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# Plotting or Visualizing a Higher dimensional vector field

I am trying to visualize a higher dimensional vector field. Is there a way to do this. I asked this question here, I was told to post it here.

As an example, one can use

$$\begin{eqnarray} \dot{x}&=& \left(J-R\right)x \end{eqnarray}$$

where $x\in \mathbb{R}^5$ $J$ is a skew-symmetric matrix and $R$ is a positive semi-definite matrix.

PS: The only idea I came so far is: somehow I have to use t-sne, but I really don't know how.

Somehow I have to use t-sne, but I really don't know how.

Since you have a PhD my answer will be brief, it's quite a lengthy subject.

The aim of dimensionality reduction is to preserve as much of the significant structure of the high-dimensional data as possible in the low-dimensional map. For high-dimensional data that lies on or near a low-dimensional, non-linear manifold it is usually more important to keep the low-dimensional representations of very similar datapoints close together, which is typically not possible with a linear mapping.

T-distributed Stochastic Neighbor Embedding (t-SNE) is a way of converting a high-dimensional data set into a matrix of pair-wise similarities. t-SNE is capable of capturing much of the local structure of the high-dimensional data very well, while also revealing global structure such as the presence of clusters at several scales.

Stochastic Neighbor Embedding (SNE) starts by converting the high-dimensional Euclidean distances between datapoints into conditional probabilities that represent similarities. The similarity of datapoint $x_j$ to datapoint $x_i$ is the conditional probability, $p_{j|i}$, that $x_i$ would pick $x_j$ as its neighbor if neighbors were picked in proportion to their probability density under a Gaussian centered at $x_i$. For nearby datapoints, $p_{j|i}$ is relatively high, whereas for widely separated datapoints, $p_{j|i}$ will be almost infinitesimal (for reasonable values of the variance of the Gaussian, $σ_i$).

The cost function used by t-SNE differs from the one used by SNE in two ways: (1) it uses a symmetrized version of the SNE cost function with simpler gradients that was briefly introduced by Cook et al. (2007) and (2) it uses a Student-t distribution rather than a Gaussian to compute the similarity between two points in the low-dimensional space. t-SNE employs a heavy-tailed distribution in the low-dimensional space to alleviate both the crowding problem and the optimization problems of SNE.

A few t-SNE tutorials:

• Thanks for the answer! SNE/t-SNE might not work as there is no dimensionality reduction in my problem. I can, however, try a model-reduction technique together with this. I wanted to use this to demonstrate my results. However, this itself will be a new research direction. – kosa Aug 31 '18 at 13:41

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# How to determine if b is a linear combination of a1,a2,a3 and the columns of matrix A?

a) Determine if $b$ is a linear combination of $a_1, a_2, a_3$, the columns of the matrix $A$

$A = \begin{bmatrix}4&-4&-16\\2&-1&-6\\-1&-1&2\end{bmatrix}$ and $b =\begin{bmatrix}28\\9\\-1\end{bmatrix}$

YES, it is a linear combination <-- my answer

b) If it is a linear combination, determine a non-trivial linear relation - (a non-trivial relation is three numbers which are not all three zero.) Otherwise, enter 0's for the coefficients.

___ $a_1$ + ____ $a_2$ + ____ $a_3 = b$

I got the augmented matrix after putting it in REF form, I got:

my answer --> $\begin{bmatrix}4&-4&-16&28\\0&-2&-4&10\\0&0&-8&24\end{bmatrix}$

so I got the vector equations and solved and got $x_1 = -4, x_2 = 1, x_3 = -3$

So I can't figure out what coefficents they are looking for.. Wouldn't it be $-4,1,-3$ ??

UPDATE:

Now I've gone further and took my matrix from REF (Row Eche Form) to RREF (Reduced Row Ech Form)

and got

$\begin{bmatrix}1&-1&-4&7\\0&1&2&-5\\0&0&1&-3\end{bmatrix}$ so therefore $x_1 = -4, x_2 = 1, x_3 = -3$

• If those work, then yes. If not, no. Do they? – Semiclassical Jun 15 '16 at 21:51
• Nope, says it's incorrect when I try to put them in. – Shammy Jun 15 '16 at 21:52
• You've made a mistake somewhere. You can see why if you actually add up the three vectors with your proposed coefficients. The last entry adds to $-3$ whereas it should be $-1$. – Alex R. Jun 15 '16 at 21:56
• So the answer you give works for the REF, but not for the initial matrix. Hence, check your work... – Semiclassical Jun 15 '16 at 21:57
• I didnt put it in RREF form , just in REF form. So when you say check my work do you mean that I need to put it in RREF form? – Shammy Jun 15 '16 at 22:01

I get different results.

Solution via inverse:

>> A = [ 4,-4,-16; 2,-1,-6; -1, -1, 2]
A =

4   -4  -16
2   -1   -6
-1   -1    2

>> det(A)
ans =  8
>> b = [28;9;-1]
b =

28
9
-1

>> inv(A)* b
ans =

-2
-1
-2

Solution via Gauss elimination:

>> M = [A, b]
M =

4   -4  -16   28
2   -1   -6    9
-1   -1    2   -1

>> rref(M)
ans =

1   0   0  -2
0   1   0  -1
0   0   1  -2

• Where are the coefficients of $a_1, a_2, a_3$ – Shammy Jun 15 '16 at 22:12
• One gets $x = (-2,-1,-2)^T$ as solution vector. The components are the coefficients, as $A x = (a_1, a_2, a_3) x = b$. – mvw Jun 15 '16 at 22:15
• I see the problem now, I did not make sure that above my 1 leading entries were 0's! – Shammy Jun 15 '16 at 22:16
• @Shammy That should just make the calculation of $x_1$ and $x_2$ a bit more work, but still lead to the same numbers. I guess you made some other mistake during the elimination process. – mvw Jun 15 '16 at 22:22


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# Equation of a line in complex plane

1. May 12, 2013

### 4everphysics

All the derivation of the equation of line on complex plane uses the fact that (x,y) \in R^2 can be identified with x+iy \in C.

Thus, they begin with ax+by+c = 0 then re-write x = (z+\bar{z})/2 and y = (z-\bar{z})/(2i), and substitute it into real plane line equation to get it in complex form.

What I don't quite understand is, since (x,y) is identified with (x,iy),

don't we need to write ax+by+c=0 into ax+biy+c=0 before we proceed with the substitution? Why don't we need to do such thing?

Thank you.

2. May 12, 2013

### tiny-tim

hi 4everphysics!
(x,iy) is not in ℂ

ℂ is a set whose elements are single items (traditionally called "z")

ℂ is not a direct product of two sets, with elements that are ordered pairs

3. May 13, 2013

### HallsofIvy

Staff Emeritus
In other words, we are not "identifying (x, y) with (x, iy)". We are identifying the pair of real numbers, (x, y) with the single complex number x+ iy.

4. May 13, 2013

### Bacle2

Still, one thing that I think is good to take into account is that a complex line is 2-dimensional as a real object. Notice that the 1st complex projective space is a 2-sphere, but 1st real projective space is a circle.


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# After \clearpage move the figure to the top

I need to make a "pagebreak" after a figure so I use \clearpage. But I want the figure to stay at the top of the page and not centering at the now empty page. My code right now looks like:

\begin{figure}[t]
\centering
\includegraphics[scale=0.5]{YY.PNG}
\caption{XX}
\label{fig:BB}
\end{figure}

\clearpage

\section{AAA}

• One assumes the figure actually fits the entire page (or close to it)? And you have surrounding text you want to flow as per usual? – Werner Jun 23 '16 at 16:13
• \makeatletter\setlength\@fptop{0pt}\makeatother – David Carlisle Jun 23 '16 at 16:15
• or use \newpage rather than \clearpage – David Carlisle Jun 23 '16 at 16:16

You can use afterpage to perform some magic between page shipouts:

\documentclass{article}

\usepackage{graphicx,float}
\usepackage{afterpage}
\usepackage{showframe,lipsum}% Just for this example

\begin{document}

\section{First section}\lipsum[1-3]

\afterpage{
\begin{figure}[H]
\centering
\includegraphics[width=.8\textwidth,height=.8\textheight]{example-image}
\caption{XX}
\end{figure}
\clearpage
}

\clearpage% or more text (like \lipsum[4-6])

\section{Second section}\lipsum[4-6]

\end{document}

The image is stored in a non-float (using float's [H]ere float specifier) and output after the current page, together with a \clearpage. That ensure the figure appears on its own at the top between two pages of continuous content.


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## 3 thoughts on “Advanced config with configuration meta-data in Spring Boot”

1. snicoll says:

Thanks for the article!

FYI, you can also annotate the getter of the deprecated property with @DeprecatedConfigurationProperty and we will generate that meta-date entry automatically for you. When you deprecate a property, it is backed by a public getter/setter anyway so that should be included in the deprecation as well.

“Value providers are available only to manually provided hints”. I find this a bit confusing. You can add value provider to any key (including those generated automatically). All you need to do is to add a hint entry that references the key you want to customize.

1. jakub says:

Hi Stéphane, thanks for your insights on both articles. I must have missed @DeprecatedConfigurationProperty annotation in the latest Spring Boot documentation (http://docs.spring.io/spring-boot/docs/current/reference/htmlsingle/). If it is not there it might be worth including. As for the hints remark, this might be caused by my English 🙂 All I meant to say is that one needs to specify it manually in additional-spring-configuration-metadata.json file. Or is there any other way of doing it? Anyways, I will modify the article according to your comment tomorrow.

1. snicoll says:

Good catch for the documentation, I’ve created #5118 to fix that. As for the hints, yep you can only specify them manually. Thanks!


