Need sources for pot with planetary drive

i remember these used to have slippage problems at the ball to shaft (after a few hundred rotations) and required feedback looped mechanical drives to accurately position anything they were used to control.

try something new!

at

Wow, I had seen those in the past but not for many many years.

If you are truly updating some rotators, there are some articles in QST, CQ, 73, etc. from the past 20-30 years on doing it using something other than 360-degree pots. Some of them use essentially homebrew shaft encoders.

Tim.

I remember them, also, and the slippage. I used to lube them to make them work better. They didn't.

I now know they must never be lubed and must be kept really clean. They work because of the friction between the shaft, the ball and the housing. Our pick-and place machine, used to place SMT components on circuit boards, uses the same technology to convert rotary motion to very accurate linear motion. The wheels and shaft must be kept very clean, or we get positioning errors.

I have no clue about the current source of the devices, but probably have a few in the junk box!

Paul, KD7HB

agreed, i have used pc mouse encoders with a hacked mouse circuit board for various projects.

2 axis, three butttons , plenty of easy to use software, even write your own visual basic

can even determine direction and speeds!

its all simple enough and uses standard rs232 ports.

mouse click buttons can be used to determine limit positions at extreme points

the newer mice with some 6 or 7 variables can do lotsa stuff!

at

try ebay search "ball potentiometer"

msg wrote in news:139f22h91ghhna7 @corp.supernews.com:

Do you mean the rotator or the controller?

For the rotator, you can maybe use a standard pot with gearing. I made an old 3 wire rotator a 5 wire with a pot an a 4 to 3 gear. One gear was fixed to the output shaft, the other to the pot, so that one full rotation (360 degrees) of the rotator shaft turned the pot its full path, or therabouts (270 degrees). The controller just had a typical 270 degree pot and a 3/4 scale that rotated N-E-S-W-N.

at

Interesting. I've never seen a pot like this.

So the turns ratio is determined by the shaft diameter to ball diameter ratio? And the wiper arm is connected to the ball carrier? So this assumes the metal to metal, shaft to ball won't slip, but the ball will turn in the carrier, causing the carrier to have 1/3 the rotation of the shaft?

Sorry, this doesn't help you at all... Alan Nishioka

I had considered gearing but the available space and the necessary additional fabrication was more involved than the solution I chose.

Here are photos of the modified rotor:

This rotor is common to a number of makes and models; the controller that came with it was for the Cornell-Dubilier AR22R, an interrupter based spring and escapement controller version.

The modifications permit continuous rotation and rely on the controller to manage limits.

Regards,

Michael msg _at_ cybertheque _dot_ org

Thanks to all for replies so far.

Regards,

Michael msg _at_ cybertheque _dot_ org

Jackson Brothers

Makes some excellent planetary drives. There are dealers in the USA.

Bill W0IYH

The ratio, is determined by the shaft diameter, to the internal diameter of the casing in which the balls sit. If (for instance), you have a 3mm shaft, and 10mm balls, the casing has to have an internal diameter of 26mm, and the ratio is 26/3 = 8.66:1. The drive can be made remarkably good (the same system is used on some micro focussers for telescopes, which carry significantly more torque than needed for a pot). Units like the 'Williams optics feathertouch focusser', use exactly this drive.

Best Wishes

================= Sorry ,but I don't understand the above . If the shaft has a diameter of 3mm and the balls a diameter of 10mm, the inner diameter of the casing should be 10 + 10 + 3 equals 23 mm

When the shaft makes 1 revolution the balls will make 3/10 revolution . When the balls make 1 revolution the casing makes 10/23 revolution . Hence when the shaft makes 1 revolution the casing will make

3/10 * 10/23 equals 3/23 equals 0.13 revolution , hence reduction factor is 7.7 Please correct me if I made an error.

Frank GM0CSZ / KN6WH

I could be missing something, but at first glance I read the radius as the diameter of the 10mm ball (10mm) plus half the diameter of the 3mm shaft (1.5mm) and get 23mm internal diameter, ratio of 23:3 or 7.66:1.

Just in case this turns out to be a corecktion I've included speling erors to comply with tradishun.

--
Postulate a group whose intent is to destroy the United States from within
via anarchy and bankruptcy. The actions of the United States Congress are
completely consistent with the actions one would predict from such a group.

Yes. Sorry, I just added up the internal diameter wrong....

It is fairly simple to visualise what is happening, if you realise that the balls are rolling round the internal diameter of the casing, driven by the small shaft at the centre, and for the assembly of three balls to complete one complete turn, their surfaces, must travel the internal diameter of the outer casing. Since the surfaces of the balls are driven by the internal shaft, you get the ratio from the ratio of these two diameters.

Best Wishes