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# Vectors  (Page 2/10)

 Page 2 / 10

## Bearing

The final method of expressing direction is to use a bearing . A bearing is a direction relative to a fixed point.

Given just an angle, the convention is to define the angle with respect to the North. So, a vector with a direction of $110{}^{\circ }$ has been rotated clockwise $110{}^{\circ }$ relative to the North. A bearing is always written as a three digit number, for example $275{}^{\circ }$ or $080{}^{\circ }$ (for $80{}^{\circ }$ ).

## Scalars and vectors

1. Classify the following quantities as scalars or vectors:
1. 12 km
2. 1 m south
3. $2\phantom{\rule{2pt}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}$ , $45{}^{\circ }$
4. $075{}^{\circ }$ , 2 cm
5. $100\phantom{\rule{2pt}{0ex}}\mathrm{k}·{\mathrm{h}}^{-1}$ , $0{}^{\circ }$
2. Use two different notations to write down the direction of the vector in each of the following diagrams:

## Drawing vectors

In order to draw a vector accurately we must specify a scale and include a reference direction in the diagram. A scale allows us totranslate the length of the arrow into the vector's magnitude. For instance if one chose a scale of 1 cm = 2 N (1 cm represents 2 N), aforce of 20 N towards the East would be represented as an arrow 10 cm long. A reference direction may be a line representing a horizontal surface or the points of a compass.

Method: Drawing Vectors

1. Decide upon a scale and write it down.
2. Determine the length of the arrow representing the vector, by using the scale.
3. Draw the vector as an arrow. Make sure that you fill in the arrow head.
4. Fill in the magnitude of the vector.

Represent the following vector quantities:

1. $6\phantom{\rule{2pt}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}$ north
2. 16 m east
1. $1\phantom{\rule{2pt}{0ex}}\mathrm{cm}=2\phantom{\rule{2pt}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}$
2. $1\phantom{\rule{2pt}{0ex}}\mathrm{cm}=4\phantom{\rule{2pt}{0ex}}\mathrm{m}$
1. If $1\phantom{\rule{2pt}{0ex}}\mathrm{cm}=2\phantom{\rule{2pt}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}$ , then $6\phantom{\rule{2pt}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}=3\phantom{\rule{2pt}{0ex}}\mathrm{cm}$
2. If $1\phantom{\rule{2pt}{0ex}}\mathrm{cm}=4\phantom{\rule{2pt}{0ex}}\mathrm{m}$ , then $16\phantom{\rule{2pt}{0ex}}\mathrm{m}=4\phantom{\rule{2pt}{0ex}}\mathrm{cm}$
1. Scale used: $1\phantom{\rule{2pt}{0ex}}\mathrm{cm}=2\phantom{\rule{2pt}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}$ Direction = North
2. Scale used: $1\phantom{\rule{2pt}{0ex}}\mathrm{cm}=4\phantom{\rule{2pt}{0ex}}\mathrm{m}$ Direction = East

## Drawing vectors

Draw each of the following vectors to scale. Indicate the scale that you have used:

1. 12 km south
2. 1,5 m N $45{}^{\circ }$ W
3. $1\phantom{\rule{2pt}{0ex}}\mathrm{m}·\mathrm{s}{}^{-1}$ , $20{}^{\circ }$ East of North
4. $50\phantom{\rule{2pt}{0ex}}\mathrm{km}·\mathrm{h}{}^{-1}$ , $085{}^{\circ }$
5. 5 mm, $225{}^{\circ }$

## Mathematical properties of vectors

Vectors are mathematical objects and we need to understand the mathematical properties of vectors, like adding and subtracting.

For all the examples in this section, we will use displacement as our vector quantity. Displacement was discussed in Grade 10.

Displacement is defined as the distance together with direction of the straight line joining a final point to an initial point.

Remember that displacement is just one example of a vector. We could just as well have decided to use forces or velocities to illustrate the properties of vectors.

When vectors are added, we need to add both a magnitude and a direction. For example, take 2 steps in the forward direction, stop and then take another 3 steps in the forward direction. The first 2 steps is a displacement vector and the second 3 steps is also a displacement vector. If we did not stop after the first 2 steps, we would have taken 5 steps in the forward direction in total. Therefore, if we add the displacement vectors for 2 steps and 3 steps, we should get a total of 5 steps in the forward direction. Graphically, this can be seen by first following the first vector two steps forward and then following the second one three steps forward (ie. in the same direction):

where we get a research paper on Nano chemistry....?
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
how did you get the value of 2000N.What calculations are needed to arrive at it
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# If 4l - 3m = 15 and lm = 10, then find the value of 16l^2 + 9m^2?

