Nature has an article about a brain-computer-interface hat that "can read your thoughts". You can "stroll down a virtual street" by just thinking about walking. I have to admit, if I could do something in the virtual world just by thinking about it, I doubt that it would be walking down a street.

Here's a great selection of unfortunate domain name choices. Be careful what you register. Did you mean ExpertsExchange.com or ExpertSexChange.com?

Wired has a good story about ZigBee ,"a specification for building large networks of low-power radio transmitters". Expect consumer products by the end of the year. Because of the low-power consumption, batteries will run for years. Applications include home-monitoring systems and checking up on the elderly.

MDI (Moteur Developpment International) claims that they have developed a car that uses "bi-energy" (compressed air + fuel) that has an effective driving range of 2000 km with "zero pollution in cities and considerably reduced pollution outside urban areas". They are promoting it to be used as a CAT vehicle.

There's a nice roundup of do-it-yourself robot building here. Lots of links and resources for the budding robotocist.

DVD Jon has done it again. This time, he's hacked encryption coding Microsoft's Media Player NSC files. Now you can watch Media Player files on VLC (VideoLAN Client). The Register has a story.

Linspire, everyone's favorite WalMart distro, can be downloaded for free until September 6. They claim they are doing it to remove any confusion with Andrew Betts' Freespire, which is now called Squiggle. Actually, this confused me even more. According to their site, Linspire (free or not) includes "a lot of legal and paid-for 3rd-party licenses for things like mp3, Java, Flash, Quick Time, Windows Media, Bitstream fonts, Real media, music, etc.".

Everything USB reports that Mazda has developed a concept car that abandons the traditional cylinder key in favor of a USB key. It can also store driving directions and, of course, songs. This seems so obvious, I can't believe nobody has already done this.

We interrupt this program to annoy you and make things generally irritating.

1) Spazmolytic, Skinny Puppy
2) This Charming Man, The Smiths
3) How Soon Is Now, The Smiths
4) Add It Up, Violent Femmes
5) Thieves, Ministry
6) Stepping Razor, Peter Tosh
7) You're Wondering Now, Skatalites
8) Reggae and Ska, Judge Dread
9) Nothing Natural, Lush
10) Hoosier Love, Mu330
11) 007 (Shanty Town), Desmond Dekker & the Aces
12) Long Shot Kick De Bucket, The Pioneers
13) Schpreck, Infectious Grooves
14) Repeater, Fugazi
15) Astronomy Dominae, Voivod
16) Soup is good food, Dead Kennedys
17) Astrozombies, Misfits
18) Precious Things, Tori Amos
19) Psycho Killer, Talking Heads
XX) Moon Rocks, Talking Heads
YY) Life During Wartime, Talking Heads
D3) Memories Can't Wait, Talking Heads
>2) Brendan #1, Fugazi
44) Public Bath, Shonen Knife
88) ANtibody, Spahn Ranch
22) Riding the Rocket, Shonen Knife
00) Got The Time, Joe Jackson
*8) Glamorous Glue, Morrissey
#2) Love in a Void, Souixie and the Banshees
71) Cuts you up, Peter Murphy
8@) Low Self Opinion, Rollins Band
4$) Youth Against Fascism, Sonic Youth
32) Wasted, Black Flag
73) Group Sex, Circle Jerks
91) Radio, Elvis Costello
84) Tame, Pixies
2%) Head like a hole, Nine Inch Nails
3#) Its Tommorrow, Overwhelming Colourfast
5%) Sheila na gig, PJ Harvey
40) Channel Zero, Public Enemy
%%) Reign in Blood, Slayer
3#) PRove You Wrong, Prong
8*) Frizzle Fry, Primus
6^) New Boy, Sub Humans
$5) Soundsystem, Operation Ivy
4#) Dalis Car, Dalis Car
43) interimsliebden, einstuerzende neubauten
&1) S.K.I.N.H.E.A.D., MDC
77) She gone, Mephiskapheles
9!) Ice Breaker, Skinny Puppy
etc. etc. etc.

