|Tosh McCaber||20/10/2020 10:59:29|
|223 forum posts|
I have a question relating to trimming my mini pattern plane from many years ago, removing the original Cox TD .049, and replacing it with an (overpowered) electric motor. It is set up with 0-0-0 motor wings and elevator. Wings have a semi symmetrical aerofoil. I am enclosing some photos.
I’ve managed to pretty well trim it out – it’s very fast and quite sensitive, but quite controllable, despite its top speed. Everything is pretty good upright. However when I roll inverted, I have to hold a reasonable amount of down elevator to keep it level. I require an overly more bit more down when turning inverted otherwise it tends to dip further down and tries to go out of control. I have only tried inverted at height!!
Now, I’m old enough that I should know what I should be doing to alleviate the problem (CG position too far back, forward, elevator trim or what? Cg is exactly where it was on the original plan, although that sometimes can need adjusting?
Any thoughts and advice will be appreciated.
|Alan Gorham_||20/10/2020 11:02:34|
1364 forum posts
If your incidences truly are 0-0 then your model is nose heavy I'd say.
2391 forum posts
Yep, most likely nose heavy. It looks like a fairly small model, so grab a 5-10g weight, put it an inch or so behind the CG, then go fly. Keep edging it back until you get the handling you want, mark the final CG then rebalance for that point without the trim weight.
You can also try a (power off) dive test to help understand the CG - from your description I would expect it to self correct quite strongly (path A)!
EDIT - This trimming guide from the F3A guys may also be useful as an alternative approach to get the handling you want...
Edited By MattyB on 20/10/2020 11:56:42
|John Wagg||20/10/2020 11:46:51|
|136 forum posts|
If the wing is semi-symmetrical then won't that have less lift when inverted so will need more elevator ? Or is that irrelevant ?
|Frank Skilbeck||20/10/2020 12:03:26|
4868 forum posts
A Cox TD049 would weigh very little, has the electrification caused any weight gain which would be compensated for by flying either faster or at a slightly increased angle of attack (i.e. more up trim) and the reverse would apply when inverted, especially with the semi symmetrical wing section as John notes. If weight is the same then adjusting the c of g is probably your best bet.
|Rob Ashley||20/10/2020 13:26:06|
263 forum posts
As you have a semi-symmetrical wing section, you will not have an equal amount of lift from upright to inverted flight for the same pitch attitude and speed. Therefore, you will need to increase the angle of attack (AoA) when flying inverted, generally by holding a little down elevator. Moving the CG will not remove this tendency completely, but may reduce the effects.
As you state that you currently have to hold in a 'reasonable amount of down' in when inverted, so you could move the long CG aft a little and retest using the upright dive method shown above, (don't use the inverted method as this really only works for symmetrical sections). I would recommend 5 mm at a time as you are fine tuning - but I will leave that to you.
Hope this helps
|Jonathan M||20/10/2020 15:14:27|
870 forum posts
Agree with all of the above: its the wing section most probably combined with too nose-heavy a model.
My Boomerang is a bigger, heavier unit than your mini-beast, but the need for lots of down-elevator when inverted is directly attributable to the semi-symmetrical wing section, certainly compared to my Wot4 which needs hardly any. However, once I'd got 60g (2oz) of lead onto the tail (in my case to also compensate for the heavy .46 up front on an airframe weighing a total of 6.5lbs!), less down was needed.
If moving the CG progressively rearwards doesn't solve the problem enough or produces too unstable a model in the process, then (assuming you've got separate servos for each aileron) how about adding a bit of reflex to the wings via a snap-flap mix? This is a bit of a crude way of 'symmetrifying' an old-fashioned semi-symmetrical wing, but I've done this on my Middle Phase sloper to reasonable effect and can maintain inverted for longer without excessive, draggy elevator.
|Martin Harris||20/10/2020 15:40:48|
9594 forum posts
If it has a single aileron servo or basic transmitter then mechanical adjustment would do the same.
|Tosh McCaber||20/10/2020 16:16:17|
|223 forum posts|
Thanks for the excellent replies and discussion so far.
The weight, 21 oz/ 600g, is the same as when the TD.049 was on board- slightly lighter than the originally specified weight for the model, which could be up to 26oz (heavy!)
Wing area 200 sq in/ 0.13 sq.m- that's giving a wing loading of about 15oz/sq. ft.
I already did try the dive test upright, with engine in high cruising throttle, as opposed to low throttle. Cut the engine, and it gradually pulled out of the dive. I hadn't got round to try it inverted (due to my inverted problems!) But I'm now seeing your comments Rob re not using the dive inverted to trim out- sop maybe not.
