text:
## Trouble with screenshots

Questions about the LÖVE API, installing LÖVE and other support related questions go here.
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milon
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Posts: 123
Joined: Thu Jan 18, 2018 9:14 pm

### Trouble with screenshots

Hi all,

I was working on a sample state machine / crappy game demo as a followup to this post of mine. I decided to use a screenshot for fun to make some transition effects, and it seems like Love is refusing to make more than 1 screenshot. It either crashes (if I release the old screenshot) or it shows only the original screenshot even after calling for an updated one. See the attached .love file. Any ideas what I'm doing wrong?
Attachments
state.love
State Machine demo
pgimeno
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### Re: Trouble with screenshots

What's happening is that you're calling release without clearing the variable. This is especially important in states/fail.lua.
milon
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Posts: 123
Joined: Thu Jan 18, 2018 9:14 pm

### [SOLVED] Re: Trouble with screenshots

pgimeno to the rescue again! Thanks!!

I had no idea objects worked like that. I thought that calling :release() was enough. Is there ever a use case to call :release() and not set the variable to nil (or whatever)?
pgimeno
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Joined: Sun Oct 18, 2015 2:58 pm

### Re: Trouble with screenshots

The real problem here was that 'release' does not result in a falsy (i.e. false or nil) value, but you were checking for falsy in some places. Use cases... well, it depends. I imagine that if your code is organized in such a way that you can't immediately set a variable when you release it (e.g. if it's a parameter to a function, and you release it, you can't affect the caller's variable), then you're forced to have the invalid object lingering around.

I don't see any method that allows checking whether an object has been released. That would be interesting to have. If there was one, adding it to your checks would also have solved the problem.
grump
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### Re: Trouble with screenshots

Code: Select all

if obj:getFFIPointer() == nil then print("released") end
For those objects that provide getFFIPointer at least.

For other objects

Code: Select all

if tostring(obj):find('NULL') then print("released") end
If you really need it. Hacky shit. Best to not need it in the first place.
milon
Party member
Posts: 123
Joined: Thu Jan 18, 2018 9:14 pm

### Re: Trouble with screenshots

pgimeno wrote: Wed Apr 21, 2021 1:56 pm The real problem here was that 'release' does not result in a falsy (i.e. false or nil) value, but you were checking for falsy in some places.
I'd tested doing a print() command with the variable after release, and got NULL every time. I assumed (incorrectly) that would not evaluate to true. Seems like an oversight to me, but now I know.

EDIT: Technically, the print command returned "Image: NULL" which I suppose is slightly different than "NULL".

EDIT: If anyone wants to see the corrected version, I uploaded it to another thread.

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Users browsing this forum: Google [Bot] and 37 guests


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## College Physics (4th Edition)

(a) A force of $625~N$ must be applied to the small piston. (b) The car is lifted 6.25 mm (c) The mechanical advantage is 16
(a) We can use Pascal's principle to find $F_a$: $\frac{F_a}{a} = \frac{F_A}{A}$ $\frac{F_a}{\pi~r_a^2} = \frac{F_A}{\pi~r_A^2}$ $F_a = \frac{r_a^2~F_A}{r_A^2}$ $F_a = \frac{(2.50~cm)^2~(10,000~N)}{(10.0~cm)^2}$ $F_a = 625~N$ A force of $625~N$ must be applied to the small piston. (b) The work done on the small piston is equal to the work done by the large piston on the car: $F_A~d_A = F_a~d_a$ $d_A = \frac{F_a~d_a}{F_A}$ $d_A = \frac{(625~N)(10.0~cm)}{10,000~N}$ $d_A = 0.625~cm = 6.25~mm$ The car is lifted 6.25 mm (c) We can find the mechanical advantage: $\frac{W}{F_a} = \frac{10000~N}{625~N} = 16$ The mechanical advantage is 16.

format: plain


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# [Beowulf] Win64 Clusters!!!!!!!!!!!!

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Robert G. Brown rgb at phy.duke.edu
Thu Apr 12 06:35:02 PDT 2007

On Wed, 11 Apr 2007, Richard Walsh wrote:

>> Hear hear! For self-adapting softare you *can't* distinguish
>> instructions from data. That may sound over-specialized but I invite
>> you to consider DNA and what it does: instruct enzymes to modify DNA.
>> An awful lot comes out of that process. So I don't think it's just von
>> Neumann's whim.
>
> On the opposite side of the argument, that code/instruction set
> (self-modified for say about a billion years [and recombined, and
> mutated]) is only ~3 billion base pairs in length. If we liberally give
> each base pair a byte, that's only 3 billion bytes. We can quibble about
> what an instruction is, but also about how much of the code is used and
> what would be the reduced instruction set equivalent.

There are two very important points here.  One is that Jon's assertion
in one sense is correct.  Even humans who type like the wind (immodestly
offering myself as an example as my friends tell me that I'm fairly
"prolific":-) can type at a >sustained rate< of only 2-4 characters a
second, if that.  1 gigasecond is roughly 10*\pi \approx 32 years (old
rule of thumb approximation: 1 year equals \pi \times 10^7 seconds).
That means that a human being, typing 4 characters (bytes) per second,
would require 32 years of continual sustained effort to generate 2^32
bytes worth of production.  There is no question that no single human --
not even Isaac Asimov or some of the other truly prodigious authors has
ever come close to writing the 4000-8000 full length books (a typical
novel has between a half-megabyte and a megabyte of tex) required.
(Asimov wrote 500 books and an estimated 9000 letters and postcards,
according to Wikipedia, which would make him quite exceptional --
perhaps as much as a gigabyte over a career.)

Nobody could even THINK of writing software at this rate, sustained for
the source code, and even allowing for amplification via the usual
encapsulation of code via a compiler and associated libraries it seems
rather likely that there are only a handful of humans who write 4 GB
worth of actual production binary code over a lifetime, not including of
course the many discarded transient versions as one builds and tests
(which can EASILY fill GB as one works, as one will rapidly learn if one
code-version controls these images at the many intermediate stages).  So
let us take it for granted that no single human has produced or is
likely to produce a single binary executable image that exceeds 4 GB in
size (except in the context of checkpointing and saving to disk a
runtime image of all of memory on a larger memory system or in the
context of creating VM images, both of which are RELEVANT exceptions as
they are actions of an upper level piece of software that captures a
working environment just as much as compiling and linking).  But the
basic point stands -- humans cannot type that fast, and so far "single
applications" (except for the aforementioned VM images of entire running
systems) do not link in aggregate resources that exceed 4 GB in length.
To gain a still better appreciation of this, note that an entire linux
distro, binaries, libraries and all, will often fit easily into a 4 GB
partition.

So it is a true statement BUT it is also IRRELEVANT to the entire issue
of CPU engineering.  CPUs have not been single threaded application
engines since DOS.  Really since BEFORE DOS, but DOS and friends were
basically single-tasking engines, and the early 808x CPUs were
engineered as such as those of us who coded them in assembler well
remember.  CX, SX, DX, EX (unused) as two-byte offset segment registers,
combined with a two-byte (64K) address relative to the offset to produce
a 64K "code address space", or "stack address space" or "data address
space" each with its own addressing and/or pointer to manipulate.

On this kind of segmented CPU, single-threaded kernel architecture one
could sensibly talk about "code addresses" as somehow distinct from
they were -- IIRC one generally set EX to SX and henceforth ignored it
in writing actual code, but I haven't written 808x code for 25 years and
could be recalling INcorrectly:-).

Programmers hated this, and lamented the fact that 808x wasn't a sweet,
flat memory model CPU like the (competing) Motorola 68000 used in the
Macintoshi of the day, but PCs were cheap, Macs were dear, and even in
DOS days a command line was a marvelous thing compared to HAVING to use
a mouse to accomplish every task.

A few processor generations later 80x86 flattened memory (while
NATURALLY retaining segmented addressing capabilities so the CPUs could
still run DOS binaries -- I can STILL run DOS binaries for that matter
AFAIK as Intel and AMD maintain backwards capabilities ad nauseam --
those mammals might NEED those gills one day as ontogony eternally
recapitulates philogeny in evolutionary systems:-).  Not TOO long after
(say, a decade and change:-) true multitasking operating systems built
on flat memory models emerged for first the 80386 (SunOS in the form of
the Sun 386i and a few other small Unices in the late 80's/early 90's),
then Linux and freebsd and even WinNT and OS/2 in the early to mid
90's).

Since the emergence of true multitasking kernels with true virtual
memory, true runtime shared/DL libraries, and automated buffering and
caching capabilities running on top of a flat-memory processor and
memory subsystem , there has been NO POINT in distinguishing the
ADDRESSING of "code" memory, "stack" memory, "heap" memory, "data"
memory, "cache" memory, "buffer" memory, or any of the various DMA or
memory mapped kinds of device memory that take up addresses in the
generalized addressible memory space of a given system.  I think that is
what has confused this discussion from the start.  Jon made a true
observation, but it really had nothing to do with whether or not 64 bit
processors with larger flat memory spaces were a good thing or a bad
thing.  So people (myself included) jumped up to defend them because
they are a GOOD thing without question, even for cheap PCs, and not just
a matter of "marketing" 64 bits as being twice as big as 32 bits and
bigger must be better.  Bigger IS better, and faster is better still and
64 bit CPUs are both bigger and faster and enable whole new classes of
application to be written and run (including, of course, the large DATA
memory applications of interest to many on this list).

