text:
User talk:CarrieBerg

Hello Former Student!

I'm writing to you because you took a class from me in the past few years at the Art Institute. As of Spring, 2010, I'll not be teaching further classes at AIS for the foreseeable future. (We're parting ways on good terms!) Until now, you've been able to reach my Wikiversity educational resources from my website, but I've decided to delete that link. You can still reach all of those materials here.

If you're interested in what I'm up to in the future (which will include an album in 2010 and a novel in 2011) you can friend my facebook page or drop me an email at info@stevenarntson.info and you'll receive a periodic update about performances, classes, and publications.

If you know any current or former students of mine who might be interested in any of this, please forward this information if you have a moment.

I hope all is well for you.
 * Best! Stevenarntson

 Hello CarrieBerg, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --JWSchmidt 15:33, 16 April 2009 (UTC)

space opera
You might be interested in Science Fiction Challenge, VirileMail, [http://fiction.wikia.com/wiki/How_an_Alien_Wife_committed_Adultery_and_tricked_her_Husband. rewrite of The Miller's Tale by Chaucer but with aliens] --JWSchmidt 15:39, 16 April 2009 (UTC)

Stories
you write on your User page : "Here are some of my short stories. Please do not use without my permission!".

When you click on the Save button, you put the text under the GFDL Licence, so, you can not prohibit someone from using your writing.Crochet.david 15:06, 29 April 2009 (UTC)

"sigh," Very well, I do not like people using my work without me knowing about it, but I shall edit my sentence... I can at least ask them not to use it without citing me - I believe that falls under the licence.CarrieBerg 03:55, 30 April 2009 (UTC)

Great Stuff!
There is some great development on this page so far! I'm enjoying it. I have found, as you mention, that working with Oulipian restrictions increases my vocabulary more than just about any other kind of writing.

Additionally, it's true that Wikiversity uses a very permissive license for work, which is one of the things I enjoy about the place--everything is open to sharing and reuse, but attribution is, as you say, required. Best! Stevenarntson

.:.:.:.:.:.

 * Hmm... I had not heard of "Oulipian". A search yielded a more common phrase, "oulipian constraints" which would be a good resource at Wikiversity, imo. I also found Claude Berge who may be a kindred spirit. Interesting stuff, indeed. • CQ 05:38, 7 June 2009 (UTC)

Nine Healers
I feel good about this one doing double-duty, as you suggest. Great work on it! Stevenarntson 16:16, 12 May 2009 (UTC)

Self-Assessment
Thanks for your writing this quarter--recently, I enjoyed the story of the watch. I agree that there's an accumulative nature to oral stories, and also to written ones, when they get passed around. I'd enjoy seeing this grow. . . Best, Stevenarntson

Response
Carrie -- Nice work. I'm really impressed by the amount of writing that you've managed to do here. It seems that you have no shortage of inspiration, which I really envy! I think the piece I enjoyed the most was the Alchymyst's Dilemma, for two reasons -- first, because of your clever rewriting of Hamlet's soliloquy, and second, because (as you noted) you cunningly avoided doing a typical compare/contrast essay (which, based on other examples of your analytical and compositional skills, I have no doubt that you could do if you wanted to). So three reasons -- thanks for not boring us. Oh, and I also have to comment on your 'Senses' Oulipo -- I think it is an especially successful example of the style because for the most part you managed to create a very natural-sounding composition, rather than something that sounds like words were just shoehorned in and sentences distorted to fit your limitation. P.S. Congratulations. I think that's the last example of Oulipo that I ever want to read. It quickly becomes rather tiresome, doesn't it? ChristopherHailey

Carrie Berg: The Sorcerer’s Daughter It is indeed a beautiful image painted in words, including the dark and foreboding places described. I think it could be worked with for readers of the Valdemar series age grouping, or people like me who can’t help but get sucked into something with a fantasy air. Any time someone goes to the effort of trying to make something up completely, they often can get so involved in creating a place so different than our world that they get way laid by the detailing of the image, and forget the story element, or the fact that not everyone has the patience to obsess over the precise shape of leaves and exact hexadecimal color in the brief flash of a rainbow dress. For the most part you manage to maintain image without forgetting the journey. Unfortunately you do not continue the story. I too am guilty of stopping short. I recommend that you continue this one when you find the time. Young readers are often molly coddled away from this type of story style, and I feel you could really make something to break them free of that bubble, while simultaneously keeping sheltering parents and guardians wooed over that it is not too scary. Keep up the entertaining work! ^_^ Sethleneelliott

title: User talk:CarrieBerg


text:
User:Dnessett

My Projects

User:Dnessett/Sandboxes

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

title: User:Dnessett


text:
User:Dnessett/Sandboxes

First Sandbox

title: User:Dnessett/Sandboxes


text:
User:Egm6322.s09.mafia/HW7

 See my comments below. Egm6322.s09 09:37, 3 May 2009 (UTC)

=The Eigenvalue Problem, Cot'd=

--Egm6322.s09.xyz 19:30, 23 April 2009 (UTC)

Let $$ \mathbf{A} =

\begin{pmatrix} 2 & 5 \\ 5 & 7 \end{pmatrix}

$$

HW: Matlab to find Eigenvalues/Eigenvectors

--Egm6322.s09.xyz 19:52, 23 April 2009 (UTC)

Using Matlab, the eigenvalues are given as:

$$\lambda_1 =-1.0902$$ and $$\lambda_2 = 10.0902$$

OR $$ \mathbf{\Lambda} =

\begin{pmatrix} -1.0902 & 0 \\ 0 & 10.0902 \end{pmatrix}

$$

The corresponding eigenvectors are the columns of the following matrix, V:

$$ \mathbf{V}=

\begin{pmatrix} -0.8507 & 0.5257 \\ 0.5257 & 0.8507 \end{pmatrix}

$$

MATLAB CODE: A = [2 5 ; 5 7]; [v,d] = eig(A); %where eigenvalues are given in matrix d, and, %the eigenvectors are the columns of matrix v %such that A*v = b*v*d

HW: Use Matlab to prove that $$\mathbf{V}^{-1} = \mathbf{V}^T$$

--Egm6322.s09.xyz 02:17, 24 April 2009 (UTC)

Recall that for the matrix given above, the eigenvector matrix is given as:

$$

\mathbf{V} =

\begin{pmatrix} -0.8507 & 0.5257 \\ 0.5257 & 0.8507 \end{pmatrix}

$$

Using Matlab, the inverse of the eigenvector matrix is:

$$

\mathbf{V}^{-1} =

\begin{pmatrix} -0.8507 & 0.5257 \\ 0.5257 & 0.8507 \end{pmatrix}

$$

MATLAB CODE for the inverse of a matrix, V inv(v);

The respective transpose matrix is given as:

$$

\mathbf{V}^T =

\begin{pmatrix} -0.8507 & 0.5257 \\ 0.5257 & 0.8507 \end{pmatrix}

$$

MATLAB CODE for the transpose of a matrix, V v';

$$\therefore \mathbf{V}^{-1} = \mathbf{V}^T \surd$$ (verified)

HW: Use Matlab to show that $$\mathbf{V} \mathbf{\Lambda}\mathbf{V}^T = \mathbf{A}$$

--Egm6322.s09.xyz 02:19, 24 April 2009 (UTC)

Again, the matrix A is given as:

$$ \mathbf{A} =

\begin{pmatrix} 2 & 5 \\ 5 & 7 \end{pmatrix}

$$

From the information given above, it has already been shown that the eigenvalue (diagonal) matrix, given as $$\mathbf{\Lambda}$$ and the transpose of the eigenvector matrix, $$\mathbf{V}^T$$ are given as:

$$ \mathbf{\Lambda} =

\begin{pmatrix} -1.0902 & 0 \\ 0 & 10.0902 \end{pmatrix}

$$

and

$$ \mathbf{V}^T =

\begin{pmatrix} -0.8507 & 0.5257 \\ 0.5257 & 0.8507 \end{pmatrix}

$$

where

$$

\mathbf{V} =

\begin{pmatrix} -0.8507 & 0.5257 \\ 0.5257 & 0.8507 \end{pmatrix}

$$

Breaking the stated multiplication of the aformentioned matrices into two steps, one can see that:

$$

\mathbf{V} \mathbf{\Lambda} =

\begin{pmatrix} -0.8507 & 0.5257 \\ 0.5257 & 0.8507 \end{pmatrix}

\begin{pmatrix} -1.0902 & 0 \\ 0 & 10.0902 \end{pmatrix}

$$

$$ \therefore

\mathbf{V} \mathbf{\Lambda} =

\begin{pmatrix} 0.9274 & 5.3047 \\ -0.5731 & 8.5832 \end{pmatrix}

$$

and

$$

\left( \mathbf{V} \mathbf{\Lambda} \right) \mathbf{V}^T =

\begin{pmatrix} 0.9274 & 5.3047 \\ -0.5731 & 8.5832 \end{pmatrix}

\begin{pmatrix} -0.8507 & 0.5257 \\ 0.5257 & 0.8507 \end{pmatrix}

$$

$$ \therefore

\mathbf{V} \mathbf{\Lambda} \mathbf{V}^T = \mathbf{A} =

\begin{pmatrix} 2 & 5 \\ 5 & 7 \end{pmatrix}

\surd $$ verified

MATLAB Code for the combined matrix multiplication given above: A = [2 5; 5 7]; [v,d]=eig(A); %note: v = eigenvector matrix and d = diagonal eigenvalue matrix A_mult = v*d*v'; %Comparison of the matrix A to matrix A_mult %will yield A = A_mult

note: Given any matrix, attempt to change coordinates so to express that matrix as simple as possible. Best = diagonal form. Not all matrices are diagonalizable; best = Jordan canonical form. An important case is the real symmetric matrix

$$

\begin{bmatrix} \lambda_1     & \cdots & 0      \\ \vdots & \ddots & \vdots \\ 0     & \cdots & \lambda_n \end{bmatrix}

$$

Ref: pg. 34-1, eqn 2

$$ \left \lfloor x \; y \right \rfloor

\mathbf{V} \mathbf{\Lambda}\mathbf{V}^T

\begin{Bmatrix} x \\ y \end{Bmatrix}

+

\left \lfloor d \; e \right \rfloor

\begin{Bmatrix} x \\ y \end{Bmatrix}

+ f

= 0

$$

Definition

$$

\begin{Bmatrix} \bar{x} \\ \bar{y} \end{Bmatrix}

\mathbf{V}^T

\begin{Bmatrix} x \\ y \end{Bmatrix}

$$

\Rightarrow

\begin{Bmatrix} x \\ y \end{Bmatrix}

= \mathbf{V}

\begin{Bmatrix} \bar{x} \\ \bar{y} \end{Bmatrix}

$$

 * where $$\mathbf{V} (\mathbf{V}^T) = \mathbf{I}$$

$$ \therefore

\left \lfloor \bar{x} \; \bar{y} \right \rfloor

\mathbf{\Lambda}

\begin{Bmatrix} \bar{x} \\ \bar{y} \end{Bmatrix}

+

\left \lfloor \bar{d} \; \bar{e} \right \rfloor

\begin{Bmatrix} \bar{x} \\ \bar{y} \end{Bmatrix}

+ f

= 0

$$

HW: Prove $$\mathbf{\Lambda}= \mathbf { V^T A V}$$ --Egm6322.s09.lapetina 00:42, 24 April 2009 (UTC)

Prove:

$$\mathbf{\Lambda}= \mathbf { V^T A V}$$

where:

$$\mathbf{V \Lambda V^T}=\mathbf{A}$$

$$\mathbf{V^{-1}}=\mathbf{V^{T}}$$

and

$$\mathbf{VV^{T}}=\mathbf{I}$$

If we premultiply the statement:

$$\mathbf{V \Lambda V^T}=\mathbf{A}$$

by $$\mathbf{V^{T}}$$, we are left with:

$$\mathbf{ V^{T} V \Lambda V^T}=\mathbf{V^{T} A}$$, and if we post-multiply this by

$$\mathbf{V}$$, we are left with:

$$\mathbf{ V^{T} V \Lambda V^T V}=\mathbf{V^{T} A V}$$.

Substituting $$\mathbf{V^{-1}}$$ for $$\mathbf{V}$$, we're left with:

$$\mathbf{ V^{-1} V \Lambda V^{-1} V}=\mathbf{V^{T} A V}$$.

