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User Journal

Journal Journal: Better But Still Stressed Out 2

On Exams. I feel better though still stressed out. In the real world, I refer to documentation all the time even for basic stuff just to confirm things. Why do profs expect you to know everything from memory?

On Championships. Wow. I was afraid this might happen. Many of the people that I roll with during practice really want to prove themselves.

The first person I fought beat himself up for a few days afterward. I tried comforting him and instructing where he went wrong. Now he really wants a rematch even though I just want to have fun!!!

The guy who took third wanted to spar very hard Wednesday, because we didn't get to fight at the tournament. I think he wanted to know whether he could have beat me if he won his match. Instead I gather it clouded his judgement. During randori I tapped him twice. However, I was really sad and just played around for the rest of the session.

I grapple just to have fun and lose weight. That's it. I'm very worried that now I'm expected to do well at every future lesson and tourney, and everyone will be gunning for me! That is very stressful and not fun at all. :-/

On the other hand it was actually a relief yesterday when I went to the advanced BJJ practice, since I got decidedly crushed when rolling with the blue belts. I usually go on beginner BJJ day with most of the other white belts, and I was beginning to get nervous when my luck would run out!

Unexpected Consequences Odd that is felt better losing than winning. On the bright side, a clothing manufacturer wanted to sponsor me.... Methinks the advertiser is a tad too enthusiastic, or that he couldn't get Fedor to do endorce his product, or that he tipped the bottle too much. Still gave me a chuckle! :-D

Ya Drew! Thanks to a buddy of mine who also fought, you can download a quicktime highlight clip set to some funny music. Feel free to download more videos from our club if you wish.

I promise more Geeky stuff later...

User Journal

Journal Journal: First place yet still depressed (more non-nerd stuff) 4

I won first place in both the gi and no-gi lightweight novice divisions in the Ultimate Force 2005 grappling tournament. I fought twice with my kimono on and twice without. Therefore, I went 4-0 today. Pictures and video forthcomming.

I should feel great; however, I don't. I spent all week stressing over my weight (I was 160, now I'm 152) and having nightmares about losing in the first ten seconds. It is a relief that I don't have to worry any more!!! However, I am still a little nervous since now I must to well for now on. I will have to fight in more experienced weight divisions like intermediate or beginner in the future... I feel like everyone expects me to do good, and I keep thinking I will choke.

I even found myself rationalizing my wins. With my kimono on, the first person I defeated was a lot lighter than me and actually wanted me to go up a weight class. He is my friend and I look up him; I still feel he has better technique than me. In the championship gi match, my opponent was also a friend. I actually think fighing people I know is more fun, but he had trouble getting intense for the match. I ended up winning by scoring 12 points more than him (a technical submission).

Without the kimono I almost lost both matches; however, I won at the end partly because my opponents were wrestlers not grapplers. Their technique was very bad, and I just held out until they made a mistake. I tapped both of them; however, they were in much much better shape than me. I was very intimidated and expected to lose.

Does this take away from my victories? This is the first time I ever won first place at any sport. I want to be proud, and I bragged a little. But too much pride is often a sin (and not solely for religious reasons). I feel humbled by my opponents, yet nervous about the future. I'm even sad a little for hurting them enough to submit during the matches! (Not enough to quit, but still is worries me.)

I began studying Brazilian Jiu-Jitsu only to lose weight. Now I have, but I still feel fat. My self image is shot even though I won first place twice. Does anyone else feel not as good as they thought they would after winning?

User Journal

Journal Journal: Second Place at Grappling Tournament 5

Today, I got second place in the novice lightweight weight class at Mark Shrader's Third Annual Grappling Tournament. First, congratulations to the first place winner: Jay Morris. He is an awesome wrestler and grappler, and I wish him luck with his MMA career (no, we weren't fighting MMA).

Despite losing I shouldn't feel bad. In nearly six months of training I worked hard enough to lose 45 pounds (I'm 154 now); clothes that were too small for years are now too big. That was why I started grappling, after all. I only entered the tournament for the experience and not to place. I fought a tough opponent and still came out with a medal.

However, those are rationalizations. The only reason I got a metal is that there were only two people in my weight class: the champion and me. In other words, I lost and still received a metal. I was beaten in skill and strength. Even worse, during an exhibition later on, I tapped out to exactly the same submission from the same opponent. They weren't short matches, but they weren't especially long either. I'm angry at myself for losing in that way.

Every fight is won before anyone steps on the mat through how they prepare. Instead of dwelling on the past, I should direct my energies on the future.