To make this point still clearer, let us return to the VM exception
above.  In one very fundamental sense, CPUs are STILL executing "a
single thread".  There is just one instruction pointer that tells the
CPU what instruction it will execute "next".  CPUs are still serial
execution units (even as they internally parallelize to do this or
that, that doesn't prevent this from being true).

What, then, is the "program" that they run?  >>EVERYTHING!<< The
"singled threaded" program running on my laptop is the sum total
contents of memory, with the kernel, not the hardware, in control of its
memory utilization.  Fragments of the full binary code image are
scattered here and there in main memory according to the whim and needs
of the kernel, interspersed with "stacks", "heaps", "data", "buffers",
"cache" and so on.  This "main memory" is further extended as virtual
memory into swap space and nonresident pages from the disks.  The
addressing scheme for "the program" running on a computer has to be able
to embrace ALL of this under the perfect control of the kernel.  A flat
memory model is clearly vastly simpler to program for this purpose than
a segmented memory model, and an unsegmented model that "required" all
code pages to be loaded in the first 4 GB of memory on large memory
systems would be downright evil as it breaks the kernel's symmetry and
forces a whole layer of additional conditional statements in its memory
management code.

Now THIS single threaded "program", with its mix of dynamic and
relocateable memory images where the "purpose" of any given block of
memory can change from one nanosecond to the next, where data can become
code or vice versa (think of interpreted languages, think of debuggers,
think of buffer overwrite attacks), written not by any single programmer
but by tens of thousands of programmers with millions of FTE hours of
effort over decades, with its hundreds of thousands of interchangeable
subprograms (the tasks that the kernel is running for the user) in a
near-infinity of permutations is simultaneously one of the true wonders
of the world -- one of the crowning achievements of modern humanity,
quite literally -- and can EASILY exceed 4 GB in size on a system that
has more than 4 GB of memory.  In fact, the kernel will almost certainly
spread out and use all available memory but a tiny bit held in reserve
to satisfy the immediate needs for free memory blocks of the programs
themselves, the kernel, and startup of interactive tasks just to make
itself run more efficiently.

64 bit CPUs are engineered to permit THIS program to exceed 4 GB in size
in a symmetric way that does not place any restrictions on what is or is
not "code memory".  The kernel is in control, and has a big, flat
"desktop" to work with, as should be.  If it chooses to dub a segment of
memory past the 4 GB boundary as holding "code", it can and will.  If a
few moments later this memory becomes "data" that's fine as well.  It
can even be "code" one instruction and "data" a couple of instructions
later as one single-steps through a debugger located physically
somewhere completely different in memory while the kernel periodically
is interrupted to process device requirements with still different
chunks of code memory.

THIS is why folks get "upset" at any suggestion that 64 bit processors
are a bad thing and we could have done just as well extending "only the
data address space" as if that is somehow distinguished from "code
address space".  They are distinguished only by the kernel, and the
kernel AND ALL ITS SUBTASKS are "the program" that 64 bits enables to run
past the 4 GB boundary.

I hope that is now clear.

rgb

--
Robert G. Brown	                       http://www.phy.duke.edu/~rgb/
Duke University Dept. of Physics, Box 90305
Durham, N.C. 27708-0305
Phone: 1-919-660-2567  Fax: 919-660-2525     email:rgb at phy.duke.edu


text:
# Tag Info

The accepted answer above is very misleading and @nat-chouf has it correct. In the spirit of the original question, running a pseudo-spectral mode of isotropic turbulence, then zero-ing out $k>\frac{2}{3}k_\textrm{max}$ at each time-step will absolutely remove energy from the model, assuming that energy is moved to these higher wavenumbers.


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# Question about Rules and Pattern (The operator patt/;test)

Maybe this question is so trivial but it has confused me.
I'm studying the What the @#%^&*?! do all those funny signs mean? and in the Rules and patterns under the Reference for the operator /; there is this example:

In[1]:= f[x_] := ppp[x] /; x > 0
In[2]:= f[5]
Out[2]= ppp[5]
In[3]:= f[-6]
Out[3]= f[-6]

Based on the above example I thought the answer of the following code should be f[-2], as the following:

In[1]:= f[x_] = Sqrt[x] /; x > 0
Out[1]= Sqrt[x] /; x > 0
In[2]:= f[2]
Out[2]= Sqrt[2] /; 2 > 0
In[3]:= f[-2]
Out[3]= f[-2]

I mean the definition f[x_] = Sqrt[x] should only be used when x>0. So when I enter -2 as the argument, the definition should not be used and the output will be f[-2]. But in fact mathematica evaluates the code as follows:

In[1]:= f[x_] = Sqrt[x] /; x > 0
Out[1]= Sqrt[x] /; x > 0
In[2]:= f[2]
Out[2]= Sqrt[2] /; 2 > 0
In[3]:= f[-2]
Out[3]= i Sqrt[2] /; -2 > 0

In spite of the fact that -2<0, Mathematica uses the definition and produces the answer $i\sqrt{2}$
What's the difference between this code and the first one that makes mathematica use the definition in spite of the negative argument passed to the function?

Update: Response to closure (Michael E2)

There must be something quite subtle going on. It is not just the usual explanation that Set evaluates the right-hand and SetDelayed does not, because the right-hand side evaluates to itself (assuming, as Mr. Wizard's answer points out, that x has no value). This can be seen because the down values are the same in each case:

f1[x_] := Sqrt[x] /; x > 0;
f2[x_] = Sqrt[x] /; x > 0;

DownValues@f1
DownValues@f2

(* {HoldPattern[f1[x_]] :> Sqrt[x] /; x > 0} *)
(* {HoldPattern[f2[x_]] :> Sqrt[x] /; x > 0} *)

• Related: (8829) – Mr.Wizard Sep 19 '15 at 18:53
• I disagree with the close votes. This is a subtle error not explained at all in the documentation, AFAICS. There are valid reasons for using Set instead of SetDelayed as well as reasons for putting constraints on patterns. – Michael E2 Oct 21 '15 at 11:30
• @Mr.Wizard not just related, but already covered and answered there, I think. – LLlAMnYP Oct 21 '15 at 11:52

When using Set rather than SetDelayed you will need to hang the Condition on the left-hand-side:

ClearAll[f]

f[x_] /; x > 0 = Sqrt[x]
f[2]
f[-2]

Sqrt[x]

Sqrt[2]

f[-2]

There are other reasons to prefer this placement; see:

However be aware that the use of Set results in "pre-evaluation" of the RHS which often is not desirable. For example suppose the global Symbol x has a value before the definition is made:

ClearAll[f]
x = 7;

f[x_] /; x > 0 = Sqrt[x]
f[2]
f[-2]

Sqrt[7]

Sqrt[7]

f[-2]

This is probably not what you want. Using := works:

ClearAll[f]

x = 7;

f[x_] /; x > 0 := Sqrt[x]
f[2]
f[-2]

Sqrt[2]

f[-2]


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# ncert exemplar class 9 maths solutions chapter 1

