squareloop

My Feedback: (1)

If my attachment points at both the servo arm and the control horn are 3/4" from the pivot point maintaining a 1:1 relationship, is the torque the same or am I using 25% less torque? Thanks. Sorry if the answer is obvious. :stupid:

dhooks

My Feedback: (24)

The distance doesnt matter as long as your ratio is one ~one. Obviously you lose a little torque through the linkage but in such a miniscule amount you'll never notice Im sure. So your getting very close to 100% transfer.

mglavin

My Feedback: (31)

Servo torque will never change irregardless of the combination you utilize. The FORCE delivered will with varying arm lengths.

It's desirable to have a mechanical advantage. Meaning the control horn pivot center or distance should be greater than the servo arm pivot distance. NEVER (in a perfect world anyway) allow the control arm to be less than the servo arm pivot distance. In your example 3/4", I'd suggest a control arm pivot spacing measured from the hinge center line to the control horn pivot point of 1".

A mechanical advantage will change the FORCE realized at the control surface.

1.0" control arm and a .75" servo arm, 1.0/.75 = 1.33:1 ratio. Nets 333oz-in of FORCE on a servo delivering 250oz-in FORCE.

If the servo was rated at 188oz-in with a one inch arm and we are using a .75" arm then we can calculate the available FORCE by dividing the known FORCE of 188 by the servo arm used of .75" and arrive at a new delivered FORCE rating with this combination of 250.66ozs of FORCE at the servo arm pivot point.

As noted above available Force will change if you use other than a 1" arm. Servos are rated in ounce-inches. So one inch is the factor used to calculate the TORQUE together with the FORCE delivered by the servo. 188ozs of FORCE x one inch = 188oz-in of TORQUE.

squareloop

My Feedback: (1)

Yes, that's what I meant. So with any given oz/inch rating is the force the same as if the attachment point is 3/4" at both ends (1:1)? I think that's what dhooks is saying unless he's answering my question from a torque point of view (?).

Sorry for confusing the terms.

bgi

My Feedback: (7)

Yes. To calculate the force if the ratio isn't 1:1, simply divide.

Approximate (more accurate when the arm is at 90deg to pushrod - torque decreases at surface end as the arm rotates away from 90deg):

Torque at surface end = servo torque * (surface arm len / servo arm len)


If you want to get an excel spreadsheet which is "the business" for this type of stuff, check out post 10 in this thread and download the spreadsheet.

http://www.rcuniverse.com/showthread...88&forumid=230]

squareloop

My Feedback: (1)

Thanks for the help people!

squareloop

My Feedback: (1)

Wow that spreadsheet is interesting! How do you approximate what your average airspeed is though? That would be an important factor as to whether or not your servos are up to the task.

OK folks. Been thinking (oh no!!!) about this one last nite.

Just as an example a surface needs 75 oz of torque minimum. Say I have a servo rated at 100 oz/inch. Setup: the rod is connected at 1" on both the servo arm and the control arm maintaining a 1:1 mechanical advantage. I also get a desired surface deflection of 50 degrees with the servo pegged at both endpoints.

Now I move in the attachment points at 3/4" (.75) at both ends for an increased force value of 133 oz/inch (?). Assuming I moved them both exactly the same amount (1:1) the surface deflection should be the same (?).

My question is: why would I use (or not use) one setup over the other? What am I sacrificing in exchange for the extra force in the second setup?

bgi

My Feedback: (7)

Shortening the arms on both ends will:

1) Increase the effect of any linkage slop which may increase liklihood of flutter.

2) Increase the force on the pushrod (important with thin or flexible rods)

With the typical arm lengths we work with, longer is generally better up to a point.

rcFp

regarding control surfaces movement i'm trying to solve extra or less control surfaces movements by moving pushrod location on both horn and servo arm sides
my Q are :
1) is it the correct way to do it or should i leave both sides on their out most location and make radio ATV adjustment
2) if not what is the right location to move the pushrod (horn or servo side) and what is the effect (can i reduce output toque to a risky level )

thanks

bgi

My Feedback: (7)

1) You lose servo resolution and torque if your arms require you to cut down ATV. So try to keep ATV up to 100% or more. More ATV means more torque and more resolution but slightly longer transit time. It's a trade-off

2) Assuming you want to reduce the surface travel, move the pushrod out to the further clevis positions on the surface. If you run out of positions there, then use a longer surface arm or start coming in closer to the pivot on the servo arm. Either adjustment will give you more power/torque on the surface and vice versa.

rcFp

Thanks
Just one more thing to clarify .
if i need ito increase control movement can i use the pivot's closest horn hole or am i going to lose too much torqe ?

mglavin

My Feedback: (31)

Aerobatic models are not typically not fast models. But the forces applied can be extreme, even at slow speeds. I'd suggest an air speed factor of 75mph.