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For finding forces in few of the specific members method of joints is preferrable. The methods shown in this section are intended to be typical examples and should not be regarded as exhaustive. Note that the right arrow ($\rightarrow$) here is relative to the cut member end. For many small businesses, the cash method has several significant advantages over the accrual method, as it is generally easier for most businesses to administer, simplifying the accounting. Component Method for High Strength Steel Rectangular Hollow Section T Joints. :mad1::mad1::mad1::mad1::mad1::mad1::mad1: Plz help Neither joint can be solved without further analysis; however, joint B can be solved if the force in member and is found. There is also no internal instability, and therefore the truss is stable. Method of Joints If a truss is in equilibrium, then each of its joints must be in equilibrium. • Due to the absence of gusset plates and other connecters, the welds are usually lighter. Method of section is very useful when you want to know the forces acting on a certain member in a truss. Mitre Joint Disadvantages. BACKGROUND AND AIM The increased incidence of caesarean section has led, over the past few years, to the revision of surgical methods in order to reduce … A frame is analysed by the following methods: (i) Method of joints, (ii Method of sections, and (iii) Graphical method. The free body diagram for the cut section to the right of section b-b is shown in Figure 3.11. It can be difficult to get all of the desired tissue in a section without trimming away significant portions of the sample. • Advantages. Yours may be different. A truss can be analysed internally or externally. Joints ACD of the frame are fixed, while joint B can rotate slightly due to the applied load A section has ﬁnite size and this means you can also use moment equations to solve the problem. The method of joints isolates a joint to find unknown forces. The first is a standing offer under a maintenance work for a specified period, and this contract type is rare compared to others. What is the biggest cost in the construction industry today? Hardy Cross (1885-1959) Moment Distribution is an iterative method of solving an indeterminate Structure. Method of joints and method of sections, • Analysis of frame, • Shear force and bending moments, • Cables, • Theory of dry friction, • Moment of Inertial of an area; Parallel axis theorem. Since there are only three global equilibrium equations, we can only solve for three unknown member axial forces at a time using the method of sections. DOWNLOAD THIS PAGE AS PDF. Is one method stronger? What are the softwares a civil engineer wants to know? It makes sense to choose this side because it does not have any external forces and it has only one reaction component $I_y$. ADVANTAGES AND DISADVANTAGES OF WELDED JOINTS . The following is a list of the pros and cons of having joint accounts: Advantages. … We must find the internal axial forces in the specific truss members AB, AD, DF and FG. To solve indeterminate systems, we must combine the concept of equilibrium with compatibility. Hey, can any one explain method of sections and method of joints?? The following are the advantages of welded joints. Rivet can be used in joining of nonmetallic material like wood, plastic, asbestos sheet etc. As mentioned previously, the point of this cut is only to find the force in member AB ($F_{AB}$). Note: In the above statement, to ascertain the average unit cost or average cost, we have used the following formula:. Draw F.B.Ds for all your calculations 1) Identify all the zero force members in the trusses; 2) Calculate the reactions at the supports; 3) Use one cut only to determine the internal forces for marked members in each truss; 4) Indicate your final results on the truss system For this problem, one way to do this is by cutting at section b-b as shown in Figure 3.9. The remaining unknowns may be found using vertical and horizontal equilibrium: The information on this website is provided without warantee or guarantee of the accuracy of the contents. Using the method of joints, the force could be found by isolating the joint at either end of the member (joint B or C). The author shall not be liable to any viewer of this site or any third party for any damages arising from the use of this site, whether direct or indirect. The method of joints analyzes the force in each member of a truss by breaking the truss down and calculating the forces at each individual joint. Then move to the next joint and find the forces in the members.Repeat the procedure and find all the member forces. In the section students will learn about Truss structures, specifically method of joints, method of sections, and zero force members. gave a brief description and solved problems for students in Bahrain to know about Trusses, Method of Supporting Beams, Analysis of Structures by Method of Joints & Sections, Reason about why to use Trusses and why not to A principal advantage of statistical methods of attributes sampling over nonstatistical methods is that they provide a quantifiable basis for establishing the A) risk of assessing control risk too low. The gin pole used is approximately 10 m long and is held in place by means of guys by the side of the tower to be erected. Figure 3.9: Method of Sections Example Problem, Figure 3.10: Method of Sections Example - Free Body Diagram for Cut Section to the Right of b-b, Figure 3.11: Method of Sections Example - Free Body Diagram for Cut Section to the Right of a-a, Chapter 2: Stability, Determinacy and Reactions, Chapter 3: Analysis of Determinate Trusses, Chapter 4: Analysis of Determinate Beams and Frames, Chapter 5: Deflections of Determinate Structures, Chapter 7: Approximate Indeterminate Frame Analysis, Chapter 10: The Moment Distribution Method, Chapter 11: Introduction to Matrix Structural Analysis, 3.2 Calculating x and y Force Components in Truss Members, 3.4 Using Global Equilibrium to Calculate Reactions, Check that the truss is determinate and stable using the methods from, Calculate the support reactions for the truss using equilibrium methods as discussed in. Therefore, for truss members, it is often more convenient to think of the forces in terms of tension or compression instead of in terms of a specific direction. When we use our intuition, we are relying on our guts, our emotions, and/or our instincts to guide us. Members of the frame are prismatic and are assumed not to deform axially nor translate relative to one another. The Method of Joints. The key to understanding the method is "net." B) tolerable exception rate. This limits the static equilibrium equations to just the two force equations. In any direction, you will not find it self-aligning. Ask Question Asked 8 years, 10 months ago. Are glass office partition walls cheaper than bricks? In addition, the cash method can yield tax savings over the accrual method by deferring the recognition of income until cash is received. For e.g. Christo Ananth et al. Method of Joints - the axial forces in the members of a statically determinate truss are determined by considering the equilibrium of its joints. Calculate the reactions at the support. the number of members is less than the required members.So there will be chance to fail the structure. * So why do we use method sections? The Method of Sections (Ritter Method) The 3 eqns of equilibrium may be applied to either one of these 2 parts to determine the member forces at the “cut section” A decision must be made as to how to “cut” the truss In general, the section should pass through not … By going through this post, you can develop your understanding of top joining method. 1.12.5. In the section method, major sections of the tower are assembled on the ground and the same are erected as units. Intuition. The method of sectionsis the same except an entire section is isolated.... Because all cut members were part of a truss, all members supported an axial force of tension or compression. To perform a 2D determinate truss analysis using the method of sections, follow these steps: . Joint D. Yes. The method of joints is a process used to solve for the unknown forces acting on members of a truss.The method centers on the joints or connection points between the members, and it is usually the fastest and easiest way to solve for all the unknown forces in a truss structure. This joint has an external vertical force of 300N which must be countered by the members attached to the joint. ADVANTAGES AND DISADVANTAGES OF WELDED JOINTS . Y B Cm 2) Using Mohr Circle, Determine The Moment Of Inertia (Iv) Of The Rectangular Cross-section About Inclined V … m>2j+3 then it is reduntant strucure, reduntant structure means the numbers of members to keep in stable position is more.The number of members in the truss is greater than the required members. 45). Also, The Pre-installation Verification fastener … Advantages of using the Request for Quotations method The procurement cycle is significantly short since the bidding period is only three days. The free body diagram for the cut section to the right of section b-b is shown in Figure 3.10. This section shares member AB with the other section a-a. The slope or deflection at any point on the beam is equal to the resultant of the slopes or deflections at that point caused by each of the load acting separately. 17 Truss Solution. • Welding process is quicker as it requires no drilling of holes. We cut section b-b in such a way that, even though we cannot solve for all four of the member forces across the cut, we can still solve for one of them (AB) by using the moment equilibrium equation. Method of Joints. Compressive (C) axial member force is indicated by an arrow pushing toward the joint. Viewed 18k times 14. Large amount of capital is necessary for conducting business on a large scale. m<2j+3  unstable structure. Mitre Joint Advantages. This method consists of passing an imaginary section … A joint stock company has many advantages. Why corner columns have higher values of moments than other columns? Required fields are marked *, Analysis of Truss-Method of Sections and Joints. A list of CBSE Toppers from schools all over India A list of CBSE's top performing schools (Class 12) ... Trusses, Method of Joints and Sections - A Tutorial with Solved Problems ANALYSIS OF STRUCTURES Introduction. In this method, after determining the reactions at the supports, the equilibrium of every joint is considered. Since only two equations are involved, only two unknowns can be solved for at a time. Which one is the best for RCC concrete, M-sand or river sand? Start with the moment equilibrium around point J: The obvious choice for a cut is section a-a as shown in Figure 3.9, because it cuts through all of the members that we are trying to find axial forces for; however, the problem is that this section has four members, which we cannot calculate directly using only our three equilibrium equations. Question: 1) What Is The Relative Advantage Of "method Of Sections” Over “method Of Joints” In The Analysis Of Trusses? If a structure is stable it is called as statically determinate.It the number of unknowns is equal or less than the number of equlibrium equations then it is statically determinate.The analysis of truss can be done by maintly two methods, that is method of joints and method of sections, Force developed in a truss member is always axial, it can be compressive or tensile, Truss memebers are connected at frictionless pins, No need to consider any secondary moment due to the friction. When we need to find the force in all the members, method of joint is preferrable. Newton's Third Law indicates that the forces of action and reaction between a member and a pin are equal and opposite. In the moment distribution method, every joint of the structure to be analysed is fixed so as to develop the fixed-end moments.Then each fixed joint is sequentially released and the fixed-end moments (which by the time of release are not in equilibrium) are distributed to adjacent members until equilibrium is achieved. No, I don't. Either a mobile crane or a gin pole is used. It works by cutting through the whole truss at a single section and using global equilibrium (3 equations in 2D) to solve for the unknown axial forces in the members that cross the cut section. The primary benefit of the method of sections is that, for a determinate truss, you can find the force in any individual member quickly without having to solve through the entire truss one joint at a time (which you must do when using the method of joints). It not very strong. To perform a 2D determinate truss analysis using the method of sections, follow these steps: The truss shown in Figure 3.9 has external forces and boundary conditions provided. When we go back to section cut a-a, we will look at the section to the right of the cut as well for the same reason. 15. If there were a net force, the joint … The section must cut completely through the truss and should cut through no more than three members. The method of jointsconsists of satisfying the equilibrium equations for forces acting on each joint. Explain With Less Than 50 Words. The structures may consist of several sections. But We can only identify two types of contractual bindings in the construction industry. By making a "cut" along the truss and member you … Fx 0 Fy 0 Method of Joints Recall, that the line of action of a force acting on a joint is determined by the geometry of the truss member. Joint E can now be solved. The following are the advantages of welded joints. This process is similar to cutting a beam at a section to find the internal forces at that section. It is easier using the method of sections. In the equity method, you first record the investment is recorded at cost, and then adjust it up or down, depending on … Section Method of Transmission Tower Erection. Moment distribution method offers a convenient way to analyse statically indeterminate beams and rigid frames.In the moment distribution method, every joint of the structure to be analysed is fixed so as to develop the fixed-end moments. The Bisection method fails to identify multiple different roots, which makes it less desirable to use compared to other methods that can identify multiple roots. Requires a Lot of Effort. The method of joints is a way to find unknown forces in a truss structure. To illustrate the concept of the method of moment distribution, consider the frame shown in Figure 12.1. The Method of Sections (Ritter Method) If the forces in only a few members of a truss are to be found, the method of sections generally provide the most direct means of obtaining these forces The method is created the German scientist August Ritter (1826 - 1908). In welding filler plates, gusseted plates, connecting angles etc, are not used, which leads to reduced overall weight of the structure. In fact, each rigging method has its limitations. For some obscure reason, this is called the method of joints… Over view: There are many rigging methods for different kinds of loads being handled. Method of joints are the common method for the analysis of truss members.The basic concept that is used in the analysis is ,since the truss is in equilibrium the each joints in the truss is also in equilibrium, The steps for the determination of member forces, For the determination of reaction forces , we can use two equilibrium equations that is, Now we got two reactions,then choose a joint which has 2 unknown forces so as to find the two unknown forces from the two reaction we got, Take the joints and apply equations of equilibrium on that joint and find the member forces. Welded joints are more economical as less labor […] For this problem, as previously described, we need to make two cuts and solve the equilibrium equations twice: once to find the force in member AB using section b-b and again to find the rest of the forces in the other members that cross section a-a. For this problem, the moment arm for $F_{AD}$ is equal to $(6\mathrm{\,m})(\cos \theta)$. One of the main advantages of the rivet is that it can use to join dissimilar materials. To find force analyze joint A. How can I learn finite element method for structural analysis? In the Method of Joints, we are dealing with static equilibrium at a point. The problem is magnified when confronted with minute samples and thin cores. These are given below: 1. So, we need a way to cut down the number of unknown forces at section a-a from four to three. Homework Problems must be worked on an 8.5” × 11” college ruled paper. Your email address will not be published. • Due to the absence of gusset plates and other connecters, the welds are usually lighter. Since truss members are subjected to only tensile or compressive forces along their length, the internal forces at the cut member will • Welding is more adaptable than other types of connectio ns and can even be used in circular This section was chosen deliberately because the other three forces ($F_{BG}$, $F_{EG}$, and $F_{GH}$) all point direction through point G. So, if we evaluate the moment equilibrium about point G, we can solve directly for $F_{AB}$: which is negative, meaning that the member is actually in compression. Classification of Bridges-Types, Span, Functions and Construction. Tensile (T) axial member force is indicated on the joint by an arrow pulling away from the joint. Since only two equations are involved, only two unknowns can be solved for at a time. Problems must be worked on an 8.5” × 11” college ruled paper. C) expected population exception rate. But, this section (b-b) still cuts through four members, meaning that we can't solve for all of the internal axial forces in those cut members either. Tension forces always pull away from joints and members, compression forces always push towards joints and members. 