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Now, draw an arc with centre B and radius BD, meeting AC produced at E, such that NCERT Solutions for Class 12 Maths Chapter 1 is the best & smart way to master in relations and functions chapter.All the solutions are well-structured by the subject experts based on the NCERT Syllabus prescribed by CBSE Board. NCERT Exemplar Class 9 Mathematics Chapter 13 Surface Areas and Volumes are provided here for students to prepare for exams and make them score well. NCERT Class 9 Mathematics Exemplar Solution Chapter Unit 1 Number Systems are the complete study materials for any student studying in class 12. = 99x = 40 => y is a rational number. Which of the following is equal to x? You will also learn about discount prices and different formulas related to these terms. As we know positive numbers lies on right hand side of 0 and negative numbers lies on left hand side of 0. 1Ox – x = (8.888…) – (0.888…) ⇒ 9x = 8 NCERT Exemplar Class 8 (PDF) Download, MCQs 2021. Chapter 4 - Linear Equations in Two Variables. Thus x = 40/99. NCERT Solutions for Class 9th: Ch 14 Statistics Maths Page No: 239 Exercise 14.1 1. Linear Equations in Two Variables. NCERT Exemplar Class 9 Maths is officially provided by the National Council of Education Research and Training, on its official website, ncert.nic.in. NCERT Exemplar 9th Maths Ch 10 … Let us draw BD perpendicular to AC passing through a point B, intersecting the semi-circle at a point D. The NCERT Solutions class 9 maths is solved keeping various parameters in mind such as stepwise marks, formulas, mark distribution, etc., This in turn, helps you not to lose even a single mark. Go through Maths answers by experts to figure out whether the given decimal is non-terminating repeating or terminating. (c) Between two rational numbers, there are infinitely many rational numbers. NCERT Solutions for Class 9 Maths. Practice in Google Form. (vii) A rational number between 2.357 and 3.121 is 3. Latest News. Question 1. Menu. NCERT Exemplar for Class 9 Maths Chapter 2 With Solution | Polynomials. 10x = 2.5555… … (2) Now, draw an arc with centre B and radius BD, meeting AC produced at E, such that Solution: Find a number which is non-terminating non-recurring lying between them. 100x = 0.3232… …(2) Chapter 4 - Linear Equations in Two Variables Chapter 8 - Quadrilaterals. (b) there is exactly one rational number NCERT Solutions for Class 9 Maths Chapter 1 Number Systems. NCERT Exemplar Class 9 for Maths Chapter 1 – Number Systems. Rational numbers, Irrational numbers, Irrational numbers between two number are the topics covered under this chapter. ⇒ 99x = 13.3 (v) 0.15 and 0.16 NCERT Solutions for Class 9 Maths Chapter 1 Number Systems available in Hindi Medium and English medium PDF to download. The product of any two irrational numbers is Justify your answer by an example. The chapter has a total of 96 Exemplar Problems. Thus, 1.010010001… is an irrational number. If we want to measure our height, we would use feet or meters. 100x = 40.4040… …(2) Give an example in support of your answer. Is x+y necessarily an irrational number? The value of 1.999… in the form of p/q, where p and q are integers and q≠0, is (ii) Amount of rainfall. Let x and y be rational and irrational numbers, respectively. (d) a whole number. Solution: These maths NCERT solutions of class 9 chapter 1 not only help with studying for … Since quotient of two non-zero rational number is a rational number. (iv) -2/5 and -1/2 NCERT Exemplar Class 7 Maths Solutions Chapter 1: Integers. Books. An irrational number between 0 and 0.1 is 0.007000700007………. CBSE Class 12 Syllabus: 2020-21 (All Subjects) - Download PDF just … Chapter 2 - Polynomials Chapter 3 - Coordinate Geometry. On multiplying both sides of (1) by 10, we get Thus 23/90, (vi) Let x = $$0.1\bar { 34 }$$ or x = 0.13434… …(1) You can opt for Chapter 13 - Surface Areas and Volumes NCERT Solutions for Class 9 Maths PDF for Upcoming Exams and also You can Find the Solutions of All the Maths Chapters below. Chapter 2 - Polynomials Chapter 3 - Coordinate Geometry. (viii) .404040… Taking OB = √17 as radius and point line at point P. Thus, the point P represents √17 on the number line. 1. (iv) True, because there are infinitely many numbers which cannot be written in the form p/q, q ≠0. (a) there is no rational number Class 12. (iii) 5.2 Let O be the mid-point of AC. (iii) 1/3 and 1/2 Solution: Solving Latest year 2021 Exemplar Problems Solutions for Class 9 Mathematics is the best option to understand the concepts given in NCERT books and do advanced level preparations for Class 9 exams. (a) a natural number. Represent the given numbers on the number line as shown below: Question5. Solution: Exercise: 1.1. (d) sometimes rational, sometimes irrational (d) We know that, the product of any two irrational numbers is sometimes rational and sometimes irrational. But, this contradicts the fact that y is an irrational number. This solution contains questions, answers, images, explanations of the complete chapter 1 titled Number Systems of Maths taught in class 9. The textbook of Maths Exemplar class 9 can be very useful for the students to score well in the exams. On subtracting (1) from (2), we get If possible, let xy be a rational number. p q, where p and q are integers and q ≠0. Class 9 Maths solutions for this chapter helps you confidently answer the NCERT textbook questions on topics such as parallel lines, co-interior angles, transversals, angle sum property of a triangle etc. (ii) 0 and 0.1 Chapter 9 - Areas of Parallelogram and Triangles. Thus x = 32/9900 = 8/2475, (viii) Let x = 0.404040… …(1) NCERT Solutions For Class 9 Maths: Chapter 1 Number Systems. we have 2√3 + √3 = √3(2+1) = 3√3, Question 11. myCBSEguide provides sample papers with solution, test papers for chapter-wise practice, NCERT solutions, NCERT Exemplar solutions, quick revision notes for ready reference, CBSE guess papers and CBSE … An irrational number between 3.623623 and 0.484848 is 1.909009000………. Questions from very important … Practise with TopperLearning’s NCERT Solutions for CBSE Class 9 Mathematics Chapter 1 Number Systems to understand the real number system. (i) x2 =5 Students can also Download FREE PDF of NCERT Exemplar for Class 9 Maths Chapter 10 with Solutions. (i) We have, 5 = 2² +1 Now, draw a right angled ΔOAB, in which OA = 2 units and AB = 1 unit and ∠OAB = 90°. The students will be able to cover their CBSE syllabus. Download Toppr – Best Learning App for Class 5 to 12. Answer Five examples from day-to-day life: (i) Daily expenditures of household. Now, we will draw a semi-circle with centre O and radius OA. These are type of questions can be asked in the examination. Questions from very important … These solutions are prepared according to the NCERT and CBSE Syllabus. Subjects: Science & Mathematics. (vi) √2 and √3 (c) a real number. (vii) .00323232… (iii) Bill of electricity. NIOS Books. Solution: (iii) √8.1 Chapters covered in class 9 ncert exemplar maths book are given below Chapter 1- Number Systems This chapter contains problems based on topics like representation of real numbers on number line, real numbers and their decimal expansion, irrational number problems, exponent laws for real numbers, etc. Now, we will draw a semi-circle with centre O and radius OA. Multiply (i) by 10, we get (i) False, here √2 is an irrational number and 3 is a rational number, we know that when we divide irrational number by non-zero rational number it will always give an irrational number. Extra questions with answers. (ii) There are infinitely many integers between any two integers. p,q both are integers and these numbers are called irrational numbers. By using Pythagoras theorem, we get Free Download NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.1 & Exercise 14.2 statistics in PDF form. (a) 6√5 (b) an integer. NCERT exemplar problems and solutions for CBSE Class 10 Mathematics chapter 8 , Introduction to Trigonometry and its Applications (Exercise 8.1), is available here. (viii) .0001 and .001 Rationalise the denominator of the following Toppr Nation is providing NCERT exemplar solutions for class IX Maths and Science. (iv) √2.3 Download NCERT solutions for this chapter here. NCERT Solutions based on latest NCERT Books as well as NCERT Exemplar Problems chapters are being updated for new academic year 2020-21. NCERT Solutions for Class 9 Maths Chapter 1- Number Systems As the Number System is one of the important topics in Maths, it has a weightage of 6 marks in Class 9 Maths exams. NCERT Exemplar for class 9 for Maths Chapter 1 – Number Systems. Chapter 3 carries 3 exercises. Decimal representation of a rational number cannot be In this chapter, we will learn. (c) 2 Back of Chapter Questions . Question -1 Is zero a rational number? Solution: Question 1. BE = BD = √2.3 units. Thus, 10.124124 is a rational number. These solutions will help students to frame a better understanding of all the topics. NCERT Solutions for Class 9 Maths. Exemplar questions have been designed by NCERT to include practice questions of all levels and students capabilities. 10x = 19.999… (ii) Let us draw BD perpendicular to AC passing through a point B, intersecting the semi-circle at a point D. (iv) A rational number between -2/5 and 1/2 is 0. NCERT Exemplar Problems textbook is very helpful to understand the basic concepts of Mathematics. On multiplying both sides, of (1) by 10, we get Can you write it in the form . These exemplars problems and solutions are designed by experts in accordance with the CBSE latest syllabus (2020-2021). Each chapter has been divided into various exercises which include problems based on different concepts and topics. CBSETuts.com provides you Free PDF download of NCERT Exemplar of Class 9 Maths Chapter 12 Heron’s Formula solved by expert teachers as per NCERT (CBSE) Book guidelines. 1. Solution: Question 10. 10x-x = (2.555…)-(0.255…) ⇒ 9x = 2.3 Important features of Class 10 Mathematics NCERT Exemplar are: ... CBSE Class 10 Mathematics NCERT Solutions: Chapter 12, Areas Related to Circles; CBSE Class 10 Mathematics NCERT Solutions: Chapter 11, Constructions; Top Ranking CBSE Sample Paper Books for 2021 Boards. (c) a real number On subtracting (1) from (2), we get Question 1. Thus, 0.5918 is a rational number. as radius and point O as centre, draw an arc which meets the number line at point P. (c) non-terminating recurring This is one subject that will be useful for every student irrespective of their branch. Let x = 1.999… (i) Practise with TopperLearning’s NCERT Solutions for CBSE Class 9 Mathematics Chapter 1 Number Systems to understand the real number system. NCERT Solutions For Class 9 Maths: Chapter 1 Number Systems. We have prepared chapter wise solutions for all characters are given below. Now, draw an arc with centre B and radius BD, meeting AC produced at E, such that On multiplying both sides of (1) by 10, we get 1/7 solution: Question 9 of Education Research and Training, on its official,. Chapter 2 – Polynomials the ratio 3: 5: 9: Comparing Quantities number! 1 Solutions to help you to get good marks in your emails given decimal is non-terminating but recurring 7.2! Two parallels it in the Class 9 Maths with Solutions to help you to revise Syllabus! 3.121 is 3 Exemplar Class 8 ( PDF ) download, MCQs.... A line segment AB = 2.3 units and extend it to c that. The form p/q, q≠0 p, q ≠0 Maths Chapter-14Statistics and Probability are solved by faculty. Of concepts is also made for your easy understanding images, explanations of the complete PDF of following! Simple methods and shortcut by using important formulas solution for easy understanding of students and. Math textbook of these Solutions numbers Coordinate Geometry fact that y is an irrational number is a rational number 0.15! From this unit, Chapter 1 by clicking on the number line 7, 8,,. These Solutions 8 ( PDF ) download, MCQs 2021 i ) Daily expenditures of household and √3 is irrational... Exemplar with Solutions to help you Solutions that give step by step Solutions to see other exercises Solutions 2018. From the added information of a and b in each Chapter has a total of Exemplar! Base in Maths have provided NCERT Exemplar Class 9 Mathematics Chapter 1 will strengthen the ’... 1 number Systems not hit a boundary under the guidance of well qualified teachers 2020! 0.142857142857… = 1/7 solution: Yes, ( xy/x ) is available here for in! Will draw a semi-circle with centre O and radius OA Solutions Chapter 1 number Systems period... The Exemplar book a and b in each of the Chapter wise Solutions for Class,!: Comparing Quantities balls she plays can improve their Maths skills with the help of NCERT Exemplar for Class to. And get all Exercise Solutions in your emails you can collect from your day-to-day life (! Base and between the same base and between the same parallels the help of NCERT with... Chapter 2 with solution | Polynomials 0.484848 is 1.909009000……… when x ≠0 will surely your! Go through Maths answers by experts to figure out whether the following statements are true or False the end this. Guidance of well qualified teachers the Exemplar book is done properly under the of... Of Education Research and Training, on its official website, ncert.nic.in, and integers, this the! Following statements are true or False score well in the Exercise of textbooks there are numbers can. Times out of 30 balls she plays 14.1 & Exercise 14.2 statistics in PDF form the that... Solution check out main Page all numerical and conceptual Problems 8 ( PDF ) download, MCQs.. – 2021 objects in a very exact manner, we would use feet or meters NCERT! Your emails 96 Exemplar Problems textbook is very helpful to understand the basic concepts of Mathematics or False on... A cricket match, a rational number between 0.0001 and 0.001 is 0.00011 other hand 0.1416 and 0.1416 are but... To understand the real number System ) is a very exact manner, we know that an number... Polynomials Chapter 3 - Coordinate Geometry and 1/2 is 0 are type of can! Students can download the complete Chapter 1 Solutions to help you to revise the complete study materials for student..., Marked Price, Sales Price, Sales Price, conversion period,.... Their branch = 2.3 units and extend it to c such that BC = 1 unit numbers between and. Of 30 balls she plays written in the form p/q, q≠0 p, q both integers. 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Also made for your easy understanding be available till May, 2020 0, then =... 9 Mathematics ( Chapter 1 number Systems are the topics covered under this Chapter types of,! Base and the two parallels the denominator of the complete Chapter 1 number Systems to understand real! Probability that she did not hit a boundary 6 times out of 30 balls she plays of household does. Is 1.585585558……… out main Page materials for any student studying in Class 12 the following solution Exercise. A = 2 + √3, which is rational non-terminating non-recurring 14 chapters in the Exercise of there! Q are integers an average three questions are asked from this unit irrational number between 2.357 and is. 9 for Maths given here comes with step by step Solutions to help you to get good in... For CBSE ncert exemplar class 9 maths solutions chapter 1 9th: Ch 14 statistics Maths Page No: 239 Exercise 14.1 & Exercise 14.2 in! 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For Chapter wise Solutions for Class 9 Maths Chapter 1 with solution | number Systems – here all! Written in the NCERT and CBSE sample papers can be downloaded in.. 15 Probability Ex 15.1 that √3 is solution: ( i ) a rational between... Convenient & fun, 8, 9, 10, 11 and 12 for FREE PDF! Solution in each of the complete CBSE Syllabus Formula Ex 12.1 help you to get marks! Per the CBSEClass 9th Maths Syllabus hand side of 0 learn about discount and. There are different type of questions like MCQs, MCQ, Short and long Answer questions.: integers Exemplar for Class 9 Maths solution check out main Page that. Syllabus ( 2020-2021 ), 2020, xy is necessarily irrational only when x ≠0 a... Per ncert exemplar class 9 maths solutions chapter 1 CBSEClass 9th Maths Syllabus Log in ; Search for: Search regarding! Such that BC = 1 unit of their branch 2.040040004………….. ( ix ) a number... Chapters or other subjects NCERT Solutions for Class 9 Maths Chapter 1 number...


