Closed-loop linkages

[John Cole]

When I started modelling I learned that closed-loop linkages must have equal-length arms (at the servo and control-surface ends), and that the (short) servo arm should be linked to the (longer) forward closed-loop arm by a push-rod.  This then gives a smaller angular deflection at the control surface than at the servo.  A  servo will typically move 60 degrees either way.  The closed-loop arms AND the control surface itself will all deflect the same as each other, and as you probably want (say) 15 degrees at the control surface the used length of the servo arm would need to be connected to a point on the forward closed-loop arm about four times as this from the pivot (that is, the closed-loop control arms are four times as long as the servo arm).

Putting in this extra linkage potentially adds slop and it certainly adds complication, but if you simply connect the short servo arm to the long tail-end closed-loop arm the book says the controls will bind when the servo approaches maximum travel.

As I've seen no reference to this 'problem' recently, I built a large mock-up to test it: from softwood scraps, screws, nails and string.  And it's true!  The string tension rises as soon as the (pretend) servo arm rotates.  Turn it far enough and the nails pull out!

So I decided to calculate the size of the effect, and see what influenced it.  To do this I built a spreadsheet to reflect the x and y co-ordinates of the pivot points, 'servo' holes etc.  What I found is that the longer the control runs and the smaller the servo/control arm ratio, the smaller the effect.  But it was still there.

So I looked to see if I could find a way round it, and I did.

If you reposition both the servo arm holes angled FORWARD (you probably need a servo disc for this) OR both the holes in the closed-loop arm angled BACKWARDS then you can get the same tension at 60 degrees servo move as at zero.  In between there's a little bit of slack, more if you correctly reposition the closed-loop holes than the servo-holes .  But for equivalence (at 60 degree servo movement) you need to reposition the closed-loop arm holes by a bigger angle than the servo-arm holes: moving the servo holes has much more effect, roughly in proportion to the two arm lengths.

Better than that, if you move BOTH sets of holes FORWARDS you can get very nearly constant geometry (at least up to 60 degrees servo movement): say move the servo holes forward 5 degrees and the closed-loop holes 6 degrees.  The right amounts depend on the length of the control run.

If anyone would like a copy of the (rather poorly documented) spreadsheet then message me with your Email address.

[Bruce Richards]

John,

 If you want people to be able to PM you, You need to turn on messaging in your profile

[Guest]

the above seems a bit over complicated.

I understand that the recommendation is that  that the geometry should be symetrical at both ends,

ie the radius of the servo arms should be the same as that of the control surface arms so that everything moves the same amount without any additional stresses being set up.

if you want less movement of control surface you move the links IN at both ends and for more, you move links OUT - reducing/increasing  the operating radii at both ends.

In practice I have varied the control surface end only with no observed effect on the servo ( hunting, servo buzzing under strain  etc ) but the length between servo and control surface has been considerable 30" plus, and this bears out my original assumption that as long as each SIDE of the loops are symetrical, then if servo arm moves equivalent of eg 1/2", then the control arms cable point also moves 1/2" whether they are on same radius or not and therefore the throw of the control surface can be varied by adjusting the radius af the control surface arm only. this of course assumes that there is a decent amount of radius in use at both ends. I tend to use the outer holes on the servo arms and adjust at control end to achieve desired throw.

[John Cole]

Bruce: thanks for pointing this out to me; I've enabled messaging.

John: if you move the links IN or OUT at each end (the same at both ends) you won't get binding, but you won't alter the control surface movement either.  If the servo moves 60 degrees and the arms at each end are the same length then the control surface will also move 60 degrees.

To make the control surface move less of an angle than the servo the arms must be of different lengths: servo arm short and control arm long.  This is not a point specific to closed-loop systems.  Exactly the same applies to a single-sided link connection like pushrods or bowden-type cables.

