What is the CG and aileron relationship?

[ziomatrixacs]

Im working on a new plane with a wing kinda like an EDGE 540 and it just made me wonder about this..
The root rib is 9 inchs, and the tip rib is 6 inchs and the span per wing is 18 inchs (so its a 36" plane). I intend to have 3" at the root and 3" at the tip be aileron. When I calculated the CG that I wanted , it came out to 3.56". I usually place my spar on the CG to bare the weight better but if you do the maths, my tip is 6" and 3 of it will be aileron, so the spar would end up going though my aileron which just wont work right.
Then I got to thinking... If the aileron crosses the line of CG (yes, I know cg is a point, but just think of how my spar is layed out as a line across the wing) how would that effect the roll rate? My thinking would be that the aileron infront of the CG would produce a negative torque to the spar and fight the rotation.. yet the CG is of course a point , not a line, so the distance is still positive (positive being one side of the wing and negative being the other wing)so it technically couldnt make a negative torque..?
Would it begin to behave like a very odd canard since it it forward of the CG?

Should I go on and experiment or should I redesign my wing?

[Brandano]

Are you talking about ailerons or flaperons? As ailerons, the pitching moment of one aileron is at least partly counterbalanced by the one on the opposite wing. In the case of flaperons you'd expect to see a nose down effect when these are deployed, but you might actually get a nose up pitching moment depending on how large and effective the tail is. With your forward swept setup perhaps some part of the down pitching moment will be counteracted by the portion of the flap ahead of the CG. Mind you, dropping down the wingtips, especially on a sharply tapered wing, sounds like a bad idea.

[Cory]

You are going from ailerons that are 33% of the root chord to 50% of the tip chord. This will increase tip stalls. I'd taper the ailerons to 2" at the tip to keep the same chord %.

[BMatthews]

You're not thinking about the correct axis. Even if you build the wing like that the ailerons produce a lift variation torque that works on the longitudinal axis running down the length of the fuselage. Not the spanwise CG axis. So you're worrying about a non existent issue.

The spar should actually be placed where it supports the lift of the airfoil at any given point on the wing. Symetrical airfoils lift at the 25% point so actually that is where the spar should be located so that any flex in the wing doesn't try to twist the wing when it's lifting. To complicate your wing since you've got a straight leading edge with a strongly swept forward trailing edge the 25% chord line from root to tip actually has some forward sweet as well. And that's where the spar should be with that same swept forward line. Mind you the leading edge contributes some help and if it's a D tube leading edge the sheeting holds a lot of the load as well so it's more common to place the main spar at around the 30% chord point. But that still means a strongly swept forward spar line if you want to do the job right.

So where does the extra lift go or come from to balance the lift of the plane when the CG is behind the wing's lift line like this one is? Believe it or not it comes from teh stabilzer and elevator. When the CG is located behind this 25% MAC point where the lift of the wing is centered the stabilizer has to lift upwards as well as the wing so the model flys level. It's just that the stabilizer is angled so it lifts less than the wing and so you still get some positive pitch stability.

[ziomatrixacs]

That drawing is not the wing, its just a sketch toshow how the aileron would go past the CG point...

Exactly now... what does a tip stall look like? Is it that annoying wobble that it hard to get rid of at high angles of attack flying?

My thinking was that since the CG is a point, and there was a force being applied by the ailerons at some distance, you get a torque. (T=F*D) Since the aileron is at an angle and crossed the CG line, and it applies a force but on opposite sides of the CG line, you would end up with a less effective aileron prone to tip stalling.
In my mind, I see it as essentially as a see-saw. The CG is the center, the left side of the seesaw is the part of the aileron behind the CG and the part infront of the aileron is the other side of the seesaw.

To help with my idea, I made a rough sketch. using a dummy plane.
I know that the ailerons produce a force about the length of the fuselage, but if you have a bit of the aileron infront of the CG you get another torque..thats the one I am talking about.. Just look at the drawing..

The seesaw is on ground so it pushes back with a force equal and opposite to the total force of both of the boxes sitting on it. Now the bigger box gives more down force than the little box, so you get a torque around the axis of rotation. In this case, the ground doesnt move so the seesaw does not sink nor float , unlike the airplane case.
With the plane, there is nothing except drag and lift (and torque of the motor if you want to get technical) working on the plane. Its the fact that its in equilibrium that keeps it flying straight. So in order to make the airplane roll, you increase the lift of one wing and decrease the lift of the other wing (ailerons tend to decrease the lift into the "negative" so it pushes down) Now when it rolls and taking the aileron that points down as in my drawing, the area infront of the CG is less than the area behind the CG, therefore the force acting torsionally on the wing is greater in the behind area of the aileron and lower in the area infront of the aileron, al though they both apply a force down. This is ignoring the fact that the distance span wise increases the force that the aileron applies because I am not analyzing that axis. (Remember, all 3 axis act independently) So as with the seesaw, the force infront of the CG and the force behind the CG add to give a resultant force since they are acting in the same direction. That resultant force is the airplane starting to roll since the total lift is now greater (or less than) than the drag above the wing. But as with the see saw, there is still a torque being applied to the airplane from the side, so it would seem to cause a pitching movement as if it were an elevator. But we usually do not experience anything because the other aileron works in the opposite direction, and assuming its equal, it cancels out.

However, there is still the force there,(as if the seesaw were made from wood, the wood will still warp from there being forces on it) so I am asking, would the wing tips warp, how would this effect flight, could it induce tip stall(which is yes according to what others have said)

Either way, im still going to take your advice and make it the same % chord instead of constant chord. I just found this to be yet another curiosity of model airplane flight.. :rolleyes: