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Author Topic: Self-Centering Ciphering  (Read 571 times)

Offline Howard Rush

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Self-Centering Ciphering
« on: January 06, 2023, 06:39:41 PM »
I recently revised a control geometry program I've been messing with for awhile and wondered if I'm missing anything by not including bellcranks that don't have the leadout-attach (input) holes in line with the pivot.  Such bent bellcranks may have some effect on the relationship between leadout travel and control deflection, but they mainly contribute some differential tug on the lines proportional to "centrifugal" and aerodynamic forces on the airplane.  I decided to cipher the latter effect using Impact numbers (available upon request), and assuming both leadouts exit the wing at the same point.  The first picture shows differential line tug vs. bellcrank angle.  The second shows differential line tug vs. leadout travel, which is what you'd actually feel.  We had a nice discussion on "self-centering" bellcranks here in 2017.  As folks noted then, the straight bellcrank with the holes in line is slightly destabilizing.  Having bellcrank arms perpendicular to the leadouts at neutral does eliminate the effect of forces other than control deflection near neutral, as Brett said (I think), but the effect goes nuts toward the ends of control throw. The third picture shows why. 

I figure it's bad enough having to deal with springy lines when trying to move the control surfaces precisely; I don't want to add another springy-line effect that doesn't do anything to move the control surfaces.  I tried to compare the magnitudes of these two effects of pulling on the lines, but failed because of the onset of dementia. 

Conclusions so far:
1. Putting bent-bellcrank calculation in my control geometry program is not worth the bother.
2. If your bellcrank is bent, it better be big.

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Offline BillLee

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Re: Self-Centering Ciphering
« Reply #1 on: January 07, 2023, 08:11:04 AM »
And, of course, if you're looking for in-flight stability instead of doing tricks, your bellcrank will be bent big.
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Online Brett Buck

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Re: Self-Centering Ciphering
« Reply #2 on: January 07, 2023, 10:40:28 PM »
1. Putting bent-bellcrank calculation in my control geometry program is not worth the bother.
2. If your bellcrank is bent, it better be big.

     Everything goes nuts when it is that far over, you cannot afford to have that much movement. I try to get it under 30 degrees for normal flight, so my controls end up a bit faster than you might think. But still very slow by traditional standards. I am in the midst of building an airplane the old fashioned way, and the 3" bellcrank looks comically "flippy" after all these years.

      Brett

Offline Howard Rush

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Re: Self-Centering Ciphering
« Reply #3 on: January 08, 2023, 12:26:41 AM »
30 degrees of bellcrank?  That amount of bellcrank movement and the arms 90 degrees to the leadouts would be the best setup to keep the centrifugal , etc. force evenly distributed between the lines.   I reckon it depends on how far you get into line elasticity to react the flap hinge moment. Your people are famous for keeping hinge moment down by using sensible flap chord.  I've been shooting for more leadout travel per degree of flap.  One can get about 9 degrees of flap per inch of delta leadout travel with a 4" bellcrank in an Impact and keep control response pretty linear. 

Igor's flap mechanism lets you get your money's worth out of bellcrank travel: mine's almost +/- 90 degrees.  Degrees of flap per delta leadout travel is about 9.9 at neutral and 4 at 30 degrees of flap. 
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Offline Howard Rush

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Re: Self-Centering Ciphering
« Reply #4 on: January 14, 2023, 03:47:29 PM »
Here’s a comparison of relative magnitudes of control surface hinge moment and “self-centering” bellcrank moment.  First the control surface hinge moment of an Impact with flaps at 30 degrees flying a 5.25-second lap on 70-foot lines: 
Hf=(ρV2/2)*Sf*c ̅f*Chf
       =(ρV2/2)*Sf*c ̅f*(∂Chf/∂δff +∂Chα/∂α*α),
where
Hf  is hinge moment, the torque reqired to deflect the flap
ρ is air density, currently .002246 slugs/ cubic ft. at my house
V is true airspeed = 70 ft. * 2π / 5.25 sec. = 83.8 ft./sec. for a 5.25-second lap on 70-ft. lines, neglecting the variation in airspeed from the inside tip to the outside tip
Sf is flap area, call it 144 sq. inches = 1 sq. ft.
c ̅f is flap mean aerodynamic chord, call it 2.4 inches = 0.2 ft.
Chf is flap hinge moment coefficient, which is nondimensionalized hinge moment, expanded as shown and probably not as scientific as it looks.
∂Chf/∂δf  is rate of change of hinge moment with flap deflection at constant angle of attack: = -.011/degree*
δf is flap deflection, 30 degrees for this example
∂Chα/∂α is rate of change of hinge moment with angle of attack at constant flap deflection: = -.005/degree*
α is angle of attack, degrees.  Assume 6 degrees for this example.
Hf=((.002246 slugs/ cubic ft.) *83.8 ft./sec.2)/2*1  sq.ft.*0.2 ft.*(-.011/degree*30°-.005/degree*6°)
=-1.14 ft.lb.
I figure an Impact with conventional control mechanism should have -8.8 degrees of flap per inch of differential line pull (ciphering upon request).  That gives 2.1 lb. of differential line tension, I think, at -30 degrees of flap.  If total bellcrank travel is restricted to 60 degrees, leverage is -14.7 degrees of flap per inch of differential line pull, which gives 3.5. lb. of differential line tension at -30 degrees of flap.  Hope I came close on all those signs.