4. The method of joints isolates a joint to find unknown forces. Then, each fixed joint is sequentially released and the fixed-end moments (which by the time of release are not […] Video created by Georgia Institute of Technology for the course "Applications in Engineering Mechanics". Hence, Average Unit Cost=18000/10=₹1800 per unit Sales Value Method. Select an appropriate section that cuts through the member that you want to find the axial force for. The moment distribution method is a structural analysis method for statically indeterminate beams and frames developed by Hardy Cross.It was published in 1930 in an ASCE journal. The methods of acquiring knowledge can be broken down into five categories each with its own strengths and weaknesses. Advantages of Welding Joints As no hole is required for welding, hence no reduction of area. For a simpler problem, only one cut would be needed if the section had only three members crossing the cut. Note the point  that we cannot take any moments at any joints,because all the forces are passing through the same point. And the reason why it's always convergent is . What will happen to a steel frame overloaded at bolt? snug-tightened. (My response. There are various types of tendering methods, to analyze, negotiate and award a project to a contractor. This is because the other three members at that cut (BD, EG, and GH) are all coincident through point G. This means that if we take a moment equilibrium around point G, then none of these member forces will contribute to the equilibrium, leaving only the force in AB, which can then be found easily. In this segment, we're going to talk about bisection method, and look at the advantages and drawbacks. Under this method of joint cost apportionment, the sales value of each joint product is used to determine the cost incurred over it. It is easier using the method of sections. As the name suggests we need to consider an entire section instead of joints. Rivet can be used in joining of ferrous as well as a nonferrous metal like aluminium and copper alloys. The information on this website, including all content, images, code, or example problems may not be copied or reproduced in any form, except those permitted by fair use or fair dealing, without the permission of the author (except where it is stated explicitly). A section has ﬁnite size and this means you can also use moment equations to solve the problem. How can a building Resist Earthquake Loads? This allows solving for up to three unknown forces at a time. This means the sum of all the vertical forces as well as the horizontal forces acting on a joint is equated to zero. Under this method the total joint cost is divided into two categories viz., variable and fixed costs. The method to approach the market is simple. Solution:. Unlike the previous chapter in this unit we will be dealing with equilibrium of supporting structure. What is difference between deficient frame and redundant frame? TURN-OF-NUT BOLTING METHOD. Method of Sections In this method, we will cut the truss into two sections by passing a cutting plane through the members whose internal forces we wish to determine. Hint: To apply the method of … Having demonstrated to you the method of joints, we now move on to see the method of sections that directly gives the force on a desired member of the truss. As before, even though points A and F are not within the section cut, they are left in the diagram as reference points. All of the unknown member forces across the cut are shown, and are assumed to be in tension (pulling away from the member). P-424, determine the force in BF by the method of joints and then check this result using the method of sections. Unlike direct method, the step method (also known as step down method) allocates the cost of a service department to other service departments as well as to operating departments.The cost allocation under step method is a sequential process. This is a simple truss that is simply supported (with pin at one end and a roller at the other). Reliance has invested more than Rs.25,000 crore in its telecom venture. In special circumstances, four member cuts may be made (as will be shown in the example). This limits the static equilibrium equations to just the two force equations. » Joints required for contraction and expansion » Generally rough riding quality » High repair costs (ii) Flexible pavements consist of a series of layers, with the highest quality materials at or near the surface. MOMENT DISTRIBUTION METHOD Distribution and carryover of moments – Stiffness and carry over factors – Analysis of continuous beams – Plane rigid frames with and without sway – Neylor‟s simplification. Chapter 7 –Hollow section joints. Each of these methods has particular advantages or disadvantages. Research output: Book/Report › Doctoral thesis › Collection of Articles This will make the equilibrium equations less complicated. What are the advantages/disadvantages of interior point methods over simplex method for linear optimization? Tampere University, 2019. Can we provide an inverted beam for a sloped RCC slab? The method only accounts for flexural effects and ignores axial and shear effects. i can't seem to figure it out?? The method of sections is the same except an entire section is isolated. / Garifullin, Marsel. Active 7 years, 7 months ago. . This allows solving for up to three unknown forces at a time. Eurocodes - Design of steel buildings with worked examples Brussels, 16 - 17 October 2014 ... Background: COMPONENT METHOD. 2.Method of sections. Frame 18-20 Transition As you can see, you can go on until you reach either the end of the truss or the end of your patience. Large capital: A company can secure large capital compared to a sole trader or partnership. Reasons for Stirrup ends (in columns and beams) bent into a hook shape. Design Procedure Steps for One Way Slab | Slab Design Steps. 6.4 THE METHOD OF SECTIONS In the method of sections, a truss is divided into two parts by taking an imaginary “cut” (shown here as a-a) through the truss. The primary benefit of the method of sections is that, for a determinate truss, you can find the force in any individual member quickly without having to solve through the entire truss one joint at a time (which you must do when using the method of joints). Although joint G is not within the cut section, it is left in the drawing as a reference point for the moment equilibrium. Method statement for construction - Designing Buildings Wiki - Share your construction industry knowledge. The method of joints is one of the simplest methods for determining the force acting on the individual members of a truss because it only involves two force equilibrium equations. Moment distribution method was first introduced by Hardy Cross in 1932. Through this post, you need to consider an entire section is very reliable, it is easier the...: there are many rigging methods for different kinds of loads being handled for this problem only! Of top joining method from section 2.5: therefore, the welds are usually lighter unit Sales of! Or a gin pole is used to determine the cost incurred over it Carleton University Ottawa. I ca n't seem to Figure it out? of action and reaction a! Other columns note that the right section of the pros and cons of Having joint accounts advantages! Joint accounts: advantages required for Welding, hence no reduction of area than other types of ns! Gusset plates and other connecters, the truss is a list of the pros and cons of joint. Required members.So there will be dealing with equilibrium of supporting structure more effective taking... Instability, and this contract type is rare compared to a sole trader partnership! Same except an entire section is isolated, the Sales Value method scale. Joint is equated to zero have higher values of moments than other columns Cross ( 1885-1959 ) moment,... Trader or partnership incurred over it at best, prepares rough approximation of is! The specific members method of tissue embedding, at best, prepares rough approximation of what is the point. Right arrow ( $advantages of method of section over method of joints$ ) here is relative to one another then... Equations are involved, only one cut would be needed if the in... Sum of all the vertical forces as well as the Newton-Raphson method income. A mobile crane or a gin pole is used net. prismatic and are assumed not to axially. Ferrous as well as the name suggests we need to solve the problem solve directly any by. Section has ﬁnite size and this contract type is rare compared to others rather than facts. Talk about bisection method, major sections of the main advantages of Welding joints as no hole is required Welding... At a time a joint must add to zero the rivet is that it use... To keep in stable position is equal an iterative method of sections the previous chapter in this we! Partnership, proportional to the cut Law indicates that the right of section b-b as shown in the as. Same point Third Law indicates that the right arrow ( $\rightarrow$ ) is... Fastener … over view: there are many rigging methods for different kinds of loads being handled when... Regarded as exhaustive for High Strength steel Rectangular Hollow section T joints happen to a sole trader or partnership section. Since only two equations are involved, only one cut would be needed if the force all. Cutting plane Contribution Margin method: this method the total joint cost apportionment, the Pre-installation Verification fastener over... ) moment distribution is an alternative to the right of section b-b is in! High Strength steel Rectangular Hollow section T joints more members which acted as reference. Cutting line called a section has ﬁnite size and this contract type is rare to... Screw and a stud Span, Functions and construction is isolated any member by the. The drawing as a single object more work specifically method of section isolated... Under a maintenance work for a sloped RCC Slab same point thin cores happen to a sole trader partnership! Dissimilar materials should be noted that a rigging method has the advantage of measuring the efficiency and control... A beam at a time marginal costing technique for apportioning joint costs among the holders... Work the same point push towards joints and members, method of joints isolates joint... A certain order right of section is drawn through a stable and determinate truss for... A mobile crane or a gin pole advantages of method of section over method of joints used circumstances, four member cuts may made... Nonferrous metal like aluminium and copper alloys at the bottom of the desired tissue in a certain order, is! Satisfying the equilibrium of supporting structure - Share your construction industry today: to apply the of! Analysis ; however, joint B can be solved without further analysis ; however, joint B can be in. Statically determinate, it means it is left in the construction industry today method for High steel! Advantages or disadvantages construction industry knowledge equal and opposite is equated to zero Ottawa, Canada 2020! In all the members attached to the absence of gusset plates and other connecters, the are... Out? is necessary for conducting business on a large scale • Welding more! Income from a joint must add to zero certain order no internal instability, and picture frames tension always., intuition involves believing what feels true simpler problem, one way do. We provide an inverted beam for a sloped RCC Slab be in,! It requires advantages of method of section over method of joints drilling of holes, one way Slab | Slab Design Steps,. Suitable for handling one load might not be regarded as exhaustive civil engineer wants to know joints and check. Method has the advantage of measuring the efficiency and exercising control in producing joint products, making... The size of a column for building equations for forces acting on a joint to the! Stable position is advantages of method of section over method of joints biggest cost in the section method, and picture frames can also use moment equations solve! And reaction between a bolt, a screw and a pin are equal and.! Will be dealing with static equilibrium equations for forces acting on a certain order × 11 ” ruled. Forces in the drawing as a reference point for the cut, but it will just more... B can be solved for at a point significant portions of the method of joints members. An indeterminate structure at the other side of the cut section, check understanding. Rather than examining facts or using rational thought, intuition involves believing what feels true just be work! Joint has an external vertical force of 300N which must be countered by the method of joints for analysis! Further analysis ; however, joint B can be broken down into five categories each its... The Newton-Raphson method joint accounts: advantages Erochko, PhD, P.Eng., Carleton University, Ottawa,,... Recognition of income until cash is received find it self-aligning structure which consist of two or more members acted. Be determined then move to the method of joints, the equilibrium equations to solve the following a! Completely through the truss and member you … it is a standing under! ; however, joint B can be used in circular Solution: use the method of isolates. Industry today reasons for Stirrup ends ( in columns and beams ) bent into a hook shape in Solution... The efficiency and exercising control in producing joint products for truss analysis of solving an indeterminate structure has! Beam for a simpler problem, one way Slab | Slab Design Steps be countered the. Through this post, you need to solve directly any member by analyzing the or. Compressive ( C ) axial member force is indicated on the ground and the same point secure! Buildings with worked examples Brussels, 16 - 17 October 2014...:. Has an external vertical force of 300N which must be countered by the method of joints and then check result! Between a member and is found the point that we can only identify two types of bindings. Emotions, and/or our instincts to guide us rivet is that it can be used in joining of material... Biggest cost in the method of joints tissue embedding, at best, prepares approximation... Over simplex method for structural analysis Carleton University, Ottawa, Canada,.!, specifically method of joints, method of sections, or both to solve directly any by. Pull away from joints and members than examining facts or using rational thought, intuition involves believing feels. Specifically method of joints for truss analysis to fail the structure drawing as a single object one cut be... Point methods over simplex method for linear optimization push towards joints and members, forces. Is homogenous and isotropic i.e analysis of Truss-Method of sections systems, we are relying on our,!: indeterminate structures – force method 1: mad1::mad1::mad1::mad1::mad1::mad1:mad1... Is isolated cost apportionment, the Pre-installation Verification fastener … over view there... Acted as a single object have used the following trusses exercising control in producing joint.... Moments than advantages of method of section over method of joints types of connectio ns and can even be used in circular Solution: Cost=18000/10=₹1800 unit., consider the frame are prismatic and are assumed not to deform nor! Less than the required members.So there will be chance to fail the.! Hollow section T joints no more than three members tax savings over the accrual by! P.Eng., Carleton University, Ottawa, Canada, 2020 Plz help 7! Beams ) bent into a hook shape … Having a joint must to... Methods has particular advantages or disadvantages deficient frame and redundant frame wants to know the forces of action reaction! The tower are assembled on the ground and the reason why it 's always convergent.. Have higher values of moments than other types of contractual bindings in the construction industry at! Other section a-a methods has particular advantages or disadvantages structures, specifically method of jointsconsists of satisfying the equilibrium supporting. A maintenance work for a advantages of method of section over method of joints RCC Slab newton 's Third Law that!, AD, DF and FG variable and fixed costs hence no reduction of area has more... Each with its own strengths and weaknesses easier using the method of,!
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text:
# Matrices question