The difference with closed-loop connections is that they are by definition doubly-connected with both sides working in tension.  If the arms at both ends are at right-angles to the centre-line the geometry of the links means that when the servo moves the combined length of both links MUST increase, causing binding UNLESS the servo arm and control arm are of identical lengths,  This causes binding, as this increase in length can only come from increased tension (stretching the links and perhaps compressing the fuselage),

[Lew Weaver]

I always introduce a bit of Ackerman ( rotary to linear differential) which helps to eliminate binding problems with closed loop linkages

An explanation can be found here -

http://members.cox.net/bdfelice/Ackerman/ackerman.htm

[Guest]

John,

we are both saying the same thing re your 1 st 2 para's.

re the 3rd para - I disagree and my experience to date is that there is no binding if the control arms are of different lengths to the servo arms.   the rotary motion of the servo arm gets translated into a linear push/ pull on the cables and this amount of p/p gets translated back to rotary at the control surface end.

the amount the control surface moves will be proportionate to the postion of the cable links on the control arms and as both sides move equally ( unless one is daft enough to have different postions on either side of the control arm ) there will be no additional stress from binding as the movement at both ends and on both sides of the push/pull will all be identical.

My comments are based on the use of cabling to connect servo to control surface.

[John Cole]

Lew: good stuff.  But I'm not sure I entirely agree with your point about slack being unimportant.  The slack starts to develop as soon as the servo rotates.  That means the trim-set point (where the servo is un-centred by the trim control) will have slack, and I don't think that's what you want.  I agree slack is otherwise not such an issue, except in extreme manoevres such as tail-slides where the aerodynamic forces on the control surface may be reversed.

In your example you have looked at backwards-angular-offset at the control surface horn.  What my calculations showed (and what surprised me) is that the same type of change at the servo arm (but reversed: forwards-angular-offset) is more powerful but creates essentially the same amount of slack per degree.  That leads to the surprising conclusion that you can combine servo-arm-forward-angular-offset (which releases binding at extreme servo travel at the expense of a little slack at e.g. half-travel) with some control-arm-forward-angular-offset.  The latter would normally create additional binding forces, but if you have an excess of servo-end offset there is exactly no binding at full travel.  Just as the control-backward-offset produces more slack at mid-point, the control-forward-offset REMOVES more slack at the midpoint.  This combination of forward-angular-offset on both horns can therefore result in essentially slack-free travel over +/- 60 degree servo travel with no binding at either zero or full travel.

And no: I don't like your using the term Ackermann to describe this: for me, Ackermann steering is all about the differential angles of the forward (steering) stub axles on a road vehicle and the way the projections of those axles intersect on the projected line of he rear axles. So when steered to make a turn, all four wheels describe circles around this single intersection point, with no scrub.

[John Cole]

John: your post says

'if you want less movement of control surface you move the links IN at both ends and for more, you move links OUT - reducing/increasing  the operating radii at both ends.'

I disagree!  Moving the links in (or indeed out) at BOTH ends (doing the same at each end) has no effect whatsoever if the 'amount' of movement is the same, such as halving the arm length at each end.

As to your comments that you have experienced no additional binding stress with unequal arms (i.e. short servo arm and long control-surface arm) then I must tell you it IS there even if you've not spotted it.  The shorter the control run and the more unequal the (servo vs. control) arms, the bigger the effect.  As I said in my opening note, in my test-rig I managed to pull the 'control'-wire fixing point (nails) right out just by rotating the (pretend) servo.  Lew's diagrams show the differential movement clearly.

[Former Member]

[This posting has been removed]

[John Cole]

Depends on what the fuselage is made of! 

However the sensible answer to your question is that the coefficient of expansion of steel is about 7*10^-6 / degree F so a delta of 20 deg F from when it was tensioned (65 shed to 85 warm day, or 65 shed to 45 cool mountain top) will result in a length change of about 140*10^-6 (140 parts per million).  This will be offset by the expansion of the fuselage, which mey even be greater:  bottle glass has a coefficient of about 5*10^-6 / degree F; I think E glass would probably be similar, so that would offset 5/7 of the expansion, giving a net contraction or expansion of 40*10^-6.

The strain on the closed loop system with runs 40 times the (single) servo arm length and control arm 4 times the servo arm at 60 degrees servo deflection is about 800*10^-6, or twenty times the net contraction / expansion figure calculated above.

[Guest]

John,

I agree with the moving in and out - I only move the rear end  to adjust throw. 

