text:
Welcome Guest

### NetConnect

#### Internet Domains, simple and cheap

##### Find a domain name:

Home

The time is now 12:02 am
You last visited December 11, 2013, 12:02 am
All times shown are
Eastern Time (GMT-5:00)

# Has anyone tried the Robert Walsh Pick 3 or 4 Lottery System?

Topic closed. 293 replies. Last post 1 year ago by mcginnin56.

 Page 3 of 20

Belgium
Member #128223
May 19, 2012
96 Posts
Offline
 Posted: June 9, 2012, 4:31 am - IP Logged

Hey, it's going to sound crazy, but his strategies give you a point of reference.  They do work, but you have to figure out which two work the most in your particular state.  Every state shows patters.  So it will require a lot of work in the beginning, but it will pay off.  Got back 6 full months and go through each drawing one by one and figure out how each number was created.  Once you find the two most dominant strategies, apply them to every number that is drawn from here on on out and you will narrow down the winning number groups significantly.  For example a young man in NC found that the A1 and E5 strategies are the most dominant in his state so he uses only those two.

Here in DC I found that for his Pick 4 system, that the Family Trend, 7-11 and TOC strategies are the most dominant (all three combined accounts for 60% of the winning numbers in DC).  So I only apply these numbers to every drawn number. It give me a 60% chance of winning.  I hope this helps.

You know, each found base number plays a set of 8 numbers.
Using the two most dominant strategies costs 16$per day. Not cheap... Belgium Member #128223 May 19, 2012 96 Posts Offline  Posted: June 9, 2012, 5:14 am - IP Logged You know, each found base number plays a set of 8 numbers. Using the two most dominant strategies costs 16$ per day.
Not cheap...

English:

Are there people from Belgium who bought "Winning The Pick 3 Lottery System" of Robert Walsh? If so, can you send me a private message?  Thanks in advance.

People from other countries can send me a private message too.

Dutch (Nederlands):

Zijn er mensen in België die "Winning The Pick 3 Lottery System" van Robert Walsh gekocht hebben? Indien ja, stuur me even een privébericht.  Bedankt.

Texas
United States
Member #86155
January 30, 2010
1180 Posts
Offline
 Posted: June 9, 2012, 1:09 pm - IP Logged

Like lots of others, I've been around here for a little while...roughly about five years. I changed my handle, though, to put some "wildfires" out which were burning pretty good due to all the controversy around me and my style of playing. Anyhow, during all this time, I've witnessed many a systems introduced by regular players here as well as those purchased on the internet and posted here for review. If my memory serves, most of these systems are simply "recirculated" by new members looking for the best way to match numbers and win. All the veterans here have already been exposed to probably 90% of the famous name lottery systems which all provide very similar results in the end.

These systems are directly dependant upon the user's level of comprehension. The developer of any said system is basing all of their ideas and findings soley on how THEY INTERPRET THE GAME/NUMBERS, and in turn, put together their own system for marketing. See, in terms of these games, a lottery approach is purely a personal interpretation for a given game...and every player thinks differently because they all see numbers differently for the most part. The reason I say that is because Pick 3 is the same game with the same numbers all across the states but, everyone's brain and train of thought is different.

There is no one best way to play the game because all the numbers/combinations are good and can be drawn at any time. The pre-draws have a huge influence on the end results, too. In my honest opinion, the best way to play is with consistency and reasonable coverage of a complete combination because the combination must be drawn at some point...I've said this many times already. This comes with patience and changing numbers every draw will net a loss lots of times. Players win during the pre-draws and lose on the official draw every single day. Some players get very, very lucky playing just a handful of numbers but, how long it took for that number to hit must be taken into account.

If a player has spent $200 total over a period of say two weeks, and, finally hits on a$140 box, they're still $60 in the hole. It usually takes a str8 hit to even think of recovering all monies spent over a period of time...let alone making a profit in the process. Every time a loss is incured, be it boxed or str8, the level of money needed to win is simply raised in order to recover previous expenses. Do not be fooled by tickets posted with str8 or boxed wins for only$1 or so, and, how it can be done, okay. When I see this, the first thing I wonder is how long it took for that win to happen and how much money has been spent already. More than likely, it's a simple offset and a wash because it just doesn't happen this way for anyone on a regular basis. I'm not knocking or discrediting anyone's wins here...I just believe in telling it like it is. I like to see players win but, I believe they have an obligation to disclose how long it took to win on what I consider a "skeleton" set of numbers...THE WHOLE TRUTH.

The same goes for the Walsh, Player, and any other systems. They should provide players with realistic win loss ratios along with true profit margins over a period of time because that's what's most important if the player is serious. If it's only for entertainment, then the integrity of the system and money is of no concern at all. This isn't to say that a said system cannot help a given player but, rather a player is really trying to compile someone else's thought process into their own...which can cause confusion. I hope this helps, and, I'll be discussing how I approach the game at some point.

L.L.

Belgium
Member #128223
May 19, 2012
96 Posts
Offline
 Posted: June 9, 2012, 1:15 pm - IP Logged

Thanks L.L.

Texas
United States
Member #86155
January 30, 2010
1180 Posts
Offline
 Posted: June 9, 2012, 1:53 pm - IP Logged

Thanks L.L.

You're very welcome. I hope I didn't come off as rough or arrogant because I'm really not. Lots of people get this confused with simply calling it like it is and being honest. Luck definitely happens on $1 str8 or boxed...but how how often and consistent is it? Internet systems, to me, are a ripoff with all the sophisticated formulas and terminologies...it's a bunch of BS. All we have is 0-9 in three positions...five numbers are odd and five are even. There are only six total combinations which can be formed from this array of numbers...which can also be broken down into sums. A player can simply pick one combination, filter it down to where it's reasonably affordable and play it consistently until it hits. Cost will depend on how much is filtered out and the player's budget. You must begin with a complete combination, though, as it contains all the single digits for that combo to have the best chance of winning. All the rest of this PI=3.14, circumferences, etc. etc. is all just like the little decorations on a plate of food. They do nothing...meat and potatoes is what get's the job done. Pick 3 isn't rocket science. The only thing which really makes Pick 3 challenging is the pre-draws because they allow anything to happen again, and again...especially what even the sharpest players have taken out as low odds of re-occurring. Pick 3 simply draws a combination every mid-day and evening...that's it. If you play one or two 'em, and your filtering works out, you win consistently. L.L. Texas United States Member #86155 January 30, 2010 1180 Posts Offline  Posted: June 9, 2012, 2:56 pm - IP Logged Every single system, pattern, method, and approach seen on this site will still boil down to the basics which I've described...even the vtracs deal. It seems to be very popular which is good. In the end, all the winning combinations must still end up as OOE, EEO OEE, EOO, EOE, OEO, EEE, and OOO. This is all the combinations, and, I left out two of 'em on my previous post because the str8 odd and str8 even are the least drawn of the eight. Take a look around at what players are posting in terms of "near misses", followers, mirror numbers, one up/one down etc. and they all still arrive at the above mentioned locations in terms of combinations. No way around it and it's pure basics. If the last draw was 2-9-7, this is a combination of EOO and a sum of (18). A vtrac version of the next combo to be anticipated could be 1-7-2 which is OOE, and, a product of one up/one down along with a sum of (10). This is only an example but, definitely illustrates a basic approach to a simple game. Mirror numbers such as 8/3, 0/9, 6/1 etc. all follow suite. There are only a handful of players here that truly understand this game, and, refuse to make it overly complicated. Odd, even high, low...this is Pick 3. L.L. Texas United States Member #86155 January 30, 2010 1180 Posts Offline  Posted: June 9, 2012, 10:09 pm - IP Logged ...My sincere apologies for hijacking this thread. I thought I was helping but, it seems I've caused more problems. Won't post here anymore. L.L. My Pick 4 goal. Win$1,500 a week.
bel air maryland
United States
Member #90253
April 24, 2010
1574 Posts
Online
 Posted: June 9, 2012, 10:53 pm - IP Logged

...My sincere apologies for hijacking this thread. I thought I was helping but, it seems I've caused more problems. Won't post here anymore.

