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VSPAero Coeff CDo

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Nicolas Gonthier

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Jun 15, 2015, 10:02:19 AM6/15/15
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Hi,

How is compute the CDo coefficient in VSPAero ?

Thanks

All the best

Nicolas

Rob McDonald

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Jun 15, 2015, 11:27:42 AM6/15/15
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Nicolas,

The CDo coefficient in VSPAERO is calculated with an _extremely_
simple wetted area form factor buildup.

It is an easy first cut, and it comes at almost no computational cost,
so it is great to have it included in the file. However, it is very
simple, and it is quite likely that you will want to perform your own
separate form drag buildup. This will give you a lot more control.

Probably most significantly, it is likely that you would want to use
different component Sets for thin surface aero vs. your drag buildup.
For example, if you have sensors, antennae, landing gear, etc. you
probably don't want them in the VSPAERO lift / induced drag
calculation, but you probably do want to keep track of the drag they
cause.

For reference, here is the VSPAERO code that does form drag.

Wings:
https://github.com/OpenVSP/OpenVSP/blob/master/src/vsp_aero/solver/VSP_Solver.C#L2620

Bodys:
https://github.com/OpenVSP/OpenVSP/blob/master/src/vsp_aero/solver/VSP_Solver.C#L2668

Rob
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Nicolas Gonthier

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Jun 17, 2015, 4:56:35 AM6/17/15
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Rob,

Thanks for your answer.

I have an other question about VPSAero.
When I plot the characteristics of my aircraft (or just my wing) : Cl function of the angle of incidence or Cl function of Cd, I find very accurate results.

But when I plot the spanwise variation of lift coefficient (thanks to the .lod file), I find a very strange result. The curve increases instead of decreasing and some points have very high value (Cl > 13 ). These strange points are located in the extremity of my wing.

I don't understand why I have these kinds of results.

All the best

Nicolas

Rob McDonald

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Jun 17, 2015, 10:59:41 AM6/17/15
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Nicolas,

Can you post your input file, a picture of your wing, some results, or
something for us to help you with?

Do the Cp results look reasonable?

What angle of attack are you looking at the load distribution?

What do you mean by 'The curve increases instead of decreasing ...'
In what context?

Rob

Nicolas Gonthier

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Jun 18, 2015, 4:50:28 AM6/18/15
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Rob,

I join the wing, the vspaero input file, an image of VSPAERO and the spanwise variation of lift coefficient : Sivells wing at Mach 0.17, angle of incidence 8.0° in the image but I found the same type of result at 0° or 13° (but the amplitude increases with alpha).

The Cp results look reasonable.

I can't share (for the moment) the wing which presents the most aberrant results. For this wing, I am waiting for a decreasing of the Cl with Yavg but I observed the opposite. (see Cl_function_y_AlbatrosWing). The case is really simplistic, Mach = 0.75, angle of incidence = 1.5°.

Thanks

Nicolas
sivells1947_wing_DegenGeom.vspaero
sivells1947_wing.vsp3
Sivells_image_vspaero.JPG
sivells_cl_function_y.PNG
Cl_function_y_AlbatrosWing.PNG

Rob McDonald

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Jun 18, 2015, 10:34:51 AM6/18/15
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Nicolas,

I'm not sure about the spike at the wingtip. I'll try to get Dave to
check it out.

At the root of the wing, looking at the Cp distribution, it is clear
that you are only modeling half of the wing (with no symmetry
applied). To use a half-wing, you need to set the Symmetry flag to Y
(instead of No). Or, you can include the other half-wing in VSP and
model both sides.

Note that a half model will need a matching Sref to get appropriate
coefficients. Also, you will need to watch the reference span - used
to calculate AR and thereby e.

Rob

Rob McDonald

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Jun 18, 2015, 10:37:54 AM6/18/15
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Also,

I would not expect you to need anywhere near that many panels in the
chordwise direction (spanwise too, but that isn't so over the top). A
vortex lattice model should converge to the best it can do with far
fewer panels -- it will run faster that way too.

Rob

Nicolas Gonthier

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Jun 18, 2015, 11:07:54 AM6/18/15
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Rob,

I compute only a half wing because I decided to compare the VSPAero results with experimental one realized in a wind tunnel for a half wing. 

But I noticed something, I have less panels, I found a Cl value equal to 1.12 instead of the 0.8 expected (at 8° of incidence). 

For the other chart : Cl_function_y_AlbatrosWing, I have a complete wing.


Nicolas

Rob McDonald

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Jun 18, 2015, 11:15:06 AM6/18/15
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I would expect a half-model in a wind tunnel to be mounted along the
tunnel wall -- thereby simulating a symmetry boundary condition. You
might want to check how the experiment was conducted more closely.

Rob

Rob McDonald

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Jun 18, 2015, 8:02:29 PM6/18/15
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Nicolas,

I emailed some with Dave offline -- he ran into the same problem last
week as well. You've found a bug.

He sent me a fix, it'll go into the next release. It will also have a
slight effect on the force coefficients.

Rob


On Thu, Jun 18, 2015 at 7:34 AM, Rob McDonald <rob.a.m...@gmail.com> wrote:

Nicolas Gonthier

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Jun 19, 2015, 6:21:05 AM6/19/15
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Rob,

Thanks for your help. 

I have an other question about the number of panels. As you can see on the joined file, with a low tesselation I found a "quadratic" evolution of Cl(alpha) and with a high tesselation I found a linear evolution. It is with the same half wing. Do you want more informations ?

Thank you

Nicolas
Comparison Tesselation.PNG

Rob McDonald

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Jun 19, 2015, 10:30:43 AM6/19/15
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Thanks a lot for this observation. I think this should be enough to
help track this down.

This may be related to the same load integration bug.

Rob

Nicolas Gonthier

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Jun 23, 2015, 10:56:24 AM6/23/15
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Rob,

Do you think that this bug can be the reason of the quadratic behaviour of the Cl function of alpha, that one can see on the joined chart : Cl_function_alpha_comparisonNASA. On the chart, one can see the AVL result on the same wing and the experimental NASA results in a wind tunnel.
It is the Sivells Wing mentionned above, with a high tesselation on the wing (see image Sivells_image_vspaero.JPG) and the same flying conditions (Mach 0.17, Re 4.410^6).

Nicolas
Cl_function_alpha_comparisonNASA.PNG

Rob McDonald

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Jun 23, 2015, 4:20:59 PM6/23/15
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It wouldn't surprise me at all.

Rob
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