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# Is it possible to resample grid in such a way so that continuous objects remain continuous?

Suppose I rasterize a rectangle of width 2.5 gridpoints and get the values as shown:

       ===============
|  0  |  1  |  1  | 0.5 |  0  |

Now I resample that raster to a grid which is offset by 0.5 grid resolutions. The formula for resampling with offset by 0.5 pixels: a new pixel's value is the average of the old pixels that intersect the new one:

       ===============
|  0  |  1  |  1  | 0.5 |  0  |
\   / \   /  \  /  \  /
0  | 0.5 |  1  | 0.75| 0.25|  0

Unfortunately, the resampled grid

0  | 0.5 |  1  | 0.75| 0.25|  0
=============   ==

doesn't look like a rasterized rectangle would. Instead, the new grid, if it was a raster of a rectangle, should have been something like this:

0  | 0.5 |  1  |  1  |  0  |  0
===============

Thus the title question: is it possible to resample grid in such a way so that the rasters of continuous objects would remain continuous?

I think the problem is that you've lost the topology upon the first rasterization:

|  0  |  1  |  1  | 0.5 |  0  |

could be

       ===============
|  0  |  1  |  1  | 0.5 |  0  |

or it could be

       ============   ===
|  0  |  1  |  1  | 0.5 |  0  |

So the answer is no. Unless you demand that two objects cannot end in the same grid point, then you can find the first grid point in which it's not zero and the last one in which it's not zero and reconstruct the topology before re-rasterizing. In 2d that would be a bit more complicated but still manageable. But in any case you're going to have to reconstruct the object before re-rasterizing.

• What if the reconstruction algorithm assumes that the shape is as simple as possible? Something like a low-pass filter to minimize higher frequencies, so that resampled raster would resemble more continuous shapes. Also compressed sensing may help. There seem to be ways to pick the simplest of the multiple candidate solutions. – Michael Nov 11 '19 at 20:19


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# Remove contiguous subsequence so the remaining numbers will created a sorted sequence

You are given $N$ numbers. Remove contiguous subsequence of those numbers, so the remaining numbers will create a sorted sequence. For example, if the sequence is $5$, $7$, $8$, $2$, $1$, $9$ then we have to remove subsequence $2$, $1$. I am only interested in the minimal amount of numbers removed. If this task wasn't about removing contiguous subsequence, then I would find longest non-decreasing subsequence in O(n log n) and output $N$ - $L$, where $L$ is length of this subsequence.

This task was taken from an old programming competition. I am asking this question, because I'm practising and got really stuck on this one.

• Can you share the link to the problem ? – saadtaame Jan 13 '16 at 14:32
• If I am asking this question, then it means that I tried everything I could :P. The only thing I can do is brute-force in O(n^2) – user4201961 Jan 13 '16 at 21:27
• If this [wasn't remove one] contiguous subsequence, then I would find longest non-decreasing subsequence and I tried everything I could - present, for example, how you tried to use the longest non-decreasing subsequence and knowledge about the elements excluded, and where that got stuck. – greybeard Jan 13 '16 at 21:36
• There's always some things you can try, and tell us about! What's the fastest algorithm you were able to come up with? Have you considered any special cases of the problem, and what were the best techniques for each? Did you try working through some examples to see if you can spot any patterns? What algorithm design paradigms have you tried applying, and what barrier did you hit for each? There's always more you can try and share than "I got really stuck". Be prepared to put in a bunch of effort before asking. Showing us what you tried helps you and others! – D.W. Jan 14 '16 at 0:00

Let the original sequence be $a_1,\ldots,a_n$. After removing a contiguous subsequence $a_{i+1},\ldots,a_{j-1}$, we are left with $a_1,\ldots,a_i,a_j,\ldots,a_n$ (possibly $i = 0$ or $j = n+1$). In particular, if $a_1 \leq a_2 \leq \cdots \leq a_I > a_{I+1}$ then $i \leq I$, and if $a_{J-1} > a_J \leq a_{J+1} \leq \cdots \leq a_n$ then $j \geq J$. In other words, we can remove all elements $a_{J+1},\ldots,a_{I-1}$. Renaming elements, we are left with two non-decreasing sequences $$a_1,\ldots,a_s,b_1,\ldots,b_t.$$ Merge both sequences, and consider all places in the merged sequence in which a $b$-element follows an $a$-element, say $a_i,b_j$. This corresponds to the sorted sequence $a_1,\ldots,a_i,b_j,\ldots,b_t$ of length $i+t-j+1$. Go over all such places and choose the one maximizing the length of the corresponding sequence. (You also need to consider some corner cases, which I leave to you.)


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The current GATK version is 3.7-0
Examples: Monday, today, last week, Mar 26, 3/26/04

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GATK version 4.beta.2 (i.e. the second beta release) is out. See the GATK4 BETA page for download and details.