L.L.

No need to apologize. You brought up some very valid points. I agree, the pick 3 isn't as complicated as we make it to be. It

all depends on how well you track everything and how well you read the trends. Your tracking charts will always tell you

what is going to happen. In my opinion that is all you really need and if you read them correctly the wins will happen.

Good luck!!!

"You can observe a lot just by watching."  Yogi Berra, Hall of Fame baseball player.

To win the lottery, watch the #'s.

Texas
United States
Member #86155
January 30, 2010
1180 Posts
Offline
 Posted: June 9, 2012, 11:34 pm - IP Logged

No need to apologize. You brought up some very valid points. I agree, the pick 3 isn't as complicated as we make it to be. It

all depends on how well you track everything and how well you read the trends. Your tracking charts will always tell you

what is going to happen. In my opinion that is all you really need and if you read them correctly the wins will happen.

Good luck!!!

Thanks, I really appreciate that. I try very hard to just be honest with players instead of sugar coating things because sugar coating creates a false sense of accomplishment. I mean, we're talking about spending money on numbers on a regular basis here, you know. It adds up over time if you're not able to win. You're the only person that was willing to respond to the truth. It seems everyone else wants to hear a lie...that a person can win consistently spending a couple of dollars daily and make all the money.

Well, people do get lucky as I've mentioned already and lucky is good. But, it's not dependable at all. The lapse between wins is far and few between when only playing a couple of numbers or so. Odds don't lie and you get what you pay for just like anything else in life...within reason.

Your assessment is very correct and wins aren't very hard at all. It does cost money to make money, though. A very, very, very good player here has already told the people here at LP that they're never going to win like they're supposed to because they don't spend enough money on their numbers. This took place years ago and nothing has changed since then.

You can't win all the money...a player must be willing to accept a profit off of spending enough to get the hit with reasonable consistence. I'm out of breath...again. Thank you for your response!!!!!!

L.L.

My Pick 4 goal. Win $1,500 a week. bel air maryland United States Member #90253 April 24, 2010 1574 Posts Online  Posted: June 10, 2012, 12:27 am - IP Logged Thanks, I really appreciate that. I try very hard to just be honest with players instead of sugar coating things because sugar coating creates a false sense of accomplishment. I mean, we're talking about spending money on numbers on a regular basis here, you know. It adds up over time if you're not able to win. You're the only person that was willing to respond to the truth. It seems everyone else wants to hear a lie...that a person can win consistently spending a couple of dollars daily and make all the money. Well, people do get lucky as I've mentioned already and lucky is good. But, it's not dependable at all. The lapse between wins is far and few between when only playing a couple of numbers or so. Odds don't lie and you get what you pay for just like anything else in life...within reason. Your assessment is very correct and wins aren't very hard at all. It does cost money to make money, though. A very, very, very good player here has already told the people here at LP that they're never going to win like they're supposed to because they don't spend enough money on their numbers. This took place years ago and nothing has changed since then. You can't win all the money...a player must be willing to accept a profit off of spending enough to get the hit with reasonable consistence. I'm out of breath...again. Thank you for your response!!!!!! L.L. You're absolutely right. As the saying goes, scared money doesn't make money.There are plenty of people who might spend$20 a week and wonder how come I don't win. There are also a lot of people that will play the same few numbers for months

at a time. That is CRAZY. The numbers they are using could be on a long cold streak, which means no chance of a hit. But

they will all give the same reason for doing it ."The last time I played a number a long time and stopped it came in right after

that." Well that happens to everyone, I've been through it so many times it doesn't even bother me in the slightest. But I very

seldom play a number that long. Six days maybe, and even that is rare for me. I basically change every day because the

trends change every day and I try to keep up with it. If a number I played Monday, hits on Wednesday, oh well. Forget it,

and try again tomorrow. What else can you do?

That being said, I'm not going to play 50 or 100 numbers on each drawing either. I try to keep it to 15 to 20 generally and

if  you've done your homework that should be enough to get a win. I always play straight and box too.

You also have to realize when one strategy is no longer working, and be willing to switch to something else. You can't just

keep doing the same old, same old or else you're just wasting your money.

"You can observe a lot just by watching."  Yogi Berra, Hall of Fame baseball player.

To win the lottery, watch the #'s.

Texas
United States
Member #86155
January 30, 2010
1180 Posts
Offline
 Posted: June 10, 2012, 2:10 am - IP Logged

You're absolutely right. As the saying goes, scared money doesn't make money.There are plenty of people who might spend

$20 a week and wonder how come I don't win. There are also a lot of people that will play the same few numbers for months at a time. That is CRAZY. The numbers they are using could be on a long cold streak, which means no chance of a hit. But they will all give the same reason for doing it ."The last time I played a number a long time and stopped it came in right after that." Well that happens to everyone, I've been through it so many times it doesn't even bother me in the slightest. But I very seldom play a number that long. Six days maybe, and even that is rare for me. I basically change every day because the trends change every day and I try to keep up with it. If a number I played Monday, hits on Wednesday, oh well. Forget it, and try again tomorrow. What else can you do? That being said, I'm not going to play 50 or 100 numbers on each drawing either. I try to keep it to 15 to 20 generally and if you've done your homework that should be enough to get a win. I always play straight and box too. You also have to realize when one strategy is no longer working, and be willing to switch to something else. You can't just keep doing the same old, same old or else you're just wasting your money. Well, I definitely agree with most of what you've said here, but I must disagree on the idea of strategy no longer working...at least when it comes to what I've outlined. What I've described in my previous posts must happen with every single Pick 3/4 mid-day and evening draw. Those combinations absolutely cannot be changed, and, must be drawn. They cannot draw the same combination over and over and over...even with pre-draws figured in which is why I've said that playing in "cycles" is best. I don't care what numbers are selected to play on any given draw...it still boils down to OOE, EEO, EOO, OEE, EEE, OOO, EOE, EOE. If a player will just select one format to play like (OOE), and include all the numbers of that format, they have a much better chance of winning be it str8 or boxed. Your suggestion of at least$15 is a very good starting point but, you need to be patient and willing to have small bankroll to do this. The formats will usually change from one draw to the next. If the same format comes back from a mid-day to an evening draw, then it's "ripe" for playing most of them time.

Take a look at any str8 odd or str8 even combo and tell me what you see after those most of the time...there are basically only (6) options left which can still be narrowed down much farther. This is providing that the draw doesn't go from str8 odd to str8 even, or, vice-versa. If you have a complete format, then you'll have all the numbers for a given win on a combo.

I had to wonder on the part about people waiting for "a number" to fall. Let's take a number like 6-7-4 and how a person is only playing that combo daily. This number is of the Even-Odd-Even combination... there are a total of (125) str8 EOE combinations and this person is only playing (1) of them Even a boxed play is tough here as one would still need those three numbers to hit. If a person is gonna switch anything, then switch the combination to be played. Pick 3 is all about the combination of odd and even numbers. I look around and see terms like "the mirror" of this number or that number...the follower etc. and I find myself using those terms when I post.

A mirror shows an exact reflection of a given object, right? So, how are 8/3, 6/1, 0/9, 7/2, etc. mirrors of each other? These are odd/even, high/low numbers. In my opinion, the mirror of (6) is going to be (6) and not (1).  I'm struggling with all this stuff and how it works here, man. Am I over simplifying this game or what?

L.L.

My Pick 4 goal. Win $1,500 a week. bel air maryland United States Member #90253 April 24, 2010 1574 Posts Online  Posted: June 10, 2012, 4:06 am - IP Logged When I mentioned strategies I was just saying that a lot of people use different systems to pick their numbers and they go hot and cold too. You also have to consider the LHH, HLH, HHL, LLH, LHL, HLL, LLL, HHH, combos and be aware of how they are running also. Both combos can trend differently at any time. Sometimes they alternate back and forth, other times they may run several days in a row. Keeping track of both can definitely help to reduce the amount of numbers to play. Same thing with sums and root sums. Remember every thing balances itself out. If for example you know that you have been getting a lot of high numbers for the past week then the low numbers are going to start a good run very soon. If you have had singles for the past 2 weeks, you know doubles are going to be coming, since doubles hit every 4 draws statistically and they are overdue. As I said earlier if you track every thing your charts will tell you what will be coming. You will know what to play and what not to play and when. This is how you can really reduce the amount of numbers to play. You can play 10 numbers, 25, 50 or 150, only ONE is going to win. Think about it, if you play 100 numbers how many really have a good chance of hitting and how many are just guess work or hunches? You have to pick the numbers intelligently, by studying whats happening and trying to put the odds or percentages in your favor as much as possible. As another member has said, play the probabilities, not the possibilities. "You can observe a lot just by watching." Yogi Berra, Hall of Fame baseball player. To win the lottery, watch the #'s. Vancouver BC Canada Member #96080 August 22, 2010 92 Posts Offline  Posted: June 11, 2012, 1:20 am - IP Logged I posted a strategy for winning in Pick 3, the way I see it, on my LP blog titled "7 principles for being successful in Pick 3 lottery". Have you seen it, folks? If not, take a look, or sent me a note for more info. My strategy agrees with many things you say in your posts. I'm a rationalist and like seeing thing as the are, not what I would like them to be. Many people loose money in lottery because they do not want to use logic and prefer to go by emotions. I backtracked my strategy on several occasions and it works every time. Aquarius ...... My Pick 4 goal. Win$1,500 a week.
bel air maryland
United States
Member #90253
April 24, 2010
1574 Posts
Online
 Posted: June 11, 2012, 7:23 pm - IP Logged

I posted a strategy for winning in Pick 3, the way I see it, on my LP blog titled "7 principles for being successful in Pick 3 lottery". Have you seen it, folks? If not, take a look, or sent me a note for more info.

My strategy agrees with many things you say in your posts. I'm a rationalist and like seeing thing as the are, not what I would like them to be. Many people loose money in lottery because they do not want to use logic and prefer to go by emotions. I backtracked my strategy on several occasions and it works every time.

Aquarius

I took a look. Good sensible advise.

"You can observe a lot just by watching."  Yogi Berra, Hall of Fame baseball player.

To win the lottery, watch the #'s.

Station C:\somewhere\in\space\time
Fiji
Member #123705
February 27, 2012
2203 Posts
Offline
 Posted: June 30, 2012, 3:49 am - IP Logged

I've recently made a small investment with this "How to Play and Win the Pick 3 Lottery" system. Seems very intriguing, doesn't appear to be full

of gimmicks or magic wands to be effective.

Uses a base "5" number conversion method, along with several other tactic's to predict both straight and boxed pick 3 winners. The author is

Robert Walsh, who according to it's history, has been developing this system with success since the late 80's.

Anyone have  experience or feedback with this guy?

It is MOD 5, not base 5.

The modus after transformation is 1 repetition, second in frequency is 2 repetitions (positional).

The up and down is limited to 2, plus or minus, due to overflow. 9-1=8. 8>5. In base 5: 14-1=13, but you only have one digit. Decimal with Mod 5: 4-1=3 (9/5=1 rest|mod 4).

The expectancy still is 50% loss.

For pick 3 you can only use mod 5, 2, 1 to get equal parts.

.

 Page 3 of 20


text:
# NAG Library Function Document

## 1Purpose

nag_pde_interp_1d_coll (d03pyc) may be used in conjunction with either nag_pde_parab_1d_coll (d03pdc) or nag_pde_parab_1d_coll_ode (d03pjc). It computes the solution and its first derivative at user-specified points in the spatial coordinate.

## 2Specification

 #include #include
 void nag_pde_interp_1d_coll (Integer npde, const double u[], Integer nbkpts, const double xbkpts[], Integer npoly, Integer npts, const double xp[], Integer intpts, Integer itype, double up[], double rsave[], Integer lrsave, NagError *fail)

## 3Description

nag_pde_interp_1d_coll (d03pyc) is an interpolation function for evaluating the solution of a system of partial differential equations (PDEs), or the PDE components of a system of PDEs with coupled ordinary differential equations (ODEs), at a set of user-specified points. The solution of a system of equations can be computed using nag_pde_parab_1d_coll (d03pdc) or nag_pde_parab_1d_coll_ode (d03pjc) on a set of mesh points; nag_pde_interp_1d_coll (d03pyc) can then be employed to compute the solution at a set of points other than those originally used in nag_pde_parab_1d_coll (d03pdc) or nag_pde_parab_1d_coll_ode (d03pjc). It can also evaluate the first derivative of the solution. Polynomial interpolation is used between each of the break-points ${\mathbf{xbkpts}}\left[\mathit{i}-1\right]$, for $\mathit{i}=1,2,\dots ,{\mathbf{nbkpts}}$. When the derivative is needed (${\mathbf{itype}}=2$), the array ${\mathbf{xp}}\left[{\mathbf{intpts}}-1\right]$ must not contain any of the break-points, as the method, and consequently the interpolation scheme, assumes that only the solution is continuous at these points.

None.

