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Linear Algebra/Span of a set

Definition
Let V be a vector space over a field F. Choose n vectors x1, x2, x3, ..., xn from the vector space V. The linear manifold spanned by x1, x2, x3, ..., xn is defined to be all elements of V of the form a1x1+a2x2+a3x3+...+anxn where a1, a2, a3, ..., an are all elements of the field F, and shall be denoted S(x1, x2, x3, ..., xn). This is obviously a linear subspace of the vector space V. Since every linear subspace of V contains x1, x2, x3, ..., xn and their linear combinations, S(x1, x2, x3, ..., xn) is the smallest subspace containing x1, x2, x3, ..., xn.

Theorem
If y1, y2, xy, ..., ym are elements of S(x1, x2, x3, ..., xn), then S(y1, y2, y3, ..., ym) is contained within S(x1, x2, x3, ..., xn)

Proof
All linear combinations of vectors which belong to a linear manifold also belong in the linear manifold (since a linear combination of linear combinations of vectors is also a linear combination of those vectors), and since any element of S(y1, y2, y3, ..., ym) is a linear combinations of vectors within the manifold, it too is within the set, thus proving that S(y1, y2, y3, ..., ym) is contained within S(y1, y2, y3, ..., ym).

Theorem
If x is linearly dependent on other vectors upon other vectors x1, x2, x3, ..., xn, then it belongs to S(x, x1, x2, x3, ..., xn) also

Proof
x, x1, x2, x3, ..., xn all belong to S(x1, x2, x3, ..., xn), then S(x, x1, x2, x3, ..., xn) must be contained within S(x, x1, x2, x3, ..., xn). Therefore, if x is linearly dependent upon x1, x2, x3, ..., xn, then S(x, x1, x2, x3, ..., xn) is equal to S(x1, x2, x3, ..., xn).

Theorem
The maximum number of linearly independent vectors of a set of vectors is equal to the dimension of the span of the set.

Proof
Suppose that there are d linearly independent vectors among x1, x2, x3, ..., xn with all other vectors being a linear combination of those d linearly independent vectors. This number d is the maximum number of linearly independent vectors among x1, x2, x3, ..., xn. Then any element of S(x1, x2, x3, ..., xn) must be a linear combination of those d linearly independent vectors, so they form a basis, and so d is the dimension of S(x1, x2, x3, ..., xn), which is equal to the maximum number of linearly independent vectors among x1, x2, x3, ..., xn.

title: Linear Algebra/Span of a set


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General Genetics/Gene Expression

Gene expression is basically the control of the cell over which genes to make and which not to make at a specific time. Now why would a cell want to do this? The example we'll use to explore prokaryotic gene expression is the lac operon. The lac operon controls the transcription of genes to make an enzyme to catalyze lactose. It turns lactose into glucose and galactose, the glucose is used in cell respiration to make energy for the cell. The cell takes lactose from the environment and catalyzes it. But if there is already glucose in the environment whats the point of taking lactose when it requires an extra step to make it useful? The cell takes actions to conserve energy and protein by not making the enzyme required for lactose metabolism.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: General Genetics/Gene Expression


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User:Michaelwayw2

Hi!

title: User:Michaelwayw2


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User:Sesquialtera II

Greetings,

I'm an organist primarily working on the Organ subject.

Please visit my userpage on en.Wikipedia for more information.

Links for my own use: /Sandbox

title: User:Sesquialtera II


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LEGO Design/Software

During the last couple of years, the LEGO company and community have developed several software packages to improve the builders' experience with LEGO. Some of these software packages are required to program certain LEGO robotics kits (Mindstorms etc).

Lego Digital Designer (LDD)
Lego Digital Designer is a freeware download available at Lego.com. LDD is mainly used to virtually construct and animate your own home-made models and create a set of printable instructions for your model. After your model is completed, you can check the price of your model if you wish to buy the pieces required to construct it, as well as uploading it onto LEGO gallery for other Lego fans to see.

Lego Building Instructions
Lego Building Instructions is an official mobile app available for in Google Play Store and App Store for viewing instructions of lego sets in a 3d environment where you can zoom, rotate and track your progress.

Lego Mindstorms EV3
Lego Mindstorms EV3 is a software package that comes with the Lego Mindstorms kit. It can also be bought from the Lego store. You can use this software to program your Mindstorms creation with an easy drag and drop, no programming skills needed interface.

Lego Boost
Lego Boost is a mobile app to program your Boost robot with a simple drag and drop environment just like its predecessors.

Other Software
There are many other unofficial software packages available for download on the internet. Some of this software allows the user to program the EV3 beyond using the official software package by using ev3dev (a Debian Linux-based operating system).

LeoCAD
Leocad is a free and open source Lego CAD software available at leocad.org. Just like LDD, its used to create Lego models with the rich up-to-date LDraw library. The models can be converted to parts lists, instructions, obj files and rendered with built-in POV-ray.

Stud.io
Stud.io is another freeware CAD software developed by Bricklink that allows purchasing the build you made in the software from Bricklink store.

Mecablocks
Mecablocks is a web-based online Lego CAD software with an extensive parts list and Blender-based online rendering that gives professional results. While its free to use, the site requires inserting credits for faster and better rendering results.

ev3dev
Ev3dev is a Debian Linux-based operating system for your EV3 brick to fully use its capabilities in a unix-like programming environment. Its mostly used by programmers to extensively use the Mindstorms environment.

title: LEGO Design/Software


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User:I Heart Taylor Guitars

Hello
As you can see from my name, I like Taylor Guitars. I have a Big Baby guitar.

title: User:I Heart Taylor Guitars


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User:Jacques Bergeron/T1

This is a test template - please do not delete

title: User:Jacques Bergeron/T1


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Physics Exercises/Vibrations and Sound

Waves, basic
The distance between successive crests in the waves in the ocean is 7.5 m. The crests travel 12 m in 4.0 s.  What is the frequency of a buoy bobbing up and down in this water?

Solution
First, we calculate the velocity of water wave: $$v = \frac{\Delta S}{\Delta t} = \frac{12~\mathrm{m}}{4.0~\mathrm{s}} = 3.0~\mathrm{m/s}$$. We are given information about wavelength, so we can solve the problem using the following relationship between velocity, wavelength, and frequency: $$f = \frac{v}{\lambda} = \frac{3.0~\mathrm{m/s}}{7.5~\mathrm{m}} = 0.40~\mathrm{Hz}$$.

title: Physics Exercises/Vibrations and Sound


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Physics Exercises/Magnetostatics

 * 1) Earth is essentially a magnet. Why then does the Earth's magnetic field have little effect on the weight of ferromagnetic objects?  (Hint: What is the net force on an induced dipole due to a (more or less) constant magnetic field?)

title: Physics Exercises/Magnetostatics


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Physics Exercises/Blackbody radiation

 * 1) What is the maximum efficiency of the black body as a source of visible light? Assume "visible light" means EM radiation between 380 nm and 750 nm.  (Hint: First work out the optimal surface temperature (it should be close to the surface temperature of the sun), and then work out the efficiency. Integral using Planck's blackbody radiation formula required.)
 * 2) What is the maximum temperature of the light side of the moon? Assume that the amount of sunlight incident on the moon is also given by the solar constant.

title: Physics Exercises/Blackbody radiation


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Physics Exercises/Circuits

 * 1) How does the change in resistance of an incandescent lamp and the thermal inertia of the filament affect its reactance?
 * 2) A simple lead-acid battery charger consists of a variable transformer, rectifier, moving coil ammeter and a fuse. Why the fuse blows well before the current, according to the ammeter, reaches the rated value of the fuse.

title: Physics Exercises/Circuits


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Physics Exercises/Electromagnetic Waves

 * 1) Consider an oil film on water. If you see green light (532 nm) reflecting back when sunlight is normally incident, what is the minimum thickness of the film?
 * 2) An astronaut on board the International Space Station is giving a presentation using his 5 milliwatt green laser pointer (at 532 nm). The pointer projects a bright dot of 1 millimeter diameter on a screen. During the presentation he accidentally shines his laser pointer out the window toward Earth. The Space Station is in the Earth's shadow at this moment and a person on Earth happens to be in the beam. If the Space Station is at an angle $$\alpha$$ above the horizon, how bright will it appear to this person? Express the brightness in terms of intensity (power per area) and assume that the collimation of laser pointer is diffraction-limited (in reality, collimation of laser pointer beam is not as good as that).

title: Physics Exercises/Electromagnetic Waves


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LEGO Design/Power Functions

Power Functions (PF) are a LEGO Product that allows the user to motorize their design. Unlike the NXT (See LEGO Mindstorms), the Power functions range cannot be programmed and does not have any sensors. The Power Functions equipment includes:

Large Motor - A motor smaller than the Mindstorms Servo Motor, but equally as powerful.

Medium Motor - A motor slightly smaller and less powerful than the Large Motor

Battery Pack - Requires six AA batteries. Powers the Power Functions

IR Receiver - Picks up Infra-Red signals from the remote, has 4 channels

IR Transmitter Remote - Allows the user to control their model via Infra-Red. Has 4 channels

Sound Block - Acts like a Touch Sensor only plays a sound when triggered. Included with the LEGO CREATOR Dinosaur.

 Lights - 2x LED lights connected together by a "y" shaped wire - gives your model lights ^^

 Switch - enables forward/off/reverse control of a motor ^^

^^ = available in LEGO kit 8293 - supplementary motor kit

title: LEGO Design/Power Functions


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Physics Exercises/Forces

 * 1) Suppose you are in a moving train. Assume it is moving with a constant acceleration. You have just come from the mall, and have with you only a ball of string, a long ruler, and a massive paperweight. Describe how to find the acceleration of the train using these items. Assuming that the train accelerates at 3 m/s2, what would you measure?
 * 2) [[Image:Soyuz rocket engines.jpg|200px|thumb|The booster and 2nd stage engines of the Soyuz rocket.]] Why are the booster engines of the Soyuz rocket inclined at a small angle to the 2nd (central) stage? What are the advantages and disadvantages of this design?

title: Physics Exercises/Forces


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Physics Exercises/Ideal Gas Law

 * 1) An air bubble at the bottom of a lake 50.0 m deep has a volume of 10 mL. If the temperature at the bottom is 4°C and at the top 27°C, what is the volume of the bubble just before it reaches the surface?

title: Physics Exercises/Ideal Gas Law


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Physics Exercises/Heat Engines

 * 1) At a steam power plant, steam engines work in pairs, the heat output of the first one being the approximate heat input of the second. The operating temperatures of the first are 800 deg C, and 500 deg C, and of the second 485 deg C and 300 deg C. If the heat of combustion of coal is 2.8 x 107 J/kg, at what rate must coal be burned if the plant is to put out 1 GW of power? Assume the efficiency of engines is 70 percent of the Carnot efficiency operating between the said temperatures.

title: Physics Exercises/Heat Engines


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Physics Exercises/Electrostatics

Electric field superposition principle
1. In words, and by equation, state the principle of superposition for electric fields.

In physics, the electric field superposition principle postulates that, for all linear systems, the net response at a given place and time caused by two or more charges is the sum of the responses which would have been caused by each charge individually.

Mathematically that can be written:

 * $$\vec E=\vec E_1+\vec E_2+\vec E_3+...+\vec E_n=\sum_{i=1}^n\vec E_i=\sum_{i=1}^n\frac{1}{4 \pi \epsilon_0 r^2}\vec {u_r}_i$$

Note that this is a vector equation. Hence one must add the components using vector rules.

Finding the electric field associated with a dipole
1. An electric dipole is matched, equal but opposite, charges separated by a fixed distance. Find the electric field associated with an electric dipole. Perform this for three cases:
 * a. Find the electric field perpendicular to the axis between dipole charges. Find the far field approximation.
 * b. Find the electric field along the axis of the dipole charges. Compare the far field approximation for this case with that of a) above.
 * c. Find the electric field perpendicular in a plane which includes the dipole charges

In physics, the electric dipole moment is a measure of the separation of positive and negative electrical charges in a system of charges, that is, a measure of the charge system's overall polarity. It is important in understanding practical matters like the construction of a capacitor.

