Model aircraft on Antiques Roadshow.

[Peter Miller]

Did anyone see the 1/6th scale control line Short Stirling plus otther models on last night's Antiques Roadshow.

The chap had a truck full of some nice looking models including a Canberra.

The Stirling was massive and was flown on 120 foot control lines. I wonder where he flew it.

Not really sure about the £10,000 plus valuation though.

[Erfolg]

Saw it Peter, just before switching over to the football.

Could anybody safely fly that thing on wires? I just wonder about line strength, tension. Do you pre book AE to have your dislocated shoulder put back in?

[John Muir]

He described it as a 'tethered' model. Sort of giant RTP?

[Martin Harris - Moderator]

As a control line expert (I flew one for the first time in over 40 years last week!) I wonder if there's actually much tension if the turn radius is large enough?

I saw the programme but I don't remember them saying how many functional engines it had so it may well have weighed in at less than 40 pounds. I suppose I should do the sums properly but (if my mental calculations are correct) we're looking at something like about 0.5g at 30 mph so maybe no more than 20 or so pounds of pull?

[Peter Miller]

I strongly suspect that a model that size would pull quite a bit more. I know that I would want a very firm pull as any tendancy for the lines to go slack on the up wind side of the circle would be very dangerous.

I know the bloke said that he flew on 120 foot lines so there is a lot of line drag there. especially as they will be fairly heavy lines.

The model will be 200" span so it is going to be heavy.

Oh, and I flew my first control liner in 1954

[Erfolg]

Calculation you say Martin

F = mv^2/r

If we assume mass to be 40lb or approx 20kg.

The speed at 30 mph approx 15 mps

Take the radius as 120ft approx 40m

then F = 20 * 15^2/40

F = 112.5N

or in common language 11 kg

given there is a v^2 , increasing the speed has a proportionally bigger impact say 40mph or 18 mps. This would not be unreasonable in one rev, as the model is not in a static body of air (as many like to argue with RC)

then we would get about 16 kg.

So if there is a difference of going down wind to upwind of 10mph we would feel a difference in pull of 5 kg.

To think that when I sat in that class room in my yuof, that there was a good reason to be interested.

Forces seem very low to me now, having corrected  my original mistake.

 

Edited By Erfolg on 24/06/2013 21:08:35

Edited By Erfolg on 24/06/2013 21:10:12

Edited By Erfolg on 24/06/2013 21:24:20

[Mike Etheridge 1]

Here it is!

Full description is on the internet.

MJE

[SDF]

Erflog,

(Looks like you spotted the error whilst I was typing)

I think there is and error in your calculation. You appear to have multiplied the mass by the speed then squared the result making your estimate of the force a factor of ten too big.

40 lb = 18.1kg

30mph = 13.4m/s

120ft = 36.6m

18.1*(13.4^2) / 36.6 = 88.8N

or just over 9kg

or at 40mph just over 16kg

But I think 40 mph might be too low a speed for such a machine to fly so it would probably still need to be tethered to the preverbial immovable object.

 

Edited By SDF on 24/06/2013 21:53:12

[Tim Hooper]

I can't comment on the maths involved, but I do agree with Peter that £10000 seems way over the top.

tim

[John Olsen 1]

The centrifugal force * for a control line model is actually quite low, which is why it was desirable to do things like putting offset on the rudder and sometimes sidethrust on the engine(s) too. The line attachment point makes a difference too, further aft will give more line tension, further forward will give less. A multi engine model like that one would probably be set up so that the outer engines run out of fuel first. Sometimes they were arranged so that they all cut together.

* Yes, I know there is no such thing, unless we adopt a rotating frame of reference. Since the pilot actually is rotating, it does make a certain amount of sense in this case.

John

[Martin Harris - Moderator]

I'd think that the rudder offset/out-thrust etc. basically slightly overbalances the line drag but surely if the model is turning on a radius then the lateral g force must be in proportion with the speed?

I see from the article on the 'net that it weighs 95 pounds and is due to be attached to the pilot and a friend but hasn't flown yet.

Edited By Martin Harris on 24/06/2013 22:32:04

[Erfolg]

SDF

I got a decimal point out, dumpy finger syndrome, I think.

As a calculation it does ignore line drag, which would have some impact on the line tension. Although the calc in principal is dead easy, getting my mind around how it actually impacts on line tension is another matter.

