text:
#### Question 1

Find the midpoint of the following pairs of coordinates:

 a b A(−2, −3) and B(−2, 5) (−3, 1) (−2, 1) (−2, 0) A(−2, 2) and B(5, 2) (3, 2) (2, 3) (1$$\dfrac{1}{2}$$, 2)

#### Question 2

Find the coordinates of the midpoint of the line interval joining the following points:

 a b A(−3, 2) and B(6, 8) (7, $$\dfrac{1}{2}$$) (1$$\dfrac{1}{2}$$, 5) (4$$\dfrac{1}{2}$$, 2) A(−2, 8) and B(1, −1) (−$$\dfrac{1}{2}$$, 3$$\dfrac{1}{2}$$) (3, 0) (3$$\dfrac{1}{2}$$, $$\dfrac{1}{2}$$)

#### Question 3

The coordinates of the midpoint of the interval joining the points (−2, −5) and (3, 5) are:

 ($$-3\dfrac{1}{2}$$, 4) ($$\dfrac{1}{2}$$, 0) (2$$\dfrac{1}{2}$$, 5)


text:
.

# Bayesian Inference

In the classical approach, a probability is regarded as a long run frequency/propensity. We take random samples from a population of infinite size, and parameter $\theta$ is fixed.

In the Bayesian approach, a probability is a subjective degree of belief. Parameters themselves are regarded as random, and beliefs are updated based on data.

## Ingredients

• Model for the data given some known parameters, $f_{\left.X\right|p}\left(\left.x\right|p\right)$.
• Prior distribution of parameters, $f_{p}\left(p\right)$.
• Bayes Theorem: $\underset{\text{posterior distribution}}{\underbrace{f_{\left.p\right|X}\left(\left.p\right|x\right)}}=\frac{f_{X,p}\left(X,p\right)}{f_{X}\left(x\right)}=\frac{\overset{\text{likelihood function}}{\overbrace{f_{\left.X\right|p}\left(\left.x\right|p\right)}}.\overset{\text{prior distribution}}{\overbrace{f_{p}\left(p\right)}}}{\int f_{\left.X\right|p}\left(\left.x\right|p\right)f_{p}\left(p\right)dp}$

In this framework, estimating a parameter means finding $f_{\left.p\right|X}$ for some data $x_{1}..x_{n}$. Unlike in the classical approach, parameters are now distributed according to function $f_{p}\left(\cdot\right)$, although we’ll keep writing them in lowercase. In actual estimation, this function may encompass the researcher’s past experience or data from previous experiments. In practice, the function is often chosen to be relatively uninformative.

The posterior distribution is the updated distribution of the parameters, conditional on data. We may use the distribution to generate point estimates (by taking expectations of the parameters, for example).

format: plain


text:
# Small Signal Analysis

 Question 1
For an ideal MOSFET biased in saturation, the magnitude of the small signal current gain for a common drain amplifier is
 A 0 B 1 C 100 D infinite
GATE EE 2022   Analog Electronics
Question 1 Explanation:
For ideal MOSFET, $i_G=0$
Therefore, Current gain, $A_I=\frac{i_s}{i_G}=\infty$
 Question 2
The magnitude of the mid-band voltage gain of the circuit shown in figure is (assuming $h_{fe}$ of the transistor to be 100)
 A 1 B 10 C 20 D 100
GATE EE 2014-SET-1   Analog Electronics
Question 2 Explanation:
AC model,

$\; \; \; \; Z_{i}=10\, k\Omega$
Mid band voltage gain $\; \; \; \; A_{v}=\frac{A_{I}R_{L}}{Z_{i}}$
$\; \; \; \; A_{v}=\frac{-h_{fe}R_{L}}{Z_{i}} =\frac{-100\times 10\, k\Omega }{10\, k\Omega }=-100$
$\; \; \; \; \left | A_{v} \right |=100$
 Question 3
In the single-stage transistor amplifier circuit shown in figure, the capacitor $C_{E}$ is removed. Then, the ac small-signal midband voltage gain of the amplifier
 A increases B decreases C is unaffected D drops to zero
GATE EE 2001   Analog Electronics
Question 3 Explanation:
$\frac{A_{V_1}}{A_{V_1}}=1+\left ( \frac{1+h_{fe}}{h_{ie}} \right )R_{e}$
$\frac{A_{V_{1}}}{A_{V_{2}}}> 1$
$\Rightarrow \; A_{V_{2}} \lt A_{V_{1}}$
There are 3 questions to complete.


text:
AN-VI

This article was considered for deletion at Wikipedia on October 17 2016. This is a backup of Wikipedia:AN-VI. All of its AfDs can be found at Wikipedia:Special:PrefixIndex/Wikipedia:Articles_for_deletion/AN-VI, the first at Wikipedia:Wikipedia:Articles_for_deletion/AN-VI. Purge

The topic of this article may not meet Wikipedia's general notability guideline. But, that doesn't mean someone has to… establish notability by citing reliable secondary sources that are independent of the topic and provide significant coverage of it beyond its mere trivial mention. (October 2012)

AN-VI is a member of the rubber family, with characteristics different from those of traditional elastomers. It has the elasticity of a good rubber but it also has unusual damping capacities: by deformation it absorbs energy from an elastic material and as a damper it dissipates the energy absorbed. One part of the energy is converted into heat while the other part is returned so slowly as to have practically no dynamic effects. In solid form AN-VI can be used to reduce shock and impact stress effects as a vibration damper.Foamed, it finds use with not too heavy loads.

AN-VI is the industrial evolution of a product, Noene, used for years in the medical field, where the material has also shown a very low tendency to aging.

AN-VI foamed material, magnified
AN-VI foamed material, magnified

Mechanical features

Shock absorption

Shock absorption

The image shows the reduction of shock effects and the attenuation of impact vibrations generated by repeated blows of metal on metal. The graphs illustrate the force transmitted by a free-falling steel ball on a load cell. Between the ball and the instrument there are three test pieces consisting alternatively of natural rubber, ethyl-vinylacetate (EVA) and AN-VI. The test was performed under equal conditions (highness of the fall and size of the pieces): AN-VI absorbs the shock energy completely (there is no second or rebound shock) and this reduces the transmitted force.

Rebound elasticity

Rbound elasticity

One of the major characteristics which sets AN-VI apart from other good quality elastomers is its rebound elasticity. This quantity is a parameter expressing the capacity of the material to dissipate mechanical energy. It is measured by means of an instrument called the Rebound Pendulum which measures how much mechanical energy is put back into a steel ball falling from a specific height on to a test piece of the material under examination. With AN-VI, for temperature between 0° and 30°, less than 5% of the ball’s prefall potential energy is put back into the ball. The remaining 95% is dissipated by the AN-VI.

On the picture on the right the AN-VI’s rebound elasticity is illustrated in comparison with that of other common elastomers, as a function of temperature. In a range of temperature from -20° to +60 °C, which includes most normal applications, the rebound elasticity of An-VI is lower than that of all other elastomers.

Loss factor

The picture on the left shows the behavior of AN-VI according to the frequency of vibration, compared with other different elastomers.

The chart was defined in 2012 from the Politecnico di Torino - Interdisciplinary Mechatronics Laboratory during an experiment carried out using AN-VI to stabilize electrodynamics rotor-bearings.

Damping properties

AN-VI’s capacity in damping impulse and transitory vibrations as well as stationary and pseudostationary dynamic actions is apparent in its hysteresis behaviour.

Referring to the symbol table and to the oscillator model below,

Oscillation model
Parameter Symbol Unit
Time t S
Frequency f Hz
Pulsation w = 6,28 f rad/s
Displacement s(t) m
Speed v(t) = ds(t) / dt m/s
Acceleration a(t) = dv (t) / dt (m/s)/s
Force F (t) N
Mass M Kg
Elastic stiffness K N/m
Viscous damping B N /(m/s)
Hysteresis stiffness C N/m
Imaginary unit i i
Viscous damping factor nv = B / (2 Sqr (K/M)) ad
Hysteresis damping factor ni= C/2K ad
Natural pulsation wo= Sqr (K/M) rad/s
Natural frequency fo= (1/6,28) Sqr (K/M) Hz
Natural period To= 1/fo s
Natural damping factor no=nv + ni ad
$F (t) = Ma (t) + Bv (t) + (K + iC) s (t) \!$

Take test pieces of materials and subject them to a squash s (t) of unit amplitude (1 cm),changing according to harmonic law (with pulsation w). Then:

$s (t) = (1 cm) sin (wt) \!$

Graph the relationship between the squash (s) and the force (F) required to produce the desired displacement

The behaviour shown in the first figure refers to a perfectly elastic material, such as steel.

The relation between (s) and (F) is a straight line whose slope (K) represents the elastic stiffness of the sample.

$F (t) = K s (t) = K ((1 cm) sin (wt)) \!$

The elastic energy (Ee) required to cover the load-phase (half circle)

$Ee = \frac{1}{2}K (2 cm)^2 \!$

Is given back in total in the upload phase.

The behaviour of the elastomers, shown in the other figures on next page, diverges from that of the steel in that the straight lines of steel becomes an s/F closed curve and represents the hysteresis circle of the material.

The area enclosed by the curve is a measure of the hysteresis Energy (E) expended in one circle of the system.

The ratio (Ri) between hysteresis energy (Ei) and the energy associated with the elastic half-cycle (Ee) is an index of the material’s hysteretic dissipative capacities:

$Ri = \frac{Ei}{Ee} \!$

The hysteresis properties of materials are expressed in terms of hysteretic stiffness $(C)$, or hysteretic damping coefficient, which has the same dimensions as elastic stiffness $(K)$, and therefore the total force $(F)$ to be applied to the sample to produce the cycle is given by:

$F(t) = (K + iC) x (t) = K (1 cm) sin (wt) + C (1 cm) cos (wt) \!$

The ratio (2 nl) between the two stiffnesses represents the loss factor or hysteresis damping factor or hysteresis damping factor.