By definition, $$\mathbf{V^{-1} V}=\mathbf{VV^{-1}}=\mathbf{I}$$, and

$$\mathbf{ I \Lambda I}=\mathbf{V^{T} A V}$$, equivalent to:

$$\mathbf{ \Lambda}=\mathbf{V^{T} A V}$$

Note that $$\mathbf{\Lambda} $$ is a diagonal matrix

$$

\begin{bmatrix} \lambda_1     & \cdots & 0      \\ \vdots & \ddots & \vdots \\ 0     & \cdots & \lambda_n \end{bmatrix}

$$

OR

$$

\mathbf{\Lambda} =

\begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}

\Rightarrow

\left[ \lambda_1 \bar{x}^2 + \lambda_2 \bar{y}^2 \right]

+\bar{d} \bar{x} + \bar{e}\bar{y}

+ f = 0

$$

Consider the cases where:

Case#1: $$\lambda_1 \lambda_2 < 0$$ (hyperbola) or $$\lambda_1 \lambda_2 > 0$$ (ellipses)

Case#2: $$\lambda_1 \lambda_2 = 0$$(parabolas)

HW: Show case#1 and case#2

Egm6322.s09.bit.sahin 16:44, 24 April 2009 (UTC)

We know that Report 6. $$\lambda^2-\lambda(a+c)+(ac-b^2)=0$$

and

$$\left (\lambda-\lambda_1  \right )\left (\lambda-\lambda_2   \right )=\lambda^{2}-\left ( \lambda_{1}+ \lambda_{2}\right )\lambda+\lambda_{1}\lambda_{2}=0$$

therefore

$$\lambda_{1}\lambda_{2}=ac-b^{2}$$

So,

Case#1: $$\lambda_1 \lambda_2 < 0$$ (hyperbola) or $$\lambda_1 \lambda_2 > 0$$ (ellipses)

Case#2: $$\lambda_1 \lambda_2 = 0$$(parabolas)

Hyperbolic Case
Complete the details to show how we arrive at:

$$\xi ^2- \eta^2= \pm 1$$, the canonical form of hyperbolae,

from the equation:

$$\lambda_1 (\bar\bar x)^2+ \lambda_2 (\bar\bar x)^2=g$$,

where

$$g=- \left [ \lambda_1 r^2 + \lambda_2 s^2 -\bar d r -\bar e s +f \right ]$$

and

$$\lambda_1 \lambda_2 < 0$$. We will assume $$\lambda_1 > 0$$, $$ \lambda_2 < 0$$ and $$g>0$$.

Elliptic Case
--Egm6322.s09.xyz 16:46, 24 April 2009 (UTC)

Cases $$\lambda_1 \lambda_2 >0$$ and $$\lambda_1 \lambda_2 <0$$,

$$\lambda_1 \lambda_2 >0$$is the case of ellipse

$$\lambda_1 \lambda_2 <0$$is the case of hyperbola

In both of these two cases:$$\lambda_1 \lambda_2 \neq0$$

Geometric interpretation:

Eigenvalue problem $$\Leftrightarrow$$ Rotation of $$(x,y)$$to$$(\overline {x},\overline{y})$$which are parallel to principal axes of conics. Orthogonal of $$\underline {V}$$ $$(\underline {v} \ \underline {v}^T$$ $$=\underline {I})$$

Next Step: Get rid of linear terms in

$$\left \lfloor \overline {x} \ \overline {y} \right \rfloor$$ $$\underline {\Lambda}$$ $$	\binom{\overline {x}}{\overline {y}}   $$ + $$\left \lfloor \overline {d} \ \overline {e} \right \rfloor$$ $$	\binom{\overline {x}}{\overline {y}}   $$ +$$f=0$$ Equation(1)

Translation of Coordinate System.

Define the new coordinate system $$(\overline{\overline {x}} \ \overline{\overline {y}}) $$as:

$$	\binom{\overline{\overline {x}}}{\overline{\overline {y}}}$$ = $$\binom {\overline {x}+r}{\overline {y}+s}$$

(r,s)are unknowns.It's the "center" of ellipse or hyperbola.

$$\binom {\overline {x}}{\overline {y}}$$ = $$	\binom{\overline{\overline {x}}-r}{\overline{\overline {y}}-s}$$

Plugging it into Equation (1), we obtain:

$$\lambda_1 (\overline{\overline{x}}-r)^2+\lambda_2 (\overline{\overline{y}}-s)^2$$ $$+\overline {d} (\overline{\overline{x}}-r)+\overline {e} (\overline{\overline{y}}-s) +f =0$$

--EGM6322.S09.TIAN 21:36, 23 April 2009 (UTC)

HW:Find Out r and s Expanding

$$\lambda_1 (\overline{\overline{x}}-r)^2+\lambda_2 (\overline{\overline{y}}-s)^2$$ $$+\overline {d} (\overline{\overline{x}}-r)+\overline {e} (\overline{\overline{y}}-s) +f =0$$

We obtain:

$$\lambda_1 \overline {\overline {x}}^2 + \lambda_2 \overline {\overline {y}}^2$$ $$+(\overline {d}- 2 \lambda_1 r )\overline {\overline {x}}$$ $$+(\overline {e}- 2 \lambda_2 s )\overline {\overline {y}}$$ $$+(\lambda_1 r^2 + \lambda_2 s^2 - \overline {d} r -\overline {e} s +f)=0$$

Because we need to get rid of linerar terms,

$$ \overline {d} - 2 \lambda_1 r =0$$

$$ \overline {e} - 2 \lambda_2 s =0$$,

Finally we have:

$$r= \overline {d} / 2 \lambda_1$$

$$s= \overline {e} / 2 \lambda_2$$

--EGM6322.S09.TIAN 21:36, 23 April 2009 (UTC)

HW:Find Out g $$\lambda_1 (\overline{\overline{x}}-r)^2+\lambda_2 (\overline{\overline{y}}-s)^2=g$$

Comparing it to

$$\lambda_1 \overline {\overline {x}}^2 + \lambda_2 \overline {\overline {y}}^2$$ $$+(\lambda_1 r^2 + \lambda_2 s^2 - \overline {d} r -\overline {e} s +f)=0$$

we get,

$$g=-(\lambda_1 r^2 + \lambda_2 s^2 - \overline {d} r -\overline {e} s +f)$$

--EGM6322.S09.TIAN 21:36, 23 April 2009 (UTC)

=The Geometric Interpretation of Classification= =The Canonical Form of Conics=

Egm6322.s09.Three.nav 13:49, 24 April 2009 (UTC)

The modified form of the Equation of Conics is derived at the end of the section on The EigenValue Problem. This standard form on further simplification reduces to

$$\lambda_{1}\bar{\bar{x}}^2+\lambda_{2}\bar{\bar{y}}^2+f-\frac{d^2}{4\lambda_{1}}-\frac{e^2}{4\lambda_{2}}=0$$ ---(1)

Define $$ g= -(f-\frac{d^2}{4\lambda_{1}}-\frac{e^2}{4\lambda_{2}})$$. Then (1) reduces to the form

$$\lambda_{1}\bar{\bar{x}}^2+\lambda_{2}\bar{\bar{y}}^2= g$$--(2)

Here $$\lambda_{1}$$ and $$\lambda_{2}$$ denote the eigenvalues obtained by solving the eigenvalue problem for the original equation of conics: $$Ax^2+Bxy+Cy^2+\bar{d}x+\bar{e}y+\bar{f}=0$$

Equation (2) can be further reduced to equations representing the canonical forms for an ellipse, hyperbola and parabola.

Case (i):Ellipse: $$\lambda_{1}$$ and $$\lambda_{2}$$ >0 $$\Rightarrow$$ g>0

The canonical form representing the ellipse then reduces to: $$\xi^2+\eta^2=1$$ --(3)

Case (ii): Hyperbola: $$\lambda_{1}$$ and $$\lambda_{2}$$ <0 $$\Rightarrow g\neq 0 $$

The canonical form representing the hyperbola takes two different forms depending on the sign of 'g'.

Case ii(a): g>0 $$\to \xi^2-\eta^2=+1$$ ---(4a) (This is also true for $$\lambda_{1} < 0$$,$$\lambda_{2} > 0$$ and g>0)

Case ii(b): g<0 $$\to \xi^2-\eta^2=-1$$ ---(4b) (This is also true for $$\lambda_{1} < 0$$,$$\lambda_{2} > 0$$ and g<0) Case ii(c) $$\bar{\xi}\bar{\eta}=\pm 1$$--(4c)

This is a different representation of the canonical form for hyperbolas. It is derived from either of the equations (4a) or (4b)

Note that $$[\xi,\eta]$$ represent a different coordinate system using which (2) has been transformed into the various canonical forms shown. The derivation of forms(3),(4a) and (4b) are shown in separate sections. The derivation of form (4c) from (4a)or (4b) is shown below.

Define a new coordinate system $$[\bar{\xi}, \bar{\eta}]$$ in terms of the old coordinate system $$[\xi, \eta]$$ such that $$ \begin{Bmatrix}\bar{\xi} \\ \bar{\eta}\end{Bmatrix}= \begin{Bmatrix}\xi-\eta \\ \xi+\eta\end{Bmatrix}$$

$$ \Rightarrow \begin{Bmatrix}\bar{\xi}\\ \bar{\eta}\end{Bmatrix}= \begin{bmatrix}\ \ 1 \ -1 \\ -1 \ -1\end{bmatrix}\begin{Bmatrix}\xi-\eta \\ \xi+\eta\end{Bmatrix}$$

$$\Rightarrow \begin{Bmatrix}\xi \\ \eta \end{Bmatrix}= \begin{bmatrix}\ \ 1 \ -1 \\ -1 \ -1\end{bmatrix}^{-1}\begin{Bmatrix} \bar{\xi} \\ \bar{\eta} \end{Bmatrix}= \frac {1}{2}\begin{bmatrix}\ \ 1 \ -1 \\ -1 \ -1\end{bmatrix}\begin{Bmatrix}\bar{\xi} \\ \bar{\eta} \end{Bmatrix}$$

Eqn.(4a) or (4b) can be re-written in the form: $$\begin{Bmatrix}\xi \ \eta \end{Bmatrix}\begin{bmatrix}1 \ \ \ \ 0 \\ 0 \ -1 \end{bmatrix}\begin{Bmatrix}\xi \\ \eta \end{Bmatrix}= \pm 1$$--(5)

Rewriting (5) in the new coordinate system $$[\bar{\xi}, \bar{\eta}]$$, we get:

$$ \begin{Bmatrix}\bar{\xi} \ \bar{\eta} \end{Bmatrix}\frac{1}{2}\begin{bmatrix} \ \ 1 \  -1 \\ -1 \ -1 \end{bmatrix}\begin{bmatrix}1 \ \ \ \  0 \\ 0 \ -1 \end{bmatrix}\frac{1}{2}\begin{bmatrix} \ \ 1 \   -1 \\ -1 \ -1 \end{bmatrix}\begin{Bmatrix}\bar{\xi} \\ \bar{\eta} \end{Bmatrix}= \pm 1$$

$$ \Rightarrow \frac{1}{4} \begin{Bmatrix}\bar{\xi} \ \bar{\eta} \end{Bmatrix}\begin{bmatrix} \ \ 1 \  -1 \\ -1 \ -1 \end{bmatrix}\begin{bmatrix}1 \ \ \ \  0 \\ 0 \ -1 \end{bmatrix}\begin{bmatrix} \ \ 1 \   -1 \\ -1 \ -1 \end{bmatrix}\begin{Bmatrix}\bar{\xi} \\ \bar{\eta} \end{Bmatrix}= \pm 1$$

$$\Rightarrow \frac{1}{4}\begin{Bmatrix}\bar{\xi} \ \bar{\eta} \end{Bmatrix}\begin{bmatrix}\ \ \ 0 \ -2 \\ -2 \ \ \ \ 0 \end{bmatrix}\begin{Bmatrix}\bar{\xi} \\ \bar{\eta} \end{Bmatrix}= \pm 1$$

$$\Rightarrow \bar{\xi}\bar{\eta}=\pm 1$$ -(4c)

Thus we see that by a rotation of coordinates, form (4a) or (4b), which is the standard canonical form representing a hyperbola reduces to form (4c). This simplified form is used mostly in algebraic calculations and even in high-school geometry lessons.

Canonical Form of a Parabola
Egm6322.s09.Three.nav 13:49, 24 April 2009 (UTC)

Given the general,modified form of a PDE: $$\lambda_{1}\bar{x}^2+\lambda_{2}\bar{y}^2+d\bar{x}+e\bar{y}+f=0$$--(1)

Derive the canonical form representing a parabola from (1).

The canonical form representing a parabola is given by the following equation

$$\bar{\xi}^{2}- \bar{\eta} = 0$$ --(2)

where $$\bar{\xi}$$ and $$\bar{\eta}$$ represent coordinates in a different coordinate system.

So how do we derive (2) from (1)? We apply rotation once and translation of coordinates twice to $$\bar{x}$$ and $$\bar{y}$$.

