Martin Harris - Moderator

I wouldn't advise charging at more than  1C and then not on a regular basis.  1/10C is ideal but you probably can't go below 100mA which is what I'd use on a regular basis. For discharge but most of these chargers don't do less than 500mA - so that's what I'd use.

Profilm removes without leaving anything other than a small amount of glue residue (uncoloured) in the wood fibres and more or less ready to recover. Solarfilm leaves the coloured residue.

All I can say is that making best guess alterations to RF systems is at best risky and at worst could result in prosecution or even more dire consequenses if an accident were to occur. Using an aerial which boosts ERP would definitely be illegal although I'm not qualified to say if a directional aerial actually does so (Tony)

Posted by David the flyer on 03/10/2010 22:32:45:  I spoke to a club member and he said that was ok, it's a preference if you want the engine to point slight left and down, his yak, he has the engine in a slight up position, he said it's better for landing?  

Up thrust usually leads to rather nasty trim changes with power so I don't see how adding it would help with a landing. As you reduce power for the flare, it will tend to pitch the model nose down - probably resulting in a rather undignified arrival!

Don't Spektrum supply an aerial that will attach to the module?  I assume this is what you want to do and looking at various adverts they all seem to come with remote aerials.  I think this query would be better addressed direct to Horizon as there might be legal repercussions should a non-standard aerial create RF anomalies. It might well be that the aerial is a "standard" part but only the manufacturer/designer can confirm it with any authority.

Why do you ask?

(continued) wt * 40 Example 2 Windspeed 5 m/sec, model airspeed 20 m/sec. Initial state: Groundspeed(1) = airspeed(1) - windspeed = 20 - 5 or 15 m/sec End state: Groundspeed(2) = airspeed(2) - windspeed = -20 - 5 or -25 m/sec Change in momentum = wt * groundspeed(1) - wt * groundspeed(2) which is wt * 15 - (wt * -25) which becomes wt * 40 same as before. Example 3 Windspeed 10 m/sec, model airspeed 20 m/sec. Initial state: Groundspeed(1) = airspeed(1) - windspeed = 20 - 10 or 10 m/sec End state: Groundspeed(2) = airspeed(2) - windspeed = -20 - 10 or -30 m/sec Change in momentum = wt * groundspeed(1) - wt * groundspeed(2) which is wt * 10 - (wt * -30) which again becomes wt * 40 Example 4 Windspeed' 570 mph, or 250 m/sec (well, almost). Inside the transport aircraft within which the model is to fly, it has to be launched towards the tail, so that the mass of moving air will be coming in the opposite direction to the model's flight and thus constituting a headwind. The speed of the big aircraft therefore has to be entered as -250 m/sec. Initial state: Groundspeed(1) = airspeed(1) - windspeed = 20 - 250 or -230 m/sec End state: Groundspeed(2) = airspeed(2) - windspeed = -20 - 250 or -270 m/sec Change in momentum = wt * groundspeed(1) - wt * groundspeed(2) which is wt * -230 - (wt * -270) which once more becomes wt * 40 Conclusion No matter how you cut it the calculation, done with proper respect for vector quantities, brings the same result. The values of windspeed, however great they may be, cancel out and only the aerodynamic flight effects contribute to the change in momentum. Edited By Martin Harris on 04/10/2010 00:14:06