## 5Arguments

Note: the arguments u, npts, npde, xbkpts, nbkpts, rsave and lrsave must be supplied unchanged from either nag_pde_parab_1d_coll (d03pdc) or nag_pde_parab_1d_coll_ode (d03pjc).
1:    $\mathbf{npde}$IntegerInput
On entry: the number of PDEs.
Constraint: ${\mathbf{npde}}\ge 1$.
2:    $\mathbf{u}\left[{\mathbf{npde}}×{\mathbf{npts}}\right]$const doubleInput
Note: the $\left(i,j\right)$th element of the matrix $U$ is stored in ${\mathbf{u}}\left[\left(j-1\right)×{\mathbf{npde}}+i-1\right]$.
On entry: the PDE part of the original solution returned in the argument u by the function nag_pde_parab_1d_coll (d03pdc) or nag_pde_parab_1d_coll_ode (d03pjc).
3:    $\mathbf{nbkpts}$IntegerInput
On entry: the number of break-points.
Constraint: ${\mathbf{nbkpts}}\ge 2$.
4:    $\mathbf{xbkpts}\left[{\mathbf{nbkpts}}\right]$const doubleInput
On entry: ${\mathbf{xbkpts}}\left[\mathit{i}-1\right]$, for $\mathit{i}=1,2,\dots ,{\mathbf{nbkpts}}$, must contain the break-points as used by nag_pde_parab_1d_coll (d03pdc) or nag_pde_parab_1d_coll_ode (d03pjc).
Constraint: ${\mathbf{xbkpts}}\left[0\right]<{\mathbf{xbkpts}}\left[1\right]<\cdots <{\mathbf{xbkpts}}\left[{\mathbf{nbkpts}}-1\right]$.
5:    $\mathbf{npoly}$IntegerInput
On entry: the degree of the Chebyshev polynomial used for approximation as used by nag_pde_parab_1d_coll (d03pdc) or nag_pde_parab_1d_coll_ode (d03pjc).
Constraint: $1\le {\mathbf{npoly}}\le 49$.
6:    $\mathbf{npts}$IntegerInput
On entry: the number of mesh points as used by nag_pde_parab_1d_coll (d03pdc) or nag_pde_parab_1d_coll_ode (d03pjc).
Constraint: ${\mathbf{npts}}=\left({\mathbf{nbkpts}}-1\right)×{\mathbf{npoly}}+1$.
7:    $\mathbf{xp}\left[{\mathbf{intpts}}\right]$const doubleInput
On entry: ${\mathbf{xp}}\left[\mathit{i}-1\right]$, for $\mathit{i}=1,2,\dots ,{\mathbf{intpts}}$, must contain the spatial interpolation points.
Constraints:
• ${\mathbf{xbkpts}}\left[0\right]\le {\mathbf{xp}}\left[0\right]<{\mathbf{xp}}\left[1\right]<\cdots <{\mathbf{xp}}\left[{\mathbf{intpts}}-1\right]\le {\mathbf{xbkpts}}\left[{\mathbf{nbkpts}}-1\right]$;
• if ${\mathbf{itype}}=2$, ${\mathbf{xp}}\left[\mathit{i}-1\right]\ne {\mathbf{xbkpts}}\left[\mathit{j}-1\right]$, for $\mathit{i}=1,2,\dots ,{\mathbf{intpts}}$ and $\mathit{j}=2,3,\dots ,{\mathbf{nbkpts}}-1$.
8:    $\mathbf{intpts}$IntegerInput
On entry: the number of interpolation points.
Constraint: ${\mathbf{intpts}}\ge 1$.
9:    $\mathbf{itype}$IntegerInput
On entry: specifies the interpolation to be performed.
${\mathbf{itype}}=1$
The solution at the interpolation points are computed.
${\mathbf{itype}}=2$
Both the solution and the first derivative at the interpolation points are computed.
Constraint: ${\mathbf{itype}}=1$ or $2$.
10:  $\mathbf{up}\left[\mathit{dim}\right]$doubleOutput
Note: the dimension, dim, of the array up must be at least ${\mathbf{npde}}×{\mathbf{intpts}}×{\mathbf{itype}}$.
The element ${\mathbf{UP}}\left(i,j,k\right)$ is stored in the array element ${\mathbf{up}}\left[\left(k-1\right)×{\mathbf{npde}}×{\mathbf{intpts}}+\left(j-1\right)×{\mathbf{npde}}+i-1\right]$.
On exit: if ${\mathbf{itype}}=1$, ${\mathbf{UP}}\left(\mathit{i},\mathit{j},1\right)$, contains the value of the solution ${U}_{\mathit{i}}\left({x}_{\mathit{j}},{t}_{\mathrm{out}}\right)$, at the interpolation points ${x}_{\mathit{j}}={\mathbf{xp}}\left[\mathit{j}-1\right]$, for $\mathit{j}=1,2,\dots ,{\mathbf{intpts}}$ and $\mathit{i}=1,2,\dots ,{\mathbf{npde}}$.
If ${\mathbf{itype}}=2$, ${\mathbf{UP}}\left(\mathit{i},\mathit{j},1\right)$ contains ${U}_{\mathit{i}}\left({x}_{\mathit{j}},{t}_{\mathrm{out}}\right)$ and ${\mathbf{UP}}\left(\mathit{i},\mathit{j},2\right)$ contains $\frac{\partial {U}_{\mathit{i}}}{\partial x}$ at these points.
11:  $\mathbf{rsave}\left[{\mathbf{lrsave}}\right]$doubleCommunication Array
The array rsave contains information required by nag_pde_interp_1d_coll (d03pyc) as returned by nag_pde_parab_1d_coll (d03pdc) or nag_pde_parab_1d_coll_ode (d03pjc). The contents of rsave must not be changed from the call to nag_pde_parab_1d_coll (d03pdc) or nag_pde_parab_1d_coll_ode (d03pjc). Some elements of this array are overwritten on exit.
12:  $\mathbf{lrsave}$IntegerInput
On entry: the size of the workspace rsave, as in nag_pde_parab_1d_coll (d03pdc) or nag_pde_parab_1d_coll_ode (d03pjc).
13:  $\mathbf{fail}$NagError *Input/Output
The NAG error argument (see Section 3.7 in How to Use the NAG Library and its Documentation).

## 6Error Indicators and Warnings

NE_ALLOC_FAIL
Dynamic memory allocation failed.
See Section 2.3.1.2 in How to Use the NAG Library and its Documentation for further information.
On entry, argument $〈\mathit{\text{value}}〉$ had an illegal value.
NE_EXTRAPOLATION
Extrapolation is not allowed.
NE_INCOMPAT_PARAM
On entry, ${\mathbf{itype}}=2$ and at least one interpolation point coincides with a break-point, i.e., interpolation point no $〈\mathit{\text{value}}〉$ with value $〈\mathit{\text{value}}〉$ is close to break-point $〈\mathit{\text{value}}〉$ with value $〈\mathit{\text{value}}〉$.
NE_INT
On entry, ${\mathbf{intpts}}\le 0$: ${\mathbf{intpts}}=〈\mathit{\text{value}}〉$.
On entry, ${\mathbf{itype}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{itype}}=1$ or $2$.
On entry, ${\mathbf{nbkpts}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{nbkpts}}\ge 2$.
On entry, ${\mathbf{npde}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{npde}}>0$.
On entry, ${\mathbf{npoly}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{npoly}}>0$.
NE_INT_3
On entry, ${\mathbf{npts}}=〈\mathit{\text{value}}〉$, ${\mathbf{nbkpts}}=〈\mathit{\text{value}}〉$ and ${\mathbf{npoly}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{npts}}=\left({\mathbf{nbkpts}}-1\right)×{\mathbf{npoly}}+1$.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
See Section 2.7.6 in How to Use the NAG Library and its Documentation for further information.
NE_NO_LICENCE
Your licence key may have expired or may not have been installed correctly.
See Section 2.7.5 in How to Use the NAG Library and its Documentation for further information.
NE_NOT_STRICTLY_INCREASING
On entry, break-points xbkpts badly ordered: $\mathit{I}=〈\mathit{\text{value}}〉$, ${\mathbf{xbkpts}}\left[\mathit{I}-1\right]=〈\mathit{\text{value}}〉$, $\mathit{J}=〈\mathit{\text{value}}〉$ and ${\mathbf{xbkpts}}\left[\mathit{J}-1\right]=〈\mathit{\text{value}}〉$.
On entry, interpolation points xp badly ordered: $\mathit{I}=〈\mathit{\text{value}}〉$, ${\mathbf{xp}}\left[\mathit{I}-1\right]=〈\mathit{\text{value}}〉$, $\mathit{J}=〈\mathit{\text{value}}〉$ and ${\mathbf{xp}}\left[\mathit{J}-1\right]=〈\mathit{\text{value}}〉$.