Step(1)- First Translation

Define $$\begin{bmatrix}\bar{\bar{x}}\\ \bar{\bar{y}}\end{bmatrix}= \begin{bmatrix}\bar{x}+r \\ \bar{y}\end{bmatrix}$$

Substituting $$\bar{\bar{x}}$$ and $$\bar{\bar{y}}$$ in (1), we get

$$\lambda_{1}(\bar{\bar{x}}-r)^2+\lambda_{2}\bar{\bar{y}}^2+d(\bar{\bar{x}}-r)+e\bar{\bar{y}}+f=0$$

$$\Rightarrow \lambda_{1}\bar{\bar{x}}^{2}+\lambda_{2}\bar{\bar{y}}^2+\bar{\bar{x}}(d-2\lambda_{1}r)+e\bar{\bar{y}}+\lambda_{1}r^{2}+f=0$$--(3)

For equation (3) to resemble (2), we need the terms: $$\bar{\bar{y}}^2, \bar{\bar{x}} \ and \ \lambda_{1}r^{2}+f$$ to vanish.

Note that for a parabola, the equation of conics: $$Ax^2+Bxy+Cy^2+\bar{d}x+\bar{y}+\bar{f}=0$$, $$B^2-4AC=0$$ i.e only one eigenvalue exists. Thus assume $$\lambda_{2}$$= 0.

Also in order for the $$\bar{\bar{x}}$$ term to vanish, we need its coefficient to be equal to 0 $$\Rightarrow d- 2\lambda_{1}r=0$$

$$\Rightarrow r= \frac{d}{2\lambda_{1}r}$$

Substituting $$\lambda_{2}$$ and r in (3), the equation reduces to $$\lambda_{1}\bar{\bar{x}}^{2}+e\bar{\bar{y}}+\frac{d}{4\lambda_{1}^{2}}+f=0$$-(4)

Denote g= $$\frac{d}{4\lambda_{1}^{2}}+f$$. Thus (4) reduces to

$$\Rightarrow \lambda_{1}\bar{\bar{x}}^{2}+e\bar{\bar{y}}+g=0$$-(5)

Step(2)- Rotation

Eqn.(4) can be rewritten as $$\left \lfloor \bar{\bar{x}} \ \bar{\bar{y}} \right \rfloor\begin{Bmatrix} \lambda_{1} \ 0 \\ 0 \ \ 0 \end{Bmatrix}\begin{bmatrix}\bar{\bar{x}}\\ \bar{\bar{y}}\end{bmatrix}+\left \lfloor 0 \ e \right \rfloor\begin{bmatrix}\bar{\bar{x}}\\ \bar{\bar{y}}\end{bmatrix}+g=0$$ (6)

Now apply rotation of coordinates by defining a new coordinate system $$(\xi, \eta)$$ where $$\begin{bmatrix} \xi \\ \eta \end{bmatrix}= \mathbf{J_{\beta}}\begin{bmatrix}\bar{\bar{x}}\\ \bar{\bar{y}}\end{bmatrix}$$

where $$\mathbf{J_{\beta}}= \begin{bmatrix}\frac{1}{\sqrt{\lambda_{1}}} \ \ \ 0 \\ \ \ \ 0 \ \ \ \frac{1}{e}\end{bmatrix}$$

Due to $$\mathbf{J_{\beta}}$$ being an orthogonal matrix, $$\mathbf{J_{\beta}}^{-1}= \mathbf{J_{\beta}}^{T}= \mathbf{J_{\beta}}$$

$$\Rightarrow \begin{bmatrix}\bar{\bar{x}}\\\bar{\bar{y}} \end{bmatrix}= \mathbf{J_{\beta}}^{-1}\begin{bmatrix}\xi\\\eta \end{bmatrix}= \mathbf{J_{\beta}}\begin{bmatrix}\xi\\\eta \end{bmatrix}$$

Substituting this rotation of coordinates in (6), we get

$$\begin{bmatrix}\xi\ \eta \end{bmatrix} \mathbf{J_{\beta}} \begin{Bmatrix}\lambda_{1} \ 0 \\ 0 \ \ 0 \end{Bmatrix} \mathbf{J_{\beta}}\begin{bmatrix}\xi\\\eta \end{bmatrix}+\left \lfloor 0 \ e \right \rfloor \mathbf{J_{\beta}}\begin{bmatrix}\xi\\\eta \end{bmatrix}+g=0$$

$$\Rightarrow \begin{bmatrix}\xi\ \eta \end{bmatrix} \begin{bmatrix}\frac{1}{\sqrt{\lambda_{1}}} \ \ \ 0 \\ \ \ \ 0 \ \ \ \frac{1}{e}\end{bmatrix} \begin{Bmatrix}\lambda_{1} \ 0 \\ 0 \ \ 0 \end{Bmatrix} \begin{bmatrix}\frac{1}{\sqrt{\lambda_{1}}} \ \ \ 0 \\ \ \ \ 0 \ \ \ \frac{1}{e}\end{bmatrix} \begin{bmatrix}\xi\\\eta \end{bmatrix}+\left \lfloor 0 \ e \right \rfloor \begin{bmatrix}\frac{1}{\sqrt{\lambda_{1}}} \ \ \ 0 \\ \ \ \ 0 \ \ \ \frac{1}{e}\end{bmatrix} \begin{bmatrix}\xi\\\eta \end{bmatrix}+g=0$$

Simplifying the above equation further, we obtain

$$\xi^2- \eta+g=0$$ ---(7)

Step(3)- Second Translation

Now define $$\begin{bmatrix}\bar{\xi}\\ \bar{\eta}\end{bmatrix}= \begin{bmatrix}\xi \\ \eta-g \end{bmatrix}$$

This is a valid translation because $$\lambda_{1}$$ which is a function of the coefficients A,C (as defined in the equation of conics) is a constant value. Hence g which is a function of $$\lambda_{1}$$ and f is also aconstant. Thus all we are doing in the second translation is translate the $$\bar{\bar{y}}$$ by a constant. Hence making this substitution in (7) gives us the final canonical form (2)

$$\bar{\xi}^{2}- \bar{\eta} = 0$$

Egm6322.s09.Three.nav 03:39, 24 April 2009 (UTC)

Canonical Form of a Hyperbola
HW:Deriving the Canonical Form for a Hyperbola

--Egm6322.s09.lapetina 01:18, 24 April 2009 (UTC)

Complete the details to show how we arrive at:

$$\xi ^2- \eta^2= \pm 1$$, the canonical form of hyperbolae, where $$\xi$$ and $$\eta$$ represent axes in coordinate system that coincide with the principle axes of the ellipse.

We begin with the equation:

$$\lambda_1 (\bar\bar x)^2+ \lambda_2 (\bar\bar x)^2=g$$,

where

$$g=- \left [ \lambda_1 r^2 + \lambda_2 s^2 -\bar d r -\bar e s +f \right ]$$.

This has been derived in other sections of the report here.

We will assume that $$\lambda_1 \lambda_2 < 0$$, $$\lambda_1 > 0$$, $$ \lambda_2 < 0$$ and $$g>0$$.

The eigenvalue equation can be re-expressed in matrix form as:

$$

\left \lfloor \bar{\bar{x}} \ \bar{\bar{y}} \right \rfloor

\begin{bmatrix} \lambda_{1} \ 0 \\ 0 \ \lambda_2 \end{bmatrix}

\begin{bmatrix}

\bar{\bar{x}}\\ \bar{\bar{y}}

\end{bmatrix}

=g $$

Here we want to apply a rotation transformation of the form:

$$\begin{Bmatrix}

\xi \\ \eta

\end{Bmatrix}

=

\mathbf{J_\beta}

\begin{Bmatrix}

\bar{\bar{x}}\\ \bar{\bar{y}}

\end{Bmatrix}$$.

We need to determine the values of

$$\mathbf{J_\beta}$$.

For a hyperbola of this type:

$$

\mathbf{J_\beta}=

\begin{bmatrix} \frac{1}{d_x} & 0 \\ 0 & \frac{1}{d_y} \end{bmatrix}

$$

where $$d_x = \frac{\sqrt{g}}{\sqrt{|\lambda_1|}}$$

and

$$d_y = \frac{\sqrt{g}}{\sqrt{|\lambda_2|}}$$

The inverse of this Jacobian matrix is :

$$

\mathbf{J_\beta}=

\begin{bmatrix} d_x & 0 \\ 0 & d_y \end{bmatrix}

$$,

and

$$ \begin{Bmatrix}

\bar{\bar{x}}\\ \bar{\bar{y}}

\end{Bmatrix}

= \begin{bmatrix} d_x & 0 \\ 0 & d_y \end{bmatrix}

\begin{Bmatrix}

\xi \\ \eta

\end{Bmatrix} $$

and

$$ \left \lfloor \bar{\bar{x}} \ \bar{\bar{y}} \right \rfloor

=

\left \lfloor \xi \ \eta \right \rfloor

\begin{bmatrix} d_x & 0 \\ 0 & d_y \end{bmatrix}$$.

Substituting this directly into the eigenvalue equation yields:

$$

\left \lfloor \xi \ \eta \right \rfloor

\begin{bmatrix} d_x & 0 \\ 0 & d_y \end{bmatrix}

\begin{bmatrix} \lambda_{1} \ 0 \\ 0 \ \lambda_2 \end{bmatrix}

\begin{bmatrix} d_x & 0 \\ 0 & d_y \end{bmatrix}

\begin{Bmatrix}

\xi \\ \eta

\end{Bmatrix}

=g $$

which simplifies to a polynomial form of:

$$d_x^2 \lambda_1 (\xi)^2- d_y^2 \lambda_2 (\eta)^2=g$$.

The sign in the middle is negative because $$\lambda_2<0$$.

This in turn simplifies to:

$$(\frac{\sqrt{g}}{\sqrt{|\lambda_1|}})^2 \lambda_1 (\xi)^2- (\frac{\sqrt{g}}{\sqrt{|\lambda_2|}})^2 \lambda_2 (\eta)^2=g$$.

We only divide the terms on the left by the square of the root of the absolute values of $$\lambda_1$$ and $$\lambda_2$$, preserving their signs. Because g is a function of constants, as described in the above reference, it retains its sign.

The result is the canonical form:

$$\xi ^2- \eta^2= 1$$

for the case where $$\lambda_1>0$$, $$\lambda_2<0$$, and $$g>0$$. To arrive at the form:

$$\xi ^2- \eta^2= -1$$ for the case where $$\lambda_1>0$$, $$\lambda_2<0$$, and $$g<0$$ is trivial. Since we divide by $$g$$ by its absolute value, it becomes a $$-1$$ using the same Jacobian.

However, for the case where:

$$\lambda_1<0$$, $$\lambda_2>0$$, and $$g<0$$, to arrive at:

$$\xi ^2- \eta^2= 1$$, we need to use different values of $$d_x$$ and $$d_y$$:

$$d_x = \frac{i \sqrt{g}}{ \sqrt{|\lambda_1|}}$$

and

$$d_y = \frac{i \sqrt{g}}{\sqrt{|\lambda_2|}}$$.

with a Jacobian of the form

$$

\mathbf{J_\beta}=

\begin{bmatrix} 0 & d_y \\ d_x & 0 \end{bmatrix}

$$.

The inverse of this is:

$$

\mathbf{J_\beta^{-1}}=

\begin{bmatrix} 0 & \frac{1}{d_y} \\ \frac{1}{d_x} & 0 \end{bmatrix}

$$

By substitution, the matrix operator form of the conic then becomes:

$$

\left \lfloor \xi \ \eta \right \rfloor

\begin{bmatrix} 0 & \frac{1}{d_y} \\ \frac{1}{d_x} & 0 \end{bmatrix}

\begin{bmatrix} \lambda_{1} \ 0 \\ 0 \ \lambda_2 \end{bmatrix}

\begin{bmatrix} 0 & \frac{1}{d_y} \\ \frac{1}{d_x} & 0 \end{bmatrix}

\begin{Bmatrix}

\xi \\ \eta

\end{Bmatrix}

=g $$

This leaves us with the polynomial equation:

$$ -1 \left [ \lambda_2 \frac{\xi^2}{d_x^2}+ \lambda_1 \frac{\eta^2}{d_y^2} \right ]=g$$.

Upon substitution, this becomes

$$(\frac{\sqrt{g}}{\sqrt{|\lambda_2|}})^2 \lambda_2 (\xi)^2- (\frac{\sqrt{g}}{\sqrt{|\lambda_1|}})^2 \lambda_1 (\eta)^2=-g$$.

Constant values cancel, leaving:

$$\xi ^2- \eta^2= 1$$ for the case where:

$$\lambda_1<0$$, $$\lambda_2>0$$, and $$g<0$$.

For the case where

$$\lambda_1<0$$, $$\lambda_2>0$$, and $$g>0$$,

to arrive at the form $$\xi ^2- \eta^2= -1$$,

we can use the same Jacobian, because the sign of $$g$$ will change the sign on the right side.

