Finding minimum pressure with the help of Newton Raphson method

Given below is the python code for solving cushion pressure for a particular thickness of ice with the help of Newton Raphson method

import numpy as np
from matplotlib import pyplot as plt


b = .5; r=40; s=150
def f(p,h):
	return pow(p,3)*(1-pow(b,2))  + (.4*h*pow(b,2)-(s*pow(h,2))/pow(r,2))*pow(p,2) + ((pow(s,2)*pow(h,4))/3*pow(r,2))*p - pow(((s*pow(h,2))/3*pow(r,2)),3)
def f_prime(p,h):
	return 3*pow(p,2)*(1-pow(b,2))  + (.4*h*pow(b,2)-(s*pow(h,2))/pow(r,2))*pow(p,1)*2 + ((pow(s,2)*pow(h,4))/3*pow(r,2))
h = np.linspace(.6,4.2,7)
#print(h)
p_min = []
itr = 1
for x in h:

	def f_1(p):
		return f(p,x)
	def f_prime_1(p):
		return f_prime(p,x)
	p_guess = 160
	while(abs(f_1(p_guess))>.001):
		p_new  = p_guess - .5*(f_1(p_guess)/f_prime_1(p_guess))
		if abs(p_new - p_guess) == 0:
			break
		p_guess = p_new
		#print(p_guess)
	p_min.append(p_guess)
	
#print(p_min)
[H,P] = [h,p_min]
print(H,P)
plt.plot(H,P)
plt.xlabel('Thickness of ice')
plt.ylabel('Pressure')
plt.show()

Results:

given below is the relationship between thickness of ice and cushion pressure

Thickness	cushion pressure
.6	31676.68448708497
1.2	126706.79124494307
1.8	285090.3202735967
2.4	506827.271573046
3.0	791917.6451432907
3.6	1140361.4409843315
4.2	1552158.6590961672