[QUOTE]OK folks. Been thinking (oh no!!!) about this one last nite

No mechanical advantage here 1:1 is an equal advantage.

133oz-in of Force is correct, this is realized at the end of the servo horn. No the surface deflection will be less. I would only move the servo horn pivot point to 3/4" this will increase your FORCE and provide a mechanical advantage. .75:1.

Short servo arms are desirable as are long control arms otherwise known as lever arms.

Short servo arms increase the Force realized but sacrifice the travel arc.

Short control arms will provide more travel movement but will require considerable more FORCE to do so.

Long servo arms provide more travel arc but reduce the FORCE delivered.

Long control arms provide a longer lever, thus requiring less FORCE to operate the surface but inherently reduce the travel arc.

Best setup is MAX ATV or End-point at TX, as short a control arm as will provide the desired surface deflection. Generally were limited to an absolute minimum with regard to control horns, this is measured from the hinge center line to the control horn pivot point. Always maintain at least a 1:1 ratio between both arms, preferably a mechanical advantage would be present (smaller servo arm than control arm).

mglavin

My Feedback: (31)

How does this happen? Slop is slop. If its present the control arm lengths don't realize it moving or not, it simply exists and is not a variable of arm length, me thinks.

If you have thin or flexible rods and you increase the FORCE it will exacerbate an existing problem, FLEX!

I don't agree unless your specifically talking control arms, a long control arm is always better, but a long servo arm is always undesirable, but sometimes required to achieve wanted throw.

mglavin

My Feedback: (31)

Best setup is MAX ATV or End-point at TX, as short a control arm as will provide the desired surface deflection. Generally were limited to an absolute minimum with regard to control horns, this is measured from the hinge center line to the control horn pivot point. Always maintain at least a 1:1 ratio between both arms, preferably a mechanical advantage would be present (smaller servo arm than control arm).

What I would first move would be dependent on the problem in front of me. Binding would lead me to shorten the servo arm. Inadequate throw or deflection would bring my attention to the control arm first, but I would first have to consider the overall length of the control arm from hinge center line to pivot point. If it's possible I would shorten this first, keeping in mind I want a mechanical advantage. As a secondary consideration I would lenghten the servo arm.

You cannot ever reduce the servos rated TORQUE this is a fixed value. You can reduce the FORCE by lengthening or shortening the servo arm.

chimchim

My 2 cents.

Two points of interest here:
1. Control surface throw and
2. Slop

Number one, the length of the servo arm and control arm only effect the control surface throw when the ratio changes. To visualize this, think about a belt running around two pulleys. As long as the two pulleys are the same diameter, they will both turn at the same rate. ie. Move one pulley 45 degrees, and the other will move 45 degree as well. But if you make one larger than the other, they will no longer turn at the same rate. A side note regarding torque, provided that you don't change the servo/control horn ratio it doesn't matter what length they are, the same amount of force will be applied to the control surface. In the previous example of 100 oz/in @ 1 inch causing 133 oz/in @ .75 inch is correct, therefore the force applied to the control rod is increased. But at the other end of the control rod that higher force now has less advantage on the control surface due to the shorter horn.

Now as for number two, the entire control setup is about motion translation. Rotary to linear and back to rotary. Any slop involved in the rotary component is related to its turning axis, and doesn't change regardless of radius of movement. ie a berring is going to have the sazme amount of play when installed in a two inch wheel as when it is installed in a 2 foot wheel. But the linear components slop is relational to its size. The linear slop comes from the size difference between the hole in the servo arm or control horn, and the pin in the clevis used. This slop can be measured, it is small though but consistent with all lengths of servo and control horns. Now we need to consider the amount of distance the control rod will travel. The longer the servo and control horns are, the further the rod wil move, and vise versa. Therefore, the more travel you can get in the control rod, the less effect the slop will have.

Sorry for going so long and confusing, but this proves to be a hard topic to explain without pictures and hand gestures.

Note: I would put in a lol, no wait LOL but it appears to be a fopaw to laugh at ones self here. ( please see lol ettiquete thread )

Steve

mglavin

My Feedback: (31)

How does servo resolution effect TORQUE?

Lets see some numbers. I have a 100oz-in torque rated servo with a 1" arm.