Canonical Form of an Ellipse
Egm6322.s09.Three.ge 20:30, 23 April 2009 (UTC)

HW:Deriving the Canonical Form for an Ellipse

Given the general,modified form of a PDE: $$\lambda_{1}\bar{x}^2+\lambda_{2}\bar{y}^2+d\bar{x}+e\bar{y}+f=0 \qquad \ldots (1)$$

Derive the canonical form representing an ellipse from (1).

The canonical form representing an ellipse is given by the following equation

$$\xi^{2}+ \eta^2 = 1 \qquad \ldots (2)$$

where $$\xi$$ and $$\eta$$ represent axes in coordinate system that coincide with the principle axes of the ellipse.

In order to show equation (1) in the form of equation (2)the coordinate system $$(\bar{x},\bar{y})$$ must be translated and then rotated.

Step(1)- Translation

Define $$\begin{bmatrix} \bar{\bar{x}}\\ \bar{\bar{y}}\end{bmatrix}= \begin{bmatrix}\bar{x}+r \\ \bar{y}+s\end{bmatrix}$$

Substituting $$\bar{\bar{x}}$$ and $$\bar{\bar{y}}$$ in (1) yields:

$$\lambda_{1}(\bar{\bar{x}}-r)^2+\lambda_{2}(\bar{\bar{y}}-s)^2+d(\bar{\bar{x}}-r)+e(\bar{\bar{y}}-s)+f=0$$

Expanding the squared terms gives

$$\begin{matrix} \Rightarrow \lambda_{1}\bar{\bar{x}}^{2}+\lambda_{2}\bar{\bar{y}}^2+\bar{\bar{x}}(d-2\lambda_{1}r)+\bar{\bar{y}}(e-2\lambda_{2}s)+ f=0 \\ \therefore \\ r= \frac{d}{2\lambda_{1}} \qquad s= \frac{e}{2\lambda_{2}} \end{matrix}\qquad \ldots (3)$$

Where the values for $$r$$ and $$s$$ are set so that equation (3) resembles (2). Specifically, $$r$$ and $$s$$ are found by solving the following equations.

$$\begin{matrix}(d-2\lambda_{1}r)=0 \\ \\ (e-2\lambda_{2}s)=0 \end{matrix}$$

Substituting $$r$$ and $$s$$ into equation (3), the equation reduces to:

$$\begin{matrix}\lambda_{1}\bar{\bar{x}}^2+\lambda_{2}\bar{\bar{y}}^2-\frac{d}{4\lambda_{1}^{2}}-\frac{e}{4\lambda_{2}^{2}}+f=0 \\ \textrm{Let}\quad g=\left[-\frac{d}{4\lambda_{1}^{2}}-\frac{e}{4\lambda_{2}^{2}}+f \right ] \\ \end{matrix}$$

Thus,

$$ \lambda_{1}\bar{\bar{x}}^2+\lambda_{2}\bar{\bar{y}}^2=g \quad\ldots (4)$$

Step(2)- Rotation

Equation (4) can be rewritten as $$\left \lfloor \bar{\bar{x}} \ \bar{\bar{y}} \right \rfloor

\begin{bmatrix} \lambda_{1} & 0 \\ 0 & \lambda_{2} \\ \end{bmatrix}

\begin{Bmatrix} \bar{\bar{x}}\\ \bar{\bar{y}} \end{Bmatrix}=g \quad \ldots (5)$$

Now one applies rotation of coordinates by defining to arrive at the new coordinate system $$(\xi, \eta)$$.

Define the rotation as:

$$\begin{bmatrix} \xi \\ \eta \end{bmatrix}=

\mathbf{J_{\beta}} \begin{bmatrix} \bar{\bar{x}}\\ \bar{\bar{y}} \end{bmatrix}$$

where $$\mathbf{J_{\beta}}= \begin{bmatrix} \sqrt{\frac{g}{\lambda_{1}}} & 0 \\ 0 & \sqrt{\frac{g}{\lambda_{2}}} \end{bmatrix}$$

Since $$\mathbf{J_{\beta}}$$ being an orthogonal matrix, (i.e. $$\mathbf{J_{\beta}}^{-1}= \mathbf{J_{\beta}}^{T}= \mathbf{J_{\beta}}$$) it can be shown that:

$$\begin{bmatrix} \bar{\bar{x}}\\ \bar{\bar{y}} \end{bmatrix}=

\mathbf{J_{\beta}}^{-1} \begin{bmatrix}\xi \\ \eta \end{bmatrix}=

\mathbf{J_{\beta}} \begin{bmatrix}\xi \\ \eta \end{bmatrix}$$

Substituting this rotation of coordinates into equation (5) results in:

$$\left \lfloor \xi \quad \eta \right \rfloor

\mathbf{J_{\beta}} \begin{bmatrix} \lambda_{1} & 0 \\ 0 & \lambda_{2} \end{bmatrix}

\mathbf{J_{\beta}} \begin{Bmatrix} \xi \\ \eta \end{Bmatrix} =g$$

Plugging in the value of $$\mathbf{J_{\beta}}$$ yields the following expression.

$$\left\lfloor \xi \quad \eta \right\rfloor

\begin{bmatrix} \sqrt{\frac{g}{\lambda_{1}}} & 0 \\ 0 & \sqrt{\frac{g}{\lambda_{2}}} \end{bmatrix}

\begin{bmatrix} \lambda_{1} & 0 \\ 0 & \lambda_{2} \end{bmatrix}

\begin{bmatrix} \sqrt{\frac{g}{\lambda_{1}}} & 0 \\ 0 & \sqrt{\frac{g}{\lambda_{2}}} \end{bmatrix}

\begin{Bmatrix} \xi \\ \eta \end{Bmatrix}

=g$$

Simplifying the above equation further, we obtain

$$g\xi^{2}+ g\eta^{2} = g$$

Which clearly reduces to:

$$\xi^{2}+ \eta^{2} = 1$$

 For the ellipse case, $$\displaystyle \lambda_1 \lambda_2 > 0$$. Assume that $$\displaystyle \lambda_1 > 0$$ and $$\displaystyle \lambda_2 > 0$$, so that $$\displaystyle \lambda_1 \lambda_2 > 0$$, it follows from Eq.(4) in the collapsible box that $$\displaystyle g > 0$$, and thus $$\displaystyle \sqrt(g)$$ is a real number. For the case in which $$\displaystyle \lambda_1 < 0$$ and $$\displaystyle \lambda_2 < 0$$, a similar argument can be made with $$\displaystyle - \lambda_1 > 0$$, $$\displaystyle - \lambda_2 > 0$$, and $$\displaystyle - g > 0$$. Egm6322.s09 10:09, 3 May 2009 (UTC)

=The Relationship to Transformations of Variables=

Homework: Get the Canonical form for 2nd order linear Elliptic PDE Egm6322.s09.bit.la 19:58, 24 April 2009 (UTC) Find

$$u_{\xi \xi }+u_{\eta \eta }=g(\xi ,\eta ,u,u_{\xi },u_{ \eta })$$

Using the PDE notation: $$\lambda_1u_{xx}+\lambda_2u_{yy}=g$$

Where the Jacobian = $$J_B=\begin{bmatrix} \frac {1}{\sqrt {\lambda_1}} & 0 \\ 0 & \frac {1}{\sqrt {\lambda_2}}\end{bmatrix}$$

The Jacobian is an orthogonal matrix and then:

$$\left \lfloor \partial_{\xi} \partial_{\eta} \right \rfloor J_B \begin{bmatrix} u \lambda_1 & 0 \\  0 & u \lambda_2\end{bmatrix} J_B\begin{bmatrix} \partial_{\xi} \\ \partial_{\eta}\end{bmatrix}=g $$

substituting in the equation above:

$$ \left \lfloor \partial_{\xi} \partial_{\eta} \right \rfloor \begin{bmatrix} \frac {1}{\sqrt {\lambda_1}} & 0 \\ 0 & \frac {1}{\sqrt {\lambda_2}}\end{bmatrix} \begin{bmatrix} u \lambda_1 & 0 \\  0 & u \lambda_2\end{bmatrix} \begin{bmatrix} \frac {1}{\sqrt {\lambda_1}} & 0 \\ 0 & \frac {1}{\sqrt {\lambda_2}}\end{bmatrix}\begin{bmatrix} \partial_{\xi} \\ \partial_{\eta}\end{bmatrix}=g$$

Solving it we find the expected result:

$$u_{\xi \xi }+u_{\eta \eta }=g$$

 For PDEs, it is better to put the function $$\displaystyle u$$ at the end of the matrix operator, and even though $$\displaystyle \mathbf J_B$$ is diagonal, it is still better to specify the transpose in the general formula for transformation of coordinates, i.e., $$\displaystyle \left \lfloor \partial_{\xi} \partial_{\eta} \right \rfloor \mathbf J_B \begin{bmatrix} \lambda_1 & 0 \\  0 & \lambda_2\end{bmatrix} \mathbf J_B^T \left\{ \begin{matrix} \partial_{\xi} \\ \partial_{\eta}\end{matrix} \right\} (u) = g $$. Egm6322.s09 10:09, 3 May 2009 (UTC)

Homework: Get the Canonical form for 2nd order linear Hyperbolic PDE For hyperbolic PDE: $$\lambda_1 u_{xx}-\lambda_2 u_{yy}=g$$

$$\begin{Bmatrix} \partial_{x} \\ \partial_{y} \end{Bmatrix} =J_B \begin{Bmatrix} \partial_{\xi} \\ \partial_{\eta} \end{Bmatrix}$$  ... (1)

The Jacobian matrix is:

$$ J_B= \begin{bmatrix} \frac {1}{\sqrt {\lambda_1}} & 0 \\ 0 & \frac {1}{\sqrt {\lambda_2}} \end{bmatrix}$$

so the hyperbolic PDE can be written as:

$$\left \lfloor \partial_x \partial_y \right \rfloor \begin{bmatrix} u \lambda_1 & 0 \\ 0 & -u \lambda_2 \end{bmatrix} \begin{bmatrix} \partial_x \\ \partial_y \end{bmatrix}=g$$

substituting (1) into the above equation gives:

$$\left \lfloor \partial_{\xi} \partial_{\eta} \right \rfloor \begin{bmatrix} \frac {1}{\sqrt {\lambda_1}} & 0 \\ 0 & \frac {1}{\sqrt {\lambda_2}} \end{bmatrix} \begin{bmatrix} u \lambda_1 & 0 \\ 0 & -u \lambda_2 \end{bmatrix} \begin{bmatrix} \frac {1}{\sqrt {\lambda_1}} & 0 \\ 0 & \frac {1}{\sqrt {\lambda_2}} \end{bmatrix} \begin{bmatrix} \partial_{\xi} \\ \partial_{\eta} \end{bmatrix}=g$$

So the canonical form for this hyperbolic PDE is

$$u_{\xi \xi}-u_{\eta \eta}=g$$

Egm6322.s09.three.liu 16:21, 24 April 2009 (UTC)

HW:Getting the Canonical Form for Parabola PDEs Now we have $$\lambda_1 u_{xx} + e u_y =h$$ (1)

and we want $$u_{\xi \xi} - u_{\eta} =g$$ (2)

Rewriting (1) in matrix form, we have:

$$\left \lfloor \partial {x} \ \partial {y} \right \rfloor$$ $$\begin{bmatrix} \lambda_1     &  0      \\ 0             & 0 \end{bmatrix} $$ $$\begin{bmatrix} \partial_x u           \\ \partial_y u            \end{bmatrix} $$ $$+ \left \lfloor 0 \ e \right \rfloor$$ $$\begin{bmatrix} \partial_x u           \\ \partial_y u            \end{bmatrix}=g $$(3)

We have already derived that,

$$\begin{bmatrix} \partial_x            \\ \partial_y \end{bmatrix} $$ = $$J_{\beta} \begin{bmatrix} \partial_{\xi}            \\ \partial_{\eta} \end{bmatrix} $$ (4)

$$J_{\beta}= \begin{bmatrix} 1/ \sqrt{\lambda_1}     &  0      \\ 0             & 0 \end{bmatrix} $$(5)

Plugging (4) and (5) into (3), we obtain:

$$\left \lfloor \partial {\xi} \ \partial {\eta} \right \rfloor J_{\beta}^T$$ $$\begin{bmatrix} \lambda_1     &  0      \\ 0             & 0 \end{bmatrix} J_{\beta}$$ $$\begin{bmatrix} \partial_{\xi} u           \\ \partial_{\eta} u            \end{bmatrix} $$ $$+ \left \lfloor 0 \ e \right \rfloor J_{\beta}$$ $$\begin{bmatrix} \partial_{\xi} u           \\ \partial_{\eta} u            \end{bmatrix}=g $$

$$\left \lfloor \partial {\xi} \ \partial {\eta} \right \rfloor \begin{bmatrix} 1/ \sqrt{\lambda_1}     &  0      \\ 0             & 0 \end{bmatrix}$$ $$\begin{bmatrix} \lambda_1     &  0      \\ 0             & 0 \end{bmatrix} \begin{bmatrix} 1/ \sqrt{\lambda_1}     &  0      \\ 0             & 0 \end{bmatrix}$$ $$\begin{bmatrix} \partial_{\xi} u           \\ \partial_{\eta} u            \end{bmatrix} $$ $$+ \left \lfloor 0 \ e \right \rfloor J_{\beta}$$ $$\begin{bmatrix} \partial_{\xi} u           \\ \partial_{\eta} u            \end{bmatrix}=g $$

The above equation is:

$$u_{\xi \xi} - u_{\eta} =g$$ , which is exactly we want.