Finding the electric field perpendicular to the axis between dipole charges
To determine the electric field which result from the dipole charges in Figure 1 that are separated by a distance 2a, we simply sum the fields from each charge individually. Note that the fields are vectors.

 * $${\vec E}={\vec E_1}+{\vec E_2}$$

The distance of each charge (r) from the point where the electric field is measured is:

 * $$ r =\sqrt{a^2+x^2}$$

The magnitude of the electric field (E1) due to the first charge (q1) and the magnitude of the electric field (E2) due to the second charge (q2) are the same although the directions are different.

$$E_1 = E_2 = \frac{1}{4\pi \epsilon_0} \frac{q}{r^2} =  \frac{1}{4\pi \epsilon_0} \frac{q}{a^2+x^2} $$

The components of the electric field along the x axis $$E_{1_x}$$ and $$E_{2_x}$$ have the same magnitude but opposite directions and cancel each other out.

 * $$E_{1_x} + E_{2_x} = 0$$

As a result of this the horizontal components of the electric field vectors cancelling is that summing $${\vec E_1}$$ and $${\vec E_2}$$ reduces to summing $$E_{1_y}$$ and $$E_{2_y}$$. Further, the magnitude of the vertical components are identical, $$E_{1_y}=E_{2_y}$$. Since $$\cos\theta = \frac {adjacent} {hypotenuse} $$ finding the vertical component reduces to:

 * $$E_{1_y} = E_{2_y} = E_1 \cos\theta\,\!$$ or :$$E_y = 2E_1 \cos\theta\,\!$$

But we also know that:

$$\cos\theta = \frac {a}{\sqrt{a^2 + x^2}}$$

So we can substitute this into the equation for $$E\,\!$$ which yields:

$$E = 2E_1 \cos\theta\,\! = \frac{2}{4\pi \epsilon_o}\frac{q}{a^2+x^2}\frac{a}{\sqrt{a^2+x^2}} =\frac{1}{4\pi \epsilon_o}\frac{2aq}{(a^2+x^2)^{\frac{3}{2}}} $$

In the region distant from the dipole x is much greater than a (x>>a). This is commonly referred to as the "far field" region. We can find how the field behaves in the far field region by factoring out x.

$$E =\frac{1}{4\pi \epsilon_o}\frac{2qa}{(a^2+x^2)^{\frac{3}{2}}} =\frac{1}{4\pi \epsilon_o}{\frac {2qa}{x^3}}\frac{1}{(1+{(\frac {a}{x})}^2)^{\frac{3}{2}}} $$

By inspection we can conclude that when x>>a then this approaches

$$E \approx \frac{1}{4\pi \epsilon_o}\frac{2qa}{x^3} $$

An important question for some analyses might be: over what range is the far field approximation valid? We can find out by using a binomial expansion (review of binomial expansion immediately below this section if you need it).

$$E =\frac{1}{4\pi \epsilon_o}\frac{2qa}{(a^2+x^2)^{\frac{3}{2}}} =\frac{1}{4\pi \epsilon_o}{\frac {2qa}{x^3}}\frac{1}{(1+{(\frac {a}{x})}^2)^{\frac{3}{2}}} $$

Where we will expand

 * $$\frac{1}{(1+{(\frac {a}{x})}^2)^{\frac{3}{2}}} $$$$= (1+{(\frac {a}{x})}^2)^{-\frac{3}{2}} = 1 -\frac{3}{2} {(\frac {a}{x})}^2 + ... $$

If you want to use the approximation only when it introduces an error less than 1%, then - since the first term is unity, you want the second term to be less than 1/100th of unity, or:

 * $$ \frac {1}{100} > \frac{3}{2} {\left (\frac {a}{x} \right )}^2  $$

 * $$ \frac {2}{300} > { \left(\frac {a}{x} \right)}^2  $$

 * $$ \sqrt {\frac {2}{300}} > {\left (\frac {a}{x} \right )}  $$

 * $$ x > \sqrt {150} \cdot a $$

So if x is 12.3 times larger than a, this approximation is valid within 1%.

Binomial expansion review
Review of the binomial expansion:

 * $$\mathbf{ (1 + a)^n} = 1 + \frac {n}{1!} \cdot a + \frac {n(n-1)}{2!} \cdot a^2 +  \frac {n(n-1)(n-2)}{3!} \cdot a^3 + ...$$

 * $$\mathbf{ (1 + a)^{- \frac {3}{2} }} = 1 + \frac {1}{1!} \cdot \frac {-3}{2} \cdot a + \frac {1}{2!} \cdot \frac {-3}{2} \cdot \frac {-5}{2} \cdot a^2 +  \frac {1}{3!} \cdot \frac {-3}{2} \cdot \frac {-5}{2} \cdot \frac {-7}{2} \cdot a^3 + ...$$

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Finding the electric field along the axis formed by dipole charges

The net electric field is the sum of the contributions to the electric field from the dipole charges.

$${\vec E}={\vec E_1}+{\vec E_2}$$

The first thing one always should do when solving a problem is to consider what you know by simple inspection. In this case the magnitude and layout of the charges tells use several things. Since they are equal and we know that the electric field from a point charge falls off as 1 over r2 then we know that the nearest charge will always dominate. Further we also know that along the axis that we're calculating the electric field there will be no vertical component of the electric field.

$$E_{1_y}= E_{2_y}=0$$

So now we will calculate the magnitude of the electric field from each in the x direction: $$E_{1_x}$$ and $$E_{2_x}$$.

$$E_1 = \frac{1}{4\pi \epsilon_0} \frac{q}{{(r-a)}^2}\qquad\qquad E_2 =  \frac{1}{4\pi \epsilon_0} \frac{-q}{{(r + a)}^2}$$

Adding the vector magnitudes $$E_{1_x}$$ and $$E_{2_x}$$

$$E=E_{1_x} + E_{1_x}\,\!$$

And substituting in the equations above yields:

$$E=\frac{1}{4\pi \epsilon_0} \frac{q}{{(r-a)}^2} + \frac{1}{4\pi \epsilon_0} \frac{-q}{{(r + a)}^2}= \frac{1}{4\pi \epsilon_0}\left [ \frac{q}{{(r-a)}^2}- \frac{q}{{(r + a)}^2}\right ]=\frac{1}{4\pi \epsilon_0}\ 4aq\ \frac{r}{{(r^2 - a^2)}^2}$$

The quantity 2qa is defined as the dipole moment (p) and so we can substitute:

$$p=2qa\,\!$$

With that substitution we now get:

$$E=\frac{1}{2\pi \epsilon_0} \frac{pr}{{(r^2 - a^2)}^2}$$

Consider the far field case (i.e., the case where $$r\,\!$$ >> $$a\,\!$$)

Then one can see that $$r^2\,\!$$ >> $$a^2\,\!$$ and that $$r^2 - a^2\,\!$$ approaches $$r^2\,\!$$, which allows us to simplify the equation in the far field.

$$E=\frac{1}{2\pi \epsilon_0} \frac{p}{r^3}$$

Note that one can use the binomial theorem just as in the section above to find out where this far field approximation (x>>a) is valid.

Finding the electric field associated with a dipole anywhere on a plane
Now that you've solved for two special cases and determined that the dipole field falls off as $$ \frac {1} {r^3} $$ at great distances (referred to as the far field), it is instructive to find the field's behavior off the axis.

Consider the case of an electric dipole's field at a point $$P\,\!$$ with the coordinates $$(x,y)\,\!$$ as represented in the figure. Note that we can represent the coordinates either as Cartesian, or as cylindrical.

 * $$r_1=\sqrt{{(y+a)}^2+x^2}\quad\cos\alpha=\frac{y+a}{\sqrt{{(y+a)}^2+x^2}}\quad\sin\alpha=\frac{x}{\sqrt{{(y+a)}^2+x^2}}$$

 * $$r_2=\sqrt{{(y-a)}^2+x^2}\quad\cos\beta=\frac{y-a}{\sqrt{{(y-a)}^2+x^2}}\quad\sin\beta=\frac{x}{\sqrt{{(y-a)}^2+x^2}}$$

In Cartesian coordinates the electric field at any point in the plane from each charge can be found as follows:

 * $${E_r}_1= \frac{1}{4\pi\epsilon_0}\frac{+q}{{(y+a)}^2+x^2}\quad{E_r}_2= \frac{1}{4\pi\epsilon_0}\frac{-q}{{(y-a)}^2+x^2}$$

We can apply the principle of superposition and add the fields from each of the charges, remembering that they are vectors so that the vertical components add to vertical and horizontal add to horizontal.

First finding the horizontal coordinates labeled as $$E_x\,\!$$:

 * $${{E_r}_1}_x= \frac{1}{4\pi\epsilon_0}\frac{+q}{{(y+a)}^2+x^2}\sin\alpha=\frac{1}{4\pi\epsilon_0}\frac{+q}{{(y+a)}^2+x^2}\frac{x}{\sqrt{{(y+a)}^2+x^2}}=\frac{1}{4\pi\epsilon_0}\frac{+qx}{{\left [ {(y+a)}^2+x^2\right ]}^{\frac{3}{2}}}

$$

{{E_r}_2}_x= \frac{1}{4\pi\epsilon_0}\frac{-q}{{(y-a)}^2+x^2}\sin\beta=\frac{1}{4\pi\epsilon_0}\frac{-q}{{(y-a)}^2+x^2}\frac{x}{\sqrt{{y-a)}^2+x^2}}=\frac{1}{4\pi\epsilon_0}\frac{-qx}{{\left [ {(y-a)}^2+x^2\right ]}^{\frac{3}{2}}} $$

Next finding the vertical components for each of the two charges labeled as $$E_y\,\!$$ we obtain:

 * $${{E_r}_1}_y= \frac{1}{4\pi\epsilon_0}\frac{+q}{{(y+a)}^2+x^2}\cos\alpha=\frac{1}{4\pi\epsilon_0}\frac{+q}{{(y+a)}^2+x^2}\frac{y+a}{\sqrt{{(y+a)}^2+x^2}}=\frac{1}{4\pi\epsilon_0}\frac{+q(y+a)}{{\left [ {(y+a)}^2+x^2\right ]}^{\frac{3}{2}}}

$$

{{E_r}_2}_y= \frac{1}{4\pi\epsilon_0}\frac{-q}{{(y-a)}^2+x^2}\cos\beta=\frac{1}{4\pi\epsilon_0}\frac{-q}{{(y-a)}^2+x^2}\frac{y-a}{\sqrt{{y-a)}^2+x^2}}=\frac{1}{4\pi\epsilon_0}\frac{-q(y-a)}{{\left [ {(y-a)}^2+x^2\right ]}^{\frac{3}{2}}} $$

Summing the horizontal components yields:

 * $$E_x=\frac{1}{4\pi\epsilon_0}q \left \lbrack \frac{x} {{\left [{(y+a)}^2+x^2 \right ] }^{\frac{3}{2}}}-\frac{x} {{\left [{(y-a)}^2+x^2 \right ] }^{\frac{3}{2}}}\right \rbrack

$$

And for large r (either large x, large y, or their combination) this becomes:

 * $$E_x=\frac{qa}{2\pi\epsilon_0}\frac{1}{r^3}

$$

Summing the vertical components yields:

 * $$E_y=\frac{1}{4\pi\epsilon_0}q \left \lbrack \frac{y+a} {{\left [{(y+a)}^2+x^2 \right ] }^{\frac{3}{2}}}-\frac{y-a} {{\left [{(y-a)}^2+x^2 \right ] }^{\frac{3}{2}}}\right \rbrack

$$

Finding the electric field associated with an electric quadrupole

The electric field measured at point along the vertical axis due to any single charge along the horizontal axis (shown in Figure 3) is:

 * $$E_q=\frac {q}{4\pi \epsilon_o r^2}$$

So the center charges contribute:

 * $$E_{2q}=\frac {2q}{4\pi \epsilon_o r^2}$$

While the left and right charges contribute a total field of magnitude:

 * $$E_{-q}=\frac {-q}{4\pi \epsilon_o (r^2+d^2)}$$

By inspection we can see that the horizontal components of the electric field cancel out. Only the vertical components are left. So we multiply by

$$ sin \theta = \frac {r}{\sqrt{r^2+d^2}} $$

To get the vertical component of the electric field for a single negative charge:

 * $$E_{-q} =\frac {-q}{4\pi \epsilon_o (r^2+d^2)}\frac {r}{\sqrt{r^2+d^2}}$$

And for the two negative charges together we get:

 * $$E_{-2q} =\frac {-2q}{4\pi \epsilon_o (r^2+d^2)}\frac {r}{\sqrt{r^2+d^2}}$$

Now we can add to obtain the total vertical component of the electric field.

