Yet another bellcrank/flap spreadsheet.

[TomLaw]

This demonstrates the effect of bellcrank and flap horn cant. For this particular model, the differences are the handle sensitivity and the ratio of flap to elevator movement.

[Howard Rush]

What are the rest of the numbers?

[TomLaw]

Hinge to hinge: 24 "
Bellcrank to flap (center to center): 4"
Flap horn height from bellcrank: 1"
Flap horn to elevator height: 75"
Elevator horn height: .5"
Bellcrank width: 4"
Bellcrank arm: 1"
Stabilizer offset: 1"
Line length: 70'

[Howard Rush]

There are some more numbers.  It's a 3D problem.  If you are interested (not many people are), I can send you my program. 

[TomLaw]

I already included the cant of the bellcrank and flap horn. I convinced myself that the lateral offset of the horns doesn't matter. If there are any more parameters I would be interested. I used the distance formula on the 3D points in question.

[Howard Rush]

See http://stunthanger.com/smf/index.php?topic=19644.0 .

[TomLaw]

Very interesting. I see that I left out some parameters by design and failed to consider others. I'll keep working. I assumed the builder would eliminate the necessity for some of the decisions you made.

[Howard Rush]

You don't need to have all those variables, but the combination of assumptions and stuff you can vary has to suffice to specify the three-dimensional location of all the parts.  It was interesting to fiddle with parameters such as spanwise bellcrank location and bellcrank output arm angle.  They make more difference than a lot of things that people do worry about. 

I think that probably the most important thing one can calculate about control geometry is the slope of control deflection per unit of leadout travel.  If you are better at high school math than I, you can tell me how to calculate flap control horn deflection knowing the flap control horn geometry, the 3D position of the bellcrank output, and the pushrod length:

Pick a point on the bellcrank end of the rod: Call it (X1, Y1, Z1).  I chose X = 0, Z = 0 as the flap control horn joint, and the Y axis is the flap hinge line, so the flap control horn joint with the pushrod, (X2, Y2, Z2), is on a circle: r^2 = X2^2 +  Z2^2 .  Y2 is constant and known.  For pushrod length L, L = sqrt((X2 – X1)^2 + (Y2 – Y1)^2 +(Z2 - Z1)^2).  So there are two equations in two unknowns, X2 and Z2. Gimme a closed-form solution.  There are maybe two or four solutions.  I want the one (or two) with X2>0 and Z2<0 (forward and above the flap hinge.  I said Z2>0 in the original post by mistake).

I did this by a crude iterative method.  It gives OK plots of control deflection vs. leadout position, but a plot of the abovementioned slope is ugly.

[TomLaw]

This is what I did:

Given a bellcrank pivot at (0, 0, 0), the bellcrank arm at (0, d, 0), and the flap horn pivot at (a0, d, 0), with d and r0 being the bellcrank arm and flap horn lengths, respectively. Let the control handle and bellcrank arm be displaced by an angle of δ and the flap horn by α. Let θ = α - α0 be measured relative to the vertical. Both the flap horn and the plane of the bellcrank are initially canted forward α0 and α1 from the vertical, respectively. In general the bellcrank pushrod hole is located at (dSinδCosα1, dCosδ, dSinδSinα1) and the flap pushrod hole at (a0 + r0Sinθ, d, r0Cosθ). There is no influence from the lateral offsets of bellcrank and horn, even if they differ. This is easily visualized by imagining the pushrod to parallel the model’s centerline with an L shaped extension at the aft end to the horn. Let the pushrod length be

L = {[a0 + r0Sin(- α0)]^2 + [r0Cosα0]^2}^1/2 = [a0^2 - 2a0r0Sinα0 + r0^2]^1/2, evaluated at 
α = δ = 0 (when θ = - α0), then

L^2 = a0^2 - 2a0r0Sinα0 + r0^2

After displacement

L^2 = [(a0 + r0Sinθ) - (dSinδCosα1)]^2 + (d - dCosδ)^2 + (r0Cosθ - dSinδSinα1)^2 . Expand , set equal and solve for δ. You'll have to read between the lines on subscripts.

[Howard Rush]

How do you type a delta?  I tried those Alt-abc things and couldn't find it.  

You sure the lateral offset doesn't matter?  That L shape would either have a varying-length base if the pushrod stays in the X-Z plane or a varying angle if it doesn't.  You'd just add the offset to d in the Y distance in the last equation, I think.  I'd look at what I did, but I gotta get my stunt plane flying.    

I think that's what I got, but I couldn't figure out how to solve for delta (or theta if you start with delta).  I just iterated until I came within an RCH.  That's crude.  It worked, but it's not amenable to differentiation.

[TomLaw]

If I replace d by d1 in (dSinδCosα1, dCosδ, dSinδSinα1) that should take care of it.

I typed in Word and pasted. Insert symbol in Word. Click Here

Square the terms , use trig identities  and factor out (Cosδ)^2 and Cosδ. Three pages later this gives a quadratic equation of the form

ACos^2 +BCos + C = 0

Solve the equation for Cosδ. Find δ by taking the ArcCosine.

[Howard Rush]

That looks familiar.  I think after two pages I gave up, figuring I was on the wrong track, which I may well have been.  I guess I'd better stick with iteration.

If I replace d by d1 in (dSinδCosα1, dCosδ, dSinδSinα1) that should take care of it

Wouldn't that just make the bellcrank output arm length d1?  You could add the lateral offset to the dCosδ or to the control horn Y value, depending on what you want to call Y=0.

ξ  Aha, there's the Worm.  I am now a man to be reckoned with.  Thanks.

[TomLaw]

I found it easier to adjust the flap coordinates. You were correct.

[Dennis Toth]

Tom,
This is a very interesting discussion but a little hard to work through without the spreadsheet. Can you post the working Excel sheet?

Best,        DennisT