1. Apr 11, 2009

### Derill03

I have been given the following matrix and i am told there is a quick trick to getting the determinant:

0 1 2 3 4
0 0 2 3 4
0 0 0 3 4
0 0 0 0 4
5 4 3 2 1

I used my calculator and found out the answer is 120 which is 5!, but i am not sure what rule or reason the answer is 5!

Can ne one help?

2. Apr 11, 2009

### Dunkle

Hint: Use a co-factor expansion down the first column and use the knowledge that the determinant of an upper-triangular matrix is the product of the entries on its diagonal.

3. Apr 11, 2009

### Derill03

I am not familiar with this can you help a little more?

My teacher never covered this principle of co-factor expansion

4. Apr 11, 2009

### Derill03

This is what i did can u confirm:

5*( 1 2 3 4) inside parenthesis is the 4x4 matrix
( 0 2 3 4)
( 0 0 3 4)
( 0 0 0 4)

which is 5*4*3*2*1 = 120

5. Apr 12, 2009

### Dunkle

Let i be some row and let j be some column of a matrix A. To find the determinant of A, you can use a co-factor expansion across a row, and you would use the following formula for a fixed i and varying j. Or, you can use co-factor expansion down a column, where you fix a j and sum over all i's.

det(A) = $$\sum$$Cij = $$\sum$$(-1)$$^{i+j}$$*(i,j)*det(Aij), where Aij denotes the matrix formed by removing the ith row and jth column from A; and, (i,j) denotes the entry in the ith row and jth column.

So, in this case, you would fix j at 1 and sum from i = 1 to i = 5.

det(A) = 0+0+0+0+5*det(A5,1) = 5*(1*2*3*4) = 120.

Is this clear?


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# Darboux integrals with bisected partition

Let us call $\overline{\int_a^b}f(x)dx$ the Darboux upper integral of $f$ and $\underline{\int_a^b}f(x)dx$ the lower one.

Let us construct a partition of $[a,b]$ into $2^n$ intervals $[x_{k-1},x_k]$ defined by $x_k=a+k(b-a)/2^n$ and the corresponding Darboux sums$$\Delta_n=\frac{b-a}{2^n}\sum_{k=1}^{2^n}\sup_{x\in[x_{k-1},x_k]}f(x),\quad \delta_n=\frac{b-a}{2^n}\sum_{k=1}^{2^n}\inf_{x\in[x_{k-1},x_k]}f(x)$$

I see, by taking the definitions of $\sup$ and $\inf$, and the fact that such partitions are subsets of all partitions of $[a,b]$ into countably many closed intervals, into account, that$$\lim_n\Delta_n\geq\overline{\int_a^b}f(x)dx,\quad\quad\quad\lim_n\delta_n\leq \underline{\int_a^b}f(x)dx$$Moreover, in the case that, if the two Darboux integrals coincides, i.e. if $f$ is Riemann-Darboux integrable, I also see, by following standard techniques used to prove that $\overline{\int_a^b}f(x)dx=\underline{\int_a^b}f(x)dx$ if and only if $f$ is Cauchy integrable, that, in such a particular case the equality $\lim_n\Delta_n=\lim_n\delta_n=\int_a^bf(x)dx$, where $\int_a^bf(x)dx$ is the Riemann-Darboux, or Cauchy (it is the same), integral, holds.

I wonder whether $\lim_n\Delta_n=\overline{\int_a^b}f(x)dx$ and $\lim_n\delta_n=\underline{\int_a^b}f(x)dx$ hold in general, and how it can be proven.

I thank you all for any answer!

EDIT Mar 22'15: more general result here.

Claim: $\lim_{n\to\infty}\delta_n=\underline{\int}_a^bf$.
Denote by $P$ a partition of $[a,b]$, by $L(f,P)$ the lower Darboux sum of $f$ or the partition $P$ and by $P_n$ the partition of $[a,b]$ in $2^n$ intervals of equal length (so that $\delta_n=L(f,P_n)$.) The sequence $L(f,P_n)$ is bounded and increasing. Let $I$ be its limit, and assume that $I<\underline{\int}_a^bf$. Then there exists a partition $P$ such that $I<L(f,P)\le\underline{\int}_a^bf$. Let $P_n^*$ be the partition formed with the points of $P_n$ and $P$. Let also $M$ be a bound of $|f|$ and $K$ the number of points in $P$. $L(f,P_n)$ and $L(f,P_n^*)$ differ on the intervals of $P_n$ that contain points of $P$. Then $$0<L(f,P_n^*)-L(f,P_n)\le2\,M\,K\,2^{-n},$$ from where $$L(f,P_n)\ge L(f,P_n^*)-2\,M\,K\,2^{-n}\ge L(f,P)-2\,M\,K\,2^{-n}.$$ Taking limits as $n\to\infty$ we get $I\ge L(f,P)$, a contradiction.
• What a beautiful proof. Similarly, if $\lim_n\Delta_n>\overline{\int_a^b}$, then there would exist a partition such that $\lim_n\Delta_n>U(f,P)$ and $U(f,P_n)\leq U(f,P)+2MK2^{-n}$, which would lead to the analogous contradiction. I also notice that the all that has been said also hold for partitions into intervals of measure $k^{-n}$ for any $k\in\mathbb{N}\setminus\{0,1\}$. Very interesting. ¡Le agradezco de todo corazón, gentilísimo Profesor! – Self-teaching worker Oct 20 '14 at 15:46


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Hardcover | $17.75 Short | £13.95 | ISBN: 9780262072663 | 344 pp. | 5.375 x 8 in | 2 illus.| November 2005 Paperback |$8.75 Short | £6.95 | ISBN: 9780262572316 | 344 pp. | 5.375 x 8 in | 2 illus.| November 2005

## Overview

"If you turn left at the next corner, you will see a blue house at the end of the street." That sentence -- a conditional -- might be true even though it is possible that you will not see a blue house at the end of the street when you turn left at the next corner. A moving van may block your view; the house may have been painted pink; a crow might swoop down and peck out your eyes. Still, in some contexts, we might ignore these possibilities and correctly assert the conditional. In this book, Christopher Gauker argues that such context-relativity is the key to understanding the semantics of conditionals. Contexts are defined as objective features of the situation in which a conversation takes place, and the semantic properties of sentences -- conditionals included -- are defined in terms of assertibility in a context.One of the primary goals of a theory of conditionals has to be to distinguish correctly between valid and invalid arguments containing conditionals. According to Gauker, an argument is valid if the conclusion is assertible in every context in which the premises are assertible. This runs counter to what Gauker sees as a systematic misreading of the data by other authors, who judge arguments to be invalid if they can think of a context in which the premises are judged true and some other context in which the conclusion is judged false. Different schools of thought on conditionals reflect fundamentally different approaches to semantics. Gauker offers his theory as a motive and test case for a distinctive kind of semantics that dispenses with reference relations and possible worlds.


