A servo has three basic identities, i.e. its weight,its speed (time it takes to turn at extremes), and the torque.(as far as i know)
So how to calculate the type of servo needed for a particular model?
Does it have any relation with the flying weight or we are just concerned about the time it will take to make the required turn?

The torque required depends on the model weight and the speed of the craft. For a newbie I would begin with about 2 x all up weight.

About controls.

it has 'Force', 'Deflection' and 'Harmony'. what we are interested is in the force/unit deflection. The force that servo will feel will be force exerted on the control surface. this can be calculated by applying simple formula 1/2 x 1.225 (Air density at sea level) x square of max speed that you will envisage that your model will travel (Keep the units same) x the surface area. this will give the max force that your servo will feel from an unbalanced surface (??!! ) add 1/2 kg for friction and bends, your servo should have twice the the torgue needed. why twice ? reason 1. in its life time the servo will continue to loose its torgue. reason 2. after 75% of its rated torgue, servo can experience something akin to 'Jack Stall'.

Quote from: PankajC on December 23, 2010, 10:27:14 AM
The torque required depends on the model weight and the speed of the craft. For a newbie I would begin with about 2 x all up weight.

Theoretically it has nothing to do with weight. in fact in a non-powered controlled aircraft, there is a protrusion ahead of the hinge line (Refer image) this is called the horn balance (Not the control horn we know of), purpose of this horn balance is to reduce the control force. in an RC aircraft if this horn is long enough even a micro servo would do (In an aircraft called Devon, sometimes the balancing force was more than the opposing force and once you moved the stick the control surface used to travel to it max deflection of his own accord, this is called overbalance). if you are designing an aeroplane and want to knock weight and money, carefully design long enough horn. (on the lighter side, Make it more Horney).

More to come in Basic Aerodynamics for RC flying Post.

Good work there mate. I am sure this would help lot of people

But to slightly correct what you mentioned
"there is a protrusion ahead of the hinge line (Refer image) this is called the horn balance (Not the control horn we know of), "
These are correctly referred asaerodynamic balances. Not horn balances. As you would know the same effect is also acheived by placing appropriate mass (commonly what we call weight) forward of the hinge line for control surfaces and is termed as mass balances. The function of these is exactly as you have stated.

Now as you adviced fellow flyers " square of max speed that you will envisage that your model will travel" it will help them if you also put in the method to calcualte the max speed of the model (under given conditions)  

You may also need to put in a discalimer about your calculation and advice up there , as the stated torque of the servo is generally at the innermost hole on the horn (mostly 10mm from centre) and as one moves out the tourque reduces but the throws acheived increases. So the calcualtion of tourque required should take into account which hole the builder/assembler is intending to use.     

Capt, I thought about those and the pull pull system,  and its maths, how to use Sine and cosine of the horn set to control throttle curve etc, then it was sort of getting overboard. As regards Horn balance, thank you for pointing it out, it's correct nomenclature is 'Horn aerodynamic balance' (there is Horn Mass Balance as well, (Refer Image)) others are inset hinge, tab balance etc, no use for RC (I guess).

As regards mass balance, a good topic you started and i guess it will help people who design models, or to understand the same (Foamies incl). to keep it simple, the CG of the control surface should be ahead of the hinge line, if not it will flutter (at high speeds and at max deflection angles of the surface) and the control loss can take place, because, the centre of pressure of the control surface should not go ahead of the CG of the control surface. so any (Preferably there should be) reinforcement to the control surface should be ahead of the hinge line.

also
readers, Kindle your brains more, the answers are in there, in aviation if any info, if it  can't be put to use it is of no use.

Disclaimer ? of course, these info is just to start the discussion, must do your own research, because, these are tailored to keep it simple and avoid intimidation from the complexities of maths.

ah! one other thing
horn mass balance serves to keep the CG of the control surface ahead of the hinge line, especially when the hinge is very close to the leading edge of the control surface, or if it is highly swept back. it does not help in reducing the control force, (Meaning it is not a Aerodynamic balance). This type is used even in supersonic aircraft like MiG 21, (To avoid flutter at supersonic speeds) (refer image)

Quote from: augustinev on December 23, 2010, 07:03:31 PM
About controls.

Theoretically it has nothing to do with weight.