At the simplest level, it is an extra force that the engines need to overcome. At a practical level it cause the lines to adopt a parabolic type shape in plan. Probably needing some turning force via the rudder to turn the model out of the circle, which will increase the line tension, to eliminate or reduce the parabola.

As with most calcs, what I have done is a crude model, in that so much is missed out.

I had thought about the model doing an overhead, with the calc done, it would indicate that the lines would go slack, as the mass force is lower than the line tension. I do recognise that such a model is unlikely to attempt an overhead manoeuvre. Yet it does indicate that the model needs to travel faster than indicated, or a turning out force introduced, if the model is ever to go above the horizontal.

Edited By Erfolg on 24/06/2013 22:36:06

[PatMc]

I have an old Aeromodeller (late 1940s or early '50s) that had a feature on c/l models being flown for full size research in the US.

The line were fixed to a rotating centre pylon that had the "handle" fixed to it. The pilot was seated outside the circle using a joystick with cables running to the pylon operating the handle. All the models pictured were pretty big & IIRC one of the models was a large scale B17. There may have been a third line for motor control.

[Peter Miller]

Like Tim, I can't comment on the maths but based on my control line experience, a 200" span, 95 lb model is going to need to fly fairly fast and it is going to pull very hard and you do NOT want to see that coming in on the lines and heading for you.

I do know that a 2 lb model on 60 foot lines doing over 130 mph pulls like Billy Oh!. I know that I could not keep my wrist in the yoke and my arm was pulled through up to the elbow in spite of using all my strength.

[Erfolg]

Increasing the airspeed to approx 40 mph and the mass to 95 lb, gives us a new pull of 33kg. Again if we consider that CL do not fly in still air, this value would be a minimum.

We both get approx the same values.

I changed/edited my calc, whilst SDF posted his. I had a decimal point out, on the calc. I spotted the error when doing a mental check, SQ 10 then dividing by 2, which gave 50N.

What we need to recognise these values are only approx, representing the minimum ball park forces. There is so much we do not know about the model and under what weather conditions it will operate.

I did wake up this morning and thought, that probably a significant force and its consequences are being ignored, that is the effect of upwind forces on the model, blowing it into the circle. Again some calcs could be done, to determine if these are greater than the line tension forces from just flying. I suspect that you would want to turn the model quite strongly out of the circle, just to make sure the lines stay tight, rather than the model drifting in, when upwind.

Having read Peters practical experience, it seems that the simple calc, is just that, not representing the real world model.

Edited By Erfolg on 25/06/2013 08:59:14

[Bob Cotsford]

Don't forget that as Peter said, you also need offset rudder and thrustline to provide enough pull to counteract the drag of the lines. Not 'just enough' but enough to get the worst of the bellying out of them. The faster the model, the more the lines will drag and the more you need the model to turn right to keep the lines taught. Even my little Veron FW190 put enough strain on the lines to test the structure, at least it did when I stuck a racing .29 in it!

[Peter Miller]

Just had a thought! The BMFA Safety ADvice, Page 42 of the 2010 handbook says that the lines should bet tested to 10 (ten) times the model weight.

Did someone say that the Stirling weighs 95 lbs?

Er, Yes. Can I watch the pull test please!

[Erfolg]

If we take the weight at 95 ib, then apply the factor of 10 we get 950 lb pull.

Using rule of thumb values often used for ball park assessments of a max permissible stress level of 10 tons in^-2. Although standard mild steel is about 32 tons in^2 and that some steels are rated circa 60 tons in^2.

We then assume the that there are two wires, so each takes 475 lb. or 0.21 tons

then

Using stress = force / area resiting

sigma = f * 4/ Pi * D^2

rearranging

D = (f * 4/Pi * sigma)^1/2

= (0.21 *4 / Pi * 10)^1/2

= 0.2 " each (approx)

I have had some difficulty with this one, my calculator does not consistently register 0, it took at least 5 attempts.

I guess that drawn wire will be rated higher, if a suitable cable can be found.

This is based on the wire also having a rating of the pull test, but I guess, this is not done, something like 1.5 > 2 the anticipated loads?

However you look at the issue, line drag could be a very dominant factor in the power required.

The line pull test, could have safety issues.

I have to go out now, but I am starting to think of a simple calc for drag arising from the cross wind. A reasonable Cd being the big unknown.