text: Close. save hide report. Fonts.com Web Fonts is Now the Monotype Library Subscription. 4 days ago. æ¶ä»£çä»ç»ï¼ Every font in the Palatino family produced through 1962 â¦ Two italics: the foundry version is narrower than Linotypeâs and includes swash alternates, which in 1953 were extended to a full set of swash caps. Palatino is a classic serif typeface designed by German typeface designer Hermann Zapf. 03/11/2020; 2 minutes to read; B; P; W; M; In this article Overview. 31 comments. It was the kind of book you start and have to read through without putting it down. ÙÙÙØ±Ø¯ ÙÙØªØµÙÙÙ alfont_com_AlFont_com_PalatinoSansArabic-Bold.ttf ØªØÙÙÙ ÙØªÙØ²ÙÙ Palatino Sans Arabic Bold Century font family. ¶ The original Palatino for photocomposing machines (Diatype and Diatronic, produced by the Berthold AG) was widely used in Germany until the early nineties. 2 years ago. Adobe Fonts is the easiest way to bring great type into your workflow, wherever you are. The font family includes Cyrillic and Greek letters as well as accented Latin ones, titling fonts, and many other typographic niceties. palatino the natural history of a typeface Sep 16, 2020 Posted By Patricia Cornwell Publishing TEXT ID b42441be Online PDF Ebook Epub Library typeface by robert bringhurst hardcover barnes nobler hermann zapf was one of the great practitioners of the graphic arts and palatino is probably the most widely known This page was last edited on 8 November 2020, at 01:39. The five boxing wizards jump quickly. 2006-10-27 13:51 GearedBull 198×234× (81709 bytes) Specimen of the typeface Palatino. Adobe Fonts is the easiest way to bring great type into your workflow, wherever you are. See more ideas about Poster fonts, Type posters, Typography poster. . Oct 31, 2016 - Explore Drew Dietz's board "palatino", followed by 129 people on Pinterest. share. Palatino Linotype Font Family. Herman Zapf (The Wizard of Fonts) is probably the best known of modern type designers, and Palatino is certainly one of the best known fonts in the world. the fonts themselves: the color and feel of copy produced by different means are quite different. Palatino history-Out of many of the old style typefaces, Palatino has become one of the most common fonts used to this day, having five other versions of it within the palatino family. It is wider and heavier than Century Expanded, there is also less contrast between thick and thin strokes. Sep 25, 2020 palatino the natural history of a typeface Posted By Jeffrey ArcherLibrary TEXT ID 842925aa Online PDF Ebook Epub Library publication date 2016 title variation natural history of a typeface series book club of california publication number 235 note this book was designed and except for the letterpress pages set in type by robert Making the web more beautiful, fast, and open through great typography The minimal example \documentclass[12pt,oneside,english,titlepage, a4paper, fontsize=12pt]{scrbook} \usepackage{mathpazo} \usepackage{geometry} \geometry{letterpaper} â¦ Subscriptions are a big part of what we do here at Fonts.com, and weâve had a quite a bit of activity on the subscription front as of late. Text serif fonts. The subtle changes in each redesign of Palatino show the care and skill of the great master Zapf. I am using Palatino for the content and I have a problem displaying \chi in math mode. Brawny gods just flocked up to quiz. There are 2 types of font families which you can use â Specific Font-Family â This is a specific type of font like Arial, Verdana, Tahoma; Generic Font-Family â This is a All structured data from the file and property namespaces is available under the Creative Commons CC0 License; all unstructured text is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply. Palatino sans arabic mostra.png 1,042 × 1,034; 81 KB Palatino specimen sheet.jpg 1,123 × 816; 103 KB Palatino Swash Capitals Type Specimen (8033811913).jpg 1,123 × 816; 75 KB Windows 8: See the Windows 8 page. Designed to fulfill the need for a solid, legible face for printing schoolbooks. The ten most popular text serif fonts, as selected by Identifont users over the past seven days. Jul 28, 2015 - Explore gabrielle mccourt's board "Palatino" on Pinterest. ¸ë¬ë¤ ë³´ë©´ ì´ì ë ì§¬ë½ì´ ëì´ ì ì ê¸°ë³¸í°í¸ê° ë¬´ìì´ ìëì§ë ëª¨ë¥¼ ëê° ììµ.. Windows 7: See the Windows 7 page. The problem happens when I have a roman formatting in math, e.g., vector x, X, and Ï; the Ï turns to some other character.. A serif typeface with 6 styles, available from Adobe Fonts for sync and web use. Palatino nova was released in 2005, while the others were released in 2006. Product name Font version; Windows 10: See the Windows 10 page. u/kylpere. Adobe Fonts is the easiest way to bring great type into your workflow, wherever you are. palatino the natural history of a typeface Sep 01, 2020 Posted By Frank G. Slaughter Media TEXT ID 842925aa Online PDF Ebook Epub Library Palatino The Natural History Of A Typeface INTRODUCTION : #1 Palatino The Natural ~~ Last Version Palatino The Natural History Of A Typeface ~~ Uploaded By Frank G. Slaughter, palatino is more than a type nerds delight or an ode to a great â¦ We recently introduced some major enhancements to the Monotype Library Subscription, including web font support for all fonts in the . 1. Herman Zapf (The Wizard of Fonts) is probably the best known of modern type designers, and Palatino is certainly one of the best known fonts in the world. \documentclass[10pt,letterpaper]{memoir} \usepackage{amsmath} \usepackage{amssymb} \usepackage{mathpple} \usepackage{upgreek} \usepackage{mathpazo} \usepackage{lipsum} â¦ Posted by. {{message.actionLabel}} Browse Fonts 327. Released in 1950 and adapted for the Linotype machine. See more ideas about Hermann zapf, Typeface, Typography poster design. Posted by. The last book I read was a hardback novel about the story of an estranged Sri Lankan family with plenty of drama and emotional upheaval. Itâs proven methods have been expanded over the years, adapting to new technologies. Another member of the Century family which was based on Century Expanded. Designed by Hermann Zapf in 1948 and cut in metal by August Rosenberger at D. Stempel AG. At the end of â¦ And I would like a good font for the maths. ... had a strong urge to create a font, so 2 hours later this is what I came up with.. 385. Fonts classification - Bembo fontâ¦ Products that supply this font. This book is basically a history of the conception, design, creation, and production of Palatino faces and the many, many variations thereof. Fonts history. The picture was photographed from "Berthold Types" (1985), volume 2, page 1106. For over a century, Linotype has successfully produced, marketed, and licensed quality fonts. Looking for Earthquake Palatino fonts? Palatino Named after 16th century Italian master of calligraphy Giambattista Palatino, Palatino is based on the humanist fonts of the Italian Renaissance, which mirror the letters formed by a broadnib pen; this gives a calligraphic grace. . Palatino in use. palatino the natural history of a typeface Sep 27, 2020 Posted By Yasuo Uchida Media Publishing TEXT ID b42441be Online PDF Ebook Epub Library typeface 96 letterpressphotographic every font in the palatino family produced through 1962 was three dimensional intended for use in a letterpress every font in the family Files are available under licenses specified on their description page. CSS font-family defines the priority for the browser to choose the font from multiple fonts. Click to find the best 1 free fonts in the Earthquake Palatino style. ... FONTS HISTORY - The Sumer letter - Sabon Font - Typographical fonts. It was a high-quality font, identical in design with the original Palatino for hot-metal hand typesetting. Futura is a versatile, high-quality Font foundry in Germany with 5,925 fonts. u/WillAdams. You can change text for preview: Change text. Palatino: The Natural History of a Typeface by Robert Bringhurst now available in a non-limited edition. I would like to use Zapf's Palatino font for the text in my document, which also includes a lot of maths. This book is basically a history of the conception, design, creation, and production of Palatino faces and the many, many variations thereof. Windows 8.1: See the Windows 8.1 page. This is a snapshot of what I am doing. Adobe Fonts is the easiest way to bring great type into your workflow, wherever you are. Geometric san-serifs are in vogue this year in web and logo design, so it makes sense to talk about the one that started it all, Futura. Palatino Linotype Regular Font. A script typeface with 3 styles, available from Adobe Fonts for sync and web use. File history Click on a date/time to view the file as it appeared at that time. Every font is free to download! This is what I am doing new technologies for sync and web use skill of the great Zapf. In 1950 and adapted for the maths font family a strong urge create... 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# Outlier/anomaly detection on histograms