Jul 7, 2016

$l = \frac{10}{m}$

$4 \left(\frac{10}{m}\right) - 3 m = 15$

$\frac{40}{m} - 3 m = 15$

$40 - 3 {m}^{2} = 15 m$

$0 = 3 {m}^{2} + 15 m - 40$

$m = \frac{- 15 \pm \sqrt{{15}^{2} - \left(4 \times 3 \times - 40\right)}}{2 \times 3}$

$m = \frac{- 15 \pm \sqrt{705}}{6}$

$l = \frac{10}{\frac{- 15 \pm \sqrt{705}}{6}}$

$l = \frac{60}{- 15 \pm \sqrt{705}}$

Now, we can do our calculation:

$16 {\left(\frac{60}{- 15 \pm \sqrt{705}}\right)}^{2} + 9 {\left(\frac{- 15 \pm \sqrt{705}}{6}\right)}^{2}$

I will leave the rest of the simplification up to you.

Hopefully this helps!


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undesirable/buggy change in behavior of Limit in Mma 11.2?

The behavior of Limit changed in Mma 11.2, and I think it is in some ways undesirable or possibly buggy. Consider

Limit[Sin[t Pi], t -> Infinity, Assumptions -> Element[t, Integers]]

This used to (11.1) produce a correct result and no longer does (11.3). Is this intentional or a bug? (More understandable but still slightly disappointing is that Limit no longer returns limiting sets of values. E.g., removing the assumptions from the above expression, it used to return an interval, while it now returns the less informative Indeterminate.)

I am aware than I can change to DiscreteLimit, which is a nice addition, or use other tricks (e.g., Sum of differences).

• v11.2 on my Mac also evaluates to 0 so although DiscreteLimit was introduced with v11.2, the change observed was not fully implemented until v11.3 – Bob Hanlon Aug 31 '18 at 14:36

As the documentation told me, the interval arithmetic can be restored with

Limit[Sin[x Pi], x -> ∞, Method -> {"AllowIndeterminateOutput" -> False}]

Interval[{-1, 1}]

I guess that this change was made for performance reasons, speeding up Limit with the default options.

• While this is useful for my subsidiary question, it does not address my main question. Is it a bug to apparently ignore the Assumptions? – Alan Sep 1 '18 at 13:27
• Well, I know. There a pros and cons for calling that a bug. I am rather indifferent in that matter. – Henrik Schumacher Sep 1 '18 at 13:32
• What are the cons? Is there a motivation in the changes for the new behavior, which unlike the old behavior fails to appropriately use this assumption? (I'm not seeing it.) Just to be clear, it fails even if you change the Method as above. – Alan Sep 1 '18 at 18:43
• I submitted a report and received this answer: "The assumptions in Limit and related functions are only supposed to apply to parameters and not the limit variable." – Alan Sep 7 '18 at 18:44


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### Author Topic: [Bug] v111.0  (Read 3402 times)

0 Members and 1 Guest are viewing this topic.

#### transporter

• Posts: 136
##### [Bug] v111.0
« on: January 02, 2012, 02:35:13 AM »
Found a crash for the game. No error box popped up and all the Terminal program said was world destroyed after the game load-up text. Gamer version 111.0 r4911M, pak128.britain v1.1 on Intel OSX. Steps: Load up the save and pause the game. Move the view to Southampton and delete the bus stop and terminal then rebuild the same models so they read as the airport stop. Build a garage on the end of road directly right (north) of the airfield stop by the soccer field. Make a bus route between the 2 stops then move to Hartlepool and build a garage next to the closest soccer field. Make a route between the bus stops and select the garage. Assign 4 or 6 Guy Specials (crashed with both) to that line. Go back to the Southampton garage and assign 3 Guy Specials to that line. Go to the Farnham airport and assign 1 BAC 111-500 for both routes 1 and 2. Then fast-forward and it should crash before the next year. I have no idea what combination or single event makes it crash. I tried to see if it was the save itself by fast-forwarding but got past the timeline for the crash so that wasn't it. It seemed to have crashed the whole program I think, because it caused the Apple Crash Report to open.

#### prissi

• Developer
• Posts: 10135
• Languages: De,EN,JP
##### Re: [Bug] v111.0
« Reply #1 on: January 02, 2012, 09:42:12 AM »
Please try this again in a nightly as several cause for such crashes might have been removed in the meantime. You will find the nightlies here: http://nightly.simutrans-germany.com/

#### transporter

• Posts: 136
##### Re: [Bug] v111.0
« Reply #2 on: January 02, 2012, 05:51:32 PM »
Same crash, replicated exactly as above with r5070. With the length of time before crash, I'm pretty sure it has to do with the monthly bankruptcy report. I've tried with only aircraft and an account of -$1M in the hole and crashes before it reaches February #### Dwachs • DevTeam, Coder/patcher • Administrator • Posts: 4739 • Languages: EN, DE, AT ##### Re: [Bug] v111.0 « Reply #3 on: January 02, 2012, 06:33:38 PM » still could not reproduce it. Can you upload the savegame right before the crash? And can you start simutrans with "-log 1 -debug 3" and post the simu.log file after the crash? #### prissi • Developer • Administrator • Posts: 10135 • Languages: De,EN,JP ##### Re: [Bug] v111.0 « Reply #4 on: January 02, 2012, 10:44:18 PM » Maybe a wrong translation. What langugae did you set it too? #### transporter • Posts: 136 ##### Re: [Bug] v111.0 « Reply #5 on: January 03, 2012, 02:51:37 AM » It's in English. It's been that the whole time. Also, I opened up a completely new game to see if I get a bankruptcy report at all, ran up a debt of -$2M and no report at all.

