"This answer does not explain how the slip-skid ball can be centered in a normal stabilized "coordinated" turn..."

Ball centered IS the definition of a coordinated turn.

How can it be that way? Simple: The combination of forces acting on the ball is centered on the aircraft's vertical axis.

@MichaelHall Combination of AERODYNAMIC forces acting on the ball is centered on the aircraft's vertical axis, but combination of ALL forces acting on the ball is not centered on the aircraft's vertical axis. First line of answer says ball measures ALL forces, yet the ball is centered.

@RobertDiGiovanni "In a loop the magnitude and direction of the resultant vector is a combination of gravity and acceleration G forces." -- the NET vector is what you describe but the net FELT vector is not. The new instrument I described, or the panel mounted G-meter, or the pilot's own body, all feel the FELT vector, which is entirely determined by the net aerodynamic force, nothing else.

@RobertDiGiovanni My new answer aviation.stackexchange.com/questions/77275/…, and the internally linked other answer within, address the situation of negative G's in a loop in more detail. The wing is generating skyward rather than earthward aerodynamic force.

@RobertDiGiovanni -- re "If you somehow did your loop at constant speed, would you not feel 2G at the bottom?" -- if you some how did it at constant speed and constant angle-of-attack, and therefore the same lift vector, you would feel the same G-load all the way around, whatever it was. I've done loops that were 4G all the way around but of course airspeed varied, so angle-of-attack must have also. Gravity adds to the actual net force but not to the "felt" force.

@RobertDiGiovanni -- edit-- I've done loops that were 4G all the way around but of course airspeed varied, so angle-of-attack must have also. Yet the aerodynamic lift force was undoubtedly 4G all way around. Gravity adds to the actual net force but not to the "felt" force. Not sure why the instructor selected this odd way of teaching loops in a sailplane, obviously they were not round, I learned better ways later, guess he wanted to keep it simple.

@RobertDiGiovanni -- In that particular case the actual net force, not the 4G on the meter, would have varied from 6G at bottom to 2 G at top. Would be interesting to see from outside exactly what flight path looked like but surely not round; you'd have to know the starting airspeed to conclusively know what it looked like I think.

If the combination of forces acting on the ball is not centered on the airplane's vertical axis, then the ball will not be centered. Period.

@MichaelHall No, in coordinated turn where sideforce is zero, the only forces are lift and gravity (ignore thrust and drag because they cancel each other). The vector sum of lift and gravity is the centripetal (horizontal) force that drives the turn. Force is not centered, but ball is. Key is that AERODYNAMIC force is centered.

@RobertDiGiovanni -- re "'If you somehow did your loop at constant speed, would you not feel 2G at the bottom?'" -- if you some how did it at constant speed and constant angle-of-attack, and therefore the same lift vector, you would feel the same G-load all the way around, whatever it was." -I'm assuming loop is not constrained to be round. If constrained to be round, and speed is constant, then angle-of-attack cannot be constant, so that's a different question-- hang on--

@RobertDiGiovanni -- in that latter case you need constant NET G all the way around. Meaning the FELT G i.e. the aerodynamically created-G i.e. the G-load on the G-meter must be 2G larger at the bottom than at the top. You could do it with 2G's reading at bottom and 0 at top, or 4G's reading at bottom and 2 at top, etc.

In a coordinated level turn what force is "not centered"? The lift vector is always perpendicular to the aircraft wings. In other words, aligned with the vertical axis of the aircraft.

In a banked turn this can be broken into vertical and horizontal components. This is "What makes an airplane turn 101".(Please don't make me draw the sketch for you...) If the sum of the vectors create a force acting in parallel to the vertical axis of the aircraft, and there is zero side slip, then the ball will be centered.

@RobertDiGiovanni -- sorry you are having trouble replying in chat-- So many possible permutations. You want a round circle AND constant "felt" G (i.e. aerodynamic-caused G i.e. what you read on the G-meter) -- that surely requires variable airspeed since ACTUAL force will be 2 G more on top than on bottom. Radius of curvature must stay constant and is proportional to V squared divided by ACTUAL force.

The centripedal force is not a sum, but is the horizontal component of lift remaining when the vertical component is subtracted.