 * $$E = E_{2q}+E_{-2q} =\frac {2q}{4\pi \epsilon_o r^2}+ \frac {-2q}{4\pi \epsilon_o (r^2+d^2)}\frac {r}{\sqrt{r^2+d^2}}$$

Which simplifies to:

 * $$E=\frac {q}{2\pi \epsilon_o}\left [ \frac {1}{r^2}-\frac{r}{{(r^2+d^2)}^{\frac{3}{2}}}\right ]$$

Factoring out r3 from the denominator's final term yields:

 * $$E=\frac {q}{2\pi \epsilon_o}\left [ \frac {1}{r^2}-\frac{r}{{r^3 (1+(\frac {d}{r})^2)}^{\frac{3}{2}}}\right ]$$

Which reduces to:

 * $$E=\frac {q}{2\pi \epsilon_o r^2}\left [ 1-\frac{1}{{ (1+(\frac {d}{r})^2)}^{\frac{3}{2}}}\right ]$$

So now we have a perfectly good equation for calculating the field. An engineer might think that sufficient since they can calculate the value when they need it, but a physicist would like to know how this behaves in the "far field" where r >> d. Fortunately we have a tool - our friend the binomial expansion - which allows us to answer the far field question, as follows:

 * $$\left [ 1-\frac{1}{{ (1+(\frac {d}{r})^2)}^{\frac{3}{2}}}\right ] = \left [ 1- {{ \left (1+ \left ( \frac {d}{r} \right )^2 \right )}^{-\frac{3}{2}}} \right ] = \left [ 1- \left ( 1 -\frac {3}{2} \left ( \frac {d}{r} \right )^2 \right ) + ... \right ] = \frac {3}{2} \left ( \frac {d}{r} \right )^2 $$

Substituting back into the original equation:

 * $$E=\frac {q}{2\pi \epsilon_o r^2}\frac {3}{2} \left ( \frac {d}{r} \right )^2$$

Simplifying, we find that in the far field when r >> d the electric field from a quadrupole falls off as $$\frac {1}{ r^4} $$

 * $$E=\frac {3qd^2}{4\pi \epsilon_o r^4}$$

Electric field associated with a line charge

 * 1) A very long solid nonconducting cylinder of radius R0 and length L (R0 << L) possesses a uniform volume charge density ρE (C/m3). For points far from the ends and for which r << L,
 * 2) Find electric fields for all r, distance from axis of cylinder. You may need separate expressions for r < R0 and r > R0. Show that electric field is continuous at r = R0, i.e. two expressions for electric fields agree at r = R0.
 * 3) Describe the motion of a small free charge Q of opposite sign (i.e. Q ρE < 0) placed inside the cylinder. Find the frequency of oscillation $$\nu$$ if the charge oscillates about the axis of cylinder. (Note: $$\nu$$, frequency, has units of cycles / second, while $$\omega$$, angular frequency, has units of radians / second.)
 * 4) Let's change the problem and say we have a cylindrical shell of charge (of radius R0, and σE so that the total charge remains the same). How do the expressions for electric fields change for either regions r < R0 or r > R0? Is the electric field still continuous? Give a plausible argument why we might not expect electric field to be continuous.

Find the electric field for a finite line with uniform unit charge of λ coulombs per unit length

The figure shows an finitely long line chage with linear charge density of $$\lambda \,\!=\frac{dq}{dx}$$. The charge extends from x = -a to x = +b which are arbitrary (although not infinite) and are not shown on the figure.

Just as it the derivation for the infinite lint, the magnitude of the electric field which results from a short section dx of this line charge as measured at point P is a result of the charge $$dq=\lambda dx \,\!$$:

And with that charge we can calculate the field contribution (dE) from dq as:

 * $$dE=\frac{1}{4\pi {\epsilon}_0}\frac{dq}{r^2}=\frac{1}{4\pi {\epsilon}_0}\frac{\lambda dx}{x^2+y^2}$$          (1)

We can no longer argue by symmetry that the electric field will be the same at any point of equal separation from the infinite line charge, so we can solve this in detail, which we will do using 2 dimensions. Once we have a solution for 2 dimensions, simple rotation around the line charge will allow us to determine the field in the 3rd dimension.

The electric field is a vector which we'll represent as $$d\vec E$$. This we'll separate into 2-dimensional Cartesian components:

 * $$dE_x=-dE \sin \theta\,\!$$     and      $$dE_y=dE \cos \theta\,\!$$

The sign of the electric field at $$dE_x\,\!$$ indicates the $$d{\vec E}_x$$ location of the point with respect to x.

We find the total electric field for the horizontal $$E_x$$ and vertical $$E_y$$ components to determine $$\vec E$$ at the point P by integrating. Note that $$\sin \theta$$ is used to find the components of the field horizontally, while $$\cos \theta$$ is used to find them vertically.

 * $$E_x=\int dE_x=-\int_{x=-a}^{x=+b} \sin \theta \, dE$$

and

 * $$E_y=\int dE_y =\int_{x=-a}^{x=+b} \cos \theta \, dE$$

Or in full vectorial form:

 * $$\vec{E}=\int_{L} \, d\vec{E}=\int_{L}\frac{1}{4 \pi \epsilon_0}\frac{dq}{r^2}\vec {u_r}=\int_{L}\frac{1}{4 \pi \epsilon_0}\frac{\lambda dx}{r^2}\vec {u_r}$$

In this example the distance, r, from the charge to point P can be expressed in terms of Cartesian coordinates x and y.

$$r = \sqrt {x^2+y^2}$$.

which allows us to calculation the magnitude of the electric field at point P from the charge dx at r using the equation. $$dE=\frac{1}{4\pi {\epsilon}_0}=\frac{1}{4\pi {\epsilon}_0}\frac{\lambda dx}{x^2+y^2}$$

 * $$\sin \theta = \frac {opposite}{hypotenuse} = \frac {x}{\sqrt{x^2 + y^2}}$$         and         $$\cos \theta = \frac {adjacent}{hypotenuse} = \frac {y}{\sqrt{x^2 + y^2}}$$

However we must recall that that magnitude corresponds to the vector $$d\vec E$$, which changes as one moves along the line. So we must find the component of $$d\vec E$$ which points in the x or in the y directions.

To find the component in the x direction $$dE_x=-dE \sin \theta\,\!$$ and to find it in the y direction apply $$dE_y=dE \cos \theta\,\!$$.

The x component of the field $$d{\vec E}_x$$ can be found by integrating from one end of the line source at -a to the other at a. If the line source is equally long to left and right, the net field in the x direction will cancel as below.

 * $$E_x=\int dE_x=-\int_{x=-a}^{x=+b} \sin \theta \, dE =-\int_{x=-a}^{x=+b} \frac {x}{\sqrt{x^2 + y^2}} \, dE $$

 * $$E_x=\int_{x=-a}^{x=+b} \frac {x}{\sqrt{x^2 + y^2}}, \frac{1}{4\pi {\epsilon}_0}\frac{\lambda dx}{x^2+y^2}$$       (2)

 * $$E=\frac {\lambda }{4\pi {\epsilon}_0}\int_{x=-a}^{x=+b} \frac {xdx} {\left[{x^2+y^2}\right ]^\frac {3}{2} }$$           (3)

 * $$E_x=\frac {\lambda }{4\pi {\epsilon}_0} \left [ - \frac{1}{ \left[a^2+y^2 \right ]^\frac {1}{2}} + \frac{1}{ \left[b^2+y^2 \right ]^ \frac {1}{2}}\right ]$$        (4)

hold 1
Similarly the y component of the field $$d{\vec E}_y$$ can be found by integrating from one end of the line source at -a to the other at a.

 * $$E_y=\int dE_y =\int_{x=-a}^{x=+b} \cos \theta \, dE =\int_{x=-a}^{x=+b} \frac {y}{\sqrt{x^2 + y^2}} \, dE $$

And substituting in the equation above we get:

 * $$E_y=\int_{x=-a}^{x=+b} \frac {y}{\sqrt{x^2 + y^2}}\, \frac{1}{4\pi {\epsilon}_0}\frac{\lambda dx}{x^2+y^2}=\frac {\lambda y}{4\pi {\epsilon}_0}\int_{x=-a}^{x=+b} \frac{dx}{\left[x^2+y^2 \right]^\frac{3}{2}}$$       (5)

 * $$E_y=\frac {\lambda y}{4\pi {\epsilon}_0}\int_{x=-a}^{x=+b} \frac{dx}{\left[x^2+y^2 \right]^\frac{3}{2}}$$       (6)

 * $$E_y=\frac {\lambda }{4\pi {\epsilon}_0 y} \left[\frac{x}{\left[x^2+y^2 \right]^\frac{1}{2}} \right] _{x=-a}^{x=+b}

$$       (7)

Find the electric field for an infinite line with uniform unit charge of λ coulombs per unit length

The figure shows an infinitely long line chage with linear charge density of $$\lambda \,\!=\frac{dq}{dx}$$.

The magnitude of the electric field which results from a short section dx of this line charge as measured at point P is a result of the charge $$dq=\lambda dx \,\!$$:

And with that charge we can calculate the field contribution (dE) from dq as:

 * $$dE=\frac{1}{4\pi {\epsilon}_0}\frac{dq}{r^2}=\frac{1}{4\pi {\epsilon}_0}\frac{\lambda dx}{x^2+y^2}$$          (1)

Since we can argue by symmetry that the electric field will be the same at any point of equal separation from the infinite line charge, we can solve this in using 2 dimensions. Once we have a solution for 2 dimensions, simple rotation around the line charge will provide the field in the 3rd dimension.

The electric field is a vector which we'll represent as $$d\vec E$$. This we'll separate into 2-dimensional Cartesian components:

 * $$dE_x=-dE \sin \theta\,\!$$     and      $$dE_y=dE \cos \theta\,\!$$

The sign of the electric field at $$dE_x\,\!$$ indicates the $$d{\vec E}_x$$ location of the point with respect to x.

We find the total electric field for the horizontal $$E_x$$ and vertical $$E_y$$ components to determine $$\vec E$$ at the point P by integrating. Note that $$\sin \theta$$ is used to find the components of the field horizontally, while $$\cos \theta$$ is used to find them vertically.