Here's the savegame for the original scenario posted. Just let it run and the crash happens (for me) before 1:00h in February. https://docs.google.com/open?id=0B4x-6whHppW7NTcwODYwZTctNTU4Mi00ZmZlLTllYjQtOTZmOTk5NTQzNjcz

Remind me how to run the debug code. I did it once before, but forgot how

#### prissi

• Developer
• Posts: 10135
• Languages: De,EN,JP
##### Re: [Bug] v111.0
« Reply #6 on: January 03, 2012, 10:27:17 AM »
at the command line add "-debug 3 -log 1"

#### Dwachs

• DevTeam, Coder/patcher
• Posts: 4739
• Languages: EN, DE, AT
##### Re: [Bug] v111.0
« Reply #7 on: January 03, 2012, 11:39:46 AM »
Could not reproduce it, also valgrind did not show any suspicious things :(

#### transporter

• Posts: 136
##### Re: [Bug] v111.0
« Reply #8 on: January 03, 2012, 04:42:52 PM »
Hmmm....I'll start another scenario with a similar setup using airports to connect the cities and see if it will crash again. It could just be a isolated event for that specific game. Do you at least get a bankruptcy report?

#### transporter

• Posts: 136
##### Re: [Bug] v111.0
« Reply #9 on: January 04, 2012, 05:17:12 PM »
I just tried with pak128 and the same thing happens. If I start a new map with airports and aircraft, the game crashes after a month of running. This time I had a positive account though

#### prissi

• Developer
• Posts: 10135
• Languages: De,EN,JP
##### Re: [Bug] v111.0
« Reply #10 on: January 04, 2012, 08:42:32 PM »
Looks like you translations are broken. Either reinstall the executable from sourceforge or download new tranlations from simuranslator.

#### transporter

• Posts: 136
##### Re: [Bug] v111.0
« Reply #11 on: January 04, 2012, 11:20:03 PM »
Downloaded translations for the executable and the paks, downloaded a new nightly, pakset, and the package from sourceforge and still get the same crash.

#### Milko

• Devotee
• Posts: 868
##### Re: [Bug] v111.0
« Reply #12 on: January 05, 2012, 08:23:13 AM »
Hello

Have you tried deleting the settings.xml file?

In windows the file is in documents\simutrans folder.

Giuseppe

#### transporter

• Posts: 136
##### Re: [Bug] v111.0
« Reply #13 on: January 05, 2012, 07:36:34 PM »
I found a settings.txt file in the simutrans\text\en, and I run a Mac. I also looked in the library folder and didn't find that file.

#### Milko

• Devotee
• Posts: 868
##### Re: [Bug] v111.0
« Reply #14 on: January 07, 2012, 04:38:10 PM »
Hello

It is a file "xml", contains the configuration of simutrans. I do not know where he is in the mac ...

Maybe someone who uses the mac can help.

Giuseppe

#### transporter

• Posts: 136
##### Re: [Bug] v111.0
« Reply #15 on: January 08, 2012, 05:52:12 AM »
There's a settings.xml in the Library folder, and the simuconf.tab in the main folder.


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## anonymous 3 years ago 4 = the square root of p, -2?

• This Question is Open
1. zepdrix

$\huge 4=\sqrt p$To solve for p, square both sides.$\huge 4^2=\sqrt{p}^2$$\huge p=4^2$Is this what you were trying to do? Solve for p? :D

2. zepdrix

os was there a minus 2 in the problem also? :O i couldn't tell based on the comma, i thought maybe that was a guess :D

3. anonymous

Yes there is a minus 2 . So does it just become p = 2^2 ?

4. zepdrix

$\huge 4=\sqrt p - 2$So this is the problem? :D

5. zepdrix

We'll add 2 to both sides, giving us:$\huge 6=\sqrt p$Then we'll do the same step we did before, square both sides, that will eliminate the square root.$\huge 6^2=\sqrt p ^2 \qquad \rightarrow \qquad 6^2=p$

6. anonymous

Thanks so much !!

7. anonymous

I understand how to do the rest of the problems now (:


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# How to test that the difference between two sample means is greater that a given value (in Matlab)

I (unfortunately) only have an elementary knowledge of statistics, and so please excuse me for my following basic question.

I have two independent samples (with sample size of 60 and 765 respectively) and I need to test if the mean of first sample is greater than the mean of second sample by a non-zero given value (the two samples have unequal variance). Since, Matlab is the tool that is used in my workplace, I need to do this in Matlab (I had looked at the "ttest2" function in Matlab, but from the help document,I could not figure out how to use ttest2 to test whether the difference between two sample means is greater than a given non-zero value).