I didn't listen to radio. EVAR.
I don't think it was ever good.
And MTV has been a sell-out since day 1 (not wanting to play Michael Jackson? OMGWTFBBQ!)

Pleasure pool at the beach!

Thank you for the precious palpitation that anyone is unable to give me.

I myself, Mr. Gurbik, half shark half man, skin like an alligator, carrying a dead walrus. Check it.

My new case mod website.

Okay, perhaps thats a bit misleading.

As you all remember my computers power supply blew up. So this is a photo-journal of my attempts at replacement.
Cower in fear as Mekka bends metal and plastic to his will.

Oh, and PS- verizon "automagic website builder" is stupid, turgid, and hysterical. Enjoy!

Every wonder why E = mc^2? It has always bugged me why it was so . Hardly any college introductory physics courses go into why. In fact, hardly any college courses derive the formula and just assume it is correct. I hardly ever accept anything on face value: I like running experiments, confirming observations, compiling source code, and deriving formulas. By doing so, I get a better understanding of the stuff I use. That is a worthwhile endeavor.

Now I finally figured out why E = mc^2, all by hand. It doesn't take much more than what you learn after a single year of a single college physics courses (Physics I + II) and a year and a half of calculus (Calc I + II + III). I highly encourage anyone who has taken those classes to try it out. It is very enlightening!

The Speed of Light Is a Constant

To start, you have to accept the following axiom.

The laws of the universe are valid in all inertial reference frames.

That seems super obvious, but there are interesting concequences. You see, one law of the universe is that the strength of electric fields is a certain constant in space. This is represented by the variable e - AKA the "Permittivity of free space" - and the higher e is the weaker the electric field.

Another property of the universe is the strength of magnetic fields. This term was mostly subsumed into the permittivity constant since magnetic fields are created by electric fields, so you don't really need to consider it. This magnetic constant was should be a geometric factor (4*pi) but is usually multiplied by 10^-7 for practical reasons.

The reason this is important is that light is a wave. If you solve Maxwell's equations for an electromagnetic field (in space), you find that the speed of light only depends on the permittivity of space, e, and the strength of magnetic fields in space. In fact, the final result describing the speed of light is simple:

c^2 = 1/(e*u)

c

Speed of light

e

Strength of Electric fields (Permittivity)

u

Strength of Magnetic fields (Permeability)

Such a simple result is rarely a fluke. It also makes little sense. For many years, scientists tried to disprove this result with many experiments. Their skepticism was based on the prevailing theory of the day: that light is not only a wave but a particle too. It is like Jello: while hot it looks like a liquid and waves can be seen. While cold, Jello looks like a solid and can hold its shape. Light is similar - in certain circumstances is acts like a wave and others it acts a particle. This violates the principle of relativity.

The Principle of Relativity

Relativity is a basic observation. Say you are in a bus, and you walk from one end to the other. You don't feel like you are traveling very fast... only a few miles per hour. And to the rest of the people in the bus, you are moving that slowly. However, to people on the street, you are moving very fast. Your speed, to them, is your velocity (relative to the people on the bus) plus the velocity of the bus (relative to the people on the street). Take this example:

Look at this picture (from this page).

O

Represents the "stationary" reference frame with respect to us (you and me).

O'

Represents the "moving" reference frame wrt O (and us).

v

Velocity of O' relative to O.

x, y, z

Coordinates according to O.

x', y', z'

Coordinates according to O'.

t

Time in O (not shown in diagram) since some start event.

t'

Time in O' (same as O).

