Specific activity

2012-08-22T12:45:16Z

194.138.39.56: /* Formulation */

When you compare registry products there are a amount of details to look out for. Because of the sheer amount of for registry products available on the Internet at when it might be quite effortless to be scammed. Something frequently overlooked is that a few of these products might in actual fact end up damaging your PC. And the registry they state they have cleaned will just cause more problems with the computer than the ones you started with. Whenever you registry gets cluttered up with a great deal of junk you don't employ, a PC will run slower. Therefore it is very prudent that we frequently receive a registry cleaned. Perfect Optimizer also has to remove junk files and is completely Windows Vista compatible. Many registry product simply don't have the time plus money to analysis Windows Vista mistakes. Because best optimizer has a large customer base, they do have the time, income plus factors to support fully support Windows Vista. The issue with most of the persons is that they do not desire to spend cash. In the cracked adaptation one refuses to have to pay anything plus can download it from internet easily. It is easy to install as well. However, the issue comes whenever it happens to be not able to identify all possible viruses, spyware plus malware in the program. This is because it happens to be obsolete inside nature plus does not get any standard updates from the site downloaded. Thus, the program is accessible to problems like hacking. Another way when arresting the 1328 error is to wash out a PC's registry. The registry is especially important as it is actually where settings plus files chosen by Windows for operating are stored. As it is frequently utilized, breakdowns and situations of files getting corrupted are not uncommon. Additionally considering of the technique it's configured, the "registry" gets saved in the wrong fashion continually, which makes the system run slow, ultimately causing your PC to suffer from a series of errors. The best method one will use inside cleaning out registries is to use a reliable [http://bestregistrycleanerfix.com/tune-up-utilities tuneup utilities] system. A registry cleaner could find out and repair corrupted registry files and settings permitting one's computer to run normally again. The initial thing you should do is to reinstall any system which shows the error. It's typical for countless computers to have specific programs which require this DLL to show the error whenever we try plus load it up. If you see a particular system show the error, you need to initially uninstall that system, restart a PC plus then resinstall the program again. This must substitute the damaged ac1st16.dll file and remedy the error. We require an option to automatically delete unwanted registry keys. This can protect you hours of laborious checking by a registry keys. Automatic deletion is a key element when you compare registry products. There is a lot a superior registry cleaner can do for your computer. It can check for plus download changes for Windows, Java and Adobe. Keeping updates present is an significant part of advantageous computer health. It will furthermore protect a personal and company confidentiality and the online safety.

Voltage droop

2011-08-17T14:13:19Z

194.138.39.56: /* References */

[[Image:Line-Line Intersection.png|400px|thumb|right|The intersection of lines.]]
In [[Euclidean geometry]], the intersection of a [[line (mathematics)|line]] and a line can be the [[empty set]], a [[point (geometry)|point]], or a line. Distinguishing these cases and finding the intersection point have use, for example, in [[computer graphics]], [[motion planning]], and [[collision detection]].

The number and locations of possible intersections between two lines and the number of possible lines with no intersections ([[parallel (geometry)|parallel]]) with a given line are the distinguishing features of [[non-Euclidean geometry]].

== Mathematics ==
First we consider the intersection of two lines <math>L_1\,</math> and <math>L_2\,</math> in 2-dimensional space, with line <math>L_1\,</math> being defined by two distinct points <math>(x_1,y_1)\,</math> and <math>(x_2,y_2)\,</math>, and line <math>L_2\,</math> being defined by two distinct points <math>(x_3,y_3)\,</math> and <math>(x_4,y_4)\,</math>.<ref name="Wolfram">
{{cite web | title=Weisstein, Eric W. "Line-Line Intersection." From MathWorld | work=A Wolfram Web Resource | url=http://mathworld.wolfram.com/Line-LineIntersection.html| accessdate=2008-01-10}}
</ref>

The intersection <math>P\,</math> of line <math>L_1\,</math> and <math>L_2\,</math> can be defined using [[determinant]]s.

