We have a LogLog Graph and need some Values out of it

Leistung-E2-180.csv

Leistung-E2-180.png

Leistungsdiagramm_E2-180.png

unLogLog.ipynb

unLogLog.py

# -*- coding: utf-8 -*-
# <nbformat>3.0</nbformat>

# <headingcell level=1>

# We have a LogLog diagram and need the values

# <markdowncell>

# ![Diag](Leistungsdiagramm_E2-180.png)

# <headingcell level=3>

# So, let's Un-LogLog it

# <codecell>

import numpy as np
%matplotlib inline
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit

# <markdowncell>

# Get some Datapoints, e.g. with [DigitizeIT](http://www.digitizeit.de/)

# <codecell>

# for the E2-180 curve
logx = np.array([250.0, 1650.0, 4320.0, 10000.0])  # rpm
logy = np.array([3.0, 70.0, 180.0, 180.0]) # power

# <markdowncell>

# Because the diagram is [LogLog](http://en.wikipedia.org/wiki/Log-log_plot) and has only straight lines, that means we can fit a function of the form $P(n)=c \cdot n^m$ to find the parameters `c` (constant) and `m` (slope).

# <codecell>

def loglog(n, c, m):
    return c * n**m

# <markdowncell>

# Now lets fit the three parts of the LogLog graph to find the parameters.

# <codecell>

var={} # Dictionary to save the optimal parameters
for p in range(3):
    # do the curve fitting
    popt, pcov = curve_fit(loglog, logx[p:p+2], logy[p:p+2])
    
    # Save the optimal parameters
    var[p] = [popt[0], popt[1]]
    
    # Print them
    print('Constant: %.6f, Slope: %.3f for %i. part of the graph' % (popt[0], popt[1], p+1))

# <markdowncell>

# So now we have the variables for the function $P(n)=c \cdot n^m$, we can calc a continuous function

# <codecell>

n=[]
P=[]
for p in range(3):
    x=np.arange(logx[p], logx[p+1]+1, 10)
    n.extend(x)
    P.extend(loglog(x, var[p][0], var[p][1]))

# <markdowncell>

# Let's plot it and see, if the original datapoints and the found curve are fitting

# <codecell>

plt.loglog(n, P, label='Curve Fit', ls='--')
plt.loglog(logx, logy, label='Original', alpha=0.6)

plt.xlabel('n / 1/min')
plt.ylabel('P / kW')
plt.title('Leistungsdiagramm E2-180')
plt.ylim(0, 200)
plt.legend(loc='best')

# <markdowncell>

# You see, exactly!

# <headingcell level=3>

# Plot it with normal scale axes

# <codecell>

plt.figure(figsize=(10,6))
plt.plot(n, P)

plt.xlabel('n / 1/min')
plt.ylabel('P / kW')
plt.title('Leistungsdiagramm E2-180')
plt.ylim(0, 200)
plt.savefig('Leistung-E2-180.png', dpi=300, bbox_inches='tight')

# <markdowncell>

# Now dump it to csv for use with `Excel` or something you guys are using...

# <codecell>

data = np.array([n, P])
np.savetxt('Leistung-E2-180.csv', data.T, fmt='%.1f', header='n [1/min], P [kW]', delimiter=',')

# <markdowncell>

# Remember:

# <markdowncell>

# ![XKCD Python](http://imgs.xkcd.com/comics/python.png)