We are only going to consider one dimention, so y, z, y', z' can be ignored for now. Time (t) and accelerations are the same in each frame (O and O') if V is a constant - i.e. O' doesn't accelerate and moves with a constant speed in a constant direction (wrt O.) Say we have a point called P with coordinate x' in O' and is not moving. Relative to O it is moving with speed v (i.e. the same as O'). To figure out what point P has in O - called x - you use:

x = x' + v*t'

y = y'

z = z'

t = t'

Velocity times time is the distance O' moved, and x' is the distance P is in O', so added together you get the distance P is from O.

From the vantage point of O', things are just the opposite. You come up with the following equations, which can be derived from the equations above:

x' = x - v*t

y' = y

z' = z

t' = t

Just use algebra to figure out those. That second set (and the following set) seems almost trivial, but there are important effects later. There is one more set of equations to note:

ux = ux' + v

uy = uy'

uz = uz'

These are the velocities of a particle at the point if it was moving instead of stationary wrt O'. We used u to differentiate it from the speed of the reference frame, v.

Those are the Galilean Transoformation Equations and are the main result of the Principle of Relativity. See if you can understand that before going on. It is pretty standard and makes sense if you think about it.

How the Speed of Light Mucks Everything Up

The speed of light depends on the strength of electric fields in space, as shown above. But in two reference frames, the speeds observed have to be different according to relativity. If c is the speed of light, and it travels only in the x-direction:

c = c' + v (Not true as explained below!)

Thus, each frame must see a different speed of a light beam. But Maxwell says the speed is a property of the universe (as I said before), so c = c'. One of them is wrong. After lots of experiments, it seems Galileo's Relativity is WRONG! The speed of light, no matter how it was measured, was the same for all frames of reference. So we have the following axiom:

The speed of light is 186,282.397 miles/second for ALL reference frames.

So how do we fix Relativity, since it seems to work in most cases? There must be a correction factor that has to be added in. Let's call that factor, gamma or g. That means the transformation equations will look something like:

x = g*(x' + v*t')

x' = g*(x - v*t)

Those are the same equations as above, just with the correction factor added. But what is the correction factor? Let's perform an experiment to find out!

Now, even though O' is moving, lets start it at the same place as O. So this start occurs where t = t' = 0. At this time, a light pulse is emitted from the origin of the frames (remember, they are at the same place right now) and moves in the x-axis direction. According to O it moves a distance of:

x = c*t

But according to O' it moves a distance of (note c'=c):

x' = c*t'

Plugging these into the equations above:

c*t = g*(c*t' + v*t')

c*t' = g*(c*t - v*t)

Subsitituting:

c*t = g*(c + v)*t'

c*t' = g*(c - v)*t

This is a system of two equations. Solving one for t' (say the second one) and substituting into the first one can help you solve for g:

t' = g*(c - v)*t/c

and

c*t = g*(c + v)*t'

c*t = g*(c + v)*g*(c - v)*t/c

c = g^2*(c + v)*(c - v)/c

c^2 = g^2*(c^2 - v^2)

1 = g^2*(1 - v^2/c^2)

g^2 = 1 / (1 - v^2/c^2)

g = 1 / (1 - v^2/c^2)^.5

That is what g actually is. For small values of v, it is about zero. But if you travel facter than light, it grows to infinity! That was an early indication you couldn't travel faster than light.

Another result is that, since the space direction (x) changes, the time direction must also change. This makes sense, since speed is distance over time. If the speed is constant, but the space dimention changes, then time should do. You can solve this by using the two equations above again:

x = g*(x' + v*t')

x' = g*(x - v*t)

But because we know what g is, we can solve for t now:

x = g*(x' + v*t')

and

x' = g*(x - v*t)

x' = g*(g*(x' + v*t') - v*t)

x'/g = g*(x' + v*t') - v*t

v*t = g*(x' + v*t') - x'/g

t = g*(x'/v + t') - x'/(v*g)

t = g*(x'/v + t') - g*x'/(v*g^2)

t = g*(x'/v + t' - x'/(v*g^2)

t = g*(t' + x'/v*(1 - (1 - v^2/c^2)))

t = g*(t' + x'/v*v^2/c^2)

t = g*(t' + v*x'/c^2)