I still disagree with you in respect of a closed loop set up where the servo arm starts from neutral straight across the servo and at the control end the control arm at neutral is also straight across and in line with the hinge line . This I have always assumed to be the desirable set up ( closed loop or push/pull )  In such a case then movement of the arms on both sides and at both ends will always be equal with no additional stress being put on the servo, other than the load from the control surface.

binding will occur if these conditions are not met, as Lew's differntial setting shows.

John

[John Cole]

John: you say the movement at both ends will be equal, and if the upper link were a rigid pushrod and the other a piece of elastic then  clearly the upper link-holes MUST move the same amount, and as the arms are rigid the lower link holes must also move the same amount (in reversed directions).  However, the movements at the servo end are in DIFFERENT DIRECTIONS from the movements at the control surface end, and that's what causes the tendency to bind.  Take the following worked example:

Take the (half) length of the servo arm as 1 (I use this as the basis for measurement); the (half) length of the control arm is 4.  The distance between the pivot points (servo arm pivot and control arm hinge) is 20.  Both servo and control arms have the two 'link holes' and the pivot hole in a straight line (no angular offset).

Start with them both at right-angles to the line joining the two pivot points (i.e. at neutral).  Let's call this the vertical position, assuming the pivot-point joining line is horizontal.   We now calculate the length of the control link.  We then rotate the servo 60 degrees anti-clockwise and calculate the position of the servo upper control-link hole.  From the assumed-unchanged length of the upper control link we than calculate the position of the upper control surface link hole.  This then lets us calculate the angle of rotation of the control arm, the positions of the two lower link holes, and the distance between them.  It turns out that in this worked example this lower length is now about 0.123 units longer than the upper one (servo arm half lengths), meaning that in practice in a real closed-loop system both links stretch by about half this.  So your pre-conception is wrong.  To prevent binding you must offset the control-link holes in either the servo arm or control arm, and this can be done so there is essentially no change in control lengths during 60 degrees of servo travel.

What follows in the next posting is a simple but extremely boring piece of x,y trigonometry that demonstrates this.  Only read it if you must!

[John Cole]

THE CALCULATION 

Let's follow geometric convention and call the horizontal direction the x axis, and the vertical one the y axis.  In x,y co-ordinates the position of the (upper) servo link hole is 0,1 and the (upper) control link hole is 20, 4.  The x separation of the holes is 20 and the y separation is 4-1 = 3.  Following the rules for calculating the diagonal side of a right-angled triangle, the distance between the link holes is the square root of (delta x) squared plus (delta y) squared which is 20 squared plus 3 squared =root 409.  The numerical value of this is about 20.22375. This is the length of the upper link, By symmetry the length of the lower link is exactly the same.

Now let's pretend the upper link is rigid and the lower link is elastic.  We now deflect the servo anticlockwise by 60 degrees, and the upper link moves forwards.  As we've said the upper link is rigid, so when the servo moves the control arm will move to keep that length exactly the same.

In fact the servo arm will move so its link hole goes left and slightly down.  By trigonometry the upper link hole will be at x= -0.866 (minus means left) y=0.5.  The lower link hole will be at the opposite position: x= 0.866, y=-0.5 (minus means down).

The upper control link hole moves the same distance and will also move left and slightly down.  However, it will move different x and y amounts from the servo hole as its radius of movement is greater: it will move in a different direction.  It moves anti-clockwise approximately 1/4 of the servo's 60 degrees but not exactly: it actually moves 13.423 degrees, and the x, y of the control surface hole become x= 20 minus 0.92855 = 19.07145, y= 3.891.  This means the x-separation of the link holes is 19.93745 and the y separation is 3.391.  The distance between the link holes is (again: right-angled triangle) root of (delta x) squared plus (delta y) squared = root 409 as above, confirming I've correctly calculated the position of the upper link hole.  If you check my arithmetic based on the numbers I've typed in , you'll find it comes to root 409.0007935.  That's because in this note I'm only using up to five decimal places.  My real calculations are more exact.

So how did I calculate this angle of control-arm movement?  By using a calculation that makes a guess (60 / 4 = 15 degrees), calculates the length error, makes an adjustment to the guessed angle  and recalculates (25 times): brute force!