*Gary Letsinger said, “Hinge moment calculations are easy to do, but hard to believe.”  These ∂Chf/∂δf  and ∂Chα/∂α data come from an NACA report shown in two classic textbooks as an example not to be used in general.  Bill Netzeband in his Control Line Aerodynamics Made Painless series, https://miniatureaeronautics.com/Dave_Day/netze/control.htm , used it in general, though, and if it’s good enough for Wild Bill, it’s good enough for me. 



Now the differential line tension from a “self-centering” bellcrank.  Force on bellcrank from “centrifugal” force only = mV2/r,
where
m= airplane mass.  We usually refer to airplane weight: 73 oz. for my massive Impact.  To convert to mass in slugs, one converts ounces to pounds and divides by the acceleration of gravity, about 32.2 ft./sec.2.  1 lb. = 1 slug ft./sec.2
m=73 oz.*(slug ft./sec.2)/(16 oz.*32.2ft./sec.2)=.142 slug
V is inertial speed = true airspeed if there’s no wind = 83.8 ft./sec. from above calculation
r is circle radius = 70 ft.
mV2/r=(.142 slug*83.8 ft./sec.2)/70 ft.=14.2 lb.

Plugging this number into the second graph of the first post, one gets pounds of differential line tension caused by bellcrank hole arrangement.  The red and orange lines on the plot show the contribution of hinge moment to differential line tension.  I’ll assume that flap hinge moment varies linearly with flap deflection from -30 degrees to +30 degrees.


Differential line tension is the sum of the contributions of bellcrank bend angle and hinge moment.  I recommend lots of bellcrank movement (yellow line) so’s to shy away from the Netzeband wall.  That gets one into the perverted part of the bent-bellcrank contribution.  If bellcrank movement is limited to 60 degrees total (red line), bent-bellcrank contribution is better behaved. 
The Jive Combat Team
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Online 944_Jim

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Re: Self-Centering Ciphering
« Reply #5 on: January 14, 2023, 04:55:33 PM »
I try to learn new things regularly...but this one is making me feel stupid! I can't find some of those symbols on my phone's "keyboard." HB~>
Mr.Howard's post looks an awful lot like signage outside Bagdad to me...and I mean the one in Iraq. This is more proof that these models are more than "just silly toys."

Online Lauri Malila

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Re: Self-Centering Ciphering
« Reply #6 on: January 15, 2023, 04:08:59 AM »
Hi Howard,

Maybe I don't know what I'm writing about, so please correct if necessary.
Could you add also the handle geometry to your cipherings? I think the bellcrank alone tells only half of the story. As an example;

-With the Yatsenko stuff there is a big bell crank with quite a lot of this "self-centering" overhang, and a handle with very little overhang (line pivot point is about in line with knuckles).

-If you look at the Chinese school models, they use a tiny (even down to 70mm line spacing, like in 2-time world champion Zhang's Skywriter) bellcrank with no overhang, and quite a bit of overhang in the handle. Chinese models are very light.

To me the minimum overhang in the handle makes more sense, as the "self centering" feeling changes less between different wind conditions.

Cheers, Lauri

Offline Howard Rush

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Re: Self-Centering Ciphering
« Reply #7 on: January 16, 2023, 05:09:54 PM »
Could you add also the handle geometry to your cipherings?

I'll try.  The handle is on the other side of the spring, so it's different.  It's difficult to compensate for airplane problems with the handle. I remember thinking about handle overhang awhile back, but I forgot what I thought.  I think one might approach the handle-lines-airplane combination like Dolby noise reduction: what can you do with the handle and airplane to diminish the line spring effect?
The Jive Combat Team
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Offline Howard Rush

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Re: Self-Centering Ciphering
« Reply #8 on: January 16, 2023, 05:12:26 PM »
This is more proof that these models are more than "just silly toys."

More likely this is more proof that I can't write very clearly.
The Jive Combat Team
Making combat and stunt great again

Offline Air Ministry .

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Re: Self-Centering Ciphering
« Reply #9 on: January 18, 2023, 09:42:08 PM »
The handle overhang seems to be a sort of ' dampner ' ( Shock absorber in U.S.Speak . ) zero  overhang , no resistance . 12 in. overhang , to much resistance . so damps hand  Tremmers ! .


« Last Edit: January 18, 2023, 10:06:30 PM by Air Ministry . »


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