Aerohead

My Feedback: (37)

I'm getting ready to start on my H9 Sukhoi and have several Hitec 5945 digitals to use on the control surfaces. I'm wondering what you guys are using for the servo arms, the blue anodized ones that come in the box or something else? What I'd really like is something longer than the blue ones, and something that has threaded holes for 4/40 hardware, like the H9 gold arms, but I don't think they make them that fit these servos, do they? I know that they make them for both Futaba and JR servos, but what about Hitecs? I really am not interested in using the plastic(nylon?) ones from Dubro, I want metal arms, and something long enough to use for 3D throws. Any ideas out there?

mglavin

My Feedback: (31)

Airwild and SWB Mfg. offer 4-40 1" and 1.25" aluminum arms for Hitwc servos.

Mike Rojas

My Feedback: (2)

SWB arms are really nice,I have them on some 8411's and some 5945's.They are red and the best fitting arms around.Check them out at www.swbmfg.com
Mike

Stick Jammer

Ditto on the Airwild arms. The 1" singles are plenty long enough even for 3D.

www.airwildpilotshop.com/store.asp?title=product

Aerohead

My Feedback: (37)

Thanks alot guys, I'll check those out.

Shortman

My Feedback: (21)

Nelson makes some really heavy duty arms

xp8103

My Feedback: (16)

Aerohead,
I am using the Dubro HD arms on all but throttle and rudder. Standard arm on the throttle and a 3" SWB arm on the pull pull rudder.

Mike Rojas

My Feedback: (2)

I wouldn't use the 1" arm for 3d.I don't think they are long enough,and you stand the risk of overextendind the servo and locking it up.I have the 1" arms on the ele. of my H-9 cap,and it needs a little more,JMO.The advantage of the SWB arms is that they have 4 threaded holes so you can move the ball/clevis in or out.The Airwild ams aonly have one hole on the end,FWIW.
Mike

Stick Jammer

lv2rcav8,

If you can shorten up the length of the control horns on your elevators you'll get more throw with the 1" arms. If the control horn length (measured from the center of the hinge line to the pivot point) matches the length of the servo arm, then the servo has to travel 45 degrees in each direction to get 45 degrees of elevator throw. At 45 degrees of servo travel you won't even come close to over extending it.

Mike Rojas

My Feedback: (2)

That sounds good,but I already have the center of trhe control horn 1"up from the surface.I don't want to get any closer to the surface because of flutter possiblity.Stick,I sure hope this doesn't turn into one of those "Bellcrank"threads,I don't think my mind is up to it yet.I will check out those measurements though.Thanks

Stick Jammer

No, I don't think it will! LOL! As long as the control horn length isn't shorter than the servo arm length it won't in itself cause any flutter. All things being equal, putting on a 1/4" longer servo arm has the same mechanical effect as shortening the control horn by 1/4". Maximum available servo force is decreased by the same amount either way. If you can't shorten the control horn enough to get the throw you desire then you'll will need a longer arm.

CrashMeister

My Feedback: (4)

I used Airwild 1.25" arms on every surface on my H9 CAP. I also found the 1" ones a bit short for sufficient throw. Great arms, very tight.

Craig.

mglavin

My Feedback: (31)

1"servo arm (SA) x 1" control arm (CA) nets a 1:1 arm and torque ratio. I'll use a servo rated at 155oz-in for my example.

155ozs of FORCE x 1" control arm = 155oz-in TORQUE realized by the control surface.

1" servo arm delivers 155ozs of FORCE
1.25" servo arm delivers 124ozs of FORCE
1.5" servo arm delivers 103.33ozs of FORCE

oz-in = TORQUE or twisting force at the control surface. The data below is with respect to the arm ratio of the control surface.

