coade.wikia.com/wiki/Children_of_a_Dead_Earth_WikiWith the new Wiki just set up, that may be a better place for this. But I wanted to set up something
direct and to the point to help out those who like myself don't have much background in the sciences
or may have forgot them. This is a quick and dirty guide which will pay huge dividends to those who take
the time to make sure they have these concepts down cold. I included the Member who shared the info
where appropriate.
This is the comments eluda was classy enough to private message me to help with some of the basics I was missing.
Many who took physics long ago as I did may benefit.
Further down the page are the basics of Railguns.
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By @eluda in response to some of my misconceptionsOk, some basic terms and units. I (and most of scientific world) works with SI units (Système international d'unités - International System of Units). This is the metric-based system of units, and comes down to 7 base units. All other units are can be expressed as these. A 'unit' in this context refers to a unit like a kilogram, which happens to be a 'base' unit, or something like the Joule, which is not a base unit - it can be expressed in base units as kg·m2/s2. Units like this are known as 'derived' units.
Note the '·' used here is to denote multiplicative units, like the Watt-hour (expressed as W·h), or as other non-base SI units, such as N·m, or W·s, or C·V. you brought up in the thread. I'm using it to make sure that they're split up correctly and easy to read. '/' is used the same way, except for divisive units (sometimes this will be shown using the '·', but with a negative power unit). Thus the 'Joule' above is read as kilogram metres-squared, per second-squared.
You can also express a unit, like the Joule, as other non-base SI units, such as N·m (newton-metres), C·V (coulomb-volts) or W·s (watt-seconds). What's important is that ultimately, these must 'reduce' down to the same kg·m2/s2 in terms of base units.
A useful concept to understand is that the units on both sides of an equation must be the same, and hence you can use equations to show what the units should be (and in reverse, you can sanity check equations by checking that the units on both sides are the same - if they are not, something is wrong, as it is not physically possible). For example, if I said that A = B x C, and B = 5kg, and C = 2m, then A would be 10 kg·m. If this was not true (say you knew the units for A had to be kg·s instead), then you instantly know something is wrong. This is called Dimensional Analysis - if you study physics or engineering you will probably learn about this at some point, but its such a useful tool it worth knowing about anyway.
More on this in both the wiki article on SI units (linked above) and here;
physics.nist.gov/cuu/Units/units.html
hyperphysics.phy-astr.gsu.edu/hbase/units.html#uni1
www.npl.co.uk/reference/measurement-units/So, we come back to the concept of Energy and Power.
Energy has a unit of Joules, which is a derived unit, with base units of kg·m2/s2, as shown above. This is easy to show with dimensional analysis using the very common equation of
Kinetic Energy = 1/2 x Mass x Velocity x Velocity (or written as KE=1/2 mv2)
in terms of units this is
Joule = Kilograms x Metres/Second x Metres/Second = kg·(m/s)·(m/s) = kg·m2/s2
Note we lost the 1/2 term because it is dimensionless, and has no units
This same approach is true for every single equation involving energy, and will reduce down to the same base units (kg·m2/s2). This includes things like U=kQq/r (electric potential energy), E=mgh (gravitational potential energy), W=Fd (force-distance), E=mc2 (energy-mass equivalence).
Power is defined as Energy / Time. It has a unit of Watts, which is also a derived unit, in the form of J/s, or in base units kg·m2/s3. Therefore Power is called a 'rate' unit, in this case the rate of Energy transfer (or work done, same thing). Again, this is true regardless of if we are talking Watts in an electrical sense (P=VI, or P=RI2), a 'kinetic' sense. This is easy to see with dimensional analysis again, using P=RI2 this time;
Power = Resistance x Current x Current
in units
Watts = Ohms x Amperes x Amperes = Ω·A·A = Ω·A2
We need to change the Ohms into base units, so
Resistance = Voltage / Current
in units
Ohms = Volts / Amperes
So using this we get
Watts = (Volts / Amperes) x Amperes x Amperes = Volts x Amperes = V·A
Hey, we seem to have hit P=VI accidentally on the way. Great. But we must go further, Volts are be defined as
Voltage = Potential Energy / Charge
so
Volts = Joules / Coulombs
Using this above, we get
Watts = (Joules / Coulombs) x Amperes
Lastly, we need to turn Amperes into another unit. An Ampere is the 'rate unit' for charge, much like Watt is for Energy, so
Current = Charge / Time
so
Ampere = Coulomb / Second
Putting this into the above
Watts = (Joules / Coulombs) x (Coulombs / Second) = Joules / Second = J/s
You can do this for any equation so long as you know the units.
How a Rail guns works.... by @eluda in response to some of my misconceptions1. The railgun is 'connected', as the power draw at this point is practically zero. After the armature is inserted it closes the circuit and draws current and generates a magnetic field, and propels the armature. After the armature exits the rails, the current drops to zero again.
2. Each shot gets whatever energy is imparted by the current and magnetic field, which is dependent on the current, which is again dependent on the energy source. If the source can provide the same current as for the first shot, then yes, each shot got the same, if not, then they did not get the same.
3. The current (acts on the armature), and it draws energy from the source equivalent (in a 100% efficient system) equal to the kinetic energy of the round. The 0.2MW is a result of this energy draw, but is impossible to determine without knowing how often the system draws that energy.
4. "So as fast as you can load rounds into the railgun (flip that lamp switch), the railgun will fire provided the barrel or
round does not melt or blow up from the pressure and you don't put more than one round in the rail gun at once."
This is also correct, with the added condition that the energy source can provide the current to the rails for the increased number of shots. If you add more shots beyond that, the current will drop, and each projectile will receive less energy.
Every additional projectile you fire, draws more energy. Power is Energy / Time (from above), so if you draw 2x more Energy in the same time (for the sake of convenience, during one second), then you must provide 2x more Power to the system. Alternatively, if you do not, then the Energy per shot must decrease (lower current and lower energy draw per shot).
The railgun does not 'use' 0.2MW on a shot. If the system is drawing 0.2MW, then it is capable of providing 0.2MJ in energy to its shots per second. This could be one shot with 0.2MJ, or 10 shots with 0.02MJ.
32kJ per shot (or 0.032MJ per shot) is the energy of the projectile, and thus the energy the system draws (assuming 100% efficiency) from its energy source. If it were to only fire one round per second at this energy, its power draw would be 32kW, or 0.032MW. If the system can draw 0.2MW as mentioned above, then at best it could fire 6.25 projectiles with 32kJ of energy per second. If more projectiles were fired, then either the power draw must go up, or the energy per projectile must go down.
You cannot look at just energy per shot. You can look at energy per shot vs energy drawn by the current to check for physics compliance on that part of it, but that's a fairly complicated bit of calculus if you want to do it accurately. If you want to look at the railgun as a 'system', then the energy out (in terms of kinetic energy of the projectiles) must be equal to the energy in (in terms of electrical energy drawn into the system) over a period of time (and assuming 100% efficiency again). Or, in other words, power out must equal power in.
Any result where the power out (or kinetic energy out / time) is higher than the power in (or electrical energy in / time), violates the first law of thermodynamics, as the system as a whole would be creating energy. This makes it by definition, physically impossible.