 * $$E_x=\int dE_x=-\int_{x=-\infty}^{x=+\infty} \sin \theta \, dE$$

and

 * $$E_y=\int dE_y =\int_{x=-\infty}^{x=+\infty} \cos \theta \, dE$$

In these expressions $$E_x\,\!$$> must integrate to zero because every element of charge to the left of the perpendicular point P has a corresponding element on the right so that their contributions to the field in the horizontal direction cancel each other. So $$\vec E$$ points in the vertical direction only. As for the electric field contributions to $$E_y\,\!$$, the right half and left half of the charged lines will make an equal contribution, so we can write:

 * $$E=E_y=2\int_{x=0}^{x=+\infty} \cos \theta \, dE$$

Substituting the expression (1) above into this equation we get:

 * $$E=2\int_{x=0}^{x=+\infty} \cos \theta \, \frac{1}{4\pi {\epsilon}_0}\frac{\lambda dx}{x^2+y^2}=\frac {\lambda }{2\pi {\epsilon}_0}\int_{x=0}^{x=+\infty} \cos \theta \, \frac{dx}{x^2+y^2}$$         (2)

By definition $$\tan \theta = \frac {x}{y}$$, which we can write as $$x=y\tan \theta \,\!$$. Differentiating this expression yields: $$dx=y{\sec }^2 \theta d \theta\,\!$$. Substitute this into (2) and you will get:

 * $$E=\frac {\lambda }{2\pi {\epsilon}_0}\int_{x=0}^{x=+\infty} \cos \theta \, \frac{y{\sec }^2 \theta d \theta}{x^2+y^2}=\frac {\lambda }{2\pi {\epsilon}_0}\int_{x=0}^{x=+\infty} \frac {y}{x^2+y^2}\cos \theta {\sec }^2 \theta d \theta$$     (3)

Applying further trigonometric relationships: $$\cos \theta = \frac{y}{r}$$,   $$\sec \theta = \frac{r}{y}\quad $$    and    $$\quad x^2+y^2=r^2\,\!$$,     we can establish equivalencies:

 * $$\frac {y}{x^2+y^2}\cos \theta {\sec }^2 \theta d \theta=\frac {y}{r^2}\frac {y}{r} \frac {r^2}{y^2}=\frac {y}{ry}= \frac {\cos \theta}{y}$$

Which can the be substituted into equation (3) to yield a significantly simplified integral:

 * $$E=\frac {\lambda }{2\pi {\epsilon}_0 y}\int_{x=0}^{x=+\infty} \cos \theta d \theta$$

To evaluate this we need only observe that $$x \rightarrow 0 \,\!$$, $$\theta \rightarrow 0 \,\!$$ and similarly $$x \rightarrow +\infty \,\!$$, $$\theta \rightarrow \frac {\pi }{2} \,\!$$, to simplify our range of integration to:

 * $$E=\frac {\lambda }{2\pi {\epsilon}_0 y}\int_{\theta=0 }^{\theta =\frac {\pi }{2}} \cos \theta d \theta =\frac {\lambda }{2\pi {\epsilon}_0 y}\Bigg[ \sin \frac {\pi }{2}- \sin 0\Bigg] = \frac {\lambda }{2\pi {\epsilon}_0 y}$$

Which yields a final solution:

 * $$E=\frac {\lambda }{2\pi {\epsilon}_0 y}$$

And this can be generalized for any point at distance r from the line charge to:

title: Physics Exercises/Electrostatics


text:
Physics Exercises/Kinematics in One Dimension/Solutions

Without Calculus
The speed of Cinderella,$$ v_c $$ is 12mph The time used by Cinderella,$$ t_c $$ is 0.25h The speed of Prince,$$ v_p $$ is unknown The time used by Prince,$$ t_p $$ is 0.16666h or $$1/6h$$ Since they have to meet, their distance traveled will be equal, then $$ v_c t_c = v_p t_p$$ $$ (12)(0.25) = v_p(1/6)$$ $$ v_p = 18 $$
 * 1) 3.1 s.  Use the kinematics equation $$y = y_0 + v_0 t + (1/2) a t^2$$.  Let the rock's initial height $$y_0 = 0$$. Then equation simplifies to $$y = t(v_0 + (1/2) a t)$$. So, two solutions for t are: t = 0 s and $$t = - 2 v_0/a $$. Plugging in $$v_0 = 15~\mathrm{m/s}$$ and $$a = - 9.8~\mathrm{m/s^2}$$ gives t = 3.1 s.
 * 2) Let L be the length of the escalator, v1 be the speed of the escalator, and v2 be the speed the man walks on solid ground. It takes t1 for a man standing on the escalator to travel its length, so $$L = v_1 t_1$$. If the man walks against the escalator, his net speed is $$v_2 - v_1$$; it takes him t2 to travel the length of the escalator, so $$L = (v_2 - v_1) t_2$$.  Since $$L = v_1 t_1 = (v_2 - v_1) t_2$$, we find that $$v_1 (t_1 + t_2) = t_2 v_2$$ or $$v_2 = \{(t_1 + t_2)/t_2\} v_1$$.  Now, when the man walks down the escalator, his net speed is $$v_1 + v_2$$.  If it takes him t seconds this time, then $$L = (v_1 + v_2) t = \{1 + (t_1 + t_2)/t_2\} v_1 t$$. Rewriting $$v_1$$ in terms of given quantities, $$v_1 = L / t_1$$, and solving for t, we get: $$t = (t_1 t_2)/(t_1 + 2 t_2) = 3.75~\mathrm{s}$$. Note that by carrying out the calculation in symbols as far as we can, we actually discover that the final answer is independent of L. This is one of many benefits of not substituting in numbers until the very end. (As with any rule, this rule has exceptions, but start by following the rule—you will notice exceptions by yourself when you become sufficiently skilled.)
 * 3) 14.4 s and 200 m  Let xb be the position of the bike and xc be the position of the car. We know from our kinematic equations that $$ x_f = x_i + v_i t + (1/2) a t^2 $$.  Because we are assuming that both the bike and the car are moving at a constant velocity (25 km/hr and 50 km/hr respectively) we can say that the acceleration (a) in this equation is zero (in both cases).  Thus we can write two equations of motion :  The equation for the bike is $$ x_b = x_{bi} + v_b t $$ and the equation for the car is $$ x_c = x_{ci} + v_c t $$.  Because we are interested in the point at which the two meet - we can safely say that we are looking for the point at which $$ x_b = x_c $$.  Solving this expression gives us $$ x_{bi} + v_b t = x_{ci} + v_c t $$ and thus the time which it will take for the two objects to reach the same location can be found to be $$ t = ( x_{bi} - x_{ci} ) / ( v_c - v_b ) $$.  Before we can plug in any values we need to make sure that they share the same units, so 25 km/hr becomes 6.944 m/s and 50 km/hr becomes 13.888 m/s.  Plugging in these values gives us $$ t = (100 m - 0 m) / (13.888 m/s - 6.944 m/s) = 14.4 s$$  Finally, in order to find the location at which they meet, simply plug this value for time into either of the two equations of motion, and we get $$ x = 0 + ( 13.888 m/s ) (14.4 s) = 200 m $$
 * 4) 18 mph

With Calculus

 * 1) The basic equations are these: $$a = \frac{d v}{d t}$$, $$v = \frac{d x}{d t}$$. Since our final goal is x as a function of t, we should solve for v(t) first by integrating the first expression: $$v(t) = \int_0^t dv = \int_0^t a~dt' = a t + v_0$$  And integrating this a second time yields the final expression for x(t) $$x(t) = \int_0^t dx = \int_0^t v~dt' = \int_0^t  (a t + v_0)~dt' = (1/2)a t^2 + v_0 t + x_0$$.
 * 2) e cm.  The positions as a function of time can be written down as the infinite series: $$x = x_0 + v_0 t + (1/2) a_0 t^2 + (1/3\cdot 2) j_0 t^3 + \cdots$$. Once you let $$x_0 = v_0 = a_0 = j_0 = \cdots = 1$$, you should recognize the resulting series as the Taylor series of $$e^t$$ (times 1 cm): $$e^t = 1 + t + (1/2)t^2 + (1/3!)t^3 + \cdots$$.  If you plug in $$t = 1$$, you get $$x = e^1 = e~\mathrm{cm}$$.  Observant reader should notice that we suddenly dropped units in the middle. To be thorough, we would factor out "1 s" from each written expression of t and absorb it into the coefficient (that way, t becomes unitless and all coefficients have unit of cm).


text:
Introspector/GccCpp

Intro
The motivation for compiling the gcc with g++ is to allow for a fine tuned capture of information about the trees directly from the compiler. Currently, the access to the tree data structures are not wrapped as tightly as they could be. By using the c++ compiler we can monitor all read and write accesses to the tree nodes. that way we can isolate how they are used.

then we should be able to recreate a better and more well documented tree interface.

I have tried using java to compile the gcc source code, but it does not support operator overloading and makes you rework all types of things.

Here I have introduced a smart pointer for all the members. and a integer wrapper for the direct values.

At this point the bitfields are no longer optimized.

the next step would be to include the original tree definitions as the internal storage and use the wrapper for the public interface.

I have also replaced the central tree union as a base class.

The idea behind using classes instead of unions is this :

Later when we have reworked the parts, there should be no need for unions. The usage of virtual base classes and inheritance will remove the need for them.

we will need to do multiple iterations over the model before we get to the point that the gcc will run with a c++ compiled back end. It is sure to meet some critics, so we we keep the pages here until it is ready.

here is the consolidated header Introspector/GccCpp/mytypes.h]

Here is the edited tree header Introspector/GccCpp/tree.h

Here is the edited rtl header Introspector/GccCpp/rtl.h

Here is my current patch : Introspector/GccCpp/patch.diff

/tree-dump.c/char class switch/

/tree-dump.c/test rtl/

title: Introspector/GccCpp


text:
Radiation Oncology/CNS/Cancer pain

 * St. Petersburg; 2006 (Russia) PMID 16715713 -- "[Proton irradiation of the pituitary gland for alleviating pain in patients with disseminated prostate cancer] -- [Article in Russian]" (Kondrate'ev BV, Vopr Onkol. 2006;52(1):92-4.)
 * Retrospective. 80 prostate cancer patients with pain from bony mets. Stereotactic ablation of frontal lob of pituitary. Narrow beams of 1,000 MeV protons.
 * Outcome: 70% suspended pain meds, remainder decreased
 * Toxicity: No untoward side-effects

title: Radiation Oncology/CNS/Cancer pain


text:
User:Jacques Bergeron/T2

Personal test template - please do not delete /T3

title: User:Jacques Bergeron/T2


text:
Radiation Oncology/Corporate/Protons

Proton Therapy

 * ''Please see the Protons page for links to clinical information

Associations

 * Particle Therapy Cooperative Group (PTCOG)
 * "An organization for those interested in proton, light ion, and heavy charged particle radiotherapy"

 * National Association for Proton Therapy (NAPT)
 * U.S. based, independent, non-profit, public benefit corporation
 * Promotes the clinical benefits of proton beam radiation therapy for cancer patients and their families

Existing Proton Facilities
North America
 * Canada
 * TRIUMF - ocular diseases only
 * United States
 * Loma Linda
 * Harvard University (Massachusetts General Hospital)
 * Midwest Proton Radiotherapy Institute
 * MD Anderson Cancer Center
 * University of California San Francisco/University of California Davis - ocular diseases only
 * University of Florida
 * ProCure Oklahoma
 * University of Pennsylvania
 * Hampton University, VA
 * CDH Proton Center ProCure, IL

Europe
 * Belgium
 * Louvain
 * France
 * Centre Antoine Lacassagne (Nice)
 * Institut Curie (Orsay)
 * Germany
 * Hahn-Meitner Institut (HMI/Charite) - ocular diseases only
 * Rinecker PTC
 * Heidelberg HIT
 * Italy
 * Catana - ocular diseases only
 * CNAO
 * Russia
 * Institute of Theoretical and Experimental Physics (ITEP, Moscow)
 * Joint Institute for Nuclear Research (JINR, Dubna), YouTube link
 * Petersburg Nuclear Physics Institute (PNPI)
 * Sweden
 * Uppsala Universitet
 * Switzerland
 * Paul Scherrer Institut (PSI)
 * United Kingdom
 * Clatterbridge Center for Oncology - ocular diseases only

Asia
 * China
 * Wanjie Proton Therapy Center
 * Japan
 * National Cancer Center Hospital East
 * Kobe Univeristy (Hyogo Medical Center)
 * National Institute of Radiological Sciences, Chiba
 * Shizuoka Cancer Center
 * University of Tsukuba
 * Southern Tohoku Proton Therapy Cancer Center, Fukushima
 * Wakasa Bay
 * Korea
 * Korea National Cancer Center