  • Deluding myself, I though I was improving against the people I usually train against. I guess my pride was hurt a little. I need to train against different people so I don't get used to anyone.
  • The submission that got me was a guillotine. I need to learn to master, avoid, and counter them.
  • I also have to work on initiating things from the guard. My confidence with guard work was virtually nill even before the tournament.
  • I was out wrestled, so improving my stand-up is also a priority.
  • I did not feel winded or tired at all during the match, instead I was simply overpowered. I need a better lifting routine to get stronger. However, I should maintain this weight at least (154). Therefore, I still need to do more jogging and cardio as well to lose the dead weight.

I was debating throwing the silver metal away (which says made in China on the back). However, I'll keep it to motivate me to work on these goals.

User Journal

Journal Journal: Secrets of the Universe (Or Why E = mc^2) 6

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

User Journal

Journal Journal: Juvenile cerebellar astrocytoma (repost for posterity)

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! :-)

User Journal

Journal Journal: 1-2 5

Well, I'm back from the second grappling tournament entered so far. Last time I lost my only match, but this time I won once!

I fought in two divisions. In the beginner division (1 year experience), I beat my opponent by points and it really wasn't close. :-) Unfortunately in the next fight I got caught in an armbar and lost even though I was ahead by points again. In the novice division (6 months experience) my opponent submitted me from a guillotine. I hate those. :-(

So my record at NAGA is 1 win 2 losses. Another grappler told me that is good since 50% lose their first match but I'm still unsatisfied. Next year I'll improve my record. Good news is that I didn't lose due to conditioning (I never got really tired), so I just have to learn the counters to those submissions... :-)

The tourney was also a very humbling experience. There was another fighter the also in the Beginner division with the same experience as me who got third place out of 20 people! There was also the college wrestler who, which almost no grappling experience, placed third in the Intermediate division. Props to both fighters, of course. I got a long and interesting road ahead to compete with them. Here's to the journey!

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Journal Journal: Juvenile fun 7

I don't like memes and chain letters, but this sounds cool in a juvenile way. I'm feeling juvenile and silly tonight, so reply and... I'll respond with something random about you. (source)

  1. I'll tell you what song/movie reminds me of you.
  2. I'll pick a flavor of jello to wrestle with you in. (I perfer girls. ;-)
  3. I'll say something that only makes sense to you and me.
  4. I'll tell you my first memory of you.
  5. I'll tell you what animal you remind me of.
  6. I'll ask you something that I've always wondered about you.

I don't like chain letters, so you may or may not post this on your journal. Your choice. It is written somewhere, probably here. ;-)

User Journal

Journal Journal: This Person Does Not Know What He Is Talking About 3

Check out this comment in a story about Design Patterns. That person makes no sense whatsoever! I have moderator points, but I won't waste them on his post. Instead I'll respond here. However if he gets modded up I will be very upset!

The quite possibly most useless book in the history of computer science gets an award. Somehow I am not that surprised, considering that everybody hails it as the end all of object oriented design and everything.

Nobody seriously thinks Design Patterns is "end all" of OOD. Design patterns are an active research topic and much new material is published about them both in academic journals and your local bookstore every year.

To be honest, modern computer science curriculum seems to be wasting a lot of bright young potential on buzzwords. Patterns, paradigms, bleh. People somehow manage to get masters degrees in CS from Berkeley without even knowing what "turing complete", "Karnaugh map", "Rice's theorem", "Goedel's completeness theorem", "planar graph", "functional language", "church-turing thesis" are. But you ask them about a singleton, model-view controller or Java's security model in reflection and they're the fucking expert.

This is not true. I know what every one of those terms are except of "Rice's theorem," and I'm not even a computer science major! At my school they are covered in much detail. I know from taking the Digital Logic, Intro to Programming, and other courses offered there. Our computer science curriculum really does iterate though those topics, as does our computer engineering school.

Well that's barely computer science, that's just OO banging. Just because it uses paradigms and object oriented terms doesn't make it anything other than advanced code banging.

That is nonsence. He keeps using that word - paradigm - but I don't think he knows what it means. Then there is "code banging." Computer science is not just advanced mathematics even though the two fields are related. I guess actually writing programs is beyond him.

Really... I pity him for taking the time to write up such nonsense. I - on the other hand - wrote this reply to prevent misinformation and to vet some frustrations. Edit: A good reply was posted by an anonymous contributor. That's what should be modded up. Why do people intentionally try to sabotage communities like /.?