So, the idea is that I have many histograms, each one representing results for something. So, I have histogram_1 for object_1, histogram_2 for object_2,...,histogram_20 for object_20. How can throw out an outlier histogram? I mean, I am searching for an entirely outlier histogram here in comparison to the mainstream of the other histograms, not the outliers number inside a histogram.

• It strongly depends on how you define the outlier, or how it differs from the rest. Is it different because of the distribution, the value of its values or is it something different? Jul 22, 2022 at 14:45
• @just_learning I don't understand the question, why don't you just pool all the data together and use a box plot? Jul 22, 2022 at 15:02
• @deps_stats: I think OP wants to detect abnormal histograms among a sample of 20 different histograms, not abnormal data points. Jul 22, 2022 at 15:13
• got it! Two more questions, are all histograms defined over the same support? and, are the bin widths equal in all histograms? Those might be critical Jul 22, 2022 at 15:26
• What is the relevance of the histograms? You can compute distances between empirical distributions (in terms of the Komogorov-Smirnov statistics) without binning the distributions in a histogram. Many CV posts explain that binning a continuous distribution means loss of power. Jul 23, 2022 at 17:13

Outlier or anomaly detection methods always rely on some notion of distance between the "data points" to be subjected to the detection algorithm. So your first step needs to be to decide on a distance metric between your "data points" - which in your case are your histograms.

There are various ways of doing this. If your histograms all contain the same number of points, and all have the same breaks, you can simply take the average of the squared difference in bin counts. If the breaks are the same, but the counts differ, you can normalize first. Alternatively, you can use the Earth Mover's Distance, which is a general distance between distributions - you can estimate this even on the raw data, before binning into histograms.

Once you have a distance matrix between your histograms, one way forward would be to cluster your histograms, e.g., with a DBSCAN method, which explicitly allows for treating some data points (i.e., histograms) as "noise". You would need to fiddle around with the tuning parameters until you get results you are comfortable with. They will depend on the bumpiness and bin counts of your histograms.

As an example, here are 20 histograms, which one is the outlier?

Our approach correctly identifies the one at the bottom right as "noise", i.e., as an outlier.

R code:

library(dbscan)

set.seed(1)
n_obs <- 2e3
sims <- cbind(replicate(19,runif(n_obs)),rbeta(n_obs,2,2))
histograms <- matrix(NA,nrow=20,ncol=10)
opar <- par(mfrow=c(4,5),las=1,mai=c(.3,.3,0,0))
for ( ii in 1:20 ) {
histograms[ii,] <- hist(sims[,ii],xlab="",ylab="",
breaks=seq(0,1,by=0.1),main="")$counts } distances <- matrix(NA,20,20) for ( xx in 1:20 ) { for ( yy in 1:20 ) { distances[xx,yy] <- mean((histograms[xx,]-histograms[yy,])^2) } } clustering <- dbscan(distances,eps=10000,minPts=2) clustering$cluster

Alternatively, since you have no more than 20 histograms, you could use an "inter-ocular trauma test for significance". Something like that might be a good idea for calibrating the clustering-based approach above, in any case.

• You can use the test statistic of the two-sample K-S test as a distance measure between each pair of histograms, instead of the squared difference in bin counts I used above or the EMD. Then you can again use DBSCAN or similar. Jul 23, 2022 at 8:44
• You shouldn't use the Kolmogorov-Smirnov statistics as a distance measure between histograms as K-S makes sense for continuous distributions only. And histograms are definitely not continuous. Jul 23, 2022 at 17:50
• @dipetkov: good point. I should have written that you can use the K-S statistic for a measure of distance of the underlying data, before binning them into histograms, analogously to the EMD. Jul 23, 2022 at 17:51
• This is exactly what's bothering me with this question -- the OP hasn't said why the histograms in the first place. Is there a good reason to bin the samples and then compute distances? Maybe the OP will explain. Jul 23, 2022 at 17:54
• If you have access to the original data (before they were binned into histograms), you could calculate distances between those, as @dipetkov and I are suggesting. Taking means per image may also be possible. The distribution/histogram approach would detect a different kind of "outlier" (e.g., same mean as other images, but different distribution). What works best will depend on your context. Jul 24, 2022 at 11:57

format: plain


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# DHE key exchange. Is it really secure?

## Context

I recently came across a video on the net showing the process of key exchange in Diffie Hellmann.

Alice and Bob want to communicate. Eve is the (wo)man in the middle.

This video is the basis of my question http://youtu.be/YEBfamv-_do?t=6m22s

## Question

Alice sends the modulo and the prefix to bob. Eve gets that.

What prevents Eve to have a giant table of all the calculations possible given a prefix and a modulo and the public value and intersect thos possibilities with the public value from bob ?

Yes the table would be very big but you can then iterate over the intersecting possibilities and try the numbers on the message up to the point to get something right.

i may miss something, but it is not that hard, even with very big numbers.

So what did I miss ?

