# Swept Volume Theory

Discussion in 'Hydrodynamics and Aerodynamics' started by Sailor Al, Aug 2, 2022.

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### Sailor AlSenior Member

Do want to think about that?

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### Doug HalseySenior Member

It's a large gradient. Not too uncommon.

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### Sailor AlSenior Member

Like this?

That's pretty large, don't you think?

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### Sailor AlSenior Member

Or this:

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### Doug HalseySenior Member

At the leading edge, the slope of the surface is infinite. Moving along the surface, the Y coordinate is changing rapidly there, but the X coordinate is not.

If you plot the pressure with respect to the X coordinate, you will grossly overestimate the gradients. That's why I'm telling you (for the third time!) that the arclength is the parameter to use when focusing on the leading edge (not the X coordinate).

Arc length - Wikipedia

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### Sailor AlSenior Member

Here are a couple of graphics to divert your attention from the infinite gradient at the leading edge to highlight the problems with the NASA graph.

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### Doug HalseySenior Member

I think you're pointing out two things you think are wrong with the NASA graph:
1-You say it shows the stagnation point to be at the leading edge, but I don't think that's correct. The graph appears to be drawn from a sparse set of data points (judging by the width of the plateau at the max-pressure point), making it hard to be sure, but I would estimate the stagnation point to be near the red line that I've added to the graph. That's not far from the leading edge, but it's clearly on the lower surface.
View attachment 181331

2-You make the claim that the minimum pressure point should be at about X/C = 0.25 on the upper surface, but that's not true at all. It can be almost anywhere on the upper surface, depending on the airfoil shape & angle of attack.

You might be thinking of the classical solution for the loading on a flat plate, which has its moment center at 25% chord. However, its minimum pressure is right at the leading edge (& actually infinite in the inviscid idealization.)

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### mc_rashSenior Member

1. normally, the leading edge experiences negative pressure since the stagnation point (velocity = 0) moves further aft on the lower surface with increased AoA. At low AoA (~0°) and dependend on the foil the stagnation point lies on/ near the leading edge which led us to 2...

...2. stagnation point does not experience the highest pressure. At this point the velocity is 0, with Bernoulli you get:
P1 + 1/2*rho*v1^2 = P2 + 1/2*rho*v2^2
P1 = static pressure of freestream
v1 = velocity free stream (1/2*rho*v1^2 = dynamic pressure free stream)
P2 = pressure stagnation point
v2 = velocity stagnation point = 0
You notice the pressure is higher compared to the STATIC free stream pressure, but this is not the MAX pressure the foil experiences

3. The 25% procent chord length with lowest pressure.. you totally misunderstand what textbooks mean with something like "the center of pressure should be somewhere around 25% chordlength".

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### Sailor AlSenior Member

No, I didn't say that at all. But I agree, the stagnation point is on the lower surface a bit aft of the leading edge. I show that clearly in the diagrams. Not sure why you are quibbling?
Yes, "about" 25% based on the aerofoil section and AoA provided in the NASA graphic. The point is that the NASA graph shows the minimum pressure at X= 0! which is the leading edge!
No, I wasn't. Please don't misquote my thoughts!

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### Sailor AlSenior Member

OK, so maybe it does, maybe it doesn't, depends on the AoA.
My point is that the NASA graphic shows both the Maximum pressure and the Minimum pressure at (or, given the sparseness of the data, within ~2-3% of) the leading edge!
That's a bit harsh. Totally?? What's wrong with that - based on the limited data attached to the NASA graphic?

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### Sailor AlSenior Member

What value do you use for rho (air density)?

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### mc_rashSenior Member

Search for "airfoil pressure distribution" and you will find many graphs with a similar distribution as your's from NASA.
What is your problem with this graph, why shouldn't it be possible that the maximum and minimum pressure are in the area of the leading edge? As you said 2-3%, on a foil with 1000 mm chord this would be 20 - 30 mm. Even 1% would still be 10 mm. At the leading edge there is the biggest change, from free stream to a stream over a closed surface so why can't the biggest change in pressure be in this region?
"All the textbooks show pressure falling from the stagnation pt. to a minimum at around 25%" I still bet you misunderstand "centre of pressure"

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### Doug HalseySenior Member

What is the source of this graph anyway? Is it experimental data or the result of some calculation? You haven't really given much context to it.

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### mc_rashSenior Member

It's from NASA's website "Beginners guide to Aerodynamics" and is just an example picture. Aerodynamic Forces - Glenn Research Center | NASA https://www1.grc.nasa.gov/beginners-guide-to-aeronautics/aerodynamic-forces/

Yes, it will be experimental/scientific/proven data but there is no additional information on that specific graph on this site. Anyway, there are foils with similiar pressure distribution.

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### Sailor AlSenior Member

I didn't say "centre of pressure".
My graphic clearly states "All the textbooks show pressure falling from the stagnation pt. to a minimum at around 25%".
"Minimum", not "Centre of pressure".