CIM motor + Electrical outlet = ?

[Ryutso]

I just wanted to know if the CIM motor could be hooked to the wall outlet natively or if something could be rigged to run the CIM motor from a regular 120V AC 20A household electrical outlet?

I'm planning to invest in some CIMs for a home made project.

[MrForbes]

No!

And a power supply to provide 12v DC at enough current to do something useful with the motors, will be kind of expensive.

When you ask a question like this, it helps to provide a lot more info about what exactly you want to do, so others here can help you figure out some other possible ways to do it.....

[artdutra04]

Originally Posted by Ryutso

I just wanted to know if the CIM motor could be hooked to the wall outlet natively or if something could be rigged to run the CIM motor from a regular 120V AC 20A household electrical outlet?

I'm planning to invest in some CIMs for a home made project.

A CIM motor is rated for 12V DC (direct current). Electrical mains voltage is 60Hz 120V AC (alternating current).

To run a 12V DC motor off of 120V AC mains, you'd need an electrical transformer/power supply to both convert alternating current to direct current and to step down the voltage from 120V to 12V.

The stall current of a single CIM motor is 133 amps at 12v, which means it's using 1596 Watts of electrical power (but at this point is outputting zero watts of useful mechanical power, at the expense of becoming an excellent space heater). This means the power supply you choose to run the CIM motor must be capable of sinking at least 1596 Watts of output power. Buying a 120VAC->12VDC power supply capable of outputting at least 1596W will put a several hundred dollar dent into your wallet.

[Chris is me]

Originally Posted by artdutra04

The stall current of a single CIM motor is 133 amps at 12v, which means it's using 1596 Watts of electrical power (but at this point is outputting zero watts of useful mechanical power, at the expense of becoming an excellent space heater). This means the power supply you choose to run the CIM motor must be capable of sinking at least 1596 Watts of output power. Buying a 120VAC->12VDC power supply capable of outputting at least 1596W will put a several hundred dollar dent into your wallet.

Can't you put an circuit breaker in between the DC converter and the CIM?

[Ether]

Originally Posted by Ryutso

I just wanted to know if the CIM motor could be hooked to the wall outlet natively

No, but you could use a battery from a car, motorcycle, lawn tractor, golf cart, or boat. Or you could visit several HVAC shops in your area and ask if they would allow you to salvage blower motors from some junk furnaces. They run on house current.




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[AdamHeard]

Originally Posted by Chris is me

Can't you put an circuit breaker in between the DC converter and the CIM?

I don't follow what the point of that would be.

[Chris is me]

Originally Posted by AdamHeard

I don't follow what the point of that would be.

If the circuit breaks at 40 amps, you wouldn't need a AC-DC converter that could output 1600 watts.

[sanddrag]

I have run a CIM from a high quality 500W computer power supply. You need to look at the specs though, as most computer power supplies provide more current on the 5V side than on the 12V side.

[Ryutso]

Originally Posted by squirrel

When you ask a question like this, it helps to provide a lot more info about what exactly you want to do, so others here can help you figure out some other possible ways to do it.....

Well what I planned to do was use 4 similar motors to drive a winch of sorts, that would lift my LCD TV up and down a stationary pole that I have in my room.

I asked around and found that the CIM's would lift the appropriate 25 pounds that my TV and 4 CIM's weigh. The only problem is, I don't want to have a car battery sitting at the foot of my bed.

Is there another electric motor that I can cluster together to lift the 20 pounds of my TV and 3 others of itself just running from 120V AC current?

EDIT: I also plan to control them with DPDT switches and would like to know if this (http://www.radioshack.com/product/index.jsp?productId=2062516&numProdsPerPage=60&y=1 0&x=13&retainProdsInSession=1) is the right switch for the job.

EDIT2: I also plan to wrap the whole controller in a project box so I'll need some tips on soldering the wire to a power cord.

[PAR_WIG1350]

how are you attaching this TV to the pole and what kind of pole is it?

Also, is this worth the cost of 4 of any decent motors? why not use pulleys and a cleat from a flagpole?

EDIT:to solder a wire to a power cord it might be best to put some type of breaker/ fuse block in between

[Ryutso]

Originally Posted by PAR_WIG1350

how are you attaching this TV to the pole and what kind of pole is it?

Steel pole and I was thinking of designing something like the pole locking mechanisms from this years game, only instead of being one pneumatics, it's just permanent.

Originally Posted by PAR_WIG1350

Also, is this worth the cost of 4 of any decent motors? why not use pulleys and a cleat from a flagpole?

Cause it'd be cooler to use motors to move it up and down the pole.

[Vikesrock]

Originally Posted by Ryutso

Well what I planned to do was use 4 similar motors to drive a winch of sorts, that would lift my LCD TV up and down a stationary pole that I have in my room.

I asked around and found that the CIM's would lift the appropriate 25 pounds that my TV and 4 CIM's weigh. The only problem is, I don't want to have a car battery sitting at the foot of my bed.

CIM peak power = 321 W
321W * 4 = 1284 W
Assuming gearbox efficiency of something extremely conservative like .5
1284W*.5 = 642W

25 lbs * 10 ft (tallest most bedroom ceilings would be) = 339 joules
Power = Work /Time
Time = Work / Power
339 Joules / 642 W = .53 seconds

This means that at half of peak power 4 CIMS will lift a TV of 0 height from your floor to your ceiling in .53 seconds. I guess my point is, how did you decide you needed 4 CIMs for this job?

[Ryutso]

Originally Posted by Vikesrock

CIM peak power = 321 W
321W * 4 = 1284 W
Assuming gearbox efficiency of something extremely conservative like .5
1284W*.5 = 642W

25 lbs * 10 ft (tallest most bedroom ceilings would be) = 271.16 joules
Power = Work /Time
Time = Work / Power
271.16 Joules / 642 W = .42 seconds

This means that at half of peak power 4 CIMS will lift a TV of 0 height from your floor to your roof in .42 seconds. I guess my point is, how did you decide you needed 4 CIMs for this job?

Oh sorry. It's not 4 CIMs doing all the lifting. I'm going to arrange them diagonally like:

\ /
/ \

around a rubber ball so that I can move it up and down, but also swing it around the pole to differ in direction. So something like this:

Bold Red means it's powered:

\ /
/ \
Moves up

\ /
/ \
Moves down

\ /
/ \
Moves right

\ /
/ \
Moves left

[Ether]

Originally Posted by Vikesrock

25 lbs * 10 ft (tallest most bedroom ceilings would be) = 271.16 joules

250 ft-lb = 339 joules


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[Vikesrock]

Originally Posted by Ether

250 ft-lb = 339 joules


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Oops

I used 8 ft, then thought it would be better to use 10 and apparently didn't go back and fix the math.