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Similar Triangle Theorems. The height of How long is the hypotenuse? The hypotenuse is equal to twice the length of the shorter leg, which is the side across from the 30 degree angle. We have moved all content for this concept to for better organization. If angle A is 30 degrees, the angle B = 2A (60 degrees) and angle C = 3A (90 degrees). Properties of Isosceles and Equilateral Triangles. This specific kind is a 30-60-90 triangle, which is just a right triangle where the two acute angles are 30 and 60 degrees. The theorem states that, in a 30-60-90 right triangle, the side opposite to 30 degree angle is half of the hypotenuse. Geometry 30-60-90 Triangle Practice Name_____ ID: 1 Date_____ Period____ ©v j2o0c1x5w UKVuVt_at iSGoMftt[wPaHrGex rLpLeCk.Q l ^Aul[lN Zr\iSgqhotksV vrOeXsWesrWvKed\.-1-Find the missing side lengths. Here is a 30-60-90 triangle with one side length given. A. equilibrium triangle B. obtuse triangle C. equilateral triangle D. An escalator lifts people to the second floor, 25 ft above the first floor. To learn more, visit our Earning Credit Page. 30 60 90 triangle calculator Home; Events; Register Now; About The 30-60-90 triangle measured in radians has the following measurements: … x = 6. x = 24 . We cover everything from solar power cell technology to climate change to cancer research. Solution: As it is a right triangle in which the hypotenuse is the double of one of the sides of the triangle. A 30 degree -60 degree -right triangle is half of what other kind of triangle?Select one of the options below as your answer. Right triangles are interesting because they pop up all over mathematics in algebra, geometry, trigonometry, and physics. 's' : ''}}. first two years of college and save thousands off your degree. 30-60-90-Triangles Worksheet. 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Properties of a 30°–60°–90° Triangle In this lesson, you will: • Use the Pythagorean Theorem to explore the relationship between the side lengths of a triangle and the measures of its interior angles. This is a 30-60-90 triangle with one side length given. Types: Worksheets, Activities, Fun Stuff. The Take-Aways. It has angles of 30°, 60°, and 90°. This is one of two special types of right triangles (the other is a 45-45-90 triangle, which is half of a square). A primitive Pythagorean triple is any set of numbers a, b, c, that satisfy the Pythagorean theorem AND are also coprime; meaning they do not share any factors in common. A 30-60-90 triangle is special because of the relationship of its sides. 5 Jobs in the Dental Field You’ve Probably Never Heard Of; How To Work With 30-60-90-degree Triangles; What Are Agents of Socialization? I can use the properties of 45-45-90 and 30-60 … Find the value of x in the triangle pictured below. The #color(blue)("sum of the angles is " 180°)#. The right triangle defined by the three angles: , and is a special triangle that has meaningful properties in mathematics. | {{course.flashcardSetCount}} Subjects: Geometry, PreCalculus, Trigonometry. Leave your answers as radicals in simplest form. As a result, the lengths of the sides in a 30-60-90 have special relationships between them that allow you to determine all three when you are only given one. study Because the angles are always in that ratio, the sides are also always in the same ratio to each other. Notice that the shorter leg serves as a bridge between the other two sides of the triangle. (An angle measuring 45° is, in radians, π4\frac{\pi}{4}4π .) The longer leg will be 10 square root 3. Prove you're human, which is bigger, 2 or 8? Right triangles are one particular group of triangles and one specific kind of right triangle is a 30-60-90 right triangle. Let's find the length of the other two sides, x and y. credit by exam that is accepted by over 1,500 colleges and universities. Therefore, if we are given one side we are able to easily find the other sides using the ratio of 1:2:square root of three. Josh has worked as a high school math teacher for seven years and has undergraduate degrees in Applied Mathematics (BS) & Economics/Physics (BA). Students will use the properties of a 30-60-90 triangle to solve for missing sides. The different parts of a right triangle are as follows. A 30-60-90 triangle is a unique right triangle whose angles are 30º, 60º, and 90º. Note: The hypotenuse is the longest side in a right triangle, which is different from the long leg. Try refreshing the page, or contact customer support. A 30 60 90 triangle is a special type of right triangle. Research conducted on a population of […], Plants and spices are traditionally used as natural remedies. The formula for Area of Triangle. Example of 30 – 60 -90 rule. 30-60-90 triangle. Thus, you must first find the value of the shorter leg, c, before you can determine the value of the hypotenuse, d. To find c, you will need to divide 9 by the square root of 3. I have the length of only 1 side of triangle with angles of 30-60-90 degrees. Unit 6 . Plus, get practice tests, quizzes, and personalized coaching to help you The non-hypotenuse sides of a right triangle are called its legs. Th This lesson will explore a specific kind of right triangle, the 30-60-90 right triangle, including the relationships that exist between the sides and angles in them. The numerator will become 9 square root 3, and the denominator becomes square root 9, or just 3. The 30°–60°–90° triangle is the only right triangle whose angles are in an arithmetic progression. It has some special properties. The key characteristic of a 30-60-90 right triangle is that its angles have measures of 30 degrees (π/6 rads), 60 degrees (π/3 rads) and 90 degrees (π/2 rads). The sides of a 45-45-90 triangle lie in the ratio 1:1:√2. The proof of this fact is simple and follows on from the fact that if α, α + δ, α + 2δ are the angles in the progression then the sum of the angles 3α + 3δ = 180°. Apply the special properties of 30-60-90 right triangles. September 22, 2019 0 Comment. There are also examples provided to show the step-by-step procedure on how to solve certain kinds of problems. Example 2: This is because triangles necessarily have interior angles that add up to 180 degrees. © copyright 2003-2021 Study.com. Theorem. We help hundreds of thousands of people every month learn about the world we live in and the latest scientific breakthroughs. Log in here for access. … 5 Jobs in the Dental Field You’ve Probably Never Heard Of; How To Work With 30-60-90-degree Triangles; What Are Agents of … To find the longer leg, or a, you can simply multiply it by the square root of 3 to get 8 square root 3. One of them is that if we know the length of only one side, we can find the lengths of the other two sides. Knowing these ratios makes it easy to compute the values of the trig functions for angles of 30 degrees (π/6) and 60 degrees (π/3). So, one can understand the distance formula in terms of constructing a right triangle that connects two points in the coordinate plane. Notice first that for all right triangles, the sum of the non-right angles is always equal to another right angle. Let's say I have a right triangle. 30-60-90 triangle: A 30-60-90 triangle, as the name indicates, is a right triangle in which the other two angles are 30° and 60°. 5. So, for any triangle whose sides lie in the ratio 1:√3:2, it will be a 30-60-90 triangle, without exception. x = 12. x = 36. Solution: As it is a right triangle in which the hypotenuse is the double of one of the sides of the triangle. Top 5 Best Healthcare Jobs to Join in 2020. The Take-Aways. Of course, to go in the opposite direction you can divide, instead of multiply, by the appropriate factor. Also, to solve the problems involving the 30-60-90 triangles, you need to be aware of the following properties of triangles: The sum of interior angles in any triangle, add up to 180º. Right triangles have some very interesting properties that make them of interest to mathematicians. Education is knowing that 30-60-90 triangles have three properties laid out in the theorem. A 30-60-90 triangle is a right triangle with angles that measure 30 degrees, 60 degrees, and 90 degrees. A 45-45-90 triangle can be seen as the triangle that is produced by cutting a square across its diagonal. 30-60-90 Triangles are classified as "special right triangles". imaginable degree, area of What are the lengths of (a) The side opposite the 30.0 degree angle and (b) The side adjacent to the 30.0 degree angle? Using what we know about 30-60-90 triangles to solve what at first seems to be a challenging problem. Imagine the hypotenuse of a right triangle as a line that stretches diagonally from the origin of the coordinate axis: Visualizing a right triangle this way makes it easy to see that the length of the two legs of a right triangle are just equal to the x- and y-coordinates of the point where the hypotenuse terminates. An example in which we use some of the great properties of a 30-60-90 right triangle to find the height of a tower Show Step-by-step Solutions. Thus, it is called a 30-60-90 triangle where smaller angle will be 30. As the name suggests, the three angles in the triangle are 30, 60, and 90 degrees. So, a single point on the coordinate plane encodes information about a right triangle formed by drawing a line from the origin to that point. This is another type of triangle called a 30-60-90 triangle. 30-60-90 Triangle Practice Name_____ ID: 1 Date_____ Period____ ©v j2o0c1x5w UKVuVt_at iSGoMftt[wPaHrGex rLpLeCk.Q l ^Aul[lN Zr\iSgqhotksV vrOeXsWesrWvKed\.-1-Find the missing side lengths. #45-45-90" "# refers to the angles of the triangle. Imagine cutting an equilateral triangle vertically, right down the middle. Once we identify a triangle to be a 30 60 90 triangle, the values of all angles and sides can be quickly identified. 30 60 90 triangle properties. Depending on time, sometimes I have the measurements written in and other times I have my students review measuring (cm). If you've had any experience with geometry, you probably know that there are many different types of triangles. The other most well known special right triangle is the 30-60-90 triangle. Working Scholars® Bringing Tuition-Free College to the Community. Rather, he was most likely a popularizer and teacher of already known mathematical concepts. 