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Sailplane polar coefficients

Posted 3 days ago

Hi
I want to find the coefficients a & b for the equation (a*V^3+b = Ws). I have a hand-drawn graph, which is an empirically-derived polar of sink rate vs speed for a sailplane polar

The graph can also be described by the quadratic (aV^2+bV+c = Ws), and I am able to derive these coefficients, using formulae given in Reichmann (p105) with 3 sets of coordinates read from the physical graph. This gives a pretty good fit.

However, it's reported that the cubic form a*V^3 + b is more accurate. I don't know, but would like to try and derive the a & b coefficients. However, I don't have handy algebraic solutions. I have tried iterating over a range of possible solutions in Python but, at 4+ billion loops, it's a rather clumsy process.

Can anyone suggest a Wolfram function that would factorize or expand this cubic and thus simplify the muscling required, or alternatively know of an algebraic method?

Attachments:
POSTED BY: Denis Buckley
3 Replies

It does not look like a polynomial curve to me. I would try with a rational function instead. Here is my attempt. IĀ imported the picture into Mathematica, traced the black curve with the mouse, rescaled the values so that they have the right range, and then I fit the points with a rational curve. The plot seems to confirm that it is a good approximation:

originalPts = {{122.51588254168202`, 
    204.24489892928017`}, {127.02236509049423`, 
    221.52865750038615`}, {130.42278577285526`, 
    239.81157233003358`}, {135.19447646436674`, 
    261.8608539614255}, {139.27744833569022`, 
    279.5372850538925}, {123.52326230852417`, 
    201.8950314323402}, {129.08852155105825`, 
    233.34995068271752`}, {134.78330104932917`, 
    259.49659532495923`}, {141.40733698558506`, 
    281.2148807472459}, {147.6387004004801, 
    298.1696989863218}, {156.52831287359328`, 
    308.0934176282546}, {169.37754459351862`, 
    311.48150304816454`}, {187.2821780413314, 
    314.0369582526232}, {200.62482025930188`, 
    314.7030624249842}, {215.92260157577624`, 
    313.8087558972775}, {234.60024480385977`, 
    311.19984788886376`}, {252.22322309237188`, 
    308.04613245552525`}, {274.14504034510213`, 
    304.00222224863035`}, {297.7362297828851, 
    297.8798203187203}, {320.3611569953296, 
    292.55098693983285`}, {341.4092264911051, 
    282.98087915483245`}, {359.90800841364717`, 
    276.7104540755089}, {378.6863896184149, 
    267.1115640114558}, {399.77557665569407`, 
    258.7133061593126}, {422.3676098349356, 
    248.06180703276328`}, {445.5969649074855, 
    234.01811073215362`}, {467.67091706377965`, 
    220.77003885964177`}, {485.83870277721655`, 
    205.78680673567118`}, {500.4004801007735, 
    193.5317234904752}, {535.1591938109781, 
    166.49282819760185`}, {566.0960320384079, 
    138.57812927069835`}, {596.705985810883, 
    109.16469595598278`}, {620.1717667470796, 
    84.05626923671105}, {641.7481966511782, 58.68674612889038}};
rx[{x_, y_}] = {Rescale[x, {48.84, 642.8069733449}, {60, 240}], 
   Rescale[y, {56.5, 384.1}, {3, 0}]};
pts = Map[rx, nera];
model = (a + b  x + c  x^2 + f  x^3)/(d + e  x);
ft = model /. FindFit[pts, model, {a, b, c, d, e, f}, x]
Plot[ft, {x, 50, 240}, Epilog -> Point[pts]]
POSTED BY: Gianluca Gorni
Posted 2 days ago

Apologies. I should clarify. The area of interest is to the right of the curve's maximum, ie, ~110+ on the black line. At speed below this, the aircraft is entering stall, which is why the curve drops away sharply. A function describing the right hand side is useful for predicting the aircraft performance, in particular how fast to fly in given atmospheric conditions. That said, I do not have access to Mathematica, indeed, my math skills generally are fairly rudimentary. Could you please provide the rational function you have derived? It would be most interesting to be able to map the entire curve. Incidentally, the coefficients I found for the form ax^2+bx+c are as follows (for the black curve x>110): a= +0.0015337999101090623 b= -0.08509049632639594 c= +1.7970110804340482

POSTED BY: Denis Buckley

Ok, here is a rational function that approximates the whole black curve:

(-2021.93 + 66.0922 x - 0.71923 x^2 + 0.00306214 x^3)/(-2393.81 + 
 30.8477 x)

If you need to fit only the part to the right, this polynomial of degree 2 gives a good approximation:

2.01015 - 0.0260691 x + 0.000125399 x^2

I also tried the degree 3, but the improvement is very small, and not worth the complication, however, here it is:

1.87598 - 0.0233908 x + 0.000108471 x^2 + 3.40822*10^-8 x^3
Attachments:
POSTED BY: Gianluca Gorni
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