Power equation in railgun operation

In summary: I don't know what the right name is.If you haven't studied basic EM (like Faraday's law, etc.), basic differential equations, and LCR circuit analysis, I think you might not really be ready for the "how does it work question".I could understand how motors, generators, transformer, capacitor, faraday's law, etc. work.As for railguns, since there cannot be a simplified "layman" answer, I don't think I am ready to ask the question.I don't think it's too different than a DC motor. If you really understand motors that spin, I'm sure you can figure out linear motors and railguns. The basic concepts are mostly the same
  • #1
avicenna
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In a capacitor discharge through a plain resistor, the capacitor power supplied at any instant is VI; the power dissipated in the resistor is I²R. So VI = I²R.

Consider a railgun operated with a capacitor bank. At any instant of capacitor discharge, the power supplied is VI. The total power supplied for ohmic loss is sum I²R for two rails plus the resistance of the armature.

Question: Since VI = total I²R, how can the power equation include the kinetic energy supplied to the armature?
 
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  • #2
avicenna said:
In a capacitor discharge through a plain resistor, the capacitor power supplied at any instant is VI; the power dissipated in the resistor is I²R. So VI = I²R.

Consider a railgun operated with a capacitor bank. At any instant of capacitor discharge, the power supplied is VI. The total power supplied for ohmic loss is sum I²R for two rails plus the resistance of the armature.

Question: Since VI = total I²R, how can the power equation include the kinetic energy supplied to the armature?
The short answer is that in electro-mechanical machines, like motors or railguns, there is a significant inductance term in the equations and additional losses associated with the energy transferred. Your schematic may look like an RC discharge, but there are other effects that need to be included in your model.

Beware of using circuit analysis before you've worked out the fundamental physics in the system. The circuits are just a simplified, standardized, representation of the real world. They are just a way of telling other EEs "you don't have to worry about physics, I've already done that for you."*

Personally, I'm a big fan of google searches. Here's (literally) the first result for "railgun electrical model". Look at figure 2.

* This concept, BTW, is something many physicist don't understand. But I'll spare y'all my diatribe on that.
 
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  • #3
DaveE said:
The short answer is that in electro-mechanical machines, like motors or railguns, there is a significant inductance term in the equations and additional losses associated with the energy transferred. Your schematic may look like an RC discharge, but there are other effects that need to be included in your model.

Beware of using circuit analysis before you've worked out the fundamental physics in the system. The circuits are just a simplified, standardized, representation of the real world. They are just a way of telling other EEs "you don't have to worry about physics, I've already done that for you."*

Personally, I'm a big fan of google searches. Here's (literally) the first result for "railgun electrical model". Look at figure 2.

* This concept, BTW, is something many physicist don't understand. But I'll spare y'all my diatribe on that.
The link you gave involves a mathematical modeling with mathematics beyond me.

I would be interested if anyone could provide a simplified explanation as to how some of the power of the capacitor bank is transferred to kinetic energy in the armature. And what factors determined how much power are wasted as ohmic loss.
 
  • #5
DaveE said:
Still, I don't know that we can do better than what's already out there.
https://en.wikipedia.org/wiki/Railgun
https://science.howstuffworks.com/rail-gun1.htm

If you haven't studied basic EM (like Faraday's law, etc.), basic differential equations, and LCR circuit analysis, I think you might not really be ready for the "how does it work question".
I could understand how motors, generators, transformer, capacitor, faraday's law, etc. work.

As for railguns, since there cannot be a simplified "layman" answer, I don't think I am ready to ask the question.
 
  • #6
I don't think it's too different than a DC motor. If you really understand motors that spin, I'm sure you can figure out linear motors and railguns. The basic concepts are mostly the same.
 
  • #7
The key point with railguns is that when the current starts to flow it creates an increase in the magnetic flux. But "nature resists a change in magnetic flux", so that creates either (or both) a back emf (voltage) to oppose the current increase (this is inductance) and a force on the conductors to increase the area enclosed by the current. This is because flux is inversely proportional to area; if the current increases by 10% but the area also increases by 10%, then the flux doesn't change. Since one of the conductor in the rail gun loop is the arc between the rails, then this makes a force to push the arc away along with whatever projectile is in the way. I think it's Lenz's law, but I get confused about the names of these things.

Edit: not Lenz, Lorentz force is a better name.
 
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  • #8
 

FAQ: Power equation in railgun operation

What is the power equation in railgun operation?

The power equation in railgun operation is used to calculate the amount of electrical power required to operate a railgun. It is represented as P = VI, where P is power in watts, V is voltage in volts, and I is current in amps.

How is the power equation used in railgun design?

The power equation is an important factor in railgun design as it helps determine the amount of power needed to accelerate the projectile to the desired velocity. It is also used to calculate the size and capacity of the power source required for the railgun.

What factors influence the power equation in railgun operation?

The power equation in railgun operation is influenced by several factors, including the mass and velocity of the projectile, the length and resistance of the rails, and the efficiency of the power supply. Any changes in these factors can affect the amount of power needed to operate the railgun.

How does the power equation affect the efficiency of a railgun?

The power equation plays a crucial role in the efficiency of a railgun. A higher power input can result in a more efficient acceleration of the projectile, while a lower power input may result in a slower and less efficient acceleration. Therefore, the power equation must be carefully considered in order to achieve optimal efficiency in railgun operation.

Are there any limitations to the power equation in railgun operation?

While the power equation is a useful tool in railgun design, it does have some limitations. It assumes ideal conditions and does not account for factors such as friction, heat loss, and electromagnetic forces. Additionally, the power equation does not take into account the effects of the projectile's shape and size on its acceleration. Therefore, it should be used as a guide and not as an exact representation of the railgun's performance.

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