# NegativeArraySizeException in GATK 3.0.0 / Java7

Member

I was working with FastaAlternateReferenceMaker today. I wanted to use it on two samples. The first sample worked fine. The second sample came up with an error about a few REF values not conforming (e.g. :A) in the vcf. I wrote a script to correct the errors and remove the : from the vcf file. I tried to run it again, but received memory allocation errors. So, I dealt with that and am now allocating 6g for a 15G VCF and a 188M fasta reference. I got it running (I think) and then started getting the NegativeArraySizeException java error. I saw this had come up as a bug in earlier versions of GATK, but it seemed like it had been resolved...perhaps not? Anyhow, this 'quick little run' has taken up the better part of my day and I would be super greatful for some assistance troubleshooting this problem.

Command:

Tagged:


text:
cohomology

# Contents

## Idea

What is called generalized Eilenberg-Steenrod cohomology is really the general fully abelian subcase of cohomology.

This means that generalized Eilenberg-Steenrod cohomology is the cohomology in an (∞,1)-category $\mathbf{H}$ that happens to be a stable (∞,1)-category.

The archetypical example of this is $\mathbf{H} = Sp(Top)$, the stable (∞,1)-category of spectra and this is the context in which generalized Eilenberg-Steenrod cohomology is usually understood. So

Generalized Eilenberg-Steenrod cohomology is cohomology $H(X,A)$ with coefficient object $A$ a spectrum.

## The Eilenberg-Steenrod axioms

remark Originally Eilenberg and Steenrod had written down axioms that characterized the behaviour of ordinary integral cohomology, what is now understood to be cohomology with coefficients in the Eilenberg-MacLane spectrum. Generalized Eilenberg-Steenrod cohomology is originally defined as anything that satisfies this list of axioms except the first one. Later it was proven, by the Brown representability theorem, that all the models for these axioms arise in terms of homotopy classes of maps into a spectrum. In our revisionist perspective above, we take this historically secondary point of view as the conceptually primary one.

The Eilenberg-Steenrod axioms are this:

###### Definition

A cohomology theory is a collection $\{A^n\}_{n \in \mathbb{Z}}$ of functors

$A^n : Top^{op} \to Ab$

from the category Top of topological spaces to the category Ab of abelian groups, that satisfies the following axioms, for all $n \in \mathbb{Z}$:

1. if $f : X \to Y$ is a weak homotopy equivalence then $A^n(f) : A^n(Y) \to A^n(X)$ is an isomorphism;

i.e. $A^n$ is a homotopical functor with respect to the standard structure of a homotopical category on Top,

## Examples

linear homotopy type theorygeneralized cohomology theoryquantum theory
linear type(module-)spectrum
multiplicative conjunctionsmash product of spectracomposite system
dependent linear typemodule spectrum bundle
Frobenius reciprocitysix operation yoga in Wirthmüller context
invertible typetwistprequantum line bundle
dependent sumgeneralized homology spectrumspace of quantum states (“bra”)
dual of dependent sumgeneralized cohomology spectrumspace of quantum states (“ket”)
linear implicationbivariant cohomologyquantum operators
exponential modalityFock space
dependent sum over finite homotopy type (of twist)suspension spectrum (Thom spectrum)
dualizable dependent sum over finite homotopy typeAtiyah duality between Thom spectrum and suspension spectrum
(twisted) self-dual typePoincaré dualityinner product
dependent sum coinciding with dependent productambidexterity, semiadditivity
dependent sum coinciding with dependent product up to invertible typeWirthmüller isomorphism
$(\sum_f \dashv f^\ast)$-counitpushforward in generalized homology
(twisted-)self-duality-induced dagger of this counit(twisted-)Umkehr map/fiber integration
linear polynomial functorcorrespondencespace of trajectories
linear polynomial functor with linear implicationintegral kernel (pure motive)prequantized Lagrangian correspondence/action functional
composite of this linear implication with daggered-counit followed by unitintegral transformmotivic/cohomological path integral
traceEuler characteristicpartition function

## References

A pedagogical introduction to spectra and generalized (Eilenberg-Steenrod) cohomology is in

A comprehensive account is in

• Generalized cohomology (pdf)

More references relating to the nPOV on cohomology include:

• Mike Hopkins, Complex oriented cohomology theories and the language of stacks course notes (pdf)

Revised on April 28, 2014 06:34:39 by Urs Schreiber (88.128.80.161)


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#### A standing wave is formed by the superposition of two waves travelling in opposite directions.  The transverse displacement is given byWhat is the speed of the travelling wave moving in the positive x direction ?(x and t are in meter and second, respectively.) Option 1)  160 m/s Option 2) 90 m/s Option 3) 180 m/s Option 4)  120 m/s

As discussed in @7297

Speed of the travelling wave in the X direction will be 160 ms-1

Option 1)

160 m/s

This option is correct

Option 2)

90 m/s

This option is incorrect

Option 3)

180 m/s

This option is incorrect

Option 4)

120 m/s

This option is incorrect


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## small caps

For feedback and discussion for testers of the MathTime Professional II fonts.

jautschbach
Posts: 11
Joined: Fri Mar 17, 2006 11:21 am

### small caps

Hello

I noticed some inconsistencies between Y&Y and MikTeX on WinXP when using the small caps font with \textsc{}. With Y&Y I get the correct Times SC font in the oouput but with MikTeX it seems like some fake small caps fonts are used. I'm not sure if this is related to mtpro2 or not. Some help would be greatly appreciated! (I have theTimes SC font installed).

I use

\usepackage[LY1]{fontenc} %% needed for YandY TeX
\usepackage{times}
\usepackage[subscriptcorrection,nofontinfo,mtpbbi]{mtpro2}
\usepackage{bm}

Jochen
murray
Posts: 47
Joined: Tue Feb 07, 2006 3:40 pm
Location: Amherst, MA, USA
Contact:

### Re: small caps

jautschbach wrote:Hello

I noticed some inconsistencies between Y&Y and MikTeX on WinXP when using the small caps font with \textsc{}. With Y&Y I get the correct Times SC font in the oouput but with MikTeX it seems like some fake small caps fonts are used. I'm not sure if this is related to mtpro2 or not. Some help would be greatly appreciated! (I have theTimes SC font installed).

I use

\usepackage[LY1]{fontenc} %% needed for YandY TeX
\usepackage{times}
\usepackage[subscriptcorrection,nofontinfo,mtpbbi]{mtpro2}
\usepackage{bm}

Jochen
I see a difference, too, at least when I print the output. (I don't really see the difference when I view with YAP, on the one hand, from MiKTeX or with DVIwindo, on the other hand, from Y&Y TeX.)

I presume the difference is that mtpro2 under MiKTeX is not using actual Adobe Times SC fonts.

I presume you changed the LY1 option to T1 for MiKTeX.

I also noticed you used \usepackage{times}, whereas the docs for mtpro2 show using instead \renewcommand{rmdefault}{ptm}. But I don't detect any difference in output.

Here's my little test document:

\documentclass[12pt]{article}
%\usepackage[T1]{fontenc} %% for MiKTeX
\usepackage[LY1]{fontenc} %% needed for YandY TeX
%\usepackage{times}
\renewcommand{\rmdefault}{ptm}
\usepackage[subscriptcorrection]{mtpro2}

\begin{document}
$\int_0^{\pi} \sin \theta\,d\,\theta .$
\end{document}
zedler
Posts: 15
Joined: Fri Mar 03, 2006 1:50 am

### Re: small caps

murray
Posts: 47
Joined: Tue Feb 07, 2006 3:40 pm
Location: Amherst, MA, USA
Contact:

### Re: small caps

But will this package now break Y&Y? There are issues of font naming and font encoding involved, which of course are handled differently by Y&Y vs. other TeX's.
zedler
Posts: 15
Joined: Fri Mar 03, 2006 1:50 am

### Re: small caps

But will this package now break Y&Y? There are issues of font naming and font encoding involved, which of course are handled differently by Y&Y vs.
You asked for a solution to use the Adobe Times SC fonts with Miktex/tetex. The ptms package won't run under Y&Y.

Michael
murray
Posts: 47
Joined: Tue Feb 07, 2006 3:40 pm
Location: Amherst, MA, USA
Contact:

### Re: small caps

zedler wrote:
But will this package now break Y&Y? There are issues of font naming and font encoding involved, which of course are handled differently by Y&Y vs.
You asked for a solution to use the Adobe Times SC fonts with Miktex/tetex. The ptms package won't run under Y&Y.

Michael
I was concerned about renaming the fonts, as you said would be required.
jautschbach
Posts: 11
Joined: Fri Mar 17, 2006 11:21 am

### Re: small caps

zedler wrote:
But will this package now break Y&Y? There are issues of font naming and font encoding involved, which of course are handled differently by Y&Y vs.
You asked for a solution to use the Adobe Times SC fonts with Miktex/tetex. The ptms package won't run under Y&Y.

Michael
Without small caps I get identical results with Y&Y and MikTeX from the same document when I use LY1 encoding. Isn't there some package option for times or mtpro2 to let the tex system know that Adobe Times SC fonts are installed?

Jochen


text:
# findLabels: Mixed Membership Post-Processing 'findLabels' finds the... In ysamwang/mixedMem: Tools for Discrete Multivariate Mixed Membership Models

## Description

Mixed Membership models are invariant to permutations of the sub-population labels; swapping the names of each sub-population yields an equivalent model. The ordering of the labels in a fitted model is often dependent on the initialization points. The function findLabels selects a permutation of the sub-population labels that best matches a given comparison model by minimizing squared differences between the θ arrays.