:<math>
P_x = \frac{\begin{vmatrix} \begin{vmatrix} x_1 & y_1\\x_2 & y_2\end{vmatrix} & \begin{vmatrix} x_1 & 1\\x_2 & 1\end{vmatrix} \\\\ \begin{vmatrix} x_3 & y_3\\x_4 & y_4\end{vmatrix} & \begin{vmatrix} x_3 & 1\\x_4 & 1\end{vmatrix} \end{vmatrix} }
{\begin{vmatrix} \begin{vmatrix} x_1 & 1\\x_2 & 1\end{vmatrix} & \begin{vmatrix} y_1 & 1\\y_2 & 1\end{vmatrix} \\\\ \begin{vmatrix} x_3 & 1\\x_4 & 1\end{vmatrix} & \begin{vmatrix} y_3 & 1\\y_4 & 1\end{vmatrix} \end{vmatrix}}\,\!
\qquad
P_y = \frac{\begin{vmatrix} \begin{vmatrix} x_1 & y_1\\x_2 & y_2\end{vmatrix} & \begin{vmatrix} y_1 & 1\\y_2 & 1\end{vmatrix} \\\\ \begin{vmatrix} x_3 & y_3\\x_4 & y_4\end{vmatrix} & \begin{vmatrix} y_3 & 1\\y_4 & 1\end{vmatrix} \end{vmatrix} }
{\begin{vmatrix} \begin{vmatrix} x_1 & 1\\x_2 & 1\end{vmatrix} & \begin{vmatrix} y_1 & 1\\y_2 & 1\end{vmatrix} \\\\ \begin{vmatrix} x_3 & 1\\x_4 & 1\end{vmatrix} & \begin{vmatrix} y_3 & 1\\y_4 & 1\end{vmatrix} \end{vmatrix}}\,\!
</math>

The determinants can be written out as:

: <math>\begin{align}
(P_x, P_y)= \bigg(&\frac{(x_1 y_2-y_1 x_2)(x_3-x_4)-(x_1-x_2)(x_3 y_4-y_3 x_4)}{(x_1-x_2)(y_3-y_4)-(y_1-y_2)(x_3-x_4)}, \\
&\frac{(x_1 y_2-y_1 x_2)(y_3-y_4)-(y_1-y_2)(x_3 y_4-y_3 x_4)}{(x_1-x_2)(y_3-y_4)-(y_1-y_2)(x_3-x_4)}\bigg)
\end{align}</math>

Note that the intersection point is for the infinitely long lines defined by the points, rather than the [[line segment]]s between the points, and can produce an intersection point beyond the lengths of the line segments. If (rather than solving for the point in a single step), the solution in terms of first degree [[Bézier curve#Linear curves|Bézier]] parameters is first found, then this intermediate result can be checked for 0.0 ≤ ''t'' ≤ 1.0 and 0.0 ≤ ''u'' ≤ 1.0 (where ''t'' and ''u'' are the driving variables).

When the two lines are parallel or coincident the denominator term is zero:

: <math>
(x_1-x_2)(y_3-y_4)-(y_1-y_2)(x_3-x_4)=0\text{ if the lines are parallel}
</math>

If the lines are very close to being parallel, then a computer solution may encounter numeric problems in the solution described above, and so recognition of this condition may require an appropriately "fuzzy" test in practical application. A more robust and general solution may be obtained by rotation of the line segments to drive one of them horizontal, whence the solution of the rotated parametric form of the second line is easily obtained. Careful discussion of the special cases is required (parallel lines/coincident lines, overlapping/non-overlapping intervals).

== ''n''-line intersection ==
In two dimensions, more than two lines [[almost certainly]] do not intersect at a single point. Similarly, in three or more dimensions, even two lines almost certainly do not intersect; pairs of lines that do not intersect are called [[skew lines]]. However, in two or more dimensions, we can usually find a point that is mutually closest to two or more lines in a [[least-squares]] sense.

In the two-dimensional case, first, represent line ''i'' as a point, <math>p_i</math>, on the line and a [[unit vector|unit]] [[normal vector]], <math>\hat n_i</math>, perpendicular to that line. That is, if <math>x_1</math> and <math>x_2</math> are points on line 1, then let <math>p_1 = x_1</math> and let

:<math>\hat n_1:= \begin{bmatrix}0&-1\\1&0\end{bmatrix} (x_2-x_1) / \|x_2-x_1\|</math>

which is the unit vector along the line, rotated by 90 degrees.

Note that the distance from a point, ''x'' to the line <math>(p, \hat n)</math> is given by

:<math>d(x,(p,n))=\|(x-p)\cdot \hat n\| = \|(x-p)^\top \hat n\| = \sqrt{(x-p)^\top \hat n \hat n^\top (x-p)}.</math>

And so the squared distance from a point, ''x'', to a line is

:<math>d(x,(p,n))^2=(x-p)^\top (\hat n \hat n^\top) (x-p).</math>

the sum of squared distances to many lines is the [[Loss function|cost function]]:

:<math>E(x) = \sum_i (x-p_i)^\top (\hat n_i \hat n_i^\top) (x-p_i).</math>

This can be rearranged:

:<math>
\begin{align}
E(x) & = \sum_i x^\top \hat n_i \hat n_i^\top x - x^\top \hat n_i \hat n_i^\top p_i - p_i \hat n_i \hat n_i^\top x + p_i^\top \hat n_i \hat n_i^\top p \\
& = x^\top \left(\sum_i \hat n_i \hat n_i^\top\right) x - 2 x^\top \left(\sum_i \hat n_i \hat n_i^\top p_i\right) + \sum_i p_i^\top \hat n_i \hat n_i^\top p_i.
\end{align}
</math>