Together with the distance-equation, these form the new relativity equations, called the Lorentz Transformations:

x = g*(x' + v*t')

y = y'

z = z'

t = g*(t' + v*x'/c^2)

g = 1 / (1 - v^2/c^2)^.5

And from the other point-of-view:

x' = g*(x - v*t)

y' = y

z' = z

t' = g*(t - v*x/c^2)

g = 1 / (1 - v^2/c^2)^.5

Lorentz Velocity Transformations

But wait: there's more! Velocity is the derivative of distance with respect to time. Time and distance are distorted between the two reference frames, so velocities measured between the two should also be distorted. Say a particle moves with a velocity u (components ux uy uz) in O, and say that it moves with a velocity u' (components ux' uy' uz') in O'. It is the same particle looked at from two vantage points O and O'. How is the velocity in O' related to the velocity in O? First note the definitions:

ux = dx/dt

uy = dy/dt

uz = dz/dt

and

ux' = dx'/dt'

uy' = dy'/dt'

uz' = dz'/dt'

Let's begin by looking at the x-direction velocities. Note the following identity from the chain rule:

dx/dt' = dx/dt * dt/dt'

or

dx/dt = dx/dt' / dt/dt'

ux = dx/dt' / dt/dt'

We want the derivatives of the varibles in O wrt variables in O', since that's how the Lorentz transforms are defined. The velocity ux is dx/dt of course. Now taking the derivatives of the transforms and plugging them into the equation yeilds:

dx/dt' = d/dt' (g*(x' + v*t'))

dx/dt' = g * d/dt' (x' + v*t')

dx/dt' = g * (dx'/dt' + v*dt'/dt')

dx/dt' = g * (ux' + v)

and

dt/dt' = d/dt' (g*(t' + v*x'/c^2))

dt/dt' = g * d/dt' (t' + v*x'/c^2)

dt/dt' = g * (dt'/dt' + v*dx'/dt'/c^2)

dt/dt' = g * (1 + v*ux'/c^2)

Therefore

ux = dx/dt' / dt/dt'

ux = (ux' + v) / (1 + v*ux'/c^2)

Compare with the Galilean result, ux = (ux' + v). There is a correction term, and it is caused by the time distortion of Relativity. What about y-direction and z-direction velocities? Well, the two will have the same form, since anything in one direction perpendicular to the direction of motion - x-direction - isn't special in any other direction perpendicular to the direction of motion. (Can you see why?) So let's solve for the y-direction, and the z-direction follows the same logic:

dy/dt' = dy/dt * dt/dt'

or

dy/dt = dy/dt' / dt/dt'

uy = dy/dt' / dt/dt'

also

dy/dt' = d/dt' (y')

dy/dt' = dy'/dt'

dy/dt' = uy'

and remember

dt/dt' = g * (1 + v*ux'/c^2)

Therefore

uy = uy' / g / (1 + v*ux'/c^2)

And in the z-direction

uz = uz' / g / (1 + v*ux'/c^2)

It is interesting that the changes to the Galilean result in the y/z-directions depend on the distortion from the x-direction (and the ux' velocity). This is complete counter-intuitive at first glance, but makes sense after thinking about it. The distortion is from movement in the x-direction of the O' frame, so that is what the change depends on. In summary (from both points of view):

ux = (ux' + v) / (1 + v*ux'/c^2)

uy = uy' / g / (1 + v*ux'/c^2)

uz = uz' / g / (1 + v*ux'/c^2)

and

ux' = (ux - v) / (1 - v*ux/c^2)

uy' = uy / g / (1 - v*ux/c^2)

uz' = uz / g / (1 - v*ux/c^2)

Momentum changes

Momentum is highly depended on velocity. So does it change too? Let's perform an experiment and find out! :-) Here's the skinny:

Look at this picture in the O frame and this picture in the O' frame from this page.

a

Ball thrown by O

uax

x-velocity of "ball a" wrt O before the collision

uay

y-velocity of "ball a" wrt O before the collision

uax'

x-velocity of "ball a" wrt O' before the collision

uay'

y-velocity of "ball a" wrt O' before the collision

wax

x-velocity of "ball a" wrt O after the collision

way

y-velocity of "ball a" wrt O after the collision

wax'

x-velocity of "ball a" wrt O' after the collision

way'

y-velocity of "ball a" wrt O' after the collision

and

b

Ball' thrown by O'

ubx

x-velocity of "ball b" wrt O before the collision

uby

y-velocity of "ball b" wrt O before the collision

ubx'

x-velocity of "ball b" wrt O' before the collision

uby'

y-velocity of "ball b" wrt O' before the collision

wbx

x-velocity of "ball b" wrt O after the collision

wby

y-velocity of "ball b" wrt O after the collision

wbx'

x-velocity of "ball b" wrt O' after the collision

wby'

y-velocity of "ball b" wrt O' after the collision

also

uy

Velocity each person measures throwing their ball in their reference frame (x-componet=0)

My notation is slightly different from the picture. Note the differences!

Here is what the situation is. Say the person at O throws a baseball straight out (y direction) relative to her. Say the person' at O' also throws a baseball' straight out (-y' direction) relative to her'. Each ball has constant velocity (no gravity), and each person throws the ball with the same velocity as measured in their reference frame. Well, relative to the person at O, the baseball' moves in a diagonal line.

(Think of it this way. If you throw a ball up in a car, it seems to go straight. To a person on the sidewalk, it is moving in a diagonal. You just don't notice any horizonal direction because you are moving at the same speed in that direction.)

Well, the velocity of the "ball a" thrown by O is:

before the collision

uax = 0

uay = uy

and after the collision

wax = 0

way = -uy

Using the classical definition of momentum p = m * u then the change of momentum observed by O is:

before the collision

pax = 0

pay = m * uy

and after the collision

qax = 0

qay = m * (-uy)

So the net momentum change

Pax = qax - pax = 0 - 0 = 0

Pay = qay - pay

Pay = m * (-uy) - m * uy

Pay = -2 * m * uy

Likewise for "ball b" thrown by O':

before the collision (using the velocity transforms)

ubx = v

uby = -uy / g

and after the collision

wbx = v

wby = uy / g

and the momentums:

before the collision

pbx = m * v

pby = - m * uy / g

and after the collision

qbx = m * v

qby = m * uy / g

So the net momentum change

Pbx = qbx - pbx = m * v - m * v = 0

Pby = qby - pby

Pby = m * uy / g + m * uy / g

Pby = 2 * m * uy / g

This makes no sense! The momentum from one side of the collision is not balanced by momentum on the other side:

Should be zero, but isn't:

Pynet = Pay + Pby

Pynet = -2 * m * uy + 2 * m * uy / g

Pynet = 2 * m * uy * ( 1 / g - 1)

Pynet != 0

Since the net is not zero, momentum was NOT conserved using Lorentz Transforms! To preserve the law of conservation of momentum the definition of momentum must be changed. How should it be adjusted? Well the problems occured when we calculated the momentum of the particle in the O' frame:

Remember this?

Pby = 2 * m * uy / g

The 1/g factor came from the Lorentz velocity transformations. To fix it, we adjust the definition of momentum from:

p = m * u

To:

p = m * u * g

This doesn't affect the "ball a" result since for the O frame, relative to itself, g is zero. Plugging it into the above result allows momentum to be conserved. Cool!