Africa
 * South Africa
 * IThemba LABS, South Afrcia

Chronologic List Overview

Proposed Proton Facilities
United States
 * Northern Illinois University, IL
 * ProCure
 * ProCure: CentraState Health, NJ - IBA - target 2012
 * ProCure: Beaumont Hospital, MI
 * ProCure: Boca Raton, FL
 * ProCure: Seattle Cancer Care Alliance: Seattle, WA
 * American Shared Hospital Services
 * American Shared Hospital Services: Tufts University, MA
 * American Shared Hospital Services: MD Andersen Orlando, FL
 * American Shared Hospital Services: Long Beach Memorial, CA
 * American Shared Hospital Services: San Francisco Bay area, CA
 * American Shared Hospital Services: Kettering Medical Center, Dayton, OH
 * Lehigh Valley Hospital, PA
 * South Florida Proton Center, FL
 * Broward Medical Center, FL
 * Washington University, MO
 * First Coast Oncology, FL
 * McLaren Health Care, MI - ProTom
 * Dayton, OH - Optivus
 * University of Iowa, IA
 * Scripps Health, CA - Varian
 * Wright State University, OH
 * Knoxville, TN
 * University of Maryland, MD
 * Mayo Clinics, MN and AZ
 * Emory University, GA
 * Atlantic Health System, NJ - ProTom
 * Texas Oncology, Dallas, TX
 * Willis-Knighton, Shreveport, LA (Summer 2014)

Europe
 * MedAustron, Austria - proton and carbon ion
 * Abstract 2007
 * WPE Essen, Germany
 * Prague, Czech Republic
 * Rozumberok, Slovakia
 * Krakow, Poland
 * Caen, France - carbon ion
 * Dimitrovgrad, Russia

Middle East
 * Qatar

Asia
 * National Taiwan University Hospital, Taiwan
 * Linkou Chang Gung Hospital, Taiwan
 * Shanghai Proton and Heavy Ion Hospital

Vendors

 * Hitachi and Hitachi North America
 * Proton facilities: Tsukuba, MD Anderson
 * IBA
 * Proton facilities: Massachusetts General Hospital, National Cancer Center Japan, National Cancer Center Korea, Wanjie Proton Therapy Center, Midwest Proton Radiotherapy Institute
 * Optivus Proton Therapy
 * Proton facilities: Loma Linda
 * Varian Medical Systems (ACCEL subsidiary)
 * Proton facilities: Paul Scherrer Institute
 * Siemens AG
 * Still River Systems
 * ProTom International
 * Compact Particle Acceleration Corporation & TomoTherapy

Treatment Planning

 * PerMedics Medical Systems - Odyssey
 * Elekta via CMS software - XiO and Focal

Patient Positioning

 * Brindisi, Italy; 2010 PMID 20465816 -- "Patient positioning in the proton radiotherapy era." (Devicienti S, J Exp Clin Cancer Res. 2010 May 13;29(1):47. [Epub ahead of print])
 * Review of various robotic couch applications

Overview by Institution

 * Heidelberg, Germany; 2010 PMID 20831505 -- "Heidelberg Ion Therapy Center (HIT): Initial clinical experience in the first 80 patients." (Combs SE, Acta Oncol. 2010 Oct;49(7):1132-40.)

 * CNAO Italy; 2009 PMID 19579862 -- "The National Center for Oncological Hadron Therapy: status of the project and future clinical use of the facility." (Orecchia R, Tumori. 2009 Mar-Apr;95(2):169-76.)
 * Overview of plans for the facility

 * Hyogo Japan; 2009 PMID 19920379 -- "[Current status and future of particle radiotherapy at the Hyogo Ion Beam Medical Center]" [Article in Japanese] (Murakami M, Gan To Kagaku Ryoho. 2009 Nov;36(11):1791-4.)
 * Overview. 2,639 patients treated with prostate, H&N, liver, lung and sarcomas (total 87%). Proton beam 2,122 (80%), carbon ion 517 (20%)

 * European ENLIGHT network
 * 2008 PMID 18208851 -- "COST-ENLIGHT strategic workshop on hadron (particle) therapy, CERN, Geneva, 3-4 May 2007." (Taylor RE, Br J Radiol. 2008 Mar;81(963):169-71. Epub 2008 Jan 21.)
 * Overview of meeting to review current status of hadron therapy in Europe
 *  2007 PMID 17513898 -- "European developments in radiotherapy with beams of large radiobiological effectiveness." (Amaldi U, J Radiat Res (Tokyo). 2007;48 Suppl A:A27-41.)
 * Overview of European experience, including development of facilities in Heidelberg (HIT) and Pavia (CNAO). Also review of Marburg, Wiener Neustadt, Lyon and Uppsala and the development of ENLIGHT network

 * Paul Scherrer Institut; 2007 PMID 17878101 -- "[Proton beam therapy: clinical indications and summary of the Swiss experience] - [Article in French]" (Weber DC, Bull Cancer. 2007 Sep 1;94(9):807-15.)

 * Harvard
 * 2007 PMID 17668954 -- "Clinical proton radiation therapy research at the Francis H. Burr Proton Therapy Center." (DeLaney TF, Technol Cancer Res Treat. 2007 Aug;6(4 Suppl):61-6.)
 * Review of MGH current clinical research goals
 * 1980 PMID 6245096 -- "Clinical experience with proton beam radiation therapy." (Suit HD, J Can Assoc Radiol. 1980 Mar;31(1):35-9.)
 * Cannot yet assess efficacy, but results warrant further investigation

 * Loma Linda; 2007
 * 2007 PMID 17668955 -- "Development and operation of the Loma Linda University Medical Center proton facility." (Slater JD, Technol Cancer Res Treat. 2007 Aug;6(4 Suppl):67-72.)
 * Overview of a hospital-based center development
 * 2006 PMID 16551128 -- "Clinical applications of proton radiation treatment at Loma Linda University: review of a fifteen-year experience." (Slater JD, Technol Cancer Res Treat. 2006 Apr;5(2):81-9.)
 * Review of 15 year experience using protons

Other Entities

 * GSI, Germany
 * ProBeam Oncology, USA
 * Partnership between MD Anderson Cancer Center, Sanders Morris Harris, and The Styles Company to manage the Proton Therapy Center at MD Anderson
 * Proton Brotherhood of the Balloon, USA
 * ProCure Treatment Centers, USA
 * TERA Foundation, Italy
 * Proton Therapy Global Management
 * SAH Global

title: Radiation Oncology/Corporate/Protons


text:
User:Jacques Bergeron/T3

Test template please do not delete  This is the long description of that section

title: User:Jacques Bergeron/T3


text:
A-level Physics (Advancing Physics)/Digital Processing

As we have already seen, a digital image consists of pixels, with each pixel having a value which represents its colour. For the purposes of understanding how digital images are manipulated, we are going to consider an 8-bit grayscale image, with pixel values ranging from 0 to 255, giving us 256 (28) levels of grey. 0 represents white, and 255 represents black. This is the image we are going to consider:

The image consists of an edge, and some random noise. There are two methods of smoothing this image (i.e. removing noise) that you need to know about:

Mean Smoothing
In order to attempt to remove noise, we can take the mean average of all the pixels surrounding each pixel (and the pixel itself) as the value of the pixel in the smoothed image, as follows:

This does remove the noise, but it blurs the image with means crucial anomalies and points may be missed.

Median Smoothing
A far better method is, instead of taking the mean, to take the median, as follows:

For this image, this gives a perfect result. In more complicated images, however, data will still be lost, although, in general, less data will be lost by taking the median than by taking the mean.

Edge Detection
We can detect the positioning of edges in an image using the 'Laplace rule', or 'Laplace kernel'. For each pixel in the image, we multiply its value by 4, and then subtract the values of the pixels above and below it, and on either side of it. If the result is negative, we treat it as 0. So, taking the median-smoothed image above, edge detection gives the following result:

Questions
1. How could the above methods be applied to a digital sound sample?

2. Which of the above methods would be suitable for smoothing sharp edges? Why?

3. Use median smoothing to remove noise from the following image of a white cat in a snowstorm (the black pixels have a value of 255):

4. Why would mean sampling not be appropriate for smoothing the image given in question 3?

5. Use mean smoothing to remove noise from the following image of a black cat in a coal cellar:

/Worked Solutions/


text:
Talk:Circuit Idea/Deboo Integrator

I would like to start the discussion about the famous Deboo integrator by inserting my comment to the EDN's article Consider The "Deboo" Integrator For Unipolar Noninverting Designs. In this story that I wrote in May 2007, I showed how we can convert, in two different ways (Miller's and Deboo's), the passive RC integrator into an active op-amp one. To my much surprise, six months later I received an email from the very inventor of this clever circuit - Gordon Deboo (see the two emails below). In my reply I asked him to tell how he has invented this clever circuit. Circuit-fantasist (talk) 15:35, 24 December 2007 (UTC)

Reader's comment to a related EDN's article
Consider The "Deboo" Integrator For Unipolar Noninverting Designs

The secret of an active integrator is simple: it is just an improved passive integrator. Then, how do we convert the passive integrator into an active one? Here is a possible "scenario".

RC-INTEGRATOR. The bare capacitor is the simplest integrator. It is perfect, if only we drive it by a constant current. Most frequently, we need to drive the capacitor by a voltage. For this purpose, we connect before a resistor R acting as a voltage-to-current converter. Only, a problem appears - the voltage Vc across the capacitor disturbs the input voltage Vin. As a result, the input current decreases: Ic = (Vin - Vc)/R = Vin/R - Vc/R. What do we do to compensate the losses in the capacitor? We may borrow the remedy from the human routine: connect an additional power source, which "helps" the input source. From the Kirchoff's times, we can do this connection in two ways - series or parallel.

INVERTING (MILLER) INTEGRATOR. Here, we connect in series with the capacitor an additional following voltage source (op-amp voltage "inverter"). It produces a compensating "mirror" voltage -Vc equal and inverse to the voltage drop Vc across the capacitor. In this way, the op-amp adds as much voltage to the input voltage source as it loses across the capacitor; thus, it "helps by voltage" the input source to create the right current Ic = (Vin - Vc + Vc)/R = Vin/R. In this arrangement, the capacitor becomes "flying" but the compensating voltage source is grounded. That is why, we use its "mirror" voltage -Vc as an output that gives two advantages: first, the load is connected to the common ground; second, it consumes energy from the "helping" voltage source (the power supply) instead from the input source (so, the load may be low-resistive enough). A disadvantage is the need of a bipolar power supply. All the circuits with parallel negative feedback (i.e., all the inverting op-amp circuits) exploit this great idea.

NON-INVERTING (DEBOO) INTEGRATOR. Here, we connect in parallel to the capacitor an additional "helping" current source. It produces a compensating current Vc/R, which is proportional to the voltage drop Vc across the capacitor. In this way, the source adds the "missing" current Vc/R to the insufficient input current Iin = Vin/R - Vc/R; thus, it "helps by current" the input source to create the right current Ic = Vin/R - Vc/R + Vc/R = Vin/R. Thus we obtain the famous Howland circuit, which acts as a perfect voltage-to-current converter (if we include the input voltage source, we name it Howland current source).

A negative resistor -R can serve as the "proportional helping current source" needed. It does this magic by "pushing" a current I = Vc/R through the capacitor instead by "sucking" the same current (if it was an ordinary "positive" resistor R). In order to create a negative resistor -R, we may convert a positive resistance R into a negative one -R ; thus we will create the famous negative impedance converter (NIC). For this purpose, we connect in series a "positive" resistor R and a proportional voltage source, which produces a compensating voltage 2Vc (i.e, an amplifier with K = 2). Half the voltage (Vc) vanishes in the capacitor; the rest half (Vc) remains applied across the resistor R, which acts as a voltage-to-current converter. As a result, the whole combination produces the "helping" current needed I = (2Vc - Vc)/R = Vc/R. We may implement this idea into a practical NIC circuit by using a resistor R (R3 on Fig. 1) and a classical op-amp non-inverting amplifier with K = 2 (the op-amp and the resistors R1, R2 on Fig. 1).