User Journal

Journal Journal: DOCTYPE declarations for versioning information

Over at Anne's journal there is a debate about using DOCTYPE declarations as versioning information. For example, the external subset for HTML 3.2 is different from the external subset for the HTML 4.01 family. There are also different external subsets for each "subversion" in the HTML 4.01 family. i.e. Transitional, Strict, and Frameset versions. Some people think this doesn't work. My opinion is that DOCTYPE declarations can be used to specify version information for the following reason:

  1. Say you have two documents with identical DOCTYPE decarations. Both the internal, external, and root element declarations are the same. Then you can say their syntactic doctypes are identical. The structure of each document must conform to the same SGML rules.

  2. The two documents could have different semantic doctypes though. That is, the meaning content in one document could mean something totally different in the other one even though they both conform to the same syntax rules in the DTD. For example, say the "rel" attribute is defined with character data content in the DTD. One document may specify that the rel attribute specifies a relationship and should be a character string; the other could specify that those attributes specify links and should be URI references.

  3. There could also be syntatic differences not captured by the syntatic doctypes. For example one spec may indicate that "name" and "id" attributes MUST have identical content if both are present; however, this is impossible to specify in a DTD.

  4. Therefore, the content of the external and internal subsets can not be used to differentiate between languages or versions of a single lanugage like HTML. There could be different syntatic or semantic meaning not captured by the doctype.

  5. External subsets are specified with either a Formal Public Identifier (FPI) or a URI reference to a DTD. However in practical applications a FPI uniquely identifies a resource just like a URI, so I'm going to assume both are the same for this line of reasoning, and I'll call both DTD names. DTD names have the following property: the owner of the name gets to determine what it means. For HTML's DTD names specified by the W3C, only the W3C gets to say what they mean, for example.

  6. Therefore, it is legal for a DTD names to indicate the doctype for a single language including semantic and syntatic requirements not captured in the content of the DTD. The meaning of the HTML 4.01 Strict doctype string is unambiguous even though the DTD's content may not specify all of the semantic or syntatic requirements of HTML 4.01 Strict.

  7. If you change the external subset's DTD name, it may or may not refer to the same language. Even if the external subset contains the same content, other requirements not encoded by the DTD could be different. Therefore, you can ONLY use the cannonical DTD names for unambiguously identifing the resource with third parties.

  8. Even when using cannonical DTD names, if you change the root element then you might no longer conform with the specification. For example, changing the root element while using HTML 4.01 strict's DTD name violates the global structure semantics of HTML 4.01 strict. Therefore, the document is not valid HTML even though it is syntatically valid SGML. Note that the DTD name still specifies the particular language you are using even though the content of resolving the DTD name is not enough to validate the document as HTML.

  9. If you add an internal subset, then the meaning of those changes is undefined even though the syntax of those changes specified as well as can be by the content of the subset. The content of the internal subset simply cannot capture the semantic or all the possible syntatic requirments you specify. Therefore if anything in the internal subset conflicts with the HTML 4.01 strict's external subset, or additional elements or attributes or attribute lists are defined, then the resulting lanugae is not HTML even though it is valid SGML for example.

  10. There is an exception to the above point. If the internal subset contains entity declarations with valid HTML content as its content (even though the entity by itself may not be valid HTML content), and those entity declarations don't interfere with the HTML DTD, then the meaning of those entities is clear (it is defined by SGML) and the specified syntax of the external subset is unchanged. Therefore, it is still HTML of the specified version for example.

Thus if you use the same internal subset content (with an exception), the same external subset declaration, and same root element declaration as the HTML language version you are declaring, then your document is HTML. If you change anything (with an exception), then it can never be unambiguously determined to be HTML by a computer.

If you can specify different languages using the above rules, then you can specify different versions of a language family using the above rules. Every version of a language family is a different language. They may share certain semantics, but they are not compatible except as explicitly defined by the language family's specification.

Therefore, you can use DOCTYPE declarations to specify version information. Q.E.D.

User Journal

Journal Journal: URI References and data: URI strings

Not to overload sw@w3, I'm going to propose answers to my questions on URI References in RDF. Any comments are greatly welcomed, since I don't know if I am correct!

1) RDF describes URI references - not URI strings themselves - in anticipation of IRIs (AKA a 'URIRef') URIRefs are always encoded in UTF-8 too. Correct?

Correct. (of course ;-)

2) My main question concerns converting a URIRef into a URI. Say we have the URIRef:

<data:,Hello, World>

Is that legal? Would that be converted into the URI:

<data:,Hello%2C%20World>

Since the comma is illegal in the URI after the first one?