1"SA x .75"CA = 116.25oz-in
1"SA x 1"CA = 155oz-in
1"SA x 1.25"CA = 193.75oz-in
1"SA x 1.5CA = 232.5oz-in

1.25"SA x .75CA = 93ox-in
1.25"SA x 1"CA = 124oz-in
1.25"SA x 1.25CA = 155oz-in
1.25"SA x 1.5"CA = 186oz-in

1.5"SA x .75"CA = 77.5oz-in
1.5"SA x 1"CA = 103.33oz-in
1.5"SA x 1.25"CA = 129.75oz-in
1.5"SA x 1.5"CA = 154.99oz-in

Simply changing the arm length's 1/4" in either case does not always provide a 1:1 change in FORCE and or TORQUE as evidenced above.

An ideal set-up is one that would provide "mechanical advantage". As such an arm ratio that nets a CA that is longer than the SA is preferred.

Yakdude

My Feedback: (2)

Mglavin is correct.

I always try to run as long a control horn on the surface as possible that will still net the required control surface throw.


Lenny.

Stick Jammer

I agree with mglavin's numbers and DIESEL's post in regards to always using the longest control horn possible, But if lv2rcav8 has 1" servo arms on his elevators and lets say (just for example) 1.25" control horns. If he put on 1.25" servo arms and left the control horns at 1.25" his ratio would be at 1:1
If he simply shortened the control horn by .25" and left the 1" arm on the servo the result is the same, a 1:1 ratio. Same force available in either case as is shown by mglavin's numbers. I think the last line of numbers is off though:

1.5"SA x 1.5"CA = 199.75oz-in ?

This should be the same value as the other 1:1 examples - 155oz-in

mglavin

My Feedback: (31)

OK, I fixed the typo for 1.5 x 1.5 arms.

lv2rcav8 example consists of a 1" servo arm and 1" control arm.

Adding 1/4" servo arm nets 124oz-in
Subtracting a 1/4" from the control arm nets 116.35oz-in

The assertion that it always remained constant when adding or subtracting the 1/4" is what I had exception to. As I mentioned it is not an ALWAYS thing.

Stick Jammer

No, lv2rcav8's example was a 1" servo arm and he said his control horn was 1" from the SURFACE of the elevator, not the hinge line. I think I mentioned earlier that the control horn should never be shorter than the servo arm or it will create some problems for sure. At any rate, I'm not sure how you are coming up with these numbers but in both of the above examples the servo arm is .25" longer than the control horn. Ratio is the same, torque doesn't change, = same amount of available force in either case.

mglavin

My Feedback: (31)

Stick, your right 1" from the surface is noted. I mistook the 1" as from the hinge line.

If the servo arm and control arm were 1" long. My notes suggest the data above to be correct.

1"SA x 1"CA = 155oz-in, arm ratio = 1:1=1

subtract 1/4" from the control arm, 1"SA x .75"CA : 155x.75 =116.25oz-in, arm ratio = 1:.75=1.33

add 1/4" to the servo arm, 1.25"SA x 1"CA = 124x1=124oz-in, arm ratio = 1.25:1=1.25

All things being equal, putting on a 1/4" longer servo arm has the same mechanical effect as shortening the control horn by 1/4". Maximum available servo force is decreased by the same amount either way.

The above statement is only true if you begin with an arm ratio of 1.25 or 1.25:1. TORQUE will remain constant in this example. The available servo FORCE used to calculate TORQUE realized by the control surface will change with the servo arms length.

TORQUE and FORCE are two different entities.

TORQUE: The quantitative measure of the tendency of a force to cause rotational motion.

FORCE: Power made operative against resistance.
(Physics) Any action between two bodies which changes, or tends to change, their relative condition as to rest or motion.

There are two TORQUE specifications to consider with this info.

1] Servo TORQUE (remains constant no matter what the circumstances). FORCE available at the end of the servo arm changes dependent on servo arm length, but TORQUE remains constant.
2] TORQUE generated by the arm ratio in use to deflect the control surface. This TORQUE is factored by the sum of the servo arms FORCE multiplied by the control arm length and can be ever changing.

There is no absolute here, it all depends on what you have to work with.

Mike Rojas

My Feedback: (2)

See,I told you this was going to happen stick!

