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I'm trying to build a sounding rocket to reach the 100km as my senior project next year. To do this I need an equation to approximate the atmospheric density at any given altitude in the rocket's ascent, so that I can better understand the engine performance requirements of the rocket. I got my atmospheric data from https://www.engineeringtoolbox.com/standard-atmosphere-d_604.html

Here is some python code with outputs:

## Atmospheric density curve fit from Engineering Toolbox US standard Atmosphere
## Design by whisperingShiba

from scipy import optimize

import numpy as np
import matplotlib.pyplot as plt



def fxn(x, a, b):
    return a*np.exp(-b*x)
def polyFxn(h, a0, a1, a2, a3, a4, a5, a6, a7, a8, a9):
    return  ((((((((a9*h + a8)*h + a7)*h + a6)*h + a5)*h + a4)*h + a3)*h + a2)*h + a1)*h + a0


altitudes = np.array([0,5000,10000,15000,20000,25000,30000,35000,40000,45000,50000,60000,70000,80000,90000,100000,150000,200000,250000])

density1 = np.array([23.8,20.48,17.56,14.96,12.67,10.66,8.91,7.38,5.87,4.62,3.64,2.26,1.39,.86,.56,.33,.037,.0053,.00065])

altitudeLinSpace = np.linspace(0,500000,500000)

#Popt returns a array containing constants a,b,c... etc for function 'fxn'
popt, pcov = optimize.curve_fit(fxn, altitudes, density1, p0=(1, 1e-5))
popt2, pcov2 = optimize.curve_fit(polyFxn, altitudes, density1)

print(popt)
print(popt2)



#Plots data
plt.grid(True)
plt.ylim((0,25))
plt.ylabel('Density of atmospher in slug/ft^3 * 10^-4')
plt.xlabel('Altitude, feet')
#Data from engineering toolbox
plt.plot(altitudes, density1)
#curve_fit for negative exponential
plt.plot(altitudeLinSpace, fxn(altitudeLinSpace, popt[0], popt[1]))
plt.plot(altitudeLinSpace, polyFxn(altitudeLinSpace, popt2[0],popt2[1],popt2[2],popt2[3],popt2[4],popt2[5],popt2[6],popt2[7],popt2[8],popt2[9]))
plt.show()

Here's the output of the python script, where blue is the data, orange is the negative exponential, and green is the 9th order polynomial fit:

enter image description here

As can be seen in the image, the 9th order polynomial fit is really good for the first part, but deviates massively past 100000 feet. The negative exponential fit has the opposite problem, having good performance as altitude approaches infinity, but having poor performance from 50000 ft to 100000 feet.

How could I best fit this data? I want to avoid super long polynomials if possible (I only have 19 data points, and the curve is already trying too hard to 'hit' all the points). I have already put this data into excel and its even worse than python's curve fit. Any advice would be greatly appreciated.

Thanks to Claude for his work. Look at this beautiful graph: Beautiful

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  • $\begingroup$ First workout how good you need the fit. The atmosphere is pretty thin above 50k ft, does it make that much difference to the engine above this? If not, consider leaving out the high altitude points, or at least give them a lesser weight then refit the exponential. $\endgroup$ – user121049 Nov 22 '18 at 22:52
  • $\begingroup$ @user121049 Its actually shockingly sensitive because the rocket could be going trans-sonic at any point in the flight, which amplifies the effect of any error tremendously. I'll give the weights a try, Ill edit my post with the results. $\endgroup$ – WhisperingShiba Nov 22 '18 at 23:02
  • $\begingroup$ I would stick with something exponential like because that is what the physics suggests. Either use some form of perturbed exponential like $(A+\alpha h)e^{-\beta h}$ or split the region into two and fit two different curves but be sure to join them up smoothly. $\endgroup$ – user121049 Nov 23 '18 at 8:15
  • $\begingroup$ Could you plot it on a log scale ? Thanks. $\endgroup$ – Claude Leibovici Nov 24 '18 at 11:43
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I should not use a strict polynomial regression.

Looking at the data $(h_i,\rho_i)$, I first had a look at the plot of $\log \left(\frac{\rho_i}{\rho_1}\right)$ as a function of $h$. This is quite nice and using a model $$\log \left(\frac{\rho}{\rho_1}\right)=\sum_{k=1}^n a_k \left(\frac {h}{10^4}\right)^k$$ I obtained a very good fit $(R^2=0.999977)$ using $n=4$. For this model, the parameters are all significant as shown below $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a_1 & -0.2457039080 & 0.006144797 & \{-0.25909227,-0.23231555\} \\ a_2 & -0.0351850842 & 0.001530225 & \{-0.03851916,-0.03185101\} \\ a_3 & +0.0021044026 & 0.000110197 & \{+0.00186430,+0.00234450\} \\ a_4 & -0.0000390562 & 0.000002385 & \{-0.00004425,-0.00003386\} \end{array}$$ Using $n>4$ leads to non-significant parameters.

So, the model would be $$\rho=23.77\, \exp \left(\sum_{k=1}^4 a_k \left(\frac {h}{10^4}\right)^k \right)$$

Below are compared the experimental values and the predicted values from this model. $$\left( \begin{array}{ccc} h & \text{measured} & \text{predicted} \\ 0 & 23.77 & 23.770 \\ 5000 & 20.48 & 20.843 \\ 10000 & 17.56 & 17.986 \\ 15000 & 14.96 & 15.296 \\ 20000 & 12.67 & 12.839 \\ 25000 & 10.66 & 10.651 \\ 30000 & 8.91 & 8.743 \\ 35000 & 7.38 & 7.112 \\ 40000 & 5.87 & 5.739 \\ 45000 & 4.62 & 4.600 \\ 50000 & 3.64 & 3.665 \\ 60000 & 2.26 & 2.297 \\ 70000 & 1.39 & 1.423 \\ 80000 & 0.86 & 0.877 \\ 90000 & 0.56 & 0.541 \\ 100000 & 0.33 & 0.335 \\ 150000 & 0.037 & 0.0366 \\ 200000 & 0.0053 & 0.00534 \\ 250000 & 0.00065 & 0.000649 \end{array} \right)$$

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  • $\begingroup$ Wow that looks like exactly what I needed... One thing though. I'm trying to program this into python, and I am realizing I misunderstand something here. What number is $ a_i $ supposed to be? I initially took it to be $ a_k$ but I get an overflow in my code with that number, so thats clearly wrong. $\endgroup$ – WhisperingShiba Nov 24 '18 at 4:13
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    $\begingroup$ Wait it was just a bug!!! Its perfect thanks you a million times! What are good resources to learn how to analyze these kinds of problems myself? $\endgroup$ – WhisperingShiba Nov 24 '18 at 4:41
  • $\begingroup$ @WhisperingShiba. Sorry for the typo's ! By the way, I divided the $h$'s by $10^4$ in order to avoid overflows if working with the exponential function and also to have parameters of "human" size ! Glad to know that you liked it. $\endgroup$ – Claude Leibovici Nov 24 '18 at 4:43

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