LX A-10 - trim issue

Hi guys,

Not sure if this is the right thread to post this in. If not feel free to move it.

I need help and so does my LX A-10 twin ducted hair drier thing. I maidened her yesterday but after trimming her out I needed to have about 6 degrees of UP elevator. Flew a dream but that amount of elevator trim is not right. She is absolutely fine in pitch and roll. The CoG is very close as tested by climbing at 45 degrees and inverting her. Needs a slight touch of down elevator. Just right.

So am I correct in thinking that the tailplane needs reduced incidence, i.e. reduce the angle of attack? I don't want to mess with the wing incidence as it is a major piece of surgery. However the tailplane is an easy fix.

Suggestions please, preferably polite ones!!

Hi Ralph,

I'm happy to give you an opinion, but I'd like more info please:

- A picture of the side and plan view of the model
- Model span and length
- Position of the C of G
- Angles of the wing and tailplane relative to the fuselage datum
- thrust lines relative to fuselage datum
- wing and tail sections - symmetrical, lifting, flat bottom, flat plate, etc etc

Thanks,
Andy

My own view is that your CofG 'test 'is not necessarily proof of it being correct, just that the combination of aerodynamic forces and thrust lines produced the result you were expecting.
If the wing/tailplane incidences are incorrect it is still possible to fly quite well simply by a compensation by the CofG and elevator positions.

Wow, thanks for the responses.

I will take some measurements but won't be able to do that until the weekend.

The LX refers to Shenzhen Lanxiang models which I couldn't spell so went for LX.

I have managed to take some dimensions and hopefully got the picture loaded.





- Model span and length = w/span = 1534mm; length = 1400mm
- Position of the C of G - 90mm from wing root LE
- Angles of the wing and tailplane relative to the fuselage datum

wing = +2^o; tailplane = -4^o

- thrust lines relative to fuselage datum = 0^o

- wing and tail sections - wing is semi symmetrical (similar to Clark) tailplane is symmetrical

wing root = 260mm; tailplane root = 160mm; distance from wing LE to t/plane LE = 615mm

Interesting!
Are those wing and tail incidences actually measured?
My thinking goes like this:-
You say it needs just a touch of 'down' to fly inverted but you are already trimmed 6 degrees up so depending on how much 'a touch' really is you could be flying inverted with not needing any 'applied' down. The tailplane is already applying it.
It follows that the tailplane itself is not applying little or no down for normal flight so you are having to make it up with 'up' elevator.
If the CofG is where it is supposed to be then it is already reasonably far back for a power plane at close to 35% so the 6 degrees up is not required to compensate for a forward CofG but simply to overcome the tailplane. It looks like the tailplane incidence may be incorrect with its leading edge too high, hence my initial question about its incidence measurement.
Of course I could be completely wrong.

Thanks for the feedback.

I have rechecked my incidence measurements by distance measurement and trig calculation and the tailplane is -3^o not -4^o as per my previous post. The original measurement was done using an incidence meter.

Quorneng wrote: Interesting!

If the CofG is where it is supposed to be then it is already reasonably far back for a power plane at close to 35% so the 6 degrees up is not required to compensate for a forward CofG but simply to overcome the tailplane. It looks like the tailplane incidence may be incorrect with its leading edge too high, hence my initial question about its incidence measurement.
Of course I could be completely wrong.

Just a clarification point - when you say the tailplane is -ve incidence that means that the tailplane LE is lower than the TE (referenced to the datum) - effectively up elevator - pushing the tail down). This is conventional nomenclature for tailplane (and wing) incidence. If that is the case and you are adding up elevator to hold it in level flight then , yet the same amount (or slightly less) is required for inverted flight suggests that the CG is about right - I am assuming a semi symmetrical wing section.

However, if the tailplane incidence is actually +3 (as revised) (LE higher than TE reference the datum), you will be flying with a geometric decalage of -1 and with a semi symmetrical section would give a aerodynamic decalage of about +3. This would probably need the additional up elevator that you have stated.

Can you just confirm what you are actually measuring please?

Ta

Hi Martyn,

The -ve incidence on the tailplane is as you say leading edge lower that trailing edge or -ve angle of attack with the fuselage datum horizontal. On the wing the reverse, leading edge is higher than trailing edge, i.e. +ve angle of attack.

I am measuring the leading edge and trailing edge at the root of the tailplane and main wing against the same horizontal datum on the fuselage. Then taking the chord of the wing/tailplane at the root as one length of a triangle with the difference between leading edge and trailing edge gives the other side of the triangle then solving for the angle.

Hi Ralph,

Thanks for providing he data. I believe the short answer to your question is yes, reduce the incidence of the tailplane.

However, there's a few things that appear to be contradictory in the analysis:

- In an aircraft with such a short tail moment, I would expect a lot less longitudinal dihedral* with a C of G at around 35%. You currently have a difference of angle between the wing and tail of 5 deg and the fix we are considering here would increase that angle.

(* Longitudinal dihedral or decalage is the angle between wing and tailplane incidence)

- You say that when inverting the aircraft, only 'slight down elevator' trims the machine. As it's stable in upright flight with a longitudinal dihedral of 5 deg plus up elevator, I would expect a change of tailplane angle of at least 10 degrees plus at least the same down elevator as you now have up when inverted, even with a semi-symmetrical section. However, if she's stable with only a small amount of down elevator it would indicate that the model is tending towards neutral stability, which would require longitudinal dihedral of around 0 deg.

So, in a nutshell, I'm stuck!

Is the model an own design or is it commercial. If the latter, what's the recommended C of G position?

Thanks for the very comprehensive reply Andy.

The model is a commercial design and the CoG is stated as 90 - 95mm. I set it up to 90mm for the first flights and it flies perfectly. To be honest I have not tried moving the CoG back but that is something I will do to get the best CoG position.

Just for clarity when you say reduce the tailplane incidence I understand you to mean increase the negative angle of attack, i.e. go from -3⁰ to say -5^o. I am going to make the tailplane adjustable to fine tune it.

Apart from this high elevator trim the plane fly's almost like a trainer. So precise to control inputs, doesn't balloon when deploying flaps, trim doesn't change with power or with u/carriage up or down. Hopefully I will get to the fly-in and bring it with me.

Thanks again to everyone for taking the time to make all the comments. I have learnt a lot from what I thought was a simple question. I will make the change and let you know the result.