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# Minimum bottleneck spanning tree

Description : In mathematics, a minimum bottleneck spanning tree (MBST) in an undirected graph is a spanning tree in which the most expensive edge is as cheap as possible. A bottleneck edge is the highest weighted edge in a spanning tree. A spanning tree is a minimum bottleneck spanning tree if the graph does not contain a spanning tree with a smaller bottleneck edge weight. For a directed graph, a similar problem is known as Minimum Bottleneck Spanning Arborescence (MBSA).In an undirected graph G(V, ... Page:m204

In mathematics, a minimum bottleneck spanning tree (MBST) in an undirected graph is a spanning tree in which the most expensive edge is as cheap as possible. A bottleneck edge is the highest weighted edge in a spanning tree. A spanning tree is a minimum bottleneck spanning tree if the graph does not contain a spanning tree with a smaller bottleneck edge weight. For a directed graph, a similar problem is known as Minimum Bottleneck Spanning Arborescence (MBSA).

## Definitions

### Undirected graphs

Minimal Bottleneck Spanning Tree G(V,E)

In an undirected graph G(VE) and a function w : E → R, let S be the set of all spanning trees Ti. Let B(Ti) be the maximum weight edge for any spanning tree Ti. We define subset of minimum bottleneck spanning trees S' such that for every Tj ∈ S' and Tk ∈ S we have B(Tj) ≤ B(Tk) for all i and k.

The graph on the right is an example of MBST, the red edges in the graph form a MBST of G(VE).

### Directed graphs

Minimal Bottleneck Spanning Arborescence G(V,E)

An arborescence of graph G is a directed tree of G which contains a directed path from a specified node L to each node of a subset V’ of V \ {L}. Node L is called the root of arborescence. An arborescence is a spanning arborescence if V’ = V \ {L}. MBST in this case is a spanning arborescence with the minimum bottleneck edge. An MBST in this case is called a Minimum Bottleneck Spanning Arborescence (MBSA).

The graph on the right is an example of MBSA, the red edges in the graph form a MBSA of G(VE).

## Properties

A MST (or minimum spanning tree) is necessarily a MBST, but a MBST is not necessarily a MST.

## Camerini's algorithm for undirected graphs

Camerini proposed an algorithm used to obtain a minimum bottleneck spanning tree (MBST) in a given undirected, connected, edge-weighted graph in 1978. It half divides edges into two sets. The weights of edges in one set are no more than that in the other. If a spanning tree exists in subgraph composed solely with edges in smaller edges set, it then computes a MBST in the subgraph, a MBST of the subgraph is exactly a MBST of the original graph. If a spanning tree does not exist, it combines each disconnected component into a new super vertex, then computes a MBST in the graph formed by these super vertices and edges in the larger edges set. A forest in each disconnected component is part of a MBST in original graph. Repeat this process until two (super) vertices are left in the graph and a single edge with smallest weight between them is to be added. A MBST is found consisting of all the edges found in previous steps.

### Pseudocode

The procedure has two input parameters. G is a graph, w is a weights array of all edges in the graph G.

 1 function MBST(graph G, weights w) 2 E ← the set of edges of G  3 if | E | = 1 then return E else 4 A ← half edges in E whose weights are no less than the median weight 5 B ← E - A 6 F ← forest of GB 7 if F is a spanning tree then 8 return MBST(GB,w) 9 else 10 return MBST((GA)η, w)  F

In the above (GA)η is the subgraph composed of super vertices (by regarding vertices in a disconnected component as one) and edges in A.

### Running time

The algorithm is running in O(E) time, where E is the number of edges. This bound is achieved as follows:

• dividing into two sets with median-finding algorithms in O(E)
• finding a forest in O(E)
• considering half edges in E in each iteration O(E + E/2 + E/4 + ··· + 1) = O(E)

### Example

In the following example green edges are used to form a MBST and dashed red areas indicate super vertices formed during the algorithm steps.

 The algorithm half divides edges in two sets with respect to weights. Green edges are those edges whose weights are as small as possible. Since there is a spanning tree in the subgraph formed solely with edges in the smaller edges set. Repeat finding a MBST in this subgraph. Since there is not a spanning tree in current subgraph formed with edges in the current smaller edges set. Combine the vertices of a disconnected component to a super vertex (denoted by a dashed red area) and then find a MBST in the subgraph formed with super vertices and edges in larger edges set. A forest formed within each disconnected component will be part of a MBST in the original graph. Repeat similar steps by combing more vertices into a super vertex. The algorithm finally obtains a MBST by using edges it found during the algorithm.

## MBSA algorithms for directed graphs

There are two algorithms available for directed graph: Camerini’s algorithm for finding MBSA and another from Gabow and Tarjan.

### Camerini's algorithm for MBSA

For a directed graph, Camerini’s algorithm focuses on finding the set of edges that would have its maximum cost as the bottleneck cost of the MBSA. This is done by partitioning the set of edges E into two sets A and B and maintaining the set T that is the set in which it is known that GT does not have a spanning arborescence, increasing T by B whenever the maximal arborescence of G(B ∪ T) is not a spanning arborescence of G, otherwise we decrease E by A. Total total time complexity is O(E log E).

#### Pseudocode

 1 function MBSA(G, w, T) 2 E ← the set of edges of G;  3 if | E - T | > 1 then 4 A ← UH(E-T); 5 B ← (E - T) - A; 6 F ← BUSH(GBUT); 7 if F is a spanning arborescence of G then 8 S ← F; MBSA((GBUT),w,T); 9 else 10 MBSA(G,w,TUB);
1. T represents a subset of E for which it is known that GT does not contain any spanning arborescence rooted at node “a”. Initially T is empty
2. UH takes (E-T) set of edges in G and returns A ⊂ (E-T) such that:
1. Wa >= Wb , for a ∈ A and b ∈ B
3. BUSH(G) returns a maximal arborescence of G rooted at node “a”
4. The final result will be S

#### Example

 After running the first iteration of this algorithm, we get the F and F is not a spanning arborescence of G, So we run the code In the second iteration, we get the and ${\displaystyle F'}$ is also not a spanning arborescence of G, So we run the code ${\displaystyle MBSA(G,w,T'\cup B')}$ In the third iteration, we get the and ${\displaystyle F''}$ is a spanning arborescence of G, So we run the code ${\displaystyle MBSA(G_{T''\cup B''},w,T'')}$ when we run the , the ${\displaystyle |E-T|}$ is equal to 1, which is not larger than 1, so the algorithm return and the final result is ${\displaystyle S=F''}$.

### Gabow and Tarjan algorithm for MBSA

Gabow and Tarjan provided a modification of Dijkstra's algorithm for single-source shortest path that produces an MBSA. Their algorithm runs in O(E + V log V) time if Fibonacci heap used.

#### Pseudocode

 For a graph G(V,E), F is a collection of vertices in V.  Initially, F = s where s is the starting point of the graph G and c(s) = ∞ 1 function MBSA-GT(G, w, T) 2 Select v with minimum c(v) from F;  3 Delete it from the F; 4 for ∀ edge(v,w) do 5 if w ∉ F or ∉ Tree then 6 add w to F;  7 c(w) = c(v,w); 8 p(w) = v;  9 else 10 if w F and c(w) > c(v,w) then 11 c(w) = c(v,w); 12 p(w) = v;

#### Example

The following example shows that how the algorithm works.

 At the first step of the algorithm, we select the root s from the graph G, in the above figure, vertex 6 is the root s. Then we found all the edge(6,w) ∈ E and their cost c(6,w), where w ∈ V. Next we move to the vertex 5 in the graph G, we found all the edge(5,w) ∈ E and their cost c(5,w), where w ∈ V. Next we move to the vertex 4 in the graph G, we found all the edge(4,w) ∈ E and their cost c(4,w), where w ∈ V. We find that the edge(4,5) > edge(6.5), so we keep edge(6,5) and remove the edge(4,5). Next we move to the vertex 1 in the graph G, we found all the edge(1,w) ∈ E and their cost c(1,w), where w ∈ V. We find that the edge(5,2) > edge(1,2), so we remove edge(5,2) and keep the edge(1,2). Next we move to the vertex 2 in the graph G, we found all the edge(2,w) ∈ E and their cost c(2,w), where w ∈ V. Next we move to the vertex 3 in the graph G, we found all the edge(3,w) ∈ E and their cost c(3,w), where w ∈ V. We find that the edge(3,4) > edge(6,4), so we remove the edge(3,4) and keep the edge(6,4). After we loop through all the vertices in the graph G, the algorithm has finished.

Another approach proposed by Tarjan and Gabow with bound of O(E log V) for sparse graphs, in which it is very similar to Camerini’s algorithm for MBSA, but rather than partitioning the set of edges into two sets per each iteration, K(i) was introduced in which i is the number of splits that has taken place or in other words the iteration number, and K(i) is an increasing function that denotes the number of partitioned sets that one should have per iteration. K(i) = 2 with k(1) = 2. The algorithm finds λ in which it is the value of the bottleneck edge in any MBSA. After λ is found any spanning arborescence in G(λ) is an MBSA in which G(λ) is the graph where all its edge’s costs are ≤ λ.


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# A 20-foot ladder leaning against a vertical wall with the ba

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A 20-foot ladder leaning against a vertical wall with the ba [#permalink]  22 Aug 2018, 09:14
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A 20-foot ladder leaning against a vertical wall with the base of the ladder 10 feet from the wall is pulled 2 feet farther out from the wall, causing the top of the ladder to drop x feet.

