right electric motor for this...?



No, it is 2 or 3 linear strokes per hour, for example once up and once down. You are locked into one mechanical design where strokes = revolutions, but there are others. Obviously you ignored my reference to a lead screw.
For example... How many revolutions of a 1/4-20 threaded rod would it take to move a nut 36 inches? To do that in 30 minutes (for one complete up-down cycle per hour), at what speed would you rotate the threaded rod? Can an induction gearmotor rotate at that speed?
Now couple the rod to the gearbox shaft, put limit switches at each end, attach the nut to a cable, hang his rod from the cable over a pulley, etc. Of course, in actual practice I might use a leadscrew assembly with ball bearings.
Ben Miller
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Benjamin D. Miller, PE
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Phil Scott wrote:

For that method. But the op did not specify that method. Run a cord over a pulley at the ceiling. One end of the cord attaches to the pole. The other end attaches to a drum, with a circumference of 1/2". Rotate the drum at 1 rpM(inute). In one hour, the drum will reel in (or let out) 30" of cord, meaning the pole will travel 30".
Or drive a 16 TPI linear actuator with an 8 rpM motor. Same distance - 30" in an hour.
Ed
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If the pole weight is a constant, then a counter-weight would seem in order.
daestrom
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message

good idea... I hadnt thought of a counter weight.. that would allow the use of a small timer motor to do the job.
Phil Scott

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snipped-for-privacy@gmail.com wrote:

You need a speed reduction mechanism, not a super slow motor. Go here to get a sense of the various types of gearing: http://www.bostongear.com/products/enclosed/faqs.html
Take a look at the sites below. You didn't specify the weight of the pole - so maybe something small is all you need to raise/lower it.
http://www.hobby-lobby.com/gearbox.htm http://www.imagesco.com/catalog/motors/index.html http://www.hobbyplace.com/robotics/gearbox.html http://www.scientificsonline.com/product.asp_Q_pn_E_3041860 http://www.hobbyengineering.com/H2031.html
Here's a circuit to drive the motor (assumes 12V motor):
| <|Ry1-1 N/O | | +12---+----+----o +--------+ | | | | | <| MS U |>| MS D | | N/o | N/C | | | | +-----------+---[Ry1]---+--- Gnd | | +------------+ +-------------+ | | | | | N/C|>| <| Ry1-2 | | | | | [Motor] | | | | | N/O|> |<| Ry1-3 | | | | | +-----------+ +--------------+
MS = MicroSwitch (U = Up, D = Down)
When the follower reaches the up position, MS-U closes its N/O contact, which energizes the relay. That reverses the polarity to the motor, so the follower begins to move down. The relay remains energized through the Ry1-1 N/O contact and the MS-D N/C contact, until the follower reaches MS-D and opens its N/C contact. The relay drops out, the motor reverses, and the cycle repeats.
Ed
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