Thanks

• You're underestimating what "very big" means. No conventional computer will ever be able to count to $2^{256}$ and that doesn't even consider storing that much. – CodesInChaos Nov 19 '14 at 8:55
• the "giant table" would be as large or larger than the observable universe for the key sizes currently used (4096-bits), and take more energy to search than was released by the big bang – Richie Frame Nov 19 '14 at 9:09
• What "prefix" does Alice send to Bob? $\;$ – user991 Nov 19 '14 at 10:30

A lot of modern cryptography is based on some mathematical assumptions and aims to achieve what is called Computational Security. That means that the adversary (Eve) could get some information about the plaintext with a negligible probability and the adversary is modeled as someone with bounded computational power, storage and bounded time. So all the (encryption) schemes are dimensioned in sort to achieve such level of security: that means that all mathematical elements are chosen to be large enough to make the best known attack feasible in a reasonable amount of time.

Even if this is a weaker notion of security, this assumption on the adversary is easy to accept in practice.

Other approach exists: Information theoretic Security or Perfect Security

So coming back to your solution based on a big table, if the scheme is correctly dimensioned the table you are supposed to use to attack the scheme would be so large that it cannot be stored in a (very very) large disk or to be (too) long to be generated.

• I think it is time for me to write a paper. – Larry Nov 19 '14 at 8:33
• What do you mean? – ddddavidee Nov 19 '14 at 8:34
• @Larry I look forward to your paper where you ridicule the entire cryptographic community. Please jot me down as the idiot that ridiculed you before you proved that the crypto community is a bunch of asses :) – Maarten Bodewes Nov 19 '14 at 10:10
• @Larry if you are somehow able to find a solution to storing $4.4*10^{598}$ times the total amount of storage in the world then I believe you will become far too rich to worry about breaking Diffie-Hellman. – flashbang Nov 19 '14 at 21:32
• Humans are so bad at understanding exponential growth... Best example is the question "How often can you fold a paper, and guess how thick it will be" -> typical are 6 or 7, record is at 12 atm – tylo Nov 24 '14 at 14:54

Guess the catch in the video is in how the participants exchange details 'publicly'.

If the Man-In-The-Middle can intercept and manipulate what is being 'publicly' shared, then the attempt to eavesdrop would still be successful.


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What theories solve the problem of weight dispersion to minimize damage?

I assume there's some simple link to give me to answer this, but I don't know the terminology for it and so haven't been able to solve this.

The use case it to avoid having my TV put divots into my IKEA KALLAX.

Here's what my intuition tells me:

1. The best shape to distribute the weight is a conical frustum, with the larger circle on the KALLAX, and the TV legs sitting on the smaller one.
2. The frustum should have increasing density as its diameter increases. (The density at the bottom should be high to "avoid bending", and low at the top to minimize added weight.)

Are the intuitions correct? I have the sense that a dense substance would distribute weight much better than a less dense one, but why?

My current solution is to use dense foam, which is an incredibly crude approximation of the frustum I'm talking about. How can I tell if it's useful?

Again, my intuition is telling me that there's a falloff curve such that, while nowhere along the foam, is the dispersed weight going to be zero, there's going to be a threshold such that there's no point in having the foam be as wide as it is without trimming.

• Since you already put an aesthetically rather unpleasing (in my opinion :P) foam on top of the shelf, would you be willing to rest the TV legs on some large steel washers with soft padding underneath? This is all about distributing the force of the TV legs on a larger area of the Kallax top board, look up surface pressure to read a bit about it. – OpticalResonator Aug 25 '20 at 7:52
• It looks fine at normal light levels, but sure, I'd do what you're asking if someone could validate why it's a good solution. "Surface pressure" didn't turn up anything good for me. – Jessy Aug 25 '20 at 8:15
• I think it's a clumsy translation on my end, contact mechanics maybe comes closer. – OpticalResonator Aug 25 '20 at 8:35
• @Jessy, a better way is use Self Stick Felt Furniture Pads – Mahendra Gunawardena Aug 25 '20 at 12:00
• I think as long as the spacer is strong enough to withstand the TV's weight, it doesn't really matter what shape it is. – user253751 Aug 25 '20 at 16:56

Assume that your TV weighs $$100 kg$$, then each leg exerts a force of $$0.25 \cdot 100 kg \cdot 9.81 m/s^2 = 245.25 N$$ on the top board of the shelf. The contact area of the foot I can only guess, but let's be pessimistic and approximate it as a patch of $$5 mm \times 5mm = 25 mm^2$$. Pressure is calculated as $$\text{Pressure} = \frac{\text{Force}}{\text{Area}},$$ so every leg exerts a surface pressure of $$245.25N/25mm^2 = 9.81 N/mm^2$$ on the shelf top.
It's a bit difficult to find how much stress wood, especially the thin top layer of IKEA furniture, can handle. For pine wood we can find values of approx. $$2 N/mm^2$$, so let's take this as conservative goal. Then the TV legs on their own would indeed damage the top of the shelf.
If you now take a large M20 washer, with an inner diameter of $$22 mm$$ and an outer diameter of $$60 mm$$, you get a significantly larger area, at about $$2450 mm^2$$. The surface pressure is now at about $$0,1 N/mm^2$$, something the top of the shelf should be able to handle.


text:
OpenStudy (gamer56):

Select the mathematical sentence that is true. A. 2 • (8 – 5) + 3 < 7 + 3 • 2 B. 2 • (8 – 5) + 3 > 7 + 3 • 2 C. 2 • (8 – 5) + 3 = 7 + 3 • 2

3 years ago
OpenStudy (gamer56):

@shamil98 @shamil98 @kitteh♥ @ztp5031

3 years ago
OpenStudy (anonymous):

C'mon mate, this is basic math.

3 years ago
OpenStudy (anonymous):

PEMDAS

3 years ago
OpenStudy (gamer56):

Im redoing some stuff can u tell me how or guide me in some way???

3 years ago
OpenStudy (anonymous):

8-5+3?

3 years ago
OpenStudy (gamer56):

8-5=3 3+3=6

3 years ago
OpenStudy (anonymous):

7+3×2?

3 years ago
OpenStudy (gamer56):

7+3=10 10 x 2=20

3 years ago
OpenStudy (anonymous):

is 6<20?

3 years ago
OpenStudy (gamer56):

3 years ago
OpenStudy (anonymous):

ok so statement 1 is true

3 years ago
OpenStudy (gamer56):

ITS A !!!

3 years ago
OpenStudy (gamer56):

o

3 years ago
OpenStudy (gamer56):

can you help with 3 more?

3 years ago
OpenStudy (anonymous):

statement 2 is false because it's saying 6 is greater than 20 and statement 3 is false because it's saying 6 is equal to 20

3 years ago
OpenStudy (anonymous):

i can try

3 years ago
OpenStudy (gamer56):

Find the solution set for the given inequality if the replacement set for x is the set {1, 2, 3, 4}. 4x – 3 ≥ 5

3 years ago
OpenStudy (anonymous):

4(1)-3?

3 years ago
OpenStudy (gamer56):

Assuming that i do 4 x 1 = 4 - 3=1 Correct??

