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Compare d.S and Heron's formula for face area
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#!/usr/bin/env python3 | |
import math | |
import numpy as np | |
import numpy.linalg as la | |
def herons_area(v1, v2, v3): | |
a = la.norm(v2 - v1) | |
b = la.norm(v3 - v2) | |
c = la.norm(v1 - v3) | |
p = 0.5*(a+b+c) | |
return math.sqrt(p*(p-a)*(p-b)*(p-c)) | |
x1 = 1 | |
y1 = 1 | |
x2 = 3 | |
y2 = 0.5 | |
x3 = x1 # assume this is always true! | |
y3 = 6 | |
v0 = np.array([0, 0]) | |
v1 = np.array([x1, y1]) | |
v2 = np.array([x2, y2]) | |
v3 = np.array([x3, y3]) | |
area1 = herons_area(v0, v1, v3) | |
area2 = herons_area(v1, v2, v3) | |
print("area", area1+area2) | |
d = v2 | |
S = np.array([y3 - y1, 0]) # should really use cross product | |
print("0.5*S.d", 0.5*np.dot(S,d)) |
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