# Homework Help: How to find drag polar equation of Cl vs Cd graph

1. Mar 7, 2017

### MattH150197

1. The problem statement, all variables and given/known data
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. Ive plotted the graph which ive included.

2. Relevant equations
CD=CDO+K*CL^2

3. 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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2. Mar 9, 2017

### FactChecker

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.

Last edited: Mar 9, 2017
3. Mar 9, 2017

### MattH150197

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?

4. Mar 9, 2017

### FactChecker

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

5. Mar 10, 2017

### FactChecker

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.