3 years ago
OpenStudy (anonymous):

yes always follow this rule Pemdas Parenthesis Exponents Multiply Divide Add Subtract

3 years ago
OpenStudy (gamer56):

:D

3 years ago
OpenStudy (anonymous):

is 1 greater than or equal to 5?

3 years ago
OpenStudy (gamer56):

neither 5 is greater then 1 correct?

3 years ago
OpenStudy (anonymous):

yes correct so 1 wont be in the function I'm in and out because i have to catch up in my math class can you do these by yourself and tell me what you get 4(2)-3= 4(3)-3= 4(4)-3=

3 years ago
OpenStudy (gamer56):

4 x 2= 8-3= 3 4 x 3=12-3= 9 4 x 4=16-3=13

3 years ago
OpenStudy (anonymous):

8-3=3?

3 years ago
OpenStudy (anonymous):

8-3=5 $\ge$ this means a number is greater than or equal to another number 5,9,12 are greater than or equal to 5 so your solution set will be 2,3,4

3 years ago
OpenStudy (gamer56):

Determine which value is a solution to the open sentence. 2 + 4x = 18

3 years ago
OpenStudy (gamer56):

i think i got this

3 years ago
OpenStudy (anonymous):

subtract 2 by 18 then divide that solution by 4

3 years ago
OpenStudy (gamer56):

16 correct?

3 years ago
OpenStudy (gamer56):

wait i didnt divide

3 years ago
OpenStudy (anonymous):

18-2 is 16 now divide by 4

3 years ago
OpenStudy (gamer56):

4?

3 years ago
OpenStudy (anonymous):

yes

3 years ago
OpenStudy (gamer56):

Which is a word phrase for the variable expression? 3(n – 1)

3 years ago
OpenStudy (anonymous):

whats another name for multiplying

3 years ago
OpenStudy (gamer56):

dividing??

3 years ago
OpenStudy (anonymous):

no mathmatical terms for multiplying dividing would be the inverse

3 years ago
OpenStudy (gamer56):

oh

3 years ago
OpenStudy (gamer56):

so what do i do?

3 years ago
OpenStudy (anonymous):

well what would be a mathmatical term for multiplication?

3 years ago
OpenStudy (gamer56):

inverse

3 years ago
OpenStudy (anonymous):

no thats the opposite of a sign

3 years ago
OpenStudy (anonymous):

product is one mathmatical term

3 years ago
OpenStudy (anonymous):

and what is n and 1 doing

3 years ago
OpenStudy (gamer56):

subtraction

3 years ago
OpenStudy (gamer56):

@texaschic101 help someone

3 years ago
OpenStudy (anonymous):

now a mathmatical term for subtraction?

3 years ago
OpenStudy (gamer56):

3 years ago
OpenStudy (anonymous):

no thats the inverse

3 years ago
OpenStudy (anonymous):

mathmatical term would be another way of saying subtraction more advance

3 years ago
OpenStudy (gamer56):

can you tell me?

3 years ago
OpenStudy (anonymous):

when you're subtracting you're finding the difference of two numbers

3 years ago
OpenStudy (anonymous):

soo in that equation you're finding the product of the difference of n and 1

3 years ago
OpenStudy (gamer56):

yes

3 years ago
OpenStudy (anonymous):

yes that is what the word phrase would be or similar

3 years ago
OpenStudy (gamer56):

So my answer would be?? @KlOwNlOvE

3 years ago
OpenStudy (anonymous):

well are you given options

3 years ago
OpenStudy (gamer56):

whoops i thought i posted them

3 years ago
OpenStudy (gamer56):

A. three times the quotient of a number n and 1 B. three more than the difference of a number n and 1 C. three times the difference of a number n and 1 D. one less than the product 3 and a number n

3 years ago
OpenStudy (anonymous):

quotient means divide and theres no dividing so A is no more means adding theres no adding so B is no C you are multiplying 3 by the difference of n and 1 D would be 3n-1

3 years ago
OpenStudy (anonymous):

so C is right

3 years ago


text:
# GATE1991-5-c

9.4k views

Find the maximum clock frequency at which the counter in the figure below can be operated. Assume that the propagation delay through each flip flop and each AND gate is $10$ $\text{ns}$. Also assume that the setup time for the $JK$ inputs of the flip flops is negligible.

edited
1

What I know is Total time (Propagation) = T(flip flop) + T(combinational circuit)

So from the above Total time= 10(Flip Flop) + 30(combinational AND gate);

f=1/T and we can solve easily . But here why only one AND gate delay is considered ?

Expalin plz.

1
@bikram sir

Answer not should be 30(3 And Gate ) + 10 ( Synch Circuit) = 40 ns

1/40 ns= 25 Mhz ?

Why here they have taken it only for 1 And and 1 FF ?
0
This might be silly but I can't see any combinational circuit such that output of flip flops is again given to input of flip flops. plz help.
0
Can You check K1, K2?
3

[KINDLY IGNORE THE YELLOW PATH WITHIN FF2]

I understood it in this way . I’m adding my answer just in case someone finds it easier to assimilate.

we determine the current state of each FF based on the previous state of FF.

In the diagram , lets say the current state values are at Aout , Bout and Cout . It will take effect once Ain , Bin and Cin is available from the previous state.

For FF1 to change state (Follow the green path):

t1 = Time for Ain to generate + time for FF1 to propagate = 0 (it’s always high)+ 10 = 10ns

For FF2 to change state (follow the blue path):

t2 = Time to generate Bin + time for FF2 to propagate

= Time to generate Aout + Time for logic 1 + time for AND gate + time for FF2 to propagate

= 0(Aout is already there from prev state) + 0(always high) + 10 + 10

= 20 ns

For FF3 to change state(follow the yellow path):

t3 = Time to generate Cin + time for FF3 to propagate

= time for Bout + time for Bin + time for AND gate + time for FF3 to propagate

= 0 + time for Bin + time for AND gate + time for FF3 to propagate (Bout is already there from prev state , hence time is 0)

=  Time to generate Aout + Time for logic 1 + time for AND gate + time for AND gate + time for FF3 to propagate

= 0 + 0 + 10 + 10 + 10 (like Bout , Aout is already there from previous state)

= 30ns

So time taken for FF1 , FF2 and FF3 to change their respective states are 10 , 20 and 30 ns . We take the maximum time among it i.e 30 ns as it will ensure that no FF is left out.

In a JK flip flop the output toggles when both J and K inputs are $1.$ So, we must ensure that with each clock the output from the previous stage reaches the current stage.

Setup time is defined as the minimum amount of time before the clock's active edge that the data must be stable for it to be latched correctly

It is given that set up time is negligible - means as soon as data is stable, next clock can be given.