--EGM6322.S09.TIAN 21:40, 23 April 2009 (UTC)

Homework: General classification of 2nd order linear PDEs to 2nd order nonlinear PDEs The general form of a linear second order PDE in the matrix form is give as

$$\left[{\begin{matrix} \frac{\partial }{\partial x} &  \frac{\partial }{\partial x}  \\ \end{matrix}}\right]\left[{\begin{matrix} A& B \\ B & C \\ \end{matrix}}\right]\left[{\begin{matrix}

\frac{\partial u }{\partial x}   \\

\frac{\partial u}{\partial y}    \\ \end{matrix}}\right] +\left[{\begin{matrix} D &  E  \\ \end{matrix}}\right]\left[{\begin{matrix}

\frac{\partial u }{\partial x}   \\

\frac{\partial u}{\partial y}    \\ \end{matrix}}\right]+F(x,y)u+G(x,y)=0 $$

and for a second order non linear PDE the general equation in two independent variables is given as

$$F\left ( u,x,y,\frac{\partial u}{\partial x} ,\frac{\partial u}{\partial y},\frac{\partial^2 u}{\partial x^2},\frac{\partial^2 u}{\partial x\partial y},\frac{\partial^2 u}{\partial y^2}\right )=0$$

Let, $$p=\frac{\partial^2 u}{\partial x^2}$$; $$q=\frac{\partial^2 u}{\partial x\partial y}$$; $$r=\frac{\partial^2 u}{\partial y^2}$$

and denote

$$a=\frac{\partial F}{\partial p}$$

$$b=\frac{1}{2}\frac{\partial F}{\partial q}$$

$$c=\frac{\partial F}{\partial r}$$

Let the specific solution $$u=u(x,y)$$ be considered at any point (x,y)and the values of a,b,c be calculated and then let the determinant $$\delta $$ be calculated using $$ac-b^2$$

if

$$\delta < 0$$ then hyperbolic

$$\delta > 0$$ then elliptic

$$\delta = 0$$ then parabolic

In a nonlinear PDE the coefficients are not only dependent on the point$$(x,y)$$but also on the selection of the solution.Therefore it is not possible to know the nature of the equation without knowing the specific solution.The type of the equation will be changing with the specific solution at the same point (x,y)

Reference from Andreĭ Dmitrievich Poli͡a︡nin, Aleksandr Vladimirovich Manzhirov, 2006 p.653,

Egm6322.s09.bit.gk 19:43, 24 April 2009 (UTC)

 The above collapsible box described the linearization of a nonlinear PDE with respect to the second order derivatives to do the classification. On the other hand, note also a PDE like $$u_{\xi \xi }+u_{\eta \eta }=g$$ is also a nonlinear PDE if $$\displaystyle g(\xi, \eta, u, u_\xi, u_\eta)$$ is nonlinear with respect to its arguments. Egm6322.s09 10:09, 3 May 2009 (UTC)

Homework: Verify (2) in Solution of (1). Derivation of d'Alembert solution for 1-D wave equation Given that

$$U_{\xi\eta}=0$$

and the solution is obtained by

$$U\left ( \xi,\eta \right )=F\left ( \xi \right )+G\left ( \eta \right )$$

To verify this ,we have to differentiate twice,Therefore

$$\frac{\partial u}{\partial \xi}={F}'$$

$$\Rightarrow \frac{\partial^2 u}{\partial \xi\partial \eta}=0$$

$$\Rightarrow U_{\xi\eta}=0$$

Hence the solution.

Egm6322.s09.bit.gk 20:11, 24 April 2009 (UTC)

Homework: Finding $$\bar{a}$$, $$\bar{b}$$, $$\bar{c}$$

Egm6322.s09.bit.sahin 16:52, 24 April 2009 (UTC)

--Egm6322.s09.lapetina 18:01, 24 April 2009 (UTC)

$$\mathbf{J}=\begin{bmatrix} \phi _{x} &\phi _{y} \\ \psi _{x}&\psi_{y} \end{bmatrix}$$, $$\mathbf{J^{T}}=\begin{bmatrix} \phi _{x} &\psi _{x} \\ \phi _{y}&\psi_{y} \end{bmatrix}$$ and $$\mathbf{A}=\begin{bmatrix} a &b \\ b &c \end{bmatrix}$$

$$\mathbf{\bar{A}}=\mathbf{JAJ^T}=\begin{bmatrix} \bar{a} &\bar{b} \\ \bar{b} &\bar{c} \end{bmatrix} $$

Substituting matrices into the expression

$$\mathbf{\bar{A}}=\begin{bmatrix} \phi _{x} &\phi _{y} \\ \psi _{x}&\psi_{y} \end{bmatrix} \begin{bmatrix} a &b \\ b &c \end{bmatrix} \begin{bmatrix} \phi _{x} &\psi _{x} \\ \phi _{y}&\psi_{y} \end{bmatrix}$$

$$\mathbf{\bar{A}}=\begin{bmatrix} a\left (\phi _{x} \right )^{2}+2b\phi _{x}\phi _{y}+c\left (\phi _{y}  \right )^{2} & a\phi _{x}\psi _{x}+b\phi _{x}\psi _{y}+b\psi _{x}\phi _{y}+c\phi _{y}\psi _{y} \\ a\phi _{x}\psi _{x}+b\phi _{x}\psi _{y}+b\psi _{x}\phi _{y}+c\phi _{y}\psi _{y} & a\left ( \psi _{x} \right )^{2}+2b\psi _{x}\psi _{y}+c\left ( \psi _{y} \right )^{2} \end{bmatrix}$$

Therefore,

$$\bar{a}=a\left (\phi _{x} \right )^{2}+2b\phi _{x}\phi _{y}+c\left (\phi _{y}  \right )^{2}$$

$$\bar{b}=a\phi _{x}\psi _{x}+b\phi _{x}\psi _{y}+b\psi _{x}\phi _{y}+c\phi _{y}\psi _{y}$$

$$\bar{c}=a\left ( \psi _{x} \right )^{2}+2b\psi _{x}\psi _{y}+c\left ( \psi _{y} \right )^{2}$$

The sign of $$det{\underline A}$$ would not change under only change of coordinates, including nonlinear change of coordinates.

Therefore, the classification of 2nd order PDEs includes the canonical forms and the invariance of $$det{\underline A}$$.

=Summary : Differences Between conics and PDE's= Egm6322.s09.bit.la 17:31, 24 April 2009 (UTC)

Transformation of Coordinates

$$\begin{Bmatrix}\bar{x}\\ \bar{y}\end{Bmatrix} = \mathbf{J} \begin{Bmatrix}x\\ y\end{Bmatrix},\begin{Bmatrix}x\\ y\end{Bmatrix}=J^{-1} \begin{Bmatrix}\bar{x}\\ \bar{y}\end{Bmatrix}$$

Conics : $$\mathbf{\bar{A}}=\mathbf{J}^{-T}\mathbf{A}\mathbf{J}^{-1}$$

PDE's : $$\mathbf{\bar{A}}=\mathbf{J}\mathbf{A}\mathbf{J}^{T}$$

Invariance:

For Conics : $$det\mathbf{\bar{A}}=(det\mathbf{A})(\frac{1}{det\mathbf{J}})^2$$

For PDE's: $$det\mathbf{\bar{A}}=(det\mathbf{A})(det\mathbf{J})^2$$

Conics - $$\mathbf{A}=V\mathbf{\Lambda}V^T $$ Eigenvalue problem : Rotation

And the transformation of coordinates in this case is:

$$\begin{Bmatrix}\bar{x}\\ \bar{y}\end{Bmatrix} = \mathbf{V}^T$$

where:

$$=\mathbf{J_\alpha }\mathbf{V}^{-1}=\mathbf{V}^T$$

$$\mathbf{\bar{A}}=\mathbf{\Lambda }=\mathbf{V^T}A\mathbf{V}$$

where:

$$\mathbf{V}=\mathbf{J_\alpha }^T=\mathbf{J_\alpha }^{-1}$$

And Translating $$\mathbf{J_\alpha }^{-T}=\mathbf{J_\alpha }$$

$$\bar{\bar{x}}=\bar{x}-r$$

$$\bar{\bar{y}}=\bar{y}-s$$

After that Scaling:

$$\xi =\bar{\bar{x}}/dx$$

$$\eta=\bar{\bar{y}}/dy$$

where:

$$dx=\sqrt{\left |\lambda _1 \right |}$$

$$dy=\sqrt{\left |\lambda _2 \right |}$$

Applying the Form 1 - Assuming hyperbola

$$\xi^2-\eta ^2=\pm 1$$

First Step:

$$\begin{Bmatrix}\bar{\xi} \\ \bar{\eta}\end{Bmatrix}=\mathbf{J}_\gamma \begin{Bmatrix}\xi \\ \eta\end{Bmatrix}$$

$$\begin{Bmatrix}\xi \\ \eta\end{Bmatrix}=\mathbf{J}_\gamma^{-1} \begin{Bmatrix}\bar{\xi} \\ \bar{\eta}\end{Bmatrix}$$

Second Step:

$$\mathbf{\bar{B}}=\mathbf{J}_\gamma^{-T}\mathbf{B}\mathbf{J}_\gamma^{-1}$$

where: $$\mathbf{B}=\begin{bmatrix} 1 &0 \\ 0 &-1 \end{bmatrix}$$

This is a rotation in hyperbola cases

Going to Form 2 (Canonical Form):

$$\pm \bar{\xi} \bar{\eta}=1$$

Scaling again as the same processes described above:

$$\begin{Bmatrix}\xi \\ \eta\end{Bmatrix}=\begin{bmatrix}1/dx &0 \\ 0 &1/dy \end{bmatrix}\begin{Bmatrix}\bar{\bar{x}} \\ \bar{\bar{y}}\end{Bmatrix}$$

For the 2nd order linear PDE's:

General Transformation of Coordinates:

$$\mathbf{\bar{A}}=\mathbf{J}_\alpha\mathbf{A}\mathbf{J}_\alpha^{T}$$

$$\begin{Bmatrix}\partial x \\ \partial y\end{Bmatrix}=\mathbf{J}_\alpha^{T} \begin{Bmatrix}\bar{\partial x} \\ \bar{\partial y}\end{Bmatrix}$$

Particularize to Eigenvalue problem:

$$\mathbf{\bar{A}}=\mathbf{\Lambda }=\mathbf{V^T}A\mathbf{V}$$

Again where: $$\mathbf{V}=\mathbf{J_\alpha }^{T}$$

$$\lambda _1u_{\bar{x}\bar{x}}+\lambda _2u_{\bar{y}\bar{y}}=h(\bar{x},\bar{y},u,u_{\bar{x}},u_{\bar{y}})$$

where :

$$\lambda _1=\lambda _x;\lambda _2=\lambda _y$$

Assuming a hyperbolic PDE's:

$$\lambda _1>0;\lambda _2<0$$

$$\begin{Bmatrix}\xi \\ \eta\end{Bmatrix}=\begin{bmatrix}1/\sqrt{\left |\lambda _1 \right |} &0 \\ 0 &1/\sqrt{\left |\lambda _2  \right |} \end{bmatrix}\begin{Bmatrix}\bar{x} \\ \bar{y}\end{Bmatrix}$$

Where:

$$\mathbf{J}_\beta =\begin{bmatrix}1/\sqrt{\left |\lambda _1 \right |} &0 \\ 0 &1/\sqrt{\left |\lambda _2  \right |} \end{bmatrix}$$

And Scaling:

$$u_{\xi \xi }-u_{\eta \eta}=h$$ Form 1

Find: $$\begin{Bmatrix}\bar{\xi} \\ \bar{\eta}\end{Bmatrix}=\mathbf{J_\gamma }\begin{Bmatrix}\xi \\ \eta\end{Bmatrix}$$