## 7Accuracy

See the documents for nag_pde_parab_1d_coll (d03pdc) or nag_pde_parab_1d_coll_ode (d03pjc).

## 8Parallelism and Performance

nag_pde_interp_1d_coll (d03pyc) is not threaded in any implementation.

format: plain


text:
Check that object names conform to a naming style. The default naming styles are "snake_case" and "symbols".

## Usage

object_name_linter(styles = c("snake_case", "symbols"))

## Arguments

styles

A subset of ‘symbols’, ‘CamelCase’, ‘camelCase’, ‘snake_case’, ‘SNAKE_CASE’, ‘dotted.case’, ‘lowercase’, ‘UPPERCASE’ . A name should match at least one of these styles.

## Details

Note when used in a package, in order to ignore objects imported from other namespaces, this linter will attempt getNamespaceExports() whenever an import(PKG) or importFrom(PKG, ...) statement is found in your NAMESPACE file. If requireNamespace() fails (e.g., the package is not yet installed), the linter won't be able to ignore some usages that would otherwise be allowed.

Suppose, for example, you have import(upstream) in your NAMESPACE, which makes available its exported S3 generic function a_really_quite_long_function_name that you then extend in your package by defining a corresponding method for your class my_class. Then, if upstream is not installed when this linter runs, a lint will be thrown on this object (even though you don't "own" its full name).

The best way to get lintr to work correctly is to install the package so that it's available in the session where this linter is running.


text:
# Lie group quotient bundle with image of normalizer as structure group

In Glen Bredon "Topology and Geometry", Ch. II-13, I am stuck on the following

"Problem 1. Finish the proof at the end of this section that $G\rightarrow G/H$ is a bundle. Also show that this is a fiber bundle with structure group $N(H)/H$ where $N(H)$ is the normalizer of $H$ in $G$."

Here $H$ is a closed subgroup of the Lie group $G$ (which the book assumes to be again a closed subgroup of $Gl(n, \mathbb{C})$."

I was able to compute that this is a fiber bundle with structure group $H$, using the trivializations given in the text. Obviously, some other trivializations of the same bundle will be required to get $N(H)/H$ as structure group. I have now been stuck on this for a while, and any help would be greatly appreciated.

-


text:
# How can I calculate knee current for this foldback current limiter

So as the title says, I'm trying to calculate the knee current for this schematic. I know Veb(Q2)/R1 is approximately the maximum current through the emitter of Q1 if R4 wasn't there and this was a normal current limiting scheme. I also know R2 is there for limiting the current through the base of Q2. R3 is the current limiting resistor for the base of Q1. but how can I calculate the knee current point? (Rload is just a dummy value)

simulate this circuit – Schematic created using CircuitLab

• What is Vin and Vout under normal conditions? Apr 13 '17 at 15:51
• Vin and Vout should both be around 27V in normal condition Apr 14 '17 at 8:12
• Who misled you into thinking this design will work? Impossible with R3/R1 = 560 even if you had hFE=100 to saturate this power transistor which only has hFE =30 at Vce(sat) .... NG design for 700mA and depends on V+/700mA thus Rload/Re>> hfe(sat) Apr 14 '17 at 17:24
• Just added an analysis for you. I didn't pull any details from models for the 2N2955 or BC807. But you can do that and recompute results. The approach I show is solid.
– jonk
Apr 30 '17 at 19:42

## 4 Answers

I've drawn two versions of the same schematic below. These diagrams are arranged so that you can focus on their operation during two different phases. Diagram 1 shows the arrangement where the load current hasn't yet risen to the point where $Q_2$'s collector current is significant and where the current in $R_2$ is pulling up on $Q_2$'s base, keeping it out of a foldback behavior. Diagram 2 shows the arrangement where the load current is moving sufficiently towards foldback that $R_2$ is now supplementing $Q_2$'s base current (along with current through $R_4$) and foldback is near. I've used a small green arrow to point out this detail: if you carefully look, you can see that the direction through $R_2$ is diametrically opposite in these two cases.

simulate this circuit – Schematic created using CircuitLab

DIAGRAM 1

$Q_1$ is fully saturated and the current in $R_1$ is almost the same as the current in $R_3$. $Q_1$'s base-emitter junction is mostly just a diode junction here.

With the load current effectively zero, $R_2$ and $R_4$ act as an uncontrolled (but small) collector bypass current that pulls up on $Q_1$'s collector enough to likely yield an almost zero $^{Q_1}V_{CE_{SAT}}$ and there will be nearly zero current through $R_2$ and $R_4$, as well.

(Technically, since $Q_1$'s base has "access" to a lower voltage through $R_3$, $Q_1$'s collector is allowed to be a few millivolts below its emitter and there will be a negligible base current into $Q_2$: it's likely that $V_{OUT}$ will be perhaps $100-200\:\textrm{mV}$ below $V_{IN}$. That's with exactly zero load current.)

As the load current ($I_L$) increases from zero, $Q_1$'s emitter current increases to provide that collector current, causing an increasing drop through $R_1$. This will lower $Q_1$'s emitter ($V_A$) and $Q_1$'s base ($V_B$) in paired motion.

As the load pulls down more on $Q_1$'s collector, $^{Q_1}V_{CE_{SAT}}$ opens up a bit and this begins to increase the still negligible current through $R_2$ and $R_4$. The important bit here is that $^{Q_2}V_{BE}$ is increasing and $Q_2$'s collector is starting to supply some of the current into $R_3$, reducing pressure on and gradually supplanting more of $Q_1$'s currently saturated base current.

Since $V_A$ and $V_B$ track together tightly ($^{Q_1}V_{BE}$ guarantees this), the load will effectively see $R_1\vert\vert R_3$ as the source impedance.

(This isn't quite true, because $^{Q_1}V_{CE_{SAT}}$ is also opening up due to the relaxing pressure on its base current and because the load is also pulling down. However, $Q_1$ is also still quite saturated and this effect is gradual. While measurable, the added complication is not important for this analysis.)

During this phase (and keeping in mind that some details are held as fixed values although they do vary slightly), $V_{OUT}$ will closely follow:

$$V_{OUT}=V_{IN}-^{Q_1}\!\!V_{CE_{SAT}}-\frac{V_{IN}-^{Q_1}\!\!V_{BE}}{R_1+R_3}\cdot R_1-\left(R_1\vert\vert R_3\right)\cdot I_L$$

(The last term in the above equation shows the effect of load current on the output voltage [discounting variation in $^{Q_1}V_{CE_{SAT}}$.])

