More bad physics - turns

[Bede]

Sorry, pet peave of mine, but why is nearly every diagram of forces in a turn incorrect?

This plane won't turn:


This plane will turn the wrong way - like an outside loop:


This diagram is correct:

[digits_]

Your last diagram won't just turn, it will keep accelerating forever. There needs to be an equilibrium of forces.

The confusion is caused by not defining which coordinate system the writer is using. For the turning airplane, the use of the cylindircal coordinate system (https://en.wikipedia.org/wiki/Cylindric ... ate_system) feels more natural. One assumes that is the one being used.

In the cylindrical coordinate system, the first picture is correct. There is no acceleration in the display, because the coordinate system itself "takes care of that movement". (I don't know the correct english terminology, sorry). The rho value is constant, and the phi value is changing at a constant speed.

We are used to working in a cartesian coordinate system. X,Y,Z. In that case, the first picture is misleading, because it does not show the acceleration the airplane is making. The airplane is subject to accelerations/decelerations in the X and Y plane.

You are basically drawing the forces while slicing through a plane that is perpendicular to the airplane. If you want to study the full motion of the airplane in a constant turn, you take a slice every second, you'll see that in a cylindrical plane, the phi value will change by, let's say 2 degrees every second. The plane is defined by phi = [certain degrees]

If you do this with a cartesian coordinate system, you'll have to define your plane with a more complicated x and y relationship, with values that move more 'irratic'. Those values explain the hidden acceleration during the turn.


Either way, in neither coordinate system is your last drawing correct, IMO.


Now we'll wait for photofly to explain it properly!

[PilotY]

Is it wrong? I thought that centrifugal force was an apparent outward force due to inertia. Take for example a ball on a string that you swing around your head, it feels a force outward, because if it would suddenly become detached, its path would extend straight outwards from that point. But in the case of the turn, the horizontal component of lift acts as the string (which is the centripetal force) since it is directed inwards towards the turn.

To me this makes sense, as in a coordinated steep turn, you feel an increased load straight down in your seat (which in a banked turn, is actually on an angle to the horizontal, not straight down). Remember load factor?

[Bede]

Your last diagram won't just turn, it will keep accelerating forever. There needs to be an equilibrium of forces.

Well yes it's accelerating- towards the center of the turn- that's what happens in a turn. No there doesn't need to be an equilibrium of forces- only in unaccelerated flight.

I think you're confusing cylindrical coordinate systems with rotating frame of reference.

[Bede]

To me this makes sense, as in a coordinated steep turn, you feel an increased load straight down in your seat (which in a banked turn, is actually on an angle to the horizontal, not straight down). Remember load factor?

What you feel is the effect of inertia, hence why centrifugal force is an apparent force in the rotating frame of reference.

Look at the first picture. Is the plane climbing or descending? Why/why not?
Now ask yourself- given your previous reasoning, is the plane turning? (Hint: add up your vectors)

[PilotY]

Your last diagram won't just turn, it will keep accelerating forever. There needs to be an equilibrium of forces.

Well yes it's accelerating- towards the center of the turn- that's what happens in a turn. No there doesn't need to be an equilibrium of forces- only in unaccelerated flight.

I think you're confusing cylindrical coordinate systems with rotating frame of reference.

You can apply the first diagram to a ball on a string. The centrifugal force is an apparent force, but we all know the ball continues in a circle because of the string. In this case, the Horizontal component of lift is the string is it not?

[digits_]

Your last diagram won't just turn, it will keep accelerating forever. There needs to be an equilibrium of forces.

Well yes it's accelerating- towards the center of the turn- that's what happens in a turn. No there doesn't need to be an equilibrium of forces- only in unaccelerated flight.

True, but that's not displayed in your last pic. If those forces are correct, the plane would indefinitely slide sideways and keep accelerating. That's not what happens in a turn either. There is nothing in there that would eventually reverse it's course.

