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Member postings for John Cole

Here is a list of all the postings John Cole has made in our forums. Click on a thread name to jump to the thread.

Thread: Irvine Tutor 40 Wing Bolts
25/08/2008 14:25:00
I would say the same applies to the U/C for a first trainer, though I appreciate that for e.g. a torsion wire arrangement this would not work.
Thread: Closed-loop linkages
24/08/2008 20:53:00

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).

24/08/2008 16:49:00
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.
24/08/2008 16:48:00
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.
http://www.modelflying.co.uk/sites/3/images/member_albums/26009/Closed_Loop_Diagram.JPG

23/08/2008 14:54:00

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.

23/08/2008 12:10:00

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.

23/08/2008 12:03:00

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!

22/08/2008 12:35:00

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.

21/08/2008 14:39:00

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.

21/08/2008 14:30:00

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.

21/08/2008 10:08:00

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),

20/08/2008 16:09:00

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.

Thread: Prop failures
08/08/2008 17:25:00

Brian: No you're not.  What you say is exactly what I thought before I did the sums.  The prop is a bit of plastic; it has no brains - and in particular no memory! 

In my example the prop fails when in the vertical position.  Your argument (and my predjudice) is that the separated blade will go off vertically upwards (if it's pointing up).

My considered argument is different: that the instant after failure the blade "does not know" it was ever connected to the hub - so the blade (and all parts of it) will try to continue moving at the same speed and direction as before failure, subject only to the after-failure aerodynamic forces.

At the instant of failure, if the prop is in the vertical position then all parts of the (failed) blade will be travelling HORIZONTALLY.  The complication is that all the bits of the blade are travelling at different speeds.  In my example, the tip is travelling at 160 mps (about 360 Mph) and the base much slower.  The CofG is moving at 65 mps (145 Mph) and that's the speed at which the blade moves off (HORIZONTALLY).

However, the tip will try to move faster than this (and the base slower) and as a result the blade will TRY to continue to rotate STILL at 12k - but now round the CofG rather than the hub.

See my original note which shows why the blade then tumbles.

Thread: Futaba 6EX exchange programme
08/08/2008 09:17:00
Rob Crewe: my trims had not changed at all!  I've flown with exactly the same settings and no need to re-trim.
Thread: The August Grand Prize Draw
05/08/2008 17:13:00
Maybe I'll be first-time lucky!
Thread: Prop failures
03/08/2008 18:08:00
Yes there's a letter in All Write, but I was not convinced by what it said.  So I did some sums and came to different conclusions.
03/08/2008 13:46:00

......"ordinary" revs.

So what's the damage that causes failure at "ordinary" revs?  I don't know but I would guess that both the backwards bending force of a dead-stick prop touching the ground on landing, and the rotational bending force of a fast-revving prop being stopped dead by ground contact, might fracture the glass filaments inside the prop (on the side put in tension by these actions).  T

his would not be visible but would weaken it in proportion to the amount of glass fractured.  Repeated impacts would cause additional damage and would leave it prone to failure at "ordinary" revs.  Note that the damage caused by the backwards bending force will weaken the FRONT of the blade-root - the part which the aerodynamic force will tend to put into compression, offsetting the centripetal tension.  This reduces the importance of the aerodynamic effect.So what happens when the prop fails? 

The forces that WERE acting on the blade (and caused it to fail) are now not relevant.  What will happen is that all parts of the blade will try to continue to move at unchanged speed along the straight line they were following at the instant of failure; these paths will be subsequently modified by the aerodynamic forces acting on them AFTER failure.  If the blade is upright at the instant of failure then it will fly off (initially) horizontally (NOT vertically outwards!) in the plane of the blade disc.  

The blade will fly off at the rotational speed of the C of G of the blade. For my Master prop I estimate the C of G to be at 5 cm from the hub, so rotating as 12k RPM the fractured blade will fly off at 65 metres per second horizontally, or about 145 mph.  But what's the effect of the aerodynamic forces?The blade as a whole will fly off at the speed of the C of G of the blade.  But every part of the blade will try to continue at its own speed.  So the tip will seek to continue horizontally at about 160 mps, while the fracture point will only move horizontally at about 10 mps. 

What this means may seem surprising, but the single blade will therefore initially continue to rotate at 12k rpm!  But now it's rotating about the blade's C of G rather than the prop hub.  This means that while the tip of the blade continues to generate thrust forward (that is, in the direction of the plane's axis) the part of the blade on the hub side of the blade's C of G is now generating BACKWARDS thrust!  Hold a prop at the blade's mid-point (with your fingertip on the blade front-face and your thumb on the blade rear-face) then rotate it and you'll see this is true).  So the blade instantly tumbles (tip forward and root backwards) and there is no significant net aerodynamic force to throw the blade off its course (in this example, horizontally in the prop-disc plane).Well, that's what I think!

Regards,  John   

03/08/2008 13:46:00

I've been thinking about prop failures.......

In looking at what happens when a prop fractures at speed you need to look at what the forces are (both at the instant of failure and immediately afterwards) and get a feel for the relative magnitudes of these.  It's raining today so I thought I would do a few sums.  I've based my sums on an old Master 10*6 prop.  I chose it as it is of relatively constant chord, which makes the sums easier.  I understand these props are made from nylon with chopped glass fibres added for strength.My prop is 25 cm in diameter and weighs 23 gms.  I guess each blade weighs about 10 gms and the hub 3 gms. 

My 40 engine will turn it at 12k rpm, or 200 revs per second.  It will just about hold a 2 Kgm model's weight so each blade is delivering about 1000 gms aerodynamic force forwards. The centripetal force is calculated from the acceleration of each element of the blade, converted to a force and summed. This is made arithmetically difficult by the way the blade section thins from root to tip, but again the Master prop is not too hard to approximate. 

The acceleration of each element is calculated as the square of the speed divided by the radius of the curved path, so for the tip (which is travelling at about 160 metres per second (mps) and pulled round a radius of 0.125 m) the acceleration is a little under 200,000 mpsps.  Divide that by 9.81 mpsps (the acceleration due to gravity) and you see that the acceleration at the tip is about 20,000 g.  So one gm at the tip would produce an outward pull of 20 kg.I calculate that the centripetal force at the blade hub totalling along the blade is about 84 kg (less than you might guess from the tip figure of 20 kg per gm as the inward pull decreases with the square of the radius as you move in towards the hub, and the mass is predominantly near the hub). 

This is a LOT more than the aerodynamic force and is the preponderant force in causing fracture, even allowing for the section-moment effect of leverage at the hub.  As well as being much greater than the aerodynamic force it is also much greater than the engine torque even allowing for the positive-negative variation in this during the engine's rotational cycle (which makes the peak instantaneous torque just after firing TDC much greater than the torque averaged round the engine's rotational cycle).But how does this force compare with the nominal strength of the prop? 

 I've seen the tensile strength of roving-reinforced GRP quoted as 1000 MPa, approximately 10,000 Kg / sq cm.  The material of a prop will be considerably weaker (chopped glass not roving, lower glass content, poor fibre alignment near the hub) but even if it's 10 times weaker this is much greater than the centripetally-caused stress: my Master prop has a cross-section near the hub of a bit under 1 sq cm and so at low estimate tensile strength of 1,000 Kg / sq cm would fail at over 36k revs, not 12k.  And we know, undamaged props don't fail at ...........(ctd next post)

Thread: Futaba 6EX exchange programme
24/07/2008 13:55:00
Nick Rigg: I got Ripmax to exchange mine before the recall was announced.
24/07/2008 09:47:00

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