I've read quite a few posts here on RCU where folks use the term "proper mechanical advantage" for setting up controls.
I think I have a basic understanding of what it is but does anyone know what the rules are or perhaps what some good rules of thumb for achieving proper mechanical advantage are.

Thanks in advance
Clm

Proper mechanical advantage means a 1:1 length ratio, of the servo arm relative to the distance from the hinge line to where the pushrod attaches to the control horn. Therefore, if you have 1.5" from hingeline to pushrod connection, you will need a 1.5" servo arm.
Also, for proper geometry, you need to have the angle of the theoretical line created by the point at the hingeline to the point where the pushrod attaches be parallel to the angle of the servo arm.
In addition, for perfect geometry, it is nice for the theoretical line of the control surface hingeline to the hingepoint of the servo arm to be parallel to the pushrod.
These are just a few qualities of proper mechanical advantage, as there are more details for pull-pull systems and such, but I have to go to work, so perhaps someone else will chime in...

I see "proper mechanical advantage" to mean simply that the leverage for adequate control surface motion is such that the servo and linkage can handle the load without failure. This could mean using long control arms with a standard servo, to using high torque servos with a concealed linkage where the moment arm, or leverage force, is greater. The linkages may need to be heavier as well, to prevent flex or buckling.

Hear is my rule of thumb. Use the minimum amount of control surface travel to accomplish the desired result. Use the maximum amount of servo travel (rotation) to accomplish this. This does not apply to the rudder in a pull-pull set up however. The idea is to give the servo the maximum mechanical advantage. and yet have enough travel to get the desired roll rate, loop size etc. The same applys to 3D set up's but often requires the use of larger servo arms to achieve this. Hope I havent confused you.
Z

zx has the right idea. Mechanical advantage is the same thing as leverage, which is all about moment arms. Mechanical advantage is achieved by linkage closer to the hub on the servo arm and out on the end of the control arm.

I pick up a Hitec standard short servo arm, and measure the distance from the rotor hub to the linkage holes: ~7, 10, 13 mm. I pick up a plastic ARF control horn and measure from the base to linkage holes: ~10, 15, 20, 25 & 30 mm. Add half the thickness of the control surface to the distances on the control horn--call it another 5 mm for convenience. So my choices at the control surface from hinge line to linkage connecting point are 15, 20, 25, 30 & 35 mm.

If I connect my linkage from the 10 mm hole (middle hole) on the servo arm, to the 30 mm hole on the control horn, then I have a mechanical advantage of 3 to 1, or 3:1 as I was taught in high school. The servo arm will rotate three times further in degrees of arc than the control horn will. So 45Ëš of rotation at the servo will give 15Ëš of rotation at the control surface. But the mechanical advantage will convert 40 inch-ounces of torque from the servo to 120 inch-ounces of torque at the control arm. This is a pretty good setup for sport aerobatic flying.

A 3D flyer would be happier with a 1:1 ratio, 45Ëš servo rotation giving 45Ëš control surface rotation--which is no mechanical advantage at all. Maybe an extreme setup would be 45Ëš servo rotation giving 90Ëš control surface rotation, in which case the servo would be on the short end of the power ratio, such that 40 in-oz torque in the servo would produce only 20 in-oz at the control surface. If you wanted more control torque than that, you'd need a high-torque servo.

You can, of course, get more rotation out of a servo than 45Ëš, so you can get a 4:1 mechanical advantage, or even 5:1 (7 mm at the servo, 35 mm at the control horn). I've had pattern planes where 9Ëš deflection on ailerons and elevators was as much as I would want. This gives a somewhat slow but very controllable roll rate, and more than adequately tight loop diameter. So with standard servo I could link from the 7 mm hole on the servo arm to the 35 mm hole on the aileron/elevator control arm and multiply my 40 in-oz torque to 200 in-oz. That, by jiminy, is real mechanical advantage!

Along with the power, you also gain resolution--which for precision aerobatics is just as important as the higher torque. You do lose speed of the control surface reaction, since the servo has to travel farther to get the same deflection, but that has never been a problem for me. If it is, you can spend a bit more for a high-speed servo.

Thank you, MajorTom! Clear, concise and correct! I have had a terrible time trying to explain mechanical advantage and the evils of ATV. You did it just right.

Jim

what ever jrf said... I second that thank you to majortom... ow its crystal clear...

If I may paraphrase what MajorTom said, for those who may have gotten lost in the physics:

1-Set the ATV at maximum.
2-Connect the pushrod to the outermost hole in the control horn.
3-Connect the pushrod to the innermost hole you can use on the servo arm and still get the travel you want.

Jim

What happens if...
1) You set your ATV at maximum.
2) Set your control horn linkage to the outermost..lets say 1.25" from the hinge line.
3) Set your servo horn linkage to the outermost..lets say 1.25"
This part always confuses me. Can someone shed some light.

Thanks
Smokey

If your linkage is the same distance from hinge line to connector on control horn, and from servo rotor to connector on servo arm, you have a 1:1 ratio, therefore no mechanical advantage. Whatever your servo torque is, you'll have the same effective torque at the control horn.

Setting your ATV to maximum will give you more servo rotation, maybe +/- 60Ëš instead of 45Ëš, but no change in mechanical advantage. It really doesn't matter whether your linkage is 1.25:1.25, or any other distance, if they are equal distances.

Say you have an L-shaped lever, long arm 5' and short arm 1', which can be rotated at its corner. Say you fix a 100 pound weight at the end of the short arm. If you now pull on the long arm, the further out to the end you pull, the easier it will be to lift the 100 pounds. Pull at one foot, and you need 100 pounds of pull to move the weight. Pull at 2' and you can move it with just 50 pounds of pull. Pull at 4' and it moves with 25 pounds of pull.

The same principle operates the servo itself. A very small electric motor turns a small diameter spur gear, of let's say 10 teeth around its circumference. This spur gear engages a larger gear, with let's say 50 teeth in its circumference. The motor spins at a rapid rate, but the second gear turns only 1/5 of a revolution for each full revolution of the spur gear. However the torque of the drive motor is multiplied 5 times on the axle of the second gear.

If you then fix another spur gear on the axle of the second gear, you can get another multiplication on a third gear, and go on multiplying the torque of a tiny high-speed low-torque motor almost indefinitely. With every additional gear you add, you lose rotational speed but gain torque. If the drive motor spins at 10,000 rpm then your output gear may turn at only 100 rpm--a hundred times slower, but with 100 times the torque of the drive motor.

Thus the same motor can be geared as a high-speed or a high torque servo, all depending on the number of intermediate gears and their ratios.

Setting the ATV at maximum gives more mechanical advantage indirectly by allowing you to use a hole closer to the center of the servo arm to get the same surface travel. Maximizing the mechanical advantage of the control linkage requires that the servo travel be set to maximum before you select the position of the pushrod in the servo arm to get the surface travel you want.

Jim

I agree. Point well taken.

I am now wiser beyond my years..

Thanks
Smokey