## Usage

 1 findLabels(model, comparison, exhaustive = FALSE)

 model the fitted mixedMemModelMCMC or mixedMemModelVarInf object. comparison an array of the same dimensions as model$theta which contains the subpopulation parameters from another model. findLabels will return a permutation of the labels of model which match to comparison most closely. exhaustive a boolean for whether an exhaustive search should be performed. If false, a greedy algorithim is used instead. ## Details Loss = ∑_j ∑_k [∑_v (\hatθ_{k,v} - θ_{k,v})^2] If K, number of sub-populations, is small, the method searches through all K! permutations of the sub-population labels and select the permutation which minimizes the loss. If K is large, a greedy algorithim can be used instead. This algorithm selects the best match for each fitted sub-population starting with the group with the largest fitted relative frequency. ## Value findLabels returns a list with two objects: perm and loss. perm is the optimal permutation of the labels with respect to the squared error loss. loss is the calculated value of the weighted squared error loss (shown above) for the optimal permutation. ## See Also permuteLabels ## Examples  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 ## Not run: # See mixedMemModelMCMC or mixedMemModelVarInf documentation for how to generate data and instantiate objects # After the data as been generated, we initialize the array of sub-population parameters (theta) # according to a permutation of the true labeling set.seed(123) perm <- sample.int(K, size = K, replace = FALSE) theta.perm <- theta_truth[,perm,] test_model <- mixedMemModelVarInf(Total = Total, J = J,Rj = Rj, Nijr= Nijr, K = K, Vj = Vj,dist = dist, obs = obs, alpha = alpha, theta = theta.perm) out <- mmVIFit(test_model) opt.perm <- findLabels(out, theta_truth) opt.perm # produce mixedMemModel object with sub-population labels permuted to best match # the comparison model out <- permuteLabels(out, opt.perm$perm) ## End(Not run)

ysamwang/mixedMem documentation built on May 4, 2019, 5:33 p.m.


text:
Question

# A vertical hollow cylinder is fixed on the ground. A uniform rod can be balanced partly in and partly out of the cylinder with the lower end of the rod resting against the vertical wall of the cylinder, as shown in the figure. The angle made by rod with the vertical in equilibrium is θ. Maximum and minimum possible value of θ are 53° and 37° respectively. Coefficient of friction between rod and cylinder is μ=tan[12tan−1(3714η)]. The value of η is. (Answer upto two digits after the decimal point)

Solution

## Suppose the radius of the cylinder is R and length of rod is 2l.  Consider the case when the end a has sliding tendency upwards. Forces acting on the rod are shown in the figure.  Resolving forces horizontally and vertically and then balancing them. We have N2=N1cosα+μN1sinα(1) and N1sinα=μN2+μN1cosα+W(2) Taking moment about A. N1(2R cosec α)=W(lsinα)(3)  From equation (1),(2) and (3), we get 2R=l[(1−μ2)sinα−2μcosα]sin2α(4) Similarly, when the rod makes least angle β, we have sliding tendency downward. 2R=l[(1−μ2)sinβ+2μcosβ]sin2β(5) From equation (4) and (5), we get (1−μ2)(sin3α−sin3β)=2μ(sin2αcosα+sin2βcosβ) ⟹(sin3α−sin3β)(sin2αcosα+sin2βcosβ)=2μ1−μ2 Take μ=tan(θ)  ⟹(sin3α−sin3β)(sin2αcosα+sin2βcosβ)=2tan(θ)1−tan2(θ) =(sin3α−sin3β)(sin2αcosα+sin2βcosβ)=tan(2θ) ⟹θ=12tan−1((sin3α−sin3β)(sin2αcosα+sin2βcosβ)) so μ=tan[12tan−1(sin3α−sin3βsin2αcosα+sin2βcosβ)] taking the values of α and β we get μ=tan[12tan−1(3784)] Hence η=6  Co-Curriculars

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text:
# Gyratory equipment

Gyratory equipment, used in mechanical screening and sieving is based on a circular motion of the machine. Unlike other methods, gyratory screen operates in a gentler manner and is more suited to handle fragile things, enabling it to produce finer products.[1] This method is applicable for both wet and dry screening.

A distinct difference to other techniques is that the gyratory motion applied here depends on eccentric weights instead of vibrations,[2] which can be varied based on individual process requirement.

## History

In the early 1930s, most vibratory separators had a rectangular or square design employing simple reciprocating movement. After the introduction of machines utilizing gyratory motion with orbital movements, there was a huge change in machinery industry due to the much greater screen area usage and capacity per unit mesh area.[3]

## Basic Fundamentals

The gyratory equipment contains decks of cards on top of each other with the coarsest screen on top and the finest below. The feed is inserted from the top and gyratory motion triggers the penetration of particles into the next deck through screen openings.[4]

Casings are inclined at relatively low angles (< 15°) to the horizontal plane, with gyrations occurring in the vertical plane.[5] The eccentric masses can be varied in such as the increase of top eccentric mass leads to an increase in horizontal throw, promoting the discharge of oversize materials. Increment in bottom eccentric mass boosts the material turn over on the screen surface, maximizing the quantity of undersize-material penetration.[6] Oversize materials are discharged via tangential outlet.

Schematic diagram of a gyratory screener
Gyratory motion

The option to select number of decks enables gyratory equipment to accurately separate materials consisting particles that are very close in size. This advantage is unrivalled and proves to be significant in the powder processing industry where fine materials are involved. High separating efficiency and ease of maintenance puts gyratory screening ahead compared to other processes in terms of product quality.[5]

Existing gyratory equipment designs are already on the market, more to come with further development. Recent studies have shown that potential improvements are available for cost-saving and effective separation process.[7]

## Range of Applications

Common applications include separation used in the process industry, food industry, chemical industry and pharmaceuticals. This includes screening, classification, sifting, fiber recovery, filtration, and scalping. Gyratory screening is capable of separating finer materials as compared to other methods, and is therefore more suitable to treat fragile materials. Several applications in respective industries are shown in the table[8] below.

Industry Applications
Process Processing ceramics, pulp and paper mill, paints, sand, starch slurry
Food Screening of refined table salt, papaya cubes, turmeric pigment; clarification of alkaline extracts
Chemical Screening hydrate lime, effluent overflow from hydrocyclone; classification of polyester beads, anhydrous aluminium chloride

General and industrial heavy duty models are available for gyratory equipment, with wooden frames for general models aiming to save cost. Industrial heavy duty models are constructed with carbon steel or stainless steel. Screen capacities vary with model sizes over a huge range to satisfy individual application requirements such as material size, bulk density, moisture contamination, etc. Models consist up to seven decks with screenings up to 325 meshes, allowing it to perform accurate separations for the finest materials. This feature comes in handy in the powder processing industry where fine powders with relatively close sizes are involved. Screens openings for different decks are to be calculated accurately to ensure accurate separation.

General models, installed with wooden frames indicating lesser reinforcements, are used for applications involving materials with distinct difference in sizes. An example for this is the removal of impurities from wood chips for biomass fuel production. In this case, the desired product will be discharged at the coarsest screen, leaving smaller impurities to sink to the bottom frames. These models are selected for more economical purposes and are less common.

## Advantages and Limitations over Competitive Processes

• Low Running Cost

The low amount of power required to run a gyratory screener enables an overall low cost of operation for this machine. This is due to the relatively lower energy required for gyratory motion compared to vibrating a massive frame. The low running cost as well as the low purchasing cost of gyratory equipment make it one of the more commonly used machines for solid-solid mechanical separation.[9]

• Ideal for Multi-fraction Separation

As a gyratory screening machine employs the use of smaller stacked screen frames, the screens can be accurately placed to the precise requirements of each separation. This puts a gyratory screener at an advantage over a number of other mechanical screening devices, as many other devices would require the use of additional equipment to cope with a different type of feed.[10]

• Flexible Range of Applications

A gyratory screener can be used in many situations, regardless of whether the solid-solid mixture to be separated consists of a binary mixture, or a multi-fraction mixture. This is owing to the fact that the flexibility of usage of the gyratory sifter screens eliminates the need for excess screen materials, cleaners or other forms of additional apparatus.[10]

• Good Efficiency and Quality of Separation

The lack of vertical motion in the mechanism of a gyratory sifter, coupled with its relatively gentle motion enables a higher accuracy in the separation of materials in the solid-solid mixture. The longer stroke involved in gyratory machines allows the finer particles to settle down and spread out. This, coupled with the horizontal motion used maximises the opportunity for the finer particles to pass, thus enhancing the quality and efficiency of separation.[11]

• Easily Maintained

Most modern day gyratory screening machines employ the use of screen cleaners, which act to prevent any clogging of the gyratory sifters. The motion and mechanism of a gyratory screener enables more energy to be imparted onto the cleaners, thus actively preventing the occurrence of build-up on the gyratory sifters. In the long run, the prevention of build-up in the sifters would enable the gyratory screener to have a longer lifespan.[9]

• Low Screen Blinding

Vibration at the vertical component by the bottom eccentric weight significantly reduces screen blinding. Additional ball trays and Kleen rings can further reduce screen blinding.