The parameter:

$nl = \frac{C}{2K} \!$

is connected to the Energy ratio $(Rl)$ by the formula:

$Rl = (3,14) nl \!$

The greater the hysteresis capacity for dissipating mechanical Energy, the fatter is the cycle shape. With the test, the difference between the behaviours of AN-VI and other elastomers can be quantified. Given that all good quality rubbers have respectable hysteresis damping factor (ni), AN-VI’s particularly high.

Parameter Rubber AN-VI
Rebound elasticity at 20 °C

Loss factor 2ni = C / K

Hysteresis factor ni = C / (2K)

Energy ratio Ri = Ei / Ee

Dal 20% al 40%

From 10% to 30%

From 5% to 15%

From 16% to 47%

From 2% to 5%

From 120% to 80%

From 60% to 90%

From 188% to 283%

Transitory phenomena

Transitory phenomena have their origin in events which last only for a very short time. An example is the rectangular impulse illustrated below in the figure, resulting from the application of a load (F), which remains constant for a period of time (T). In fact the response of the mechanical system is usually characterized by complex oscillatory phenomena during the loading phase, in passing from (0) to (F).

The higher the natural damping factor (no) of the system, the quicker the phenomenon disappears. The diagrams give evidence of the differences in behaviour between an ordinary rubber (no = 30%) and AN-VI (no = 60%). The extent of displacement depends upon the stiffness (K) of the system. The response oscillation amplitude a (t) is inversely proportional to the mass (M). The diagrams confirm that significant effects are obtained through the use of AN-VI insofar as the transitory phenomena caused by the system’s free oscillations are concerned, both in the loading phase and in the unloading phase.

The shorter oscillation damping time also reduces the risk of dangerous synchronizations due to the superimposing of return waves on the outgoing waves.

The pictures show the transitory for the different materials.

Transitory phenomena

Transitory phenomena have their origin in events which last only for a very short time.

A theoretical example, schematically very simple, is the rectangular impulse illustrated below in the figure, resulting from the application of a load ( F ), which remains constant for a period of time ( T ). In fact the response of the mechanical system is usually characterized by complex oscillatory phenomena during the loading phase, in passing from ( 0 ) to ( F ).

Higher is the natural damping factor ( no ) of the system quicker is the phenomena disappearing.

The diagrams give evidence of the differences in behaviour between an ordinary rubber ( no = 30% ) and AN-VI ( no = 60% ).

The extent of displacement depends upon the stiffness ( K ) of the system.

Response oscillation amplitude a ( t ) is inversely proportional to the mass ( M ). The diagrams confirm that significant effects are obtained through the use of AN-VI insofar as the transitory phenomena caused by the system’s free oscillations are concerned, both in the loading phase and in the unloading phase.

The shorter oscillation damping time also reduces the risk of dangerous synchronizations due to the superimposing of return waves on the outgoing waves.

The pictures show the transitory for the different materials.

$Force(N)\!$

RUBBER (on the left) Vs. AN-VI (on the right) - Politecnico di Milano

$Displacement(cm)\!$

$Acceleration (cm/s^2)\!$

Electrical features

AN-VI is classified, from an electrical point of view as an insulating material.

Anyway, even if it cannot be classified as material with ESD protection, the quantity of the charge generated by triboelectric effect is limited: undergoing to a rubbing process, AN-VI features an accumulation inferior to 500V, while normal insulating plastics gets tens of thousands Volt.

According to the above features, the material results to be particularly suitable to be used in electrical and electronics field in direct contact with circuitry without any particular risk of conduction or electrostatic discharges.

Use

Industry

Vibration deadening in rotatory and reciprocating machines, linings for hoppers and containers suffering shock from falling materials, vibratory separators, metal panel deadening, handgrip covers, portable machine tool uncouplers

Automotive, Rail, Aereospace, Naval e Yachts

Lining and covers, rebound bumpers, racing car interiors

Railwais superstructure

Rail chairs, superstructure platform supports, signlling and shunting box connections

Building

Slab supports, bearings, joints, floating floors

Electronics and precision instrumentation

Coverings, equipment supports and fixings

Hi-Fi

Coverings, supports and fixings for soundboxes and record-players

Hydraulics’

Piping anchorage, pump uncouplers, joint seals

Office machines

Supports and base plates for printers and computers

Aeronautics

Helicopter landing platforms


text:
In 4th grade numbers worksheets we will solve how to read and write large numbers, how to represent 5 and 6-digit numbers on abacus, use of place value chart to write a number in expanded form, represent the large number on the abacus, write the number in standard form, how to write a large number in words and figures, place value of a number, standard form and extended form of a number, compare with another number and arrange numbers in ascending and descending order, successor and predecessor of a number and rounding off the numbers.

1. Who am I?

(i) I am the smallest 5-digit number: _______________

(ii) I am the predecessor of 8,25,481: _______________

(iii) I am one more than 99,999: _______________

(iv) My place value in 9,83,351 is ten thousand: _______________

(v) I am the largest 6 – digit number: _______________

2. Complete the given table.

 S.No. Numeral Number name (i) 24,998 (ii) Three lakh twenty three thousand four hundred two (iii) 1,84,305 (iv) Six lakh ten thousand three hundred seven (v) 3,10,706

3. Form the largest and smallest 6–digit numbers using the given digits.

 S.No. Digits Greatest Number Smallest Number (i) 3,  9,  4,  1,  7 (ii) 9,  6,  1,  5,  2 (iii) 3,  0,  8,  2,  7,  1

4. Write the place value of underlined digits in the given blanks.

(i) 2103

(ii) 4,00,493

(iii) 9,14,825

(iv) 2,13,921

5. Represent the given numbers on the abacus:

(i) 67,751

(ii) 5,44,301

(iii) 8,54,206

6. Write the given numbers in expanded form.

(i) 63,498                             _______________

(ii) 7,70,052                         _______________

(iii) 9,72,855                        _______________

(iv) 3,88,941                        _______________

(v) 8,00,527                         _______________

7. Write the given number in standard form.

(i) 80,000 + 5000 + 200 + 90 + 6                     _______________

(ii) 60,000 + 7000 + 200 + 8                            _______________

(iii) 4,00,000 + 70,000 + 5000 + 50 + 1            _______________

8. Compare the given numbers. Put > , < or = sign.

(i) 35,493      _____          34,395

(ii) 2,90,018    _____      8,10,092

(iii) 7,44,582    _____     7,12,298

(iv) 8,17,736   _____      8,30,559

9. If in the number 2,39,987; 3 is replaced by 8 and the hundred place is replaced by 1, then what number is formed? Also give its number name.

_______________; _______________

10. Fill in the blanks:

(i) One lakh is a ……………………………… digit number.

(ii) The difference between the greatest 4-digit number and the greatest 3-digit number is ………………………………

(iii) A five digit number begins with ……………………………… place.

(iv) The sum of the place value of two 5s in 3,58,654 is ………………………………

(v) The successor of five lakh, fifty thousand, five hundred fifty is ………………………………

(vi) 1 lakh = ……………………………… thousand

(vii) 7 lakh = ……………………………… hundred

(viii) Form the largest 6-digit number with 9 at the tens place and without repeating any digits. ………………………………

(ix) Form the smallest 5-digit number with 0 at the thousands place and without repeating any digits. ………………………………

(x) The predecessor of 7,85,000 in words will be ………………………………

(xi) The next number in the series: 7,46,952, 7,57,952, 7,68,952, ………………………………

(xii) The number 100 more than 6,66,666 is ………………………………

(xiii) 1000 less than 5,98,456 is ………………………………

11. Given below is the area of 5 states of South India.

 S.NO. City Area (in Square km) (i) Telangana 1,12,077 (ii) Andhra Pradesh 1,62,969 (iii) Kerala 38,763 (iv) Karnataka 1,91,791 (v) Tamil Nadu 1,30,058

(i) Arrange the states in increasing order to their area.

(ii) Which state has digit 9 at the place value of hundred and ones.

(iii) Give the number name for the area of Andhra Pradesh and Karnataka.

(iv) Represent the area of Telangana on the babacus.

(v) Write the area of Tamil Nadu in expanded form.

(vi) Give the place value of digit 8 in area of kerala.

12. Write the predecessor and successor of the given numbers.

 S.NO. Predecessor Number Successor (i) 7,56,993 (ii) 8,99,999 (iii) 6,30,050 (iv) 9,82,311

Answers on 4th Grade Numbers Worksheets are given below to check the exact answers of the questions.

1. (i) 10,000

(ii) 8,25,480

(iii) 1,00,000

(iv) 8

(v) 9,99,999

2. (i) Twenty-four thousand, nine hundred ninety eight

(ii) 3,23,402

(iii) One lakh, eighty-four thousand, three hundred five.

(iv) 6,10,307

(v) Three lakh, ten thousand, seven hundred six.

3. (i) 97,431; 13,479

(ii) 96,521; 12,569

(iii) 8,73,210; 1,02,378

4. (i) 3

(ii) 4,00,000

(iii) 20

(iv) 2,00,000

6. Write the given numbers in expanded form.

(i) 60,000 3,000 + 400 + 90 8

(ii) 7,00,000 + 70,000 + 0 + 50 + 2

(iii) 9,00,000 + 70,000 + 2,000 + 800 + 50 + 5

(iv) 3,00,000 + 80,000 + 8,000 + 900 + 40 + 1

(v) 8,00,000 + 0 + 0 + 500 + 20 + 7

7. (i) 85,296

(ii) 67,208

(iii) 4,75,051

8. (i) 35,493 > 34,395

(ii) 2,90,018 < 8,10,092

(iii) 7,44,582 > 7,12,298

(iv) 8,17,736 < 8,30,559

9. 2,89,187; Two lakh, eighty-nine thousand, one hundred eighty-seven.

10. (i) 6

(ii) 9000

(iii) ten thousands

(iv) 50,050

(v) 5,50,551

(vi) 100

(vii) 7,000

(viii) 8,76,590

(ix) 10,234

(x) Seven lakh, eighty-four thousand, nine hundred ninety-nine

(xi) 7,79,952

(xii) 6,66,766

(xiii) 5,97,456

(iii) Andhra Pradesh – One lakh, sixty-two thousand, nine hundred sixty-nine,

Karnataka - One lakh, ninety-one thousand, seven hundred ninety-one,

(iv)

(v) 1,00,000 + 30,000 + 0 + 0 + 50 + 8

(vi) 8000

12.