It would also mean that during the turn, the pilot would be feeling a sideways movement, which doesn't happen -assuming a coordinated turn-.

I think you're confusing cylindrical coordinate systems with rotating frame of reference.

That's possible. I'll look into it a bit more.

[digits_]

You're right, I did confuse coordinate systems with rotating frame of reference.

I was writing a bunch of text, but I noticed that wikipedia actually describes it pretty good.

https://en.wikipedia.org/wiki/Centrifugal_force

Stone on a string
If a stone is whirled round on a string, in a horizontal plane, the only real force acting on the stone in the horizontal plane is applied by the string (gravity acts vertically). There is a net force on the stone in the horizontal plane which acts toward the center.

In an inertial frame of reference, were it not for this net force acting on the stone, the stone would travel in a straight line, according to Newton's first law of motion. In order to keep the stone moving in a circular path, a centripetal force, in this case provided by the string, must be continuously applied to the stone. As soon as it is removed (for example if the string breaks) the stone moves in a straight line. In this inertial frame, the concept of centrifugal force is not required as all motion can be properly described using only real forces and Newton's laws of motion.

In a frame of reference rotating with the stone around the same axis as the stone, the stone is stationary. However, the force applied by the string is still acting on the stone. If one were to apply Newton's laws in their usual (inertial frame) form, one would conclude that the stone should accelerate in the direction of the net applied force—towards the axis of rotation—which it does not do. The centrifugal force and other fictitious forces must be included along with the real forces in order to apply Newton's laws of motion in the rotating frame.

So what's missing is the definition of the frame of reference, both in your first picture, and in your last picture.

[Bede]

True, but that's not displayed in your last pic. If those forces are correct, the plane would indefinitely slide sideways and keep accelerating. That's not what happens in a turn either. There is nothing in there that would eventually reverse it's course.

It won't because the plane is travelling forward (make a diagram from the top). Remember acceleration is a change in velocity over time (dV/dt). Velocity has both a magnitude and direction. In the case of a turn, the magnitude is the same, but the acceleration acts to change the direction.

It would also mean that during the turn, the pilot would be feeling a sideways movement, which doesn't happen -assuming a coordinated turn-.

What the pilot feels is irrelevant. The load factor is a result of lift being greater than weight (which results in a vertical and horizontal component).

Edit: digits, sorry, this posted after your last post

[digits_]

To add, I would like to postulate that it makes more sense to make the diagram based on the frame of reference rotating with the plane. That way, the diagram is valid for every position in the turn.

If we use your last picture, from an inertial frame of reference, the picture is only valid for one particular position in the turn. You'd either have to redraw your forces, or change position, which defeats the purpose of using an inertial frame.

[photofly]

Choice of rotating or inertial frame is a personal preference (or equivalently a choice of whichever is easier for the analysis at hand).

If you want to “watch” the plane turn from the outside - my preference - use an inertial frame (plus gravity) and there is no centrifugal force. Then the forces are out of balance, and there is acceleration.

A situation where a rotating frame makes more sense to me is looking at weather patterns, and because we are standing on a rotating planet a rotating frame (turning with the Earth) makes more sense. Then we must introduce Coriolis forces to make things work.

As Denker points out, things like Coriolis forces, centrifugal forces and even gravity itself are actually pseudo-forces that arise from having to “fix” a non inertial frame to obey the same equations as a true inertial frame. The give-away is that pseudo forces are always proportional to the mass of the object in question, because they actually represent accelerations.

[Bede]

As Denker points out, things like Coriolis forces, centrifugal forces and even gravity itself are actually pseudo-forces that arise from having to “fix” a non inertial frame to obey the same equations as a true inertial frame. The give-away is that pseudo forces are always proportional to the mass of the object in question, because they actually represent accelerations.

Bring on the Lorenz transforms!

Speaking of Denker, probably the best book on aviation. www.av8n.com

[digits_]

Wait, did we just conclude a physics discussion by post 11?

