How to find drag polar equation of Cl vs Cd graph

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    Drag Graph Polar
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Discussion Overview

The discussion revolves around determining the drag polar equation from a graph of drag coefficient (Cd) versus lift coefficient (Cl) for an aircraft. Participants explore methods to extract values for [C][/D0] and k, considering both graphical analysis and regression techniques.

Discussion Character

  • Homework-related
  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • The original poster (OP) identifies that the drag coefficient at zero lift (CD0) is found at the x-axis intersection, but questions the meaning of k and other methods for analysis.
  • One participant suggests that the data fits better to a curve of the form Cl = Cl0 + k(Cd-0.4)² rather than the standard drag polar equation.
  • Another participant proposes determining the tangent line from the origin to the curve to find the optimal lift-to-drag ratio.
  • There is a discussion about the correct formulation of the tangent line equation, with a correction that it should involve (Cl - Cl0)² instead of Cl².
  • One participant expresses uncertainty about the regression results, questioning the choice of Cl0 and its impact on the fitting of the model to the data.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to derive the drag polar equation, with multiple competing views on the appropriate model and the interpretation of the data.

Contextual Notes

There are unresolved issues regarding the choice of parameters such as Cl0 and the sensitivity of the regression results to these choices. The discussion also highlights potential errors in the initial plotting of the graph.

MattH150197
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Homework Statement


I plotted the drag polar graph (Drag coefficient vs. lift coefficient) for an aircraft and are required to find the equation of the drag polar to determine values for [C][/D0] and k, using the graph or any other method. I've plotted the graph which I've included.

Homework Equations


CD=CDO+K*CL^2 [/B]

The Attempt at a Solution


I know that CD0 is the point at which the graph intersects the x-axis but what is k is it the gradient of the linear part of the graph? And what is the other method of analysing it, just out of curiosity. Thanks
 

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There is no good fit of the data to an equation of the form Cd = Cd0 + k(Cl2).
There is a better fit of the data to a curve of the form Cl = Cl0 + k(Cd-0.4)2.
Approximately Cl = 0.184 + 0.4×(Cd-0.4)2

I wonder if you are being asked to determine the tangent line from (0,0) to the curve. That gives the greatest lift-to-drag ratio. That would be the line from (0,0) to about (0.275, 0.82).

EDIT: The OP noticed that the chart had the lift and drag axis switched. So all the lift and drag coefficients in this should have been switched.
 
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What about if like you said I draw a tangent line and then simply allied y=mx + c so y = Cd m = k, x = Cl^2, c = Cdo do you think that would be correct?
 
MattH150197 said:
What about if like you said I draw a tangent line and then simply allied y=mx + c so y = Cd m = k, x = Cl^2, c = Cdo do you think that would be correct?
Not exactly. Instead of Cl2, it must be (Cl - Cl0)2, where Cl0 is the minimal lift value. See the equation in https://en.wikipedia.org/wiki/Drag_polar
 
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Just eyeballing the original data, I don't see why the regression didn't give (approximately) Cd0 = .25, Cl0 = 0.4 and k = 0.3. I wonder if I did the regression correctly. Or maybe it is sensitive to the original choice of Cl0 = 0.4 (which was eyeballed). Maybe Cl0 = 0.45 would have fit the upper Cl values better. The lower Cl values do not fit the model well. They are too low.
 

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