Physics of a Turning Aircraft

[HocBao]

I remember from elementary physics that a body traveling in a circular path has a velocity vector tangential to the circular path and an acceleration vector perpendicular to the velocity vector and pointed toward the center of the circle. A plane rolled 90 degrees from level has its lift vector pointed toward the center of the circle and therefore it turns in a circular path.

If I now pull back on the stick (i.e., add a little elevator) the plane turns tighter. Intuitively, this makes sense. But how do you explain the addition of elevator in physical terms? How does the elevator increase the acceleration of the plane toward the center of the circle? Does the elevator increase the lift of the plane? Or, does the elevator change the direction of the lift vector (i.e., moves the center of the circle). Or, does the elevator change the pitch of the plane such that a portion of the thrust vector is pointed toward the center of the circle?

Thanks for helping a novice understand aerodynamics just a little bit more. (I am in way over my head and this may be the one that drowns me!)

Hoc

[Apar]

HocBao,

This online course might help. The link refers to your specific question:

link

Apar

[bozon]

here is the short-short explenation in terms of elementry physics:

yes, pulling the stick increases the lift (and also drag) produced by the plane. it does so by increasing the "angle of attack".
the lift (call it "F") is a force perpendicular to the area of the wing. by rolling the plane at an angle "a" you get a component of the force that acts against gravity -  F*cos(a) = mg   and a radial component that creates the centripetal force needed for the turn F*sin(a) and that is the component of the force you were talking about.

mind you that if you roll your plane 90 deg. like you said, your vertical component F*cos(90) = 0 ! therefor you'll be free falling while turning (in a spiral) but will get the fastest turning rate for a given speed (this will change as speed changes).
the G figure mentioned when refering to turnes is the ratio betwin that "F" force and the weight "mg". since in a level turn F*cos(a) = mg you get G = F/mg = 1/cos(a).
by pulling on the stick you increase "F" therefor also "G" therefor you can turn at a larger bank "a" without loosing alt. (not refering to drag and pull issues).

as for the turn radius, in a circular motion the centripetal component of the force must satisfy: F*sin(a) = m*V*V/r .
therefor if you increase "F" (or "a") and keep a constant speed ("V"), the turn radius "r" will be smaller.
note that pulling hard on the stick will also make your speed drop, and "r" must be even smaller to compensate. this is how a p-47 can "outturn" a spit in the short term.

if you are interested in turn-rate, substitue V=wr and you get: w^2 = Fsin(a)/(m*r) where "w" is [radian/sec] turnrate.

hope this answered your question. wasn't so short after all, sorry

Bozon

[Andy Bush]

Minor clarification...turn rate varies with Radial G...that being the resultant vector of gravity factored along with cockpit G. The greatest value for Radial G occurs when the cockpit G vector is aligned with gravity...i.e., the aircraft is inverted relative to the ground.

Andy

[FLS]

Here's a nice site that explains flying.

http://www.monmouth.com/~jsd/how/


--)-FLS----
Musketeers

[HocBao]

Thanks all for the great responses. Very helpful.

Hoc