From this negative resistance viewpoint, we may present the Howland current source as two parallel connected resistors - a "positive" R and a negative -R. The result of this connection is that the negative resistor has "neutralized" (absorbed) the positive one and the effective differential resistance (the internal source's resistance) is infinite. In the circuit of a Deboo integrator, both the capacitor and the compensating source are grounded; the power supply is unipolar. The voltage drop across the capacitor can serve as an "original" output; only, this is not a good idea because the load will affect the output (we have to isolate the load). Then, where to take a perfect output from? As above, we may use the compensating voltage 2Vc (which creates the compensating current) as a buffered output. In order to enlarge the active region, we may scale the negative resistance by the ratio R1/R2 > 1 (see the article). Note that all the resistances and CMRR affect the accuracy (Miller integrator does not have these problems). For the first time, I encountered the negative resistance phenomenon and its implementations - the exotic circuits of negative impedance converter, Howland current source and Deboo integrator - in the French book "L’amplificateur opérationnel et ses applications Masson, Paris, 1971". Then I was a student and was trying without success to grasp the ideas behind these famous circuits. I was asking the simple questions, "What did the op-amp actually do in these circuits? What problem did it solve? Why was it connected there?"; only, they were not giving me reasonable answers. Until many years later, I managed finally to penetrate into the philosophy of negative resistance circuits. Up to now, I continue revealing the mystery of the negative resistance phenomenon to my students following the "scenario" above.

I have decided to expose my viewpoint about these legendary circuits because, looking at this article (and the entire web), I see that the idea behind them is still hidden. Although the article is useful from a practical point of view, it does not tell us anything new about root of the matter. It only shows another possible application of MAX4250 op-amp; obviously, it is its purpose.

The sentence "...ground drives the capacitor through a negative resistor..." does not sound well as a ground is just a piece of wire; it is not a source. The negative resistor is the source here. So, more correctly is just to say "...a negative resistor drives the capacitor..." Of course, I see that the author would like to show a fine symmetrical "T" configuration on Fig. 3. Really, the configuration is symmetrical - two voltage sources, Vin,Vout, drive the common capacitor through corresponding "positive" resistors,R,R3. The entire problem is that a "positive" resistor contains only a resistor while a negative resistor contains a resistor and a voltage source. Actually, a negative resistor is not a resistor, it is a varying source. IMO, it deserves a special "source" name; for example, "1-port voltage-controlled current source" or "2-terminal voltage-to-current active converter" can serve as full descriptive names:) At worst, we may name it "antiresistor" or "reverse resistor"; "negative resistor" is the most inappropriate word:)

Cyril Mechkov

email: cyril@circuit-fantasia.com

A reply from Gordon Deboo
From: Gordon Deboo To: cyril@circuit-fantasia.com Sent: Thursday, October 04, 2007 8:27 PM Subject: Deboo Integrator

I notice from http://electronicdesign.com/analog/consider-deboo-integrator-unipolar-noninverting-designs   that you expressed an interest in one of the circuits I designed in my old NASA days, viz. the Deboo Integrator. There is another more complicated circuit, which serves a different purpose, that includes a Deboo Integrator along with other components. It can be found in one of my now obsolete patents:

1. Patent # 3493901 - A GYRATOR TYPE CIRCUIT 1968

A "gyrator" makes an effective inductor using an capacitor, which is important because capacitors are easy to make with integrated circuits, while inductors are not. Therefore a gyrator enables large value inductors to be made in integrated without using physically large coils of wire, because the coil can be replaced with a capacitor.

I have been retired from NASA for many years now and I am interested and surprised when some of my old circuits are mentioned in current literature.

It's great to see younger people still working on and improving on what we old fogies did. Keep up the good work!

Gordon Deboo California

Another reply from Gordon Deboo
From: Gordon Deboo To: "Cyril"  Sent: Tuesday, October 09, 2007 3:34 PM Subject: Re: Deboo Integrator

...As you can see from the dates of my publications, I did most of my creative work in the 1960s and early 1970s working for NASA (The  USA’s National Aeronautics and Space Administration). After that I got into management, which wasn’t nearly as much fun! So I have been away from circuit design for 30 years and most of what I did has been improved upon by the many younger people who have followed me. Also I haven’t kept up with developments in the field. A brief look at what has happened tells me that the part of the work I did that is  still somewhat applicable is in circuits using operation amplifiers.

I have looked at what you have on the Internet and am impressed with your ability to explain many complicated circuits. I think your students are lucky to have a teacher with your approach.

I am of course now retired and health and other issues do not permit me to give detailed answers to the many interesting questions you raise. However, let me respond to some of your points...

''...It is interesting to know, if you had known about Howland idea when you invented Deboo integrator; if you applied directly the Howland circuit just adding a capacitor to it or you walked the whole way, etc... Circuit-fantasist''

... I was not aware of the Howland current source when I designed my integrator. I conceived my integrator a couple of years before it was published, so I don’t which came first...

...I’m sorry I’m not able to give you a more detailed reply, but already this is fairly lengthy. I can’t imagine you would want to use anything from this note in your various publications but, if you should wish to, feel free to do so...

Best wishes,

Gordon

A copy from the Wikipedia talk page about negative resistance
I have written the text below in Howland circuit section of the Wikipedia talk page about negative resistance. Circuit-fantasist (talk) 15:42, 3 March 2009 (UTC)

The second version of Howland current source (see Fig. 1 of MAXIM's AN1155) consists of an input voltage source VIN, a positive resistor R, a load and a negative impedance converter INIC (R1 = R2 = R3 = R and the op-amp). The input voltage source and the resistor R constitute an imperfect current source with positive internal resistance R that is connected in parallel to the negative resistor INIC and the load. The role of the negative resistance -R is to neutralize the positive internal resistance R of the source. As a result, the load is driven again by a constant current source with infinite internal resistance. Well, let's include some algebra to be more cogent.

The imperfect current source produces a current IL = (VIN - VL)/R = VIN/R - VL/R. As you can see, it differs from the ideal result IL = VIN/R by the term VL/R. It is more than obvious that we may compensate this error by adding the same term; then, IL = VIN/R - VL/R + VL/R = VIN/R. That means to inject an additional "helping" current IH = VL/R that is proportional to the load voltage. What is this circuit that can produce such a current IH? Of course, it is a voltage driven negative resistor with resistance -R; it is a negative impedance converter with current inversion (INIC) that is connected in parallel to the load.

As a conclusion, Howland voltage-to-current converter consists of two connected in parallel resistors with equivalent but opposite (positive and negative) resistances; Howland voltage-to-current converter = INIC + resistor. The "neutralization" between them produces an infinite resistance and, as a result, a constant current. The circuit is stable since the positive load resistance remains after the "neutralization" (another wisdom: the positive resistance has to dominate over the negative resistance to have stability).

Deboo integrator (see again Fig. 1 of MAXIM's AN1155) is just a Howland voltage-to-current converter driving a capacitive load; Deboo integrator = Howland voltage-to-current converter + capacitor. It has two advantages over the dual inverting integrator: it is a non-inverting circuit and the capacitor is grounded. But it has two disadvantages as well: it requires an op-amp with differential input and it has a bigger error depending on the resistor tolerances (here, the op-amp does not monitor the result of the compensation as it monitors the virtual ground at the inverting circuit).

It is interesting that although there was nothing new in a Deboo integrator (actually, it was just a Howland pump driving a capacitor) Deboo was not aware of the Howland current source when he invented his integrator. I found out this fact in a personal correspondence (see above) between me and Deboo a year ago (Deboo was impressed by my comments to an EDN article about his famous circuit and emailed me). By the way, then I also exchanged interesting thoughts with Deboo about the unique feature of negative feedback to reverse the causality in circuits (to swap circuit inputs and outputs). For example, the negative inductance circuit is based on this idea.

Today, I found an extremely interesting source about the invention of the Howland circuit. Although it is nameless, it seems it is written by the very Howland. Circuit-fantasist (talk) 15:42, 3 March 2009 (UTC)

title: Talk:Circuit Idea/Deboo Integrator


text:
Chess Opening Theory/1. e4/1...Nf6/2. e5/2...Nd5/3. d4/3...d6/4. c4/4...Nb6/5. f4/5...dxe5/6. fxe5/6...Nc6

=Alekhine Defence= The so-called Alekhine Knights come out to attack White's center, with d4 under attack. White must now defend the pawn. The main line is 7.Be3. 7.Nf3 is strongly met by 7...Bg4 with strong pressure on the center.

Theory table
.

'''1.e4 Nf6 2.e5 Nd5 3.d4 d6 4. c4 Nb6 5. f4 dxe5 6.fxe6 Nc6'''


text:
User talk:Hayson1991

Welcome!
Come introduce yourself at the new users page. If you have any questions, you can ask there or contact me personally.--Whiteknight (Page) (Talk) 16:16, 24 December 2007 (UTC)

Japanese Wikibook discussion
Hi Hayson1991,

I just added a comment on Talk:Japanese which I wanted to draw your attention to as you've been involved in this book. --Swift (talk) 14:51, 10 June 2008 (UTC)

title: User talk:Hayson1991


text:
Expert Systems/Refraction

Refraction
Refraction is the act of preventing a rule from firing multiple times in succession. Without refraction, at each loop iteration the same rules can be added to the agenda repeatedly. This is because the same conditions are satisfied. To prevent a single rule from firing repeatedly, a refraction condition is implemented. There are many types of refraction rules.

Once Only
Some systems use a refraction condition that each rule may only be fired once. Once a rule has been fired, it may never be fired again until the system is reset.

Intermittant Rules
In an intermittant rule refraction condition, a rule may never fire twice consecutively, and may not be added to the agenda again until a different rule fires first.

Changing Antecedent
In a changing antecedent refraction condition, a rule may only fire if the attributes in the antecedent have changed.

title: Expert Systems/Refraction


text:
User:Whiteknight/Notes/Algebra

< Book Foundry | All Books Several new contributors have been helping with this book, so it's actually doing better then it was. May still need to be merged/combined with Linear Algebra, Abstract Algebra, and some other related books.

title: User:Whiteknight/Notes/Algebra


text:
VHDL for FPGA Design/4-Bit Adder

library ieee; use ieee.std_logic_1164.all; use ieee.std_logic_unsigned.all use ieee.std_logic_arith.all;

Entity Adder4 Is port ( a,b:in integer Range -8 to 7; sum:out integer Range -16 to 15 ); end Adder4;

Architecture Adder4arch of Adder4 Is Begin sum<=a+b;

end Adder4arch;

Simulation Waveform

VHDL-kieli FPGA-suunnittelussa/4-bittinen summain

title: VHDL for FPGA Design/4-Bit Adder


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User:Whiteknight/Notes/Analog and Digital Conversion

< Book Foundry | All Books
 * 9 Nov 2008: Did a read-through today. Book is a lousy stub but has attracted a scant few contributions from other contributors. Desperately needs some images to illustrate some of it's concepts, and has a lot of empty pages and empty headings, especially in later sections/chapters.


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User:Whiteknight/Notes/Arithmetic

< Book Foundry | All Books Book has a lot of redlinks, too little content, no solid structure, etc. Needs a lot of work.

title: User:Whiteknight/Notes/Arithmetic


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User:Whiteknight/Notes/Assembly Language

< Book Foundry | All Books Is now a subject page for assembly language books.


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User:Whiteknight/Notes/Artificial Neural Networks

< Book Foundry | All Books Has good structure, navigation. Needs to be expanded.
 * 9 Nov 2008:Several pages are stubbish, and have no headers structure. Lots of equations but not a lot of explanation and no examples. A few redlinks in later parts of the TOC, and a definite need for massive expansion in later chapters.


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User:Whiteknight/Notes/Astrodynamics

< Book Foundry | All Books
 * 9 Nov 2008: Several images are not explained or integrated into the text. Existing content needs to be expanded with explanations and examples. Several redlinks on TOC. Later chapters/sections need to be written/expanded.

title: User:Whiteknight/Notes/Astrodynamics


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User:Whiteknight/Notes/Digital Circuits

< Book Foundry | All Books
 * 15 November 2008:Book is a stub with very little content and an incomplete TOC. What this book needs most is a full-time author to adopt it, because it's not far enough along yet for occasional contributors to be able to latch onto it. Needs everything.

title: User:Whiteknight/Notes/Digital Circuits


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User:Whiteknight/Notes/Digital Signal Processing

< Book Foundry | All Books
 * 17 November 2008:The whole second half of the book hasn't been written yet. Needs more content, images, explanations. Especially images, explanations of most materials will suffer without them. Not enough written yet to support a PDF or a collection.