Time to get output from $FF$ once input (and clock) is given  = $10ns.$ (Propagation Delay)

Time for inputs to reach $FF_1 = 0.$ (Zero AND gate)

Time for inputs to reach $FF_2 = 10.$ (One AND gate)

Time for inputs to reach $FF_3 = 20.$ (Two AND gates)

So, minimum time period needed for clock is $10 + \max(0,10,20) = 10 + 20 = 30ns$ which would mean a maximum clock frequency of $1/30 GHz = 33.33 MHz$

selected by
0
Is the clock frequency only dependent on no. of AND gates and not on no. of flip flops? Why can't we take no. of flip flops as the the multiple for propagation delay?
6
We just have to ensure the output from a previous flipflop reaches the next flipflop (A flipflop can store a bit) within the time period of the operating clock. So, we just need to consider the stages between these. In the question there is only one AND gate in this stage. If there were 2, we should have multiplied the propagation delay by 2.
0
sir , if just ensure that o/p of previous one reaches to next flip flop then ... for last flip flop , we have to consider 2nd flip flop delay also ...??
0
yes, for FF 3, we consider delay of FF 2 and the AND gate delay.
2
then why only 20 ns .. is it 30ns .. 2 and gate for ff2 and ff3 , and i ff2 before ff3 ??
0
we are only considering the time from input of FF 2 to input to FF 3.
0
sorry sir , i am not getting... :(
4
When a clock comes it activates all the FFs simultaneously. So, we want to ensure the output of one stage reaches the input of next before the next clock. So, between two clocks, input must produce output and this output should reach next FF.
0
yes , but why only we are considering FF 2 TO FF3, not FF1 to FF2 ... Here i am confuse  ..
0
thats the same delay rt?
1
suppose the circuit was not the same, we should have considered all these stages, and taken the maximum time period (min. frequency).
1
okkk.... they have same delay ... and as clock given at same time so ..that delay consider only once ....

sorry .. i am considering for each ...thank you :)
0
0
it should be 1/30 as...after 20 3rd jk has it's inputs & it takes another 10 to produce the output....so 30
0

In a synchronous counter, at the end of a clock cycle the output should be present at each point that will further be used as a input to flip flops. Am i right ?

And if input for flip flops is present before giving clock. Then definitely $Tdelay = 20ns$

Otherwise if input of AND gates depends on other AND gate, then $T = 10 +(10+10+10) = 40ns$

0
what would have happened in asynchronous counter??
0

@Arjun Sir,

Is this Right ? https://gateoverflow.in/86195/me-test

This is the same related question.

1
@Arjun Sir

Thanks !!
0
@sushmita in asynchronous case there is no clock rt?
0
arjun sir clock is there in asynchronous according to me.
0
yes, it is there to the first FF. The maximum clock frquency won't be more than 50 MHz in that case also, but might be even lower. Have to see Morris Mano to tell more..
0
sir what i know is that the clock time is equal to the delay of all flipflops and combinational circuit in aynchronous counter. Its not right??
3

In the series carry scheme, the time to propagate the change in Q0 must take into account the propagation delays of the 3 AND gates (A, B, C)

so it is T+A+B+C here rt??

1
Arjun sir, in asynchronous counter I think we should consider the delay of all the flip-flops because the output of one flip-flop serves as the input to the next one. So I think delay should be 30ns for 3 ff and 20 n's for intermediate and gates. Plz plz check.
0
how can you say that it is asynchronous clock ? same clock is attached to

all flip flops here , so it should be synchronous.plz correct me.
0

But in question https://gateoverflow.in/86195/me-test , the delay for each and gate is considered.Whats the difference?

The input to 3rd flip flop is depending is two and gate delays as per this link

0
what if JK was not on high and given different inputs like 01 or 10 then ??
1
thank u arjun sir.....
0
0

ignoring the carry output of the circuit (the third AND gate). The question is only about the maximum clock frequency of the counter itself. If that output needs to be valid, then the clock period will have to be even longer.

0
@Arjun Sir , how the time for input to reach the second FF is 10ns, it should be 20ns, i.e propragation time of first FF and the AND gate , please explain
0
What if propagation delay for each flip flop was different? Let's say 1ns, 2ns, 3ns.
1

Will not the final output from ff3 have 1 AND  gate more delay before we use it finally. If yes then

minimum time period needed for the clock is

10+max(0,10,20,30) =>  10+30= 40 ns which would mean a maximum clock frequency of 1/40ns

= 25 MHz

Please correct if thinking wrong. Thank you.

0

My doubts,

Pd of FFs are 10ns. clock is simultaneously given in each FF.

J0K0 inputs are given,so when cock is active we can get this immediately,but to get the inputs of J1K1 we've to spend 20ns( 10ns for FF & 10ns for OR gate).

likewise to get the inputs at J2K2 we've to spend more 20ns(10ns for FF & 10ns for OR)

so we'll get o/p at Q2 at the time of 20 + 20 = 40ns.

again if I consider the last OR gate, then finally the total time is 50ns.

Don't know where I'm lagging.

see I'm taking this in the ay like - for J1K1 & J2K2 FF can't operate untill it gets the output from it's previous AND gate which is total 20ns.

1
0
Here I think we can approach in the following way:-

10ns (at 10ns all the FFs will give their respective o/ps) + 10ns (after 10ns o/p of FF1 will be available for FF2 (after AND op) and o/p of FF2 will be available for i/p at FF3) + 10ns(FF1 o/p will be available at FF3 so that the AND operation can be done) = 30ns

Is the above understanding correct?

Can anyone kindly confirm ?

Thank you
0

Let’s say, we are starting the FFs now. They will be in some initial state, eg Q0Q1Q2=000. Now these are outputs from those different FFs which are available and can be used wherever they are needed as inputs. Similarly, at each clock pulse, we will have the previous state value, which we will use to generate the next state (which is basically how FFs work)

We start with the first clock. Let us first calculate the input delays for each FF and nothing else.

The delay to get inputs available at:-

• J0K0 is 0, which is quite clear.
• J1K1 is 10 ns. Why? Because J1 = 1 AND Q0 (On a different note, you might think this is not a very optimal way of implementation, I agree with that, but here 1 is acting like a switch and we can make 1 as 0 and control the inputs at J1K1, just extra info). So a delay of ! AND is involved
• J2K2 is 20 ns. Why? Because J2 =  (1 AND Q0) AND Q1. So here a delay of 2 AND gates are involved. Note that the inputs to 1st AND gate are 1, Q0 which are already available (Q0 is the output from previous state and not current state so we don’t have to consider the FF delay for getting Q0), and for 2nd AND gate Q1 is readily available but it is also using the output of 1st AND gate, and hence the total delay becomes 20ns for J2K2

So now we have the input delays for each FF as 0ns, 10ns, 20ns. All we have to do is add FF delay, so the total delays would become 10ns, 20ns, 30ns. Maximum is 30 ns and so the minimum time period is 30 ns.

