How do I calculate how much power/current etc. is required?

[KevinTOC]

(Title is actually supposed to say: How do I calculate how much power/current etc. is required to propel an object forwards at a certain speed using electromagnetism? but it's too long for the title tab, apparently.)

So, I'm wondering how you calculate the amount of power/current etc. is required to propel an object forwards.

so say I have a 10kg object, and I have a 2m long barrel with electromagnetic coils, and I want to propel this object to 10m/s. (Assuming it's in a vacuum, and there's no air resistance, and assuming the coils can withstand the heat)

How would I calculate how much electric power and current (maybe voltage?) I'd need to make it go that fast through the length of that barrel?

 

[CWatters]

Simplest way to ball park the numbers would be to assume constant acceleration. Work out the acceleration required using equations of motion (eg SUVAT), Then apply Newton's law F=ma to get the required force. Then power = force*velocity. Using that method the power required will be at a maximum as it leaves the barrel because that's where the velocity is highest.

However it's likely your power source will have some limiting value. If you assume constant power the acceleration won't be constant and the equations of motion are slightly more complicated but I think they will give you a lower value for the power required. I suggest you Google equations of motion under constant power. I think the relevant equation would be...

P=(mv^3)/3s

Where
s= length of barrel
v=exit velocity
m=mass

 

[sophiecentaur]

I’d be inclined to think in terms of required Kinetic Energy and the required Electrical Energy would be that amount plus inefficiency due to mismatch and friction losses. Assume say 25% efficiency perhaps so four times the final KE.

 

[Baluncore]

Welcome to PF.
I assume the shot is horizontal and not influenced by gravity.
I assume this problem is not homework.

Projectile energy is kinetic; E = ½⋅m⋅v² = 500 joule.

The coil makes a linear motor so for a constant motor current we can expect a constant force.
The work is done on the projectile over a barrel length of, s = 2 m; What force?
E = F⋅s; F = E/s; ∴ F = 500 / 2 = 250 Newton.

What is the acceleration?
F = m⋅a; a = F/m; ∴ a = 250 / 10 = 25 m/s²

How long does it take? S = ½⋅a⋅t²; ∴ t = √( 2⋅s/a )
t = √( 4 / 25 ) = 2 / 5 = 0.4 sec.

Now we have energy and time; Power, W = joules / sec = watt.
W = E/t = 500 / 0.4 = 1250 watt.

It is not possible to specify the coil current without coil and power supply details.

 

[Babadag]

As usually I am late. I prepared the same answer as Baluncore.
length=2 m; v=10 m/sec; m=10 kg; length=a*t^2/2; a=v/t; length=v*t/2; t=2*length/v=2*2/10=0.4 sec; a=v/t=10/0.4=25 m/sec^2
F=m*a=10*25=250 N;P=F*length/t=250*2/0.4=1250 W

 

[sophiecentaur]

The Power / Energy needed by the 'gun' will be more than the 1250 / 500 figures because there will be energy used up in creating the magnetic field and also in the resistance of the coil. That's what @Baluncore was referring to in his comments about the power supply details. That's the crucial thing. You really need to be looking at some of the many RailGun links you will find on Google.
PS these things are a bit of a safety hazard and you need to be careful of possible high voltage shocks, high current burns and shooting yourself or someone else. They can be serious weapons.

 

[Baluncore]

You really need to be looking at some of the many RailGun links you will find on Google.

https://en.wikipedia.org/wiki/Railgun

I have a 2m long barrel with electromagnetic coils

The particular device being discussed here seems to be a coilgun rather than a railgun.
https://en.wikipedia.org/wiki/Coilgun

I would discharge a separate capacitor through each stage coil.
The current would have a half-sinewave profile because of the inductance and capacitance.
For a capacitor, energy, E = ½⋅C⋅V².
The capacitor voltage, V, would be partially reversed by the current pulse, the lost voltage representing the sum of projectile energy, resistive energy losses and radiated magnetic field energy.

Designing a coilgun is educational. Making a coilgun can be dangerous and is not recommended.

 

[KevinTOC]

Welcome to PF.
I assume the shot is horizontal and not influenced by gravity.
I assume this problem is not homework.

Projectile energy is kinetic; E = ½⋅m⋅v² = 500 joule.

The coil makes a linear motor so for a constant motor current we can expect a constant force.
The work is done on the projectile over a barrel length of, s = 2 m; What force?
E = F⋅s; F = E/s; ∴ F = 500 / 2 = 250 Newton.

What is the acceleration?
F = m⋅a; a = F/m; ∴ a = 250 / 10 = 25 m/s²

How long does it take? S = ½⋅a⋅t²; ∴ t = √( 2⋅s/a )
t = √( 4 / 25 ) = 2 / 5 = 0.4 sec.

Now we have energy and time; Power, W = joules / sec = watt.
W = E/t = 500 / 0.4 = 1250 watt.

So, according to that, in order to propel a 10kg object at 10m/s, through a horizontal 2m long barrel in a vacuum, you'd need 1250 Watts of power to propel it at that speed?

And no, this isn't homework.

It is not possible to specify the coil current without coil and power supply details.

Right, I forgot about that when writing the post. Say it's uses 400V (U), then I (Current) Would be I=P/U 1250/400 = 3.125A

Designing a coilgun is educational. Making a coilgun can be dangerous and is not recommended.

I'm not building one, I'm just interested in the technology behind it, and how to design one.

 

[Baluncore]

So, according to that, in order to propel a 10kg object at 10m/s, through a horizontal 2m long barrel in a vacuum, you'd need 1250 Watts of power to propel it at that speed?

That is correct. Each projectile will carry away 500 joule. Power is the rate of flow, or rate of conversion of energy. When you launch the projectile, energy must be converted from capacitor storage to the projectile at a rate of 1250 joules per second. If it was perfectly efficient it would require 1250 W for only 0.4 sec to launch each projectile. It will not be perfect, so it will require more energy input, while the coils and electronics will get hot.

If you charge the energy storage capacitors, say in 40 seconds, at 1% of the rate you need to discharge them, then minimum input power will only need to be 12.5 watt, but with a lower possible rate of fire.

 

[KevinTOC]

That is correct. Each projectile will carry away 500 joule. Power is the rate of flow, or rate of conversion of energy. When you launch the projectile, energy must be converted from capacitor storage to the projectile at a rate of 1250 joules per second. If it was perfectly efficient it would require 1250 W for only 0.4 sec to launch each projectile. It will not be perfect, so it will require more energy input, while the coils and electronics will get hot.

If you charge the energy storage capacitors, say in 40 seconds, at 1% of the rate you need to discharge them, then minimum input power will only need to be 12.5 watt, but with a lower possible rate of fire.

And I suppose the equation gets much more complicated when you start to factor in things like air resistance, friction between the barrel and projectile, etc.?

Anyways, thank you all for your answers. :)