 S.NO. Predecessor Number Successor (i) 7,56,992 7,56,993 7,56,994 (ii) 8,99,998 8,99,999 9,00,000 (iii) 6,30,049 6,30,050 6,30,051 (iv) 9,82,310 9,82,311 9,82,312

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# Concurrent Game Structures for Temporal STIT Logic

Boudou, Joseph and Lorini, Emiliano Concurrent Game Structures for Temporal STIT Logic. (2018) In: 17th International Joint Conference on Autonomous Agents and Multiagent Systems (AAMAS 2018), 11 July 2018 - 13 July 2018 (Stockholm, Sweden).

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## Abstract

The paper introduces a new semantics for temporal STIT logic (the logic of seeing to it that ) based on concurrent game structures (CGSs), thereby strengthening the connection between temporal STIT and existing logics for MAS including coalition logic, alternating-time temporal logic and strategy logic whose language are usually interpreted over CGSs. Moreover, it provides a complexity result for a rich temporal STIT language interpreted over these structures. The language extends that of full computation tree logic (CTL*) by individual agency operators, allowing to express sentences of the form "agent i sees to it that $\varphi$ is true, as a consequence of her choice''.

Item Type: Conference or Workshop Item (Paper) AAMAS'18 : Proceedings of the 17th International Conference on Autonomous Agents and MultiAgent Systems ISBN: 978-1-4503-5649-7 ISSN: 2523-5699 https://dl.acm.org/citation.cfm?id=3237443 hal-02382734 International conference proceedings Temporal STIT logic Université de Toulouse > Institut National Polytechnique de Toulouse - Toulouse INP (FRANCE)French research institutions > Centre National de la Recherche Scientifique - CNRS (FRANCE)Université de Toulouse > Université Toulouse III - Paul Sabatier - UT3 (FRANCE)Université de Toulouse > Université Toulouse - Jean Jaurès - UT2J (FRANCE)Université de Toulouse > Université Toulouse 1 Capitole - UT1 (FRANCE) download 21 Nov 2019 14:44

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# Definitions of Dedekind cuts in Geometry

When defining $\mathbb{R}$ from the rationals they use Dedekind cuts and there are basically two definitions:

Version 1:

A Dedekind cut is any set $A\subseteq \mathbb{Q}$ with the following three properties:

(I) $A$ is not empty, and $A\neq \mathbb{Q}$.

(II) If $p\in A, q\in \mathbb{Q}$, and $q< p$, then $q\in A$.

(III) If $p\in A$, then $p<r$ for some $r\in A$

Version 2:

A Dedekind cut is an ordered pair of subsets of $\mathbb{Q}$, $(A,B)$, satisfying:

(I) $\{A,B\}$ is a partition of $\mathbb{Q}$

(II) $p<q$ for all $p\in A$, $q\in B$

(III) $A$ does not have a greatest element

The main difference between them is that the first one names the set $A$ as a cut while in the second a cut is the pair $(A,B)$.

Now, in Geometry (From Greenberg's book) there is what I think is the analogous definition for the second version:

Dedekind cut in Geometry: A Dedekind cut of a line $l$ is a partition of $l$ into two non-empty subsets $A$ and $B$ such that no point of either subset is between two points of the other.

Question: I haven't seen any place that uses the first version of the Dedekind cuts for Geometry, and I wonder why. Is there a reason? What is the analogous (standard?) definition of Dedekid cut in Geometry for the first version?.

Thanks

• The second definition doesn't really work, because according to it there are two different cuts that intuitively represent each rational, depending on whether you put that rational in $A$ or in $B$. – Henning Makholm Nov 6 '13 at 13:31
• Well, you can use betweenness to induce a total order on the line, and just carbon copy the definition. Also, to correct the mistake @HenningMakholm hints to, people usually assume $A$ to be open (that is, it has no maximal element. – Arthur Nov 6 '13 at 13:33
• @Arthur Thanks. I just did the correction. I think your idea of inducing a total order works. It would be nice though if it could be done just like in Greenberg's definition, without appealing to extra definitions. – Daniela Diaz Nov 6 '13 at 13:52
• @DanielaDiaz: But now your geometric definition is not analogous to your Version 2, because the mistake was not fixed there. (And without having a preferred direction on the line in question, it seems to be hard do fix elegantly). (Also, where did the requirement that $A$ and $B$ are nonempty go?) – Henning Makholm Nov 6 '13 at 13:55
• @DanielaDiaz: Geometically, "unbounded" can be phrased as "not contained inside any circle". – Henning Makholm Nov 6 '13 at 16:25


text:
# Can't we obtain observational values for scale factor at different periods in the history of the universe?

This may be a stupid question, but it seems to me that we could calculate the value of the scale factor at any stage in the history of the universe by analysing the relationship between recessional velocity and distance in faraway galaxy clusters. Basically, use the same method Hubble used to get to his Hubble's Law: take a base point and then look at how distance to other galaxies varies with their velocities relative to that point? Why can't we do that?

Edit: I realised we can measure the scale factor directly using redshift, but it requires that we know the light travel distance.

• Is your question then: "does Hubble law hold for other galaxies?" Mar 18 '18 at 23:37
• @caverac Well, I assume it does, seeing as the universe is isotropic
– Max
Mar 18 '18 at 23:57
• @caverac alright, if that is my question, can you give me an answer? Any information would be appreciated, because this question is bothering me a lot
– Max
Mar 19 '18 at 22:01
• If you start assuming the universe is isotropic and homogeneous, then Hubble's law will hold everywhere, right? Are you looking for an experimental proof of that? Mar 19 '18 at 22:03
• @caverac Well, if that makes it easier for you to answer, then, yes, let's say that's my question.
– Max
Mar 20 '18 at 19:00

The question of recession velocity vs distance for distant galaxies that the question is posing is not very different than the continuing uncertainties in measuring the current Hubble constant, $H_0$. The former (velocity vs distance) is problematic because the distances measurements to far off galaxies depend on the distance ladder, i.e., determining cosmological distances accurately, whereas the second depends on distance measurements as well, but also on the cosmological model as to how the acceleration/deceleration changed over cosmological times, and the dependence of that on the relative mass and energy densities - i.e. they have some model dependence.

What is fairly clear though is that the dark energy started having enough of a noticeable effect about 5 or 6 billion years ago, based on distance measurements using the distance ladder (cepheid so to supernova and so on). So it is known, from measurements which are semi model-independent (somewhat because little in astrophysics and cosmology can be, you still depend on 'known physics' to get the distance ladder, and sometimes on simplifying assumptions), that the recession velocities are indeed time, or z dependent, and it is modeled by how the scale factor grows when the universe was radiation dominated, matter dominated, or more recently becoming dark energy dominated.

So the answer comes back to how non-model dependent can one measure cosmological distances. The fact is that there are still significant uncertianTies in the measurements of the Hubble constant using the astronomical ladder distances, and using distances estimated by the CMB-related/estimated acoustic density perturbations, and the time delay measurements of gravitationally focused multiple paths in CMB observations. The first measures about 72 Kms/sec/Mpsec, while the latter is around 67 Kms/sec/Mpsec.

Two interesting articles are on the Hubble constant measurement, the first more historical, the second more on the current Hubble constant uncertianTies, where both elucidate how measuring cosmological distances are at the heart of the issue, and the model dependencies. I've seen papers on the issues/uncertianTies of the distance and velocity measurements for different cosmological epochs, and don't remember them to be extremely more accurate, nor to be totally model independent. I don't see that, and neither do cosmologists/astrophysicists, as a big weakness, but more as the normal interplay of theory and measurement in physics. Still, while it is true that we still don't a totally accurate and certain picture of the expansion, any more accuracy in the measurements will refine the models, or find some new physics.

The two references are:

-A quick reference to the standard cosmology equations showing the model dependencies, at https://ned.ipac.caltech.edu/level5/Peacock/Peacock3_2.html

-A reference to the distance ladder, but not much on the most recent attempts to go further, in https://en.m.wikipedia.org/wiki/Cosmic_distance_ladder

• I thought we didn't have to use any cosmological models to measure the Hubble constant? For small distances, i.e z<0.5, proper distance is roughly equal to light travel distance - regardless of model chosen (this is true for all models). So where does the uncertainty come from?
– Max
Mar 22 '18 at 18:55
• They at least go out to distances where supernovas are used as standard candles, so there's astrophysics assumptions on the physics providing the inherent luminosity. See some of it at nytimes.com/2017/02/20/science/…. Even at a z of 0.5 the model makes a difference, and further out it makes more of one. See the Hubble article witH the history of how much it's varied, and some of the assumptions/methods. See cfa.harvard.edu/~dfabricant/huchra/hubble Mar 23 '18 at 3:02
• alright, you convinced me that there are large uncertainties when measuring distances to faraway objects. However, they are still a fair approximation for actual values of distance. Are the errors so high that when we try to establish the light travel distance, the error bars are so large that uncertainty range includes most cosmological models? It can't be THAT large, right? I mean, the value of the Hubble Constant is more or less agreed upon to be around 72mpckm^-1s^-1 for standard calibrations and around 67mpckm^-1s^-1 for CMB calibrations, right?
– Max
Mar 23 '18 at 21:28
• And, actually, the value of the Hubble Constant doesn't even depend on distance measures. We can measure it to a sufficiently good degree of accuracy using CMB.
– Max
Mar 23 '18 at 21:52
• So, the fact that the CMB number is significantly different than the distance based measurements is actually a point of controversy. Nobody knows which is the right one, it can't be both, something is wrong in one or both. The CMB numbers and the distance ladder numbers not agreeing is a bad, unexplained, problem. So you see, you do need distance measures, if that can lead to a number. The wrong one may depend on which wrong assumption was made in the physics modeling of one or the other - they both depend on physics models. As I explained in my answer. Mar 27 '18 at 23:58


text:
# Fitting of continuous transfer function

I measure a network's transfer function using a network analyzer and therefore have a set of complex data points, representing the network's gain and phase. I wish to fit a continuous, rational transfer function to my measured data to create a model of my network. To find a discrete-time model, there is the "Frequency Domain Least Squares" algorithm:

http://robotics.itee.uq.edu.au/~elec3004/2014/lectures/Precise%20Filter%20Design%20(chapter).pdf

Several months ago, I found a PDF which presented a very similar method to estimate a continuous transfer function, but I don't remember where it is nor do I remember its name. I only remember that it was also a simple least squares algorithm as in the above link. Does someone have a hint for this?