-Tinku

• Is there a reason that you cannot subtract, the given non-zero value you want to test against, from the appropriate sample and then test if their means are equal? You do assume that the distributions are invariant under translation right? – usεr11852 Nov 20 '15 at 18:12
• Thank you for your suggestion to subtract the given non-zero value. I followed your suggestion today, but I am getting results that prima facie look wrong. I understand that if the results are indeed wrong then the error is not in subtracting the non-zero given value but in something else that I am doing wrong. I will investigate more and if I am unable to figure out where I am going wrong, I will let you know (so that, in case you have some spare time, you can look at it and kindly let me know where I am going wrong). Thanks again. – TInku Nov 22 '15 at 3:22
• Hi, I tried to investigate all that I could but I am still getting unexpected results. As there is limit on number of characters here in the Comments section, I put down the details of what I am doing in a google doc Difference between sample means If you happen to get some free time, I will be grateful if you can shed light on where I am going wrong. Many Thanks @usεr11852 – TInku Dec 2 '15 at 19:48
• i am very busy, I will probably look this up during the weekend. – usεr11852 Dec 2 '15 at 20:21
• Hi, it will help me immensely if you can have a look at the issue I am facing when you get some free time. Many thanks for your help. @usεr11852 – TInku Dec 11 '15 at 17:05

I am not a Matlab expert, but looking at the documentation for ttest2 it looks like there is not an option to change the null value (what the difference is under the null hypothesis).
Another option is that if you subtract the null value difference from each value in one set (the one you expect to be higher) then that adjusts the mean appropriately and does not change the variances or other values. So you can pass the new vectors (one with the subtraction done) to the ttest2 function and it will give you the appropriate p-value.


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# Totally disconnected space

In topology and related branches of mathematics, a totally disconnected space is a topological space that is maximally disconnected, in the sense that it has no non-trivial connected subsets. In every topological space the empty set and the one-point sets are connected; in a totally disconnected space these are the only connected subsets.

An important example of a totally disconnected space is the Cantor set. Another example, playing a key role in algebraic number theory, is the field Qp of p-adic numbers.

## Definition

A topological space X is totally disconnected if the connected components in X are the one-point sets. Analogously, a topological space X is totally path-disconnected if all path-components in X are the one-point sets.

## Examples

The following are examples of totally disconnected spaces:

## Constructing a disconnected space

Let be an arbitrary topological space. Let if and only if (where denotes the largest connected subset containing ). This is obviously an equivalence relation. Endow with the quotient topology, i.e. the coarsest topology making the map continuous. With a little bit of effort we can see that is totally disconnected. We also have the following universal property: if a continuous map to a totally disconnected space, then it uniquely factors into where is continuous.


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View Single Post
Math
Emeritus
Thanks
PF Gold
P: 39,304
 Quote by JameB Sorry, I don't mean to be rude but I'm just kinda lost. This is my first course in linear algebra and that's why I'm pretty confused. What I was trying to do was to find a basis.
I don't mean to be rude but this is your 6th post in this thread and that is the first mention of finding a basis!

 The theorem states that vectors are said to form basis for V if a) if they span V b) they are linearly independent
That's not a theorem, that is the definition of "basis". You can add a "c" to this- the number of vectors in the set is the dimension of the space. And then there is a theorem that says any two implies the third.

 I know that to prove b) I need to put it in a matrix, reduce and if I get a matrix with trivial solution meaning everything equals to 0 then it's a trivial solution and it's linearly independant.
Well, no, you don't need to do that- that's one method. What you need to do is think about the definition: a set of vectors spans a space if and only if, any vector, y, in the space can be written as a linear combination of the vectors in the set. If the set is $\{v_1, v_2, \cdot\cdot\cdot, v_n\}$ then there exist scalars, $a_1, a_2, \cdot\cdot\cdot, a_n$ such that $y= a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n$.

 But I'm having trouble with checking if they span or not... that's the whole point of this post :) The examples in my book don't show step by step solution so that's why I'm lost. e.g. Show that S = {v1, v2, v3, v4} where v1 = [1 0 0 1], v2 = [0 1 -1 2], v3 = [0 2 2 1], v4 = [1 0 0 1] is a basis for R4. (R4 is written as R subscript 4 meaning it's referring to the row spaces?...) to show that S is linearly independant, I formed the eqn a1v1 + a2v2 + a3v3 + a4v4 = 0 turned it into a augmented matrix and got its RREF. This means that it's linearly independent.
Then you have a problem- either you did the problem wrong or you gave the vectors incorrectly here. Those vector are NOT independent because $v_1$ and $v_2$ are the same: $(1)v_1+ (-1)v_2= 0$.

 NOW, to show that S spans R4, I let x = [a, b, c, d] be a vector in R4 what should I do next? thanks!
If those vectors were actually the ones you were given, you do nothing now. Since there are 4 vectors and the are NOT indpendent, they cannot span R4. If you gave the fouth vector incorrectly and the ones you were given were independent, you still would not have to do anything more- four independent vectors must span R4.