Giving us a maximum frequency of  = 1/30 GHz

= 33.333333333333333333333333333333333333333333333333333333333 MHz     (:P)

0

Your approach is partially correct. here’s why

I think there are two possible approaches:-

1. At the start of clock pulse, only outputs from previous states are available. They will have to go through the combinational circuits, and then the results can be used as inputs for the FFs to calculate the output for current state (which is what i have explained in the above comment)
2. At the start of clock pulse, the outputs from previous states, as well as the required inputs for FFs for current state are also available. So we can immediately calculate the output for current state using those inputs. After that, we will have to use this current state output, feed them to the combinational circuits, and make sure that the inputs are available for the respective FFs before clock ends. These inputs will be used by the FFs in the next clock pulse

From your comment, it seemed that you were trying to use the current state o/p of one FF into another FF, which is not how it works. We always use the previous state o/p to produce the next state.

A clock cycle time should be large enough that all flip-flops can come in to a stable state before the next clock cycle starts
(or) from the time we provide the input and till the moment output is available will be a cycle.
Suppose cycle 1 is just completed and cycle 2 is just about to start:
1. to reach the result of q0Λ1 to J1K1 it needs $10 ns$
2. to reach the result of (q0Λ1)ΛQ1 to J2K2, other 10 nanoseconds are needed.
3. Now, we need other 10  nanoseconds to operate flipflops as the propagation delay of each flip flop is 10ns.
4. Finally, we need 10 more nanoseconds to get the output from the last AND gate.

For better understanding we can enclose this circuit in a box with one input-pin and one output-pin out, when we provide input to the box, we need at least 40ns to get the output.

Answer should be (1/(2*10ns + 10ns + 10ns) = $\frac{1}{40ns} = 25MHZ$.

edited
0

I took think it should be 40ns.

By the time of next clock pulse. All current changes should be settled.

https://electronics.stackexchange.com/questions/328294/maximum-frequency-of-the-synchronous-counter

0
@rahul that's what i said, right?
0
Yes.I was trying to say that this answer seems to be correct than the selected one
1
Each FF along with the AND gate is an independent module since the clock is given synchronously to each FF.  Each module takes 20ns to produce next output. So the output is available after every 20ns. So cycle time is 20ns and thus freq. is 50MHz
0

@Manu Thakur Why clock is required for last Nand Gate, it can work even without clock pulse being active

0
Since, it's a synchronous counter, all the flip flops will get clock at the same time. Also, All AND gates will get the input simultaneously. I think you shouldn't consider it in serial manner.

$T_{CLK}>= T_{FF}+T_{AND}$

$T_{CLK}>=10+10$

$T_{CLK}>=20\ ns$

$f=\frac{1}{T_{CLK}}=\frac{1}{20\ ns}=50\ Mhz$
So at all I have concluded , it had asked maximum clock frequency .

And as freq=1/Time period

So we will take best case .

As it is synchronous circuit than all flip flop activate simultaneously and here key is

"Maximum clock frequency at which the counter in the figure below can be operated."

So 1FF+1AND GATE =20 NS

1/20=50 MHZ.
0
I also thought the same way, please confirm if it is right answer.
As all flip flop are connected to common clock so it is Synchronous Counter
all flip flop changes simultaneously
and there are 3 AND gates
For last FF input will be avail after 2(AND gate)*10

So Frequency = $\frac{1}{t_{ff} + 2t_{and}}$
= $\frac{1}{10 + 20}$
= 0.033 = 33.33MHz

edited
0
yes . this is according to the formula. right . one confusion. i think 20 second time period will be enough. if all will get the time of 20 why the output will not be generated. ??
2
1
The answer would be 1/20 because 2nd AND GATE do not have to wait for getting the output of first AND GATTE.
–3
why ? it will depend .
0

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Introducing Textbook Solutions. benzoic acid dissociates in water as shown in the equation. $$\ce{RNH_2}$$), and neutral components to be purified through a series of extractions, as summarized in Figure 4.59 (which uses an organic solvent less dense than water). 4-amoniumbenzoate + NaOH---->Ethyl 4-aminobenzoate + H2O.

Für nähere Informationen zur Nutzung Ihrer Daten lesen Sie bitte unsere Datenschutzerklärung und Cookie-Richtlinie. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. The. This reaction is a exothermic reaction.

Benzoic Acid + Water = Benzoate + Hydronium Cr + O2 = CrO3 MnO4{-} + SO3{2-} + H{+} = Mn{2+} + SO4{2-} + H2O MgO + HNO3 = Mg(NO3)2 + H2O Mg + AgNO3 = Mg(NO3)2 + Ag Br2 + OH{-} = Br{-} + … $$\ce{RCO_2H}$$), basic (e.g. benzoic acid + NaOH ---> water + sodium benzoate HC7H5O2(aq) + NaOH(aq) ---> H2O(l) + NaC7H5O2 You can use parenthesis () or brackets []. … Sodium salicylate is roughly 350 times more soluble in water than salicylic acid due to its ionic character (Figure 4.55), and it is rather insoluble in organic solvents such as diethyl ether. Due to its acidic nature, benzoic acid can undergo a reaction with $$\ce{NaOH}$$ as follows, resulting in the carboxylate salt sodium benzoate.

For a limited time, find answers and explanations to over 1.2 million textbook exercises for FREE! Separation of a mixture of benzoic acid and cyclohexane is however possible using a wash with a base such as $$\ce{NaOH}$$.

benzoic acid reacts with NaOH to make sodium benzoate, (sodium salts are very soluble: C6H5CO2H & NaOH(aq) --> Na C6H5CO2(aq) & H2O(l) Na C6H5CO2(aq) & HCl(aq) --> C6H5CO2H & NaCl(aq) 0 0. To isolate, wash with brine ($$\times 1$$) if using diethyl ether or ethyl acetate, dry with a drying agent, and remove the solvent via rotary evaporator to leave the pure neutral component. To enable Verizon Media and our partners to process your personal data select 'I agree', or select 'Manage settings' for more information and to manage your choices.