Update: Meanwhile I was able to recover the paper I mentioned. It is called 'Practical Transfer Function Estimation and its Application to Wide Frequency Range Representation of Transformers'. In my application, I am not dealing with a transformer but with some other passive electrical network involving RLC components, and I try to fit a transfer function to that. The network analyzer gives me some complex data for logarithmically spaced frequencies. I tried to use the method presented in the paper, but since my frequencies range from several Hz to several 100 MHz, the algorithm is unable to converge because the matrix A involves $$\\omega\$$, $$\\omega^2\$$, $$\\omega^3\$$ and so on - which leads to an ill-conditioned equation system, and therefore it cannot be solved. I wonder whether there are better algorithms than this, or whether this algorithm could be used, though, perhaps with some pre-processing of my frequency data.

Link to the paper I mentioned: https://ieeexplore.ieee.org/document/252689

• Have you tried normalizing the frequency? Feb 27 '19 at 8:37

You'll be jumping down the rabbit hole of system identification and modelling. Generally speaking you will need two things:

The first is a so-called estimator (wiki), that you can use to find the best "fit" of a general rational transfer function. The ones I've encountered most are

• Least-Squares Estimator
• Maximum-Likelihood Estimator (wiki)

They work by defining a cost function that you try to minimize using the unknown tranfer function coefficients. For the least-squares estimator, the cost function is simply the sum of the squared errors.

The second one is a model selection criterion (wiki). In order to avoid modelling the noise of your data points, you can use this criterion to formally find the optimal model complexity (order of the system).

• yes, I know we are talking about system identification here - and since I have already dealt a bit with this topic I know it is quite a rabbit hole :-) meanwhile I found the paper I was talking about - question updated. Feb 26 '19 at 17:07

The goal of Matlab function invfreqs is to identify continuous-time filter parameters from frequency response data. The idea is based on Levi, E. C. “Complex-Curve Fitting.” IRE Trans. on Automatic Control. Vol. AC-4, 1959, pp. 37–44. The main problem is to adjust n and m parameters (the orders from numerator and denominator).


text:
## Zero-divisors in polynomial rings

Posted: November 1, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: , , ,

Problem. (McCoy) Let $R$ be a commutative ring with identity and let $f(x)= \sum_{i=0}^n a_ix^i \in R[x].$

1) Prove that $f(x)$ is a zero-divisor if and only if there exists some $0 \neq c \in R$ such that $cf(x) = 0.$

2) Prove that if $R$ is reduced and $f(x)g(x)=0$ for some $g(x)=\sum_{i=0}^m b_ix^i \in R[x],$ then $a_ib_j=0$ for all $0 \leq i \leq n$ and $0 \leq j \leq m.$

Solution. 1) If there exists $0 \neq c \in R$ such that $cf(x)=0,$ then clearly $f(x)$ is a zero-divisor of $R[x].$ For the converse, let $g(x)=\sum_{i=0}^m b_ix^i, \ b_m \neq 0,$ be a polynomial with minimum degree such that $f(x)g(x)=0.$ I will show that $m = 0.$ So, suppose to the contrary, that $m \geq 1.$ If $a_jg(x)=0,$ for all $j,$ then $a_jb_m=0,$ for all $j,$ and so $b_mf(x)=0$ contradicting the minimality of $m$ because $\deg b_m = 0 < m$. So we may assume that the set $\{j: \ a_jg(x) \neq 0 \}$ is non-empty and so we can let

$\ell=\max \{j : \ a_jg(x) \neq 0 \}.$

Then

$0=f(x)g(x)=(a_{\ell}x^{\ell} + \cdots + a_0)(b_mx^m + \cdots + b_0).$

Thus

$a_{\ell}b_m=0$ and so $a_{\ell}g(x)=a_{\ell}b_{m-1}x^{m-1} + \cdots + a_{\ell}b_0.$

Hence $\deg a_{\ell}g(x) < m=\deg g.$ But we have $f(x)(a_{\ell}g(x))=a_{\ell}f(x)g(x)=0,$ which is impossible because $g(x)$ was supposed to be a polynomial with minimum degree satisfying $f(x)g(x)=0.$

2) The proof of this part is by induction over $i+j.$ It is obvious from $f(x)g(x)=0$ that $a_0b_0=0.$ Now let $0 < \ell \leq m+n$ and suppose that $a_rb_s=0$ whenever $0 \leq r+s < \ell.$ We need to show that $a_rb_s=0$ whenever $r+s=\ell.$ So suppose that $r+s=\ell.$ The coefficient of $x^{\ell}$ in $f(x)g(x)$ is clearly

$0=\sum_{i < r, \ i+j=\ell}a_ib_j+ a_rb_s + \sum_{i > r, \ i+j=\ell} a_ib_j,$

which after multiplying both sides by $a_rb_s$ gives us

$\sum_{i < r, \ i+j=\ell} a_rb_sa_ib_j+ (a_rb_s)^2 + \sum_{i > r, \ i+j=\ell} a_rb_sa_ib_j=0.$

Call this (1). Now in the first sum in (1), since $i < r,$ we have $i+s < r+s=\ell$ and hence by the induction hypothesis $a_ib_s=0.$ Thus $a_rb_sa_ib_j=0.$ So the first sum in (1) is $0.$ In the second sum in (1), since $i > r$ and $i+j=r+s=\ell,$ we have $j < s.$ Therefore by the induction hypothesis $a_rb_j=0$ and hence $a_rb_sa_ib_j=0.$ So the second sum in (1) is also $0.$ Thus (1) becomes $(a_rb_s)^2=0$ and so, since $R$ is reduced, $a_rb_s=0. \ \Box$


text:
시간 제한메모리 제한제출정답맞힌 사람정답 비율
1 초 (추가 시간 없음) 256 MB222100.000%

## 문제

Unfortunately, the boy finally got bankrupt. Looking at his receding figure, Rikka understood more about the world --- but she is still herself. Though, she still sat on the tower and spent time with the low gravity, feeling lost.

She looked at the deep dark sky, where the blue round Earth is shining overhead. Rikka thought about her families, her friends, and her homeland. Is she in her dream or a "real" world? The girl of chunibyo felt afraid for the first time since the journey began.

She saw a bright star traveling fast around the earth --- maybe a geostationary space station. How could she get there? Daydream started again.

In other words, Rikka wonders the minimum distance she needs to travel from her position $s$ to the position of the star $t$, while a sphere --- the earth staying there as an obstacle. They are in a 3-dimensional Euclidean space. $s$ and $t$ may be at the same position.

## 입력

The first line contains an integer $T (1 \leq T \leq 1000)$, the number of test cases. Then $T$ test cases follow.

Each test case contains two lines.

The first line contains four integers $o_x, o_y, o_z, r$, the positions in each dimension of the center of the sphere $o$ and its radius.

The second line contains six integers $s_x, s_y, s_z, t_x, t_y, t_z$, the positions in each dimension of the starting point $s$ and the terminal point $t$, respectively. Notice that $s$ and $t$ may be at the same position.

It is guaranteed that neither $s$ nor $t$ is strictly inside the obstacle, and each value of a position or a radius in the input belongs to $[1, 1000]$.

## 출력

Output one number, the minimum distance from $s$ to $t$ without entering the sphere obstacle. Your answer is considered correct if the absolute or relative error doesn't exceed $10^{-6}$.

## 예제 입력 1

2
2 1 1 1
1 1 1 3 1 1
2 1 1 1
1 2 2 3 1 1

## 예제 출력 1

3.14159265
2.64517298


text:
# Homomorphism from Z into a nonabelian group

1. Feb 7, 2009

### quasar987

What would be an example of a (nontrivial) homomorphism from Z into a nonabelian group??

More specifically, I am looking for two homomorphisms f, g: Z-->W such that for some n,m in Z f(n)g(m) does not commute.

Last edited: Feb 7, 2009
2. Feb 7, 2009

### quasar987

Ah! I found out that the smallest nonabelian group is the diedral group of order 6, D_3. And from its multiplication table, we see that embedded in it is a copy of Z_2 and a copy of Z_3 that contain elements that do not commute.

3. Feb 7, 2009

### StatusX

A homomorphism from Z into any group G is completely specified by picking f(1), because if f(1)=g, then f(n)=gn. And you can pick f(1) to be any element of G that you want, basically because Z is a free group.

