Ok, looks like I can barely get about 40deg of throw on my elevator after a lot of trying and 1.25" servo arms. I've found that swb makes a 1.5" servo arm but its 'not suggested for aircraft use', anybody use these, why not for aircraft, and I imagine I should use 5945's to drive them, no?

Thanks,
Sam

The 1.5" arms work BUT you have to be careful with them as you loose 50% of the FORCE the servo can drive. Therefore Scott does not recommend there use. If you have multi ganged surfaces you can likely use them. What is the application, control arm length from hinge line to control link pivot point.

Another consideration is the Hitec digital can travel farther than other servos, they have a maximum travel arc of 180 degrees compared to 120 of all the rest. BUT it's not all usable as you get deminishing returns as you get close to 80 degrres in one direction.

I rarely have a problem obtaining maximum throw with the Hitec's and 1.25" arms...

finaly got my camera working hitec 605 high torq is the best way to go and use some polly carbon methol servo arms here are some pictures. the white servo arm is an axis extender but the servo am itself is will give you the 3D throws you need to do extreem manuvers you want. good luck

Yes, its for my elevator servos in a 27% extra, one servo for each elevator half at 6v and I currently have HiTec 5925's mounted (126oz torque) though I could mount some 5945's (180oz torque). The servo arm length is the last variable I can modify to get more throw, everything else is at the most optimal position for maximum throw.

o i see its an extra your mounting it on

Yah, I have a funtana too and 1.25" arms was _plenty_ on it

Servo arm lenght is not so much of a factor.
You'd better see it this way : If the servo moves 60 degrees each way and the control surface moves 60 degrees each way, you are getting pretty close to the limits of resolution and holding torque.

The above assertion it true but there are caveats which displace the assertion.

There are other factors in play, in this case the primary factor which would dictate 60=60 is the control arm length of the elevator or surface in question. Generally speaking we do not see a 1:1 ratio in this area, in fact it’s undesirable. We'd prefer to see an uneven ratio that would promote mechanical advantage. If the servo arm length is equal to the control arm length we see 1:1 ratio if the servo arm is 1.25" and the control arm is 1.5" we realize mechanical advantage at the cost of surface deflection. This small concept plays a factor in the FORCE passed to the surface from the servo, in this example FORCE will be greater than that provided by the servo.

188in-oz servo with a 1.25" arm coupled to 1.5" control arm will net 225ozs of FORCE to the surface, TORQUE remains constant. Mechanical advantage is in play...

This maybe an unrealistic with this specific model as the control arm maybe much shorter. 1.25"/188oz-in=150ozs Force, coupled to a 1" control arm you'll see 150ozs Force at the control surface, NO mechanical advantage.

I've got 2 slightly more difficult questions for you.

1. Instead of 1 inch arms on each end 1:1 suppose you go with 1/2 inch arms on each side. Do you then increase the torque available in the 1:1 ration by 50%. Or, overall is it still the same as the oz/inch rating for the servo.

2. Doesn't the force the servo exherts lessen as it nears the end of its range due to the force vector no longer being 90 degrees to the pushrod.

If you opt for 1/2" EACH END, you get exactly the same results in terms of actual torque transferred to the control surface. The only advantages of using smaller control horns are maybe price, and the fact that you have a more compact setup, which in most cases is not needed. There are some disadvantages of using smaller control horns. The most important IMO is that the smallest slop in the linkage will be much more apparent so higher quality hardware is a must so price is no longer and advantage. Another thing to consider is the force carried by the pushrod. If you have a 100 oz.in servo with a 1/2 control horn, you get 200 oz. of FORCE in the pushrod. On the other end, you have 200 oz. of FORCE that is converted to 100 oz.in TORQUE through a 1/2" control arm.

In short, it is not the servo arm that produces torque, it is the motor and gear ratio in the servo that produce torque so no matter the servo arm you put, you will all the time get the servo rated torque.

The force does not lessen, as the control surface horn is also in the same situation. The pushrod will work harder on full deflection than neutral so it makes it even more important to use the largest servo and control arms as practically possible.

All the above is said considering a 1:1 ratio between servo and control surface.

Thanks for the technical description, now, how does one tell if they're overloading their servo (before the servo fries)? Given that even at full 140% deflection of near 90deg at the servo I only get 40deg at the surface I'm not running at 1:1 and the servo is getting the advantage, and using that ratio with the 5925's rating I get 238oz of torque (about 17lbs). NOTE: the axis of rotation at the control horn isn't over the hinge, its a fraction of an inch behind, so you can't really use the length-of-arm:lengh-of-horn ratio, you have to use the deg-of-rot:deg-of-rot as I used above. So, given that, its stands to reason that I could move up from a 1.25" arm to a 1.5" arm and the servo would still have the advantage, no?

Sam

Are the servo arm and pushrod square with each other when the servo is neutral? Same for the pushrod and control surface horn?

If not, you can loose mechanical advantage frightening fast.

Actually, they're not, the servo-rod connection is about 3/4" down from the horn-rod connection. I found some interesting Excell spreadsheets for calculating force and they take into account if the connections aren't square so I'm going to try calc'ing those out. All in all the elevator-servo setup is very not ideal on this plane, they should have located the 'servo-hole' closer to the elevator so the connections could be more square.

Sam

Unless I misunderstood the questions/assertions the numbers favor a slightly offset alignment. Whereas the servo arm and control arm are not 90 degrees to one another. If while at neutral the arms are 90 degrees to one another you realize exaggerated numbers in one direction as opposed to the opposite direction of travel.

Sizam

How are you netting 90 degrees of servo arm travel? Is this the total number of the travel arc lock to lock or in one direction?

If you'd prefer to forgo all the arm length, offset and what not math, there is a down and dirty method of calculating the realized numbers...

Simply accurately measure the degrees of rotation of the servo arm in one direction from neutral, and then measure the control surface deflection degrees of rotation. Then divide servo arm numbers by the latter and then multiply by the servos rated torque...

So if your getting 45/40=1.12, 1.12x100=112oz-in

Oh, no, it was in each direction, it turns out its 80 deg each way (or 160 total) so given your math thats 80/40 = 2 x 126oz/in = 252oz/in. But I still need to determine how much the surface is pushing back, still 2:1 seems pretty good.

What servos are you using? The only servos that can move more than the typical servos 60 degrees are Hitec's programmables.

You must have some long control arms? This meauremnet would be the cumulative total length from the hinge center line to the control arm pivot point.

Your servo arm length is?

Surface pushing backs means?

These are digital hitec HS5925's, the control arms are currently 1.25" but I'm considering 1.5" arms to try to get some more throw, the distance from the hingeline to where the control-horn->pivot-point is 1.75" and thats as small as it can be since the elevator is airfoiled. By 'Pushing back', I ment how much force is deflecting the elevator going generage and 'push back' on the servo. Its nice to know that my servo has effective torque of around 250oz/in but thats meaningless unless I can figure out how much the elevator is going to torque the servo, no?

Thanks,
Sam

What model are you working on and how heavy is it?

Its the CA 27% extra, I built it 'SX' so there are counterbalances on the elevator, about 2", should come out to 12-13lbs.

you'll be fine on this model.. I have used mid sized servo with a less torque several times without issue with these smaller models...

There are some online programs for calculating servo requirements. Try doing a serach you should come up with something.

Have fun.

"Smaller", hah!

Carbon Copy in the UK will be bringing out some carbon 3D servo arms soon. They are CNC cut from carbon sheet.