And the transition point in the load current, where $R_2$'s current goes through zero and reverses direction (and diagram 2 takes over), is:

$$I_{L_{SW}}=\frac{V_T\cdot\operatorname{ln}\left(\frac{^{Q_1}V_{CE_{SAT}}}{R_4}\cdot\frac{^{Q_2}\beta}{^{Q_2}I_S}\right)-\frac{V_{IN}-^{Q_1}V_{BE}}{R_1+R_3}\cdot R_1}{R_1\vert\vert R_3}$$

$V_T$ is, of course, the thermal voltage and is, at room temps, $\approx 26\:\textrm{mV}$. The remaining BJT values are based upon experience and/or the datasheet and only need to be approximate for this use. The saturation current can vary in parts by a factor of 2 or 3, but can be picked up from a spice model. There are techniques to use with a datasheet to extract that parameter, if the datasheet is sufficiently detailed. But I won't discuss that here. (Typical values are a few tens of femptoamps for small signal devices and an order or two of magnitude more for high current devices.)

CALCULATION: Using your resistor values and modest BJT experience, I'll assume $V_{IN}=35\:\textrm{V}$, $^{Q_1}\!V_{CE_{SAT}}=200\:\textrm{mV}$, $^{Q_1}\!V_{BE}=900\:\textrm{mV}$, $^{Q_2}\!\beta=200$, and $^{Q_2}\!I_S=10\:\textrm{fA}$. From this, I compute $I_{L_{SW}}\approx 637.4\:\textrm{mA}$.

A much simpler approach to dealing with this earlier phase is to simply guess (accurately enough) that $Q_2$ will be "turning on" about when its $V_{BE}\approx 600\:\textrm{mV}$ and will be off before that point. (Clearly, the point at which $R_2$'s current goes to zero will be the point where the voltage across $R_1$ is the same as the voltage across $Q_2$'s $V_{BE}$.) This idea leads to a much simpler equation:

$$I_{L_{SW}}=\frac{600\:\textrm{mV}}{R_1}$$

In this case, the estimate is now $I_{L_{SW}}=600\:\textrm{mA}$.

$R_1$ usually dominates $R_1\vert\vert R_3$, so that explains the simplified denominator. The last term in the numerator is typically on the order of only a few tens of millivolts, by itself. And the first term in the numerator only changes by $60\:\textrm{mV}$ for each factor of 10 change inside the logarithm. So practical BJT parameters pretty much lead to the conclusion that the numerator will be at, or slightly more than, about $600\:\textrm{mV}$ after all is said and done.

DIAGRAM 2

The current in $R_2$ has now reversed direction and is now adding to the base current driving $Q_2$. This is also the period where $Q_2$'s rising collector current will eventually supplant enough of $Q_1$'s base current through $R_3$ that $Q_1$ starts to leave saturation. The foldback behavior depends upon this replacement behavior, because it will eventually pull $Q_1$ out of saturation and into its active region. The moment this happens, $Q_1$'s collector current will be forced to decrease as $Q_2$'s collector current increases further and $^{Q_1}V_{CE_{SAT}}$ will be forced to suddenly open up (having reached its $\beta$ limitation.) As $Q_1$'s collector now rapidly declines, it only adds still more base current for $Q_2$ (through $R_4$.) This, of course, puts still more pressure on $Q_1$'s collector current as its base current declines still more due to the increased collector current of $Q_2$, making $Q_1$ drop its collector yet more.

This positive feedback rapidly acts to force the output voltage low enough so that the remaining collector current in $Q_1$ can be supported by the nearly vanished base current still remaining to $Q_1$, as $Q_2$'s collector continues to pull up harder on $R_3$ and pinches $Q_1$'s $V_{BE}$ still tighter.

This is the fold-back behavior.

We can predict the point at which this will take place:

$$I_{L_{FOLDBACK}}=\frac{\frac{V_{IN}-^{Q_1}V_{BE}}{^{Q_2}\beta\:\cdot R_3}+\frac{^{Q_2}V_{BE}}{R_2}+\frac{^{Q_2}V_{BE}-^{Q_1}V_{CE_{SAT}}}{R_4}}{\frac{1}{^{Q_1}\beta\:\cdot\:^{Q_2}\beta}+\frac{R_1}{^{Q_2}\beta\:\cdot R_3}+\frac{R_1}{R_2}+\frac{R_1}{R_4}}$$

CALCULATION: Using your resistor values and modest BJT experience, I'll assume $V_{IN}=35\:\textrm{V}$, use a slightly increased $^{Q_1}\!V_{CE_{SAT}}=400\:\textrm{mV}$ (to represent the moment when the positive feedback breaks loose), an estimated $^{Q_1}\!\beta=180$ (when the positive feedback takes over), $^{Q_1}\!V_{BE}=900\:\textrm{mV}$, a somewhat lower $^{Q_2}\!V_{BE}=800\:\textrm{mV}$, $^{Q_2}\!\beta=200$, and $^{Q_2}\!I_S=10\:\textrm{fA}$. From this, I compute $I_{L_{FOLDBACK}}\approx 919.6\:\textrm{mA}$. To the better accuracy any of this has a right to expect, the answer is $920\:\textrm{mA}$.

The above calculation does demonstrate, though, that some of the terms aren't as important as others. The above can be reduced in complexity (when the source voltage is very much larger than the saturation and diode drop voltages) to something like this:

$$I_{L_{FOLDBACK}}\approx \frac{R_4}{R_1\cdot R_2+R_1\cdot R_4}\left(\frac{V_{IN}}{^{Q_2}\beta}\cdot\frac{R_2}{R_3}+^{Q_2}V_{BE}\right)$$

The above formula yields $I_{L_{FOLDBACK}}\approx 918.5\:\textrm{mA}$, in this situation. Again, the answer is the same $920\:\textrm{mA}$.

NOTES ON DERIVING FOLDBACK CURRENT

$R_4$ supplies positive feedback. But at first that feedback is small and not enough to counter negative feedbacks. So it's ignorable, early on. The crux is to find the answer to, "When will $R_4$'s positive feedback overwhelm the system?"

It helps to recognize a simple BJT behavior: as recombination current is removed from $Q_1$'s base and it moves out of saturation and into its active region, its $V_{CE}$ widens rapidly in response. It has no other possible response. Prior to that point, $V_{CE}$ remains relatively slow moving (almost static) because the BJT is saturated. These earlier small changes do feedback through $R_4$. But if you look at diagram #1, and focus a bit on the fact that $R_2$ is supplying a significant part of the current into $R_4$, you may realize that minor changes at this time in $V_{CE}$ don't impact $Q_2$'s base current so much as they affect $R_2$'s current. The voltage at $V_C$ just hasn't gotten to the point where $Q_2$ "cares," yet.

But after the change in direction in the current in $R_2$?? Then yes, changes in $V_{CE}$ provide feedback that now directly and entirely affects $Q_2$'s base current. And it is here that we start to become more interested in changes in $Q_1$'s $V_{CE}$ and it is now that we need to focus much more closely on precisely when the foldback might happen.

But even after the change-over, when the current direction in $R_2$ has reversed, $Q_1$ is still heavily saturated (in a properly designed circuit.) $Q_2$'s collector hasn't yet replaced enough of it to push $Q_1$ out of saturation.

So what are really looking for is the exact moment when $Q_1$ comes out of saturation, because this is the point when its $V_{CE}$ will open up much more quickly (very fast) if any reduction in its base current happens. (Or, equivalently, its $V_{BE}$ is pinched any more.)