Personally, I believe BEB is proposing a situation which cannot physically occur in the real world when using aerodynamic effects to initiate the turn. He may very well be correct hypothetically but I freely admit that it's beyond my understanding to apply his reasoning to the real world.  All I know is what I've experienced and my basic and empirical understanding of the dynamics of flight.   Not my words but I think this sums up the situation as far as my limited maths skills can tell:   Please feel free to discuss!   The problem of the downwind turn is entirely associated with the effects of wind speed. Here are four simple calculated examples based on 4 different wind speeds: zero wind, wind at 1/4 airspeed, wind at 1/2 airspeed and 'wind' experienced inside a large aircraft cruising at 570 mph. Before we go any further it will be necessary to set some parameters. All values will be in metric units, i.e. metres, and seconds; there is no need to consider a numerical value for weight because at all times in a level turn it equals lift, which simplifies it a bit. It will be referred to as 'wt'. Also, we need to lay down a few definitions: Kinetic energy. This is a red herring and does not concern us. The common mistake is to assume that it affects turns as if there were a law of conservation of kinetic energy. There isn't, so we can forget about it. For the record kinetic energy is 1/2 mass * velocity squared. Momentum. This is the one that matters. There IS a law of conservation of momentum, both of linear momentum and angular momentum (rotation). The snag is that this too is not relevant to effects produced by the aerodynamic forces that cause turns. The point of the examples will be to demonstrate that the change in momentum during a turn is the result of the reversal of airspeed and that this is independent of any effects of windspeed. Momentum is defined as mass * velocity. Because velocity is a directional quantity we have to be very careful with the number values. When the direction reverses, velocity changes from a positive value to a negative one. It is probably neglect of the sign (+ or -), more than any other factor, that leads to confusion when the physics of the turn are considered. In all the examples the initial condition is flight into wind. The airspeed in all cases will be 20 m/sec, so the groundspeed in each case will be airspeed - wind speed. The term 'speed' is not as precise as 'velocity' but it will be used for convenience because of its familiarity. Example 1 Windspeed 0 m/sec, model airspeed 20 m/sec. Initial state: Groundspeed(1) = airspeed(1) - windspeed = 20 - 0 End state: Groundspeed(2) = airspeed(2) - windspeed = -20 - 0 Note, the reversal of the direction of flight has made airspeed(2) a negative value. The change in momentum is groundspeed(1)m - groundspeed(2) and again the sign of each value is vital. Change in momentum = wt * groundspeed(1) - wt * groundspeed(2) which is wt * 20 - wt * (-20) or

BEB, you'll need to excuse me if I'm confusing my energy types - physics lessons were a long time ago, but isn't the kinetic (doing) energy being used to oppose the wind speed resulting in zero potential energy left to whack my head if a model is doing the same airspeed as the windspeed opposing it?   Yes, of course you can stall the model by turning too tightly with the required amount of elevator but the case I think you're looking at is more that of a flat ruddered turn, reversing the direction the model is pointing in virtually instantaneously but I don't think this is actually possible - would there be enough inertia to carry the tail through the relative airflow past 90 degrees of rotation?   When I fall over rotating quickly (in or out of the train) it's due to my foot being in contact with the ground so I can't relate to that analogy.   Dan, I think you may be under a misapprehension over the ballooning issue - have a look at some of BEB's earlier posts in which we all seem to agree on in that the effect is pilot induced.Edited By Martin Harris on 03/10/2010 19:00:20

What sort of glowclip are you using?  Is it making good contact?  I'm assuming you've checked the plug is glowing from your battery?   While the plug's out, you could squirt a few drops of fuel in before putting it back - this often helps an engine to fire and if it does but won't continue, you've got a blockage somewhere in the carb. If so, my favourite trick is a bicycle pump (with a retract adapter - or similar) linked to some fuel tube on the inlet nipple to blow through the carb while wide open with the main needle in and out.

I'd guess they meant a shaft run leading to over revving?

...and then do the same experiment from downwind.  I think it will exhibit the exact same result. I'm not able to follow your logic this time but I don't claim a deep mathematical understanding of the subject so I'm happy to wait and see the results.   To me, the kinetic energy in relation to a point on the ground changes - i.e if the model hits me at 30 knots airspeed flying into a 29.9 knot headwind, I'd feel a lot happier than if it's just done it downwind at 59.9 knots in relation to my bonce...but as we all mostly seem to agree, in free air there is no effect from windspeed on a model's flight characteristics.   Interesting discussion, though.

Agreed - my problem is I've never been one to meekly accept things at face value. I remember arguing with a maths lecturer over 1/0 equalling infinity until he got quite excited - I see the argument and accept that it's very very close to being correct but simple maths tells me that if it were so, then infinity times nothing equals 1 and I know that can't be right! I'd guess that sometimes we need to accept something as near enough...   I'll bet John had no idea what he'd start when he posed his question! What's interesting is that everyone seems fairly comfortable with the basic principles of downwind/upwind turns which has certainly not been the case in past debates.

