Linear Actuators

Dear Simon Griffiths:

Rig it like you would have it in your sculpture. Orientation and load. Cycle it a few hundred times. If you are satisfied, he did good.

Once the structure is proved out, the next concern will be allowing the motor sufficient air flow to stay cool. Or keeping the cycle rate (how many times a day you operate it) low enough that it can conduct to ambient through your sculpture.

David A. Smith

I used to buy linear actuators with 36" stroke and rated at 1500 lb. The motor was about 2-1/2" or 3" in dia and about 6" long (it moved slow as molasses). So 12-14 lb with your actuator is certainly within the realm of possibility.

If you had a manufacturer and model number you could probably find all the info you need on the internet.

Lance

*****

The mechanical power from a linear actuator will be about 40% of the electrical power it consumes. But I suspect you are interested in the force available.

Roughly estimate it like this:

Motor electrical power (from the plate) times 0.4 = mechanical power (still in watts) Mechanical power is force times speed using force in newtons, speed in meters/second.

Here's a hypothetical example: Linear actuator rated 24 volt 6 amps Electrical power input = 24 X 6 watts = 144 watts Mechanical power available is 0.4 X 144 watts = 58 watts That could turn out to be, depending on the gearing:

58 newtons at 1 meter/second 232 newtons at 0.25 m/s or 580 newtons at 0.1 m/s

A newton of force is 2.2/9.8 pounds force = 0.22 lbs A m/s of speed is 2.2 miles/hour

So the kind of force and speed product that 144 watts of electrical input gets you are these:

58 X 0.22 lbs at 2.2 mph = 12.7 lb at 2.2 mph or 50 lb at 1/2 mph or 127 lb at 0.2 mph

Brian Whatcott Altus OK

Thanks for the explanation Brian... I've been working on a small project to swivel a TV stand and was trying to estimate the amperage requirements for my power supply. Your example is just what I needed! :)

- Siddharth Vajirkar

Brian Whatcott wrote:

You're most welcome, Sid

Brian