One could choose to write this using the Shockley equation. We want to know when the required base-emitter voltage exceeds the available voltage, or:

\begin{align*} ^{Q_1}\!V_{BE}&\ge V_A-V_B\\\\ V_T\cdot\operatorname{ln}\left(\frac{I_L}{^{Q_1}\!I_S}\right)&\ge \frac{V_{IN}-R_1\cdot I_L}{R_3}-V_B \end{align*}

If $V_B$ weren't itself a complex function of the behavior of the surrounding BJTs, that would be reasonably solvable using the LambertW function. Unfortunately, it is rather more complicated that that, now.

(If you are interested in what the LambertW function is [how it is defined] and in seeing a fully worked example on how to apply it to solving problems like these, then see: Differential and Multistage Amplifiers(BJT).)

But the added complexity really isn't needed. It is over-kill, partly because parameters aren't all that well known or consistent between devices. It is "good enough" to just use an approximation for $Q_1$'s $\beta$ and $V_{CE}$ and go with that rather than turn the whole thing into a mindless rat's nest for Spice.

(Simplify as much, but no more than, as necessary.)

A way out is that we know $Q_1$'s base current must be just at the point where it is entering its active region and its $V_{CE}$ is widening at an increasing rate. This will be right at the point where at its operating point $I_C=\beta\cdot I_B$, of course.

(Note: But BJTs vary a bit and we should probably account for that variation and look at best and worst cases here. [I invite you to try the earlier equations with smaller and larger $\beta$ values and see how the results vary.])

So we want to convert the above equation from looking at voltages to looking at currents, instead. Changing it so that are instead asking, "When is the moment when this required base current now exceeds the available base current?" Since $Q_2$'s collector is driving that process, we need a way to compute that side of the equation. It's not too hard to do, luckily. So here is what I came up with:

$$\frac{I_L}{^{Q_1}\!\beta}\ge\\\\ \frac{V_{IN}-R_1\cdot I_L - ^{Q_1}\!\!V_{BE}}{R_3}-^{Q_2}\!\!\beta\cdot\left(\frac{R_1\cdot I_L-^{Q_2}\!\!V_{BE}}{R_2}+\frac{R_1\cdot I_L-^{Q_2}\!\!V_{BE}+^{Q_1}\!\!V_{CE_{SAT}}}{R_4}\right)$$

The first term on the left side represents the total current through $R_3$, with the numerator being the assumed $V_B$ right at the point where $^{Q_1}\!V_{BE}$ is the value just at the point where $Q_1$ enters its active region. This will, of course, include the drop across $R_1$ (approximated by assuming that at this time $I_L$ is the only remaining important current causing that drop) and also the $V_{BE}$ drop for $Q_1$.

The second term on the left side represents the collector current for $Q_2$ that is in the process of supplanting $Q_1$'s base current. This is just the base current times $Q_2$'s $\beta$ value and includes the two currents: one from $R_2$ and one from $R_4$. That last term subtracts away from the total current in order to answer the question about what current remains for $Q_1$'s base.

It's just algebra after that.

• @BertVano I'm glad it may help some and I'm glad you noticed the addition. I hope you plugged in your $27\:\textrm{V}$ supply to come up with your own results. By the way, I thought it was a good question. The remaining answers were either hand-waving as "hard" (but I demonstrate can be done) or else focused on the turn-around and where it doesn't actually address the foldback behavior itself (which I show happens later and for different reasons.)
– jonk
May 2 '17 at 15:15
• @jonk Can you tell how did you derive the formula for ILfoldback. In my country very similar circuit is used as a fuse with "latch" obrazki.elektroda.pl/6621038800_1368118663.gif
– G36
May 2 '17 at 18:30
• @G36 That schematic is nice to see. Almost makes me want to add an analysis for it. Almost! It's different enough to be interesting, though. I think you should definitely attempt it. But sure, I suppose I could add that analysis here. You'll have to adapt the ideas to the new circumstances, though.
– jonk
May 2 '17 at 19:06
• @G36 I added the commentary at the bottom. I have more notes, where I stay with the Shockley equation and solve it that way. It's rather more involved and doesn't much improve the results. I ran a few tests using Spice and found that spice folded back within about 2-3% of the provided predictions and figured that meant it was "close enough." The added equations brought it closer, of course. But it was ugly.
– jonk
May 2 '17 at 20:24
• @jonk Thank you for adding this additional information.
– G36
May 3 '17 at 8:00

I think it will be very hard to find such a general equation for this circuit. Because the current through R2 will depend on Q1 Vce voltage and the transistors current gain. So try to use some simulation software.

Current limit is usually VBE / 1ohm or 700mA.

As VIN-VOUT increases, more current flows through R2 and R4. This generates a voltage over R2 that offsets the voltage limit threshold.

Assuming very high beta for the BJT, then we can assume that the Vbe of the Q2 is ~0.7V when current limit is active. This allows us to write the equation:

IOUT*1ohm = 0.7 - 561/9661*(VIN-VOUT)

If your supply is 27V as stated in the comments, then the current limit may prevent you from starting up from VOUT = 0V!

This is a simulation about two kinds of the circuit.The different is base resistor of Q1 .Above is connected to Vout, and below is connected to ground. In the simulation,the current limit is different. @jonk


text:
Surface area and total area

1. Integral0

56
What's the difference b/w surface area and total area? For example, the surface area of a cylinder is A = 2 pi rh while the total area of a cylinder is A = 2 pi rh x 2 pi r^2.

Thanks for the help =)

2. HallsofIvy

40,372
Staff Emeritus
There is no difference between "total area" and "surface area" and your assertion that the "surface area" of a cylinder is 2pi rh is incorrect. That is the "lateral surface area". The ends of a cylinder are as much "surface" as the lateral surface.

3. Integral0

56
RE

not according to my math book.

4. HBar

68
I guess by "surface area" your math book, as HallsofIvy said, means lateral surface area and by "total surface area" it means surface area. And this is not correct:
it should be: A=2rπh + 2πr2. Why would you multiply the lateral surface area by the area of the top and bottom?

It would be pretty ridiculous if your math book called lateral surface area, "surface area" because no one will know what the heck you mean if you call it that.

-HBar

5. Integral0

56
What does n represent in your areas?

6. Mr. Robin Parsons

It isn't N it is pi, done like this, & pi ; (minus the spaces) go back to General math, (the thread listing) and at the top you too can find "making math symbols" π

7. Integral0

56
RE

According to my SAT math book ->

"The surface area, A, of the side of the cylinder is the circumference of the circular base times the height: A = 2nrh. The area of the top and bottom are each nr^2, so the total area of a can is given by the formula A = 2nrh + 2nr^2"

So if you are eager to refute this, go right ahead! I am not mad or attempting to fight your opinions, instead, I am trying to see who is right and who is wrong . . . so -> I can do the problems right!

So plz, tell me what's wrong with the statement above . . . its exactly what the book says about total area and surface area.

If the author is wrong . . . talk to Sharon Weiner Green and Ira K. Wolf from Barron's.

8. Integral0

56
oops!

I just noticed from HBAR a mistake that I had made when typing in the total area (I didn't mean to multiply 2nrh x 2nr^2, in reality, I meant to put a positive sign.

Thanks HBAR

9. Hurkyl

16,090
Staff Emeritus
Emphasis mine.