This transformation was inspired from the conics form

Continuing:

$$\mathbf{\bar{B}}=\mathbf{J}_\gamma\mathbf{B}\mathbf{J}_\gamma^{T}$$

and

$$u_{\bar{\xi}\bar{\eta}}=h $$

 Hyperbolic PDEs

Egm6322.s09.bit.sahin 17:38, 24 April 2009 (UTC)

Method of characteristics applied in two different places

1) Finding canonical forms (Form 1 by hyperbolic PDEs) by setting

$$\bar{a}=\bar{c}=0$$

2) Solving Form 1 of hyperbolic PDEs

Setting $$\bar{a}=\bar{c}=0$$

Remembering that

$$\bar{a}=a\left (\phi _{x} \right )^{2}+b\phi _{x}\phi _{y}+c\left (\phi _{y}  \right )^{2}$$

and assuming that $$\phi _{x}$$ and $$\phi _{y}$$ are proportional

$$\phi _{x}=k\phi _{y}$$

substituting second equation to the first one, we have

$$\bar{a}=\left ( \phi _{y} \right )^{2}\left ( ak^{2}+bk+c \right )=0$$

the solution of $$\left ( ak^{2}+bk+c \right )=0$$ is

$$k=\frac{-b\pm \sqrt{\Delta }}{2a}$$, $$\Delta =b^{2}-4ac$$

so

$$k_{+}:=\frac{-b+ \sqrt{\Delta }}{2a}$$ and $$k_{-}:=\frac{-b- \sqrt{\Delta }}{2a}$$

it can be written that

$$d\phi =\phi _{x}dx+\phi _{y}dy$$ (2)

from the proportionality $$\phi _{x}=k\phi _{y}$$

$$0=\phi _{x}-k\phi _{y}$$ (3)

comparing Eq(2) and Eq(3) gives

$$\frac{d\phi }{0}=\frac{dx}{1}=\frac{dy}{-k}$$

Egm6322.s09.bit.gk 19:17, 24 April 2009 (UTC)

$$\frac{\mathrm{d} y}{\mathrm{d} x}=-k_{\pm }$$

Similarly,assume

$$\psi_x=k\psi_y$$

$$ak^2+bk+c=0$$

From $$\phi_x=k\phi_y$$  and  $$\psi_x=k\psi_y$$

We have

$$k=\frac{\phi_x}{\phi_y}$$

$$k=\frac{\psi_x}{\psi_y}$$

1-D Wave equation,we have ,

$$\hat{c}^2u_{xx}-u_{tt}=0$$ by looking at the units we have

$$\left [\hat{c}u_{xx} \right ]=\left [u_{tt}  \right ]=\left [\frac{L}{T^2}  \right ]$$

Therefore,

$$a=c^2,b=0,c=-1$$

$$\Rightarrow \xi =x-\hat{c}t$$ and $$\eta =\xi =x+\hat{c}t$$

By Scaling Method

$$\hat{c}^2u_{xx}-u_{tt}=0$$ $$\Rightarrow u_{xx}-u_{tt}=0$$

Coordinate Transformation

$$\begin{Bmatrix} \bar x\\ \bar y=t

\end{Bmatrix}\to \begin{Bmatrix} \xi \\ \eta

\end{Bmatrix}$$

$$\Rightarrow u_{\xi \eta }=0$$

 The derivation of d'Alembert solution from Form 2 of the hyperbolic canonical form is missing. Egm6322.s09 10:09, 3 May 2009 (UTC)

= References =

=Signatures= --Egm6322.s09.xyz 21:38, 23 April 2009 (UTC)

--EGM6322.S09.TIAN 21:40, 23 April 2009 (UTC)

Egm6322.s09.Three.nav 03:44, 24 April 2009 (UTC)

--Egm6322.s09.lapetina 14:44, 24 April 2009 (UTC)

Egm6322.s09.Three.ge 15:26, 24 April 2009 (UTC)

Egm6322.s09.three.liu 16:21, 24 April 2009 (UTC)

Egm6322.s09.bit.la 17:31, 24 April 2009 (UTC)

Egm6322.s09.bit.sahin 17:39, 24 April 2009 (UTC)

Egm6322.s09.bit.gk 20:14, 24 April 2009 (UTC)

title: User:Egm6322.s09.mafia/HW7


text:
Binary Stars and Extrasolar Planets

This learning activity utilizes text, imagery, and applet-simulations to introduce the concepts associated with Binary Star systems and the search for Extrasolar Planets (exoplanets for short). This is a rapidly developing field within Astronomy due to new technology allowing scientists to either directly image or better infer the presence of exosolar planets via gravitational pull, detection of change in visual magnitude, and other methods.

The activity is separated into three parts to contour the experience into basic, advanced, and mathematical conceptual understanding. The basic level will introduce the general ideas of what is occurring. The advanced level will further the conceptual experience to fully understanding the concepts necessary to apply mathematical analysis upon either a binary star system or exoplanet. The mathematical analysis will introduce Astrophysics equations in order to give a taste of how scientists analyze the data they collect to aid in the discovery of exoplanets. Lastly, if you still seek more there is a way that you too can aid in the search for exoplanets without the need for a degree in the field or a large telescope!

When you have completed this activity you should be able to; by level: Basic: Know terminology and have background-level knowledge of binary systems and exoplanets. Advanced: Know and understand select techniques pertaining to binary systems and how they can be applied to the search for exoplanets. Mathematical: Be able to use data to get practical information about either binary stars or exoplanets.

Basic Concepts:
This section looks into the types of binary stars, the light curve, center of mass, and a simple applet to understand how changing mass and distance causes changes in the orbits of binaries. Types of Binary Stars:
 * 1) Optical Double: This is actually better used to actually define what constitutes a binary star. This is not a binary star system and is actually just stars that appear close to each other based upon our vantage point and can be, often are, very far apart.  Thus, the definition of a binary star requires that the stars are gravitationally bound like the Earth and the other planets in our Solar System are to the Sun.
 * 2) Visual Binary: This describes two gravitationally bound stars that are one of or a combination of the following: bright enough, far enough apart, and/or near enough to be seen separately by high-powered telescopes. Albireo, Mira, and Sirius are three examples of visual binaries and have images displaying both stars.  Note that more stars can also be present, Polaris (the North Star) is actually a ternary system (three stars) with visual verification.
 * 3) Astrometric Binary: Only one star is visible through current telescopes, but the movement of the star from the gravitational pull of the other star indicates the presence of its unseen companion star. It is a system in which a visible star and a dimmer companion orbit a common centre of mass and detection of such binary by astrmetric means are called astrometric binary.
 * 4) Eclipsing Binary: Eclipsing binary stars means that the star system is oriented, from our vantage point, in such a way that one star passes in front of the other and then later passes behind the other. This is most notably recognized by a reduction in light due to the one passing in front of the other blocking some or all of the light from the one behind while when both remain visible they both show their entire light output. Algol, β Perseus, also known as the Demon star are the first eclipsing binary.
 * 5) Spectroscopic Binary: A spectrum of light at rest produces wavelengths that remain at the same wavelength under all stationary conditions. However, when moving towards or away from the observer this spectrum shifts.  This method uses the spectra received from stars to note shifts in the position of the bands.  We are then able to know when a star is moving away (shifts the spectrum towards the red end) or towards us (towards blue end).  This is aptly termed red shift and blue shift.

All of the above, with the exception of an optical double, can also be applied for exoplanet discovery although the size, mass, and light emission for exoplanets make it considerably more difficult. Optical doubles are impossible for exoplanets since the overwhelming majority of their light is reflected from the star they orbit.

The Light Curve
As shown, the light curve over the period (the length of the line) of orbit has two drops in luminosity. This would be the data generated by an eclipsing binary star system.

The first drop is far greater, indicating that it is the passing of the colder, less luminous star in front of the hotter one. This means that, for every unit area, it is in effect blocking more light. It does not matter which star, colder or hotter, is larger.

The second drop represents the hotter star passing in front of the cooler one. It is less because the light being blocked is that of the less luminous star which for every unit area sends less light towards the observer than the hot one.

The duration of the drops should be approximately the same (not perfectly reflected in this image) as the smaller star disappears at the same rate behind as it blocks the light in front. Also, the duration reflects the time spent behind or in front of the other star. The diagonal slopes in and out represent the partial concealment of the star being progressive depending on the duration it takes to fully block the other star.

Again, this can be applied to exoplanets, albeit far more difficult. Although the lack of light production by a planet assists by decreasing the luminosity to nearly nothing for the gap it makes, the gap of light it actually makes is so much smaller due to tiny radius relative to that of a star that it is nearly unnoticeable to even many modern telescopes.

Center of Mass
Center of mass is a point at which the combined mass of the two (or more) bodies involved in the rotation act as if they were concentrated at this single point. This point lies between the masses involved, and is closer to the larger masses then the smaller masses. If the system of rotating masses has a transverse velocity the motion can be represented by the motion of the center of mass with this same velocity. This can be related to the solar system in the sense that the Sun is (essentially, it too rotates some, in actuality, from the pull of the planets) the center of mass by which the planets orbit. However, when two bodies approach nearer masses this point is drawn out of being located within the heavier body and actually lies at a point in space directly between the two bodies. It remains equidistance from both stars (in the case of exactly equal mass) as the orbit in their elliptical orbits about it. The figure shows a center of mass located within the star, but note that they always remain on opposite sides of the center. More bodies makes the situation far more complex, but ultimately it is the same idea that at any given time the positional motion of all the bodies keeps the central rotation about the center of mass.

Application of Basic Ideas
We now turn to the Applet to gain an active appreciation for the above concepts. To keep things simple, the Applet for this section has limited options. Once you open it, you can see the white dot as the star (the Sun for most of the options) and the blue dot as the planet. The Applet also greatly exaggerates the movement about the center of mass to exemplify the effect of gravitational binding between the two objects making them both move. One must be aware of this as the 10 Jupiter setting demonstrates the movement of two near-equal mass bodies whilst in reality ten times the mass of Jupiter is still a very miniscule mass compared to the sun (a mere 0.95%) and would not send the Sun on a crash-course through the solar system as this Applet shows.

Now, let’s run some tests with the Applet (Open in new window if it fails to run in a new tab) and see if you can answer the questions posed correctly.

 * 1) Run the simulation for a while with the standard set up of Sun/Jupiter. Then switch to the Sun/Earth.  What do you notice about the center of mass?  Set to Sun/Jupiter again and observe then switch to two Jupiters, then five Jupiters.  What changes happen with each change in the mass ratio?
 * 2) Let us suppose this is a visual binary system and two stars orbiting instead of a star and a planet. What would we be able to determine about the masses based on watching the rotation?  If one or the other was not visible, would we still know it was a binary system?  What type of binary system would it be?
 * 3) Suppose we were looking at this as a spectroscopic binary. Would we be able to determine anything based on spectra obtained from repeatedly viewing this binary system?
 * 4) Let us suppose it’s an eclipsing binary. If our vantage point was from the bottom of the screen, at what position(s) in the orbit would we see a dip in light?  What about if our vantage point was from the right?

Answers:
 * 1) The center of the mass is inside the Sun when it is using the Earth because the difference of mass is so great, but with Jupiter it was not (remember that in reality it is always within the Sun though, regardless of the planets’ alignment). As the mass ratio increases the simulation leaves the center of mass at the center of rotation and Jupiter remains at the same location, but the Sun moves further away from the center of mass to maintain the center of mass at a proper distance relative to the mass ratio.
 * 2) We would know that the one moving less is more massive. It would be a spectroscopic binary system because we would only see one star wobbling back and forth in space.
 * No, we would not be able to determine anything because it has no motion towards or away from us and just maintains a flattened appearance. Note, however, that this situation of perfectly perpendicular is very unlikely in practicality and some shift would likely be able to be obtained if the motion was fast enough.
 * 1) At the bottom and the top when the two bodies are aligned from our vantage point. When we were observing from the right, it would be when the stars were on the right and left side of the orbit.  Note that, matter where we placed the vantage point, there would always be two spots in which the light would be reduced and both would be when they came into alignment from our vantage point.

That concludes the basics of binary stars and exoplanets. We now move on to flesh out more advanced concepts, some of which were alluded to here.

Advanced Concepts
This section looks into the more advanced concepts of: Kepler’s Laws of Planetary Motion, Newton’s version of Kepler’s Third Law, Orientation to Earth, Doppler Shift, and Proportionality.

Kepler’s Laws of Planetary Motion

Kepler’s Laws were created to explain the motion of the planets in the Solar System. They are based upon Tycho Brahe’s very accurate measuring of the heavens over many years. They center on the principle of rejecting the geocentric model in favor of the heliocentric model as was necessary to match the data without using epicycles to explain the motion.

They are as follows:
 * 1) "The orbit of every planet is an ellipse with the sun at a focus."
 * 2) "A line joining a planet and the sun sweeps out equal areas during equal intervals of time."
 * 3) "The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit."