But you can't possibly do that.  To do so would give an infinite G load and apart from the wings falling off it would stall because of that irrespective of any wind.   I'd guess that gravity originating from the centre of mass of the earth is theoretically relevent but any difference due to a change of position of the model would be infinitessably small?

I'm sure we're in general agreement and debating a very fine point here - it's just that I can't accept that the relationship with the ground has ANY physical effect (other than from ground induced turbulence) on a model in flight.   This statement: We don't notice these effects 99.99% of the time because the model has quite low inertia and has the small interval of time it needs to adapt to the new conditions - to adjust its ground speed so as to keep its airspeed constant.   ...just doesn't make sense to my empirical interpretation of the subject.Edited By Martin Harris on 02/10/2010 23:15:37

Ah, but which way do you define walking up the aisle!

OK I agree with that (BEB - Eric's post popped in while I was replying) but the point is that the motion of the medium I'm travelling in is irrelevent, whether in an aluminium tube or a block of moving air.Edited By Martin Harris on 02/10/2010 22:59:01

But if I walked down the aisle of a jumbo jet flying at 500 mph groundspeed and spun round to go back to my seat because I saw the queue for the toilet I don't think I would zoom at all - and what's happened to the inertial effect of going from 502 mph to -498 mph? I don't recall ever feeling any - and if I did I'd imagine my prospective visit to the little room would be superfluous!   I would contend that the forces you're referring to are all pilot induced as the flier on the ground attempts to make things look "right".   Perhaps I should have said that the various inertial forces in the calculation cancelled out?Edited By Martin Harris on 02/10/2010 22:55:09

Sorry - I'm getting bored not being able to do anything constructive for a week or two - I may get worse!!!

Shall eyelet you decide which is better or should we leave it as a tie?  Possibly a knotty problem - but do they have a similar wing cord as well?

I did once come across some apparently learned writings where it was proved mathematically that the inertia is cancelled out and can be discounted.  I too have thought about the DS conundrum and I believe that it is because of the sharp edged transition from the wind shadow - not a situation encountered in free air but very much related to a model doing a slow climbing turn with a dying engine downwind of a typical boundary hedge, tree, barn etc. - and I think we've all witnessed the result!

Posted by Big Bandit on 01/10/2010 19:01:57: Hi All,   Great to speak to you, and Jo your English is fine. It was great talking about Hailwood, Surtees and Ago. I'm watching the old Video's later, particularly the 1967 Senior TT when Hailwood's throttle broke and Ago's chain broke on the last lap.     Did you happen to mention Phil Read during your chat?  Many years ago, (well before my time) he was a member of our model club.   I didn't go to too many bike race meetings but I do recall watching a terrific battle between Read and Ago at the Race of the Year at Mallory Park in the early 70s (73?) .   Shame, but Thursdays are out for me as we always adjourn to the local pub for a meal and some liquid refreshment after flying in the afternoon...

To add a bit to Phil's advice, the air bleed needle will point at a tiny hole in the front of the carb body - take a squint down it and you should see the tip of the needle. Check that it is approximately half way across the hole as a starting position.   When Phil refers to the throttle stop being fully open, he means that the hole in the throttle barrel can be fully closed - if you screw this adjuster in slowly you'll see how it will open the throttle slightly...then go back until it's properly closed.   Start with the main needle 2 1/2 turns open which will almost undoubtedly be too rich but you'll tune this out once it's running.

Polyclarifier,   Your last point is particularly relevent to trying to explain the practical aspects of this old chestnut.   It's the turbulence / wind shadow effects that are the fly in the ointment to the  theoretical explanations.  These give rise to the false impression that the theory is flawed but the end result is that you need a margin of energy (airspeed) to counter these effects which certainly count in the real world but are outside the theoretical arguments.   I hesitate to argue with BEB as I'm aware of his qualifications but I really don't think that scale or the ability to pull more G influences the theory. What it would do is to dramatically increase the stalling speed and given sufficient elevator power I can quite believe that the model could flick out of a tight turn at high airspeed - I know my 1/12 scale combat models will do so at full throttle/max airspeed but I wouldn't want to be in one!Edited By Martin Harris on 02/10/2010 00:56:33