4. Feb 8, 2009

### quasar987

Thanks for that remark StatusX.


text:
# Superfield component decompositions and the scan for prepotential supermultiplets in 10D superspaces

@article{JamesGates2020SuperfieldCD,
title={Superfield component decompositions and the scan for prepotential supermultiplets in 10D superspaces},
author={S. James Gates and Yangrui Hu and S.-N. Hazel Mak},
journal={Journal of High Energy Physics},
year={2020},
volume={2020},
pages={1-63}
}
• Published 3 November 2019
• Mathematics
• Journal of High Energy Physics
The first complete and explicit SO(1,9) Lorentz descriptions of all component fields contained in the N $$\mathcal{N}$$ = 1, N $$\mathcal{N}$$ = 2A, and N $$\mathcal{N}$$ = 2B unconstrained scalar 10D superfields are presented. These are made possible by a discovery of the dependence of the superfield component expansion on the branching rules of irreducible representations in one ordinary Lie algebra into one of its Lie subalgebras. Adinkra graphs for ten dimensional superspaces are…
Adinkra foundation of component decomposition and the scan for superconformal multiplets in 11D, $$\mathcal{N}$$ = 1 superspace
• Physics
• 2020
For the first time in the physics literature, the Lorentz representations of all 2,147,483,648 bosonic degrees of freedom and 2,147,483,648 fermionic degrees of freedom in an unconstrained eleven
Weyl covariance, and proposals for superconformal prepotentials in 10D superspaces
• Mathematics
• 2020
Proposals are made to describe the Weyl scaling transformation laws of supercovariant derivatives $\nabla{}_{\underline A}$, the torsion supertensors $T{}_{{\underline A} \, {\underline Advening to Adynkrafields: Young Tableaux to Component Fields of the 10D, N = 1 Scalar Superfield • Mathematics • 2020 Starting from higher dimensional adinkras constructed with nodes referenced by Dynkin Labels, we define "adynkras." These suggest a computationally direct way to describe the component fields Component decompositions and adynkra libraries for supermultiplets in lower dimensional superspaces • Physics Journal of High Energy Physics • 2021 Abstract We present Adynkra Libraries that can be used to explore the embedding of multiplets of component field (whether on-shell or partial on-shell) within Salam-Strathdee superfields for A Mini-Introduction To Superfield Decompositions With Branching Rules This paper provides a short introduction to scalar, bosonic, and fermionic superfield component expansion based on the branching rules of irreducible representations in one Lie algebra (in our case, Positive Configuration Space • Mathematics • 2020 We define and study the totally nonnegative part of the Chow quotient of the Grassmannian, or more simply the nonnegative configuration space. This space has a natural stratification by positive Chow ## References SHOWING 1-10 OF 71 REFERENCES On Graph-Theoretic Identifications of Adinkras, Supersymmetry Representations and Superfields • Mathematics • 2007 In this paper, we discuss off-shell representations of N-extended supersymmetry in one dimension, i.e. N-extended supersymmetric quantum mechanics, and following earlier work on the subject, we Geometrization of$N$-Extended$1$-Dimensional Supersymmetry Algebras II • Mathematics • 2013 The problem of classifying off-shell representations of the$N$-extended one-dimensional super Poincare algebra is closely related to the study of a class of decorated$N$-regular,$N\$-edge colored
Topology Types of Adinkras and the Corresponding Representations of N -Extended Supersymmetry
• Computer Science, Mathematics
• 2008
A relationship between Adinkra diagrams and quotients of N-dimensional cubes is demonstrated, and how these quotient groups correspond precisely to doubly even binary linear error-correcting codes provides a means for describing equivalence classes of Adinkras and therefore supermultiplets.
On Clifford-algebraic dimensional extension and SUSY holography
• Mathematics
• 2014
We analyze the group of maximal automorphisms of the N-extended worldline supersymmetry algebra, and its action on off-shell supermultiplets. This defines a concept of "holoraumy" that extends the
Adinkra height yielding matrix numbers: Eigenvalue equivalence classes for minimal four-color adinkras
• Mathematics
International Journal of Modern Physics A
• 2019
An adinkra is a graph-theoretic representation of space–time supersymmetry. Minimal four-color valise adinkras have been extensively studied due to their relations to minimal 4D, [Formula: see text]
The cohomology of superspace, pure spinors and invariant integrals
• Mathematics
• 2008
The superform construction of supersymmetric invariants, which consists of integrating the top component of a closed superform over spacetime, is reviewed. The cohomological methods necessary for the


text:
# scipy.spatial.distance.pdist¶

scipy.spatial.distance.pdist(X, metric='euclidean', p=2, w=None, V=None, VI=None)

Computes the pairwise distances between m original observations in n-dimensional space. Returns a condensed distance matrix Y. For each and (where ), the metric dist(u=X[i], v=X[j]) is computed and stored in the :math:ijth entry.

See squareform for information on how to calculate the index of this entry or to convert the condensed distance matrix to a redundant square matrix.

The following are common calling conventions.

1. Y = pdist(X, 'euclidean')

Computes the distance between m points using Euclidean distance (2-norm) as the distance metric between the points. The points are arranged as m n-dimensional row vectors in the matrix X.

2. Y = pdist(X, 'minkowski', p)

Computes the distances using the Minkowski distance (p-norm) where .

3. Y = pdist(X, 'cityblock')

Computes the city block or Manhattan distance between the points.

4. Y = pdist(X, 'seuclidean', V=None)

Computes the standardized Euclidean distance. The standardized Euclidean distance between two n-vectors u and v is

V is the variance vector; V[i] is the variance computed over all

the i’th components of the points. If not passed, it is automatically computed.

5. Y = pdist(X, 'sqeuclidean')

Computes the squared Euclidean distance between the vectors.

6. Y = pdist(X, 'cosine')

Computes the cosine distance between vectors u and v,

where |*|_2 is the 2 norm of its argument *.

7. Y = pdist(X, 'correlation')

Computes the correlation distance between vectors u and v. This is

where is the mean of the elements of vector v.

8. Y = pdist(X, 'hamming')

Computes the normalized Hamming distance, or the proportion of those vector elements between two n-vectors u and v which disagree. To save memory, the matrix X can be of type boolean.

9. Y = pdist(X, 'jaccard')

Computes the Jaccard distance between the points. Given two vectors, u and v, the Jaccard distance is the proportion of those elements u[i] and v[i] that disagree where at least one of them is non-zero.

10. Y = pdist(X, 'chebyshev')

Computes the Chebyshev distance between the points. The Chebyshev distance between two n-vectors u and v is the maximum norm-1 distance between their respective elements. More precisely, the distance is given by

1. Y = pdist(X, 'canberra')

Computes the Canberra distance between the points. The Canberra distance between two points u and v is

1. Y = pdist(X, 'braycurtis')

Computes the Bray-Curtis distance between the points. The Bray-Curtis distance between two points u and v is

1. Y = pdist(X, 'mahalanobis', VI=None)
Computes the Mahalanobis distance between the points. The Mahalanobis distance between two points u and v is where (the VI variable) is the inverse covariance. If VI is not None, VI will be used as the inverse covariance matrix.
1. Y = pdist(X, 'yule')
Computes the Yule distance between each pair of boolean vectors. (see yule function documentation)
1. Y = pdist(X, 'matching')
Computes the matching distance between each pair of boolean vectors. (see matching function documentation)
1. Y = pdist(X, 'dice')
Computes the Dice distance between each pair of boolean vectors. (see dice function documentation)
1. Y = pdist(X, 'kulsinski')
Computes the Kulsinski distance between each pair of boolean vectors. (see kulsinski function documentation)
1. Y = pdist(X, 'rogerstanimoto')
Computes the Rogers-Tanimoto distance between each pair of boolean vectors. (see rogerstanimoto function documentation)
1. Y = pdist(X, 'russellrao')
Computes the Russell-Rao distance between each pair of boolean vectors. (see russellrao function documentation)
1. Y = pdist(X, 'sokalmichener')
Computes the Sokal-Michener distance between each pair of boolean vectors. (see sokalmichener function documentation)
1. Y = pdist(X, 'sokalsneath')
Computes the Sokal-Sneath distance between each pair of boolean vectors. (see sokalsneath function documentation)
1. Y = pdist(X, 'wminkowski')
Computes the weighted Minkowski distance between each pair of vectors. (see wminkowski function documentation)
1. Y = pdist(X, f)

Computes the distance between all pairs of vectors in X using the user supplied 2-arity function f. For example, Euclidean distance between the vectors could be computed as follows:

dm = pdist(X, (lambda u, v: np.sqrt(((u-v)*(u-v).T).sum())))

Note that you should avoid passing a reference to one of the distance functions defined in this library. For example,:

dm = pdist(X, sokalsneath)

would calculate the pair-wise distances between the vectors in X using the Python function sokalsneath. This would result in sokalsneath being called times, which is inefficient. Instead, the optimized C version is more efficient, and we call it using the following syntax.:

dm = pdist(X, 'sokalsneath')

Parameters : X : ndarray An m by n array of m original observations in an n-dimensional space. metric : string or function The distance metric to use. The distance function can be ‘braycurtis’, ‘canberra’, ‘chebyshev’, ‘cityblock’, ‘correlation’, ‘cosine’, ‘dice’, ‘euclidean’, ‘hamming’, ‘jaccard’, ‘kulsinski’, ‘mahalanobis’, ‘matching’, ‘minkowski’, ‘rogerstanimoto’, ‘russellrao’, ‘seuclidean’, ‘sokalmichener’, ‘sokalsneath’, ‘sqeuclidean’, ‘yule’. w : ndarray The weight vector (for weighted Minkowski). p : double The p-norm to apply (for Minkowski, weighted and unweighted) V : ndarray The variance vector (for standardized Euclidean). VI : ndarray The inverse of the covariance matrix (for Mahalanobis). Y : ndarray A condensed distance matrix.