As was discussed in the previous section, $$\ce{NaOH}$$ can be used to convert a carboxylic acid into its more water-soluble ionic carboxylate form. If no solid forms upon acidification (or if fine crystals or low quantity of solid forms), extract the acidic component back into an organic solvent ($$\times 3$$). An acid-base extraction can be used to extract carboxylic acids from the organic layer into the aqueous layer. Wir und unsere Partner nutzen Cookies und ähnliche Technik, um Daten auf Ihrem Gerät zu speichern und/oder darauf zuzugreifen, für folgende Zwecke: um personalisierte Werbung und Inhalte zu zeigen, zur Messung von Anzeigen und Inhalten, um mehr über die Zielgruppe zu erfahren sowie für die Entwicklung von Produkten. Information about your device and internet connection, including your IP address, Browsing and search activity while using Verizon Media websites and apps. Use uppercase for the first character in the element and lowercase for the second character. Therefore, a solution of bicarbonate can be used to separate mixtures of phenols and carboxylic acids (Figure 4.58b).
Damit Verizon Media und unsere Partner Ihre personenbezogenen Daten verarbeiten können, wählen Sie bitte 'Ich stimme zu.' Benzoic acid has a molecular formula of C 7 H 6 O 2 , a molecular weight of 122g/mol and a melting point in the range of 121-123 ° C. Acetanilide has a molecular formula of C8H9NO, a molecular weight of 135g/mol and a melting point in the range of 111-115 °C. Acidity of carboxylic acid . Benzoic acid is a white precipitate. What do you think of the answers? The product of the reaction is Ethyl 4-aminobenzoate. reach acid base equilibrium. Benzoic Acid - C 6 H 5 COOH Retardex E210 Phenylcarboxylic Acid Benzenemethanoic Acid Benzeneformic Acid Benzenemethonic Acid Diacylic Acid Carboxybenzene Molar Mass of C7H6O2 Oxidation State of C7H6O2 $\begin{array}{ccccccccc} \ce{PhCO_2H} \left( aq \right) & + & \ce{NaOH} \left( aq \right) & \rightarrow & \ce{H_2O} \left( l \right) & + & \ce{PhCO_2Na} \left( aq \right) & & \left( \text{or } \ce{PhCO_2^-} \ce{Na^+} \right) \\ \text{Benzoic acid} & & & & & & \text{Sodium benzoate} & & \end{array}$.

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In this manner, a mixture of benzoic acid and cyclohexane can be separated (Figure 4.54b). The acidic portion of benzoic.

The equation for the reaction is shown below. Yahoo is part of Verizon Media. For example, it reacts with sodium hydroxide (NaOH) to produce sodium benzoate C6H5COOH + …

The product of the reaction is Ethyl 4-amoniumbenzoate.

Source(s): benzoic acid naoh equation: https://shortly.im/zw5OM. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739.
As has been discussed previously, the acid-base properties of compounds can be utilized to selectively extract certain compounds from mixtures. If only a small amount of solid is seen compared to the theoretical quantity, it is likely the compound is quite water-soluble, and filtration would lead to low recovery. A modification of the extractions previously discussed in this chapter is to perform a chemical reaction in the separatory funnel in order to change the polarity and therefore partitioning of a compound in the aqueous and organic layers. Read our article on how to balance chemical equations or ask for help in our chat. When shaking an acidic solution with sodium bicarbonate in a separatory funnel, care should be taken to swirl gently and vent more frequently to release pressure from the gas.

Sodium bicarbonate is preferable to $$\ce{NaOH}$$ in this process, as it is a much weaker base; washing with $$\ce{NaOH}$$ could cause hydrolysis of the ester product. In following examples, we discuss some reactions of carboxylic acids and aqueous NaOH. Part D the solubility of benzoic acid organic acid and ethyl p aminobenzoiate, 13 out of 13 people found this document helpful. Benzoic acid reacts with aqueous NaOH to sodium benzoate salt. When 6.0 M, NaOH is added, lots of OH forms and makes the solution basic and the Ethyl 4-. amoniumbenzoate is deprotonated. Cyclohexane would remain in the organic layer as it has no affinity for the aqueous phase, nor can react with $$\ce{NaOH}$$ in any way. Use a similar process as the isolation of the acidic component, except basify the solution using $$2 \: \text{M} \: \ce{NaOH} \left( aq \right)$$ until it gives a pH of 9-10 as determined by pH paper. The result of their, When the benzoic acid was added to 1.M NaOH a soluble solution was observed. A common method is to perform an acid-base reaction, which can convert some compounds from neutral to ionic forms (or vice versa). In such a graph the slope would be 2 Kd^2 Km and the intercept Kd. If large quantities of acid are present such that acidification would require too great a volume of $$2 \: \text{M} \: \ce{HCl} \left( aq \right)$$, concentrated $$\ce{HCl} \left( aq \right)$$ may be instead added dropwise. Examples: Fe, Au, Co, Br, C, O, N, F. Ionic charges are not yet supported and will be ignored. Sodium carboxylate ion is given as the salt.

equation for the reaction is shown below. As a general rule of thumb, use one-third as much solvent for the extractions as the original layer (e.g. Course Hero is not sponsored or endorsed by any college or university. Table: 3 Compounds Water 1.0 M NaOH 1.0 M HCl Benzoic acid Soluble Soluble Insoluble Add 6.0M HCl insoluble Ethyl 4 aminobenzoate Insoluble Soluble Soluble Add 6.0 M NaOH When the benzoic acid was added to 1.M NaOH a soluble solution was observed.

Be sure to first cool the aqueous solution in an ice bath before extraction if the acidification created noticeable heat. But during the reaction, that precipitate dissolve in aqueous phase. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. chemistry123 Sat, 09/14/2013 - 16:12. The acidic portion of benzoic acid is the carboxyl group, and it reacts with a base to form a salt. You can change your choices at any time by visiting Your Privacy Controls. When the acidic component is in the aqueous layer in an Erlenmeyer flask, it can be converted back to the neutral component through addition of $$2 \: \text{M} \: \ce{HCl} \left( aq \right)$$ until the solution gives a pH of 3-4 (as determined by pH paper). Sarah. Lower concentrations of $$\ce{HCl} \left( aq \right)$$ are less hazardous, but increasing the volume of the aqueous layer by a large amount would affect the efficiency of subsequent extractions and filtering steps. And the concentration of benzoic acid in benzene as acquired by base titration is: Cb = [HBz]B + 2[(HBz)]B 5 Then one can use equations 1 and 3 to rearrange equation 5 into: Cb / Cw (1-alpha) = Kd + 2 Kd^2 Km Cw(1-alpha) 6 which shows linear aspects if one were to plot Cb/Cw (1-alpha) as a function of Cw(1-alpha). For example, imagine that a mixture of benzoic acid and cyclohexane is dissolved in an organic solvent like ethyl acetate in a separatory funnel. The residual carboxylic acid can be removed from the desired ester product using an acid-base extraction in a separatory funnel.

$\begin{array}{ccccccc} \ce{PhNH_2} \left( aq \right) & + & \ce{HCl} \left( aq \right) & \rightarrow & \ce{PhNH_3Cl} \left( aq \right) & & \left( \text{or } \ce{PhNH_3^+} \ce{Cl^-} \right) \\ \text{Basic amine} & & & & \text{Ammonium salt} & & \end{array}$. Part D, the solubility of benzoic acid (organic acid) and ethyl p-aminobenzoiate (organic, base) were determined in water, water/NaOH and water/HCl.