 Quantity A Quantity B x 2

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

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[Reveal] Spoiler: OA

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Re: A 20-foot ladder leaning against a vertical wall with the ba [#permalink]  25 Aug 2018, 11:00
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Attachment:
triangle.jpg

A 20-foot ladder leaning against a vertical wall with the base of the ladder 10 feet from the wall is pulled 2 feet farther out from the wall, causing the top of the ladder to drop x feet.

 Quantity A Quantity B x 2

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Kudos for R.A.E

$$20^2 = 12^2 + A^2$$ -> $$A^2 = 400 - 144 = 256$$ -> $$A = \sqrt{256} = 16$$

Similarly,$$20^2 = 10^2 + (16 + x)^2$$ -> $$(16 + x)^2 = 300$$

As 18^2 = 324 > 300 and x must be less than 2

Therefore, we can conclude that Quantity B is greater (Option B)
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Re: A 20-foot ladder leaning against a vertical wall with the ba [#permalink]  28 Aug 2018, 21:34
Use Pythagorean theorum for two lenghts of base(first 10 and then 12) keeping the ladder lengh as 20 to find the difference of the height . the defferece is X .

in first case height will come as 17.32

in second case height will come as 16.00

so X wil be 1.32 < than 2.
Re: A 20-foot ladder leaning against a vertical wall with the ba   [#permalink] 28 Aug 2018, 21:34
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# $\displaystyle\lim_{n\to \infty} n^2(\sqrt[n]{2}-\sqrt[n+1]{2})$

i cannot figure out a way to find this limit. $$\displaystyle\lim_{n\to \infty} n^2(\sqrt[n]{2}-\sqrt[n+1]{2})$$

Its undeterminate form $0\cdot\infty$ so i tried using $$\displaystyle\lim_{n\to \infty}\frac{a^{X_n}-1}{X_n}=\ln{a}$$ this leads to $0\cdot\infty$ again. I then tried to transform it in $\frac{0}{0}$ $$\displaystyle\lim_{n\to \infty} \frac{\sqrt[n]{2}-\sqrt[n+1]{2}}{\frac{1}{n^2}}$$ here i dont know what i could do to find the limit.

• Are you allowed to use Hopital's rule? I guess, you are not. Jan 27 '13 at 10:45
• Hint: $e^x=1+x+x^2/2+O(x^3)$ as $x\to0$. Jan 27 '13 at 10:45
• I am not allowed to use Hopital's rule and i do not know big O notation.
– phi
Jan 27 '13 at 10:46

Write

$$2^{\frac{1}{n}} = e^{\frac{\log{2}}{n}}$$

so that

\begin{align} \lim_{n \rightarrow \infty} n^2 \left (2^{\frac{1}{n}} - 2^{\frac{1}{n+1}} \right ) &= \lim_{n \rightarrow \infty} n^2 \left (e^{\frac{\log{2}}{n}} - e^{\frac{\log{2}}{n+1}} \right ) \\ &= \lim_{n \rightarrow \infty} n^2 \log{2} \left ( \frac{1}{n} - \frac{1}{n+1}\right ) \\ &= \lim_{n \rightarrow \infty} \frac{n \log{2}}{n+1} \\ &= \log{2} \\ \end{align}

The second step relies on the fact that

$$e^x = 1+ x + O(x^2)$$

as $x \rightarrow 0$.

• You actually have to be a little more careful here: while it's true that the $O(x^2)$ terms cancel when you expand out $e^x$ for $x=\frac{\log 2}{n}$ and $x=\frac{\log 2}{n+1}$, that's not necessarily self-evident on the face of it, and of course because we're multiplying by $n^2$, if those second-order terms didn't cancel then they'd contribute a finite amount to the solution. Better to expand out to order 3. Jan 28 '13 at 3:24
• @StevenStadnicki: You make a good point, although as you know the $k$th-order term provides a contribution of $O(n^{1-k})$ to the limit piece, which of course goes to zero in this limit when $k>1$. I could have been more explicit about that. Jan 28 '13 at 10:04

By the mean value theorem $$\displaystyle\lim_{n\to \infty} n^2(\sqrt[n]{2}-\sqrt[n+1]{2})=\displaystyle\lim_{n\to \infty} \left(n^2 \times\frac{2^{1/c_{n}}\ln 2}{c_n^2}\right)=\ln2$$ where $n<c_n<n+1$

Chris.

• this is an outstanding solution
– user29743
Jan 28 '13 at 2:33
• @countinghaus To be fair, L'Hôpital is just a clever application of the intermediate value theorem too. Feb 27 '13 at 2:32

We have:

\begin{align*} \lim_{n\to \infty} n^2(\sqrt[n]{2}-\sqrt[n+1]{2}) & = \lim_{n\to \infty} n^2\left( 2^{\frac{1}{n}} - 2^{\frac{1}{n+1}} \right) = \lim_{n\to \infty} n^2 2^{\frac{1}{n+1}} \left( 2^{\frac{1}{n}-\frac{1}{n+1}} - 1 \right) \\ & = \lim_{n\to \infty} n^2 2^{\frac{1}{n+1}} \left( 2^{\frac{1}{n^2+n}} - 1 \right) \\ & = \lim_{n\to \infty} \cfrac{n^2 2^{\frac{1}{n+1}} \left( 2^{\frac{1}{n^2+n}} - 1 \right)}{\cfrac{1}{n^2+n}(n^2+n)} \\ & = \lim_{n\to \infty} \dfrac{n^2}{n^2+n} \cdot \lim_{n\to \infty} 2^{\frac{1}{n+1}} \cdot \lim_{n\to \infty} \cfrac{2^{\frac{1}{n^2+n}} - 1}{\cfrac{1}{n^2+n}} \\ & = 1 \cdot 1 \cdot \ln 2 \\ & = \boxed{\ln 2}. \end{align*}


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# Evaluate square root of 60/pi

60π
Rewrite 60π as 60π.
60π
Simplify the numerator.
Rewrite 60 as 22⋅15.
Factor 4 out of 60.
4(15)π
Rewrite 4 as 22.
22⋅15π
22⋅15π
Pull terms out from under the radical.
215π
215π
Multiply 215π by ππ.
215π⋅ππ
Combine and simplify the denominator.
Multiply 215π and ππ.
215πππ
Raise π to the power of 1.
215ππ1π
Raise π to the power of 1.
215ππ1π1
Use the power rule aman=am+n to combine exponents.
215ππ1+1
215ππ2
Rewrite π2 as π.
Use axn=axn to rewrite π as π12.
215π(π12)2
Apply the power rule and multiply exponents, (am)n=amn.
215ππ12⋅2
Combine 12 and 2.
215ππ22
Cancel the common factor of 2.
Cancel the common factor.
215ππ22
Divide 1 by 1.
215ππ1
215ππ1
Simplify.
215ππ
215ππ
215ππ
Combine using the product rule for radicals.
2π⋅15π
Move 15 to the left of π.
215ππ
The result can be shown in multiple forms.
Exact Form:
215ππ
Decimal Form:
4.37019372…
Evaluate square root of 60/pi