text:
tilingArray- normalizeByReference
2
0
Entering edit mode
@anjan-purkayastha-3096
Last seen 8.2 years ago
hi all, a rookie tilingArray question. i'm having trouble creating the perfect match index vector. for my probe set (n= 6490), each probe is a unique, perfect match to its target. i have created a probe index file that contains the integers: 1, 2....6490 (one index per line). i scanned this file as: pm_vector <- scan("PerfectMatchIndex", 0) i used the pm_vector in a following command: vac_normalized_exp_data <- normalizeByReference(vac_exp_set, vac_hyb_set, pm= pm_vector, background= bg_vector, cutoffQuantile= 0) this gives me an error: 'pm' must be an integer vector with values between 1 and 6490. any idea what the problem could be? tia anjan -- =========================================== anjan purkayastha, phd bioinformatics analyst whitehead institute for biomedical research nine cambridge center cambridge, ma 02142 purkayas [at] wi [dot] mit [dot] edu 703.740.6939
probe tilingArray probe tilingArray • 780 views
0
Entering edit mode
@joern-toedling-1244
Last seen 8.2 years ago
Hello Anjan, well, the error message indicates that your vector pm_vector is not usable by the function. Is it really an integer vector. What is the output of str(pm_vector) ? Probably, it would be better to read in the file by something like pm_vector <- read.delim("PerfectMatchIndex", skip=<put number="" of="" header="" lines="" here!="">)$V1 Again, check the result using str(pm_vector) If you want to use other functionalities of package tilingArray, you should consider creating a 'probeAnno' environment, which holds the mapping between probes and genomic match positions, for your array. Please see the script "makeProbeAnno.R" in the package's "scripts" directory for details. Once you have such a probeAnno object, the function "PMindex" can be used to obtain a numeric vector that you can use as "pm" in the normalization function. Please see the vignette "assessNorm" of package tilingArray for more details. Best regards, Joern Anjan Purkayastha wrote: > hi all, > a rookie tilingArray question. > i'm having trouble creating the perfect match index vector. > for my probe set (n= 6490), each probe is a unique, perfect match to > its target. > i have created a probe index file that contains the integers: 1, > 2....6490 (one index per line). > i scanned this file as: pm_vector <- scan("PerfectMatchIndex", 0) > > i used the pm_vector in a following command: > vac_normalized_exp_data <- normalizeByReference(vac_exp_set, > vac_hyb_set, pm= pm_vector, background= bg_vector, cutoffQuantile= 0) > > this gives me an error: > 'pm' must be an integer vector with values between 1 and 6490. > > any idea what the problem could be? > tia > anjan > -- Joern Toedling EMBL - European Bioinformatics Institute Wellcome Trust Genome Campus Hinxton, Cambridge CB10 1SD United Kingdom Phone +44(0)1223 492566 Email toedling at ebi.ac.uk ADD COMMENT 0 Entering edit mode joern, thanks. i've solved the problem by wrapping an as. integer function around the pm_vector. so here is my command: vac_normalized_data= normalizeByReference(vac_exp_set, vac_hyb_set, pm= as.integer(pm_vector), background= as.integer(bg_vector)) however my story does not end here. i get a further set of errors after this command. Error in validObject(.Object) : invalid class "vsnInput" object: 'subsample' must be a numeric vector of length 1 with values between 0 and 6488. In addition: Warning message: In normalizeByReference(vac_exp_set, vac_hyb_set, pm = as.integer(pm_vector), : 'some strata of background contain fewer than 5000 features, are you sure this is alright? Error in exprs(vsnMatrix(xn, lts.quantile = 0.95, subsample = as.integer(2e+05), : error in evaluating the argument 'object' in selecting a method for function 'exprs' the first part of the error message: vsnInput" object: 'subsample' must be a numeric vector of length 1 with values between 0 and 6488, does not make much sense to me. if i understand the second part of the error message about background strata correctly, there are less than 5000 features in some background strata. that is understandable as i have only 1275 probes in the background set. there are a total of 6488 probes in my set. any ideas on how to troubleshoot this? thanks. anjan Joern Toedling wrote: > Hello Anjan, > > well, the error message indicates that your vector pm_vector is not > usable by the function. Is it really an integer vector. What is the > output of > str(pm_vector) > ? > Probably, it would be better to read in the file by something like > pm_vector <- read.delim("PerfectMatchIndex", skip=<put number="" of="" header=""> lines here!>)$V1 > Again, check the result using > str(pm_vector) > If you want to use other functionalities of package tilingArray, you > should consider creating a 'probeAnno' environment, which holds the > mapping between probes and genomic match positions, for your array. > Please see the script "makeProbeAnno.R" in the package's "scripts" > directory for details. Once you have such a probeAnno object, the > function "PMindex" can be used to obtain a numeric vector that you can > use as "pm" in the normalization function. > Please see the vignette "assessNorm" of package tilingArray for more > details. > > Best regards, > Joern > > Anjan Purkayastha wrote: > >> hi all, >> a rookie tilingArray question. >> i'm having trouble creating the perfect match index vector. >> for my probe set (n= 6490), each probe is a unique, perfect match to >> its target. >> i have created a probe index file that contains the integers: 1, >> 2....6490 (one index per line). >> i scanned this file as: pm_vector <- scan("PerfectMatchIndex", 0) >> >> i used the pm_vector in a following command: >> vac_normalized_exp_data <- normalizeByReference(vac_exp_set, >> vac_hyb_set, pm= pm_vector, background= bg_vector, cutoffQuantile= 0) >> >> this gives me an error: >> 'pm' must be an integer vector with values between 1 and 6490. >> >> any idea what the problem could be? >> tia >> anjan >> >> > > -- =========================================== anjan purkayastha, phd bioinformatics analyst whitehead institute for biomedical research nine cambridge center cambridge, ma 02142 purkayas [at] wi [dot] mit [dot] edu 703.740.6939
0
Entering edit mode
Anjan, looking at the source code of the function normalizeByReference, there is indeed the line vsnMatrix(xn, lts.quantile = 0.95, subsample = as.integer(2e+05)) , and since your matrix only has 6488 lines, this cannot work. Without knowing more details about your array platform and study, I cannot say whether or not the function "normalizeByReference" is really applicable to your array platform. However, we should augment the function anyway by chaning this line to something like vsnMatrix(xn, lts.quantile = 0.95, subsample = min(nrow(xn), as.integer(2e+05))) . We are going to make this change and it will be included in the development version of tilingArray within a few days. For the moment, you will have to work with a thus modified local copy of the function. Sorry for the inconvenience. Regards, Joern Anjan Purkayastha wrote: > joern, > thanks. i've solved the problem by wrapping an as. integer function > around the pm_vector. so here is my command: > > vac_normalized_data= normalizeByReference(vac_exp_set, vac_hyb_set, > pm= as.integer(pm_vector), background= as.integer(bg_vector)) > > however my story does not end here. i get a further set of errors > after this command. > Error in validObject(.Object) : > invalid class "vsnInput" object: 'subsample' must be a numeric vector > of length 1 with values between 0 and 6488. > In addition: Warning message: > In normalizeByReference(vac_exp_set, vac_hyb_set, pm = > as.integer(pm_vector), : > 'some strata of background contain fewer than 5000 features, are you > sure this is alright? > Error in exprs(vsnMatrix(xn, lts.quantile = 0.95, subsample = > as.integer(2e+05), : > error in evaluating the argument 'object' in selecting a method for > function 'exprs' > > the first part of the error message: vsnInput" object: 'subsample' > must be a numeric vector of length 1 with values between 0 and 6488, > does not make much sense to me. > if i understand the second part of the error message about background > strata correctly, there are less than 5000 features in some background > strata. that is understandable as i have only 1275 probes in the > background set. there are a total of 6488 probes in my set. > > any ideas on how to troubleshoot this? > thanks. > anjan > > > -- Joern Toedling EMBL - European Bioinformatics Institute Wellcome Trust Genome Campus Hinxton, Cambridge CB10 1SD United Kingdom Phone +44(0)1223 492566 Email toedling at ebi.ac.uk
0
Entering edit mode
@anjan-purkayastha-3096
Last seen 8.2 years ago
Anjan Purkayastha wrote: > joern, > thanks. > let me modify a local version and see the results. is it ok if i work > on this problem with you? i think dr. huber's method to genomic dna > hybridization based approach suits our purpose the best. > to give you a brief idea about our platform: we have built a whole > genome tiling array for the vaccinia virus (ds DNA virus, genome size: > ca 190kb); there are about 19000 overlapping probes, 60 bp in length, > for each strand. for the purposes of normalization i have removed > overlapping probes and am left with 6488 non-overlapping, contiguous > members. > > anyway, i'll rerun the procedure with the revised script and see what > i come up with . > thank you for your help. > warm regards, > anjan > > > > > Joern Toedling wrote: >> Anjan, >> looking at the source code of the function normalizeByReference, there >> is indeed the line >> vsnMatrix(xn, lts.quantile = 0.95, subsample = as.integer(2e+05)) , >> and since your matrix only has 6488 lines, this cannot work. >> Without knowing more details about your array platform and study, I >> cannot say whether or not the function "normalizeByReference" is really >> applicable to your array platform. However, we should augment the >> function anyway by chaning this line to something like >> vsnMatrix(xn, lts.quantile = 0.95, subsample = min(nrow(xn), >> as.integer(2e+05))) . >> We are going to make this change and it will be included in the >> development version of tilingArray within a few days. For the moment, >> you will have to work with a thus modified local copy of the function. >> Sorry for the inconvenience. >> >> Regards, >> Joern >> >> >> Anjan Purkayastha wrote: >> >>> joern, >>> thanks. i've solved the problem by wrapping an as. integer function >>> around the pm_vector. so here is my command: >>> >>> vac_normalized_data= normalizeByReference(vac_exp_set, vac_hyb_set, >>> pm= as.integer(pm_vector), background= as.integer(bg_vector)) >>> >>> however my story does not end here. i get a further set of errors >>> after this command. >>> Error in validObject(.Object) : >>> invalid class "vsnInput" object: 'subsample' must be a numeric vector >>> of length 1 with values between 0 and 6488. >>> In addition: Warning message: >>> In normalizeByReference(vac_exp_set, vac_hyb_set, pm = >>> as.integer(pm_vector), : >>> 'some strata of background contain fewer than 5000 features, are you >>> sure this is alright? >>> Error in exprs(vsnMatrix(xn, lts.quantile = 0.95, subsample = >>> as.integer(2e+05), : >>> error in evaluating the argument 'object' in selecting a method for >>> function 'exprs' >>> >>> the first part of the error message: vsnInput" object: 'subsample' >>> must be a numeric vector of length 1 with values between 0 and 6488, >>> does not make much sense to me. >>> if i understand the second part of the error message about background >>> strata correctly, there are less than 5000 features in some background >>> strata. that is understandable as i have only 1275 probes in the >>> background set. there are a total of 6488 probes in my set. >>> >>> any ideas on how to troubleshoot this? >>> thanks. >>> anjan >>> >>> >>> >>> >> >> > > -- =========================================== anjan purkayastha, phd bioinformatics analyst whitehead institute for biomedical research nine cambridge center cambridge, ma 02142 purkayas [at] wi [dot] mit [dot] edu 703.740.6939


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# Finding the change in K.E. G.P.E ignored?

Hi guys! Just a little confused on the solution provided.

## Homework Statement

A crate of mass 10.0 kg is pulled up a rough incline with an initial speed of 1.50 m/s. The
pulling force is 100 N parallel to the incline, which makes an angle of 20.0° with the horizontal.
The coefficient of kinetic friction is 0.400, and the crate is pulled 5.00 m.