Shortman

My Feedback: (21)

man, that is some detailed stuff Mike... very very interesting to see the control horn makes a big difference

Stick Jammer

mglavin,

I'm not sure that you and I are looking at the final drive ratio in the same light. The final drive ratio is calculated by comparing the degrees of servo rotation to the degrees of surface deflection it causes, not the length of the servo arm compared to the length of the control horn. I believe my statement is correct in regards to lengthening the servo arm 1/4" or shortening the control horn by 1/4", same effect. It really doesn't matter if the servo arm is 1" and the control horn is 3/4", or the servo arm is 12" and the control horn is 11 3/4". If there is 1/4" difference between the two, the drive ratio would be the same. Granted the amount of force available would change, but available force is not what's important here, the ratio is. The drive ratio is what determines how much or how little the servo will be able to overcome the flight load. In this extreme example of a 12" servo arm, obviously the maximum available force at the tip of it will be substantially less than a 1" arm but the 11 3/4" control horn is substantially easier to push or pull against the flight load.
Just like a longer wrench breaks a stubborn nut easier than a short one. The lack of force is made up by the increased lateral movement of the longer arm and horn, it all stays relevant to the ratio of servo travel to surface travel.

lv2rcav8, I guess you were right!

mglavin

My Feedback: (31)

Stick

"Final drive ratio" was not mentioned previously, however "RATIO" was and to be honest it's a new one on me and it makes sense. I am familiar with arm and torque ratio's for this application. Can you explain how we use the FD ratio to make comparison's of the ability to overcome flight loads. What units are used to factor the relationship? A little supporting math would be nice. How about we use a 155oz-in servo. Or is it as simple as looking at the FD numbers and sooting for the lowest FD ratio that works on the model?

I concur on the 1/4" difference being one and the same with regard to FD ratios. In fact this is true with any denominator, with respect to FD ratio. You did mention maximum servo force remained constant though. I assume this is in relation to the FD again? We both seem to concur the arm ratio does effect torque and force as realized by the control surface, in of itself.

Arm ratio remains a constant if the arm spacing is 1/4" apart to begin with, and you add 1/4" to the SA or subtract a 1/4" from the CA.

For comparison's I did a little calculating and came up with this FD ratio. I'm making some assumptions. I used the actual CA spacing as measured on my son's H9 Sukhoi. This is pretty typical 33% model setup, as far as SA arm length's go.

CA is 1.5" hinge center line to pivot point.

1.5CA/.75SA = 31/45 degrees of travel : FD=.69 AR=2.0
1.5CA/1.0SA = 31/45 degrees of travel : FD=.69 AR=1.5
1.5CA/1.25SA= 36/45 degrees of travel : FD= .8 AR=1.2
1.5CA/1.5SA = 45/45 degrees of travel : FD= 1.0 AR=1

Stick Jammer

The drive ratio would be the degrees of surface deflection divided by the degrees of servo rotation. Like we've said before, a 1" servo arm linked to a 1" control horn (or any setup where they are the same length) would be a 1:1 ratio because for every degree the servo rotates, the surface is deflected one degree. As we both agree, the control horn should never be shorter than the servo arm or it does create some problems in that the servo now has less ability to overcome the load. Let's use the example of a 1.25" servo arm and a 1" control horn. In this setup, 45 degrees of servo rotation creates 60 degrees of surface deflection which is a ratio of 1:1.34
This means that for every degree the servo rotates, the surface deflects 1.34 degrees.....obviously not desirable. On the other hand, if the servo arm was 1" and the control horn was 1.25" the ratio would be 1.34:1 For every degree of surface deflection, the servo now must rotate 1.34 degrees....very desirable. The second example is now able to overcome a greater load than the servo is actually rated at because it must travel farther to get the same amount of work done. I wish I had a formula to figure the ratio for any given arm lengths, but I don't. These examples were drawn on paper to come up with degree values. I'll keep working on a formula because one has to exist. You are correct to conclude that any scenario of arms that produces the same ratio between servo rotation and surface deflection will give the servo the same ability to overcome a given load. You are also correct in that all of this simply means to set up the linkage that gives you the highest ratio of servo travel to surface deflection without overdriving the servo. You want the servo to rotate more degrees than the surface is actually moving. Keep in mind all of this only applies if the control horn pivot point is on the hinge line where it should be.