![]() |
| | #1 |
| Old Skool Join Date: Dec 2005 Location: Winchestertonfieldville
Posts: 6,608
| Whats the relationship of these two with regard to aerodynamics? Here is how I teach it and I am looking for a little more detail so feel free to add any input - Overbanking - during turns the outside wing travels at a greater resultant velocity v. the inside wing. This results in an increase in lift which tends to create a rolling tendency in the direction of the turn. The pilot must use aileron somewhat opposite the turn to counteract such effects. Adverse Yaw - during a turn, the aileron which deflects downward (outside wing) creates a higher lift/drag situation and tends to yaw the nose opposite the direction of the turn. This is counteracted by rudder inputs.
__________________ The simplest answer tends to be correct. |
| |
| | #2 |
| Old Skool Join Date: Nov 2000 Location: Spokane, WA
Posts: 6,517
| You forgot to mention the coriolis effect on the horizontal stabilizer. That's the key... |
| |
| | #3 |
| Old Skool Join Date: Dec 2005 Location: Winchestertonfieldville
Posts: 6,608
| Actually, never heard of it.
__________________ The simplest answer tends to be correct. |
| |
| | #4 |
| Old Skool Join Date: Nov 2000 Location: Spokane, WA
Posts: 6,517
| That's cause you didn't go to Riddle. |
| |
| | #5 |
| Old Skool Join Date: Dec 2005 Location: Winchestertonfieldville
Posts: 6,608
| Care to share your thoughts on it?
__________________ The simplest answer tends to be correct. |
| |
| | #6 |
| Agent Smith | It was all about the "Coanda" at ERAU/PRC back in the early 1990s! But for the technical aspects other than saying "trigonometry and physics", i really don't know. It's certainly in the realm of "minutia" but if you really want to know, I'd talk to Aero_Engineer.
__________________ Doug Taylor http://76school.flyblog.com (old!) http://30west.flyblog.com (updated 11/28) |
| |
| | #7 |
| Old Skool Join Date: May 2003 Location: Denver Colorado
Posts: 2,974
| merit, They both sound fine to me for basic explanations. The only thing I do a bit extra with overbanking is talk about stability as well. Overbanking tendency exists in all turns, but you don't need to counteract it in shallow turns. That's because of the airplane's positive lateral stability. In shallow to medium turns, the stability wins. If you put the airplane into a 15° bank and let go, it will roll out itself. If you put the airplane into a steep turn and let go, overbanking wins. There is a "pivot point" where the two tendencies are equal. I like demos and this is one of the concepts I teach in the first intro lesson (mostly for the purpose of "advertising" how stable the airplane is). I don't do the steep turn, but I show a shallow turn righting itself an then do a 30° turn (retriming for level flight) and let go so the new student can see the airplane maintain it's bank (although it eventually degrades, I can usually get 2 180s out of it) |
| |
| | #8 |
| Old Skool Join Date: Sep 2006 Location: Memphis, TN
Posts: 2,242
| This concept is from the Airplane Flying Handbook and I disagree with it. It's based on the premise that an aircraft's lateral stability tends to roll an aircraft out of a bank. This isn't exactly true, because an aircraft has no way of knowing that it's banked. What it does recognize is sideslip. In an uncoordinated bank, the aircraft will be in a slight sideslip, which will activate the geometric dihedral and tend to roll the airplane until the sideslip goes away. In theory, if your turn is coordinated, there will be no sideslip and the aircraft will have no tendency to roll wings level. |
| |
| | #9 | |
| Old Skool Join Date: Sep 2006 Location: Memphis, TN
Posts: 2,242
| Quote:
According to some aerodynamics books, this is more significant than aileron drag. Also, during a steady turn, there is the effect that I mentioned in another thread, due to the yaw rate of an aircraft that will require rudder in the direction of the turn. This is probably excessive detail for a student pilot. They will do well to remember the aileron stuff. ;-) | |
| |
| | #10 |
| Old Skool Join Date: Nov 2000 Location: Spokane, WA
Posts: 6,517
| "This concept is from the Airplane Flying Handbook and I disagree with it" You're disagreeing with the FAA blessed resource on how flying machines work. Now THAT'S an interesting topic of discussion... |
| |
| | #11 | |
| Old Skool Join Date: Sep 2006 Location: Memphis, TN
Posts: 2,242
| Quote:
Take a look at the explanation in the predecessor to the AFH, Flight Training Handbook, as to how dihedral works. And until recently, they've included both centripetal and centrifugal force in their diagrams of turns. One advantage of the AFH compared to Flight Training Handbook is that it leaves out a lot of explanations as to how things work; keeps them from being wrong. ;-) | |
| |
| | #12 | |
| Old Skool Join Date: Mar 2006 Location: KRST
Posts: 1,819
| Quote:
__________________ Aircraft without engine(s) prohibited... -KMIA 10-9 | |
| |
| | #13 | |
| Old Skool Join Date: Sep 2006 Location: Memphis, TN
Posts: 2,242
| Quote:
Centripetal force is the force that accelerates the aircraft toward the center of the circle and is provided by the horizontal component of lift. A proper diagram should only represent this force, because there is no force that opposes it. A turn consists of a constant acceleration towards the center of the turn, so there are no balanced forces in the horizontal plane. What we call centrifugal force is merely the result of our inertia slamming us into the bottom of our seats as the airplane accelerates in the opposite direction. Now, if you want to conduct your analysis from inside the airplane, then centrifugal force can be considered real, but that's a frame of reference shift. Aircraft mechanics is usually analyized by an outside observer, so centrifugal force should not be considered. | |
| |
| | #14 |
| Old Skool Join Date: May 2003 Location: Denver Colorado
Posts: 2,974
| Sorry, I never read it there. For me this concept is from doing it. And I try to keep those turns coordinated. |
| |
| | #15 | |
| Old Skool Join Date: Mar 2006 Location: KRST
Posts: 1,819
| Quote:
__________________ Aircraft without engine(s) prohibited... -KMIA 10-9 | |
| |
| | #16 |
| Old Skool Join Date: Dec 2005 Location: Winchestertonfieldville
Posts: 6,608
| fish, where are you??