Proving E = m * c ^ 2

In summary so far... The effects of relativity means measurements between O and O' are different. The changes include the Lorentz Transforms of position and velocity. The law of momentum was adjusted:

p = m * u * g

Also note the original definitions of velocity, force and energy (work):

v = dx/dt

F = dp/dt

E = integral of F dx

The first step is to get a more specific equation for Force. Momentum changed, so the force is not just F=ma. Let's put a particle right at the origin of O' and see what happens. This is a modification of Young's derivation of Kinetic Energy. By evaluating the derivate you find:

Note: u=v

F = dp/dt

F = d/dt (m * v / (1 - v^2 / c^2)^.5

F = m * v * d/dt((1 - v^2 / c^2)^-.5) + m / (1 - v^2 / c^2)^.5 * dv/dt

F = m * v * d/dt((1 - v^2 / c^2)^-.5) + m * a / (1 - v^2 / c^2)^.5

F = m * v * (1 - v^2 / c^2)^-3/2 * d/dt(1 - v^2 / c^2) + m * a / (1 - v^2 / c^2)^.5

F = m * v * g^3 * (-1/2) * (-2 * v / c^2) * dv/dt + m * a * g

F = m * a * g * (1 + g^2 * v^2 / c^2)

F = m * a * g * (1 + v^2 / c^2 / (1 - v^2 / c^2))

F = m * a * g * (1 - v^2 / c^2 + v^2 / c^2) / (1 - v^2 / c^2)

F = m * a * g^3 * (1 - v^2 / c^2 + v^2 / c^2)

F = m * a * g^3

Now going back to the definition of energy:

E = integral of F dx

F = m * a * g^3

E = integral of m * a * g^3 dx

E = integral of m * g^3 * dv/dt dx

E = integral of m * g^3 * dx/dt dv

E = integral of m * g^3 * v dv

E = integral of m * v / (1 - v^2 / c^2)^3/2 dv

Note (from integral table):

integal du / (a^2 - u^2)^3/2 = u / a^2 / (a^2 - u^2)^1/2 + C

Using the product rule:

J = m * v

dJ = m dv

dK = dv / (1 - v^2 / c^2)^3/2

K = v / (1 - v^2 / c^2)^1/2

E = J * K - integral of K dJ

E = m * v^2 * g - integral of m * v / (1 - v^2 / c^2)^1/2 dv

Note (from integral table):

integal du / (a^2 - u^2)^1/2 = arcsin(u / a) + C

Using the product rule:

J = m * v

dJ = m dv

dK = dv / (1 - v^2 / c^2)^1/2

K = c * arcsin(v / c)

E = m * v^2 * g - J * K + integral of K dJ

E = m * v^2 * g - m * v * c * arcsin(v / c) + integral of m * c * arcsin(v / c) dv

E = m * v^2 * g - m * v * c * arcsin(v / c) + m * c^2 * integral of arcsin(v / c) dv / c

Note (from integral table):

integal of arcsin(u) du = arcsin(u) + (1 - u^2)^.5 + C

E = m * v^2 * g - m * v * c * arcsin(v / c) + m * c^2 * (v / c) * arcsin(v / c) + m * c^2 * (1 - v^2 / c^2)^1/2

E = m * v^2 * g - m * v * c * arcsin(v / c) + m * v * c * arcsin(v / c) + m * c^2 * (1 - v^2 / c^2) * g

E = m * v^2 * g + m * c^2 * g - m * v^2 * g

E = m * c^2 * g

Almost there! When the velocity is zero, g = 1. Therefore, when the object is at rest, it still has some energy. This is called rest energy. So what is the equation for rest enegy?

E = m * c^2

I hate bOINGbOING.

I only started looking at it because of some faux connection with Factsheet 5 of old, and I think I came away from the zine experience with the realization that "all zines suck, unless you can use 'em to get back stage at shows."

But I keep going back because its easy to load a page. But I HATE them. I REALLY REALLY hate Cory, Xeni, Mark F. and whats-his-wife's-name (the original bOINGbOINGers). Okay, maybe I don't hate frauenfelder... he actually seems to be the only cool one.