(d) What is the change in kinetic energy of the crate?

## Homework Equations

Change in word done = Fs - mgsin$\Theta$s - friction

I understand this part but I want to know why the solution key takes this change as the change in kinetic energy itself. Isn't there a gain in G.P.E. as well?

Thanks for the help!

ehild
Homework Helper
Change in word done = Fs - mgsin$\Theta$s - friction

I understand this part but I want to know why the solution key takes this change as the change in kinetic energy itself. Isn't there a gain in G.P.E. as well?

Thanks for the help!

The work done by all forces is

Fs - mgsin$\Theta$s - (work of friction)=ΔKE

The term -mgsin$\Theta$s is the work of gravity, and the opposite is the change of the gravitational potential energy, as
s*sin$\Theta$=h, the height reached. You can rewrite the equation as ΔKE+ΔPE=Fs+(work of friction)

the work done by all forces is

fs - mgsin$\theta$s - (work of friction)=Δke

the term -mgsin$\theta$s is the work of gravity, and the opposite is the change of the gravitational potential energy, as
s*sin$\theta$=h, the height reached. You can rewrite the equation as Δke+Δpe=fs+(work of friction)

Thanks! (: Haha I can't believe that I didn't realize that. :-X

ehild
Homework Helper
Thanks! (: Haha I can't believe that I didn't realize that. :-X

It is a pleasure if a student feels so.

ehild


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## Problem of the Day #41: Last Digits of an Exponentiated FactorialApril 29, 2011

Posted by Albert in : potd , trackback

Albert is mad at everyone for stipulating the so-called “Albert week” on MPOTD. To get his revenge, he makes everyone else work on the following problem:

1. Prove that there exists a function $g(k)$ such that the last $k$ digits of $f(n) = 2^{n!}$ becomes constant for all $n \geq g(k)$.
2. Find the smallest integer $m$ such that the last $32$ digits of $f(n)$ are constant for all $n \geq m$.


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# Tractrix

Jump to: navigation, search

A tractrix (from the Latin verb trahere "pull, drag"; plural: tractrices) is the curve along which an object moves, under the influence of friction, when pulled on a horizontal plane by a line segment attached to a tractor (pulling) point that moves at a right angle to the initial line between the object and the puller at an infinitesimal speed. It is therefore a curve of pursuit. It was first introduced by Claude Perrault in 1670, and later studied by Sir Isaac Newton (1676) and Christiaan Huygens (1692).

Tractrix with object initially at (4,0)

## Mathematical derivation

Suppose the object is placed at (a,0) [or (4,0) in the example shown at right], and the puller in the origin, so a is the length of the pulling thread [4 in the example at right]. Then the puller starts to move along the y axis in the positive direction. At every moment, the thread will be tangent to the curve y = y(x) described by the object, so that it becomes completely determined by the movement of the puller. Mathematically, the movement will be described then by the differential equation

${\displaystyle {\frac {dy}{dx}}=-{\frac {\sqrt {a^{2}-x^{2}}}{x}}\,\!}$

with the initial condition y(a) = 0 whose solution is

${\displaystyle y=\int _{x}^{a}{\frac {\sqrt {a^{2}-t^{2}}}{t}}\,dt=\pm \left(a\ln {\frac {a+{\sqrt {a^{2}-x^{2}}}}{x}}-{\sqrt {a^{2}-x^{2}}}\right).\,\!}$

The first term of this solution can also be written

${\displaystyle a\ \mathrm {arsech} {\frac {x}{a}},\,\!}$

where arsech is the inverse hyperbolic secant function.

The negative branch denotes the case where the puller moves in the negative direction from the origin. Both branches belong to the tractrix, meeting at the cusp point (a, 0).

## Basis of the tractrix

The essential property of the tractrix is constancy of the distance between a point P on the curve and the intersection of the tangent line at P with the asymptote of the curve.

The tractrix might be regarded in a multitude of ways:

1. It is the locus of the center of a hyperbolic spiral rolling (without skidding) on a straight line.
2. The involute of the catenary function, which describes a fully flexible, inelastic, homogeneous string attached to two points that is subjected to a gravitational field. The catenary has the equation ${\displaystyle y(x)=a\,\operatorname {cosh} (x/a)}$.
3. The trajectory determined by the middle of the back axle of a car pulled by a rope at a constant speed and with a constant direction (initially perpendicular to the vehicle).
Tractrix by dragging a pole.

The function admits a horizontal asymptote. The curve is symmetrical with respect to the y-axis. The curvature radius is ${\displaystyle r=a\,\operatorname {cot} (x/y)}$

A great implication that the tractrix had was the study of the revolution surface of it around its asymptote: the pseudosphere. Studied by Beltrami in 1868, as a surface of constant negative Gaussian curvature, the pseudosphere is a local model of non-Euclidean geometry. The idea was carried further by Kasner and Newman in their book Mathematics and the Imagination, where they show a toy train dragging a pocket watch to generate the tractrix.

## Properties

Catenary as evolute of a tractrix
• Due to the geometrical way it was defined, the tractrix has the property that the segment of its tangent, between the asymptote and the point of tangency, has constant length ${\displaystyle a}$.
• The arc length of one branch between x = x1 and x = x2 is ${\displaystyle a\ln {\frac {x_{1}}{x_{2}}}}$
• The area between the tractrix and its asymptote is ${\displaystyle \pi a^{2}/2}$ which can be found using integration or Mamikon's theorem.
• The envelope of the normals of the tractrix (that is, the evolute of the tractrix) is the catenary (or chain curve) given by ${\displaystyle y=a\cosh {\frac {x}{a}}}$.
• The surface of revolution created by revolving a tractrix about its asymptote is a pseudosphere.

## Practical application

In 1927, P.G.A.H. Voigt patented a horn loudspeaker design based on the assumption that a wave front traveling through the horn is spherical of a constant radius. The idea is to minimize distortion caused by internal reflection of sound within the horn. The resulting shape is the surface of revolution of a tractrix.[1]
An important application is in the forming technology for sheet metal. In particular a tractrix profile is used for the corner of the die on which the sheet metal is bent during deep drawing. [2]

## Drawing machines

• In October–November 1692, Huygens described three tractrice drawing machines.
• In 1693 Leibniz released to the public a machine which, in theory, could integrate any differential equation; the machine was of tractional design.
• In 1706 John Perks built a tractional machine in order to realise the hyperbolic quadrature.
• In 1729 Johann Poleni built a tractional device that enabled logarithmic functions to be drawn.

## Notes

1. ^ Horn loudspeaker design pp. 4-5. (Reprinted from Wireless World, March 1974)
2. ^ Lange, Kurt (1985). Handbook of Metal Forming. McGraw Hill Book Company. p. 20.43.


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sqrfree - Maple Programming Help

sqrfree

square-free factorization function

 Calling Sequence sqrfree(a) sqrfree(a, x)

Parameters

 a - multivariate polynomial or rational function x - name or list or set of names

Description

 • The sqrfree function computes the square-free factorization of the multivariate polynomial or the rational function a in the variable(s) x over an algebraic number field.
 • The square-free factorization is returned in the form $\left[u,\left[\left[{f}_{1},{e}_{1}\right],\mathrm{...},\left[{f}_{n},{e}_{n}\right]\right]\right]$ such that $a=u{f}_{1}^{{e}_{1}}\cdots {f}_{n}^{{e}_{n}}$ where ${f}_{i}$ is primitive and square-free, that is, $\mathrm{gcd}\left({f}_{i},\frac{\partial }{\partial x}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{f}_{i}\right)=1$ for all i and $\mathrm{gcd}\left({f}_{i},{f}_{j}\right)=1$ for all $i\ne j$ hence u is the content(a,x) times a unit.
 • In the case of two arguments, the second argument x denotes the main variable(s).  In the case of one argument, all the names in the polynomial a are used as variables.
 • Note that the ${e}_{i}$ are not necessarily distinct as partial factorizations in the input are preserved as much as possible.

Examples

 > $\mathrm{sqrfree}\left(-12{x}^{3}+9x+3\right)$
 $\left[{-3}{,}\left[\left[{x}{-}{1}{,}{1}\right]{,}\left[{1}{+}{2}{}{x}{,}{2}\right]\right]\right]$ (1)
 > $f≔{x}^{3}y-{x}^{3}-{x}^{2}{y}^{2}+{x}^{2}y$
 ${f}{≔}{{x}}^{{3}}{}{y}{-}{{x}}^{{2}}{}{{y}}^{{2}}{-}{{x}}^{{3}}{+}{{x}}^{{2}}{}{y}$ (2)
 > $\mathrm{sqrfree}\left(f,x\right)$
 $\left[{y}{-}{1}{,}\left[\left[{x}{,}{2}\right]{,}\left[{-}{y}{+}{x}{,}{1}\right]\right]\right]$ (3)
 > $\mathrm{sqrfree}\left(f,y\right)$
 $\left[{-}{{x}}^{{2}}{,}\left[\left[{-}{x}{}{y}{+}{{y}}^{{2}}{+}{x}{-}{y}{,}{1}\right]\right]\right]$ (4)
 > $\mathrm{sqrfree}\left(f,\left[x,y\right]\right)$
 $\left[{1}{,}\left[\left[{x}{,}{2}\right]{,}\left[{y}{-}{1}{,}{1}\right]{,}\left[{-}{y}{+}{x}{,}{1}\right]\right]\right]$ (5)
 > $g≔\left(x+y+1\right)\mathrm{expand}\left({\left(x+y+1\right)}^{2}\right){\left(x-y-3\right)}^{3}\left(3x+6y-21\right)$
 ${g}{≔}\left({x}{+}{y}{+}{1}\right){}\left({{x}}^{{2}}{+}{2}{}{x}{}{y}{+}{{y}}^{{2}}{+}{2}{}{x}{+}{2}{}{y}{+}{1}\right){}{\left({x}{-}{y}{-}{3}\right)}^{{3}}{}\left({3}{}{x}{+}{6}{}{y}{-}{21}\right)$ (6)
 > $\mathrm{sqrfree}\left(g,\left[x,y\right]\right)$
 $\left[{3}{,}\left[\left[{x}{+}{y}{+}{1}{,}{3}\right]{,}\left[{x}{-}{y}{-}{3}{,}{3}\right]{,}\left[{x}{+}{2}{}{y}{-}{7}{,}{1}\right]\right]\right]$ (7)
 > $h≔\frac{{\left(x+y\right)}^{3}}{\mathrm{expand}\left({\left({x}^{2}-{y}^{2}\right)}^{2}\left({x}^{2}+1\right)\right)}$
 ${h}{≔}\frac{{\left({x}{+}{y}\right)}^{{3}}}{{{x}}^{{6}}{-}{2}{}{{x}}^{{4}}{}{{y}}^{{2}}{+}{{x}}^{{2}}{}{{y}}^{{4}}{+}{{x}}^{{4}}{-}{2}{}{{x}}^{{2}}{}{{y}}^{{2}}{+}{{y}}^{{4}}}$ (8)
 > $\mathrm{sqrfree}\left(h\right)$
 $\left[{1}{,}\left[\left[{x}{+}{y}{,}{1}\right]{,}\left[{{x}}^{{2}}{+}{1}{,}{-1}\right]{,}\left[{-}{y}{+}{x}{,}{-2}\right]\right]\right]$ (9)
 > $\mathrm{sqrfree}\left(h,y\right)$
 $\left[\frac{{1}}{{{x}}^{{2}}{+}{1}}{,}\left[\left[{x}{+}{y}{,}{1}\right]{,}\left[{y}{-}{x}{,}{-2}\right]\right]\right]$ (10)