In either case, showing linear independence or spanning, you are, in effect, solving a system of equations. You are either trying to solve $a_1v_1+ a_2v_2+ a_3v_3= 0$ (for linear independence) or $a_1v_1+ a_2v_2+ a_3v_3+ a_4v_4= y$ for any y in the space (for spanning). You could do either by setting up the "augmented" matrix with the right side as the "fifth column". Of course, with all "0"s in the fifth column, no row operations will change those: $a_1= a_2= a_3= a_4= 0$ is an obvious solution. Showing that you can reduce the first four columns to the identity matrix is enough to show that is the only solution. Similarly, to show that the set spans the space you add the components of the vector y as the fifth column. You would row-reduce the matrix in exactly the same way, just doing the row operations on the fifth column on the y- components rather than "0"s. But the actual value of the coefficients is not important, only that you can find them. So the values you wind up with in the fifth column is not important, only that you can reduce the first four columns to the identity matrix- exactly what you did to show independence. They really are exactly the same thing.

When someone said you could look at the determinant of the matrix, earlier, you asked what to do if the matrix is not square. In that case, there is not work to do- such a set of vectors cannot be a basis (though it may still span the space or be independent). The number of columns of the matrix, the number of components in each vector, is the dimension of the vector space. The number of rows is the number of vectors in the set. And those must be equal in order to have a basis. If there are fewer vectors than the dimension- if there are more columns than row- they might be independent but cannot span the space. If there are more vectors than the dimension- if there are more rows than columns, they might span the space but cannot be independent.


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# how to reproduce a matching network of a vendor

I'm trying to create a board using the BGT24MTR chip from infinion. I noticed that they said that there is a Tx/Txx impedance.

To compensate for this impedance they created a matching network.

Since I'm trying to use a different board stack up, I have to redesign this matching system. I tried calculating the impedance of each of the lengths

Here's my breakdown of the major components.

Width Length  Ω     diff Ω   diff Ω single
.3    .55     68.5     82     61
1.15  1.6     31.19    22.2   16.5
.5    ∞       53

I thought that I was trying to use the smith chart to get the impedance to be purely real around 5/3 after the second rotation so that when it is renormalized it would be purely in the unit circle in the center, but every way I looked at using the smith chart led to results that are far away from being matched. Any suggestions on what they are doing to match this?

P.S. The new board stack up is FR408 1.6mm thick. 1oz copper on both sides (1.4mil,35um). Frequency is 24 GHz

• Can you show an example of the Smith chart you made? – The Photon Apr 25 '16 at 21:14
• @ThePhoton added the smith chart – Legen Diary Apr 25 '16 at 22:14

Your use of the smith chart is correct, but your numbers are a little off due to the wavelength you used (6.5mm).

In stripline, the fields are completely contained inside the substrate. The fields are partially in air around a microstrip line, resulting in an effective permittivity which is lower than the relative permittivity of the substrate. Wavelength can be calculated using the effective permittivity or the delay (Tpd). Both are available from your impedance calculator. Er effective is 3.0635. Tpd is 5.8e-9 s/m.

$Wavelength = \frac{c}{\sqrt{E_{r}} \cdot f} = \frac{1}{T_{pd} \cdot f}$
$Wavelength(w/ E_{r}) = \frac{c}{\sqrt{E_{r}} \cdot f} = 6.5 mm$
$Wavelength(w/ E_{eff}) = \frac{c}{\sqrt{E_{eff}} \cdot f} = 7.1 mm$
$Wavelength(w/ T_{pd}) = \frac{1}{T_{pd} \cdot f} = 7.1 mm$

The difference between 6.5mm and 7.1mm may seem small, but check out the difference in return loss.

$Z_{out}(\Lambda:6.5mm) = 33-j24 \Omega$ >>>> $ORL=9dB$
$Z_{out}(\Lambda:7.1mm) = 46-j14 \Omega$ >>>> $ORL=16dB$

For a sanity check, match from the load to the conjugate of the generator. In other words, start at a load of $50\Omega$ and go towards the chip. You'll end up around $20.8+j20.2\Omega$. This is typical for most output stages, since they're usually low resistance and capacitive.

Your method of matching will provide a decent ballpark, but I'd like to point out a few important sources of error.

1. Tx & Txx are intended to be 100ohm differential signals. The small spacing between them produces an additional shunt capacitance. You need to include the extra capacitance, or use a coupled line model.
2. The datasheet doesn't say how they measured the Tx/Txx load impedances. Did they measure at the PCB output port and try to de-embed everything up to the chip? I'm skeptical of the accuracy without knowing where they came from. Additionally, I think these are the impedances looking into the Tx/Txx ports. I would call the load impedance the conjugate of the printed values, or the impedance presented to the Tx/Txx ports. It's all unclear.
3. The package pad (footprint) will have a different impact on a different board stackup.
4. I don't see anything to cancel mismatch from the launchers. The launchers will perform differently on a different board stackup.

The matching network they used is simple and probably low loss. I guess you could call it a stepped-impedance match. I don't know. It's a plus to avoid shunt components or stubs, and the high Q resonances they might produce.

• A load impedance of 100 ohm is clearly stated in User's Guide to BGT24MTR11. " It is a differential output signal with a load impedance of 100 ohm, given that the off-chip compensation structures, shown in the data sheet, are in place. ". How could OP's use of smith chart be correct? – ivan Jun 1 '16 at 20:55
• Are you referring to the OP using single-ended 50ohm, instead of 100ohm differential? – curtis Jun 1 '16 at 22:14
• Disregard my previous comment, you are probably right. – ivan Jun 1 '16 at 23:24

Since you have a two-sided FR4 you won't be able to reproduce their matching network. What you can do is to use a microwave design software.

