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I am having the following parabola looking curve (blue curve), but its not exactly symmetric. The best fit that I am getting using a quadratic equation is also shown (black curve) but it is not perfectly fitting it.

Is there any simple equation for fitting such a skewed parabolic curve?

The data is from experiment. Y-axis is flow velocity in vertical direction (in m/s), and X-axis is the depth of water (in mm). The parabola vertex can be assumed to be at 'D' as shown in the 2nd figure.

Thanks a lot.

enter image description here

Edit 1: Below is a pic of the experiment set-up in the question for easy understanding. The axes are interchanged in the figures.

enter image description here

user2617526
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    Can you provide details of where the data comes from? Normally an understanding of the source of the data would help determine what model is appropriate. – Ian Miller Jun 08 '16 at 04:15
  • How did you get that curve? Was it simply drawn by hand? – SS_C4 Jun 08 '16 at 04:21
  • @IanMiller The data is from experiment. Y-axis is flow velocity in vertical direction (in m/s), and X-axis is the depth of water (in mm). I have attached a pic of the experiment set-up in the question for easy understanding. – user2617526 Jun 08 '16 at 04:47
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    Does the theory behind this experiment offer no plausible formula relating velocity and depth? I would think this is at least a well-studied section of physics. – J. M. ain't a mathematician Jun 08 '16 at 04:55
  • @J.M. Well, till now I didn't come across any such formula for this, so thought of asking here. But you are correct that such cavity flow has been studied a lot. – user2617526 Jun 08 '16 at 05:07
  • It might be worth also asking a similar question on physics.SE. Someone there may have a better understanding of the principles and hence what equation(s) to use. – Ian Miller Jun 08 '16 at 06:11
  • @IanMiller Thanks for the suggestion, just posted there too. – user2617526 Jun 08 '16 at 06:30

2 Answers2

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As far as your diagram looks, you might try some ombination of parabolas to fit your function..

Your lower part (both sides) look like more or less from a same parabola, andthe upper part from a different one. So you might make a function like $f(x)$=Parabola A when y coordinates below a certain point and Parabola B otherwise.

Now what these two (or more) parabolas are, you yourself have to find out , by brute force or something..

Qwerty
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  • ...a linear combination of parabolas is still a parabola. Probably you intended to talk about a piecewise parabolic curve. – J. M. ain't a mathematician Jun 08 '16 at 04:54
  • @J.M. Yes, I meant piecewise, but made a wrong choice of words. Thanks for helping. – Qwerty Jun 08 '16 at 05:06
  • @user2617526 Please dont leave answers unaccepted.. Just a request as a member of this site – Qwerty Jun 08 '16 at 05:24
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    and we should not hurry people to accept answers, either. The OP has only asked in the last hour, no need to rush him. – J. M. ain't a mathematician Jun 08 '16 at 05:27
  • @J.M. Sure. I just requested him.I didnt ask him to accept my answer if he didnt want to.. I have seen 50% or more questions have unaccepted answers even though the answer(s) have 3-4 votes..So I gave him a reminder since he is new to this site.. – Qwerty Jun 08 '16 at 05:30
  • @Qwerty The problem I see with a piecewise function (especially if build from parabolas) is it will have discontinuities in gradient which the physical data should not have. – Ian Miller Jun 08 '16 at 06:13
  • @IanMiller That can be solved by incorporating more and more piecewise parabolas until precision of the required level is reached.. – Qwerty Jun 08 '16 at 06:26
  • @Qwerty Thanks a lot for the answer. The 1 upvote u see is from me only. And i will also accept it if there is no better answer. Thanks again for taking the time to answer my question. – user2617526 Jun 08 '16 at 06:32
  • Which divorces the answer further from actual reality. I think without a appropriate physics phenomena there really is not correct answer to this problem. – Ian Miller Jun 08 '16 at 06:32
  • @IanMiller :Absolutely Physics will surely be of great help .But since there is no adequate physical background provided and the question has been posted on Math.SE so I couldn't think of a better solution. – Qwerty Jun 08 '16 at 06:39
  • @user2617526 You are welcome – Qwerty Jun 08 '16 at 06:39
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This is a fairly old question, but I wanted to share a nice solution offered in this thread. The solution they give was to use a Linex function, which looks like that:

$f(x) = \exp(ax) - ax - 1$

Which, when plotted, looks like that:

enter image description here

Note that I scaled the functions to compare their "skewness". Python code to reproduce the plot:

import numpy as np
from matplotlib import pyplot as plt

x = np.linspace(-2,2)
y = lambda a : np.exp(a * x) - a * x - 1
y_scaled = lambda a : (y(a) - np.min(y(a))) / (np.max(y(a)) - np.min(y(a)))


plt.plot(x, -y_scaled(1.25), x, -y_scaled(0.75), x, -y_scaled(0.25))
plt.legend(['a = 1.25', 'a = 0.75', 'a = 0.25'])
plt.show()

This, of course, doesn't at all look like a parabola (well, it is an exponential function), but did the trick for me when I tackled a similar problem:

enter image description here

liorr
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