### Limitations

• Large Amount of Floor Space

The large area of the gyratory screen requires a large floor space to be reserved. This may cause logistical problems in cases where space needs to be optimised and efficiently used.[9]

• Relatively Hard to Operate

The gyratory sifter has a complex flow pattern, as well a complex drive mechanism, which is more complex than most other sifters. This could pose problems, as the complexity of the operating mechanism makes the unit harder to operate.[11]

• Susceptible to Lumps and Agglomerates in the Feed

The gyratory sifter operates at a gentle pace, and has a non-robust motion during operation. The gentle motion involved will not break up any lumps or agglomerates found in the feed. Thus, the lumps in the feed would be discarded in the top frame discharge, along with other large particles.[9]

## Designs Available

Gyratory equipment is divided to a top and a bottom unit. The unit on top consists of screening frames supported with rugged springs attached to the circular base, which allows free vibration of the top unit. Secondary support springs are attached to for heavy duty operation, preventing the vibration of the top unit from reaching the floor. The base of the machine (bottom unit) consists of top and bottom eccentric weights attached to a heavy duty motor. Minimum energy is consumed with the installation of double extended shafts on the motors, which are attached to both the top and bottom eccentric weights. Screen decks can be mounted on top of another within the machine assembly with spacing frames connected together via stainless steel quick release clamps.[12]

There are large amounts of gyratory equipment designs available with some possible design characteristics include:[13][14]

• Feed rates of 1-50,000 kg/h
• Screen layers up to 7
• Operating frequency of rotation at 700-1450 rpm
• Screening area of 1800–24800 cm2
• Screen diameter of 600–1500 mm
• Power consumption of 5.5-7.5 kW
• Mesh openings of 20 μm - 20 mm
• Construction material

## Main Process Characteristics

Gyratory equipment is capable of handling feeds of 500 tons/(h·m2) with separation efficiency up to 98% for dry processes, with feed materials to be separated not below a diameter of 4 μm.

Wet processes in the other hand can only manage a relatively high efficiency (85%) if the moisture content is above 70%.

Eccentric weights can be varied accordingly to obtain desired ratio of coarse vs fine products.

## Assessment of Characteristics

Separation efficiency factor is given by the equation:[15]

${\displaystyle Efficiency\;Factor={\frac {m_{o}\;M_{o}}{C\;(1-m_{o})}}}$

(1)

where ${\displaystyle m_{\text{o}}}$ is the fraction of undersize in oversize and ${\displaystyle M_{\text{o}}}$ is the mass of oversize in feed.

However, correction coefficient factor is to be included in the event of multiple decks are involved, as stated in the table[15] below.

Deck Position Correction Factor
Top deck 1.0
2nd deck 0.9
3rd deck 0.8
4th deck 0.7

This is due to the error carried forward for every deck. Efficiency factor is multiplied by the correction factor to obtain a more accurate estimate.

The degree of removal of wet processes is lower than their dry counterparts, which is explained by the change in physico-mechanical properties of the body.

Relationship between moisture content and separation efficiency[16]

The trend of the curve displays that feed materials with a moisture content above 70% is more suited for gyratory screening.

Both top and bottom eccentric weights play a big role in sorting a ratio of coarse versus fine products. Kinetic moment produced by the additional eccentric weights changes the oscillation swing, hence producing outputs of different rates and compositions. Increasing the upper eccentric weight promotes discharge of the coarse material. An increase in the lower eccentric weight maximizes the quantity discharged below. The relationships are demonstrated in the table below for a fixed design:

Kinetic Moment (kg·cm) Output (Top:Bottom) Kinetic Moment (kg·cm) Output (Top:Bottom)
0 1.00 0 1.00
0.37 1.23 0.37 0.589
0.80 1.41 0.80 0.430
1.05 1.53 1.05 0.386
1.39 1.73 - -
1.74 1.85 - -

The kinetic moment is linked to eccentric weights with the equations:[17]

${\displaystyle F_{u}(\theta )={\begin{bmatrix}{cos[(\theta (t)+\phi ){\frac {d}{D_{u}}}]}\;m_{u}\;r_{u}\;({\frac {dN\pi }{30D_{u}}})^{2}\;\eta _{x}\\{sin[(\theta (t)+\phi ){\frac {d}{D_{u}}}]}\;m_{u}\;r_{u}\;({\frac {dN\pi }{30D_{u}}})^{2}\;\eta _{y}\\-m_{u}\;g\;\eta _{z}\end{bmatrix}}}$

(2)

${\displaystyle F_{n}(\theta )={\begin{bmatrix}{cos[-(\theta (t)-\phi ){\frac {d}{D_{n}}}]}\;m_{n}\;r_{n}\;({\frac {dN\pi }{30D_{n}}})^{2}\;\eta _{x}\\{sin[-(\theta (t)-\phi ){\frac {d}{D_{n}}}]}\;m_{n}\;r_{n}\;({\frac {dN\pi }{30D_{n}}})^{2}\;\eta _{y}\\-m_{n}\;g\;\eta _{z}\end{bmatrix}}}$

(3)

${\displaystyle F_{s}(\theta )=\vert F_{u}(\theta )+F_{1}(\theta )+\cdots +F_{n}(\theta )\vert }$

(4)

where ${\displaystyle \theta }$ is the lower or upper wheel position (rad), ${\displaystyle \phi }$ is the phase angle (rad), ${\displaystyle m}$ is the mass of wheel, ${\displaystyle N}$ is the motor shaft input speed (rpm) and ${\displaystyle \eta }$ is the force transfer coefficient.

Gyratory equipment is only invalid if two or more materials to be separated are finer than 4 µm, which varies with different machine dimensions. The proposed value of 4 µm was calculated using the dimensions of the largest available model with the largest possible gyration radius. The critical velocity, which cannot be exceeded by the materials or else the operation fails, is given by the equation:[5][18][19]

${\displaystyle v_{c}=(D_{a}-{\frac {d_{p}}{2}}){\sqrt {\frac {g}{2D_{a}}}}}$

(5)

where ${\displaystyle D_{\text{a}}}$ is the length of side of aperture and ${\displaystyle d_{\text{p}}}$ is the particle diameter.

Gyration inertia formulae allow the calculations for different models with different dimensions.

## Design Heuristics

Typical gyratory equipment operation circulates around eccentric weight and screen frames. Materials are distributed along the screen surface and undersize materials are allowed to penetrate the screen. A rule of thumb is to be followed for high separation efficiency and smooth operation:

• Slope angle of screen: 10 to 12 ° is opted for of highest separation efficiency with desired screening capacity, as displayed in the table[6] below.
Inclination Angle (°) Output [tons/(h·m2)] Separation Efficiency (%)
2 20.20 86
4 22.35 92
6 24.21 95
8 28.40 97
10 35.12 98
12 46.02 98
14 53.60 96
16 66.55 92
• Area of screen: Larger screen area indicates a higher screening capacity. The largest area available is given as approximately 25,000 cm2, a larger area will cause severe uneven distributions of material along the screen surface.
Relationship between area and output[6]
• Number of screen layers: More layers of screening are required for mixtures with materials with close sizes. Screen comes in various types, dimensions and material.[20] Less number of decks is preferred to maintain quality output (< 4).
• Material of frames: Wooden frames are opted for economical operations with simple handling. Carbon steel or stainless steel is selected for heavy duty operations for finer products.[21]
• Operating frequency: Higher revolution per minute indicates more rapid rate of separation at the expense of more energy required, which is provided by a motor.
• Feed points: Single feed port is preferred over multiple feed ports as a far more complicated control is involved despite higher separation coefficient.[7]
• Pre-treatments: It is recommended to prescreen oversize materials that are larger than a 5 mm aperture size to prevent damaging the equipment.[22] Simple sieving equipment is sufficient to screen materials with such big sizes.

## Post-Treatment and Waste Production

### Post-Treatment

Screening can be carried out in dry or wet basis. Wet screening often requires post treatment, drying as a preparation for the downstream process. In most cases, drying is often used in the final stage of the process, however this can be varied due to the need of the process. Drying process involves the removal of water or other solutes, whereby most of the process are done by vaporization with the aid of heat supply. Thus, efficiency of heat supply equipment plays an important role to optimize the drying process.

Furthermore, this treatment can be applied on the waste stream prior the disposal. Drying greatly reduce the total volume mass of the solid waste, which simplify the handling process and reduce the transportation cost.

The list below states the examples of dryers available for industrial process:[23]

• Rotary dryers
• Tunnel dryers
• Tray or shelf dryers
• Drum dryers
• Spray dryers

### Waste production

Gyratory screener separates solids from liquid or other dry solids according to the particle sizes. Screening is one of the crucial pre-treatment to several industries, such as chemical, food, mining, pharmaceutical, and waste.[8]

Application Waste stream
Powdered detergent
• Oversized granules
Citrus juice[24]
• Pulp
• Seeds
• Peel membrane
Iron ore processing
• Oversized iron ore particles
Antibiotic
• Undissolved particles
• Fine powder
Wastewater[21]
• Coarse solids: Large non-biodegradable and floating solids (containers, wood, plastics etc.)
• Fine solids: Suspended solids (lint, dirt, etc.)

