Help - 24VDC linear actuator

Discussion in 'The Projects Forum' started by ubuntu34520, Apr 8, 2013.

  1. ubuntu34520

    Thread Starter New Member

    Apr 8, 2013
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    Hi Everyone,

    I am new to electronics and have a question about a linear actuator I was thinking of using in a project. The specs for the actuator that I could find are below:

    Stroke Nominal 18"
    Push/Pull Rating 665 lbs. max
    Voltage 24 DC
    Amps 0.5 no load
    Duty Intermittent
    Avg. Extension Speed 0.3"/sec.

    This is all I can find. I've determined that I need about 60W of power. Is the power output of linear actuators controlled by maintaining a constant voltage and changing the current? So I could input 24V and 2.5A and get the power I need? Also, does anyone know what the push/pull rating means exactly? Is this an indication of the maximum force the motor can supply? Could I multiply the push/pull rating by the extension speed to get the maximum power output with the right unit conversions? I need this actuator to lift about 150 pounds, would it be capable of doing so assuming I have an appropriate power supply?I'm sorry if any of these are dumb questions, but I have very little experience with actual electronics and pretty much only know the theory.

    Answers to any/all of these questions would be greatly appreciated! Thanks!
     
  2. Brownout

    Well-Known Member

    Jan 10, 2012
    2,375
    998
    You can limit the current to control the output power, but I can't say if your numbers add up. Once you start limiting the current, the voltage will change. Do you really need to limit the power?

    Yup, it's exactly that.

    That would yield power in units of lbs*in/sec. Watts can be expressed in lbs*ft/min, so you would need to convert by dividing by 12 and multiplying by 60 ( or else just multiplying by 5 ). The conversion I've used is 1HP = 22,000lb*ft/sec ~= 750W.

    However, the speed might be considerably less at maximum lift. You might need to experiment to get the real numbers.

    It should, since it can pull 665 lb. Just be carful to not get any sideways pull. It's made to push/pull straight on.

    No need to apologize. You should read some of the dumb questions I ask:D
     
    Last edited: Apr 8, 2013
  3. ubuntu34520

    Thread Starter New Member

    Apr 8, 2013
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    Thanks a lot Brownout, your response was very helpful. For the first question, I guess I don't necessarily need to limit the power. I just want to know exactly what the voltage and current draw are going to be so I don't mess up the microcontroller I'm using to control it. As you can see from the specs, the only current value given is at no load. Since I am going to be putting a load on the actuator, I will need more than .5 A of current. Will the power output of the actuator increase linearly with increased current (since power = voltage * current), or is the voltage not always going to be 24V?
     
  4. Brownout

    Well-Known Member

    Jan 10, 2012
    2,375
    998
    First of alll, I had to make corrections to my first post, so look it over again if you're planning on using any of my information. You should be able to design a system to hold the voltage at 24V, so current should increase linearly.

    Youre requirements of 60W equates to 60/750 = .08HP. But I calculated the power available to be somewhere in the neighborhood of .045HP. (665*.3*5/22,000 = .045)
     
  5. ubuntu34520

    Thread Starter New Member

    Apr 8, 2013
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    Can you explain how you calculated that power figure? Where did you get the 5 and 22000 from? I would think it would be 665 lb * .3 in/sec *(1 foot/12 inches)*(1 HP*sec/550 ft*lb). The 60W is not a hard requirement; I need to be able to lift the weight but the speed could potentially be lower.
     
  6. Brownout

    Well-Known Member

    Jan 10, 2012
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    The '5' I explained in post #2.
    The 22,000 I also showed in post #2. 1HP = 1ft*22,000lb/1min.

    EDIT: I may have that wrong, 1HP = 33,000ft*lb/sec. Or by your calc: 665lb*.3in/sec*12 * 1HP/550ft*lb/sec = .03HP. Your calculation converts HP to ft*lb/sec where mine converted in*lb/sec to ft*lb/min. Same result by different path, though my original calculation was off by a factor of 22/33.
     
    Last edited: Apr 9, 2013
  7. ubuntu34520

    Thread Starter New Member

    Apr 8, 2013
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    Oh I see thanks for clarifying. So if I increase the power by increasing current, will that alter the speed or force output? Or would both increase slightly?
     
  8. Brownout

    Well-Known Member

    Jan 10, 2012
    2,375
    998
    Basically, you increase the torque, which in turn increases the speed, so you're doing both. Just make sure you don't increase the current to the point of burning up you moter.
     
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