So if the control arm moves by 13.423 degrees, the lower control arm link hole will be "opposite" the upper one, at x = 20.92855, y = -3.891 (minus means down).  Take the servo / control difference in x and y: delta x = 20.06255 and delta y = -3.3907.  Use the right-angled triangle rule to calculate the new length of the lower (elastic) link and it comes out as the square root of 414.003, or 20.34706.  This compares with the initial length of 20.22375 and is 0.12331 longer.

[Bob Cotsford]

I'm not going to attempt to follow your calculations, as a)  it makes my head hurt and b) you appear to assume the axis of servo and surface are not parallel or at right angles as you speak of connection points movingin two relative planes.  Are you  assuming a (considerable) degree of sweeback on the control axis?  I can only speak from practical exprience using closed loop systems on rudders for more than twenty years,  but I've never had a setup where slack or binding was a problem.  As long as connection points line up with hinges/bearings and you have equal radius either side, you won't have a problem, even with moderate amounts of sweep, eg on a soarer's rudder.  You could have a potential problem with extreme angular movements where the driving bar has a greater radius than the control hrns, but if the bar is shorter, then you can't exceed the movement on the control horns.

I usually use an intermediate tiller bar close to the servo with a short pushrod connection to the servo, just to make it easy to remove the servo without c/l cables going slack and snagging the structure or disappearing down the fuselage. 

[John Cole]

No: I am assuming that the starting point is that the servo arm and control arm (pointing in the direction of the y axis) are both initially at right-angles to the line joining the servo centre and control surface hinge (the x axis), and that servo arm and  the control arm rotate in the same plane.  I do not expect you to work through the calculations which is why I split the posting in two; I included them for completeness.

The example I chose has a relatively short control run and a high servo / control arm length ratio.  This increases the binding forces (but they are still there with low ratios, unless the ratio is exactly 1).  I built my mock-up to similar ratios, and that's why the nails were pulled straight out of the wood!  If you still don't believe me, build a mock-up to these dimenions yourself, and see how the link-wire (or string in my case) tension increases as you rotate the 'servo arm'. 

The Tiller Arm you refer to: that may be the intermediate arm I referred to in the first sentence of my first posting: if the length of the tiller arm is the same as the control arm there is no binding, just as is the case if the servo arm and control arm are the same length (with no intermediate 'tiller arm').  The two arms will then rotate through the same number of degrees when the servo moves.

[Guest]

John,

I think we may be on 2 different playing fields.

In your example 1/4/20 units a 60deg movement resulted in a 0.6% change in length or 0.3% if load spread between 2 cables as it will if both lines remain under tension.

I dont doubt your trig ( do they still teach that ?) but in my experience the amount is insignificant within normal ( normal to me anyway ) modelling precision.

Just to check, I measured the equivalent  on my 105" precedent aeronca champion ( must be over 20 years old now ) and it measured out at approx 0.5 and 0.8" for 1/2 servo/control arm length giving a ratio of 1.6 cf 4 in your example and a 40" length beteen arms.  the 40" alone reduces the proportionate change in length to 0.15% on each loop and I think the reduced ratio will also reduce the amount proportionately.

now in practise. I dont put my control loop under that tight a tension since this would reflect back onto the servo bearing ( which in early days were plain plastic/nylonn ?? ) and I still dont even altho I use ball race servos for closed loop.

In addition, the hinges on control surfaces also give a little.

both of these factors will mitigate in favour of alleviating any binding due to geometry of closed loops, particularly at the amounts we appear to be discussing

Re not noticing, I always listen to my servos when i switch on and work the controls - any significant strain on the servo bearing should be audible - as well as checking for hunting etc

Bob, - when I built the Champion, I bought one of the closed loop controls complete with spring tensioners - which from memory was simply to take the tension on tight loops off the servo. I never fitted it as I figured the ball race servos would be ok and also at end of flying I always release one link in each loop to take the tension off to avoid permanent strain on ball races and/ or stretching the fishing trace line loops ( the Champion has 3 - rudder and 2  1/2 elevators all of which go back to servos up front.

John

[John Cole]

Here's the diagram for the posting below. You can see a better copy in my gallery (when it's approved) by sorting on Title (= album name) and then look for Closed Loop Controls - it's item 73 at present, on page 10. Currently the album is there but no diagram.