[photofly]

Your last diagram won't just turn, it will keep accelerating forever.

We could talk about this bit..?

Two things happening in a "turn" - the airplane is accelerating sideways, but as soon as it picks up any sideways motion the vertical and horizontal stabilizer make it nose into the new relative wind, so it doesn't just translate in a circle but rotates as well.

At 45° bank with a level pitch attitude the rotation is half yaw, and half pitch. Turning to the right it's continuously yawing right, and pitching up. (Ever wondered why you need to pull back on the yoke in a turn ...?)

But it does both turn, and accelerate, forever....

Curious fact: if the aircraft is in level flight going very slowly (think slow flight) it has a high pitch attitude, then the turn is a combination of yaw, pitch, and roll - with the roll actually *away* from the direction of the turn. That's partly why you need out-of-turn aileron doing steep turns in slow flight.

[Bede]

Curious fact: if the aircraft is in level flight going very slowly (think slow flight) it has a high pitch attitude, then the turn is a combination of yaw, pitch, and roll - with the roll actually *away* from the direction of the turn. That's partly why you need out-of-turn aileron doing steep turns in slow flight.

Hmm. Interesting. Tell me more. Reference?

[photofly]

Don’t need a reference: this is original thought. Get a model airplane and check it out for yourself. Put the model in the correct attitude for a steep turn in slow-flight (nose up, and banked at 45 degrees), then work out what combinations of small rotations around the three aircraft axes sum to a small change in heading, but leave the pitch and bank angles unchanged.

Hint: yaw a bit into the turn, which drops the nose too, then pitch up a bit to fix the pitch, and note the bank angle has increased, so now you need to roll away from the direction of turn.

You can do it with matrices representing small rotations (in the style of 1 + δθ), or quaternions, but it’s not really necessary.

Alternatively, put the model in a steep turn in slow flight attitude, and clamp a pencil vertically right next to the plane. Then keeping the pencil and the plane fixed together bring the plane to a regular attitude and note the pencil - representing the axis of rotation - is angled across all three aircraft axes - roll, pitch and yaw.

Contrast: If you do a turn with the nose below the horizon you must add roll into the turn, or else the bank angle goes away. Descending turns, the aircraft tends to roll wings level without aileron input, and in climbing turns the aircraft tends to over bank. You knew that... but I bet you didn't know this was one reason why.

[digits_]

Have you considered the effect of anhedral/dihedral wings on this?

[photofly]

I hadn’t. Initially I would say dihedral doesn’t change things. Dihedral increases slip-yaw coupling and slip-roll coupling, but there’s no side slip angle at play here.

[tsgarp]

The physics in a turn aren't really all that complicated. The first item that must be understood about a turn is that two aspects of the aircrafts motion are changing.
1.) It's path over the ground is changing. In terms of physics, the direction of it's velocity vector is changing.
2.) The aircraft is rotating around its vertical axis; i.e. it is yawing and changing heading.

A turn is co-ordinated when the rate of change in its velocity vector is equal to the rate of change in its heading. When this happens the relative wind is kept as close to parallel as possible to the longitudinal axis.

Centripetal force, (acting towards the centre of the turn), changes the direction of the aircraft's velocity vector (in terms of physics it causes centripetal acceleration). The best way to think of centripetal force is as the force that must be supplied in order to make an object turn. In the solar system, the force of gravity supplies the centripetal force to keep the planets turning around the Sun. In a turning aircraft centripetal force is supplied by the wings (as has been shown in the diagrams the OP posted). More centripetal force means more change in direction means a tighter turn, hence more bank means a tighter turn.

The force to rotate the aircraft around it's vertical axis (to yaw it) comes from the rudders.

[RedAndWhiteBaron]

Forgive me if this is obvious - but every illustration I've seen in this thread fails to account for the force of lift (be that positive or negative) generated from the empennage. It would seem to me that any discussion of the forces involved in a turn cannot be considered to be a complete argument without that force factored in.

But then again, I know nothing.