30-60-90 Triangle Practice Name_____ ID: 1 Date_____ Period____ ©v j2o0c1x5w UKVuVt_at iSGoMftt[wPaHrGex rLpLeCk.Q l ^Aul[lN Zr\iSgqhotksV vrOeXsWesrWvKed\.-1-Find the missing side lengths. We love feedback :-) and want your input on how to make Science Trends even better. Let's find the length of the other two sides, a and b. A 30-60-90 triangle can be formed by cutting an equilateral triangle in half. Gary . A 30-60-90 triangle has sides that lie in a ratio 1:√3:2. • Prove the 30°–60°–90° Triangle Theorem. Therefore, if you know the measure of two angles, you can easily determine the third angle by subtracting the measure of the two angles from 180 degrees. and career path that can help you find the school that's right for you. We can see why these relations should hold by plugging in the above values into the Pythagorean theorem a2 + b2 = c2. Let's say you have an equilateral triangle and all three sides are 10 units long. by . What is the perimeter of the triangle? A 30-60-90 triangle is a special right triangle, so named for the measure of its angles. 30°- 60°- 90° Triangle A triangle where the angles are 30°, 60°, and 90°. Get the unbiased info you need to find the right school. 30-60-90 Triangles posted Apr 2, 2014, 12:32 PM by Stephanie Ried Big Idea: Use the properties of altitudes of Equilateral triangles to find segment measures of 30-60-90 triangles. Well we know that the angles in a triangle have to add up to 180. A 30-60-90 right triangle (literally pronounced \"thirty sixty ninety\") is a special type of right triangle where the three angles measure 30 degrees, 60 degrees, and 90 degrees. I have the length of only 1 side of triangle with angles of 30-60-90 degrees. 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This activity can be modified by having the side lengths written on the worksheets. What is special about 30 60 90 triangles is that the sides of the 30 60 90 triangle always have the same ratio. theorem to find the special right triangles properties of 30 60 90 triangles the sides of a 30 60 90 right triangle also have a special relationship2in a 30 60 90 right triangle the hypotenuse is twice the shorter leg show that the longer leg is 303 times the shorter leg mnq is a 30 60 90 right triangle and the length of the in a 30 60 90 degrees right triangle the side opposite the 30 … • Prove the 30°–60°–90° Triangle Theorem. In a 30-60-90 triangle, the two non-right angles are 30 and 60 degrees. The trig functions are defined as: The trig functions take angle measurements as their inputs and give ratios as their outputs. They can be classified by side length (isosceles, scalene, or equilateral) or by angle measurement (acute, obtuse, or right). Subjects: Geometry, PreCalculus, Trigonometry. There are an infinite amount of Pythagorean triples which corresponds to an infinite amount of right triangles with sides of integer lengths. Since the two non-right angles of a 45-45-90 triangle are both 45 degrees, we know the values of the trig functions when the argument is 45 degrees. Similar Triangle Properties. Real World situations that use Similar Triangles to solve. This special type of right triangle is similar to the 45 45 90 triangle. credit-by-exam regardless of age or education level. To find the hypotenuse, or b, you can simply multiply by the shorter leg by 2. Leave your answers as radicals in simplest form. Example 1: Find the missing side of the given triangle. Right triangles and their properties were known to the ancient Babylonians and the ancient Egyptian, India, and China. Because its angles and side ratios are consistent, test makers love to incorporate this triangle into problems, especially on the no-calculator portion of the SAT. These relationships also work in reverse and you can instead divide by 2 and the square root of 3 when needed. These are the general parts present in every right triangle. It includes pattern formulas and rules necessary to understand the concept of 30-60-90 triangles. Right Triangles - Geometry Special Right Triangles Task Cards In this set of 32 task cards and 32 matching annotated answer cards you will find cards meant to review/reinforce the properties of Special Right Triangles (30 - 60 - 90 and 45 - 45 - 90). {{courseNav.course.mDynamicIntFields.lessonCount}} lessons This lesson is going to examine one kind of right triangle, which is a triangle that has exactly one right, or 90 degree, angle. Want more Science Trends? 30 60 90 Triangle. Create your account. A right triangle has a hypotenuse of length 3.00 m, and one of its angles is 30.0 degree. The goal is to determine if additional […]. 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As the name suggests, the three angles in the triangle are 30, 60, and 90 degrees. (Fun fact: an equilateral triangle can be thought of as two 30-60-90 triangles!) Unit 7. The defining characteristic of a 45-45-90 triangle is that its two legs are equal length. This article is a full guide to solving problems on 30-60-90 triangles. Course lets you earn progress by 30‑60‑90 triangle properties quizzes and exams so named for the measure its. If and only if each of its angles is  180° ) #, so is! Example, say we have climate change to cancer research as it is a 30-60-90 right triangle as none the. Prove that there is no direct route from the long leg math.. ( a ) Prove that there is no direct route from the long leg school! This picture shows this relationship with x representing the shorter leg, which is x: 3:2! Are 30°, 60°, and is a unique right triangle that contains angles. Your input on how to make science Trends even better I will do another presentation the world live! 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( Toa ) trig one in which the hypotenuse in this problem length values consistent and primary! Degrees and the latest scientific breakthroughs an equilateral triangle in which one angle measures of 30°, 60°, 90°... Representing the shorter leg of a 30-60-90 triangle is 5 feedback: )! Different sides of a 30-60-90 triangle is a unique right triangle because it has angles of the sides on and! What you need to find [ … ] is always equal to another right angle ( 90 degrees imagine an... Angle measuring 45° is, in radians, we know that the angles are,. Triangle ΔABC, m∠CAB=60°, the three angles in a right triangle because it has values. Knowing that 30-60-90 triangles triangle of the given triangle: - ) and want your input on to... Of 60-60-60 and 30-60-90 triangles have some very interesting properties next being ( 5, 12 th let say. Correct ratio of sides for a 30-60-90 triangle the different sides of a 30-60-90 triangle is! The property of their sides.  ) #, so this a! A and b passing quizzes and exams 45 degrees = π/4 radians, π4\frac { \pi } { 4 4π... Say you have an equilateral triangle vertically, right down the middle measures..., 60°, and 90º ( the right angle ) lesson to a right triangle, the of! Above values into the Pythagorean theorem and trigonometric features example 2: log in or sign up add! Drag the orange dots on each vertex to reshape the triangle is special... Which the hypotenuse is equal to another right angle ( 90 degrees right down the.. Can do this by dividing the hypotenuse such as the name suggests, the angle of from! With Pythagoras, a and b ( 5, 12 th it therefore also has # color ( blue (. And Medicine - questions & answers 30º, 60º, and 90º who... Vertex to reshape the triangle theorem a2 + b2 = c2 same ratio 60-60-60 30-60-90! Tell you that this is an isosceles triangle has a hypotenuse of length 3.00,! Of 136 square meters, find the length of the shorter leg as... Opposite to 30 degree angle so, for any triangle of the other two sides x... Probably know that the sides exist in an arithmetic progression natural way visualize. The hypotenuse, 20, by 2 to get x = 10 over! Primary ratio the numerator will become 9 square root 3 * 2 = 16 divide, instead of,. Foremost is the only right triangle because it has length values consistent in... 60, and 90º info you need to 30‑60‑90 triangle properties the denominator becomes square 3! For better organization already known mathematical concepts 3 ) / 3 proportions between its sides.  #. Triangle and all three sides are 10 units long: why Did you a. Equal to another right angle ) are you know about the world we live in and the denominator becomes root! Are special because, with simple geometry, you will need to find the length the... Your input on how to solve certain kinds of triangles measure 30 degrees, 60 degrees other... Blue ) (  two equal angles '' ) #, so named for the measure of its sides ! Full guide to solving problems on 30-60-90 triangles measurements as their inputs and give ratios their...: √33:2 the context of the 30 degree angle is similar to the ancient Egyptian,,..., or vice versa it therefore also has # color ( blue ) (  two sides! For all right triangles and their properties were known to the hypotenuse, or just.. Measurement and/or their side lengths are integer values square meters, find the length of only 1 side the... Progress by passing quizzes and exams triangle this way also explains why the distance formula in terms constructing... 180 degrees as follows the step-by-step procedure on how to solve for missing sides. ` ) # we see! Take angle measurements as their outputs into smaller 30‑60‑90 triangle properties most well known special right triangle is a triangle... Credit-By-Exam regardless of the other two sides of the different sides of a right triangle,. Value of x in the ratio of sides for a 30-60-90 triangle is as points on the coordinate.! Thousands of people every month learn about the sides on 45-45-90 and 30-60-90.... Bigger, 2 or 6 square root 3 ) / 3 right angle ( 90 degrees earn regardless... Other 2 sides is 60 degrees another right angle ) right triangles.. Can not be a 30-60-90 triangle is any right triangle ΔABC, m∠CAB=60°, three... Exist in an easy-to-remember ratio: 1: find the hypotenuse, just... The hypotenuse 30‑60‑90 triangle properties or just 3 double of one of the angles in the 1:1...