[John Cole]

For those of you who do not want to work through the maths I suggested you build a softwood mock-up.  Here is an alternative: draw it out!  All you need are a fine pencil, a ruler and either a square or (like me) squared paper.  A protractor and pair of compasses would help, but I did not use them.  The (in) accuracy of measuring a drawing means you need to choose a configuration that exaggerates the effect, or you won’t see it.  I chose: servo arm (half-length) = 1, control arm = 4, centres separation = 10.  This allows you to measure the effect and also is the right shape to draw comfortably on a page of A4. See the drawing above which I’ve scanned in.  As before, I treat the upper (in the diagram) control-link as rigid and the lower one as elastic; this one stretches when you rotate the servo.

Theory says that the neutral-position control-lengths will be about 10.44 units, and when the servo deflects 60 degrees anti-clockwise the lower (elastic) one will stretch to about 10.67 units – a stretch of 0.23 units.

In the diagram, I have labelled key points and added key dimensions.

The servo arm has pivot point at A; the control arm at B.
The neutral servo position is shown with the control link holes at C1 and D1; the control arms ditto at E1 and F1.
When the servo is rotated anti-clockwise by 60 degrees the servo link holes move to C2 and D2.  The control arm link holes move to E2 and F2.  I treat the link C1 to E1 (or C2 to E2) as rigid and E1/E2 to F1/F2 as elastic.  So the distance C1 to E1 is the same as C2 to E2.

To draw this out, first draw the base line A to B, then the right-angle neutral positions: servo arm C1-A-D1 and the control arm E1-B-F1, all as straight lines and to the lengths shown above (scaled to fit nicely on the paper: I used 4 paper-squares per unit).  Measure the control-run length.  I made it 10.45 units, very close to the theoretical figure.

Next, draw the line of the servo arm at 60 degrees.  Use a protractor if you have one, but I ‘constructed’ the angle as follows: Extend C1 - D1 down the paper.  Mark G and H at 2.5 and 5.0 units down from A.  Draw a horizontal line to the right from G. Put the zero of your ruler on A and swing it so it intersects the new line right of G until the ruler/new-line intersection is at 5 units from A (use a pair of compasses if you have one!).  Call this point I.  Check by repeating from point H.  Now you have an equilateral triangle AHI so the line IA gives the 60 degree angle you want.  Draw it in and mark points 1 unit out from A (C2 and D2).  These are the new positions of the servo arm holes.

Now draw a faint line through B, about 15 degrees anticlockwise from E1-B-F1, and mark a point which is 4 units out from B.  What you are going to do next is place the ruler with its zero on C2 and find E2: this will be 4 units out from B and the control-run length (10.45 units according to my measurement above) along from C2 – the position of the upper control hole in the deflected position.  As you can see, my faint-line guess was not terribly accurate, but good enough.  Again, a pair of compasses makes it easier – and in this case more accurate.

Now draw E2-B-F2 with F2 4 units away from B (the position of the lower control hole).  Measure D2 to F2.  I made it 10.7 units, not too bad compared with 10.67 theoretical.  The extension demonstrated in my drawing is 10.70 - 10.45 = 0.25 units.

[John Cole]

Good news: you don't have to draw it out!  Just print out a copy from the Gallery, and you can just measure it yourself.  However, to get a good-sized print do it this way:

Open the Closed Loop Controls album and then the one item in it: Closed Loop Diagram.jpg   The album is now no. 79 if you sort on (album) Title, and is on page 10.

Then using the browser commands (in my case Microsoft Internet Explorer 7, and those are the options I'll quote) click on the picture itself, then right-click on it and choose the option Save Picture As...    and save it under the default name Closed-Loop-Diagram.jpg, in My Documents.  Then go to My Documents, select the picture and right-click Print it (using the Print Wizard), choosing Full Page Fax Print.

It will come out a different size from my A4 original (your size will depend on the printer you are using) but despite the less-than-super resolution you can (in my case, printing on a LaserJet) see that the 'fixed' control length C1/E1 or C2/E2 is 20.6 cm and the 'stretched' control length D2/F2 is 21.0 cm, 4 mm longer.

