- #1
Kalagaraz
- 28
- 0
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 converted 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?
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 converted 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?