

• By using Newton Rapson method we have to determine the height and pressure values
• plotted height vs pressure
• calculated the optimized value at the what iteration it comes
• plotted alpha and iteration