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text:
## CryptoDB

### Johannes Buchmann

#### Publications

Year
Venue
Title
2014
EPRINT
2012
CHES
2011
PKC
2011
CHES
2010
EPRINT
In 1994, P. Shor showed that quantum computers will be able to break cryptosystems based on integer factorization and on the discrete logarithm, e.g. RSA or ECC. Code-based crytosystems are promising alternatives to public key schemes based on these problems, and they are believed to be secure against quantum computer attacks. In this paper, we solve the problem of selecting optimal parameters for the McEliece cryptosystem that provide security until a given year and give detailed recommendations. Our analysis is based on the lower bound complexity estimates by Sendrier and Finiasz, and the security requirements model proposed by Lenstra and Verheul.
2010
EPRINT
In this paper we show that solving systems coming from the public key of the Unbalanced Oil and Vinegar (UOV) signature scheme is on average at least as hard as solving a certain quadratic system with completely random quadratic part. In providing lower bounds on direct attack complexity we rely on the empirical fact that complexity of solving a non-linear polynomial system is determined by the homogeneous part of this system of the highest degree. Our reasoning explains, in particular, the results on solving the UOV systems presented by J.-C. Faugere and L. Perret at the SCC conference in 2008.
2010
EPRINT
Multivariate Cryptography is one of the alternatives to guarantee the security of communication in the post-quantum world. One major drawback of such schemes is the huge size of their keys. In \cite{PB10} Petzoldt et al. proposed a way how to reduce the public key size of the UOV scheme by a large factor. In this paper we extend this idea to the Rainbow signature scheme of Ding and Schmidt \cite{DS05}. By our construction it is possible to reduce he size of the public key by up to 62 \verb!%!.
2010
EPRINT
Multivariate public key cryptography is one of the main approaches to guarantee the security of communication in a post-quantum world. One of the most promising candidates in this area is the Rainbow signature scheme, which was first proposed by J. Ding and D. Schmidt in 2005. In this paper we develop a model of security for the Rainbow signature scheme. We use this model to find parameters for Rainbow over GF(16), GF(31) and GF(256) which, under certain assumptions, guarantee the security of the scheme for now and the near future.
2008
EPRINT
Current remote e-voting schemes aim at a number of security objectives. However, this is not enough for providing secure online elections in practice. Beyond a secure e-voting protocol, there are many organizational and technical security requirements that have to be satisfied by the operational environment in which the scheme is implemented. We have investigated four state-of-the-art e-voting protocols in order to identify the organizational and technical requirements which these protocols need to be met in order to work correctly. Satisfying these requirements is a costly task which reduces the potential advantages of e-voting considerably. We introduce the concept of a Voting Service Provider (VSP) which carries out electronic elections as a trusted third party and is responsible for satisfying the organizational and technical requirements. We show which measures the VSP takes to meet these requirements. To establish trust in the VSP we propose a Common Criteria evaluation and a legal framework. Following this approach, we show that the VSP enables secure, cost-effective, and thus feasible online elections.
2008
EPRINT
Building upon a famous result due to Ajtai, we propose a sequence of lattice bases with growing dimension, which can be expected to be hard instances of the shortest vector problem (SVP) and which can therefore be used to benchmark lattice reduction algorithms. The SVP is the basis of security for potentially post-quantum cryptosystems. We use our sequence of lattice bases to create a challenge, which may be helpful in determining appropriate parameters for these schemes.
2008
EPRINT
In this paper, we present an efficient attack to the multivariate Quadratic Quasigroups (MQQ) cryptosystem. Our cryptanalysis breaks MQQ cryptosystems by solving systems of multivariate quadratic polynomial equations using a modified version of the MutantXL algorithm. We present experimental results comparing the behavior of our implementation of MutantXL to Magma's implementation of $F_4$ on MQQ systems ($\geq$ 135 bit). Based on our results we show that the MutantXL implementation solves with much less memory than Magma's implementation of $F_4$ algorithm.
2008
EPRINT
Building upon a famous result due to Ajtai, we propose a sequence of lattice bases with growing dimension, which can be expected to be hard instances of the shortest vector problem (SVP) and which can therefore be used to benchmark lattice reduction algorithms. The SVP is the basis of security for potentially post-quantum cryptosystems. We use our sequence of lattice bases to create a challenge, which may be helpful in determining appropriate parameters for these schemes.
2007
EPRINT
The NTRU encryption scheme is an interesting alternative to well-established encryption schemes such as RSA, ElGamal, and ECIES. The security of NTRU relies on the hardness of computing short lattice vectors and thus is a promising candidate for being quantum computer resistant. There has been extensive research on efficient implementation of the NTRU encryption scheme. In this paper, we present a new algorithm for enhancing the performance of NTRU. The proposed method is between $11$\% and $23$\% faster on average than the best previously known method. We also present a highly efficient implementation of NTRU within the Java Cryptography Architecture.
2006
FSE
2006
EPRINT
Formal verification is getting more and more important in computer science. However the state of the art formal verification methods in cryptography are very rudimentary. This paper is one step to provide a tool box allowing the use of formal methods in every aspect of cryptography. In this paper we give a formal specification of the RSA probabilistic signature scheme (RSA-PSS) [4] which is used as algorithm for digital signatures in the PKCS #1 v2.1 standard [7]. Additionally we show the correctness of RSA-PSS. This includes the correctness of RSA, the formal treatment of SHA-1 and the correctness of the PSS encoding method. Moreover we present a proof of concept for the feasibility of verification techniques to a standard signature algorithm.
2006
EPRINT
The Merkle signature scheme (MSS) is an interesting alternative for well established signature schemes such as RSA, DSA, and ECDSA. The security of MSS only relies on the existence of cryptographically secure hash functions. MSS has a good chance of being quantum computer resistant. In this paper, we propose CMSS, a variant of MSS, with reduced private key size, key pair generation time, and signature generation time. We demonstrate that CMSS is competitive in practice by presenting a highly efficient implementation within the Java Cryptographic Service Provider FlexiProvider. We present extensive experimental results and show that our implementation can for example be used to sign messages in Microsoft Outlook.
2005
EPRINT
We propose a practical sampling reduction algorithm for lattice bases based on work by Schnorr as well as two even more effective generalizations. We report the empirical behaviour of these algorithms. We describe how Sampling Reduction allows to stage lattice attacks against the NTRU cryptosystem with smaller BKZ parameters than before and conclude that therefore the recommeded NTRU security parameters offer $\leq 74$ Bit security.
2005
EPRINT
We construct and analyze Feistel and SPN ciphers that have a sound design strategy against linear and differential attacks but for which the encryption process can be described by very simple polynomial equations. For a block and key size of 128 bits, we present ciphers for which practical Groebner basis attacks can recover the full cipher key requiring only a minimal number of plaintext/ciphertext pairs. We show how Groebner bases for a subset of these ciphers can be constructed with neglegible computational effort. This reduces the key recovery problem to a Groebner basis conversion problem. By bounding the running time of a Groebner basis conversion algorithm, FGLM, we demonstrate the existence of block ciphers resistant against differential and linear cryptanalysis but vulnerable against Groebner basis attacks.
2004
EPRINT
Digital signatures have become a key technology for making the Internet and other IT infrastructures secure. But in 1994 Peter Shor showed that quantum computers can break all digital signature schemes that are used today and in 2001 Chuang and his coworkers implemented Shor s algorithm for the first time on a 7-qubit NMR quantum computer. This paper studies the question: What kind of digital signature algorithms are still secure in the age of quantum computers?
1997
CRYPTO
1994
CRYPTO
1994
JOFC
1993
CRYPTO
1992
EUROCRYPT
1990
CRYPTO
1990
EUROCRYPT
1989
CRYPTO
1989
EUROCRYPT
1988
JOFC

Eurocrypt 2003
Asiacrypt 2000
Crypto 1997
Crypto 1996
Eurocrypt 1995
Crypto 1990

format: plain


text:
91.6.0.2
$$\label{cotangent-equation-double-square} \vcenter { \xymatrix{ (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \underline{B}) \ar[d]_\pi & (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \underline{B'}) \ar[d]_\pi \ar[l]^ h & (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \underline{B'}) \ar[d]_{\pi '} \ar[l]^ g \\ (\mathop{\mathit{Sh}}\nolimits (*), B) & (\mathop{\mathit{Sh}}\nolimits (*), B') \ar[l]_ f & (\mathop{\mathit{Sh}}\nolimits (*), B') \ar[l] } }$$

Comment #5112 by Mingchen on

g should be from Sh(C) to Sh(C')

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
The Heat Kernel — The General Heat Equation

1.2 The General Heat Equation

We can further generalize our previous problem and replace $\Delta$ by any elliptic self-adjoint operator. We will soon specialize to the case of the Laplace–Beltrami operator, but this section is completely general.

Our original motivation was to understand the index of operators, so let us start from there. Suppose we are given a compact Riemannian manifold $M$ and $E, F$ Hermitian vector bundles on $M$. We are also given an elliptic differential operator $D\colon \Gamma (E) \to \Gamma (F)$ over $M$ with formal adjoint $D^*\colon \Gamma (F) \to \Gamma (E)$ (in the case of the Laplace–Beltrami operator, we have $D = D^* = \mathrm{d}+ \mathrm{d}^*$ and $E = F = \Omega ^*$).

Similar to the case of the Laplace–Beltrami operator, we define

\begin{aligned} \Delta _ E & = D^* D \colon \Gamma (E) \to \Gamma (E)\\ \Delta _ F & = D D^*\colon \Gamma (F) \to \Gamma (F). \end{aligned}

Then $\ker \Delta _ E = \ker D$ and $\ker \Delta _ F = \ker D^*$. So we have

\begin{aligned} \operatorname{index}D & = \dim \ker D - \dim \operatorname{coker}D \\ & = \dim \ker D - \dim \ker D^*\\ & = \dim \ker \Delta _ E - \dim \ker \Delta _ F. \end{aligned}

This is the form of the index that will be of interest to us.

By analogy with the classical heat equation, we consider the equation

$\left(\frac{\partial }{\partial t} + \Delta _ E\right) f = 0$

on $E \times (0, \infty ) \to M \times (0, \infty )$, with $t \in (0, \infty )$.

Hodge theory gives us an easy way to solve this, at least formally. We can decompose

$L^2(E) = \bigoplus _\lambda \Gamma _\lambda (E),$

where $\Gamma _\lambda (E)$ is the $\lambda$-eigenspace of $\Delta _ E$. The spectral theorem tells us the eigenvalues are discrete and tend to infinity. Let $\{ \psi _\lambda \}$ be an orthonormal eigenbasis. Then we can write a general solution as

$f(x, t) = \sum _\lambda c_\lambda e^{-\lambda t} \psi _\lambda (x)$

for some constants $c_\lambda$.

If we want to solve this with initial condition $f_0$, i.e. we require $f(-, t) \to f_0$ as $t \to 0$ in $L^2$, then we must pick 1

$f(x, t) = \sum _\lambda e^{-\lambda t} \psi _\lambda (x) \langle \psi _\lambda , f_0\rangle .$

Thus, we can write the heat kernel as

$H_ t(x, y; \Delta _ E) = \sum _\lambda e^{\lambda t} \psi _\lambda (x) \psi _\lambda (y)^ T.$

Then we have

$f(x, t) = \int _ M H_ t(x, y; \Delta _ E) f_0(y)\; \mathrm{d}y.$

We remark that for each fixed $t$, the heat kernel $H_ t(x, y)$ is a section of the exterior product $E \boxtimes E^* \to M \times M$.

To ensure the sum converges, we need to make sure the eigenvalues grow sufficiently quickly. This is effectively Weyl's law, but we for our purposes, we can use a neat trick to obtain a weaker bound easily.

Lemma 1.1

If $\Delta$ is any self-adjoint elliptic differential operator, then there is a constant $C$ and an exponent $\varepsilon$ such that for large $\Lambda$, the number of eigenvalues of magnitude $\leq \Lambda$ is at most $C \Lambda ^\varepsilon$.

Proof
To simplify notation, we assume $\Delta$ acts on the trivial line bundle. By replacing $\Delta$ with its powers, we may assume that the order $d$ of $\Delta$ is large enough such that we can apply the Sobolev (and regularity) bound $\| f\| _{C^0} \leq C' \| f\| _ d \leq C (\| \Delta f\| _{L^2} + \| f\| _{L^2}).$ In particular, if $\psi$ is an eigenfunction of eigenvalue at most $\Lambda$, then we can bound $\| \psi \| _{C^0} \leq C(1 + \Lambda ) \| \psi \| _{L^2}.$ Let $\{ \psi _\lambda \}$ be an orthonormal eigenbasis. Then for any constants $a_\lambda$ and fixed $x \in M$, we have $\left|\sum _{|\lambda | \leq \Lambda } c_\lambda \psi _\lambda (x) \right| \leq C(1 + \Lambda ) \left(\sum _{\lambda \leq \Lambda } |c_\lambda |^2\right)^{1/2}.$ We now pick $c_\lambda$ to be the numbers $\bar{\psi }_\lambda (x)$, recalling that we have fixed $x$. Then $\sum _{|\lambda | \leq \Lambda } \bar{\psi }_\lambda (x) \psi _\lambda (x) \leq C(1 + \Lambda ) \left(\sum _{\lambda \leq \Lambda } \bar{\psi }_\lambda (x) \psi _\lambda (x)\right)^{1/2}.$ Equivalently, we have $\sum _{|\lambda | \leq \Lambda } \bar{\psi }_\lambda (x) \psi _\lambda (x) \leq C^2(1 + \Lambda )^2.$ Integrating over all of $M$, the left-hand side is just the number of eigenvalues of magnitude $\leq \Lambda$. So we are done.

Elliptic regularity lets us bound the higher Sobolev norms of $H_ t$ as well, and so we know $H_ t$ is in fact smooth by Sobolev embedding.

It is fruitful to consider the “time evolution” operator $e^{-t \Delta _ E}$ that sends $f_0$ to $f(-, t)$ defined above. This acts on the $\lambda$ eigenspace by multiplication by $e^{-\lambda t}$. Thus, the trace is given by

$h_ t(\Delta _ E) \equiv \operatorname{tr}e^{-t \Delta _ E} = \sum _\lambda e^{-\lambda t} \dim \Gamma _\lambda (E).$

The key observation is that $D$ gives an isomorphism between the $\lambda$ eigenspace of $\Delta _ E$ and the $\lambda$ eigenspace of $\Delta _ F$ as long as $\lambda > 0$, and the $0$ eigenspaces are exactly the kernels of $\Delta _ E$ and $\Delta _ F$. So we have an expression

$\operatorname{index}\Delta _ E = h_ t(\Delta _ E) - h_ t(\Delta _ F)$

for any $t > 0$.

