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3 green's
29th Nov 2003, 16:47
morning all

given that acceleration is defined as a change in speed and or direction

and given that a turn is a change in direction

and given that one needs a unbalanced force to cause an acceleration

what casuses the ball in the "inclometer" to remain "centrilised" during a balanced turn ?

please excuse spelling (maths ok);)

Captain Stable
29th Nov 2003, 17:31
Think of it this way:-

You are travelling in straight and level flight. The ball is in the middle.

You put on left bank with no turn. The ball goes left due to the aircraft tipping on its left side.

Apply rudder to make the aircraft turn. Centrifugal force pushes the ball out to the right, back to the centre of the tube. You are now turning, in balance, with the ball in the middle.

john_tullamarine
29th Nov 2003, 18:28
3 Greens,

Or, to put Captain Stable's tale in a different way ..

(a) the ball reacts to the net or resultant force vector

(b) in a constant speed, level turn there are two vectors of interest - the horizontal (centripetal) force pointing to the centre of the turn, and the lift force vector pointing up. (Cast your thoughts back to private pilot licence principles of flight classes and that nice little diagram of a banked aircraft with lots of arrows going this way and that .. )

(c) the resultant of these two vectors is a third vector pointing more or less perpendicularly with respect to the aircraft .. the orientation of the resultant vector depends on whether the turn is balanced or not ... if the turn is balanced, the resultant vector will be normal (ie perpendicular to the aircraft) and the ball will be in the middle as it has no way of distinguishing between balanced flight with different angles of bank.

Or, if the above is confusing (at times I even confuse myself) ..

(a) contemplate a glass of water sitting in some sort of a string cradle and held by your good self at the other end of the string.

(b) at rest, the string hangs vertically as does the glass and the water surface (similar to a ball in its action) is horizontal. There is a vertical force vector only acting

(c) now start whirling the glass around you in a conical motion. You now have a vertical force (the glass doesn't fall down and smash on the floor) and a horizontal force (something has to be stopping the glass flying out and smashing against the wall). The resultant force is aligned with the string and the water surface is, again, horizontal with respect to the glass .. the water can't tell what angle the glass is at with respect to the real world.

... or have I just further confused the issue .... ?

3 green's
29th Nov 2003, 18:56
thank for the replys

i was under the impression that a turn was an un-equal force vector diagram ie NO SUCH THING AS CENTRIFUGAL FORCE

VECTOR DIAGRAM SEEMS TO INDICATE THE NEED FOR THE BALL TO FALL TO BOTTOM OF INCLOMETER DUE TO GRAVITY? (NO BALANCEING FORCE EXCUSE PUN.)

APPARENTLY IT MAY HAVE SOMETHING TO DO WITH "CENTRIFUGAL EFFECT" ALTHOUGH DONT KNOW WHAT THIS IS

SOMEONE KNOWS...

bookworm
30th Nov 2003, 02:55
i was under the impression that a turn was an un-equal force vector diagram ie NO SUCH THING AS CENTRIFUGAL FORCE

Centrifugal force is a fictitious force, an apparent force that needs to be there to account for the fact that you're trying to do physics in a spatial frame of reference in which Newton's Laws wouldn't otherwise work. Then again, so, according to Einstein, is gravity so let's not be too hard on it. :)

You can choose to look at the situation in one of two ways. You can either do the mechanics in an inertial reference frame fixed with respect to the earth, and look at the motion of the balance ball as it accelerates under the influence of gravity and the reaction at the turn and bank casing. Indeed those latter two forces are unequal so there is a net acceleration.

Alternatively, you can do the mechanics in a frame of reference fixed with respect to the aircraft that is rotating as it accelerates around the turn. If you do that you have to introduce a fictitious centrifugal force that acts on everything in the picture in proportion to its mass. The centrifugal force plus the gravity is opposed by the reaction at the turn and bank casing and there's no net acceleration in the frame of reference of the aircraft. Hence that's where the ball sits.

You can make your choice -- inertial frame, unbalanced forces and a net acceleration or accelerating frame, fictitious force, and (in this case) balanced forces and no net acceleration. For a case like this it doesn't make much difference, but for more complex situations it makes a lot of sense to simplify the model by using an accelerating frame and fictitious forces.

If you are prepared to combine gravity and centrifugal force (both of which are proportional to mass) into a single apparent gravity it offers another way of looking at the balance ball -- as an indicator of the direction of apparent gravity.

john_tullamarine
30th Nov 2003, 06:38
3 Greens

To add a bit to Bookworm's post ...

You may find it easier to think of forces and reactions ..