I'm going to try moving the battery pack to move the CG a little back, as suggested. CG at present may be just a tad in front of the nominal position- it's lively enough, without me having a too rearward CG!!
For discussion- if I pull both ailerons slightly up, does that not make the nominal wing incidence negative, in relation to tail and motor- just wondered what influence that has on stability?
Unfortunately, with the vagaries of the weather presently, it probably will be the end of the week before I get out again for further tests.
|Martin Harris||20/10/2020 17:52:17|
9594 forum posts
My guess would be that if the incidence was measured from the verical tangent to the leading edge, the asymmetrical wing section will have imparted a modicum of effective angle of attack. Just a guess though...
Ideally, try it on a switcheable flight mode or mix.
2391 forum posts
You often hear this as a reason for not moving the CG back, but it's not correct - if you want to get the best performance and hands off trim for aerobatic flight you have to optimise the CG for the incidence first then set the elevator to give the response you want. For some reason a lot of fliers seem to believe the rate settings in the manual are what you have to stick with, but that isn't the case - they will vary depending on your preferred CG and flying style.
As you move the CG back you should always be ready to reduce the elevator rate - I generally have 2-3 elevator rates setup during trimming, or better yet set it on a knob or slider so you can adjust to the exact response you want. Once the CG is nailed you can come back and optimise the physical linkage (for instance you might be able to go to a smaller servo horn or larger elevator horn) to minimise slop and maximise the torque and resolution of the servo.
PS - A power model like this with a traditional planform is actually pretty easy to trim, as long. A plank (unswept) flying wing is whole lot more fun, as on those the total elevator movement might only be 2mm! That's where you have to take a really rigorous approach to trimming or you'll forever be chasing your own tail...
Edited By MattyB on 20/10/2020 18:12:10
|Simon Chaddock||20/10/2020 19:13:54|
5810 forum posts
I do think you are tending to minimise the effect of your semi symmetrical wing section. A zero incidence set will not have the same aerodynamic effect when inverted and quite possibly a different CofP position when it is at the incidence angle required to create the necessary lift.
In other words it will have a different trim setting when inverted.
Unless you use a true symmetrical wing section you may reduce the inverted trim change but are unlikely to eliminate it.
|Matt Carlton||20/10/2020 23:51:31|
131 forum posts
To add to what others have said. Any wing section has a 'zero lift' angle of attack. A cambered wing like yours could potentially have a zero lift angle which is negative. Let's call that 1 degree negative. Let's also say that when trimmed in level flight, the wing is at 1 degree positive. So, when inverted, you'll need to generate the equivalent of negative 2 degrees AOA just to accomplish ZERO lift. Let's assume that the wing needs additional negative 2 degrees to generate enough lift inverted due to the positive camber. So, now you need to apply sufficient down elevator to generate negative 4 degrees compared to that required for level flight upright.
Do as you will with CG and trim, a cambered wing will always require a fair bit of down elevator unless you rig the wing at a negative incidence.
|Pete Collins||21/10/2020 09:30:20|
146 forum posts
An interesting aside: A while ago I had a model rather similar to yours. Also a semi-symmetrical section but , in my case, with the wing set at a slight positive incidence. From memory, about 1.5 degrees. To maintain inverted flight this model needed a little UP trim. I know all the experts will say that this is impossible but, like the bumble bee, nobody told the model that! In practice it needed a dab of down to establish inverted flight and then a little up to maintain level after that.
|J D 8||21/10/2020 09:44:05|
1655 forum posts
You had the wing on upside down Pete.
Only joshing but I am bored, just been allowed to drive again after illness and now in lockdown again with nowhere to go. Cheers, John.
|Tosh McCaber||21/10/2020 10:02:52|
|223 forum posts|
Thanks again for all your comments and suggestions. I agree that the semi symetrical will generate lift, upright or inverted, even at zero deg incidence. I'm going to screw in a turn or two of up aileron on each side (single aileron servo), with elevator compensation, to see how that goes. I'll let you know how it goes.
My control surfaces are presently on very low dual rate/ expo!
|Bob Cotsford||21/10/2020 11:29:29|
8845 forum posts
My MiniPanic behaves exactly as you describe if the balance point is set a bit too far back. It's still well mannered and not at all twitchy but it is unnerving!
|Robert Welford||21/10/2020 14:03:59|
|234 forum posts|
I suspect the centre of lift of the wing (s) is forward of the CofG not behind it, which is the the normal case. Thus, the restoring moment of the tailplane is in the opposite sense.
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