h a l f b a k e r y
There goes my teleportation concept.
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The rifle is heavier than you thought it would be from reading the brochure, but the foil armor plates and helmet are much less cumbersome than you'd supposed when signing the waver, so that's good.
The overhead lights begin to dim and redden allowing players in their locked cubbies time to adjust
their eyes to the near blackness which will signal the start of game play.
Attaching a crank to your VanDeGraff back-pack you slowly start the opposed flywheels within it spinning and then quickly cycle through the gears to rev them up as fast as you are able. Engaging the belt system begins charging the aluminum housing in the top of the back-pack.
This charge is then stored in a refillable Leyden jar which makes up the greater portion of your pump-action pressurized salt-water rifle, and acts as a capacitor that discharges thousands of volts to illuminate the continuous-stream of electrified water as it blindingly scracks along your opponents Faraday suit of armor plating searching/soaking for a weak spot to pack that light-show with a mild tazer action if given an accurate enough shot.
Slipping concussion sensitive capacitors into each balloon/grenade you fill them to expansion from the tap provided, tie them off, stow them in their protective compartments, and then you wait...
Red lights switch off.
||"Trick or treat ... ? MUHWHAHAHAHAH BZZZZZZZZTTTT ..."
||We'll get back to you with the conductivity figures for wet cat fur soon as we've run a few ... practical tests.
||[+]Awesome, is this a hunger games death thing or a
recreational Games center thing? Let me try and clarify
that, what are the odds of death?
||Same as every other day I would imagine barring pregnancy or a preexisting heart condition.
It's not the volts that kill you it's the amps. This would just look deadly.
||"You're thor!? - I can hardly walk"
||Interesting idea, but I'm not so sure about the use of
the Leyden jar.
You can't really store the output from a Van de
that would defeat the whole purpose. You're
technically correct that
it's the amperage, not the voltage, that's the
determining factor for a
lethal current, but that's like saying it's the heat, not
the fire that kills.
By definition (namely, Ohm's law), amperage and
voltage for a given
resistive load increase at the same rate. The
amperage is therefore
always a function of the voltage and the resistance,
and the resistance would be more or less fixed in this
||A Van de Graaff generator supplies a constant (very
low) current by
effectively adjusting its output voltage dynamically
based on the load
attached. A Leyden jar, being basically a capacitor,
would have no
means of doing so, and would basically just dump
electricity as fast as
possible into the target.
||Come to think of it, there are quite a number of ways
this idea could go
awry. A stray ground here or there causing a sudden
change in the
capacitance of the target could easily prove fatal.
There are likely
ways of achieving a similar effect safely, but the use
of a Van de Graaff
generator, as cool as it would be to do so, probably
would be be
unworkable from a safety standpoint.
||So... same as every other, other day then?
||I am trying to understand this stuff better so I appreciate your anno.
I do not understand why a charge from a Van De Graff generator can not be stored in a Leyden jar though. I don't dispute this, I just don't get why not.
||Also, if the Leyden jar can be used as a capacitor, could its design not be such that it would only hold a certain non-lethal charge and no more?
||Also also; I had pictured the flywheels in the back-pack as spinning freely until manually engauged to re-charge the capacitor for the next shot. Unless you need to 're-load' there would be no continuous charge building. I figured this would be more challenging and pretty safe, if not entirely painless.
||I'm sure that there are much better ways to achieve the same effects but I'm kinda barely scratching steam-punk tinker level here.
If nothing else it would make for a fun science project with the kids...
||Long-winded technical answer follows; for the simple explanation by analogy, just
jump down to the ***** below.
||It boils down to the fact that amperage and voltage are directly related. The
primary equation that governs electrical circuits of all types is Ohm's law, which
states that voltage equals current times resistance (commonly written as V=IR).
Understanding this equation is the key to understanding electrical circuits at their
most basic level. Current and resistance are easy to explain; voltage is a bit more
||Current is simply a measurement of how fast electrons are flowing past a certain
point. We call a certain number of electrons (about 6.2*10^18, specifically) a
coulomb. This is basically a blob of electricity. When this blob starts flowing, the
rate of flow is measured in amperes. One coulomb per second past a point is one
ampere. You may have seen batteries with reserve capacity measured in amp-hours;
amp-hours are directly convertible to coulombs. So, a coulomb is how many
electrons we have, and an ampere is a measurement of how fast those electrons are
moving. Note that static electrons don't do us any goodjust as we need a flowing
river to turn a water wheel, we need the electrons to be flowing to do work.
||Resistance simply describes the physical makeup of a substance. Is it porous, and will
let a lot of electricity through? Or is it dense, and will stop electricity from flowing?
We quantify this in ohms. A material with a low resistance is a good conductor,
which will allow electricity to flow freely. A material with a high resistance is an
insulator, which will prevent the flow of electricity.
||But what are volts, exactly? Well, the truth is that there's really no such thing as a
voltnot in the physical sense, anyway. A volt is an abstraction. If we connect a
source of electricity to a resistance, a current will begin to flow at a certain rate,
depending on what the resistance is. What determines the rate of flow? It's the
physical and electrochemical makeup of the materials involved. All of this
information is abstracted away by wrapping it up in a voltage value. The voltage
value is always measured between two points (for example, a power outlet and the
earth, or two terminals of a battery), and tells us how much current will flow
between those two points when a load of a given resistance (as measured in ohms) is
connected between them. This is why voltage is described as electrical potential.
||If our resistance is fixed, but our voltage doubles, then twice as much current will
flow. If our voltage is fixed but our resistance doubles, then half as much current will
flow. This is how most electricity sources that we are familiar with operate.