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# Spinning ball dropped on the ground

1. Feb 12, 2009

1. The problem statement, all variables and given/known data
A solid ball of mass M and radius R is spinning with angular velocity $\omega_0$ about a horizontal axis. It drops vertically onto a surface where the coefficient of kinetic friction with the ball is $\mu_k$. Find an expression for the final angular velocity once it has achieved pure rolling motion. The other part of the question asks for the time taken until its in pure rolling motion. I could solve that part easily enough once I knew the answer to the first part.
2. Relevant equations
$$\tau=I\alpha$$
$$L=I\omega$$
$$RF=i\alpha$$
$$L=RMv$$
3. The attempt at a solution
$$\omega=\frac{2}{7}\omega_0$$
$$I\omega_0=I\omega+RMv=I\omega+MR^2\omega$$


text:
## error LNK2001: unresolved external symbol ___glewBindBuffer

I use Windows 7 and 8 and Visual Studio 2010 and 2012. In all combinations I get this error message when trying to build some examples:

Code :
```1>------ Rebuild All started: Project: 6E test, Configuration: Release Win32 ------
1>  example1.cpp
1>c:\users\student\downloads\6e_example1_vc10\6e test\code\Angel.h(65): warning C4305: 'initializing' : truncation from 'double' to 'const GLfloat'
1>c:\users\student\downloads\6e_example1_vc10\6e test\code\mat.h(698): warning C4244: '=' : conversion from 'double' to 'GLfloat', possible loss of data
1>c:\users\student\downloads\6e_example1_vc10\6e test\code\mat.h(699): warning C4244: '=' : conversion from 'double' to 'GLfloat', possible loss of data
1>c:\users\student\downloads\6e_example1_vc10\6e test\code\mat.h(700): warning C4244: '=' : conversion from 'double' to 'GLfloat', possible loss of data
1>c:\users\student\downloads\6e_example1_vc10\6e test\code\mat.h(721): warning C4244: '=' : conversion from 'double' to 'GLfloat', possible loss of data
1>c:\users\student\downloads\6e_example1_vc10\6e test\code\mat.h(723): warning C4244: '=' : conversion from 'double' to 'GLfloat', possible loss of data
1>c:\users\student\downloads\6e_example1_vc10\6e test\code\mat.h(726): warning C4244: '=' : conversion from 'double' to 'GLfloat', possible loss of data
1>c:\users\student\downloads\6e_example1_vc10\6e test\code\mat.h(742): warning C4244: '=' : conversion from 'double' to 'GLfloat', possible loss of data
1>c:\users\student\downloads\6e_example1_vc10\6e test\code\Angel.h(65): warning C4305: 'initializing' : truncation from 'double' to 'const GLfloat'
1>c:\users\student\downloads\6e_example1_vc10\6e test\code\mat.h(698): warning C4244: '=' : conversion from 'double' to 'GLfloat', possible loss of data
1>c:\users\student\downloads\6e_example1_vc10\6e test\code\mat.h(699): warning C4244: '=' : conversion from 'double' to 'GLfloat', possible loss of data
1>c:\users\student\downloads\6e_example1_vc10\6e test\code\mat.h(700): warning C4244: '=' : conversion from 'double' to 'GLfloat', possible loss of data
1>c:\users\student\downloads\6e_example1_vc10\6e test\code\mat.h(721): warning C4244: '=' : conversion from 'double' to 'GLfloat', possible loss of data
1>c:\users\student\downloads\6e_example1_vc10\6e test\code\mat.h(723): warning C4244: '=' : conversion from 'double' to 'GLfloat', possible loss of data
1>c:\users\student\downloads\6e_example1_vc10\6e test\code\mat.h(726): warning C4244: '=' : conversion from 'double' to 'GLfloat', possible loss of data
1>c:\users\student\downloads\6e_example1_vc10\6e test\code\mat.h(742): warning C4244: '=' : conversion from 'double' to 'GLfloat', possible loss of data
1>..\CODE\InitShader.cpp(17): warning C4996: 'fopen': This function or variable may be unsafe. Consider using fopen_s instead. To disable deprecation, use _CRT_SECURE_NO_WARNINGS. See online help for details.
1>          C:\Program Files (x86)\Microsoft Visual Studio 11.0\VC\include\stdio.h(218) : see declaration of 'fopen'
1>example1.obj : error LNK2001: unresolved external symbol ___glewBindBuffer
1>example1.obj : error LNK2001: unresolved external symbol ___glewVertexAttribPointer
1>example1.obj : error LNK2001: unresolved external symbol ___glewGenVertexArrays
1>example1.obj : error LNK2001: unresolved external symbol ___glewUseProgram
1>example1.obj : error LNK2001: unresolved external symbol ___glewBufferData
1>example1.obj : error LNK2001: unresolved external symbol ___glewBindVertexArray
1>example1.obj : error LNK2001: unresolved external symbol ___glewGetAttribLocation
1>example1.obj : error LNK2001: unresolved external symbol ___glewGenBuffers
1>example1.obj : error LNK2001: unresolved external symbol ___glewEnableVertexAttribArray
1>InitShader.obj : error LNK2001: unresolved external symbol ___glewCreateShader
1>InitShader.obj : error LNK2001: unresolved external symbol ___glewGetShaderInfoLog
1>InitShader.obj : error LNK2001: unresolved external symbol ___glewLinkProgram
1>InitShader.obj : error LNK2001: unresolved external symbol ___glewCompileShader
1>InitShader.obj : error LNK2001: unresolved external symbol ___glewShaderSource
1>InitShader.obj : error LNK2001: unresolved external symbol ___glewGetProgramiv
1>InitShader.obj : error LNK2001: unresolved external symbol ___glewGetShaderiv
1>InitShader.obj : error LNK2001: unresolved external symbol ___glewGetProgramInfoLog
1>InitShader.obj : error LNK2001: unresolved external symbol ___glewCreateProgram
1>InitShader.obj : error LNK2001: unresolved external symbol ___glewAttachShader
1>C:\Users\student\Downloads\6E_example1_VC10\6E test\Release\6E test.exe : fatal error LNK1120: 19 unresolved externals
========== Rebuild All: 0 succeeded, 1 failed, 0 skipped ==========```

What can be done about this? I saw somebody else had a similar problem due to linking the 64 bit version of glew instead of the 32 bit verion, ho do I check that my version is the 32 bit version? Why should I use the 32 bit version and not the 64 bit version?


text:
# An optimal harvesting problem with measure valued solutions

## Giovedi' 1 Marzo 2012 - Giuseppe M. Coclite

ARGOMENTI: Seminari

SEMINARI DI EQUAZIONI DIFFERENZIALI E APPLICAZIONI
Giovedi' 1 marzo 2012 alle ore 12:15 in aula 1AD100 Giuseppe M. Coclite (Universita' di Bari) terra' un seminario dal titolo "An optimal harvesting problem with measure valued solutions".

-Abstract
In this lecture we consider a model for the harvesting of marine resources, described by an elliptic equation. Since the cost functionals have sublinear growth with respect to the pointwise intensity of fishing effort, optimal solutions are in general measure-valued. For the control problem, we prove the existence of optimal strategies. The results were obtained in collaboration with Professors Alberto Bressan and Wen Shen.

Rif. int. M. Bardi, C. Marchi, F. Ancona