That's why unequal-arm no-angular-offset closed-loop control linkages bind in tension which is created when the servo moves (and why the nails in my mock-up pulled out).

[Mike Rolls]

I'm a bit late in joining in this discussion, so please forgive me if what I am about to write is already covered and dismissed in some of the earlier postings.

I. The diagram - it shows 60 degree movement in one direction. Why? My servos only move 60 degrees in total - 30 each way.

2. Why would anyone use a horn (I assume this is the 'control arm' referred to?) which is not the same distance from pivot point (hinge) to line connection as the servo arm? Everything I can remember reading about closed loop controls stresses that the actuating lines, the servo arm and the line through the control horn line attachment points should form a rectangle, with the cotrol horn line attachment points and the hinge line forming a straight line. If  a horn dimension equal to the servo arm length doesn't give you the movement you want, you need to use an intermediary arm. Certainy this is how I have always set them up, with no problems of lines slackening, servos being stressed, etc..

3. Side loading on the servo bearings has been mentioned. This can be removed by using an intermediary arm (which sa has also been mentioned, allows the servo to be removed more easily should the need arise). There is a downside - the additional links can introduce slop. I corresponded on this aspect some years ago with Dean Pappas, the then aerobatics columnist in Flying Models, when he was advocating using an extended final drive shaft in the servo ending in a ball, with a ball link ended rod secured to the fus structure to protect the servo from such loads. I asked why go to such lengths - the reply was to avoid slop inherent in the idler bar. This was on F3A competition models - I don't know if the Americans (or anybody else) still do it, or whether more recent servos (the correspondence was in 2000) make it unnecessary.

Mike

[John Cole]

1. My servos move more than yours.

3. (yes 3 not 2) Yes an intermediate arm allows you to take the tension off, with a push-rod between the servo arm and the intermediate arm.  It also lets you have the arm on the control surface LONGER than the servo arm, by matching the length of the control surface arm to that of the intermediate arm - and maintaining the rectangular geometry you mention in 2 above ('required' to avoid binding forces).

2 (yes 2 not 3) The reason why I would LIKE to have a control surface arm (or pair of horns) LONGER than the servo arm is that otherwise the angular deflection of the control surface will exactly match that of the servo.  That may be OK for a rudder but I prefer a bit less for the elevator. 

The reason why I started all this was to establish how significant the binding issue was, and to see if there was a geometry which gave less angular deflection at the control surface than at the servo, did not introduce binding forces and was slack-free at all angles.  There is!

[Mike Rolls]

John

What servos are you using? I checked a Futaba 148 and a Hitec mini - both have a max throw (100% set on Tx)  of 30 degrees or so either side of neutral. However, to satisfy my own mind, I'll check a few more.

Point 3 (yes 3 not 2?) - exactly.

Point 2 (yes 2 not 3?) why? You are just making life unnecessariy hard for yourself, and if I read some of the comments aright - also for your servos.

Don't understand your '2 not 3' etc - the numbering is mine, not intended to refer to anything that went before.

Mike

[Mike Rolls]

John

Just checked (I was beginning to doubt my sanity) - two different Txs (Futaba FF7 abd JR 9XII) and half a dozen servos at randon - Futaba, Hitec, GWS - all gave a total movement of under 90 degrees from full deflection one way to the other. Without going to the trouble of making extended arms to measure the movement accurately, there was, however, no doubt that none of them came near to 60 degrees each side of neutral.

Mike

[Terry Whiting]

I'm one that has used closed-loop or pull-pull system for many many years, and have never knowingly been confronted with any of the problems related in this thread.   I have found to obtain a slop free control surface the tension required is reasonablly soft. 

 Last evening I tried a little experiment of my own. 

  I used a Watts meter between the RX battery  and receiver.              Operating the rudder left and right  I had a reading of .04amp.  The elevator DOWN was .03amp  and .06amp UP. I disconected the down wire and it still read .06amp. I assume the extra .03amp is due to the weight of the elevator.

These readings tell me I have no worries with my set up.  

[John Cole]

Servo movement: put it on a switched channel and you may see more, as the Tx trim and EPA use some of the potential movement.  I wanted to allow for as far as the servo could possibly travel.