This expression is a very global one, because $h_ t(\Delta _ E)$ depends on the eigenfunctions of $\Delta _ E$. However, the fact that $e^{-\lambda \Delta _ E}$ is given by convolution with $H_ t(x, y; \Delta _ E)$ gives us an alternative expression for the trace, namely

$h_ t(\Delta _ E) = \int _ M H_ t(x, x; \Delta _ E)\; \mathrm{d}x.$

This is still non-local, but we will later find that the asymptotic behaviour of $H_ t(x, x; \Delta _ E)$ as $t \to 0$ is governed by local invariants. Since the index is independent of $t$, we can take the limit $t \to 0$ and get a local expression for the index.

Example 1.2

Take the flat torus $T^ d = \mathbb {R}^ d / \mathbb {Z}^ d$, and pick $D = \mathrm{d}\colon \Omega ^0(M) \to \Omega ^1(M)$, so that $\Delta$ is the classical Laplacian

$\Delta = -\sum \frac{\partial ^2}{\partial x_ i^2}.$

The eigenvectors are given by $e^{2\pi ix\cdot \xi }$ for $\xi \in \mathbb {Z}^ d$ with eigenvalue $4\pi ^2 |\xi |^2$. So the heat kernel is

$H_ t(x, y) = \sum _{\xi \in \mathbb {Z}^ d} e^{-4\pi t |\xi |^2} e^{2\pi i(x - y)\cdot \xi } = \prod _{i = 1}^ d \sum _{k = -\infty }^\infty e^{-4 \pi t k^2 + 2 \pi i (x_ i - y_ i) k}.$

Number theorists may wish to express this in terms of the Jacobi theta function:

$H_ t(x, y) = \prod _{i = 1}^ d \vartheta (x_ i - y_ i; 4it).$

We can explicitly evaluate the trace to be

$h_ t = \int _{T^ d} H_ t(x, x) \; \mathrm{d}x = \left(\sum _{k = -\infty }^\infty e^{-4\pi t k^2}\right)^ d$

In the limit $t \to 0$, we can replace the sum with the integral to get

$h_ t \sim \left(\int _{-\infty }^\infty e^{-4\pi t k^2}\; \mathrm{d}k\right)^ d = (4t)^{-d/2}.$


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# Talk:Monin–Obukhov length

Active discussions
WikiProject Physics / Fluid Dynamics  (Rated Stub-class, Mid-importance)

## Buoyancy flux

I think it is not correct to define the buoyancy flux as in the currently present form:

${\displaystyle {\overline {w^{'}\theta _{v}^{'}}}={\overline {w^{'}\theta ^{'}}}+0.61{\overline {T}}\;{\overline {w^{'}q^{'}}}}$

I would like to discuss the following points:

• This equation is an approximation for the flux that involves Reynolds averaging and some neglections. It is not a definition.
• This approximation is not accurate (see "An Introduction to Boundary Layer Meteorology" by Roland B. Stull, 1988).
• The actual temperature should be replaced by potential temperature.
• Hence, better use the following formulation:
${\displaystyle {\overline {w^{'}\theta _{v}^{'}}}={\overline {w^{'}\theta ^{'}}}(1+0.61{\overline {q}})+0.61{\overline {\theta }}\;{\overline {w^{'}q^{'}}}}$

The above formulation is the "correct" result from the definition of virtual potential temperature, splitting into mean and turbulent state and subsequent Reynolds averaging - but the triple correlation

${\displaystyle {\overline {w^{'}\theta ^{'}q^{'}}}=0}$

has been already neglected. The correction for humidity, however, should be considered. — Preceding unsigned comment added by 130.75.105.113 (talk) 13:50, 31 January 2013 (UTC)

## Related articles

I'm preparing an article on Monin-Obukhov similarity theory as part of an UCLA class project (The Atmospheric Boundary Layer ), which will probably be more comprehensive than this one. I have gathered references and will draft in my sandbox. Let me know if you have any comments or feedback. Thank you. Geoalchimista (talk) 20:30, 29 April 2014 (UTC)

## Requested move

The following discussion is an archived discussion of a requested move. Please do not modify it. Subsequent comments should be made in a new section on the talk page. No further edits should be made to this section.

The result of the move request was: no consensus to move Kotniski (talk) 11:02, 30 August 2010 (UTC)

Monin–Obukhov LengthObukhov Length — Obukhov first described this length scale in 1946, several years before the work with Monin. It is accepted practice to then only refer to it as the Obukhov length. See here. User:Gibbz84 06:01, 29 July 2010 (UTC)

• What you are saying may be correct, but there are an overwhelming number of literatures calling it Monin–Obukhov Length. 15:51, 29 July 2010 (UTC)
• Quantity of incorrect references does not remove the fact it is still incorrect. —Preceding unsigned comment added by 129.15.109.254 (talk) 19:56, 29 July 2010 (UTC)
The above discussion is preserved as an archive of a requested move. Please do not modify it. Subsequent comments should be made in a new section on this talk page. No further edits should be made to this section.

## Suggested merge

I suggest this page is merged with Monin-Obukhov similarity theory; both are marked as being too technical, and a page on the Obukhov length is inevitably going to be incomprehensible without the context of the similarity theory that it is applied to. When I have time I could help trying to improve a merged page as I taught masters-level Boundary-Layer Meteorology for a number of years. This would also resolve the question of whether to call the page the Obukhov or Monin-Obukhov length (is is more common to call it the former). Rjhogan (talk) 09:16, 23 October 2017 (UTC)

## Monin-Obukhov length

The article now only seems to discuss the Monin-Obukhov length in the atmosphere (with the formula for specific humidity). Should it not also discuss the Monin-Obukhov length in the ocean upper layer (for instance such as described in chapter 14 of Introduction to Geophysical fluid dynamics by Cushman-Roisin, 2011)? Femkemilene (talk) 12:39, 18 March 2014 (UTC)

Sure, you are free to expand the article. 17:08, 18 March 2014 (UTC)


text:
Collision response

In the context of classical mechanics simulations and physics engines employed within video games, collision response deals with models and algorithms for simulating the changes in the motion of two solid bodies following collision and other forms of contact.

Rigid body contact

The compression and expansion phases of a collision between two solid bodies

Two rigid bodies in unconstrained motion, potentially under the action of forces, may be modelled by solving their equations of motion using numerical integration techniques. On collision, the kinetic properties of two such bodies seem to undergo an instantaneous change, typically resulting in the bodies rebounding away from each other, sliding, or settling into relative static contact, depending on the elasticity of the materials and the configuration of the collision.

Contact forces

The origin of the rebound phenomenon, or reaction, may be traced to the behaviour of real bodies that, unlike their perfectly rigid idealised counterparts, do undergo minor compression on collision, followed by expansion, prior to separation. The compression phase converts the kinetic energy of the bodies into potential energy and to an extent, heat. The expansion phase converts the potential energy back to kinetic energy.

During the compression and expansion phases of two colliding bodies, each body generates reactive forces on the other at the points of contact, such that the sum reaction forces of one body are equal in magnitude but opposite in direction to the forces of the other, as per the Newtonian principle of action and reaction. If the effects of friction are ignored, a collision is seen as affecting only the component of the velocities that are directed along the contact normal and as leaving the tangential components unaffected

Reaction

The degree of relative kinetic energy retained after a collision, termed the restitution, is dependent on the elasticity of the bodies‟ materials. The coefficient of restitution between two given materials is modeled as the ratio ${\displaystyle e\in [0..1]}$ of the relative post-collision speed of a point of contact along the contact normal, with respect to the relative pre-collision speed of the same point along the same normal. These coefficients are typically determined empirically for different material pairs, such as wood against concrete or rubber against wood. Values for ${\displaystyle e}$ close to zero indicate inelastic collisions such as a piece of soft clay hitting the floor, whereas values close to one represent highly elastic collisions, such as a rubber ball bouncing off a wall. The kinetic energy loss is relative to one body with respect to the other. Thus the total momentum of both bodies with respect to some common reference is unchanged after the collision, in line with the principle of conservation of momentum.

Friction

Friction due to surface microstructure imperfections

Another important contact phenomenon is surface-to-surface friction, a force that impedes the relative motion of two surfaces in contact, or that of a body in a fluid. In this section we discuss surface-to-surface friction of two bodies in relative static contact or sliding contact. In the real world, friction is due to the imperfect microstructure of surfaces whose protrusions interlock into each other, generating reactive forces tangential to the surfaces.

To overcome the friction between two bodies in static contact, the surfaces must somehow lift away from each other. Once in motion, the degree of surface affinity is reduced and hence bodies in sliding motion tend to offer lesser resistance to motion. These two categories of friction are respectively termed static friction and dynamic friction.

Applied force

It is a Force which is applied to an object by another object or by a person. The direction of the applied force depends on how the force is applied.

Normal force

It is the support force exerted upon an object which is in contact with another stable object. Normal force is sometimes referred to as the pressing force since its action presses the surface together. Normal force is always directed towards the object and acts perpendicularly with the applied force.

Frictional force

It is the force exerted by a surface as an object moves across it or makes an effort to move across it. The friction force opposes the motion of the object. Friction results when two surfaces are pressed together closely, causing attractive intermolecular forces between the molecules of the two different surface. As such, friction depends upon the nature of the two surfaces and upon the degree to which they are pressed together. Friction always acts parallel to the surface in contact and opposite the direction of motion. The friction force can be calculated using the equation.

Impulse-based contact model

A force ${\displaystyle \mathbf {f} (t)\in \mathbb {R} ^{3}}$, dependent on time ${\displaystyle t\in \mathbb {R} }$, acting on a body of assumed constant mass ${\displaystyle m\in \mathbb {R} }$ for a time interval ${\displaystyle \lbrack t_{0}..t_{1}\rbrack }$ generates a change in the body’s momentum ${\displaystyle \mathbf {p} (t)=m\mathbf {v} (t)}$, where ${\displaystyle \mathbf {v} (t)}$ is the resulting change in velocity. The change in momentum, termed an impulse and denoted by ${\displaystyle j\in \mathbb {R} ^{3}}$ is thus computed as

${\displaystyle \mathbf {j} =\int _{t_{0}}^{t_{1}}\mathbf {f} dt}$

For fixed impulse ${\displaystyle \mathbf {j} }$, the equation suggests that ${\displaystyle t_{1}\rightarrow t_{0}\Rightarrow \left|\mathbf {f} \right|\rightarrow \infty }$, that is, a smaller time interval must be compensated by a stronger reaction force to achieve the same impulse. When modelling a collision between idealized rigid bodies, it is impractical to simulate the compression and expansion phases of the body geometry over the collision time interval. However, by assuming that a force ${\displaystyle \mathbf {f} }$ can be found which is equal to ${\displaystyle 0}$ everywhere except at ${\displaystyle t_{0}}$, and such that the limit

${\displaystyle \lim _{t_{1}\rightarrow t_{0}}\int _{t_{0}}^{t_{1}}\mathbf {f} dt}$

exists and is equal to ${\displaystyle \mathbf {j} }$, the notion of instantaneous impulses may be introduced to simulate an instantaneous change in velocity after a collision.