I suggest awr microwave office or agilent ads. There you should create two ports for TX and TXX and assign their respective impedances. You should also create a load that you want to use.

Then use microstrip libraries to create a matching network (there's plenty of examples in both programs).

Create S21 graph and tune microstrip' width and length manually or use a built-in optimizer (set constraints and goals first).

• Thanks for the answer. I'm curious whether you have any suggestions as to why those designers chose that matching structure? Was my intuition correct or were they pursuing a different goal? – Legen Diary Apr 26 '16 at 20:53
• They chose that matching because it's the simplest one. What you was trying to calculate is called characteristic impedance of a transmission line (and you calculated it wrong - check conductor height). What you need to know is a load that you are going to use (e.g. antenna, PA etc.). – ivan Apr 26 '16 at 21:53
• Your Smith chart attempt is valid with lumped elements but these guys use microstrips. There is lumped-element matching and distributed-element matching (e.g. microstrip) which is a completely other thing. – ivan Apr 26 '16 at 21:59
• I think you are misunderstanding what I'm doing. First the main thrust of the question is to determine what they were doing because stub matching is very well documented. Second I wasn't adding lumped elements with the smith chart, but was moving away from the load and to the generator (in retrospect that is something I'm going to try, moving from the generator to the load), which is why I was rotating around a circle based upon the circle at the origin. – Legen Diary Apr 26 '16 at 23:19
• @LegenDiary, please look at page 9 paragraph 4.1 of this document link – ivan Apr 27 '16 at 5:42
1. Compensation structure in datasheet RO4350 0.254mm Estimating in QUCS / tools->line calculation (http://qucs.sourceforge.net/):

w=0.3 l=0.55 -> Z0 = 68 Ohm; Angle = 25 Deg ≈ 0.7λ
w=1.1 l=1.65 -> Z0 = 31 Ohm; Angle = 83 Deg ≈ 0.23λ

1. Tuning smith chart software (http://fritz.dellsperger.net/smith.html): Effective lengths are longer. Wide section is probably quarterwave transformer, so i used 0.25λ:
Thin section is shortened. In this particular case it is better to use EM simulation software, I think Sonnet Lite is capable of doing that.

2. If you are going to use FR4 material at 24 GHz, 1.6mm is too thick. It will result in extremely wide lines, and they may act not as expected. I recommend to use 0.6mm, maximum 1.0mm.

3. Here is my compensation structure for BGT24MTR12:

It differs from original compensation structure. Thin section is longer and act as half-wavelength impedance repeater. It will have worse bandwidth than original RO4350 compensation structure.

Check new bgt24ltr11, although it may be more difficult to solder (UPDATE 2020: bgt24ltr is not 50 Ohm matched, TX/RX port impedance is around 100 Ohm and may have some little complex part. Datasheet/appnote provides matching structure to 50Ohm).


text:
10.7 The projection matrix

Denote by $$\bm{\hat{y}}_h$$ a set of $$h$$-step-ahead base forecasts generated for each series in a hierarchical or grouped structure and stacked the same way as the data. For example for the hierarchy of Figure 10.1 $\bm{\hat{y}}_h=\begin{bmatrix} \hat{y}_h \\ \yhat{A}{h} \\ \yhat{B}{h} \\ \yhat{AA}{h} \\ \yhat{AB}{h} \\ \yhat{AC}{h} \\ \yhat{BA}{h} \\ \yhat{BB}{h} \\ \end{bmatrix}.$

In general, all forecasting approaches for either hierarchical or grouped structures can be represented as $$$\bm{\tilde{y}}_h=\bm{S}\bm{P}\bm{\hat{y}}_h \tag{10.6}$$$

where reading from right to left, $$\bm{\hat{y}}_h$$ is the set of $$h$$-step-ahead base forecasts as defined above, $$\bm{P}$$ is a matrix that projects the base forecasts into the bottom-level, and the summing matrix $$\bm{S}$$ sums these up using the aggregation structure to produce a set of coherent forecasts $$\bm{\tilde{y}}_h$$.

The $$\bm{P}$$ matrix is defined according to the approach implemented. For example if the bottom-up approach is used to forecast the hierarchy of Figure 10.1, $\bm{P}= \begin{bmatrix} 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\ \end{bmatrix}.$ Notice that the $$\bm{P}$$ comprises two partitions. The first three columns which zero out the base forecasts of the series above the bottom-level and the $$m$$-dimensional identity matrix which picks only the base forecasts of the bottom-level, to then be summed up the hierarchy by the $$\bm{S}$$ matrix.

If any of the top-down approaches were used then $\bm{P}= \begin{bmatrix} p_1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ p_2 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ p_3 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ p_4 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ p_5 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ \end{bmatrix}.$ The first column includes a set of proportions that distribute the base forecasts of the top-level to the bottom-level. These are then summed up the hierarchy by the $$\bm{S}$$ matrix. The rest of the columns zero out the base forecasts below the very top-level of aggregation.