Don't understand this. The servo will push or pull the control surface one way or another. This will mean that the control in turn offer resistance to the airflow thereby changing the direction of the aircraft. When an aircraft is flying, it will have a certain momentum (which is dependant on weight and speed). While changing the direction, it has to overcome this momentum so it has to move the weight of the aircraft doesn't it?  So why this statement?

Pankaj

the surface protruding ahead of the hinge line (Horn aerodynamic balance) is meant to assist the surface to move, at high speeds the opposing force will be more, so will the assisting force, if carefully designed, the assisting force can make flying force almost feather like.

Devon was a large transport aircraft of yester years, it had manual controls and it had a aerodynamic balance called the spring tab (Not required in RC, therefore did not delve on it) it made flying feel rather 'feather like'. even a micro servo could fly that real life aeroplane. (Max force was 4 Lbs approx), think of its weight and airspeed as compared to an RC aircraft. Thats why the statement, aim was to clarify and not to refute you.

as regards weight, another good topic for discussion, see my basic aerodynamic post, if CG is kept behind the Centre of pressure(CP), the same weight will assist the maneuver (But will be called unstable right), so technically it is the static and maneuver margin (simply, Distance between CG and CP) that will dictate the reluctance of aircraft to maneuver. this reluctance is what the controls need to overcome. More on this later in the basic aerodynamic thread (Running for midnight mass(?!!))
Good to have you thinking, any more queries ? keep it coming, makes reading interesting.  

Pnkaj,
Simply put, these balances make the work (efforts) of a servo (or pilot in the full scale) easy. With these balances lesser force (effort) is required to move the control surface for the same speed and size of control surface.

Now take the aerodynamic balances (horns projecting ahead of the hinge line). Now say you require a servo with 4 Kg torque to move the elevator surface (without the aerodynamic balances) at the given speed.
The moment you displace the elevator from centre what happens?
Let's take an example of elev being moved up.
The moment elev is moved slightly up, the projecting balance moves down of the hinge line. Now this projection meets the oncoming air stream which tries to further push it down, larger the surface of this horn higher the force generated on this projection which would translate into assisting the upward movement of elev. Thereby the same size elev surface which would have required a torque of 4kg would now require a force of (4 Kg minus x )
I am sure you would get the picture      

thanks capt

flyingboxcar,
The horn projection is just a fraction of the control surface. Lets take an RC model, the elev size would be some 150x20mm where as the control horn might be about 10x2mm. So why is this such a big consideration? The force of the incoming wind on the elevator and the horn might be same per sq cm, but the effect on the elev would be more profound wont it?

Pankaj,
Yes it just a small part of the overall control surface, but it does make the work of the servo easier by that small value.

a great example of a elevator horn of this aerobatic aircraft (Addiction X). for it's size and agility, it uses Hitec 19.0g HS-82MG Metal Gear Micro RC Servo, (Large horn ahead of the hingeline as explained by capt in Rep #8) reduces the control force.

Quote from: PankajC on December 26, 2010, 06:43:27 PM
the elev size would be some 150x20mm where as the control horn might be about 10x2mm. So why is this such a big consideration? The force of the incoming wind on the elevator and the horn might be same per sq cm, but the effect on the elev would be more profound wont it?

Great question, it is the moment arm from the hinge line (Fulcrum, load & effort), thats why horns are long (Refer image), its contribution is to reduce the control force, not to remove it completely, general rule is the horn should be not more than 1/5 of the control surface area, this is to avoid overbalancing. there are other complications as well with Horn aerodynamic balance, thats why tab, spring tab, etc.

There is another balance called 'Inset Hinge', very interesting balance, can be used, without having to use big horns. (Refer Image)

The inset hinge does the same thing without ungainly projection ahead of the control surfaces, since besides balancing these balances also play up on drag equation. So for faster (comparitively) A/c the solution was to introduce inset hinges. If you really observe though the balances are split in many parts, overall when the area is added it add ups to considerable surface (almost 20-25%).     

Dear Augustine & Flyingboxcar

Thank you very much for your valuable contribution to this topic  . It would be really nice & helpful if you could show us the calculation through an easy to understand example.....

I have understood the formula to be:
1/2 x 1.225 x Square of max. Speed (Miles/Knots /Kilometers?) x (Control?) Surface Area (In Square Inch?) = Max Force + 1/2Kg x 2 = Advised Servo 

"The force that servo will feel will be force exerted on the control surface. this can be calculated by applying simple formula 1/2 x 1.225 (Air density at sea level) x square of max speed that you will envisage that your model will travel (Keep the units same) x the surface area. this will give the max force that your servo will feel from an unbalanced surface (??!! ) add 1/2 kg for friction and bends, your servo should have twice the the torgue needed."