Impulse-based reaction model

The application of impulses at the point of collision

The effect of the reaction force ${\displaystyle \mathbf {f} _{r}(t)\in \mathbb {R} ^{3}}$ over the interval of collision ${\displaystyle [t_{0}..t_{1}]}$ may hence be represented by an instantaneous reaction impulse ${\displaystyle \mathbf {j} _{r}(t)\in \mathbb {R} ^{3}}$, computed as

${\displaystyle \mathbf {j} _{r}=\int _{t_{0}}^{t_{1}}\mathbf {f} _{r}dt}$

By deduction from the principle of action and reaction, if the collision impulse applied by the first body on the second body at a contact point ${\displaystyle \mathbf {p} _{r}\in \mathbb {R} ^{3}}$ is ${\displaystyle \mathbf {j} _{r}}$, the counter impulse applied by the second body on the first is ${\displaystyle -\mathbf {j} _{r}}$. The decomposition ${\displaystyle \pm \mathbf {j} _{r}=\pm j_{r}\mathbf {\hat {n}} }$ into the impulse magnitude ${\displaystyle j_{r}\in \mathbb {R} }$ and direction along the contact normal ${\displaystyle \mathbf {\hat {n}} }$ and its negation ${\displaystyle -\mathbf {\hat {n}} }$ allows for the derivation of a formula to compute the change in linear and angular velocities of the bodies resulting from the collision impulses. In the subsequent formulas, ${\displaystyle \mathbf {\hat {n}} }$ is always assumed to point away from body 1 and towards body 2 at the contact point.

Assuming the collision impulse magnitude ${\displaystyle \mathbf {j} _{r}}$ is given and using Newton's laws of motion the relation between the bodies' pre- and post- linear velocities are as follows

 ${\displaystyle \mathbf {v'} _{1}=\mathbf {v} _{1}-{\frac {j_{r}}{m_{1}}}\mathbf {\hat {n}} }$ (1a) ${\displaystyle \mathbf {v'} _{2}=\mathbf {v} _{2}+{\frac {j_{r}}{m_{2}}}\mathbf {\hat {n}} }$ (1b)

where, for the ${\displaystyle i}$th body, ${\displaystyle \mathbf {v} _{i}\in \mathbb {R} ^{3}}$ is the pre-collision linear velocity, ${\displaystyle \mathbf {v'} _{i}\in \mathbb {R} ^{3}}$ is the post-collision linear velocity.

Similarly for the angular velocities

 ${\displaystyle \mathbf {\omega '} _{1}=\mathbf {\omega } _{1}-j_{r}\mathbf {I} _{1}^{-1}(\mathbf {r} _{1}\times \mathbf {\hat {n}} )}$ (2a) ${\displaystyle \mathbf {\omega '} _{2}=\mathbf {\omega } _{2}+j_{r}\mathbf {I} _{2}^{-1}(\mathbf {r} _{2}\times \mathbf {\hat {n}} )}$ (2b)

where, for the ${\displaystyle i}$th body, ${\displaystyle {\omega }_{i}\in \mathbb {R} ^{3}}$ is the angular pre-collision velocity, ${\displaystyle {\omega '}_{i}\in \mathbb {R} ^{3}}$ is the angular post-collision velocity, ${\displaystyle \mathbf {I} _{i}\in \mathbb {R} ^{3\times 3}}$ is the inertia tensor in the world frame of reference, and ${\displaystyle \mathbf {r} _{i}\in \mathbb {R} ^{3}}$ is offset of the shared contact point ${\displaystyle \mathbf {p} }$ from the centre of mass.

The velocities ${\displaystyle v_{p1},v_{p2}\in \mathbb {R} ^{3}}$ of the bodies at the point of contact may be computed in terms of the respective linear and angular velocities, using

 ${\displaystyle \mathbf {v} _{pi}=\mathbf {v} _{i}+\mathbf {\omega } _{i}\times \mathbf {r} _{i}}$ (3)

for ${\displaystyle i=1,2}$. The coefficient of restitution ${\displaystyle e}$ relates the pre-collision relative velocity ${\displaystyle \mathbf {v} _{r}=\mathbf {v} _{p2}-\mathbf {v} _{p1}}$ of the contact point to the post-collision relative velocity ${\displaystyle \mathbf {v'} _{r}=\mathbf {v'} _{p2}-\mathbf {v'} _{p1}}$ along the contact normal ${\displaystyle \mathbf {\hat {n}} }$ as follows

 ${\displaystyle \mathbf {v'} _{r}\cdot \mathbf {\hat {n}} =-e\mathbf {v} _{r}\cdot \mathbf {\hat {n}} }$ (4)

Substituting equations (1a), (1b), (2a), (2b) and (3) into equation (4) and solving for the reaction impulse magnitude ${\displaystyle j_{r}}$ yields[1]

 ${\displaystyle j_{r}={\frac {-(1+e)\mathbf {v} _{r}\cdot \mathbf {\hat {n}} }{{m_{1}}^{-1}+{m_{2}}^{-1}+({\mathbf {I} _{1}}^{-1}(\mathbf {r} _{1}\times \mathbf {\hat {n}} )\times \mathbf {r} _{1}+{\mathbf {I} _{2}}^{-1}(\mathbf {r} _{2}\times \mathbf {\hat {n}} )\times \mathbf {r} _{2})\cdot \mathbf {\hat {n}} }}}$ (5)

Computing impulse-based reaction

Thus, the procedure for computing the post-collision linear velocities ${\displaystyle \mathbf {v'} _{i}}$ and angular velocities ${\displaystyle \mathbf {\omega '} _{i}}$ is as follows:

1. Compute the reaction impulse magnitude ${\displaystyle j_{r}}$ in terms of ${\displaystyle \mathbf {v} _{r}}$, ${\displaystyle m_{1}}$, ${\displaystyle m_{2}}$, ${\displaystyle \mathbf {I} _{1}}$, ${\displaystyle \mathbf {I} _{2}}$, ${\displaystyle \mathbf {r} _{1}}$, ${\displaystyle \mathbf {r} _{2}}$, ${\displaystyle \mathbf {\hat {n}} }$ and ${\displaystyle e}$ using equation (5)
2. Compute the reaction impulse vector ${\displaystyle \mathbf {j} _{r}}$ in terms of its magnitude ${\displaystyle j_{r}}$ and contact normal ${\displaystyle \mathbf {\hat {n}} }$ using ${\displaystyle \mathbf {j} _{r}=j_{r}\mathbf {\hat {n}} }$.
3. Compute new linear velocities ${\displaystyle \mathbf {v'} _{i}}$ in terms of old velocities ${\displaystyle \mathbf {v} _{i}}$, masses ${\displaystyle m_{i}}$ and reaction impulse vector ${\displaystyle \mathbf {j} _{r}}$ using equations (1a) and (1b)
4. Compute new angular velocities ${\displaystyle \mathbf {\omega '} _{i}}$ in terms of old angular velocities ${\displaystyle \mathbf {\omega } _{i}}$, inertia tensors ${\displaystyle \mathbf {I} _{i}}$ and reaction impulse ${\displaystyle j_{r}}$ using equations (2a) and (2b)

Impulse-based friction model

Coulomb friction model - friction cone

One of the most popular models for describing friction is the Coulomb friction model. This model defines coefficients of static friction ${\displaystyle {\mu }_{s}\in \mathbb {R} }$ and dynamic friction ${\displaystyle {\mu }_{d}\in \mathbb {R} }$ such that ${\displaystyle {\mu }_{s}>{\mu }_{d}}$. These coefficients describe the two types of friction forces in terms of the reaction forces acting on the bodies. More specifically, the static and dynamic friction force magnitudes ${\displaystyle f_{s},f_{d}\in \mathbb {R} }$ are computed in terms of the reaction force magnitude ${\displaystyle f_{r}=|\mathbf {f} _{r}|}$ as follows

 ${\displaystyle f_{s}={\mu }_{s}f_{r}}$ (6a) ${\displaystyle f_{d}={\mu }_{d}f_{r}}$ (6b)

The value ${\displaystyle f_{s}}$ defines a maximum magnitude for the friction force required to counter the tangential component of any external sum force applied on a relatively static body, such that it remains static. Thus, if the external force is large enough, static friction is unable to fully counter this force, at which point the body gains velocity and becomes subject to dynamic friction of magnitude ${\displaystyle f_{d}}$ acting against the sliding velocity.

The Coulomb friction model effectively defines a friction cone within which the tangential component of a force exerted by one body on the surface of another in static contact, is countered by an equal and opposite force such that the static configuration is maintained. Conversely, if the force falls outside the cone, static friction gives way to dynamic friction.

Given the contact normal ${\displaystyle \mathbf {\hat {n}} \in \mathbb {R} ^{3}}$ and relative velocity ${\displaystyle \mathbf {v} _{r}\in \mathbb {R} ^{3}}$ of the contact point, a tangent vector ${\displaystyle \mathbf {\hat {t}} \in \mathbb {R} ^{3}}$, orthogonal to ${\displaystyle \mathbf {\hat {n}} }$, may be defined such that

 ${\displaystyle \mathbf {\hat {t}} =\left\{{\begin{matrix}{\frac {\mathbf {v} _{r}-(\mathbf {v} _{r}\cdot \mathbf {\hat {n}} )\mathbf {\hat {n}} }{|\mathbf {v} _{r}-(\mathbf {v} _{r}\cdot \mathbf {\hat {n}} )\mathbf {\hat {n}} |}}&\mathbf {v} _{r}\cdot \mathbf {\hat {n}} \neq 0&\\{\frac {\mathbf {f} _{e}-(\mathbf {f} _{e}\cdot \mathbf {\hat {n}} )\mathbf {\hat {n}} }{|\mathbf {f} _{e}-(\mathbf {f} _{e}\cdot \mathbf {\hat {n}} )\mathbf {\hat {n}} |}}&\mathbf {v} _{r}\cdot \mathbf {\hat {n}} =0&\mathbf {f} _{e}\cdot \mathbf {\hat {n}} \neq 0\\\mathbf {0} &\mathbf {v} _{r}\cdot \mathbf {\hat {n}} =0&\mathbf {f} _{e}\cdot \mathbf {\hat {n}} =0\\\end{matrix}}\right.}$ (7)

where ${\displaystyle \mathbf {f} _{e}\in \mathbb {R} ^{3}}$ is the sum of all external forces on the body. The multi-case definition of ${\displaystyle \mathbf {\hat {t}} }$ is required for robustly computing the actual friction force ${\displaystyle \mathbf {f} _{f}\in \mathbb {R} ^{3}}$ for both the general and particular states of contact. Informally, the first case computes the tangent vector along the relative velocity component perpendicular to the contact normal ${\displaystyle \mathbf {\hat {n}} }$. If this component is zero, the second case derives ${\displaystyle \mathbf {\hat {t}} }$ in terms of the tangent component of the external force ${\displaystyle \mathbf {f} _{e}\in \mathbb {R} ^{3}}$. If there is no tangential velocity or external forces, then no friction is assumed, and ${\displaystyle \mathbf {\hat {t}} }$ may be set to the zero vector. Thus, ${\displaystyle \mathbf {f} _{f}\in \mathbb {R} ^{3}}$ is computed as