Batteries and wall outlets are known as constant voltage sources, meaning that
current flow increases or decreases with changing resistance.
||But what if our /current/ is fixed? Well, if our current remains constant, but our
resistance changes, then the voltage must be changing in order to keep the current
constant. If we connect a resistance of 1 ohm to a constant current device that
supplies 1 amp, then the potential across the load is 1 volt. If we then increase the
resistance of the load so it's now 2 ohms, our current remains at 1 amp, but our
voltage has now increased to 2 volts. No matter how much we vary the resistance,
the generator will increase or decrease its voltage output to keep the current flowing
at a rate of 1 amp, within the limits of its ability to do so.
||This is the case with a Van de Graaff generator, which is a constant current source
rather than a constant voltage source. It puts out a current that is, under normal
circumstances, low enough to be safe. Any load applied to it will receive a fixed
current but an unknown and variable voltage, determined from moment to moment
by the fixed current rating of the generator and the resistance applied to the
generator at that instant.
||A Leyden jar is a capacitor, however, which is a fixed voltage device rather than a
fixed current device. So, if we connect it to our Van de Graaff, we will find that it
starts storing electrons at a fairly quick rateits resistance is initially low. As more
electrons flow into it, the resistance of the Leyden jar increases, which causes the
voltage output of the generator to increase in order to maintain the same amount of
current. This cycle continues until the Van de Graaff's ability to produce more
voltage is exceeded, at which point the Leyden jar is full of electrical charge, at the
maximum voltage the Van de Graaff generator can produce.
||But when we go to discharge the Leyden jar, it doesn't care about maintaining a fixed
current. Instead, it maintains a fixed voltage (instantaneously, at leastin actual
fact the voltage drops fairly quickly, but it's that initial jolt that we're concerned
with, because that's enough to kill you). So if you were to touch a Van de Graaff
generator directly, you'd receive some voltage just high enough to get the fixed (low)
current through your body, but well below the maximum output of the generator. If
you were to touch the Leyden jar, on the other hand, you'd immediately receive a
jolt determined by the resistance of your body and the /maximum/ voltage the Van
de Graaff could generate, which is very likely to be enough to kill you many times
||So you can hopefully see now why it's not possible to store the output of a constant
current source such as a Van de Graaff generator in a constant voltage source such
as a Leyden jar. Since the Van de Graaff generator doesn't know what the resistance
of the target will be in order to generate the appropriate voltage, and the Leyden jar
doesn't care what the resistance of the target is when supplying the fixed voltage,
you end up with a potentially deadly set of circumstances when combining the two.
||***** Another way to think about it is by comparing the Leyden jar to a balloon. A
Van de Graaff generator would be a type of pump that provides a constant rate of
airflow regardless of the pressure required to do so. As our balloon gets bigger, the pressure inside it
increases, and the pump needs to increase the pressure to compensate.
Eventually, the pump can no longer increase the pressure, and the balloon is
inflated to its maximum capacity. But when we begin deflating the ballon, the
pressure will be equal to the maximum pressure of the pump. If that pressure is
greater than the pressure inside your lungs, and the capacity of the balloon is larger
than your lung capacity, well, you'd better hope your lips aren't on that balloon.
||//Remember to jiggle around a lot and shout "Ow! Ow! Stop! Stop!".// This is good advice for almost every situation.
||Actually, your balloon analogy is incorrect. Balloon pressure behavior is highly non-linear and even moderately inverse. The highest presssures occur at the earliest moments of stretching it out, which is why it's so hard to get a balloon started but rather easy after. The material is thicker at this point in the process.
Amperage = flow rate. Voltage = pressure. Easy.
||Sweet. Holy crap I think I get it. Thank you.
||I'm not claiming a balloon is a perfect analog for a
capacitor in terms of real world response curve, [Ray].
Assume an ideal balloon here, just as we assume an ideal
capacitor (i.e., one that won't blow up when five million
volts is applied to it). No real world device has a truly
linear response curve across the entire range of
conceivable operating parameters. A capacitor doesn't
have a linear response eitheryou need to take into
account the RC time constant and other factors to
determine its charge curve. But the small signal response
of a balloon actually approximates that of a capacitor quite
||The voltage=pressure analogy is only easy to understand
once you understand it. What gets lost in that explanation,
though, is the relationship between current flow and
voltage. Many people seem to think you can simply
increase the pressure somehow while regulating the flow of
electricity for a given resistancein other words, that you can adjust amperage,
voltage, and resistance for a circuit all independently of each other. It's this
that I'm trying to combat here. I think explaining volts as an abstraction of what
will happen if you connect up a given circuit helps make it clearer that, of
current, resistance, and voltage, only any two of those can be adjusted
independently and the third one will always be a calculation derived from the
||//Amperage = flow rate. Voltage = pressure. Easy.//
||Maybe. Ask Coulomb, he's in charge
||dangerous and ill advised [+]