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User:Whiteknight/Notes/Electric Circuits

< Book Foundry | All Books Is now a subject page for books on electric circuits


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User:Whiteknight/Notes/Embedded Systems

< Book Foundry | All Books
 * 20 November 2008:Poor quality mishmash of various topics. Many pages are stubs. TOC is inconsistent with subpage naming. Risks becoming an encyclopedia of hardware/software configurations. Needs more examples, more pictures, more explanations. Strongly consider and maybe start planning breaking this book into smaller, more focused volumes: RTOS, C + Assembly interaction, Embedded hardware programming, Real-Time Software, etc. Might just be better off this way.

title: User:Whiteknight/Notes/Embedded Systems


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User:Whiteknight/Notes/Engineering Analysis

< Book Foundry | All Books
 * 23 November 2008: A few pages are yet to be written, and some of the pages are stubbish. Needs more examples, badly. Also needs images to illustrate some points.


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User:Whiteknight/Notes/Engineering Tables

< Book Foundry | All Books At the moment, is just a listing of various tables for use in other books. Needs more, probably to be imported from WP.


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Talk:A-level Physics (Advancing Physics)/Digital Processing

Please could someone add edge detection to this page? I don't understand how the Laplace kernel works. --Sjlegg (talk) 18:35, 24 December 2007 (UTC)

The pixels at the edge of the array cannot be changed by the processing because they are not at the centre of a 3x3 section. Edge pixels remain unchanged - try doing mean or median processing in Scion Image it also shows the edge pixels remain unchanged. The examples on the page are incorrect.


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User:Whiteknight/Notes/Find Employment

< Book Foundry | All Books Needs more content, better organization. Possibly split large pages into smaller pages. More references, links to resources, etc. Needs a cover page and a PDF version.

title: User:Whiteknight/Notes/Find Employment


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User:Whiteknight/Notes/Floating Point

< Book Foundry | All Books
 * 26 November 2008: Book is a complete stub. Many entries in the TOC are not written, and many of the pages that do exist are just a rough outline of page headings. Book needs lots and lots of work in every way.

title: User:Whiteknight/Notes/Floating Point


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User:Whiteknight/Notes/The Grand Inquisitor

< Book Foundry | All Books
 * 29 November 2008: Book has collection, Cover (needs work), and all relevant source text from Wikisource. Needs more content and annotations, discussions, and translations.


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User:Whiteknight/Notes/GRE

< Book Foundry | All Books Book needs radical expansion, example problems, more information, etc. Needs formatting and cleanup. Is in bad shape.

title: User:Whiteknight/Notes/GRE


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User:Whiteknight/Notes/MIPS Assembly

< Book Foundry | All Books
 * 7 December 2008: Needs lots of content added, general expansion, explanation of various opcodes, etc. Many pages are stubs, or contain stubbish sections.

title: User:Whiteknight/Notes/MIPS Assembly


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User:Whiteknight/Notes/Modern Physics

< Book Foundry | All Books This large book was separated out into smaller books for each individual subject. The resultant books are listed here: Category:Modern Physics.
 * 1) The "Gravity" section needs to be separated out into it's own book.
 * 2) Go through the book, find orphaned subpages and delete them.
 * 3) Evaluate some of the appendices to see where they can be moved, or if they should be deleted.
 * 4) Redirect this book to Subject:Modern Physics
 * 5) Update all "Modern Physics" templates to point to the right places.
 * 6) Maybe improve the look of Subject:Modern Physics

title: User:Whiteknight/Notes/Modern Physics


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A-level Physics (Advancing Physics)/Digital Processing/Worked Solutions

1. How could the above methods be applied to a digital sound sample?

By taking the median or mean of the sample and the values on either side of it. This would be unreliable, as sound is a wave, so the samples need to vary quite widely. However, something similar could be used when comparing repeating patterns in a waveform.

'''2. Which of the above methods would be suitable for smoothing sharp edges? Why?'''

Mean smoothing - median smoothing would not blur the edges.

3. Use median smoothing to remove noise from the following image of a white cat in a snowstorm (the black pixels have a value of 255):

4. Why would mean sampling not be appropriate for smoothing the image given in question 3?

It would produce a really blurred mess, instead of an image, as the noise is too dense.

5. Use mean smoothing to remove noise from the following image of a black cat in a coal cellar:


text:
.NET Development Foundation/Navigation

This is a navigation template - please do not delete

title: .NET Development Foundation/Navigation


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User:Whiteknight/Notes/Programmable Logic

< Book Foundry | All Books
 * 11 December 2008: Appears that another author has converted it to book/chapter/page format. Most pages are stubs, needs massive content expansion. Some pages are redlinked


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User:Whiteknight/Notes/Reverse Engineering

< Book Foundry | All Books This is now a subject page for all software reverse engineering books.


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User:Whiteknight/Notes/Semiconductors

< Book Foundry | All Books
 * 15 December 2008:Some early pages have decent content. However, much of the TOC is unlinked or redlinked, and many of the pages in the book are stubs. needs massive content expansion, cleanup and better integration of existing content.

title: User:Whiteknight/Notes/Semiconductors


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User:Whiteknight/Notes/The Seven Wonders of the World

< Book Foundry | All Books
 * 17 December 2008: Most pages are stubs, bits of content stolen from WP and images stolen from Commons. Needs a lot more content added. Idea is good, but needs execution help.


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.NET Development Foundation/Navbox

/core/

title: .NET Development Foundation/Navbox


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User:Whiteknight/Notes/Sexual Health

< Book Foundry | All Books
 * 18 December 2008: Pages have content, but some sections are empty and some discussion points are very short. Needs a few more images in some sections, general expansion, and maybe the addition of sidebox vignettes to help with the information.

title: User:Whiteknight/Notes/Sexual Health


text: .NET Development Foundation/Navbox/core

title: .NET Development Foundation/Navbox/core


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User:Whiteknight/Notes/Signal Processing

< Book Foundry | All Books
 * 24 December 2008:Book is a stub in need of major content additions. Only about half the TOC is filled out, half those pages are redlinks, and the pages that do exist are stubs. Needs a lot of work.


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User:Whiteknight/Notes/Systems Theory

< Book Foundry | All Books Needs a lot of cleanup, improvement, and content expansion.

title: User:Whiteknight/Notes/Systems Theory


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User:Whiteknight/Notes/SPARC Assembly

< Book Foundry | All Books
 * 29 December 2008:Book is a stub with very little content. It has some good organizational structure where information could be filled in for most pages, however.

title: User:Whiteknight/Notes/SPARC Assembly


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User:Whiteknight/Notes/System Troubleshooting

< Book Foundry | All Books
 * 29 December 2008: Book is a very small stub and might need to get deleted since there isn't much here to grab on to.


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User:Whiteknight/Notes/Wikijunior:Colors

< Book Foundry | All Books
 * 2 January 2008: Book in decent condition but could use more pictures and larger/more-colorful text. I would really like to find a way to get this book faithfully published too.


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User:Whiteknight/Notes/Using Wikibooks

< Book Foundry | All Books
 * 8 December 2008:Book is rapidly expanding. Needs a little bit more content in a few places, but otherwise needs examples, pictures, and formatting work.

title: User:Whiteknight/Notes/Using Wikibooks


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User:Whiteknight/Notes/Wikibooks Community

< Book Foundry | All Books Book merged into Using Wikibooks and deleted.


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User:Whiteknight/Notes/Windows Programming

< Book Foundry | All Books
 * 02 January 2008: Book is too ambitious and needs to be split up into smaller books for particular subjects. Some subbooks to make: C/Win32API, C++/MFC/COM, Win32DDK and Win32Shell. Many sections are redlinked and many pages are stubbish.


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User:Whiteknight/Notes/x86 Assembly

< Book Foundry | All Books
 * 4 January 2009: Needs more content when it comes to the various assemblers. Needs better descriptions of some instructions and concepts, but volume of content is not too bad.

title: User:Whiteknight/Notes/x86 Assembly


text:
Microsoft SQL Server/Geospatial Data/WKT

Well-known text (WKT) is a text markup language for representing vector geometry objects on a map, spatial reference systems of spatial objects and transformations between spatial reference systems. A binary equivalent, known as well-known binary (WKB) is used to transfer and store the same information on databases, such as PostGIS. The formats are regulated by the Open Geospatial Consortium (OGC) and described in their Simple Feature Access and Coordinate Transformation Service specifications.

Geometric Objects
Geometric objects that can be represented with WKT are: points, lines, polygons, TINs and Polyhedron]. Multi geometries are also available to represent more than one geometry of the same dimension in a single object, and geometries of different dimensions can be stored in a geometry collection.

Coordinates for geometries may be 2D (x, y), 3D (x, y, z), 4D (x, y, z, m) with a m value that is part of a linear reference system or 3D with a m value (x, y, m). Three dimensional geometries are designated by a Z after the geometry type and geometries with a linear reference system have a M after the geometry type.

WKT geometries are used throughout OGC specifications and are present in applications that implement these specifications. For example, PostGIS contains functions that can convert geometries to and from a WKT representation, making them human readable.

The following are some example geometric WKT strings.

POINT(6 10) LINESTRING(3 4,10 50,20 25) POLYGON((1 1,5 1,5 5,1 5,1 1),(2 2, 3 2, 3 3, 2 3,2 2)) MULTIPOINT(3.5 5.6,4.8 10.5) MULTILINESTRING((3 4,10 50,20 25),(-5 -8,-10 -8,-15 -4)) MULTIPOLYGON(((1 1,5 1,5 5,1 5,1 1),(2 2, 3 2, 3 3, 2 3,2 2)),((3 3,6 2,6 4,3 3))) GEOMETRYCOLLECTION(POINT(4 6),LINESTRING(4 6,7 10)) POINT ZM (1 1 5 60) POINT M (1 1 80)

Spatial Reference Systems
A WKT string for a spatial reference system describes the datum, geoid, coordinate system, and map projection of the spatial objects.

Well-known text is used extensively throughout many GIS programs. ESRI uses WKT in the shapefile *.prj file.

The following is an example of a spatial reference system WKT string.

COMPD_CS["OSGB36 / British National Grid + ODN", PROJCS["OSGB 1936 / British National Grid", GEOGCS["OSGB 1936", DATUM["OSGB_1936", SPHEROID["Airy 1830",6377563.396,299.3249646,AUTHORITY["EPSG","7001"]], TOWGS84[375,-111,431,0,0,0,0], AUTHORITY["EPSG","6277"]], PRIMEM["Greenwich",0,AUTHORITY["EPSG","8901"]], UNIT["DMSH",0.0174532925199433,AUTHORITY["EPSG","9108"]], AXIS["Lat",NORTH], AXIS["Long",EAST], AUTHORITY["EPSG","4277"]], PROJECTION["Transverse_Mercator"], PARAMETER["latitude_of_origin",49], PARAMETER["central_meridian",-2], PARAMETER["scale_factor",0.999601272], PARAMETER["false_easting",400000], PARAMETER["false_northing",-100000], UNIT["metre",1,AUTHORITY["EPSG","9001"]], AXIS["E",EAST], AXIS["N",NORTH], AUTHORITY["EPSG","27700"]], VERT_CS["Newlyn", VERT_DATUM["Ordnance Datum Newlyn",2005,AUTHORITY["EPSG","5101"]], UNIT["metre",1,AUTHORITY["EPSG","9001"]], AXIS["Up",UP], AUTHORITY["EPSG","5701"]], AUTHORITY["EPSG","7405"]]

Transformations
A WKT format is defined to describe the transformation methods and parameters used to convert coordinates between two different spatial reference systems.

Below are two examples of WKT transformation descriptions.