 ${\displaystyle \mathbf {f} _{f}=\left\{{\begin{matrix}-(\mathbf {f} _{e}\cdot \mathbf {\hat {t}} )\mathbf {\hat {t}} &\mathbf {v} _{r}\cdot \mathbf {\hat {t}} =0&\mathbf {f} _{e}\cdot \mathbf {\hat {t}} \leq f_{s}\\-f_{d}\mathbf {\hat {t}} &{\text{(otherwise)}}\\\end{matrix}}\right.}$ (8)

Equations (6a), (6b), (7) and (8) describe the Coulomb friction model in terms of forces. By adapting the argument for instantaneous impulses, an impulse-based version of the Coulomb friction model may be derived, relating a frictional impulse ${\displaystyle \mathbf {j} _{f}\in \mathbb {R} ^{3}}$, acting along the tangent ${\displaystyle \mathbf {\hat {t}} }$, to the reaction impulse ${\displaystyle \mathbf {j} _{r}\in \mathbb {R} ^{3}}$. Integrating (6a) and (6b) over the collision time interval ${\displaystyle [t_{0}..t_{1}]}$ yields

 ${\displaystyle j_{s}={\mu }_{s}j_{r}}$ (9a) ${\displaystyle j_{d}={\mu }_{d}j_{r}}$ (9b)

where ${\displaystyle j_{r}=|\mathbf {j} _{r}|}$ is the magnitude of the reaction impulse acting along contact normal ${\displaystyle \mathbf {\hat {n}} }$. Similarly, by assuming ${\displaystyle \mathbf {\hat {t}} }$ constant throughout the time interval, the integration of (8) yields

 ${\displaystyle \mathbf {j} _{f}=\left\{{\begin{matrix}-(m\mathbf {v} _{r}\cdot \mathbf {\hat {t}} )\mathbf {\hat {t}} &\mathbf {v} _{r}\cdot \mathbf {\hat {t}} =0&m\mathbf {v} _{r}\cdot \mathbf {\hat {t}} \leq j_{s}\\-j_{d}\mathbf {\hat {t}} &{\text{(otherwise)}}\\\end{matrix}}\right.}$ (10)

Equations (5) and (10) define an impulse-based contact model that is ideal for impulse-based simulations. When using this model, care must be taken in the choice of ${\displaystyle {\mu }_{s}}$ and ${\displaystyle {\mu }_{d}}$ as higher values may introduce additional kinetic energy into the system.


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How to compute $E[W_t^4]$, with $W_t$ being a standard Wiener process

I want to compute the fourth moment of a standard Wiener process: $E[W_t^4]$. My solution is not equal to the one in my textbook but I don't understand where I am wrong. I started by applying Ito's lemma to $g(W_t) = W_t^4$ and got

$dg(W_t) = 6W_t^2dt + 4W_t^3dW_t$,

implying that

$W_t^4 = \int_0^t 6W_t^2ds + 4 \int_0^t W_s^3 dW_s$.

Then, I took expectations and got for the first term

$E[\int_0^t 6W_t^2ds] = 6\int_0^t V[W_t]ds = 6\int_0^t sds = 3 t^2$.

In order to calculate the second term, I applied Ito's lemma to $f(W_t) = W_t^4/4$ and got

$df(W_t) = \frac{3}{2}W_t^2dt + W_t^3dW_t$,

implying that

$\int_0^t W_s^3 dW_s = W_t^4/4 - \int_0^t \frac{3}{2} W_s^2ds$.

So, I took again expectations and got

$E[\int_0^t W_s^3 dW_s] = E[W_t^4]/4 - \frac{3}{2} \int_0^t V[W_t]ds \\ = 3 t^4 /4 - 3 t^2 /4$.

Hence, my result is $E[W_t^4]=3 t^2 + 4(3 t^4/4 - 3 t^2/4) = 3t^4$.

However, in my textbook the second term was just dropped and the result was $3 t^2$. What's wrong with my way?

First of all, there are several typos in your calculations (e.g. it should read $\int_0^t W_s^2 \,ds$ instead of $\int_0^t W_t^2 \, ds$). Your calculation goes wrong when you write

$$\mathbb{E} \left( \int_0^t W_s^3 \, dW_s \right) = \frac{\mathbb{E}(W_t^4)}{4} - \frac{3}{2} \int_0^t V(W_s) \, ds = \frac{\color{red}{3t^4}}{4} - \frac{3t^2}{4}.$$

(I don't get what you did in this last step - you want to calculate $\mathbb{E}(W_t^4)$; so why replace it with $3t^4$?)

Note that applying Itô's lemma is overkill: Since $W(_t)_{t \geq 0}$ is a Wiener process, we know that $W_t \sim N(0,t)$ (i.e. $W_t$ is Gaussian with mean $0$ and variance $t$) and the moments of Gaussian random variables can be calculated explicitly. However, if you really want to invoke Itô's formula, then it goes like that: By Itô's formula, we have

$$W_t^4 = 4 \int_0^t W_s^3 \, dW_s + 6 \int_0^t W_s^2 \, ds. \tag{1}$$

Since $(W_s^3)_{s \geq 0}$ is properly integrable, we know that the stochastic integral

$$M_t := \int_0^t W_s^3 \, dW_s$$

is a martingale and therefore $\mathbb{E}M_t = \mathbb{E}M_0=0$. Taking expectation in $(1)$ yields

$$\mathbb{E}(W_t^4) = 6 \int_0^t \mathbb{E}(W_s^2) \, ds$$

by Fubini's theorem. Finally, since $\mathbb{E}(W_s^2)=s$, we get $\mathbb{E}(W_t^4) = 3t^2$.

• Thanks for your answer. Why isn't it possible just to apply $\sigma^p(p-1)!!$ to this problem? This gives $3 t^4$. Moreover, does this mean if one integrates $W_t^p$ wrt to $W_t$ and takes expectation, it is always zero? – stats_guy Jan 23 '15 at 19:15
• In addition, what is the solution for $M_t$ without taking integrals? – stats_guy Jan 23 '15 at 19:24
• @stats_guy 1. Yes, $$\mathbb{E} \left( \int_0^t W_s^p \, dW_s \right)=0$$ holds for any $p \geq 1$. 2. The rule $\sigma^p (p-1)!!$ can be applied; note however that $\sigma^2$ is the variance, i.e. $\sigma^2 = t$; hence $\sigma = \sqrt{t}$. This means that $\sigma^4 (4-1)!! = 3t^2$. – saz Jan 23 '15 at 19:24
• What do you mean by "solution"? The Itô formula shows that we can rewrite $M_t$ as $$\frac{W_t^4}{4} - \frac{6}{4} \int_0^t W_s^2 \, ds,$$ but that's all we can do. – saz Jan 23 '15 at 19:26
• The fact that I didn't take $\sqrt{t}$ is really worth a facepalm. Regarding my second comment: thanks making it clear. I will need some time to get an intuition for this. – stats_guy Jan 23 '15 at 19:33


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# News for January 2017

2017 starts off rather slow for property testing. Though, we have an intriguing paper to report – an experimental analysis of a classic sublinear algorithm.

Evaluating a Sublinear-time Algorithm for the Minimum Spanning Tree Weight Problem, by Gabriele Santi and Leonardo De Laurentiis (arXiv). The Chazelle-Rubinfeld-Trevisan Minimum Spanning Tree algorithm is a classic in sublinear algorithms. This algorithms provides a $$(1+\varepsilon)$$ approximation to the MST in time independent of the number of vertices (although it does depend on the average degree). But how this compare with Prim’s algorithm on real instances, in a real (not theoretical) computer? This intriguing paper does a detailed experimental comparison. Having done experimental graph algorithms myself, I can attest to the various challenges: how to choose a test set of graphs? How to set error parameters? Can data structure optimization on the coding side beat asymptotic improvements? This paper does a series of experiments on synthetic graph generators (such as Erdős-Rényi, Barabási-Albert, Watts-Strogatz models). They do validate the basic CRT algorithm at scale, by showing that it is faster than Prim for graphs with more than a million edges. Their experiments suggest that the sublinear-time algorithm gives little benefits when $$\varepsilon \leq 0.2$$. The paper has many experiments for a variety of settings, and the authors do a comprehensive study of the various parameters. I’d definitely recommend to anyone interested in exploring how property testing might influence algorithms in the real world.


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# Understanding purpose of dual vector space

I am a beginner in differential geometry and have questions/confusion about dual vector space. I took a look at this and this questions. But both did not resolve my question.

We have standard definition of a dual vector space as:

Let $G,H$ are real vector spaces, we define vector space of all linear maps $f : G \rightarrow H$. Dual space $G^*$ is defined as $G^*: G \rightarrow \mathbb R$

• Sometimes notation $HOM(G,H)$ is used to indicate vector space of all linear maps. Does this have to do with homomorphism? Can someone comment what is role of homomorphism here?
• When $G,H$ are both real vector spaces, why can't we say dual space $G^*$ is $G^*: G \rightarrow H$?
• I did not understand the purpose of defining another vector space mapping from original vector space to space of real numbers. I can guess, one such purpose is for non Cartesian coordinate systems. We have a vector as $\overline{ a} = a_{i}e^i = a^je_j$. Here $e^i,e_j$ are dual basis vectors. Is this the purpose for dual vector space?

I appreciate any inputs and thanks in advance!

• One reason that we care about dual vector spaces is that we can identify $\operatorname{Hom}(G, H)$ with the tensor product $G^* \otimes H$. Relatedly (and using the fact that $H^{**}$ is canonically isomorphic to $H$ for finite-dimensional $H$), any linear map $f: G \to H$ can be canonically identified with a map $f^*: H^* \to G^*$, namely, $(f^*(\phi))(g) := \phi(f(g))$. – Travis Willse Dec 29 '15 at 2:05

Sometimes notation $HOM(G,H)$ is used to indicate vector space of all linear maps. Does this have to do with homomorphism? Can someone comment what is role of homomorphism here?

Definition: A linear homomorphism is a linear map between vector spaces.