For a middle out approach the $$\bm{P}$$ matrix will be a combination of the above two. Using a set of proportions, the base forecasts of some pre-chosen level will be disaggregated to the bottom-level to then be summed up the hierarchy with the summing matrix, with all other base forecasts being zeroed out.


text:
Asymmetric norms given by symmetrisation and specialisation order.(English)Zbl 1394.54016

A function $$d:X\times X\to [0,\infty)$$ of a set $$X$$ is called a $$T_0$$-quasi-metric if the following conditions hold for all $$x,y,z\in X$$: $d(x,x)=0,$
$d(x,z)\leq d(x,y)+d(y,z),$
$d(x,y)=0=d(y,x) \text{ implies that }x=y.$
The authors in this paper continue the investigation between $$T_0$$-quasi-metric spaces and partially ordered metric spaces. In Section $$2$$, they show that it is possible to set up a Galois connection between these two classes of spaces (see Proposition $$2.2(c)$$). Furthermore, in Section $$3$$, the authors establish a relationship between nonexpansive maps between $$T_0$$-quasi-metric spaces and nonexpansive and increasing maps between partially ordered metric spaces (see Propositions $$3.1$$ and $$3.2$$).
Finally, in the last section (Section $$4$$), they derive a representation theorem for injective asymmetrically normed spaces. This answers a question that was left open in [J. Conradie et al., Topology Appl. 231, 92–112 (2017; Zbl 1387.46050)].

MSC:

 54E35 Metric spaces, metrizability 46M10 Projective and injective objects in functional analysis 46B40 Ordered normed spaces 54F05 Linearly ordered topological spaces, generalized ordered spaces, and partially ordered spaces 06A15 Galois correspondences, closure operators (in relation to ordered sets)

Zbl 1387.46050
Full Text:

References:

 [1] Alegre, C.; Ferrando, I.; García-Raffi, L. M.; Sánchez Pérez, E. A., Compactness in asymmetric normed spaces, Topol. Appl., 155, 527-539, (2008) · Zbl 1142.46004 [2] Aliprantis, C. D.; Tourky, R., Cones and duality, Graduate Studies in Mathematics, vol. 84, (2007), American Mathematical Society Providence · Zbl 1127.46002 [3] Birkhoff, G., Lattice theory, Amer. Math. Soc. Colloq. Publ., vol. 25, (1967), American Mathematical Society Providence, RI · Zbl 0126.03801 [4] Conradie, J. J., Asymmetric norms, cones and partial orders, Topol. Appl., 193, 100-115, (2015) · Zbl 1344.46003 [5] Conradie, J. J.; Künzi, H-P. A.; Otafudu, O. O., The vector lattice structure on the isbell-convex hull of an asymmetrically normed real vector space, Topol. Appl., 231, 92-112, (2017) · Zbl 1387.46050 [6] Davie, B. A.; Priestley, H. A., Introduction to lattices and order, (2002), Cambridge University Press Cambridge · Zbl 1002.06001 [7] Gaba, Y. U.; Künzi, H-P. A., Splitting metrics by $$T_0$$-quasi-metrics, Topol. Appl., 193, 84-96, (2015) · Zbl 1345.54023 [8] Gaba, Y. U.; Künzi, H-P. A., Partially ordered metric spaces produced by $$T_0$$-quasi-metrics, Topol. Appl., 202, 366-383, (2016) · Zbl 1337.54022 [9] Kadison, R. V.; Ringrose, J. R., Fundamentals of the theory of operator algebras, vol. 1, (1983), Academic Press New York · Zbl 0518.46046 [10] Kemajou, E.; Künzi, H.-P. A.; Otafudu, O. O., The isbell-hull of a di-space, Topol. Appl., 159, 2463-2475, (2012) · Zbl 1245.54023 [11] Meyer-Nieberg, P., Banach lattices, (1991), Springer-Verlag Berlin · Zbl 0743.46015 [12] García-Raffi, L. M., Compactness and finite dimension in asymmetric normed linear spaces, Topol. Appl., 153, 844-854, (2005) · Zbl 1101.46017 [13] Riedl, J., Partially ordered locally convex vector spaces and extensions of positive continuous linear mappings, Math. Ann., 157, 95-124, (1964) · Zbl 0127.06403 [14] Schaefer, H. H., Topological vector spaces, (1971), Springer Berlin · Zbl 0217.16002
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
# Footnote - cite - Vgl. Jarchow/Rühmann (1988), S. 190

I am writing my master thesis and I need this style of citation:

Vgl. Jarchow/Rühmann (1988), S. 190.

This should be in footnotes.

I have done two variation of is. But it still not correct. The brackets are not on the place where they should be.