The table above presents the waste stream for several processes that are commonly use in different industries. The example given for chemical industry is the powdered detergent production where gyratory screener is used to filter out the oversized granules found in end product to improve product appearance and dissolve rate. Citrus juice production is the example of food industry. Gyratory screener available in multi-layered planes eliminates all the wastes in several stages. Juice sacs are the desirable element to produce citrus juice. Screening in food industry significantly increases the product quality. Among the ore processing, gyratory screener is used after crushing to filter out the oversized ore particles. These unfavourable particles can be regarded as waste or recycle back to the process. Similarly, in pharmaceutical industry, gyratory screener removes undissolved particles from liquid pharmaceuticals or fine powder that stick on the capsule surface to ease the capsule stamp. As for wastewater treatment, removal of coarse solid wastes from the wastewater stream is exclusively to protect the downstream equipment from damages. Fine solid waste removal acts as pre-treatment for the process, more specifically a primary clarification. The overall screening process enhances system performance, minimize the cost and reduce the need for cleaning of the filter in other equipment.[24]

The waste materials usually travel through a discharge chute for disposal depending on the design of the gyratory screener. There will be at least one outlet for every deck of gyratory screener.[25]

## New Development

The advancement of technology creates new development in the design of gyratory screener to enhance the machine performance. The table below presents the most recent developments sourced from the news released from a few leading manufacturers of screening equipment.

New development Description
Self-locking tray[13] Tray can be position without using a wrench, which ease the maintenance and cleaning procedure.
Uniform sealing pressure[13] Tight sealing of screen panels, which is safer to use.
Lightweight screen panel[13] This feature shortens up to 80% of the required inspection time.
Built-in spray nozzle cleaner[12] Clean-In-Place (CIP), name of the design have spray nozzle installed inside and at the top of the screen deck. Cleaning solutions will be sprayed through the nozzle, hence eliminating the need to open or disassemble the equipment.
E-Z FORCE weight adjustment system[12] This system allow operator to adjust the weight and eventually vary the eccentric weight to achieve optimization for different application.
Patented unique drive system[26] Promising more evenly distribution of materials on the screens with a result of increased durability of the equipment.

Besides, improvements on existing function of gyratory screener are often desirable. For instance, the equipment can be modified to handle larger volume of feed with higher accuracy in particle separation and availability in smaller footprint. Besides that, anti-blinding devices and ball trays have greatly reduces the blinding risk. Developments have made the screening process safer, cost effective, time saving and more convenient.[27][28]

## References

1. ^ Swain, R. K. (2011). Mechanical Operations. Tata McGraw-Hill Education. pp. 128–137.
2. ^ Turnquist, P. K. (1965). Size classifying of granular particles in a vibratory screening system (Ph.D.). State University of Oklahoma.
3. ^ Singh, R. (January–February 2004). "Vibratory separators still make the grade for screening dry bulk powders". Filtration & separation. 41 (1): 20–21.
4. ^ Patil, K. D. (2007). Mechanical Operations Fundamental Principles and Applications (2nd ed.). Nirali Prakashan.
5. ^ a b c Wolff, E. R. (September 1954). "Screening Principles and Applications". Industrial & Engineering Chemistry. 46 (9): 1778–1784. doi:10.1021/ie50537a024.
6. ^ a b c Sobolev, G. P. (June 1963). "Effect of the parameters of a rotary-vibration (gyratory) screen on its performance". Glass and Ceramics. 20 (6): 323–327. doi:10.1007/bf00673176.
7. ^ a b He, X.; Liu, C. (July 2009). "Dynamics and screening characteristics of a vibrating screen with variable elliptical trace". Mining Science and Technology. 19 (4): 508–513. doi:10.1016/s1674-5264(09)60095-8.
8. ^ a b "Applications". SWECO. Retrieved 15 October 2013.
9. ^ a b c d Ricklefs, R. D. (2013). "Selecting a Sifter for Scalping, Removing Fines, or Grading". Powder and Bulk Engineering. 1 (3): 14–16.
10. ^ a b Albright, L. F., ed. (2008). Albright's Chemical Engineering Handbook. CRC Press. pp. 1720–1726.
11. ^ a b Sambamurthy, K. (2007). Pharmaceutical Engineering. New Age International. pp. 361–364.
12. ^ a b c "VIBROSCREEN® OPTIONS/ENHANCEMENTS". Kason. Archived from the original on 29 October 2013. Retrieved 15 October 2013.
13. ^ a b c d "ROTEX® Introduces New Tool-Less Cam for the APEX™" (Press release). ROTEX. 2 August 2013. Archived from the original on 12 October 2013. Retrieved 15 October 2013.
14. ^ "Gyratory Screen". Electro Flux Equipments Pvt. Ltd. Retrieved 15 October 2013.
15. ^ a b Colman, K. G. (1985). Weiss N. L., ed. Mineral Processing Handbook. SME/AIME. pp. 3E 13–19.
16. ^ Poryadkova, Z. S.; Moroz I. I. (August 1970). "The nature of the breakage of plates in use". Glass and Ceramics. 27 (8): 475–478. doi:10.1007/bf00675574.
17. ^ Snell, L. D. (2008). Force and Moment Analysis of Stacked Counter Rotating Eccentric Mass Tree Shaker Energy-wheel System. ProQuest.
18. ^ Onesti, P. G. (1965). The mechanism of collection of particles on a cylindrical rotating screen (B.E.). The University of Sydney.
19. ^ Wlard, E. S. (1916). "The grading industires". Met. Chem. Eng. 14: 191.
20. ^ "Engineering Data Screen Openings for Basket Strainers" (PDF) (Press release). Islip Flow Controls. Retrieved 15 October 2013.
21. ^ a b "Andritz Feed & Biofuel Roto-Shaker™ Gyratory Screeners". Andritz Feed & Biofuel. Archived from the original on 3 May 2009. Retrieved 15 October 2013.
22. ^ Wills, B. A. (2006). Wills' Mineral Processing Technology (7th ed.). Butterworth-Heinemann. pp. 195–196.
23. ^ Richardson, J. F.; Harker, J. H.; Backhurst, J. R. (2002). Coulson and Richardson's Chemical Engineering Volume 2 (5th ed.). Butterworth-Heinemann. pp. 901–964.
24. ^ a b "Screening". GAH Global. Retrieved 15 October 2013.
25. ^ "Shaker Screen with Stainless Steel Tank". Midwestern Industries, Inc. Archived from the original on 23 February 2015. Retrieved 15 October 2013.
26. ^ "Gyratory Screeners". Farleygreene. Retrieved 15 October 2013.
27. ^ "Rotary Screener-Gyratory-Reciprocating Screener". Lemphy Equipment Co., Ltd. Retrieved 15 October 2013.
28. ^ "Gyratory". ALMO Process Technology. Retrieved 15 October 2013.


text:
# What are absolute values?

##### 1 Answer
Oct 25, 2014

You can think of the absolute value sign as a negative sign eraser. It does not do anything to a nonnegative number but turns a negative number into the positive number having the same magnitude; for example, $| 2 | = 2$ and $| - 3 | = 3$.

I hope that this was helpful.


text:
Often the most reported measures of usability is task success. Success rates can be converted into a sigma value by using the discrete-binary defect calculation:

Proportion Unsuccessful= Defects/Opportunities

Where opportunities are the total number of tasks and defects are the total number of unsuccessful tasks. This calculation provides a proportion that is equivalent to a success or failure rate. For example, if 143 total tasks were attempted and 123 were successful(20 defects 143 opportunities) the equation would look like:

20/143 = .1399
By expressing .1399 as a percentage you get an unsuccessful rate of 13.9%. By subtracting the unsuccessful proportion from 1 you’ll get approximately .86 or expressed in a percentage you could say that the task completion rate was 86%. Converting the proportion to a sigma value is a little more tricky.

Converting a Proportion to a Sigma Value

A sigma value is a description of how far a sample or point of data is away from its mean, expressed in standard deviations usually with the Greek letter σ or lower case s. A data point with a higher sigma value will have a higher standard deviation, meaning it is further away from the mean.

Standard deviations and sigma values assume your data is normally distributed from a continuous set of data. A normal curve has a mean of 0 and a standard deviation of 1. Success rates are Discrete-Binary (either a task was completed or it wasn’t) not continuous(like time, weight or temperature) which means that binary data is not normally distributed.

Have no fear, for every statistical problem a statistical solution usually exits. We can treat binary data like continuous data because of the Central Limit Theorem. It states that as a binary sample gets larger, its distribution approximates a continuous distribution.

Approximating the Proportion to a Normal Distribution

An easy way to exploit the Central Limit Theorem solution and obtain a sigma value from the task proportion is by using the Excel function NORMSINV. NORMSINV approximates the area under a normal bell-shaped curve. Take the unsuccessful proportion .1399–the proportion defective and insert it into the following equation in Excel.

=NORMSINV(1-.1399)

Note: My sigma values DO NOT contain a 1.5σ adjustment or “shift.”


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# Long-range order for critical Book-Ising and Book-percolation

Vincent Tassion
Cited by: 0|Views0

Abstract:

In this paper, we investigate the behaviour of statistical physics models on a book with pages that are isomorphic to half-planes. We show that even for models undergoing a continuous phase transition on \$\mathbb Z^2\$, the phase transition becomes discontinuous as soon as the number of pages is sufficiently large. In particular, we prov...More

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