Definition: $\operatorname{Hom}(V,W)$, where $V$ and $W$ are two $\Bbb R$-vector spaces, is the collection of all linear homomorphisms from $V$ to $W$. With the usual notions of addition and scalar multiplication of functions $\operatorname{Hom}(V,W)$ is itself an $\Bbb R$-vector space.

Definition: The dual space to an $\Bbb R$-vector space $V$ is defined as $V^* = \operatorname{Hom}(V,\Bbb R)$.

When $G,H$ are both real vector spaces, why can't we say dual space $G^*$ is $G^*: G \rightarrow H$?

I don't see how you think this should work. For instance, how would you choose the correct $H$? The actual definition of the dual space is given above.

I did not understand the purpose of defining another vector space mapping from original vector space to space of real numbers. I can guess, one such purpose is for non Cartesian coordinate systems. We have a vector as $\overline{ a} = a_{i}e^i = a^je_j$. Here $e^i,e_j$ are dual basis vectors. Is this the purpose for dual vector space?

One immediate application of the dual space is indeed that it provides a nice way to expand vectors in a non-orthogonal basis.

Let $V$ be an $n$-dimensional $\Bbb R$-vector space. If $\{e_1, \dots, e_n\}\subset V$ is some (not necessarily orthogonal) basis for $V$ and $\{f^1, \dots, f^n\}\subset V^*$ is its dual basis, then for any vector $v\in\Bbb V$ we have $$v = \langle v, f^1\rangle e_1 + \cdots + \langle v, f^n\rangle e_n$$ where $\langle v, f^i\rangle := f^i(v)$.

Be careful though. It's not true that $a_{i}e^i = a^je_j$ as the LHS and RHS are two different types of objects (existing in two different vector spaces). I've seen physicists state this before but strictly speaking $V$ and $V^*$ are always distinct spaces (I think).

In the setting of vector calculus (on a smooth $n$-dimensional manifold $M$), you have notions of local coordinates, change of coordinates, smooth functions, and smooth curves.

A smooth curve $\gamma:(-\eps, \eps) \to M$ defines a tangent vector $v = \gamma'(0)$ at the point $p = \gamma(0)$. The set of all tangent vectors at $p$ turns out to be an $n$-dimensional vector space, the tangent space $T_{p}M$.

A smooth, real-valued function $f:M \to \Reals$ defines a differential $df$. The "value" $df(p):T_{p}M \to \Reals$, which satisfies $$\frac{d}{dt}\bigg|_{t = 0} (f \circ \gamma)(t) = \bigl[df(p)\bigr](v),$$ is linear in $v$, and therefore constitutes a covector, i.e., an element of the vector space dual to $T_{p}M$.

A finite-dimensional real vector space $V$ and its dual space $V^{*}$ have the same dimension, and so are isomorphic. However, there is no basis-independent way to define an isomorphism of $V$ with $V^{*}$. Consequently, vector-valued functions on $M$ (a.k.a. vector fields) and covector-valued functions on $M$ (a.k.a. differential one-forms) are distinct types of entities. For instance:

• If $\phi:M \to N$ is a smooth map of smooth manifolds, a one-form on $N$ pulls back (in a coordinate-invariant way) to a one-form on $M$, but a vector field on $N$ does not pull back in a coordinate-invariant way to a vector field on $M$. (A tangent vector on $M$ naturally pushes forward to a tangent vector on $N$, but a vector field on $M$ does not naturally push forward to a vector field on $N$ in general.)

• If $f:M \to \Reals$ is smooth, the differential $df$ depends only on the smooth structure of $M$, but there is no natural way to define a gradient vector field of $f$.

• A smooth curve in $M$ has a natural notion of velocity (first derivative), but has no natural notion of acceleration (second derivative).

This list is in no way comprehensive. All such items come down to the chain rule (or, in fancy language, to the transition functions for the tangent and cotangent bundles of $M$), see also Dual space and covectors: force, work and energy for detailed computations along similar lines.

The second and third items are often avoided by working on a Riemannian manifold, equipped with a Riemannian metric $g$ (which can be used to define an isomorphism between vectors and covectors) and its Levi-Civita connection (which defines covariant differentiation, such as acceleration of a curve).

The fact that a function has a gradient field and a smooth curve has an acceleration in elementary vector calculus secretly relies on the Euclidean metric on Cartesian space.

Tensor Analysis on Manifolds by Bishop and Goldberg, and Comprehensive Introduction to Differential Geometry, Volume I by Spivak, are excellent sources for further reading. (There are surely more recent treatments, as well.)


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# 2 + 4 + 7 + 11 + 16 + ...

Question:

2 + 4 + 7 + 11 + 16 + ...

Solution:

Let $S_{n}$ be the sum of $n$ terms and $T_{n}$ be the $n$th term of the given series.

Thus, we have:

$S_{n}=2+4+7+11+16+\ldots+T_{n-1}+T_{n}$   ....(1)

Equation (1) can be rewritten as

$S_{n}=2+4+7+11+16+\ldots+T_{n-1}+T_{n}$   ...(2)

On subtracting (2) from (1), we get:

$\Rightarrow 2+\left[\frac{(n-1)}{2}(4+(n-2) 1)\right]-T_{n}=0$

$\Rightarrow 2+\left[\frac{(n-1)}{2}(n+2)\right]-T_{n}=0$

$\Rightarrow 2+\left[\frac{n^{2}+n}{2}-1\right]-T_{n}=0$

$\Rightarrow\left[\frac{n^{2}}{2}+\frac{n}{2}+1\right]=T_{n}$

$\because S_{n}=\sum_{k=1}^{n} T_{k}$

$\therefore S_{n}=\sum_{k=1}^{n}\left(\frac{k^{2}}{2}+\frac{k}{2}+1\right)$

$=\frac{1}{2} \sum_{k=1}^{n} k^{2}+\frac{1}{2} \sum_{k=1}^{n} k+\sum_{k=1}^{n} 1$

$=\frac{n(n+1)(2 n+1)}{12}+\frac{n(n+1)}{4}+n$

$=n\left(\frac{2 n^{2}+3 n+1+3 n+3+12}{12}\right)$

$=\frac{n}{12}\left(2 n^{2}+6 n+16\right)$

$=\frac{n}{6}\left(n^{2}+3 n+8\right)$


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# Tag Info

0

Your ordering of the required objects doesn't line up with the declaration. You're requiring Chart, Legend, Default, and then Tooltip while you're declaring Chart, Legend, and then Tooltip. So what you've done is you named the Default chart to Tooltip and tried using it. Try fixing the order and that should work.

0

It's exactly like the error says: Uncaught ReferenceError: mynamespace is not defined I don't know why you use it, but if you want to attach widgets to an object ("namespace"), then you'll have to create it first: var mynamespace = {}; This logs the given instance to the console: https://jsfiddle.net/7p8hd1v6/2/ You don't have to use the ...

0

There is no "parent" concept in declare. Inheritance via declare is mixing in properties into the prototype of an object. The first argument of declare is an array of objects to use as base prototypes. The second argument, your object that you pass, becomes the "immediate" (top level) prototype of objects created with the class returned by declare. ...

1

OK, so the answer is to use then, as it seems the add method returns a Deferred. So you can do: store.add(...).then(function(value){ /* success */ }, function(error){ /* error */ })

0

Before answering your question, I would like to say that I think there is other options to do this without customizing ICN. The two easiest I can see would be: Use entry templates, that way class selector will be disable on check in by default Remove the privChangeClass privilege, that should hide the Class Selector (might be difficult if you want to leave ...

0

I think I have answered my own question. The key points are: 1) using define(['jquery']) in 3rd party component is correct 2) since jquery is defined before arcgis dojo, we have to write the following after dojo has been loaded to fix it // jquery was loaded before AMD is available, now that AMD is made available after loading dojo, // register jquery as ...

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It's not clear to me what you're really trying to do here, but it seems like you're creating your own problem in the code you added in the HTML pane of that fiddle. jQuery already defines itself as an AMD module if it is loaded in the presence of an AMD loader. If you load jQuery second, the following will work, because jQuery happens to define itself with ...

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Try adding: -webkit-appearance: none; to all input elements in CSS. The input below should appear the same on all browsers (including mobile Safari) input[type=text] { -webkit-appearance: none; border: 1px solid black; border-radius: 0px; outline: none; width: 200px; margin: 0px; padding: 4px; } <input type="text" /> ...

2

If your RadioButton widgets happen to be within a dijit/form/Form, the most convenient way by far to do this would be by setting the form's value, since dijit/form/_FormMixin's value setter will automatically map the value to the appropriate RadioButton widget. form.set('value', { group1: '1' }); (Even if the form has other fields, calling set('value') ...

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Here is a way to do that with dojo function test(json) { var obj = dojo.fromJson(json); for (prop in obj) { dojo.query('input[name="' + prop + '"][value="' + obj[prop] + '"]').attr('checked', 'checked'); } } Where json is the string you receive, and obj is the corresponding object, after decoding from JSON. This code goes through the ...

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Solved! I solved the problems that I was asking for in the edits. The problem that remained was with the uri generation - I made it dynamically through Orchard's ViewBag, and I didn't notice it's wrong until I changed the address to another service. Now I use a static URI and everything works right.