Not entirely. The comma is legal in both the context and in RDF, so that URIRef would properly be converted to the URI:

<data:,Hello,%20World>

Note that I think (hope) that only the RDF semantics section applies and not the context in the scheme.

3) If there was an ambigious situation, how would it be represented as an URIRef? For example take the URI (yes, it is an unusual case where the name contains a slash - it is just an example):

<http://example.com/name%2Fslash/>

Would that be converted to the URIRef:

<http://example.com/name/slash/>

(I don't think so). But wouldn't the URIRef:

<http://example.com/name%2Fslash/>

be converted to the URI:

<http://example.com/name%252Fslash/>

That is all true, so the URI:

<http://example.com/name%2Fslash/>

Can not be represented as a URIRef. However that's not a problem, even though that means the scope of URIRefs is smaller than the scope of URI strings. RDF concerns URIRefs and not URI strings. So this is a non issue.

User Journal

Journal Journal: Howto: lose twenty pounds the wrong way 7

As of today I am 179.5 pounds - exactly twenty pounds less than I was two months ago! Yes, I was getting quite fat and I still have a lot to lose. However, I feel great and am finally down a weight class for the tourney. Realistically a target of 175 is required to improve my speed (less dead weight to carry around), but I'll try for 169 if I can. How did I do it?

Note: I am not a doctor. This is what I did and it could be harmful to you (or even me without realizing it). I am not a professional. Get professional advice and a doctor's help before attempting anything. Use at your own risk!

  1. First few weeks (lost 3-5 pounds for me):

    See a dietitian and eat according to how much you spend. You are a bank; to lose weight you must bounce your checkbook! :-p So keep track of what you eat and how much you exert through exercise (use machine or use estimates). Healthy food (fruits, vegetables) improves your "energy" so you can exercise easier. I used to go to Subway, but I stopped recently for other reasons. Limit your bread, cheese, and portions; increase your cereal/milk, fruits/vegetables, and number of meals. Eating only a few meals makes your body go into sleep-mode.

  2. Next weeks (lost 5-7 pounds more for me):

    1. Continue the previous stuff.

    2. Run run run. Do 25-35 minute workouts two-three times a week Use an elliptical if your joints hurt or you just don't running in general. You get a similar workout yet they feel much better! Use the "weight loss" workout - oscillating between high and low inclines - with some resistance (start low). Make sure you stay within your target heart rate zone. If you don't, then you will tire easily and quit.

    3. Lift weights. A strong upper body increases your metabolism and also helps make running easier. Make sure you don't hurt yourself and always use a spotter, but try to ramp up the weight to 90% or failure during reps. (Most people don't lift enough.) Try to do it two-three times a week between running days. It helps to work with a few friends, as it is safer and you push each other harder.

    4. Do a martial art one-two times a week. I joined a wrestling club at my university then did some introductionary jiu-jitsu at the one next door when the other club adjourned for the summer. You will suck at it if you never did it before (or are out of practice). That's OK!

  3. Next Weeks (lost 7-9 pounds):

    1. Continue the previous stuff, if you can. Remember to eat more as you exercise more. You need the energy, but your body will burn food more quickly rather than conserve it in fat stores. I modified my schedule to allow other stuff, but I want to add in more running/weight lifting.

    2. Train at your martial art school two days a week.

    3. Train at your martial arts club two days a week too (or go to your school). They won't be as good, but the sparring will give you a workout.

    4. Do your martial arts drills two more days a week in two minute periods. (i.e. two minutes one, two minutes rest, two minutes second, two minutes rest, etc...) Do it by yourself or with your group. This improves both your art and your conditioning.

    5. Friday is fun day!

That's what I did so far. Right now I'm working out six days a week with Friday off to party/have fun. I don't do everything as much as I can, and I'm far from perfect (missing days, not doing something, slacking off, overeating). The important thing seems to be to go back into the rhythm before falling into couch-potato mode.

Time is a problem, but I try to force exercising. I'm 10x more productive after working out, and my computer programming has increased in quality I think. Abstract algebra just doesn't kick in the fight-or-flight response! :-)

It might not work with you, and I don't think my workout is anywhere near either a scientific nor great thing. It seems to be working for me right now, so I hope it gives you ideas or inspiration to lose weight.

Note: I am not a doctor. This is what I did and it could be harmful to you (or even me without realizing it). I am not a professional. Get professional advice and a doctor's help before attempting anything. Use at your own risk!

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