If you consider the aeroplane (ball .. whatever) whizzing around in circular motion, then there has to be a force making the object keep turning toward the axis of rotation .. otherwise it would move off tangentially. This force comes from the aeroplane's lift.

To the pilot, however, the only sensation he/she feels is a pushing down into the seat .. he/she senses a "force" doing this to him .. this "force", however, is the inertial reaction to the wing lift force, the horizontal component (part) of which provides the turning force described above ...

In the horizontal plane (and if you aren't comfortable with resolution of vectors, don't sweat it too much) we talk about the inwards force (which causes the turn) being the centripetal force (which results in a centripetal acceleration) and the outwards reaction being a centrifugal reaction.

It's far easier, though, to consider the ball-on-the-end-of-a-string thing ....

As you start whizzing the ball around in a conic motion ..

(a) if you let go of the string, the ball whizzes off into the darkness somewhere .. thus, by holding onto the string you must be providing some sort of force (string tension) which causes the ball to keep going around the cone.

(b) however, you "feel" the ball trying to fly off .. this is the inertial reaction to the string tension force.

(c) now, if there were any out of balance forces at the ball, then the ball would move in the direction of the unbalanced force .. but, there it is, obediently whizzing about the cone with no other disturbing motion (unless you do something else to the string to change the ball's speed, etc ... )

(d) if you replace the ball with a glass of water, held in place by some sort of string cradle so that the glass is aligned with the string, then the water surface is at right angles to the sides of the glass. This is probably easier to see if you take a glass of water on a car trip. As the driver proceeds around a tight curve, if you hold the glass vertical to the floor of the car, the water surface inclines and eventually the water slops out of the glass. If, however, you incline the glass so the open top points somewhat to the centre of the turn, you will see the water surface (ignoring bump-induced motion) move back to being perpendicular to the side of the glass.

(e) the balance ball in the T&B instrument is doing much the same as the water in the glass .. it reacts to net (ie resultant single) force representing the various individual forces at play. If the net force is aligned with the local perpendicular (ie vertical to the aeroplane's planform), then the net reaction is opposite this and the ball goes to the bottom of the tube... no different to the case in balanced straight and level flight except that the load factor due to the turn makes everything seem a little bit heavier.


Looking at your last post ...

(a) there is no unbalanced force .. only the difficulty almost all of us had in our early days trying to grasp the idea of the radial (in toward the centre of the turn) force causing the aeroplane to keep accelerating towards the axis of rotation to maintain the turning motion

(b) the problem with the turning vector diagram is that you are looking at the (earth) vertical weight component in isolation .. if that force were the only one at play then, yes, the ball would go the the end of the tube. However, there is the resolution of the wing lift into vertical and centripetal parts to keep in mind. The vertical part of the wing lift is what is balancing the gravity influence. Try thinking either of the glass on the end of the string .. or the vector components being replaced by physical bits of string dangling over the edge of a table and having masses in proportion to the size of the force hanging on the end ... if you let the strings go, then the object on the table to which they are connected will move in a direction not aligned with either string .. rather a direction somewhere in between .. which is aligned with the net or resultant (force or reaction) of the two string tension forces. It really is worth setting this experiment up on your kitchen table ... something about seeing is believing ....

(c) I wouldn't sweat the centripetal/centrifugal thing too much .. just try thinking about whizzing a ball on the end of a bit of string around you ... or the table top experiment.

Ignition Override
30th Nov 2003, 15:06
Those are excellent descriptions. Hate to admit it, but the really strange thing for me is remembering every aspect of a banked turn with a slip, because we normally don't see it on a turbofan, unless a bit during a single engine approach in the sim (just trim the rudder a little...)and can't remember seeing it much at all on turboprops (many) years ago.:oh:

LOMCEVAK
2nd Dec 2003, 07:44
May I add one other simple way of looking at this.

Firstly, what you need to remember about a slipball is that in steady state conditions there is only one force acting upon it - gravity. However, the airframe has both gravity AND aerodynamic forces acting upon it.

In a steady turn with no sideslip, if we view the aircraft from behind (YZ plane) there are only 2 forces present, gravity and lift. (Note that the radial force which maintains the turning flightpath of the aircraft, the centripetal force, is actually just the horizontal component of lift). For the slipball to be deflected, there must be different lateral forces (along the Y or lateral axis of the aircraft) acting on the airframe and the slipball. Lift is perpendicular to the lateral axis and so has no component along it. Gravity acts on both the airframe and the slip ball. Therefore, the slip ball must remain centered. QED.

alf5071h
12th Dec 2003, 03:44
Notes on asymmetric flight here (http://www.pprune.org/forums/showthread.php?s=&postid=1099929#post1099929)