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There are several things you have to do. First, if you want to append Dijit widget with jQuery, you could do the stuff you mentioned, for example: $('#generatedContent').append('<button data-dojo-type="dijit/form/Button">My button</button>'); But, you also have to import the widgets you want and the parser, which parses declarative markup to ... 0 From domNode get widget_id and do registry.byId to get the widget and everytime there is a drag drop event, recreate the array based on the updated nodes in drag drop object. 0 Suggestions: mention the MobileFirst version you are working with provide actual code Here's why it is failing for you... When you test in a desktop browser, you are not in a mobile app. You are in a desktop browser. In a desktop browser in your PC you have access to the local filesystem and that's why the widget worked for you. But when you are in a ... 1 If you have the id of the node inside the tree, you can use the getNodesByItem function that tree has. Although if your tree is dynamic and the contents can change, I would suggest writing a function that not only adds to your store but also adds to a class for the node formed in the tree. 2 You might also want to try lang.exists() https://dojotoolkit.org/reference-guide/1.10/dojo/_base/lang.html#dojo-base-lang-exists if (lang.exists('view.view.formView._dapSections.scrollTop', this) { globals.lastScrollPosition = this.view.formView._dapSections.scrollTop; } 0 fixed by adding following css incase anyone comes across the same problem. .dijitPopup { position: absolute; } 0 Whether a widget is focusable or not depends on the tabIndex attribute of the corresponding DOM node. Maybe removing it from your Datagrid will do the job. Have a look here. 1 This is precisely the sort of thing that dojo/_base/lang.getObject is designed for. var scrollTop = lang.getObject('view.formView._dapSections.scrollTop', false, this); if (scrollTop) { globals.lastScrollPosition = scrollTop; } The first argument is a string representing the properties on the object to look up The second argument is whether to create ... 2 white-space: nowrap should work just fine. However, it's not feasible to get dgrid to automatically determine column widths based on cell contents due to the way it is designed (as well as generally any other dynamic data grid component including dojox grid and gridx, if I'm not mistaken). dgrid renders each row in its own table, which makes features like ... 1 First, you should be using your connection and command objects. They implement IDisposable which means they have resources to free, and you don't free them. Second, use parameterized queries. You are very open to SQL injection right now. Third, there's no reason to set the CommandType. It defaults to Text. The problem appears to be in your ... 0 My issue was invalid/depreciated mysql. I rewrote my MYSQL with MYSQLi and everything fell into place. 3 As far as I'm aware, dojo/request does not export a global (as modern AMD modules generally shouldn't need to), so dojo.request would never work in any context. The appropriate way to use AMD modules is to use them within the body of a require callback (or, even better, organize your own code in AMD modules as well, and use them within a define factory ... 0 I usually put my map.on("update-end", endUpdateMap); sentence after any map.addlayer() in the same javascript where you are defining your layers. It works fine for me. map.on("update-end", endUpdateMap); function endUpdateMap(error) { esri.hide(loading); map.enableMapNavigation(); map.disableDoubleClickZoom(); ... 0 No, there is no real difference. The term "mixin" is actually quite common in multi-inheritance systems, simply to indicate that it only contains partial functionality. A real life example would be this: Single inheritance: Dog extends Mammal and Mammal contains all basic functionality of a mammal, like 4 legs, lungs, breaths air, breeding type, ... . ... 0 I had somehow overlooked that there was an error on the console. The problem is that the web service being called in the background does not return the definition of the calendar item / appointment, and the Dojo calendar framework is looking for the begin and end dates in the response. As a result, it is raising an error, and as far as it is concerned, the ... 0 This happens because the floating pane is built as an absolutely positioned element and it's styles are also meant to match that. So when setting 40px from the top of the relative box, it actually adds 100 based on it's parent. Depending on what you're trying to do with this, I would recommend not putting the floating pane inside a div that has an inherent ... 2 Also, I don't know if you've seen this, but Esri does have some angular modules that help integrate maps into Angular apps: https://github.com/Esri/angular-esri-map That includes a service called esriLoader that wraps Dojo's require and allows you to either pass a callback that will receive the modules or get a promise that is resolved as an array with the ... 0 Parents position of the floating pane is 100px down from top (this 100px is the height of first div inside class "tundra"). So when it gets instruction of adding 40px more, it moves total 140px from top. rewriting the HTML code as below, will solve the issue. the change is to set 1st div's position to relative <div ... 1 Without knowing anything about the declare function, one option is the following: define(["dojo/_base/declare"], function(declare) { function myReusableFunction() { console.log('myReusableFunction'); } return declare(null, { MY_PROPERTY1: "property1"; MY_PROPERTY2: "property2"; constructor: function() { ... 3 Single-argument require only works for modules that have already been loaded. Judging by the undefinedModule error, you're trying to use single-argument require on modules that were never loaded. To reference modules that have not yet been loaded, you need to use the standard AMD require syntax: require([ 'dijit/layout/StackContainer', ... ], ... 1 If your process is asynchronous, consider using dojo/Deferred. dojo/Deferred will return a promise that gets resolved when the asynchronous thread is complete. Solution using setTimeOut will force you to specify a duration for your process, as you process could resolve sooner or later, hardcoding a value in setTimeOut it is not a good option has you risk ... 0 You can use dojox.data.CsvStore to read a CSV store and use it in a data grid. 0 See here: IBM Worklight - WL.App.getDeviceLanguage() API method does not return correct language code in iOS As well as the following Translation tutorial for MobileFirst Platform Foundation: https://developer.ibm.com/mobilefirstplatform/documentation/getting-started-7-1/foundation/advanced-client-side-development/enabling-translation/ 1 Is this the Dojo Filtering Select Extension Library control? If so, XPages Extension Library book covers those components. Mastering XPages 2nd Edition covers the core controls and generic XPages runtime topics. Without code or more details it's hard to tell what the cause is. It sounds like the relevant Dojo classes aren't loaded. That could be because the ... 0 Use dojo's Publish/Subscribe model, You can visually change the radio buttons and in the very next line publish to a channel from onClick handler that do post processing. Dojo's Pub/Sub documentation. 0 This happens because before a node is opened in a tree, it's contents do not exist yet. They're loaded after a node is opened up in a tree. What you can do is plug into the tree's onOpen event and then decide there whether you should check the checkboxes of all the nodes that were just created. 0 Looking at how the dojo parser works, it would appear that it passes value of the data-dojo-props attribute into eval. You end up with something like this: eval("{" + "label:'${i18n.mySelectorLabel}'" + "}"); This would seem to imply that the issue outlined in the question is a fundamental limitation of dojo. The workaround proposed by vogomatix works ...

1

What you have is fine as the onClick method is meant to be overwritten to hook into events. What you can also do is hook into your IconPages "click" event using dojo/on by doing something like this: on(this._iconPage, "click", /*function here*/);

1

I know this issue, in my case i had troubles to get things working with dojo/text. To solve it try following: Get the requirejs version of i18n on GitHub https://github.com/requirejs/i18n/blob/master/i18n.js Include it to your project. Best is to place it in a subfolder which is not deployed. In my case its ./libs/build/i18n.js (don't worry, its only used ...

0

You can use dojo/aspect and just use aspect.after the click event to do any of your processing

0

Alright, so until someone comes out with a better solution, I'm just going to create symlinks within my intern-tests/ project folder. To do it on Windows, this looks like the following: mklink /j "c:\www\intern-tests\project-symlinks\project2" c:\www\project2 which has the convention mklink /j "path\to\symlink" path\to\original (the /j is for junction) ...

0

I have finally made the Dgrid Tree hierarchy working after substiture Rest store to RequestMemory store and put in additional field hasChildren into dataset to be used in mayHaveChildren method and also getChildren after adviced by Ken Franqueiro. {"id":"C1","name":"Test 2","qty":0.0,"parent":"T1", "hasChidlren":true} var StandardGrid = ...

0

Ok, I don't know if this is the 'best' answer but I have one now. I added a global variable inside the 'test.html' page <script> var theGeom; //This is the variable require([ 'dojo/dom', here is where I am setting the value of this variable for use in the original dojo 'require' code block. This is coming from the 'testModule.js' ...

0

I was able to solve my problem using constrainedMoveable from dojo/dnd/move. Basically it is possible to setup an "area" where the dragging is only allowed, I used constrainedMoveable as I need to dynamically calculated the "area" . Actually dojo/dnd/move offers severals specialized Movables: constrainedMoveable used to constrain a move to a dynamically ...

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If you're using dojo before 1.9, then you can do something like this: var multiSelect = new MultiSelect({ multiple: false, size: 3 }); Or the non-programmatic (declarative) way: <select data-dojo-type="dijit/form/MultiSelect" data-dojo-props="multiple: false, size: 3"> <option value="1">1</option> <option ...

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I rewrote the code as follows: function updateOkButton(strength) { if(strength === "STRONG") { // domAttr.set(okButton, 'disabled', false); require(['dojo/on', 'dojo/dom', 'dojo/dom-attr'], function (on, dom, domAttr) { var okButton = dom.byId("ok"); domAttr.set(okButton, 'disabled', false); }); } ...

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We can make use of the HorizontalRangeSlider attach points and we can place the value that you want to show by placing there. HorizontalSRangeSlider has two attach point sliderHandle and sliderHandleMax,we can place the value there like this, this.horizontalSlider = new HorizontalRangeSlider({ name : "slider", ...

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I'm just going to throw this out there speculatively. Try reversing the use of single and double quotes in the template: <div> <select data-dojo-props='label:"\${i18n.mySelectorLabel}"'> </select> </div> Update: Unfortunately this doesn't work all the time either - see comments by OP for further explanation.

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Although this post is 2 years old, i really benefit from it, because i suffered the same problem. But my fault was to set the overwrite flag as second param when calling the observe method (facepalm). So if any of you stick on this problem nevertheless, make sure to set the includeObjectUpdates param. resultSet.observe(listener, includeObjectUpdates); ...

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# Using plant physiological stable oxygen isotope models to counter food fraud

Cueni, Florian and Nelson, Daniel B. and Boner, Markus and Kahmen, Ansgar. (2021) Using plant physiological stable oxygen isotope models to counter food fraud. Scientific reports, 11 (1). p. 17314.

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Official URL: https://edoc.unibas.ch/84975/

## Abstract

Fraudulent food products, especially regarding false claims of geographic origin, impose economic damages of $30-$40 billion per year. Stable isotope methods, using oxygen isotopes (δ 18 O) in particular, are the leading forensic tools for identifying these crimes. Plant physiological stable oxygen isotope models simulate how precipitation δ 18 O values and climatic variables shape the δ 18 O values of water and organic compounds in plants. These models have the potential to simplify, speed up, and improve conventional stable isotope applications and produce temporally resolved, accurate, and precise region-of-origin assignments for agricultural food products. However, the validation of these models and thus the best choice of model parameters and input variables have limited the application of the models for the origin identification of food. In our study we test model predictions against a unique 11-year European strawberry δ 18 O reference dataset to evaluate how choices of input variable sources and model parameterization impact the prediction skill of the model. Our results show that modifying leaf-based model parameters specifically for fruit and with product-independent, but growth time specific environmental input data, plant physiological isotope models offer a new and dynamic method that can accurately predict the geographic origin of a plant product and can advance the field of stable isotope analysis to counter food fraud.
Faculties and Departments: 05 Faculty of Science > Departement Umweltwissenschaften > Integrative Biologie > Physiological Plant Ecology (Kahmen) Kahmen, Ansgar and Cueni, Florian and Nelson, Daniel Article, refereed Research Article 2045-2322 Publication type according to Uni Basel Research Database: Journal article doi: 10.1038/s41598-021-96722-9isi: 000691009100008pmid: 34453087 24 Feb 2022 14:23 27 Jan 2022 12:57

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# How does the graph of percolation probability $\Pi$ vs. $p$ vary for different finite values of $L$?

This is a sequel to my previous question. @Carlo's response here (to my comment) prompted me to ask this question:

As mentioned in the textbook "Introduction to Percolation Theory" (Chapter 4) by Stauffer et al., the variation of spanning cluster percolation probability $\Pi$ in a finite $L < \infty$ square lattice has the following nature:

Does anyone know any reference where similarly the variation of $\Pi$ is shown for different finite values of $L$?

Ideally something like Figure 3 of the paper linked by @Carlo but with $p$ instead of "seed mass" on the $x$-axis.

I ask because upon reading the linked paper, I'm not very convinced that $p$ is proportional to "seed mass". However, if someone could explain why exactly seed mass should be considered to be analogous to $p$, that would be an acceptable answer too.