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Solenoid Force Requirements

  1. Apr 17, 2013 #1
    So I'm building a solenoid that needs to move a load of 5 ounces, 0.4 inches.

    I'm pretty sure my calculations for the force inside the coil on a iron pin are correct:

    F = (B^2 * A) / (2 * Uo). Where B is the flux density, A is cross sectional area of pin Uo is permeability of free space.

    I calculate B by finding the flux through a magnetic circuit consisiting of "housing", "pin", and "air gap" between pin and end of coil/housing. The flux I then divide the cross sectional area of pin to get B, flux density.

    While I calculated differently, I get close to same result as here (http://easycalculation.com/engineering/electrical/solenoid-force.php). Mine is probably slightly more accurate cause I take into account permeability of housing and pin and such. (I calculated 4.48 newtons, web calc said 4.473).

    Now the mechanical parts confuse the hell out of me. I need a spring that returns the load to it's original position when the solenoid is off. How much force do I actually need to move the pin? At first I assumed 5 oz load, 5 oz spring to return the load. = 10 oz required force. I now know this is wrong. I need to some how work in F = ma. but when I coverted ounces to kilogram mass and then put in an acceleration of 0.4 inches per second per second (so pin would pull in less than a second) I got an extremely low force, and the solenoid would only need 29 turns at 0.5 amps to generate that?

    Can anyone give me a push in the right direction?
     
  2. jcsd
  3. Apr 17, 2013 #2

    sophiecentaur

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    The order of your calculations is important here, imo.
    I think you need to start with the spring / pin / pin weight combination (solenoid turned off) and returning the pin fast enough This will give you the required spring constant. Then you can work out the force needed from the solenoid, to overcome the chosen spring and to lift the pin (fast enough for the purpose, of course). At least, that way round lets you work things out one at a time.
     
  4. Apr 18, 2013 #3
    Can you give me some rough guidelines as to the expected number of turns though for say moving a 1 pound load at say 0.4 inches per second? That way I can see if I am on the right track with my results?

    When I calculate for low accelerations like 0.2 inches per second per second. I get EXTREMELY low force requirements to the point where I would only need like 30 turns of wire to do it. Just doesnt seem right cause I see normal solenoids have hundreds of turns.
     
  5. Apr 18, 2013 #4

    sophiecentaur

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    Your question is really not well enough defined. This link* gets it in perspective, I think. You need to specify more than you have done if you want a specific design.
    Like I said, earlier, you need to work out the actual force needed to overcome the return spring - which first needs you to specify the constant of that spring for the particular pin mass you are planning to use. If you don't know that, you don't know the force that your solenoid needs to produce.
    The last thing you need is to have a solenoid which won't work against the spring or a spring that won't retract the spring fast enough. You need to get it right first time, if possible. So start in the right place in the design loop.
    *From a google search "solenoid design"
     
  6. Apr 18, 2013 #5
    I've never seen a store selling springs with a "spring constant" listed.

    http://www.centuryspring.com/Store/search_compression.php

    So if I figure out the spring constant I need, the length of the spring, how do I find a spring that matches these parameters?

    EDIT : I'm guessing the Rate (lbs/in) is acctually the spring constant? F = kx. Rate = F/x?

    Haven't yet recevied the material to cut the pin yet so not exactly sure what it's mass is yet.

    So to figure out the force of solenoid though, here is order or equations correct?


    1. Calculate force needed to return the pin. This would be F=ma where m is the mass of the pin and "a" my desired acceleration. I'd also add the friction force which the sliding surface is nylon so 0.25 (friction coefficient) * normal force (weight). As the spring would need to overcome the static friction before I have a net force to produce acceleration.

    2. Calculate the solenoid force to equal same acceleration as before, same static friction force, + the additional force to overcome the spring.

    3. Then calculate the turns/current needed to generate that force.
     
    Last edited: Apr 18, 2013
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