The first law works on the principle of center of mass. Kepler determined the orbits were elliptical based upon Tycho’s measurements not fitting the theory using perfect circles which is logical to abide by today when dealing with binaries and exoplanets.

The second law is also useful by relating the speed increase when the objects interacting approach each other and slow when departing. Velocity is a useful tool towards determining other information about binary stars and stars with exoplanets.

The third law was the first true astrophysical equation. Although it only applies to objects orbiting the Sun (or other approximately equal mass stars) in its current form it is still useful and becomes greatly more useful when later manipulated by Newton. Kepler’s third law has become hugely helpful in determining the masses present in binary stars or exoplanets as will be used in the mathematical concepts portion.

The third law in proportional form:

$$ {P^2} \propto {a^3} $$

 * P in Years
 * a in AU

Newton’s Law of Gravitation Applied to Kepler’s Third Law
Newton’s Law of Gravitation dictates that all objects in the universe are gravitationally bound to each other. This is drawn into Kepler’s Laws by the planet exerting a force on the star as well as the star on the planet and also separated the masses so one can apply the property to any objects that are gravitationally bound in a meaningful way (not so distant that the pull has no impact).

Newton’s revised law:

$$\left({\frac{P}{2\pi}}\right)^2 = {a^3 \over G (M+m)}$$

 * P in Seconds
 * a in Meters
 * G is the Gravitational Constant: 6.673x10-11
 * M and m in kilograms

This law is commonly used to determine the total mass of visual binaries that then allows extrapolation to large amounts of other data.

Orientation to Earth
The orientation to Earth is often known as inclination. The vast majority of stars provide an orientation of their satellites that is not eclipsing over the center of the star or perfectly upon the celestial sphere. It is for this reason that when we often are only able to extrapolate a minimum mass when viewing a star’s wobble because we do not know the inclination and, thus, are only able to detect the portion pulling the star on the plane of celestial sphere.

Doppler Shift

Doppler Shift is the basis for a Spectroscopic Binary system. It is found by either two separate shifts in spectra or a single shift generated by an unseen companion on the primary star. It is important because the shifts can be used to find the radial velocity of both stars or the visible one if only one spectrum is observed. The equation to determine radial velocity is:

$${\frac{\Delta \lambda}{\lambda_0}} = {\frac{v_r}{c}}$$

c is the speed of light in a vacuum (3x108 m/s)

λ0 is the rest wavelength of the spectra

Δλ is the change from the rest wavelength to the measured wavelength

vr is the radial velocity in m/s

If the period is known this can be paired with it to determine the semi-major axis.

Proportionality
Since all motion involving two objects revolves around a give-and-take relationship there arises the intrinsic relationship between many physical aspects of the two and their behavior with respect to each other.

All of the following relations can be derived from Kepler’s Laws and Doppler Shift and associated mathematical principles.

$${\frac{m_1}{m_2}} = {\frac{r_2}{r_1}} = {\frac{a_2}{a_1}} = {\frac{\alpha_2}{\alpha_1}} = {\frac{v_2}{v_1}} = {\frac{v_2r}{v_1r}} = {\frac{\Delta \lambda_2}{\Delta \lambda_1}}$$

Note: Units do not matter as they are ratios and the units cancel.

m is mass

r is the separation distance

a is the length of the semi-major axis

α is the angular separation

v is the velocity

vr is the radial velocity

Δλ is the change in wavelength due to Doppler Shift

Application of Advanced Ideas
This application section will use a more technical Applet that allows for more intimate manipulation of the model to better experiment with some of the ideas in this section.

 * 1) Leaving the model on the default settings, study the layout with radial velocity, the visible light spectrum, the earth view, and the privileged view.  Observe the radial velocity and spectrum.  How do they behave?  What does the negative velocity indicate?  Let the left (red end) rest wavelength be 650 nm.  Calculate the shift in wavelength when the red and blue paths cross and when red peaks in the positive.
 * 2) Experiment with the model.  First, adjust the values so that the privileged view is the same as the earth view.  What impact does this have on the Doppler Shift?  Second, adjust it to attain an eclipsing system.  Lastly, make the following changes: a = .8, e = .8, i = 30, w = -45.  Note the radial velocity curve now.  Explain this velocity curve and note the difficulty of being able to understand it if one of the curves did not exist.
 * 3) Change the three solar mass star to be .0009535, Jupiter’s mass in terms of solar mass.  Note the Doppler Shift for the star now.  Explain what two changes can be made to the variables (not changing mass) that would aid in discovering this planet.  One will be reflected in the model while the other will not readily do so.
 * 4) Formulate a way to prove the concept of center of mass will lie equidistant from both bodies under ideal conditions and test it with the model.
 * 5) Using Jupiter’s mass of 1.8986x1027 kg, the Sun's mass of 1.9891x1030 kg and average distance between them of 7.786x1011  m determine the period of Jupiter in seconds.  Verify this value by using the simplifed version (5.2 AU, ~πx107 s in a year).

Answers:

 * 1) When the velocity is going into the positive the spectrum is revealing a redshift (moving away) and when it is negative it is showing a blueshift (towards).  The negative velocity reflects the movement of the star towards the observer; this is Astronomer's customary view of the motion.  When the paths cross the change is zero because that is the point at which radial velocity from the observer's point of view is zero.  When the red is peaking in the positive (about 27 km/s) the change in wavelength is calculated to be 0.0585 nm. Work shown below.
 * 2) Making the inclination 0 makes them match.  This makes the Doppler Shift nothing because they are moving perfectly in the plane of the celestial sphere.  Eclipsing is attained by making the inclination 90 degrees.  The radial velocity curve gets rather difficult to read.  It is important to note that the intersections are still zero so we can be certain of that.  Further, the spikes in speed are when they are the closest as reflected in Kepler's Laws dictating equal area in equal time.  With only one curve it's visibly quite difficult to infer the other and we are unable to be absolutely certain of values generated about such a binary star.
 * 3) One change would be to decrease a and thereby bring the period down to a mere 0.03 years which would make observing the star for a couple days reveal an entire period worth of data so that an observer could recognize the fast, but small, wobble of the star.  The other change would be to turn the inclination to 90 degrees so that it would become eclipsing and a dip in light could be spotted from the planet transiting the star.
 * 4) Set both masses to be the same.  They will then follow the exact same path around each other.
 * 5) Answer worked out below and yes they do match up fairly closely.

Doppler Shift worked out:

$${\frac{\Delta \lambda}{\lambda_0}} = {\frac{v_r}{c}}$$

can be rewritten

$$\Delta \lambda = {\frac{\lambda_0v_r}{c}}$$

which equals

$$\Delta \lambda = {\frac{650nm*27,000m/s}{3x10^8 m/s}}$$

and yields

$$\Delta \lambda = 0.0585$$nm.

Kepler's Third Law worked out:

$$\left({\frac{P}{2\pi}}\right)^2 = {a^3 \over G (M+m)}$$

can be rewritten

$$P^2 = {4\pi^2a^3 \over G (M+m)}$$

which equals

$$P^2 = {4\pi^2(7.786x10^{11}m)^3 \over 6.673x10^{-11}Nm^2kg^{-2} (1.9891x10^{30}kg+1.8986x10^{27}kg)}$$

and yields

$$P^2 = 1.403x10^{17}$$s2 → $$P = 3.745x10^8$$s

The simplified version:

$$P^2 =a^3$$

which equals

$$P^2 =5.2^3$$

and yields

$$P^2 =140.6$$yr2 → $$P =11.86$$yr

which simplifies to

$$P = 11.86$$yr$$*\pi x10^7$$syr-1 $$= 3.725x10^8$$s.

This concludes the advanced concepts that can be associated with binary stars and exoplanets. To fully explore the nature of these entities continue onto the mathematical concepts that will utilize real stellar data to extrapolate more data about binary stars.

Mathematical Concepts:
To Be Added

Further Exploration
If you are interested in doing further exploration of exoplanets and helping the scientific community in the process you can go to visit Systemic. The Systemic site has a console to download which contains radial velocity data for stars which are suspect or known to be supporting exoplanets. You are able, with the console, to match a curve to the data by adding planets and controlling the planetary mass and distance until it is a good fit for the data accumulated so far. Upon registering, you are able to submit the curve where the system will compare it to other curves for the same data set generated by other members and, ultimately, as more data becomes available and submissions narrow down possible candidates to the stellar movement your curve fitting may contribute to the verification of the existence of one or more exoplanets orbiting a distant star. For further explanation of the software, check out their tutorials.

title: Binary Stars and Extrasolar Planets


text:
User talk:Twilane

 Hello Twilane, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Victuallers 09:35, 17 April 2009 (UTC) Victuallers 09:35, 17 April 2009 (UTC)

title: User talk:Twilane


text:
Appropriate technology designs

The following is a list of appropriate technology concepts - these are highly speculative and not actually "appropriate technology" in any commonly agreed sense.

The list only includes Completely emissionless technologies, and the technologies are especially designed to be used in villages/cities in the developed world. The designs were proposed to be placed in a global AT database.

Water supply
I suggest adding a variant of the picture Water system with cistern, where the pressure vessel is removed and the cistern itself works as a pressure vessel. This would allow the system to be build DIY (rather than needing to buy a pressure vessel or weld your own pressure vessel casing. Please include in article. Perhaps diy water towers can also be included, as they allow multiple users to take advantage of a single system (reduces cost). However, the downside would be that if the tower fails, part of or the entire village would be affected. If this happens (especially in remote arid/desert-like locations, repairs will go slow aswell (water being important to work in order to make repairs) and replacement parts may not always be available. Also, another thing is dislike in water towers is that they are outdated and somewhat inefficient (a pump is required too to pump the water up, its only after the elevation that it can works without electricity or extra pumps). 81.246.161.76 (talk) 12:21, 23 January 2009 (UTC)

Also, the waterpumping windmill described in NGC is a useful piece of material. A picture is available at http://en.wikipedia.org/wiki/File:Low_cost_wind_pump.JPG.

Water tower conversions
Water towers in the developing world may be used as a energy storage device. This can be done by adding a hydropower generator on the tube. This allows the height difference to be used as a energy source. Conversions of regular towers are theoretically also possible

Ecologic toothbrush
In order to eliminate the need of toothpaste (which is costly in the long run, and environmentally polluting; if the commercial toothpastes are used, and even if tooth powder is made at the home using baking soda and glycerin), a solution would be to design a toothbrush that keeps the middle between a "regular" commercial toothbrush and a Teeth cleaning twig. This would be done by making the handle out of a durable wood (or alternatively metal) and adding a replacable brush made from a biocidal wood species (eg Salvadora persica, Gouania lupuloides). The brush could perhaps be made with cooperation from Radius toothbrush, Recycline, Cebra Online, Gaiam, Smile Brite or Bogobrush

Telepresence robot with Multimachine
Perhaps a telepresence robot (eg Sparky) fitted with the multimachine (see Open-source robotics) could be useful for closing the technical inexpertise in workshops of developing countries. —Preceding unsigned comment added by 81.246.183.119 (talk) 13:01, 18 February 2009 (UTC)

International language and electricity system
A simplified english as the only spoken language is an appropriate technology and can be implemented in appropriate technology documents and books. Also, standards and rules to only use one type of mains electricity system (amount of volt & frequency) is best used to allow the building of internationally invariable electric devices and machines. see http://en.wikiversity.org/wiki/Juridical_national_measures_on_climate_change#Optional_digitalisation_of_city_communal_services and Thanks, 81.246.130.16 (talk) 13:29, 22 February 2009 (UTC)
 * Mains_power_systems
 * Mains_electricity
 * International_English
 * Simplified_English
 * Simple_English

 * Simplified English is useful for communications. But "simplified english as the only spoken language" is dystopian - I don't want to live in a world without linguistic diversity, and without complex language. --Chriswaterguy (talk) 00:50, 19 April 2012 (UTC)

Chainless bike with airless tires
Perhaps bikes without chains (eg shaft-driven or hydraulic bicycles) and Airless tires are more appropriate bicycles. Would reduce malfuntionings and more difficult bike repair trainings required today. Perhaps tubeless wheels can be used as a basis in which latex can be poured. —Preceding unsigned comment added by 81.246.163.71 (talk) 10:55, 27 February 2009 (UTC)

