Geek Talk

My New Laptop

2008-06-20T22:42:00.000-07:00

In a few weeks i'll finally have enough money to buy a fancy new laptop!

Features:

- Intel Core 2 Duo T9300 processor (6MB cache, 2.5MHz)
- 17" Ultrasharp screen, 1900x1200 resolution with truelife (glossy)
- 320GB SATA HDD
- 256MB Nvidia 8600 GeForce graphics card
- 3GB RAM
- All the software etc etc.
- $2200 from my bank account

Not to happy about the last part...

Once i've got my laptop I think i'll save up for a mean desktop!

The one good thing about about top of the line computers is that they are not even close to the cost of top of the line push bikes! So you really feel like your getting value for money.

Mark.

Falling Down the Rabbit Hole

2008-05-31T20:13:00.000-07:00

Many people run at the very thought of quantum physics, personally I love it.

I've always been the type that likes to try and visualise what's going on around us. I was scanning over youtube the other day and found a neat series of short clips from 'Dr. Quantum', its an animated old man in a cape, who very successfully explains some very complicated quantum topics (A simple youtube search will find the clips for you).

I've messing around with XNA for a while now, and i'm at the point where I can begin working on my final year university project, a quantum astrophysics simulation engine. The idea is that the user will be able to scroll around a 3D model of our solar system, and place all sorts of cool astro-disasters. Ever wondered what would happen if our sun had enough mass to make a black hole?, what would happen to Earth if the moon hit us?

It will also attempt to simulate a theoretical quantum perspective (only 2 of the possible infitite time dimensions though)

XNA seems realitively straight forward compared to native DirectX, so the programming side (I hope) will not take more than a year. The physics side... well I think i'm going to need some help from our superstar ex-spacerocket-builder lecturer for that one.

Anyway, it should be interesting, whether its too ambitious, well time will tell.

Mark.

Whoops

2008-05-09T16:11:00.000-07:00

Prove that sin(j*x) = j*(sinh(x)):

This is what I did:

sin(j*x) = (e^j*x - e^-j*x) / 2*j
j*(sinh(x)) = (e^x - e^-x) / 2

Therefore:

(e^j*x - e^-j*x) / 2*j = j*(e^x - e^-x) / 2

Whoops... If your not shaking your head already:

sin(j*x) = (e^jb - e^-jb) / 2j , where b = jx

sin(j*x) = (e^j*(j*x) - e^-j*(j*x)) / 2*j

sin(j*x) = (e^-x) - e^x)) / 2*j

sin(j*x) = ((e^-x) - e^x)) / 2*j ) * (j/j)

sin(j*x) = j*((e^-x - e^x)) / -2

sin(j*x) = j*(e^x) - e^-x)) / 2*j

-.-;

Can you spot the error?

2008-04-26T06:19:00.000-07:00

I was stuck on this problem for not 1, not 2 but 3 nights!

... Can you spot the stupid error I made?

Problem:
Let w = 2000
V = 6VL = 0.2H
R = 400

Given the differential equation L*(di/dt) + R*i = V*cos(w*t)

Find i(t). - The above equation is true for t > 0

NOTE: When im integrating ive used the '!' symbol

For the transient part:

-> 0.2*(di/dt) + 400*i = 6*cos(w*t)
-> (di/dt) + 2000*i = 30*cos(w*t)
-> di/dt + 2000*i = 0
-> di = -2000*i*dt
-> !(1/i)*di = !-2000*dt
-> ln(i) = -2000t + c
-> i(transient) = e^(-2000t)*e^c (Let e^c = A (a constant))
-> i(transient) = A*e^-2000t

For the steady state part:

Ohms law states that the current i = V/Z, where V is voltage, and Z is impedance.
So, V = 6>0 V. The '>' is the polar form (or phasor some might know it as) representation of the complex angle associated with the voltage, so 6V, 0 angle.

Z = R + j*w*L
Z = 400 + j*2000*0.2
Z = 400 + j400Z = 565.685>0.785 (all angles are in radians)

-> i = V/Z-> i = 6>0 / 565.685>0.785
-> i = 10.61>-0.785 mA
-> i(steady state) = 10.61*sin(w*t - 0.785) mA

To derrive i(t):

-> i(t) = i(steady state) + i(transient)
-> i(t) = A*e^-2000t + 10.61*sin(w*t - 0.785) mA

To find A make t = 0

-> i(0) = A(1) + (10.61*10^-3)*sin(-0.785)
-> 0 = A - 7.5*10^-3
-> A = 7.5*10^-3

So the final equation for i(t) is:
-7.5*e^-2000t + 10.61*sin(w*t - 0.785) mA

Trialling this solution by substitution does not satisfy the equation!

Delegates

2008-03-27T16:10:00.001-07:00

This has always been a topic that I have never quite grasped.
I learnt yet another cool trick with delegates that other day though!