To find the minimum, we differentiate with respect to ''x'' and set the result equal to the zero vector:

:<math>\frac{\partial E(x)}{\partial x} = 0 = 2 \left(\sum_i \hat n_i \hat n_i^\top\right) x - 2 \left(\sum_i \hat n_i \hat n_i^\top p_i\right) </math>

so

:<math>\left(\sum_i \hat n_i \hat n_i^\top\right) x = \sum_i \hat n_i \hat n_i^\top p_i</math>

and so

:<math>x = \left(\sum_i \hat n_i \hat n_i^\top\right)^{-1}\left(\sum_i \hat n_i \hat n_i^\top p_i\right).</math>

While <math>\hat n_i</math> is not well-defined in more than two dimensions, this can be generalized to any number of dimensions by noting that <math>\hat n_i \hat n_i^\top</math> is simply the (symmetric) matrix with all eigenvalues unity except for a zero eigenvalue in the direction along the line providing a [[seminorm]] on the distance between <math>p_i</math> and another point giving the distance to the line. In any number of dimensions, if <math>\hat v_i</math> is a unit vector ''along'' the ''i''th line, then

: <math>\hat n_i \hat n_i^\top</math> becomes <math>I - \hat v_i \hat v_i^\top</math>

where ''I'' is the identity matrix, and so

: <math> x= \left(\sum_i I-\hat v_i \hat v_i^\top\right)^{-1} \left(\sum_i (I-\hat v_i \hat v_i^\top) p_i\right).</math>
{{Citation needed|reason=The n-dimensional line-line intersection is not covered by the referenced source.|date=December 2013}}

==X and Y values of intersection on a linear curve==
The <math>X</math> and <math>Y</math> values of the point of intersection of 2 lines can easily be found using the following substitutions and rearrangements.

Suppose 2 lines with the equations <math>y=ax+c</math> and <math>y=bx+d</math> where <math>a</math> and <math>b</math> are the gradients of the lines and where <math>c</math> and <math>d</math> are the ''y''-intercepts of the lines. At the point at which the 2 lines intersect, both <math>y</math> values will be the same, hence the following equality.

:<math>ax+c=bx+d</math>

We can rearrange the latter to extract the value of <math>x</math>.

:<math>ax-bx=d-c</math>
:<math>x=\frac{d-c}{a-b}</math>

To find the ''Y'' value, all we need to do is substitute the value of ''x'' into one of the 2 line equations:

:<math>y=a\frac{d-c}{a-b}+c</math>

Hence, the point of intersection is
:<math>P\left( \frac{d-c}{a-b} ; a\frac{d-c}{a-b}+c \right)</math> 

== Using homogeneous coordinates ==

By using [[Homogeneous_coordinates|homogeneous coordinates]], the intersection point of two implicitly defined lines can be determined quite easily. In 2D, every point can be defined as a projection of a 3D point, given as the ordered triple (X,Y,W). The mapping from 3D to 2D coordinates is (x,y) = (X/W, Y/W). 2D points can be converted to homogeneous coordinates by defining them as (x,y,1). Thus, the implicit equation of a line can be given in homogeneous coordinates as:

:<math>aX+bY+cW = 0 \Rightarrow L(a,b,c) \cdot P(X,Y,W) = 0</math>

We can do a similar operation with the cross product to get the intersection of 2 lines:

:<math>L(a_1,b_1,c_1) \times L(a_2,b_2,c_2) = P(X,Y,W)</math>

This returns the intersection point in homogeneous coordinates. In the special case of W = 0, we say that the intersection point is at infinity. This means the lines are parallel. As an aside, the implicit coefficients of a line can be obtained by the cross product of two points:

:<math>P(x_1,y_1,w_1) \times P(x_2,y_2,w_2) = L(a,b,c)</math>

==See also==
*[[Line segment intersection]]
*[[Projective_plane#Lines_joining_points_and_intersection_of_lines_.28using_duality.29|Line intersection in projective space]]
*[[Distance from a point to a line]]
*[[Parallel postulate]]

== References ==
<references/>

== External links ==
* [http://softsurfer.com/Archive/algorithm_0106/algorithm_0106.htm Distance between Lines and Segments with their Closest Point of Approach], applicable to two, three, or more dimensions.

[[Category:Euclidean geometry]]
[[Category:Linear algebra]]
[[Category:Geometric algorithms]]

formulasearchengine - User contributions [en]

Specific activity

Voltage droop