Electronics
I was wondering whether anyone knows whether there are ecological and simpler alternatives to PCB boards. I was thinking about Wire_wrap, Stripboard and the like yet am not sure whether environmental and whether any difficult circuits can be made with this. Wafer-level packaging (this is I believe only used with certain components) can also sometimes? be used. Also not sure whether very environmental (presence of glues, and hazardous materials as certain plastics, ...). I also heard about the Tin-Lead ban, see leadfree soldering which has introduced the creation of alternatives (eg SnAgCu, SnCu and Sn alone). I also heard about anisothropically conductive glues (these are not environmental) and isothropically conductive glues (respectively ACA and ICA), glues for flip-chip assembly and the glue on ReUSE, ... yet not sure whether truly ecologic and whether its any durable. Perhaps Sn alone is most appropriate? This will need to be checked and further compared dough. See http://www.europeanleadfree.net, http://www.imec.be/IMECAT/, http://web.archive.org/web/20050520092018/http://www.imec.be/IMECAT/documents/16_2004_IPC_Frankfurt_Vandevelde_paper.pdf A appropriate solder and soldering method should be the following: An alternative is the use of organic materials (which are thus consistent to cradle-to-cradle design). Certain (not all!) smart plastics can be used. Examples are PEDOT:PSS. However, this approach is not yet suited today to make any durable or more advanced circuits. Also, could any other techniques as flow soldering, E-blocks, FPGA, EasyControl I/O, ... be used. E-blocks also uses flowcode as a coding language. Not sure whether this is easy to learn (aldough most coding can be simply downloaded if open-source). In addition, a suggestion on the LED-lamps. These are I believe quite expensive. In order to decrease price, I was wondering whether LED's may be made DIY (they are composed of a die, lens, cooling body and connection wires. Compounds used include aluminum, indium, gallium, phosphor, nitrogen. Especially low-power leds seem easy to build, high power ones may?? be achievable using techniques as Lumiramica (uses phosphor plates instead of powder). An alternative is the procuring of mere high-power LEDs (eg Lumileds Rebel Led, Osram Opto Semiconductors Golden Dragon, Cree X-lamp, ...)and installing them yourself with the open source LEDbus system. This may be cheaper than buying LED lamps with imbedded electronics. —Preceding unsigned comment added by 81.246.131.67 (talk) 09:59, 6 March 2009 (UTC)
 * easy assembly and disassembly of board, components and solder (so best no alloys as this reduces options with melting the parts off)
 * no etching (this would promote loss of soldering material); instead direct application of paste eg by hand soldering as in reflow soldering or automated (with workshop pick&place robots)

Synthetic biology
Synthetic biology could make cheap generators for hydrogen gas and could provide a cheap and easy way to make electronic boards. See Angela Belchers project 81.245.168.62 (talk) 14:27, 13 April 2009 (UTC)

Open-design circuit board oven and solder robot
The Elector SMT reflow oven may be used as is (or after modification) with the Sn solder (or other soldering method) noted above to solder boards ecologically and appropriate.
 * Note that this open-source project demonstrated how easy it is to implement controlling the temperature of ovens. Perhaps that a similar temperature control may be integrated in the recasting of materials. In the document made by Ingénieurs Sans Frontières it was described that casting was done "on sight". To make solder for these sensitive equipments, perhaps controlled temperature casting is better as it may allow a better purity of metal (improving the performance of these sensitive devices)

for the soldering itself, I was thinking about a open-source soldering robot (which can be probably made with a diy embedded (low-cost) microcontroller (to make a embedded computer with output devices (movable soldering syringes. This primarily as I have doubts about the local workers performing this very precise and difficult job. The local workers are probably more suited to gathering the source materials (metals, ...)

Component tester
As described above, with component boards as strip boards, easy replacing/repairs of devices are possible (allowing the repair of broken boards by replacing certain components, rather than trowing away the entire board alltogether). For this, a simple? voltage meter is needed to test the components individually (eg whether they still let the power trough). This open-source elektor meter may be used herefore

Disassembler device
Besides an oven and solder robot, an easy open-source board disassembly device would be appropriate. This would allow boards found on the junkyard (which are thus completely broken; having no more working components) of being effciently disassembled. The solder, board and components can then be easily seperated and reused (solder and metals can be recast, chemical components may be derived from large amounts of diodes, and other known components). The chemical substances and amounts are known to be in each type of component (eg diodes, leds, ...) and may therefore still be recovered if they are all seperatly detached and sorted together.

Open-design washing machine
A open-source washing machine should be created. The washing machine is to have a induction engine for the drum and a brushless DC engine for pumping the machine dry after the wash. It probably best uses (the technology of; buying the modules from the company directly may be expensive) the Motion SPM in SMD or Tuny DIP housing. see Energy Saving through Motor Control and Motor app gains efficiency with electronic control Perhaps water-saving features may be added (see Washing_machine) Cleaning agents used should be eco-friendly; see Natural cleaning product

Use of alternative currency
After some research (see updates at Alternative currency), the use of a specific alternative currency (not devaluating in time; eg no "schwundgeld") could be useful. This currency would be a currency specific for trading between villages (thus not between individuals). It would be a city gift currency; and the working would be as follows: each city is given at the start a same negative amount of credits. If a certain city requests the ordering of eg a set of windturbines, casted metals, tools, ... the product is made and delivered abd at the same time, the giving partner is given a same amount of negative points which they can use to ask another product in return (of a same value of negative points). The negative value of a product depends on the amount of workhours and the type/amount of source material used. The current amount of negative (good) or positive (bad) credits can be shown in the cities townhall (preferably on a electronic billboard together with other information; supplied via RSS) The system would encourage city trade of autonomous energy products, and other products required for ecologic, autonomous functioning of the city. As such, it would increases economy, employment and promote healthy rivalry/competitions (building new systems and trading them, ...). it would increase specialisation of each system, dividing the production of different autonomous systems (eg water harvesting, energy generation systems, ...)? This would decrease the costs as less workshops would be needed for the cities of a certain area (one area should have several cities; each specialising in the essential systems and located at a manageable/transportable distance) —Preceding unsigned comment added by 81.246.146.11 (talk) 15:17, 13 March 2009 (UTC)

Improved solar heater
I just added another design for a low-cost village. See here. This design has some new features including the seperation of rooms and making them communal (reduces cost, decreases cleaning, ...) This design makes use of a solar batch heater for heating the water for the showers. This batch heater would be a heater like, but automatic temperature control (set to eg 40°C) should be added. As such a open-source design for this temperature controller should be made. The automatic control can be created in practice by 2 tanks --> 1 with cold(er) water (eg 20° or so; ambient temperature) and 1 with warm water (this tank can be the batch heater itself). The tank with colder water needs to be shielded from the sun (eg by underground placement, ...). The 2 tanks can be connected to 1 large(r) tank that will be the primary buffer for the shower (here the water needs to be 40°). The valves of the (hot&cold) tanks will be automatically controlled by the temperature controller. The batch heater itself may be fitted with a electric coil for heating when the water cannot be heated enough by the sun; in not so sunny areas the batch heater can be improved by targeting light at it with parabolic dishes. The 40° water generated by the solar/electrical heating system can also be used for the laundry center/washing machines noted in the design. The piping may be merely extended (1 branch offcourse already going to showers; another continuing to washing machines/laundry center) and the 40° water inmediatelly enters the (open-source design; see above) laundry center. Here the water may be used as is, or be heated again in the washing machines (depends on washing program; imbedded machine thermostat will determine this) Note that instead of the batch heater, the Concentrating Solar Power plant itself (noted in the design) can also function as a batch heater. However, the CSP would then require 2 piping circuits (one for heating the shower water and 1 for the energy generation; the latter runs on chemical fluids and not water). Nonetheless the 2 circuits can be integrated into the same tower and make use of the same solar power. —Preceding unsigned comment added by 81.245.166.227 (talk) 07:54, 18 March 2009 (UTC)

Improved UDD toilet
Based on the GTZ UDD toilet (see http://www.susana.org/index.php?option=com_content&view=article&id=152&Itemid=145&lang=en )

Extra's to add: - painting the chamber black, and adding increased ventilation trough natural ventilation techniques (trought the increased warming up of the chamber). Perhaps this may be done trough the pipe (if made of metal, the metal instead can be painted black and guide the heat down to the chamber). Also, the urine collection might be inmediattly outfitted with a small hose and (non-electrical) handpump to inmediately pump the urine to the food-production plot (so that these nutrients can be inmediattely transferred and do not need to wait out the emptying; pumping may be done after each toilet visit). Finally, perhaps the heat from the faeces composting and from the solar energy may be used to further heat up the hot water system (which will already be powered by tube collectors, but which still need additional heating for certain applications (offcourse an electric boiler will need to be installed too to attain the appropriate temperature). The latter can be done by heat exchanger (to avoid contact with the faeces).

Fraction 1 protein
Does anyone know fraction 1 protein; which can be made by making a paste from tabacco leaves ? It was invented by Shuh Sheen of the Kentucky University see here I was wondering about whether this crop may be used as an easy food source (contains high amounts of protein). This would physically strengthen population and tobacco is easy to grow in tropical/subtropical areas. Not sure whether tasty, ... dough. Also not completely sure about potential toxic substances (eg nicotine). Let me know. —Preceding unsigned comment added by 81.245.166.227 (talk) 08:04, 18 March 2009 (UTC)

The agriculture simulator proposal sent before, could be made using the 3d models from Jochem Evers/Virtualplant.nl. See www.virtualplant.nl, http://www.cwe.wur.nl/UK/Staff/Dr+Jochem+B.+Evers/ —Preceding unsigned comment added by 81.246.180.114 (talk) 18:01, 29 March 2009 (UTC)

Irrigation
The Nabataeans had ceramic pipeline irrigation in the old days, which I believe could be an appropriate irrigation system. The Jacob Blaustein Institute conducts research to these nabataeans irrigation techniques. This in order to make low-cost irrigation systems to cope with the flooding periods (strange but true). It will help in preserving the rain that falls in these periods to water the crops for a great amount of time. Eg ceramic pipelines/terracotta canals can be a low cost solution see http://desert.bgu.ac.il/desert/EngSite.aspx?SiteId=3327&ItemId=4945 http://www.hackwriters.com/Nabataeans.htm http://web.archive.org/web/20110604203733/http://www.amnh.org/education/resources/rfl/pdf/petra_guide.pdf —Preceding unsigned comment added by 81.246.180.114 (talk) 18:37, 29 March 2009 (UTC)

Perhaps this terracotta irrigation should be mentioned in article ? —Preceding unsigned comment added by 81.246.171.164 (talk) 07:38, 30 March 2009 (UTC)

Finally, there is also a new system described by agrivision (see irrigation:talk), yet this probably also has the same problems as drip irrigation (eg plastic corrugates, not easily reachable/hard to repair, ...) —Preceding unsigned comment added by 81.246.171.164 (talk) 08:09, 30 March 2009 (UTC)

Improved MayaPedal
I recenently take a look at the mayapedal (MayaPedal). Seems to me that a open-sourced version of the Pedal-A-Watt (see http://www.econvergence.net/electro.htm) would be more approriate as the bike can still be used for transport. Also, should Nanosolar best be added to article, according to time magazine (50 greatest inventions) their solar panels are way cheaper made and sold. However, there are a great number of other panels (see peswiki) that may be cheap and/or have greater energy output (making price perhaps competetive if they are compared per watt produced). However, not all produced energy is always stored on the other hand in developing countries, making a cheaper panel perhaps cheaper in all. —Preceding unsigned comment added by 81.246.148.68 (talk) 14:42, 22 March 2009 (UTC)

Soil survey using satellite-images
I recently looked at a project of Agro Vision, and it seems that satellite images can be used to determine moisture, nitrogen, co²-intake, ... They are working on localized fertilisation (eg by gps-controlled fertilisermachine) were the input data (nutrient content of soil) comes from satellite data. There is also a test project which allows farmers to survey their soil using this remote sensing. The website is at http://www.mijnakker.nl/home_21.html?lang=en

If this approach allows enough minerals (nitrogen, potassium, ...) to be recorded it may be very useful. See http://www.articlesbase.com/environment-articles/agriculture-crop-management-and-production-improved-by-satellite-remote-sensing-technology-and-geographic-information-systems-gis-463274.html

http://www.utwente.nl/organisatie/stories/landbouw-vanuit-de-ruimte, http://www.itc.nl/library/papers_2011/phd/khan.pdf and the AgriVision-project at http://www.agrivision.net.au/variable-rate.php

This approach could be used to replace the more expensive sampling of the soil (by hand) in remote areas. In one of my designs sent earlier, I also mentioned a "farming-manual (telling people in developing world how they may farm more productively and without non-biodegradable pesticides). This approach should be mentioned herein.

Besides checking whether enough parameters can be recorded, it should be checked whether UAV's can also perform this "remote sensing". —Preceding unsigned comment added by 81.246.142.167 (talk) 14:16, 25 March 2009 (UTC)

Appropedia
Appropedia (note the Appropriate technology portal) is an appropriate technology wiki, with a status tagging system which can be used to tag pages as concepts, designs, prototypes or implemented technologies.

title: Appropriate technology designs


text:
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title: User talk:Kevinlee


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User talk:Troye2009

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User talk:LHSart

testing user talk.

test number 2.

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