Aerodynamic body forces and moments [Aerodynamics #3] 

Thanks! More to come for sure

It's no problem and I'm glad you're enjoying the material!

Great suggestions! Glad you enjoy!

Ah thanks so much!!

Thanks and good luck!!

You raise a good point! It's not so exact here and not entirely true to say "inviscid", you need the boundary condition to accelerate/decelerate the flow and cause the flow change due to rotation. However, it changes the flow more globally, even outside of the wall region in places we can assume the flow is safely inviscid.

Thank you!

Hi! You mean something like solving example problems? This is a common request, and something I am considering doing in the future.

@@prof.vanburen Why not a certified document, a couple of weeks course, maybe some CFD. I am just proposing what I am looking for everyday

Great questions and thanks for your interests! In the inviscid case where flow is symmetric on each side, though there may be pressure changes everything happens equally on both sides so it cancels out. For example, consider the stagnation point, where flow goes from non-zero to zero velocity---this momentum change led to a force. However, on the backside of the cylinder, there is another stagnation point where flow goes from zero velocity to non-zero, meaning there is an equal and opposite force. In the case of viscosity's role in separation, viscosity is what causes the boundary layer, where near the wall flow is slower. Separation comes from the fact that flow has reversed direction due to adverse pressure gradients (or other impediments), and in the boundary layer flow is already going slow so it is more susceptible to being reversed. I hope this helps!!

Ah good catch! Thanks! Tough to fix at this point, I will adjust in my notes. I have always been a bit more relaxed with signs/conventions, it is not precise for sure but it gives me more capability to discuss the physics efficiently.

That's a good point! It's not so exact here and not entirely true to say "inviscid", you need the boundary condition to accelerate/decelerate the flow and cause the flow change due to rotation. However, it changes the flow more globally, even outside of the wall region in places we can assume the flow is safely inviscid.

Thanks for your question! This one is a tougher one to explain in writing, unless I am mistaken this is just a matter of sign definition. Positive moment here is described as clockwise, so the terms are signed such that they either contribute (positive) or take away (negative) from this clockwise moment. Hope this helps!

I believe, A in lower surface creates CCW moment about LE and must be negative. What do you think?

Sorry to be late on this! I have to check the book to be sure, but my guess is it just boils down to a signing convention which can be counter-intuitive in aerodynamics.

@@prof.vanburen Thank you for your reply. if we look at page 21 and breakdown each force at the lower surface to x and y axis, pl sin teta will causes counterclockwise moment (The force is to the right) while Tl sin teta will cause clockwise moment (The force is downward) since the pivot is at leading edge. but both sign are positive (dMl' equation 1.10 on page 23) Pl sin teta and Tl sin teta should cause a different direction of momentum yet the sign are the same that's the only part of the equation that doesn't make sense for me Thank you

@@alvinyustiansaputra9131 i was searching exactly for this question in comments section. That makes no sense for me either. If u look at perpendicular momentum on upper surface too, u got the first term Pu*cos(theta) + Tu*sin(theta), like it has an implied counterclockwise moment, which is false

Hm, can you be more specific about where in the video? In general, when I say "use trigonometry" it refers to angle/side length systems like finding the component of one vector in another direction. For example, when finding the portion of the Normal force in the lift direction, you would need to do N*cos(alpha). There is also a component of axial force in the lift direction, which is -A*sin(alpha). I think that, for the most part, that's what I mean when say using trigonometry.

Great question, I think I am just confusing with my arrows. In that case, it's really to indicate there is a strong pulling force (low pressure region)

@@prof.vanburen Your videos are great! For this point, I also got confused, since I would expect the top/bottom pressure to be lower, since the velocity is higher. Would it make sense to add a note about that? Thanks so much for your videos.

Hey Olim Olim! Glad you're enjoying them. The spinning causes a pressure difference because of something called the Magnus Effect. The top of the cylinder accelerates the flow because the cylinder surface velocity is in the direction of rotation. On the bottom, the cylinder velocity is opposite the rotation. These velocity differences inherently lead to pressure differences through the Bernoulli relationship, leading to low pressure on top and relatively higher pressure on bottom. Regarding the edit, I'm not quite sure what you mean. Where on the airfoil specifically? Maybe you mean that the pressure difference, relative to the reference undisturbed pressure, is negative?

​@@prof.vanburen ​ sorry about that, got a little rusty with my English. while we were calculating the forces on the shaded surface ∆N= -P * ∆S... Why is P negative?

@@olimolim7230 No worries! It's just a matter of sign convention and getting direction right. You will notice that the signs are opposite for the lower surface. I'll have to double check a textbook or two when I'm back in my office, but I'm pretty sure that it's because on the top, the pressure force is acting against the conventional normal force direction. Does this help?

@@prof.vanburen yes it does help, thank you professor 🙏🏻

the forces will be experienced by the foil

@@tom_winguill Thanks!

Not a dumb question and thanks @tomwinguill2924 for replying!

Hi Fayaz, the question is a bit general to answer definitively. If you mean on a boundary then no, especially if you have a no-slip condition. If you are talking about the general velocity field around an airfoil (and we aren't strict with the no-slip condition), then yes there are possibly points where the local velocity is equal to the free-stream.

@@prof.vanburen Thanks Professor Van for the explanation. this series is amazing.. :)

Not sure where specifically you mean, but generally I draw pressure as being normal from the surface with the arrow length correlating to pressure strength.

That's a good question! I am not sure I can answer it completely but I will try. By definition, the axial force is parallel to the chord line, but the actual point the force is applied to the foil is dependent on the pressure and shear distribution. However, it is likely a decent approximation to put it on the chord line to make the math more simple. Later on in analysis we like to approximate the foil as being thin (Thin Airfoil Theory) and in this analysis we approximate things as being on the chord line. There might be a more rigorous explanation out there, but I interpret it as just being a decent assumption (because airfoils tend to be thin) that leads to more approachable calculations.

What specifically would you like explained more? I am not sure the pressure front/back is different at that timestamp.

I may have had a bit of word salad, but typically I am referring to the pressure difference with respect to the infinite case, or delta P. So when pressure decreases due to acceleration, the absolute difference gets large.

This helped. Thanks. I was really confused. @@prof.vanburen

I don't! That's a nice idea to have videos dedicated to practical applications of the tools. Please feel free to reach out and I can work with you on the integration. I'm thinking if you have discrete pressure data along the chord you can just apply the trapezoidal rule to integrate it, or possibly fit a function to the data and due a true integration (the former method is more common).

I handed in the exam paper last friday, but I did come across the trapezoidal rule and that did do the trick. I ended up taking half of the sum of the two pressures times the distance between the pressure tabs along the entire surface (which didn't include the angle of the surface). Your videos were very helpful to get me through the rest of the paper and give a good understanding on the subject.

For the segments delta N and delta A you need to multiply the stresses (Pressure and Shear) x and area. The area here is the area of the segment, delta s x l. That segment can be at an angle (theta) relative to the coordinate system, so in order to get it to align with the normal and axial force (N and A) you need some trigonometry. I hope that helps!

so theta is the angle between the shear stress and A and also between the pressure force and N ?

@@khaledfakhfakh3003 essentially yes. Shear stress and Pressure act tangential and normal to the segment surface, and that surface is angled away from A and N by angle theta.

That just defines where the moment is that you are considering. M_LE considers the leading edge moment, and M_c/4 is the moment about the quarter-chord.

Thanks for the feedback Shubham