What speed v(t) enables constant power by solenoid pull?

  • Thread starter cairoliu
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  • #1
cairoliu
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Assuming the position x = 0 if no air gap, max stroke of plunger is L.
Bordery conditions: x(t=0) = L, x(t=T)= 0.
Given magnetic force F(x) = kx(t)^-2, here k is constant.
If uncontrolled, there is big crash noise afer solenoid actuated, so I wish it pull plunger at constant power.
Help me find the dynamic position x(t), then I can get velocity.
I know Wolfram software may resolve it, but I have no money to buy it.
 

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  • #2
jack action
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Bordery conditions: x(t=0) = L, x(t=T)= 0.
Given magnetic force F(x) = kx(t)^-2, here k is constant.
If uncontrolled, there is big crash noise afer solenoid actuated, so I wish it pull plunger at constant power.
If x(t=T)= 0, then F(x(t=T)) = kx(t=T)^-2 = k(0)^-2 = ##\infty##. How do you expect not having a big crash noise at t=T ?
 
  • #3
cairoliu
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If x(t=T)= 0, then F(x(t=T)) = kx(t=T)^-2 = k(0)^-2 = ##\infty##. How do you expect not having a big crash noise at t=T ?
Now that noise not good, I can reserve a minimal air gap 𝛿, then x(t=T)= 𝛿, and still ask for the solution of constant power.
 
  • #4
jack action
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How is constant power helping you? If you have power ##P## at the end of the stroke, then if ##F = 0##, ##v## must be infinite, or if ##v = 0##, then ##F## must be infinite. Any way you choose, you will have a lot of energy to disperse rapidly when the plunger suddenly stops.

To avoid the big crash noise, you would normally remove the energy - with a damping force - before it reaches the end. Which means power would gradually go to zero.
 
  • #5
hutchphd
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Or a spring, depending upon duration.
 

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