.75" servo arm nets 133ozs of FORCE
1.0" servo arm nets 100ozs
1.25" servo arm nets 80ozs

.75"SA x .75"CA nets 100oz-in
.75"SA x 1.0"CA nets 133oz-in
.75"SA x 1.25"CA nets 166oz-in
1.0"SA x .75"CA nets 75oz-in
1.0SA x 1.0"SA nets 100oz-in
1.0"SA x 1.25"CA nets 125oz-in
1.25"SA x 1.25"CA nets 100oz-in
1.25"SA x 1.5"CA nets 120oz-in

Then there is this thing known as the Final Ratio...

mglavin

My Feedback: (31)

chimchim

Very nice explantion.

Sprink

Mglavin, servo resolution does not affect torque, it is that they are affected by the same thing.

If say your tranny has max ATV of 100%, and the servo moves 60 degrees ath that 100% setting, then reduing the ATV to 50% will mean that you only get 30 degrees travel out of the servo.

If you then set up the servo and control horn links to give the same deflection at the control surface as before you get the following:

a) 50% resolution: only half of the servo movement giving the same throw means you are missing the possible resolution of the other 50% you are not using.

b) 50% less torque: to get the same movement at the control surface, you will move from say 1:1 relationship to a 1:0.5 relationship. If you originally had 100oz-in, you will now only have 50oz-in at the control surface.

So, they do not relate to each other, but derive from the same source, which gives a misleading link between them.

Ian.W

this proberbly wont help, but visit
http://www.multiplexrc.com/calcservo.htm

Ian

mglavin

My Feedback: (31)

I agree that 50% less ATV will effectively present itself as less overall travel resolution as related to TX stick movement.

TORQUE in this case is a constant and is unchanged by the radial travel arc of the servo. A servo that develops 100oz-in of TORQUE always develops a 100oz-in of TORQUE. It matters not how long the arm is or how far the arm travels.

Torque is calculated by multiplying the arm length by the known FORCE. F(force) x L(length) = T(torque)

Servo TORQUE is greatest when the servo motor is stalled, holding a position.

The quantitative measure of the tendency of a FORCE to cause or change rotational motion is called TORQUE.

Torque (also called a moment) is the term we use when we talk about forces that act in a rotational manner. You apply a torque or moment when you turn a dial, flip a light switch, drill a hole or tighten a screw or bolt.

squareloop

My Feedback: (1)

So the net torque is what is important. I didn't realize that at .75" at both arms would net less torque. Force is gained going in on the servo arm but where it's attached on the control arm is of great significance. Now if I can only remember when to divide and when to multiply the variables in the equation!

Mike (re: surface deflection), using the spreadsheet I plug in 1" at both arms to get a certain deflection of the surface. Then I plug in .75" (or any other number) at both arms and still get the same deflection (?). I had to check because I'm picturing in my mind that the deflection shouldn't change as long as the (D)istance from the pivot points are identical. Am I missing something here? (not a first!)

T_Hill

Michael,

I think sprink is talking about the torque at the surface not at the servo. The servo torque is constant for a constant voltage. The torque at the surface varies as the length of the control horn is changed. In this case he moved the control horn to half the original distance to get back to full surface throw. This will cut the torque at the surface in half. The equation is

T = Fd

In this case the force delivered by servo the the push rod is the same. But you have changed the distance of the control horn to half the original value, so torque is also cut in half.

Tracy Hill

mglavin

My Feedback: (31)

Tracy

Your right, I read right over "at the control surface". Were talking about servo torque and or the net torque calculated by the force delivered from the servo arm in use coupled to the control arm length. FxL=T or in this case the TORQUE available to the control surface.

A 1:0.5 ratio is otherwise known as a mechanical disadvantage, highly undesirable.

mglavin

My Feedback: (31)

Square

You understand correctly, a 1:1 ratio at the arms in question will net equal travel at both ends.

Lengthening the control arm 1/4" past the 1:1 ratio will deliver about 20 degrees less surface deflection. This is a mechanical advantage. Less throw, a little faster and more Torque at the surface.

David Cutler

True, slop is slop, but if the arms are, say, twice the length, then the slop (which doesn't change) is only half the amount it was in relation to the size of the movement of the arms.

Or, to put it another way, suppose your slop is 1/100 of an inch and the arms move 1 inch. If you double the movement of the arms to 2 inches the percentage slop is not now 1 in 100 but 1 in 200 so the angle the control horn makes as it slops is half what it was before, therefore half the control surface movement,

-David C.