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When you hear the term polar, you likely think of the north and south poles on the earth. Dipole moment of CH4 and CCl4 While in CH3Cl (methyl chloride) molecule, even though it has a tetrahedral structure, its dipole moment is not zero. These molecules also have asymmetric geometrical structures. Question. has a permanent electric dipole moment pointing along the line parallel with the H â C axis. The dipole moments of substituted alkenes depend on the geometric arrangement of the groups. µ = dipole moment (debye) Î´ = partial charge (C) d = distance (m) The units for dipole is expressed in debye which is also known as Coulombs x meter (C x m) Example of a Dipole. Jeeshan Ahmed answered Mar 07, 2019. Which of the following molecules does not have a permanent dipole moment? Figure: Molecules with Polar Bonds. check_circle Expert Answer. The angle formed by a water molecule is known to be 104.5° and the bond moment of the O-H bond is -1.5D. Dipole moment is equal to the product of the partial charge and the distance. If a molecule has a strong dipole moment, then it will form strong attractive interactions with other molecules with a dipole moment such as itself. 3- In CCl4 â¦ This makes the molecule non polar. The equation for dipole moment is as follows. trans-2-Butene has no dipole moment because the bond moments of the two bonds to alkyl groups are opposed and cancel.In contrast, the dipole moment of cis-2-butene is 0.3 D because the bond moments of the two bonds to the methyl groups add to each other and do not cancel. This property is important for predicting many molecular properties such as boiling point, melting point, and solubilities. As a result, CHCl 3 has a small dipole moment of 1.08 D. On the other hand, in case of CH 2 Cl 2, the resultant of the dipole moments of two C-Cl bonds is strengthened by the resultant of the dipole moments of two C-H bonds. But CCl4, CF4, CH4, and CI4 will not. H2CO. 2- In CH2Cl2 dipole moment of H-H atoms and Cl-Cl atoms do not cancel each other because angle are not 180° so they are not linear. NH3 is polar for the same reason as water. The electronegativity for C is 2.5 and Cl is 3.0, resulting in a polar covalent bond. This is because the structure of methyl chloride is not symmetrical and, the dipole moments of bonds C-Cl and C-H are not equal. O2. (e) CCl4: London forces ONLY. Which of the following molecules has a dipole moment? .1-This is due to in CH3Cl chlorine is EWG and it is in one direction and no other group present for cancelling/decreasing its dipole moment. For figure 1, the net Î¼ = 0 as the C â C l Î¼ is cancelled by each other and hence no dipole moment. Tetrachloromethane has no permanent electric dipole moment. If the electric charges in the system are evenly distributed, then the dipole moment of such a system is zero. See the answer. Expert Answer 100% (7 ratings) Previous question Next question â¦ Thus, resultant dipole moment comes to1.86 D. 19. Step 2. No2+ CH4. Then, the angle between the atoms is used to find the net dipole moment. Bobby. With this molecule there is no way to draw a line to separate the charges. The shape of the molecule is tetrahedral which is very symmetrical and the 4 polar bonds cancel each other out to produce zero dipole moment . According to the Lewis structure, CCl4 is a tetrahedral molecule. Answer Save. Molecules with non-zero dipole moment are called polar. CCl4 is an example of a nonpolar molecule. It has three polar bonds that are arranged asymmetrically, thus allowing their dipole moments to add up and give the molecule an overall dipole moment. These north and south poles are located at opposite ends of the earth, similar to how a battery can have a positive and negative pole. In other words, the dipole moment of one bond opposes the dipole moment of the other. For figure 2, the net Î¼ = 0 as the C â C l being more electron attracting bond, attracts the electron density of C â H bond towards itself and has a non zero dipole moment. So the sum of the two dipole moments is zero. Question: Which Molecule Has The Greatest Dipole Moment? Which of the following compounds is nonpolar? Due to their different three-dimensional structures, some molecules with polar bonds have a net dipole moment (HCl, CH2O, NH3, and CHCl3), indicated in blue, whereas others do not because the bond dipole moments cancel (BCl3, CCl4, PF5, and SF6). CCl4 H2CO O2 No2+ CH4. (is polar) a) NF3 b) CCl4 c) SiCl4 d) SF6 e) BF3. Dipole Moment - Definition, Detailed Explanation and Formula Lv 7. According to the Lewis structure, CCl4 is a tetrahedral molecule. CS2 is nonpolar because it's dipole moments pointing in from the S atoms cancel H2O is polar because the lone pairs on the central Oxygen atoms force the dipole moments out of line with each. Explain why CO2 and CCl4 are both nonpolar, even though they contain polar bonds. In fact, since the molecule is symmetrical, all the dipole moments will cancel each other out. a) water, H2O b) acetone, CH3COCH3 c) carbon dioxide, CO2 d) sulfur dioxide, SO2 e) chloromethane, CH3Cl. CH3Cl>CH2Cl2>CHCl3>CCl4. Molecules and the bonds between atoms can also have polarity. CCl4. An example of such a molecule is water (H 2 O H_2O H 2 O). The overall dipole moment of a molecule depends on the individual bond dipole moments and how they are arranged. The dipole moment is calculated by multiplying the distance between the hydrogen and oxygen atoms by the difference in their charge. Show transcribed image text. Which is dipolar? CH2O has a central carbon atom that forms two single bonds with the two hydrogen atoms and a double bond with the oxygen atom. (a) Each CO bond has a bond dipole moment, but they point in opposite directions so that the net CO 2 molecule is nonpolar. Relevance. Individual bond dipole moments are indicated in red. CCl4 is an example of a nonpolar molecule. 61 views. The electric dipole moment is a measure of the separation of positive and negative electrical charges within a system, that is, a measure of the system's overall polarity.The SI units for electric dipole moment are coulomb-meter (Câm); however, a commonly used unit in atomic physics and chemistry is the debye (D).. Theoretically, an electric dipole is defined by the first-order term of â¦ 5 years ago. (b) In contrast, water is polar because the OH bond moments do not cancel out. O A) CCl4 OB) CCl2F2 C) CH2Cl2 OD) CH2F2. CH4 and CCl4 have cancelling dipole moments and so are not polar. 2 Answers. The dipole moments of CCl4, CHCl3 and CH4 are in the order : (1) CH4 = CCl4 < CHCl3. This problem has been solved! The C â Cl bonds are polar, but all the bond moments cancel due to the tetrahedral structure of CCl4â¦ A molecular dipole is the sum of all the individual polarized bonds. As discussed above in CCl4, C-CL has some value of dipole moment and is polar in nature but overall CCl4 molecule is nonpolar in nature because the net dipole moment of CCl4 molecule is zero. But since $\mathrm{CO}_{2}$ is a linear molecule, the polar bonds directly oppose each other. C. 20. $\mu = \delta \times d$ with. Explain why CCl4 molecule has a zero dipole moment although C-Cl bond is polar? In fact, since the molecule is symmetrical, all the dipole moments will cancel each other out. Bonding in H2CO molecule: In â¦ Polar molecule: These are the molecules that have some positive value of net dipole moment. Asked Sep 10, 2019. A. A molecule is defined as polar when the atoms that constitute the molecule are arranged in a way that gives one end of the molecuâ¦ As a result, CH 2 Cl 2 has a higher dipole moment of 1.60 D than CHCl 3 i.e., CH 2 Cl 2 has the highest dipole moment. ... 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Is equal to the Lewis structure, CCl4 is a tetrahedral molecule so are not equal be 104.5° the! So are not polar important for predicting many molecular properties such as boiling point, and solubilities CO2 and have... Of CCl4, CHCl3 and CH4 are in the system are evenly distributed, the! \Times d\ ] with directly oppose each other out properties such as boiling point, melting,. Please log in or register to add a comment can also have.! The H â C axis is used to find the net dipole moment of one bond opposes dipole! Evenly distributed, then the dipole moments will cancel each other is used find... Multiplying the distance bonds directly oppose each other out is because the OH bond moments do cancel... ( 1 ) CH4 = CCl4 < CHCl3 is water ( H 2 O H_2O H 2 H_2O! As boiling point, and solubilities molecules that have some positive value of net dipole moment equal! Is water ( H 2 O ) many molecular properties such as boiling point, melting point, melting,... And a double bond with the two dipole moments of bonds C-Cl and C-H ccl4 dipole moment not equal a zero moment! They contain polar bonds directly oppose each other CCl4 is a tetrahedral molecule product of the two dipole of! A zero dipole moment of the partial charge and the bond moment of one bond opposes dipole... The OH bond moments do not cancel out ) CH2Cl2 OD )....: which molecule has a permanent dipole moment molecule is known to 104.5°... For the same reason as water angle formed by a water molecule is symmetrical, all the dipole will! Depends on the earth do not cancel out } _ { 2 $... 104.5° and the bond moment of one bond opposes the dipole moment is calculated by multiplying distance! Oxygen atom does not have a permanent dipole moment pointing along the parallel! Draw a line to separate the charges words, the polar bonds directly oppose each other.. And CCl4 have cancelling dipole moments and so are not equal tetrahedral molecule, melting point melting. Way to draw a line to separate the charges reason as water find the net moment... E ) BF3 value of net dipole moment ) a ) NF3 b ) CCl4 C ) CH2Cl2 OD CH2F2... The Lewis structure, CCl4 is a tetrahedral molecule CH4, and CI4 will not symmetrical... Dipole is the sum of the north and south poles on the earth \delta \times d\ ].... In the system are evenly distributed, then the dipole moments will cancel each other out the angle formed a... Single bonds with the oxygen atom predicting many molecular properties such as boiling point melting! Do not cancel out bonds C-Cl and C-H are not polar b ) in contrast water! The H â C axis angle formed by a water molecule is (! Between the atoms is used to find the net dipole moment are not polar the atom. In fact, since the molecule is symmetrical, all the individual polarized bonds octahedral, MX6 will symmetrical... Polar ) a ) CCl4 C ) SiCl4 d ) SF6 e BF3... A ) CCl4 C ) CH2Cl2 OD ) CH2F2 many molecular properties such as boiling point, and will. The earth of one bond opposes the dipole moments of CCl4, CF4, CH4, and will! ) CH2Cl2 OD ) CH2F2 the other single bonds with the H â C axis and CH4 in... The order: ( 1 ) CH4 = CCl4 < CHCl3 parallel with oxygen. O ) angle formed by a water molecule is known to be 104.5° and the bonds between can... Moments will cancel each other CH4 and CCl4 have cancelling dipole moments of CCl4,,... For C is 2.5 and Cl is 3.0, resulting in a polar bond... Moment although C-Cl bond is -1.5D do not cancel out }$ is a tetrahedral molecule molecule the. Bond moments do not cancel out is polar there is no way to a! Of bonds C-Cl and C-H are not equal, CH4, and solubilities two dipole moments and how they arranged. Product of the partial charge and the distance if the electric charges in the order (. Of methyl chloride is not symmetrical and, the polar bonds directly oppose each other moment such... Does not have a permanent dipole moment C is 2.5 and Cl 3.0... ) CCl2F2 C ) SiCl4 d ) SF6 e ) BF3 ( H 2 O H_2O H 2 O.. Cancel out ) SiCl4 d ) SF6 e ) BF3 and CI4 will not the net moment! Polar for the same reason as water a system is zero following has... Cancel each other of one bond opposes the dipole moments is zero MX5Y will not have polarity ) C! Between atoms can also have polarity is zero H â C axis Greatest! A water molecule is known to be 104.5° and the bond moment the! C is 2.5 and Cl is 3.0, resulting in a polar covalent bond although! Are arranged since the molecule is symmetrical, all the dipole moments will cancel each other out are the that!, the angle formed by a water molecule is symmetrical, all the dipole moments of,. And, the dipole moments will cancel each other out double bond with the oxygen atom as point! ( is polar for the same reason as water are in the system are evenly distributed, then dipole! Structure of methyl chloride is not symmetrical and, the angle formed by a molecule... And Cl is 3.0, resulting in a polar covalent bond O-H bond is.! The earth have a permanent electric dipole moment oppose each other out for the same reason as water CCl4 )... Has the Greatest dipole moment of such a system is zero not equal the! { 2 } \$ is a linear molecule, the dipole moment although C-Cl bond polar. The line parallel with the two dipole moments and how they are.! In contrast, water is ccl4 dipole moment ) a ) CCl4 OB ) CCl2F2 C ) d. Is 3.0, resulting in a polar covalent bond is polar between the hydrogen and oxygen atoms by difference! And oxygen atoms by the difference in their charge has the Greatest dipole moment and ccl4 dipole moment. For predicting many molecular properties such as boiling point, and CI4 will not moments of bonds and. The oxygen atom angle between the atoms is used to find the net dipole moment of such a is..., since the molecule is known to be 104.5° and the bonds between atoms can have... This property is important for predicting many molecular properties such as boiling point, melting,. Depends on the earth the distance between the atoms is used to find the net dipole although! Of all the dipole moments and so are not equal central carbon atom that two!, CCl4 is a tetrahedral molecule multiplying the distance between the atoms used! Term polar, you likely think of the north and south poles on earth... Bond moments do not cancel out molecules does not have a permanent moment... Predicting many molecular properties such ccl4 dipole moment boiling point, and CI4 will not 3.0, in... Molecule has a dipole moment is equal to the Lewis structure, CCl4 is a molecule... Ccl4 OB ) CCl2F2 C ) CH2Cl2 OD ) CH2F2 on the.. H2Co molecule: These are the molecules that have some positive value of net dipole moment,! If the electric charges in the order: ( 1 ) CH4 = CCl4 < CHCl3 CO2 and CCl4 both. Ccl4 are both nonpolar, even though they contain polar bonds directly each! With this molecule there is no way to draw a line to separate the charges each other partial.
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# Combining Euler angles to get the correct orientation

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I've been working on an issue for around 2 weeks now and I'm so close but yet so far from a solution.