And its not because of what internet oddities they collect, but they spin they put on them. Cory and Xeni especially have a 3rd grade mentality when approaching issues. Maybe I'm not a leftist no more; maybe I'm a capitalist apologist. Perhaps I'm just a moderate republican (I did find 1 thing that Bush did that I agree with!!) and I don't know it or can't admit it.

One of the worst is:
Gmaps hack shows effects of high-yield explosive detonations
Here's a haunting Gmaps hack: "The High Yield Detonation Effects simulator maps overpressure radii generated by a ground-level detonation; these radii are an indicator of structural damage to buildings. No other effects, such as thermal damage or fallout levels, are included in this tool. Note that the displayed rings are "idealized"; that is, no account is taken of terrain, urban density, ground type, weather conditions, and so on."

Wait. Back up. This so-called-hack draws concentric rings around a point. THATS IT. Thats all it does. I was doing shit like this in LOGO back in 1983.

If they had some how tied in a diagram of how our basic freedoms of stealing copyrighted work are being eroded away, it would be the perfect crystallization of why I hate these fuckers.

Its not chilling! The FCC rulings aren't schizophrenic! They are the very obvious conclusion of a republican administration (de-reg of dsl) that wants to use technology to fight crime, never mind the lack of social freedoms (eav's dropping req for the feds). Its completely rational the police can rifle through your GARBAGE to find evidence; you threw it out on the curb! Its also logical that they can look through your windows; if you have the shades open, you are advertising! ACk!

So why do I keep going back?

There is only one conclusion; I hate myself and want to die.

I also loved the readers write-ins on why Mark Frauenfelder isn't getting Crema from his tres expensive espresso machine; he's not jerking off on coffee beans that have been shat out by lemurs. But of course! It makes me want to drink sanka.

I am not a doctor. Cerebellar astrocytoma is a form of intracranial cancer which involves brain cells call astrocytes. It is the third most common type of cancer in juveniles. There are four grades of increasing severity defined by the World Health Organization. Juvenile cerebellar astrocytoma rarely leave the cerebellum. It is a section of the brain located near the brainstem and below the occipital lobe. The cerebellum helps direct balance, attention, and complex motor control (particularly involving vision-related feedback). It also helps a person judge the passage of time and is involved in language processing too.

Astrocytes are not neurons. They are star-shaped glial cells that commonly help form the structure of the brain and provide nutrition from blood vessels. Astrocytes are the largest cells in the brain and outnumber neurons by an order of magnitude. Astrocytes help limit the spread cerain toxic neurotransmitters. Through haemodynamic regulation they can also increase blood flow to areas of intense neural activity in the brain. Functional Magnetic Resonance Imaging (fMRI) of those areas helps biologists understand which areas of the brain corrolate with certain thought patterns.

Astrocytes may also play a role in certain types of neuron-to-neuronsignal transmission by isolating or withdrawing from synapses. They can also form a second communication network within the brain by releasing neurotransmitters in response to certain stimulations. However, it is at least several orders of magnitude slower than the neuronal network.

I love google, wikipedia, and especially the library, where I first learned about these things before the world wide web even existed! :-)

After some much needed yard work, vacuuming and laundry folding it felt as if the summer heat was getting to us. My feet ached and a short nap was in order. We slept, and the house slept with us. In a half-hazy gauze of warmth and sleep I noticed how quite everything was, with only the pitter-patter of little feet to wake us (and my little feet, I mean cat paws).

What I didn't notice was that we lost power for about an hour.

Normally, I'd just shrug it off, but it killed my computer! No, not the mini. My Win XP box- an overloaded HP pavilion.
I'll try to salvage a PSU from an older 1996 era metal tower but this is a good chance to ask the subset of slashdot:
Whats the best way to consistently protect against surges?

I know that some UPS also have the side benefit of conditioning the power "signal" on the line, but whats the most cost-effective solution?

All comments welcome.