You can start a new thread as below:

Thread thread = new Thread(this.Test);

private void Test()
{
// do stuff
}

But you can also do it like this:

Thread thread = new Thread(delegate()
{
// do stuff
}
);

Cool... :-).

Dragonball - The Movie

2008-03-21T18:00:00.000-07:00

Need I say more?

http://www.imdb.com/title/tt1098327/

You can make 1 = 2... Wrong

2008-03-12T23:20:00.001-07:00

A few people may have seen this video: http://www.youtube.com/watch?v=cqj5Qrxd5M0

Just to clear it up, you cannot make 1 = 2!

The working is as follows...

a = b

a + a = b + a

2a = b + a

2a - 2b = b + a - 2b

2(a - b) = a - b ---> (it's still correct here...)

Now the idiot that wrote this working claims that you can divide both sides by (a -b), and be left with 2 = 1.

Ok great, what he failed to mention is that if a = b, a - b must equal 0. This means he is saying (2(0))/0 = (0)/(0). You can't divide by zero so this is invalid math.

I'm sure a few people's hearts skipped a beat when they first saw this though. :)

The Perfect Assignment - Or Not...

2008-03-12T23:06:00.000-07:00

After completing my math assignment the other night I remembered a simple rule that I had completely forgotten, which I'm going to post here so I don't forget again...

Most people who can do math know about the trigometric quadrant chart below:

I was given the equation v = Asin(wt) - Bcos(wt), and asked to convert it to v = Rsin(wt + 0) form.

Now I knew that R must equal the root of (A^2 + B^2), and 0 must be tan^-1(B/A).

However, most of you will know that the angle can be the calculated value above, or the above value + pi. What I overlooked was that a positive sin and negative cos in the initial equation does not mean the angle is in the second quadrant.

v = Asin(wt) - Bcos(wt) can be written as cos(wt).sin(wt) - sin(wt).cos(wt), so A must equal cos(wt), and B must equal -sin(wt).

tan^-1(B/A) can therfore be written as tan^-1(-sin(wt)/cos(wt).

This means that the sin value is infact negative and cos value is positive, placing the angle in the 4th quadrant.

There goes 15 marks :-(.

The Zerg!

2008-03-12T01:06:00.000-07:00

Blizzard have finally released the first glimpse of the Zerg for their upcoming game Starcraft 2.

For those that know me, you all know that I'm a massive Blizzard fan (except for WOW - which I hate more than anything else...).

I'm looking forward to the games' release!

Check it all out at http://www.starcraft2.com

Osbourne's Rule

2008-02-29T14:36:00.001-08:00

I came about an interesting rule the other day for converting a trigometric function into a hyperbolic.

You can take some trigometric functions such as:

i ) sin(A+B) = sinA.cosB + cosA.sinB
ii) cos(A+B) = cosA.cosB - sinA.sinB

from here, if there is a sin.sin then the sign changes. Otherwise its exactly the same.

So the above become:

i) sinh(A+B) = sinhA.coshB + coshA.sinhB
ii) cosh(A+B) = sinhA.coshB + sinhA.sinhB

This might be a simple math rule to most, but in the past I had always found the functions by substituting their exponential equivalents in, and re-arrganging.

This makes things alot simpler

AlphaC Framework

2008-02-26T20:05:00.001-08:00

Im pleased to announce that I will be building the first (known) RTS graphics engine built on the XNA framework.

I am putting alot of effort into adding a dynamic unit testing suite, which will save massive amounts of testing time for anyone wishing to build a visually appealling 3D RTS game.

More details to follow.

Reflector

2008-02-26T01:20:00.000-08:00

If you don't have this program, download it... now!

http://www.aisto.com/roeder/dotnet/

It's remarkable.

It allows you to disassemble a .NET assembly and view it's internal coding. Very handy if you want to see what makes the .NET framework tick (or any other .NET assembly).

It also allows you to view the code in several different languages. I find it very handy for translating from C# to C++.

Matrices and Transforms

2008-02-26T01:10:00.001-08:00

Over the course of the summer I have been experimenting with 3D model transforms (moving a model in 3D space). It has really been interesting, as the concept is so simple!

You can describe a model's location in space by using a series of matrices. You can have a matrix for rotation, position, scale. You can then simply multiply these all together with an identity matrix and presto!, you have the model transformation!

So all you need to do is create a series of properties in your model class which return a matrix for each of the above, then when the model render phase is called, you can multiply the returned property values together to set the transform for the model.

The best thing about this method is you can use it to control the camera movement as well.

Welcome to my Blog!

2008-02-25T17:27:00.000-08:00

In order to conform with all other geeks, I thought I might start up a blog.

I'll aim to post my experiences while building my first RTS graphics engine, and other university related topics.

Mark.