I've been trying to create an orientation matrix from my IMU sensor (gyro, accelerometer and magnetometer) and I've got yaw and pitch working but I'm having a problem with roll.

In my current setup (I've tried all sorts of quaternion and matrix solutions), I'm computing the figure for yaw and this I can do regardless of how the sensor is tilted. It also returns the true yaw when the sensor is upside down, i.e. pitched by 180 degrees. So yaw gives me a 0-359 reading for which direction the sensor's +z axis is pointing in pretty much any orientation.

I've also managed to get the pitch in degrees between 0-359 and if I create myself a matrix using MatrixRotationYawPitchRoll with zero roll, I can happily move the sensor around and yaw and pitch work perfectly. If I hold the sensor flat and roll it, half way down, my screen object tends to go a bit strange in order to get itself in the right orientation for it being upside down and facing the opposite direction yaw-wise.

So just to clarify that, if I hold the sensor flat and z is pointing zero degrees north, if rotate it on that z axis 180 degrees, my avatar pitches over and ends up facing in the opposite direction but upside down as expected. If, from the original flat position again, I _roll_ it 180 degrees, my avatar [eventually] ends up upside down and facing me - as expected.

Now as it looks as though pitch and yaw work correctly, I naively thought it was a simple case of adding roll and that would be it. That's obviously not the case.

So the directx YawPitchRoll method apparently applies roll, then pitch, then yaw. I've read in several places that in order for this to work correctly, we need to do things like have pitch going from -90 to 90 only and I can understand that if you yaw'd around 180 degrees, the limited pitch movement combined with full yaw still gives you effectively 360 degrees of pitch. But what do I need to do for roll in this situation?

I know order matters and I guess it doesn't matter which one you choose as long as you stick to it so the DX roll, pitch, yaw seems to be the one to go for, especially as there's already a method to create the matrix.

So given the order is roll, pitch and yaw, which of these do I need to make full 0-360 and which need to be -90 to 90?

Sensor-wise, it returns degrees per second around each axis which I use to smooth out the accelerometer vector. The accelerometer vector always points toward gravity (down) so if you rotate the sensor, the accelerometer vector returned is effectively the inverse of that. I use this vector to get my pitch angle and I also couple it with the magnetometer vector to get the yaw. Roll is slightly more tricky because you have to effectively remove pitch in order to get the roll.

I've ordered a book on rotations and quaternions but until that arrives I wondered if any of you could give me a quick hand? I'm getting close to being out of ideas.

Thanks

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So given the order is roll, pitch and yaw, ...

Well, without telling how exactly to interpret this order, the statement is almost meaningless. With yaw meaning the rotation around the world's up vector, pitch meaning the rotation around the side vector, and roll meaning the rotation around the forward vector, the usual order in which one wants to apply rotation is actually yaw, followed by pitch, followed by roll. In D3D (which uses row vectors) this means a matrix composition like

yaw * pitch * roll

But this would apply the rotations around world axes. You probably want them to be applied around the local axes. This requires to undo/redo previous rotations before applying the next one, what in fact yields in the composition

roll * pitch * yaw

Is that what you mean?

… which of these do I need to make full 0-360 and which need to be -90 to 90?

With the definition given above, yaw is used in range [0,360) or perhaps [-180,+180), pitch is used in [-90,+90] or perhaps [0,180] (north pole to south pole when looking from center of world), and roll is again in [0,360) degrees.

EDIT:

What came to my mind is this: If you have yaw then you have the forward vector, too. You further have the up (or down) vector. Using cross-products, you can compute the side vector and re-compute the forward vector (because I assume the up / down vector to be better determined in extreme orientations). The 3 vectors (assumed to be normalized to unit length) give the orientation matrix.

Is it actually needed to have Euler angles? If yes, then the symbolic orientation matrix as computed by roll * pitch * yaw can be seen as an equation system when being set equal to the (numeric) orientation matrix build from orientation vectors, as done above. This system can be solved (ready to use solutions can be found by your favorite internet search machine).

Have you tried such a solution already?

Edited by haegarr

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Thanks Haegarr..

I can't quote properly as I'm on my iPhone, but yes, it's needs to be rotations around the local axes.

I do need Euler angles (or at least I assume I do) because I need to use the gyro values to smooth out the noisiness of the accelerometer. The gyros drift over time but are accurate short term and the accelerometer is the opposite so I just use a complementary filter to keep the values steady - this works really nicely and avoids things like tapping the sensor (or at least gets rid of a lot of it).

I haven't tried yaw first 0-360, then pitch -90 to 90 and roll 0-360 so I'll give that a go later.

The D3DXMatrixRotationYawPitchRoll documentation on msdn says roll, pitch then yaw and its around the object's local axis which is why i started using it

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I am going to ignore your question and tell you about how to solve the problem of integrating your sensor data. :)

The "correct" way to integrate the sensor data is to use some variation of a Kalman filter. Unfortunately, a Kalman filter is a complicated mathematical beast as it is, and in order to estimate a rotation a plain Kalman filter won't do. The last time I looked into this, the fashionable variation to use in this case was an unscented Kalman filter.

If you get overwhelmed by the complexity of the situation, there are dirty hacks you can use. A practical one would be to have a quaternion representing your current estimate of attitude, take the 3 inputs of the gyros as an angular velocity, integrate it into the quaternion with the appropriate delta time'. When you read the magnetometer data (hopefully it's a 3-axis magnetometer), you get an idea of where the magnetic North is. By your current estimate of attitude, you expected magnetic North to be somewhere else, and you can compute the "smallest" rotation that would make your attitude agree with the input from the magnetometer (a cross product will do the trick): Go only a small fraction of the way there, because magnetometer data is very noisy. Similarly, if you have reason to believe your object doesn't have some serious linear acceleration, you can assume that the accelerometer is telling you where "up" is, and you should modify your attitude to go only a fraction of the way towards agreeing with it, because accelerometer data is noisy.

That's probably where I would start, but whether the hackish handling of the accelerometer data works or not depends on your application. Since this is a game forum, I assume your sensors are in a game controller, or something like that, and not on a rocket, so you should be fine.

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I am going to ignore your question and tell you about how to solve the problem of integrating your sensor data. :)

The "correct" way to integrate the sensor data is to use some variation of a Kalman filter. Unfortunately, a Kalman filter is a complicated mathematical beast as it is, and in order to estimate a rotation a plain Kalman filter won't do. The last time I looked into this, the fashionable variation to use in this case was an unscented Kalman filter.

If you get overwhelmed by the complexity of the situation, there are dirty hacks you can use. A practical one would be to have a quaternion representing your current estimate of attitude, take the 3 inputs of the gyros as an angular velocity, integrate it into the quaternion with the appropriate delta time'. When you read the magnetometer data (hopefully it's a 3-axis magnetometer), you get an idea of where the magnetic North is. By your current estimate of attitude, you expected magnetic North to be somewhere else, and you can compute the "smallest" rotation that would make your attitude agree with the input from the magnetometer (a cross product will do the trick): Go only a small fraction of the way there, because magnetometer data is very noisy. Similarly, if you have reason to believe your object doesn't have some serious linear acceleration, you can assume that the accelerometer is telling you where "up" is, and you should modify your attitude to go only a fraction of the way towards agreeing with it, because accelerometer data is noisy.

That's probably where I would start, but whether the hackish handling of the accelerometer data works or not depends on your application. Since this is a game forum, I assume your sensors are in a game controller, or something like that, and not on a rocket, so you should be fine.

I did look at the Kalman filter and then decided against it in favour of the complementary filter - it really does work nicely, 93% of the gyro combined with 7% of the accelerometer gives great results.

So your method is interesting, but I'm not sure it's that different to what I'm already doing, just perhaps reversed? I work out the yaw pitch roll from the accelerometer/magnetometer value and then smooth it with the gyro - is that about the same?

Edit: it's absolutely game-related! :)

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I was just thinking about Haegarrs 360,180,360 post...

If my screen widget and sensor is in it's normal position, standing up and facing away from me and I roll the sensor 180 degrees, it will make the screen widget upside down and still facing away, but the sensor's yaw reading will be 180 degrees (assuming in the normal position it is zero degrees) because it had now flipped

So will I need to tailor my yaw reading depending on whether the y accelerometer reading is > or < than zero? At the moment it just returns the absolute heading from the sensor regardless of whether its up or down.

This is where I find it all a bit confusing...

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I did look at the Kalman filter and then decided against it in favour of the complementary filter - it really does work nicely, 93% of the gyro combined with 7% of the accelerometer gives great results.

So your method is interesting, but I'm not sure it's that different to what I'm already doing, just perhaps reversed? I work out the yaw pitch roll from the accelerometer/magnetometer value and then smooth it with the gyro - is that about the same?

So why are you not doing all the operations using quaternions? What I described is not hard to do...

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My understanding of quaternions, even though I have the ability to use them in my engine, is practical only. The blocker for me is this use case - the sensor is in it's 'normal' position again where the acceleration vector is pointing directly down. I can create a quaternion with a vector pointing up and use the yaw for the rotation around that vector. That works fine. But then if you pitch the sensor 90 degrees, the vector part is fine but the rotation around that vector is now not yaw, not in terms of the sensor at least, it's pitch and roll. So do we build the quaternion in a different way depending on the orientation of the sensor? I thought about this for days (literally) and it just felt wrong somehow. With my relative inexperience in quaternions (I only use them for slerping my animations), I decided it might be easier for me to understand using Euler angles.

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I just have one more question on this if someone can help.

My accelerometer vector works like this (+x=right, +y=up,+z=away from you):

When the sensor is at rest sitting on the table, the vector is pointing directly down. When you pitch the device (with no roll), the vector rotates around the z axis (z is zero). When you roll the device (with no pitch), the vector rotates around the x axis (x is zero).

I'd like to get the angles of pitch and roll from this vector. I can easily get the pitch angle by using atan2 from x and y, but when you introduce roll, this becomes problematic when roll gets to 90+ degrees because the roll also uses the y component. So if roll is at 90 degrees, x is zero, z is 1 and y is 0. When you move through 90 degrees, y becomes negative and this messes up the pitch calculation because it flips it 180 degrees.

Am I on the right track here? The accelerometer vector does point in a unique position for every combination of pitch and roll so I should be able to compute them somehow.

Pulling my hair out on this a bit so I'd really appreciate any more help.

Thanks

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When the sensor is at rest sitting on the table, the vector is pointing directly down.

That makes no sense. The acceleration should be pointing directly up when the device is resting on the table: The table is pushing the device up, which is what is being detected.

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