PARAM_MT["Mercator_2SP", PARAMETER["semi_major",6370997.0], PARAMETER["semi_minor",6370997.0], PARAMETER["central_meridian",180.0], PARAMETER["false_easting",-500000.0], PARAMETER["false_northing",-1000000.0], PARAMETER["standard_parallel_1",60.0]]

PARAM_MT["Affine", PARAMETER["num_row",3], PARAMETER["num_col",3], PARAMETER["elt_0_1",1], PARAMETER["elt_0_2",2], PARAMETER["elt_1_2",3]]


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Microsoft SQL Server/Geospatial Data/GML

The Geography Markup Language (GML) is the XML grammar defined by the Open Geospatial Consortium (OGC) to express geographical features. GML serves as a modeling language for geographic systems as well as an open interchange format for geographic transactions on the Internet. Note that the concept of feature in GML is a very general one and includes not only conventional "vector" or discrete objects, but also coverages (see also GMLJP2) and sensor data. The ability to integrate all forms of geographic information is key to the utility of GML.

Standards
The OGC is an international voluntary consensus standards organization whose members maintain the Geography Markup Language standard. The OGC coordinates with the ISO TC 211 standards organization to maintain consistency between OGC and ISO standards work. GML is in the process of being adopted as an ISO standard (ISO 19136) and is expected to be released as an International Standard in 2007.

GML is the XML data standard for the GeoWeb infrastructure, enabling Internet-connected devices to access geographical information, including, for example, merchant locations and traffic conditions.

GML can also be included in version 1.0 of the United States National Information Exchange Model.

GML model
The original GML model was based on the World Wide Web Consortium's Resource Description Framework (RDF). Subsequently, the OGC introduced XML schemas into GML's structure to help connect the various existing geographic databases, whose relational structure XML schemas more easily define. The resulting XML-schema-based GML retains many features of RDF, including the idea of child elements as properties of the parent object (RDFS) and the use of remote property references.

GML contains a rich set of primitives which are used to build application specific schemas or application languages. These primitives include:

 * Feature
 * Geometry
 * Coordinate Reference System
 * Time
 * Dynamic feature
 * Coverage (including geographic images)
 * Unit of measure
 * Map presentation styling rules

Profile
GML profiles are logical restrictions to GML, and may be expressed by a document, an XML schema or both. These profiles are intended to simplify adoption of GML, to facilitate rapid adoption of the standard. The following profiles, as defined by the GML specification, have been published or proposed for public use:

 * A Point Profile for applications with point geometric data but without the need for the full GML grammar
 * A GML Simple Features profile supporting vector feature requests and transactions, e.g. with a WFS
 * A GML profile for GMJP2 (GML in JPEG 2000)
 * A GML profile for RSS

Note that Profiles are distinct from application schemas. Profiles are part of GML namespaces and define restricted subsets of GML. Application schemas are XML vocabularies defined using GML and which live in an application-defined target namespace. Application schemas can be built on specific GML profiles or use the full GML schema set.

Profiles are often created in support for GML derived languages (see application schemas) created in support of particular application domains such as commercial aviation, nautical charting or resource exploitation.

The GML Specification (Since GML v3.) contains a pair of XSLT scripts (usually referred to as the "subset tool") that can be used to construct GML profiles.

GML Simple Features Profile
The GML Simple Features Profile is a more complete profile of GML than the above Point Profile and supports a wide range of vector feature objects, including the following:

 * 1) A reduced geometry model allowing 0d, 1d and 2d linear geometric objects (all based on linear interpolation) and the corresponding aggregate geometries (gml:MultiPoint, gml:MultiCurve, etc.).
 * 2) A simplified feature model which can only be one level deep (in the general GML model, arbitrary nesting of features and feature properties is not permitted).
 * 3) All non-geometric properties must be XML Schema simple types – i.e. cannot contain nested elements.
 * 4) Remote property value references (xlink:href) just like in the main GML specification.

Since the profile aims to provide a simple entry point, it does not provide support for the following:
 * coverages
 * topology
 * observations
 * value objects (for real time sensor data)
 * nor support for dynamic features.

Nonetheless it supports a good variety of real world problems.

Subset tool
In addition, the GML specification provides a subset tool to generate GML profiles containing a user-specified list of components. The tool consists of a pair of XSLT scripts. The scripts generate a profile that a developer may extend manually or otherwise enhance through schema restriction. Note that as restrictions of the full GML specification, application schemas that a profile can generate must themselves be valid GML application schemas.

The subset tool can generate profiles for many other reasons as well. Listing the elements and attributes to include in the resultant profile schema and running the tool results in a single profile schema file containing only the user-specified items and all of the element, attribute and type declarations on which the specified items depend. Some Profile schemas created in this manner support other specifications including IHO S-57 and GML in JPEG 2000.

Application schema
In order to expose an application's geographic data with GML, a community or organization creates an XML schema specific to the application domain of interest(the application schema). This schema describes the object types whose data the community is interested in and which community applications must expose. For example, an application for tourism may define object types including monuments, places of interest, museums, road exits, and viewpoints in its application schema. Those object types in turn reference the primitive object types defined in the GML standard.

A list of known publicly available GML Application Schemas is being assembled.

Some other markup languages for geography use schema constructs, but GML builds on the existing XML schema model instead of creating a new schema language.

GML and KML
Note that the KML (Keyhole Markup Language) language made popular by Google is a complementary to GML. Whereas GML is a language to encode geographic content, by describing a spectrum of application objects and their properties (e.g. bridges, roads, buoys, vehicles etc.), KML is a language for the visualization of geographic information. KML can be used to carry GML content, and GML can be "styled" to KML for the purposes of presentation.

GML geometries
GML encodes the GML geometries, or geometric characteristics, of geographic objects as elements within GML documents. The geometries of those objects may describe, for example, roads, rivers, and bridges.

The key GML geometry object types in GML 1.0 and GML 2.0, are the following:

 * Point
 * LineString
 * Polygon

Note that this geometry model is identical to the geometry model in KML.

Features
GML defines features distinct from geometry objects. A feature is an application object that represents a physical entity, e.g. a building, a river, or a person. A feature may or may not have geometric aspects. A geometry object defines a location or region instead of a physical entity, and hence is different from a feature. The distinction between features and geometry objects in GML contrasts with models used in other geographic information systems (GIS) that make no such distinction. That is, although some other GIS define features and geometry objects interchangeably as items on a map, GML maintains them as separate entity types.

In GML, a feature can have various geometry properties that describe geometric aspects or characteristics of the feature (e.g. the feature's Point or Extent properties). GML also provides the ability for features to share a geometry property with one another by using a remote property reference on the shared geometry property. Remote properties are a general feature of GML borrowed from RDF. An xlink:href attribute on a GML geometry property means that the value of the property is the resource referenced in the link.

For example, a Building feature in a particular GML application schema might have a position given by the primitive GML geometry object type Point. However, the Building is a separate entity from the Point that defines its position. In addition, a feature may have several geometry properties (or none at all), for example an extent and a position.

Coordinates
Coordinates in GML represent the coordinates of geometry objects. Coordinates can be specified by any of the following GML elements:

 *
 *
 *

GML has multiple ways to represent coordinates. For example, the  element can be used, as follows:

 45.67, 88.56

Note that, when expressed as above, the individual coordinates (e.g. 88.56) are not separately accessible through the XML Document Object Model since the content of the  element is just a single string.

To make GML coordinates accessible through the XML DOM, GML 3.0 introduced the  and  elements. (Note that although GML versions 1 and 2 had the  element, it is treated as a defect and is not used.) Using the  element instead of the  element, the same point can be represented as follows:

 45.67 88.56</gml:pos> </gml:Point>

The coordinates of a <gml:LineString> geometry object can be represented with the <gml:coordinates> element:

<gml:LineString gml:id="p21" srsName="urn:ogc:def:crs:EPSG:6.6:4326"> <gml:coordinates>45.67, 88.56 55.56,89.44</gml:coordinates> </gml:LineString >

The <gml:posList> element is used to represent a list of coordinate tuples, as required for linear geometries:

<gml:LineString gml:id="p21" srsName="urn:ogc:def:crs:EPSG:6.6:4326"> <gml:posList dimension="2">45.67 88.56 55.56 89.44</gml:posList> </gml:LineString >

For GML data servers (WFS) and conversion tools that only support GML 1 or GML 2 (i.e. only the <gml:coordinates> element), there is no alternative to <gml:coordinates>. For GML 3 documents and later, however, <gml:pos> and <gml:posList> are preferable to <gml:coordinates>.

For more information on the srsName attribute, see Coordinate Reference System below.

Coordinate Reference System
A Coordinate Reference System (CRS) determines the geometry of each geometry element in a GML document.

Unlike KML or GeoRSS, GML does not default to a coordinate system when none is provided. Instead, the desired coordinate system must be specified explicitly with a Coordinate Reference System (CRS) or Spatial Reference System (SRS). The elements whose coordinates are interpreted with respect to such a CRS include the following:
 * <gml:coordinates>
 * <gml:pos>
 * <gml:posList>

An srsName attribute attached to a geometry object specifies the object's CRS, as shown in the following example:

<gml:Point gml:id="p1" srsName="#srs36"> <gml:coordinates>100,200</gml:coordinates> </gml:Point>

The value of the srsName attribute is a Uniform Resource Identifier (URI). It refers to a definition of the Coordinate Reference System that is used to interpret the coordinates in the geometry. The CRS definition may be in a document (i.e. a flat file) or in an online web service.

The srsName URI may also be a Uniform Resource Name (URN) for referencing a common CRS definition. The OGC has developed a URN structure and a set specific URNs to encode some common Coordinate Reference Systems. A URN resolver resolves those URNs to GML CRS definitions.

Examples
Polygons, Points, and LineString objects are encoded in GML 1.0 and 2.0 as follows:

<gml:Polygon> <gml:outerBoundaryIs> <gml:LinearRing> <gml:coordinates>0,0 100,0 100,100 0,100 0,0</gml:coordinates> </gml:LinearRing> </gml:outerBoundaryIs> </gml:Polygon> <gml:Point> <gml:coordinates>100,200</gml:coordinates> </gml:Point> <gml:LineString> <gml:coordinates>100,200 150,300</gml:coordinates> </gml:LineString>

Note that LineString objects, along with LinearRing objects, assume linear interpolation between the specified points.

Features using geometries
The following GML example illustrates the distinction between features and geometry objects. The Building feature has several geometry objects, sharing one of them (the Point with identifier p21) with the SurveyMonument feature:

<abc:Building gml:id="SearsTower"> <gml:name>Sears Tower</gml:name> <abc:height>52</abc:height> <abc:position> <gml:Point> <gml:coordinates>100,200</gml:coordinates> </gml:Point> </abc:position> <app:extent> <gml:Polygon> <gml:exterior> <gml:LinearRing> <gml:coordinates>100,200</gml:coordinates> </gml:LinearRing> </gml:exterior> </gml:Polygon> </app:extent> </abc:Building> <abc:Building gml:id="SearsTower"> <abc:position xlink:type="Simple" xlink:href="#p21"/> </abc:Building> <abc:SurveyMonument gml:id="g234"> <abc:position> <gml:Point gml:id="p21"> <gml:coordinates>100,200</gml:coordinates> </gml:Point> </abc:position> </abc:SurveyMonument>

Note that the reference is to the shared Point and not to the SurveyMonument, since any feature object can have more than one geometry object property.

Point Profile
The GML Point Profile contains a single GML geometry, namely a <gml:Point> object type. Any XML Schema can use the Point Profile by importing it and referencing the subject <gml:Point> instance:

<PhotoCollection xmlns="http://www.myphotos.org" xmlns:gml="http://www.opengis.net/gml" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.myphotos.org     MyGoodPhotos.xsd"> <Item> Lynn Valley A shot of the falls from the suspension bridge North Vancouver <gml:Point srsDimension="2" srsName="urn:ogc:def:crs:EPSG:6.6:4326"> <gml:pos>49.40 -123.26</gml:pos> </gml:Point> </Item> </PhotoCollection>

Note that when using the Point Profile, the only geometry object is the '<gml:Point>' object. The rest of the geography is defined by the photo-collection schema.