@MichaelHall "In a coordinated level turn what force is "not centered"? The lift vector is always perpendicular to the aircraft wings. In other words, aligned with the vertical axis of the aircraft." Yes, the net AERODYNAMIC force is centered, but the NET force is not, because GRAVITY is not. The resultant of lift and gravity (weight) is the horizontal centripetal force that drives the turn.

We may have overlapped responses. See my last.

And please define what you mean by "net" force...

@RobertDiGiovanni -- see my last to you above-- could be good grounds for another ASE question--

Net implies a subtraction, but the ball responds to a total of the forces. An addition of the horizontal, and vertical.

The hypotenuse of the classic banked turn vector breakdown diagram...

@MichaelHall the centripetal force is not a separate force from lift and gravity but it is the vector sum of them. It is the ACTUAL NET force on the aircraft, the result of the vector sum of EVERYTHING including all aerodynamic forces and gravity. The fact that it is non-zero explains why the flight path is accelerating (curving). And it does contain a lateral component in the aircraft's reference frame, yet the slip-skid ball is centered. NET implies a VECTOR SUM. Like you said.

NET force is the vector sum of all force.

@MichaelHall of course assuming simplified case of constant altitude and airspeed, i.e. thrust=drag.

OK, I agree with that. And I am saying that the centripetal force is acting on the aircraft's vertical axis. That's why the ball is centered.

Because the ball is responding opposite the centripetal force, in other words, centrifugal force.

@MichaelHall No, the centripetal force, which is the net force, and the resulting net acceleration, are purely horizontal. Different than the "vertical"/ "up" axis of banked aircraft. Yet the ball is centered. It's not responding opposite to anything, it's staying centered. Because it only feels the aerodynamic force, not the net force.

So the ball is feeling the actual force of the air?

@MichaelHall -- Maybe I should just say "yes" :)

It does not. The ball does not register aerodynamic forces, it only responds to centrifugal force.

And I will spot you your definition of centripetal force acting horizontally to the inside of the turn, but what you are failing to consider in defining net is the vertical component. Net is not horizontal only.

Net (total) is the combination of vertical and horizontal.

@MichaelHall How am I failing to consider vertical? I said the net force is the vector sum of lift and gravity. The resultant is horizontal but neither lift nor gravity are horizontal so how am I not considering vertical?

In a banked turn the resultant (total) lift is a combination of vertical and horizontal. That combination is NOT horizontal! It is banked from the vertical to the same extent that the wings are banked from Horizontal! The total of the two vectors is the hypotenuse of the triangle, the vector that is angled, that is neither vertical nor horizontal. That vector is the one acting on the ball.

Are you somehow surprised that the ball is centered in a coordinated turn? I really am struggling to understand how we can be miscommunicating on such a dirt simple concept...

@MichaelHall Obviously the lift vector acts "up" in the aircraft's reference frame and includes horizontal and vertical components, I never said otherwise. The vector sum of the lift vector and gravity is the NET force vector which is horizontal. That's the centripetal force vector. And yet the ball is centered. N

@MichaelHall remember this all started over my comment ""This answer does not explain how the slip-skid ball can be centered in a normal stabilized "coordinated" turn..."" Because the answer said the ball responded to unbalanced forces in the lateral direction, and gravity. I say the ball only responds to an unbalance in the AERODYNAMIC forces in the lateral direction, period. If you don't disagree then we're mostly talking about nothing.

I agree we are talking about nothing. it is time for me to move on.

Final question, in a coordinated level turn where do you think the ball should be?

@MichaellHall well, I did have response almost ready to your question about whether the ball feels the air, but feel free not to read it today.

I don't need an answer. The ball does not respond to aerodynamic force unless you can prove that air molecules strike it. See my posted answer to your question.

@MichaelHall -- re where the ball should be in a coordinated turn-- Well that's obvious, at least to the first approximation, not taking into account the small scale effects due to the sideforce generated by yaw damping as noted in the Richard Johnson linked in some of my answers. Anyway if the scale of the turn radius is large compared to the scale of the aircraft, then the ball can safely be assumed to be centered in a coordinated turn.

But why is this so? Seriously, it cannot be that way according to what you are arguing with me about. How/why is the ball centered if the forces are unbalanced?!

@MichaelHall because the ball doesn't feel gravity. It's in my long answer to the latest question.

