The Archimedean Spiral

Dan Couture
@MathYourLife
2014-12-29

This investigation was born out of work on calculating stats on running data. The Haversine formula is good for calculating distance along the great circle for two sets of latitude/longitude. The formula however does not account for elevation changes. Other solutions incorporate the Pythagorean Theorem to compensate for the additional elevation changes. For a long enough distance a linear increase in altitude actually follows a curved path known as an Archimedean spiral.

The archimedean spiral is a path that forms where the radius of the spiral increases linearly with $\theta$. This spiral occurs in situations with spooled material such as coiled wire or spools of paper.



In [1]:

    
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline

Goal: Create a function that calculates the distance for a section of an Archimedean spiral.

Create an arbitraty section of an Archimedean spiral
Calculate a numerical estimate for the total distance of the arc
Create a function that calculates the distance along an Archimedian arc
Validate the functions accuracy with the numeric simulation

Numerical Simulation

Create an Archimedean spiral starting at a radius of 1.5 increasing in radius by 2 for every revolution.



In [2]:

    
theta_0 = np.pi / 2 * 3
theta_f = np.pi / 2 * 11
r0 = 1.5
thickness = 2

thetas = np.linspace(theta_0, theta_f, 10000)
drdtheta = thickness / (2 * np.pi)
r = [drdtheta * (theta - theta_0) + r0 for theta in thetas]
pts = [(r*np.cos(theta), r*np.sin(theta)) for r, theta in zip(r, thetas)]



In [3]:

    
plt.plot(*zip(*pts))
plt.grid()
lim = np.ceil(max(r))
plt.xlim([-lim,lim])
plt.ylim([-lim,lim]);

Numerical Simulation - Length of a Archimedean Spiral

Using the points generated for the above arc determine the total length of the arc using the sum of the distances between points.

$$ \text{arc length} \approx \sum^{N-1}_{i=1} \sqrt{(x_{i+1}-x_i)^2+(y_{i+1}-y_i)^2} $$



In [4]:

    
l_est = 0
last_pt = None
for pt in pts:
    if last_pt is not None:
        dl = np.sqrt((pt[0] - last_pt[0])**2 + (pt[1] - last_pt[1])**2)
        l_est += dl
    last_pt = pt

print(l_est)









    



44.1882621017

Arc Length Formula

$$ L = \frac{a}{2} \left[ \phi \sqrt{1+\phi^2} + \ln\left( \phi + \sqrt{1 + \phi^2} \right) \right]^{\phi_2}_{\phi_1} $$

where

$ a = T / 2 \pi \\ \phi = r / a \\ r = \text{radius} \\ T = \text{change in radius for each full turn} \\ $



In [5]:

    
T = 2     # Thickness between spirals
r0 = 1.5  # Starting radius
rf = 5.5  # Ending radius

def archimedean_length(r0, rf, T):
    
    def limit(radius, thickness, a):
        phi = radius / a
        return phi * np.sqrt(1+phi**2)+np.log(phi+np.sqrt(1+phi**2))
    
    a = T / (2 * np.pi)
    return a / 2 * (limit(rf, T, a) - limit(r0, T, a))

print(archimedean_length(r0, rf, T))









    



44.1882650642

References:



In [5]: