I have data from three wind tunnel tests of a glide bomb at three different pitch angles. For each test, I have graphs showing the drag coefficient, lift coefficient, and pitching moment coefficient as a function of the angle of attack. Additionally, I have the mass, dynamic pressure, and reference area for the configuration. I was asked to determine the lift-to-drag ratio under glide conditions based on the given data. How should I find it? and why it's not simply the ratio of the coefficients founded in the experiments?
1 Answer
This feels suspiciously like a homework problem -- we generally don't like being asked to do someone's homework.
When you are having trouble with a homework problem, you can ask, but you should make clear that it is a homework, you should tell us what you've tried, and we will try to give you some direction without taking away the learning opportunity of figuring it out for yourself.
In this case, I'll go ahead and answer as if it is not a homework.
Assuming the center of gravity is in a location such that the glide bomb is statically stable, it will glide at a point where the pitching moment is zero.
That will imply a particular angle of attack -- which will correspond to a particular $C_L$ and $C_D$ -- and therefore a particular $\frac{L}{D}$. It will also (given the wing loading of the bomb) correspond to a particular dynamic pressure (equivalent airspeed).
Start by plotting $C_M$ vs. $C_L$ ($C_M$ on the vertical, $C_L$ on the horizontal). This should hopefully be close to a straight line with a negative slope.
Either fit a straight line to those points, or interpolate between the data points to find where the chart crosses $C_M=0$ -- you want to know the $C_L$ value. This is $C_L^*$ -- your cruising CL.
Next. plot $C_D$ vs. $C_L$ (Normally we would plot $C_L$ vs. $C_D$). Fit a parabola to the data $C_D=f(C_L)$. If you are doing this by hand, you will need to solve for the coefficients that match the data $C_D(C_L)=a C_L^2+b C_L + c$ -- you have three data points, so you can do it. (Here I am writing $C_D$ as a function of $C_L$, $C_D(C_L)$).
Once you have that parabola, we will treat it as a drag polar. Evaluate the drag coefficient at $C_L^*$ -- this is your cruising drag coefficient.
Then divide $\frac{C_L^*}{C_D(C_L^*)}$ -- that is your cruise L/D (again, in the denominator $C_D$ is a function of $C_L$.)
If you have the wing area and weight (or the wing loading), then you can solve for the speed. Remember
$C_L=\frac{W}{q S}$
and
$q=\frac{1}{2}\rho V^2$
or
$q=\frac{1}{2}\rho_0 V_e^2$
Ignore the dynamic pressure of the wind tunnel (assuming the data has been presented in coefficient form). The whole point of using coefficients is that you can re-use the data at a different dynamic pressure later.
You can do all of this by hand (including the plotting) or with Excel or any other appropriate computer program. If you are doing it by hand, you may be tempted to skip the steps where I say to plot the data. Don't -- go ahead and take the time to draw a plot with the data, it will help the whole process make sense in your head.
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$\begingroup$ Hi Rob, I converted the math to MathJax typesetting. Can you do a quick check to make sure I didn't make any mistakes in the conversion? $\endgroup$– DeltaLima ♦Commented Oct 28 at 10:18
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$\begingroup$ Thank you. Actually, I encountered this problem during a job interview. I'm new to this field and platform, so I hope it’s okay to ask – and please excuse my likely basic questions. Could you clarify why the solution didn’t consider the gliding angle? Also, when you mentioned calculating the cruising CL, did you mean for gliding conditions? $\endgroup$ Commented Oct 28 at 12:12
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$\begingroup$ "Please do my job interview for me" is honestly even worse than "please do my homework for me". The whole point of the exercise is to see what you can do, not what the internet can do for you. $\endgroup$ Commented Oct 28 at 15:33
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2$\begingroup$ @DarthPseudonym If he knew the interview question ahead of time, I would agree. However, since it seems the interview is over, I appreciate someone trying to learn from the experience. $\endgroup$ Commented Oct 28 at 15:46
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$\begingroup$ If you run through the math and the free body diagram, instead of $L=W$, you get $L=W cos(\theta)$. When you carry through to $L/D$, you get a sine over cosine that turns into a tangent and cancels out -- so $L/D$ is the glide slope without approximation. However, when I suggested you use $C_L=W/(q\,S)$ to get to airspeed, there is a small angle approximation embedded there which could be resolved with a little more work. $\endgroup$ Commented Oct 28 at 15:51