If you say "centrifugal force" I won't think any less of you. ;)

But seriously, it absolutely feels gravity. That's why it is centered when you are sitting on the ground!

@MichaelHall -- no, on the ground the tires are pushing up and that's what it feels.

That actually made me laugh out loud!

so if I push sideways on the tire the ball will deflect?

I think I am done here. At least for now...

@MichaelHall -- sure, if the plane is parked on ice

@MichaelHall -- hang on, I'm going to copy and paste a longer block of text. Might help you understand where I'm coming from, I'll tell you when its done-- You don't have to read it today.

1) I don't disagree with the part of your answer aviation.stackexchange.com/questions/77275/… that says we feel an apparent response to the acceleration of the aircraft. I'm not sure I disagree with any of it, actually.

2) I'm just saying that it's useful to note that the acceleration we, and the ball, "feel" always works out to be equal (or exactly opposite, if you prefer to think in terms of centrifugal rather than centripetal) to the acceleration component generated by the aerodynamic force generated by the air.

3) It is ALSO true, for the reason given in my long answer to the latest question, that this felt acceleration is equal to the net centripetal force vector minus gravity (again a vector sum), or to the net apparent centrifugal force vector plus gravity (again a vector sum). (These are just mirror images, use whichever one better fits your conception of how the ball works.)

4) So at the end of the day if you want to say the ball is feeling net centrifugal force plus gravity and nothing else, I don't actually disagree. It's just that unless someone tells you the turn rate and the airspeed and tells you that the flight path is constrained so that the turn is the only acceleration going on, you don't really know what the net centrifugal force vector is. If the flight path is bending up or down, that affects the net centrifugal/ centripetal force vector.

5) Practically speaking unless you are looking at recorded flight data, you can only find the net centrifugal/centripetal force vector by starting with the net AERODYNAMIC force created by the aircraft to get the "felt" G-loading, and then subtracting the effect of the gravity (vector math) to get the NET centripetal acceleration, (ctd)

6) (ctd) to get the NET centripetal acceleration, and then taking the mirror image to get the apparent centrifugal acceleration, and then adding in the effect of gravity (vector math again) to get the "felt" component of the centrifugal acceleration. This tells you what the ball feels. If you want to say that is a better way of explaining how the force from the wing's lift ends up affecting the ball then to say that the ball "feels" the aerodynamic forces directly, that's ok.

7) Personally I think it's more useful to just note that the ball only responds to the aerodynamic force generated by the aircraft, and not to gravity. Or if you prefer, to say that the ball only responds to the acceleration due to the aerodynamic force generated by the aircraft, and not to the acceleration due to gravity.

8) More details here aviation.stackexchange.com/questions/77275/…

@MichaelHall -- ok that's all, long block of text posted in chat. Can discuss later.

9) "Subtracting" in #5 should have been "adding". It's a vector sum. #6 is "adding" too. Maybe I need to make a step-by-step illustration of the vector math in 5 through 6 and add to my answer to latest question. Pertains to the paragraph starting with "Since gravity is already 'built in'". Anyway, the point is it's often helpful to note that what the ball (and G-meter) ends up responding to, is the net aerodynamic force generated by the aircraft.

10) Then you don't have to go through all these other permutations to figure out what the centrifugal force is and add gravity to it. But, if someone tells you the exact turn radius and velocity and bank angle and guarantees no other accelerations are happening (i.e. no up/ down bend of flight path), then the vector sum of centrifugal force and gravity is a good way to go to see where the ball will be.

11) Likewise to calculate the reading on a G-meter during a loop-- if loop radius and airspeed are definitely known, you can figure out what the G-meter will read by adding gravity to NET centripetal (or centrifugal) force. If those parameters aren't known, then you can figure the reading on G-meter out from airspeed and angle-of-attack and lift coefficient, in which case it makes no difference at all how the plane is oriented with respect to gravity, because gravity has nothing to do w/ it.

12) That last part -- #11-- may have been a pretty good summation of the value of each approach, so something useful came out of this after all. May need to add it to my answer.

@MichaelHall-- start with #11 and #12 and then read the rest if of any interest.

@RobertDiGiovanni -- #11 and #12 above may be of interest to you too.