10. HallsofIvy

40,372
Staff Emeritus
No, this is completely correct, but it is NOT what you initially said. You said " For example, the surface area of a cylinder is A = 2 pi rh while the total area of a cylinder is A = 2 pi rh x 2 pi r^2."

Your book says "the surface area of the SIDE of the cylinder" which is not the same as the "surface area of a cylinder".

Mathematics is very, very precise.

(I have friends who remark on how "anal" mathematicians are. Of course,they are generalizing from a very small sample!)

11. quantum

34
Depends on the context of the question... The surface area may be the area of one side of a cube, whereas the net area may be the sum of the four sides of the cube.

Staff: Mentor

How many sides on a cube?

Sorry, couldn't let it go following a post about how precise math is/has to be.


text:
# What is the discriminant of 7x^2+8x+1=0 and what does that mean?

Jul 25, 2015

Solve $y = 7 {x}^{2} + 8 x + 1 = 0$

Answer: -1 and $- \frac{1}{7}$

#### Explanation:

$D = {d}^{2} = {b}^{2} - 4 a c = 64 - 56 = 8 > 0$. This means there are 2 real roots (2 x-intercepts).
In this case (a - b + c = 0) we'd better use the shortcut --> two real roots: -1 and $\left(- \frac{c}{a} = - \frac{1}{7}\right)$
REMINDER OF SHORTCUT
When a + b + c = 0 -> 2 real roots: 1 and c/a
When a - b + c = 0 -> 2 real roots: -1 and -c/a


text:
# The Pay-for-Performance Myth

July 28, 2014
By

(This article was first published on Freakonometrics » R-english, and kindly contributed to R-bloggers)

Last week, Eric Chemi and Ariana Giorgi published an interesting article on “The Pay-for-Performance Myth

With all the public chatter about exorbitant executive compensation and income inequality, it’s useful to look at the relationship between chief executive officer pay and corporate performance. Typically, when the subject of their big pay packages arises, CEOs—usually through their spokespeople—say they are paid for performance. Does data back that up?

An analysis of compensation data publicly released by Equilar shows little correlation between CEO pay and company performance. Equilar ranked the salaries of 200 highly paid CEOs. When compared to metrics such as revenue, profitability, and stock return, the scattering of data looks pretty random, as though performance doesn’t matter. The comparison makes it look as if there is zero relationship between pay and performance.

In the article, they produce a copula-type plot (since ranks – only – are considered). Ariana kindly sent me the dataset (that was used in The Pay at the Top) to play with it

> base=read.table("ceo.csv",sep=";",header=TRUE)

Here I normalize (dividing by the size of the dataset) to have uniform distribution on the unit interval (instead of working with ranks, i.e. integers). If we remove that scaling factor, the scatterplot is that same as the one mentioned in  the Pay-for-performance myth.

> n=nrow(base)
> U=rank(base[,1])/(n+1)
> V=rank(base[,2])/(n+1)
> plot(U,V,xlab="Rank CEO Pay",
+ ylab="Rank Stock Return")

This is the copula type representation.

If we visualize the density of the copula (using the algorithm described in the joint paper with Gery and Davy), we get either

> library("copula")
> library("ks")
> library("MASS")
> library("locfit")
> n.res=32
> ctilde1=probtranscopkde(UVs,p=1,
+ u.out=seq(1/(2*n.res+1),1-1/(2*n.res+1),
+length=n.res),plots=TRUE)

or

> ctilde2=probtranscopkde(UVs,p=2,
+ u.out=seq(1/(2*n.res+1),1-1/(2*n.res+1),
+length=n.res),plots=TRUE)

(depending on one parameter used in our function). It looks like we can consider that the underlying copula is the independent copula (both nonparametric estimators of the density of the copula are – relatively – flat, constant, equal to one).

Since we work with copulas, a lot of things could be done. For instance, we can plot Kendall’s function

> X=cbind(U,V)
> i=rep(1:n,each=n)
> j=rep(1:n,n)
> S=((X[i,1]>X[j,1])&(X[i,2]>X[j,2]))
> Z=tapply(S,i,sum)/(n-1)
> plot(sort(Z),(1:n)/n,type="s",col="blue",lwd=2)

The line here is very close to the one we should have with independent variables (the red line). Actually, if we use monte carlo simulations to visualize a confidence interval (for a sample of size 200), it might be reasonable to assume that both variables are independent.

> M=NULL
> for(i in 1:nsim){
+ U=1:n
+ set.seed(i)
+ V=sample(1:n,size=n)
+ X=cbind(U,V)
+ i=rep(1:n,each=n)
+ j=rep(1:n,n)
+ S=((X[i,1]>X[j,1])&(X[i,2]>X[j,2]))
+ Z=tapply(S,i,sum)/(n-1)
+ lines(sort(Z),(1:n)/n,type="s",col="light blue")
+ M=cbind(M,sort(Z))
+ }
> Q1=apply(M,1,function(x) quantile(x,.05))
> Q2=apply(M,1,function(x) quantile(x,.95))
> lines(Q1,(1:n)/n,col="blue")
> lines(Q2,(1:n)/n,col="blue")

This can be confirmed by more formal tests, one based on the (rank) correlation

> cor.test(U,V)

Pearson's product-moment correlation

data:  U and V
t = 1.1178, df = 198, p-value = 0.265
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
-0.06021553  0.21555981
sample estimates:
cor
0.07918704

and one based on some contingency table if we  discretize (the chi-square test) our variables

> Uc=cut(U,seq(0,1,by=.1))
> Vc=cut(V,seq(0,1,by=.1))
> chisq.test(Uc,Vc)

Pearson's Chi-squared test

data:  Uc and Vc
X-squared = 90, df = 81, p-value = 0.2313

Warning message:
In chisq.test(Uc, Vc) : Chi-squared approximation may be incorrect

It is also possible to look at various functions to get a better understanding of what is happening in the four corners (inspired by what was done in tails of copulas)

> z=seq(0,1,by=.001)
> L00=L10=L01=L11=rep(NA,length(z))
> for(i in 1:length(z)){
+ L00[i]=sum((U<=z[i])&(V<=z[i]))/sum(U<=z[i])
+ L10[i]=sum((U>1-z[i])&(V<=z[i]))/sum(U<=z[i])
+ L01[i]=sum((U<=z[i])&(V>1-z[i]))/sum(U<=z[i])
+ L11[i]=sum((U>1-z[i])&(V>1-z[i]))/sum(U<=z[i])
+ }
> par(mfrow=c(2,2))
> plot(z,L01,type="l",xlab="",ylab="")
> segments(0,0,1,1,lty=2,col="red")
> plot(z,L11,type="l",xlab="",ylab="")
> segments(0,0,1,1,lty=2,col="red")
> plot(z,L00,type="l",xlab="",ylab="")
> segments(0,0,1,1,lty=2,col="red")
> plot(z,L10,type="l",xlab="",ylab="")
> segments(0,0,1,1,lty=2,col="red")

The four corner are respectively low pay, high return – high pay, high return – low pay, low return – high pay, low return,,

Again, the red line is the independent copula. On those graph, we can see that the null correlation is not only valid on average but also in corner : very low pays (relatively to other CEOs) is not particularly associated to low stock returns, and very high pays is not associated with very high returns.

So here again, we can confirm that this “pay for performance” principle is a myth.