Sorry but I am not a formula type (Maths was never my forte) I simply go by observation, simple common sense and suck it and see priciple. If it looks right generally it works out right too.
I would leave all that formulae and calcualtions to more learned folks

Thanks a lot buddies...

But i think that only a fraction of force can be altered by horns balancing, and even a small torque is sufficient to turn rudder or elevator even during flight, but the thing is the CG should lie at the center of wing, and elevator and rudder should be there only to balance the direction. However, lesser the torque,longer the time to make the desired turn during flight.
Though my theory may not prove to be fruitful for stunt planes or 3D , but this can be considered while making a glider or a slow flier during lack of resources.

But i agree with the fact that torque should be double the overall weight for quick and effective response.

Mathematically.

Oops ! i was dreading this, let me try and keep it as non mathematical as i can. for example if you have a model which has a surface area of say 1/2 m²(S)  and it travels at say 60 kmph and if the aerofoil's efficiency at max deflection is say 0.2 (Most symmetrical aerofoil has this value), now let us see how do we arrive at a number. 1/2 ρ v²  is the dynamic pressure. Ct 1/2 ρ v² S total force it can generate. This is what we need to see, and is the total force on the control surface. Let us see the number in SI (Système international ). (Thats what i meant by keep the units same).

0.2 ( 1, being a flat plate at 90 degs to the air flow [Aerofoil efficiency]) x 1/2 x 1.225 (ρ air density at Sea level) x 16.66² (60 kmph converted to mtrs/sec) x 1/2(Surface Area) = 17 kgs.

this is for an unbalanced (Without the aerodynamic horn balance) controls, if you aerodynamically balance it, this can be reduced to 3-7 kgs.

Practically,
Take a anemometer, connect a weighing spring to the control rod, run the engine, deflect the control to max, read the force as read on the weighing spring. this is for the measured prop wash speed as seen on your anemometer (Assuming your entire elevator is in the wash). plot this force for various speeds. now you have a graph, extrapolate it for higher speeds, simple. remember to add prop wash speed as seen on the meter at full throttle to the envisaged model speed. (In air, the prop wash will reduce, how ? later at a different time may be)

What is my models max speed ? good question.
Ans:- measure it. how ?  have two persons standing on the ground looking straight at a known distance apart, third one presses stop clock, once when the first person calls out now when the aircraft crosses his line, second time when the second one shouts similarly, what RU you doing , obviously flying the model at its highest speed. Laughing ? don't. this was how 30 years back they measured an aeroplane's speed, instead of guys they used cameras. (You can use too, no big deal).

or else fly model closer to yours in a Aerofly Simulator Delux or Real Flight simulator and read off the speed as indicated on the bottom right on the screen.

Did I make matters worse ?  

Get a speed gun and have some fun!  (measuring the models speed silly ). BTW if you have a/some relative/s in the Khakhi you dont need to buy, get one on udhaar!
But the question which you need to answer is what are you measuring ground speed or air?

Capt, there you go, IAS, CAS, TAS, GS,  

  

@ Augustine: LOL....

T O T A L L Y-----------A B S O L U T E L Y--------------C O M P L E T E L Y 

Just thought of a simple idea. Presuming its a 3D plane (Cause i think they would demand the most from the servos) could we line up servos by torque for the following:

Say: For 0.40 Size, 0.60 Size, 0.75 Size, 0.90 Size, 1.20 Size, 1.60 Size / 30cc, 55cc, 100cc and so on???

good idea, manav, my point is you can have a real size aeroplane flown by a micro servo, given the correct balance. point is well balanced ac like Addiction X requires only a micro servo, get the Drift

see the red aeroplane image above, approx 1/4 is a horn balance which ahead of the hinge line moves in the opposite direction to that of the elevator and assists the controls to move. thats why such great response from addiction, it is one of the best 3D aeroplanes ever made.

@ Augustine: that was a super explanation and understood the concept.  Many planes are built with regular control surfaces. So if we could in general identify the torque required as per plane size, i think that would help many including me. BTW did you choose theAddiction over the YAK?

its like harley and yamaha r15