__________________ The simplest answer tends to be correct. |
| |
| | #17 | |
| Old Skool Join Date: Sep 2006 Location: Memphis, TN
Posts: 2,242
| Quote:
Including centrifugal force in a diagram of a turn indicates a fundamental misunderstanding of the physics involved. Curved flight requires a net force. Centripetal and centrifugal forces are equal and opposite, resulting in no net force; no curved flight would be possible if they both were real forces in the same frame of reference. | |
| |
| | #18 |
| Old Skool Join Date: May 2003 Location: Denver Colorado
Posts: 2,974
| When I have an engineer or physicist as a student, I'll worry about it. As it is, I explain to my students that the FAA description is simplistic at best and if it doesn't fit their needs in terms of a practical understanding of aerodynamics they can apply to the airplane, I'm happy to point them in the right direction and they can deal with all the Rhos and Thetas they want. |
| |
| | #19 | |
| Old Skool Join Date: Sep 2006 Location: Memphis, TN
Posts: 2,242
| Quote:
However, some pilots at some point want to have a fuller technical understanding of what they and the airplane are doing. These needs cannot normally be met by even the above-average flight instructor. That's when they post in forums such as this. I'm glad they ask. | |
| |
| | #20 | |
| Old Skool Join Date: May 2003 Location: Denver Colorado
Posts: 2,974
| Quote:
![]() | |
| |
| | #21 | |
| Old Skool Join Date: Dec 2005 Location: Winchestertonfieldville
Posts: 6,608
| Quote:
__________________ The simplest answer tends to be correct. | |
| |
| | #22 |
| Old Skool Join Date: Sep 2006 Location: Memphis, TN
Posts: 2,242
| |
| |
| | #23 |
| Old Skool Join Date: Dec 2005 Location: Winchestertonfieldville
Posts: 6,608
| Any and/or all if possible.
__________________ The simplest answer tends to be correct. |
| |
| | #24 |
| Old Skool Join Date: Sep 2006 Location: Memphis, TN
Posts: 2,242
| Hmmm....we'll start with this. Let me know if there's something else you'd like references to. From Sears and Zemansky’s “University Physics”, 10th edition, by Young and Freedman, p. 140. Caution:>In doing problems involving uniform circular motion, you may be tempted to include an extra outward force of magnitude m(v^2/R) to “keep the body out there” or to “keep it in equilibrium”; this outward force is usually called “centrifugal force” (“centrifugal” means “fleeing from the center”). Resist this temptation, because this approach is simply wrong! First, the body does not “stay out there”; it is in constant motion around its circular path. Its velocity is constantly in direction so it accelerates and is NOT in equilibrium. Second, if there were an additional outward (“centrifugal”) force to balance the inward force, there would then be no net inward force to cause the circular motion, and the body would move in a straight line, not a circle. Third, the quantity m(v^2/R) is not a force; it corresponds to the “ma” side of Sum(F)=ma and does not appear in Sum(F). It’s certainly true that a passenger riding in a car going around a circular path on a level road tends to slide to the outside of the turn, as though responding to a “centrifugal force.” But such a passenger is in an accelerating, non-inertial frame of reference in which Newton’s first and second laws don’t apply. As we discussed in Section 4-3, what really happens is that the passenger tends to keep moving in a straight line, and the outer side of the car “runs into” the passenger as the car turns (Fig. 4-8c). In an inertial frame of reference there is no such thing as a “centrifugal force” acting on a body. We promise not to mention this term again, and we strongly advise you to avoid using it as well. http://www.amazon.com/Sears-Zemansky...e=UTF8&s=books For something more accessible, here’s the Physics FAQ: http://math.ucr.edu/home/baez/physic...al/centri.html |
| |
| | #25 | |
| Old Skool Join Date: May 2003 Location: Denver Colorado
Posts: 2,974
| Quote:
As I recall (it's been a long time) the standard elementary school "experiment" to show centrifugal force is swinging a ball around on a string. The apparent movement of the ball to the outside of the circle is actually the result of the ball's straight line momentum and the centripetal force exerted by the string. If the string broke, the ball would travel in a straight line at a tangent to the circle, not outward. If centrifugal force really existed, the ball would travel outward. There's a pretty good graphic at http://www.phy.ntnu.edu.tw/oldjava/c...cular3D_e.html. Note the string (centripetal force) and the straight ahead (not curved) velocity vector of the ball. That's why "centrifugal force" is not a "force" that acts on something. Ball or airplane, it's just a convenient shorthand to describe the net result of the centripetal force and velocity vectors. Maybe "centrifugal phenomenon" would appease those who object to the technically inaccurate use of the word "force?" tgrayson, how'd I do? | |
| |
![]() |
| Thread Tools | |
| Display Modes | |
| |