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# Tag Info

1

Nowadays, government yield curves are customarily built with only coupon bonds. Zero coupon bonds (i.e., STRIPS in the US) are much less liquid compared with coupon Treasuries, and tend to trade very differently. If you plot a zero curve implied by coupon Treasuries vs yields of STRIPS, you'll notice that they can differ quite a bit in certain parts of the ...

3

Jojo, once again the paper is about Nelson-Siegel and not Nelson-Siegel svensson (the former allows for one hump whereas the latter for two humps). Jojo, in practice people often start by fixing $\lambda$ then estimate the model by OLS and check the squared errors of the model. Then change $\lambda$ and repeat the procedure. This is highly efficient, and ...

1

Say at time $t$ , the cash flows of some bond $b$ can be described by the two vectors $\textbf{c}$ and $\textbf{t}$, containing information about the value of the nominal cash flows and cash flow times in years, respectively. Similarly, if we have a range of bonds $B = \{ b_1, ..., b_n\}$ that trade on a market, the matrices $\textbf{C}$ and $\textbf{T}$ ...

0

if I read it correctly the paper you mention is about Nelson-Siegel and not Nelson-Siegel Svensson. The easier way to cross check it is to simply start by fixing theta and estimate the remaining parameters by OLS. If that works, then the optimization algorithm should work similarly. Below I attach a simple matlab code that does it and that you can adapt for ...

2

EONIA swaps stopped trading some time in 2014. Since it stopped trading, it does not make sense to remember when it stopped trading :).

Top 50 recent answers are included


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# russia vertical cylindrical tank fire volume

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### (PDF) HOW TO CALCULATE THE VOLUMES OF PARTIALLY

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### Venting Guide for Aboveground Storage Tanks

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### Compute Fluid Volumes in Vertical Tanks

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### CYLINDRICAL STEEL TANK STANDARD SPECIFICATION

galvanized cylindrical steel bolted liquid storage tanks. These tanks are primarily used for the storage of water in the potable water, fire sprinkler and irrigation markets. The galvanized cylindrical tanks are site assembled using overlapping and bolted galvanized steel panels that are manufactured within the companys facility in the UK.Tank Calibration Chart Calculator - ODay EquipmentFiberglass Tanks. ODay Equipment provides dome end fiberglass tanks from Xerxes and Containment Solutions. The domes on fiberglass tanks vary by manufacturer. So, here are the manufacturers web sites that have calibration charts specific to their designs. Xerxes Go to the Library tab for PDF versions of their charts.How to mathematically calculate a fire water tank capacity russia vertical cylindrical tank fire volumeFeb 25, 2019 · 1. You need to determine the fire water tank demand for each area or equipment or tank in the facility 2. After determination of the individual firewater demand, you design your firewater facility based on the highest firewater demand 3. Based on russia vertical cylindrical tank fire volume

### Tank Volume Calculator - Oil Tanks

The tank size calculator on this page is designed for measuring the capacity of a variety of fuel tanks. Alternatively, you can use this tank volume calculator as a water volume calculator if you need to calculate some specific water volume. The functionality of Spherical Storage Tank Design - Chemical Engineering WorldSep 15, 2019 · Spherical Storage Tank Design:- The most common shape of a storage vessel is a cylinder with two heads which are either, hemispherical, elliptical or tori-spherical. Spherical vessels have larger surface area per unit volume.DESIGN RECOMMENDATION FOR STORAGE TANKS common use throughout Japan. The Sub-Committee first published Design Recommendation for Storage Tanks and Their Supports in 1984, and amended it in the 1990, 1996 and 2010 publications. This current revised recommendation provides bulk material pressures for silos

### Sphere Tank Volume - vCalc

Processingrussia vertical cylindrical tank fire volume russia vertical cylindrical tank fire volume russia vertical cylindrical tank fire volumeVolume and Wetted Area of Partially Filled Vertical russia vertical cylindrical tank fire volumeThe calculation of the wetted area and volume of a vertical vessel is required for engineering tasks such fire studies and the determination of level alarms and control set points. However the calculation of these parameters is complicated by the geometry of the vessel, particularly the heads. This article details formulae for calculating the wetted area and volume of these vessels for various russia vertical cylindrical tank fire volumeCalculator to Find liquid volume for vertically mounted russia vertical cylindrical tank fire volumeApr 13, 2017 · Sugar industry or other industries using vertically mounded cylindrical tanks for sealing of vacuum equipment like vacuum condensers and multiple effect evaporator lost body liquid extraction (outlet). Sometimes it has also called mound. In this cylindrical tank sometimes inside of the tank might be separated by vertical plate for easily operation purpose.

### SPCC Plan - Calculation Guidance - Asmark

Note: The volume displaced by a cylindrical vertical tank is the tank volume within the containment structure and is equal to the tank footprint multiplied by height of the concrete dike. The tank footprint is equal to D2/4, where D is the tank diameter. 1. Calculate total dike capacity: Total capacity of Cylindrical Tank - an overview | ScienceDirect TopicsIn such units the working volume is often relatively small and the mixing blades are massive in construction. 7.2.2 Baffles. To prevent gross vortexing behaviour when low-viscosity liquids are agitated in a vertical cylindrical tank with a centrally mounted impeller, baffles are often fitted to the walls of the vessel. Typically four baffles russia vertical cylindrical tank fire volumeVolume and Wetted Area of Partially Filled Vertical russia vertical cylindrical tank fire volumeThe calculation of the wetted area and volume of a vertical vessel is required for engineering tasks such fire studies and the determination of level alarms and control set points. However the calculation of these parameters is complicated by the geometry of the vessel, particularly the heads. This article details formulae for calculating the wetted area and volume of these vessels for various russia vertical cylindrical tank fire volume

### aboveground vertical tanks - Highland Tank

Our aboveground tanks are available in a variety of industry-proven cylindrical, rectangular, single-wall, double-wall and fire rated designs in standard and custom sizes from 185 to 57,500 gallons. Aboveground vertical tanks are manufactured to a variety of standards including; UL-142, API-650 Annex J, API-12F and AWWA for the gas and oil russia vertical cylindrical tank fire volumePetroleum Measurement - APICalibration of Upright Cylindrical Tanks Using the Optical-Triangulation Method (ANSI/API MPMS Ch. 2.2C-2002) Describes the calibration of vertical cylindrical tanks by means of optical triangulation using theodolites. The method is an alternative to other methods such as strapping (Ch. 2.2A) and the optical-reference-line method (Ch. 2.2B).Calculating Tank VolumeCalculating Tank Volume Saving time, increasing accuracy By Dan Jones, Ph.D., P.E. alculating fluid volume in a horizontal or vertical cylindrical or elliptical tank can be complicated, depending on fluid height and the shape of the heads (ends) of a horizontal tank or the bottom of a vertical tank.

### RESERVOIR DESIGN AND STORAGE VOLUME

volume. 9.0.1 Effective Storage . Total tank volume, as measured between the overflow and the tank outlet elevations, may not necessarily equal the effective volume available to the water system. Effective volume is equal to the total volume less any dead storage built into the reservoir. For example, a Spherical Storage Tank Design - Chemical Engineering WorldSep 15, 2019 · Spherical Storage Tank Design:- The most common shape of a storage vessel is a cylinder with two heads which are either, hemispherical, elliptical or tori-spherical. Spherical vessels have larger surface area per unit volume.BBTVI : Cylindrical storage tanks : vertical, insulated | CBSThe cylindrical storage tank for final work with purified beverage (called also serving tank, filling tank, service tank, storage tank, BBT Bright beer tank) with a vertical construction and an insulated double steel jacket.The vessel is cooling by ice water or glycol that circulates in

### * Sloped Bottom Tank

The easy part the cylindrical section above the slope, which has a volume of: (1) $\displaystyle v = \pi r^2 h$ v = volume; r = tank radius; h = cylindrical section height; More difficult the tank's sloped section, which lies between the tank's bottom and the top of the slope where the tank Cylinder | Definition of Cylinder at Dictionary, russia vertical cylindrical tank fire volumeCylinder definition, a surface or solid bounded by two parallel planes and generated by a straight line moving parallel to the given planes and tracing a curve bounded by the planes and lying in a plane perpendicular or oblique to the given planes. See more.Pressure Vessel design, Formula and Calculators russia vertical cylindrical tank fire volumeAny smooth figure of revolution if R 2 is less than infinity Uniform internal or external pressure, q force/unit area; tangential edge support Stress and Deflection Equation and Calculator.. Per. Roarks Formulas for Stress and Strain for membrane stresses and deformations in thin-walled pressure vessels.

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