That is how is looks like:

\documentclass[12pt,a4paper,bibtotocnumbered,% Literaturverzeichnis ins Inhaltsverzeichnis
liststotoc,bibliography=totoc]{scrreprt}
%Chapter weiter oben beginnen

\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage[left=2.50cm, right=2.50cm, top=2.50cm, bottom=2.50cm]{geometry}

%Deutsche Sprache
\usepackage[ngerman]{babel}
\usepackage{xcolor}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%%%%%%%%%%%%%%%%%% %BILIOGRAPHIE
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\usepackage[backend=bibtex, style=authoryear-icomp, sorting=nyvt, sortcase=false, maxnames=8, maxbibnames=8, minbibnames=8]{biblatex}
\DeclareNameAlias{sortname}{last-first}
\DeclareNameAlias{default}{last-first}

\begin{document}

Hallo\footnote{\Parencite*[Vgl. Dai/Milkman/Katherine/Riis][S.12.]{Dai2013}}

Ja\footcite[Vgl.][12.]{Goulet2017a}
\newpage
\printbibliography[title={Literaturverzeichnis}]
\end{document}

I would be grateful if someone could help me.

• What version of biblatex are you using? – moewe Oct 17 '18 at 12:33
• Excuse me where can I see which version I use? – Swanni Oct 17 '18 at 12:38
• You can find the version in the .log file. It is displayed more prominently (so you don't have to search for it) if you add \listfiles to the beginning of your document. – moewe Oct 17 '18 at 12:51

The styles of my biblatex-ext bundle make it easy to add a few parentheses that are not there in the standard styles.

Simply load ext-authoryear-comp instead of authoryear-comp and use \DeclareDelimFormat.

The MWE also uses \autocite as the go-to command and not \footcite.

\documentclass[12pt,a4paper,bibliography=totoc]{scrreprt}
\usepackage[utf8]{inputenc}
\usepackage[ngerman]{babel}
\usepackage{csquotes}
\usepackage[backend=bibtex, style=ext-authoryear-icomp,
sorting=nyvt, sortcase=false,
autocite=footnote,
maxnames=8, minbibnames=8]{biblatex}

\DeclareInnerCiteDelims{footcite}{\bibopenparen}{\bibcloseparen}

\DeclareNameAlias{default}{family-given}
\DeclareNameAlias{sortname}{default}

You are still using BibTeX instead of the more powerful Biber backend. I suggest you switch to backend=biber and tell your editor to run Biber for you (Biblatex with Biber: Configuring my editor to avoid undefined citations). You can only use the full set of biblatex's features with Biber.


text:
# CHARGED ONLY, HADROPRODUCED INSPIRE search

VALUE (MeV) EVTS DOCUMENT ID TECN CHG  COMMENT
$\bf{ 50.8 \pm0.9}$ OUR FIT
$\bf{ 50.8 \pm0.9}$ OUR AVERAGE
$49$ $\pm2$ 5840
 1984 B
HBC - 8.25 ${{\mathit K}^{-}}$ ${{\mathit p}}$ $\rightarrow$ ${{\overline{\mathit K}}^{0}}{{\mathit \pi}^{-}}{{\mathit p}}$
$56$ $\pm4$
 1984
SPEC - 200 ${{\mathit \pi}^{-}}$ ${{\mathit p}}$ $\rightarrow$ 2 ${{\mathit K}_S^0}$ X
$51$ $\pm2$ 4100
 1981
HBC - 6.5 ${{\mathit K}^{-}}$ ${{\mathit p}}$ $\rightarrow$ ${{\overline{\mathit K}}^{0}}{{\mathit \pi}^{-}}{{\mathit p}}$
$50.5$ $\pm5.6$
 1980
HBC + 32 ${{\mathit K}^{+}}$ ${{\mathit p}}$ $\rightarrow$ ${{\mathit K}^{0}}{{\mathit \pi}^{+}}$ X
$45.8$ $\pm3.6$ 1800
 1978 B
HBC $\pm{}$ 0.76 ${{\overline{\mathit p}}}$ ${{\mathit p}}$ $\rightarrow$ ${{\mathit K}^{\mp}}{{\mathit K}_S^0}$ ${{\mathit \pi}^{\pm}}$
$52.0$ $\pm2.5$ 6706 1
 1978
HBC $\pm{}$ 0.76 ${{\overline{\mathit p}}}$ ${{\mathit p}}$ $\rightarrow$ ( ${{\mathit K}}{{\mathit \pi}}$) ${}^{+-}$ X
$52.1$ $\pm2.2$ 9000 2
 1975
HBC - 14.3 ${{\mathit K}^{-}}$ ${{\mathit p}}$ $\rightarrow$ ( ${{\mathit K}}{{\mathit \pi}}$) ${}^{-}$ X
$46.3$ $\pm6.7$ 765 1
 1973
HBC - 3.13 ${{\mathit K}^{-}}$ ${{\mathit p}}$ $\rightarrow$ ${{\overline{\mathit K}}^{0}}{{\mathit \pi}^{-}}{{\mathit p}}$
$48.2$ $\pm5.7$ 1150 1, 3
 1973
HBC - 3.3 ${{\mathit K}^{-}}$ ${{\mathit p}}$ $\rightarrow$ ${{\overline{\mathit K}}^{0}}{{\mathit \pi}^{-}}{{\mathit p}}$
$54.3$ $\pm3.3$ 4404 1
 1971 B
HBC - 3.9,4.6 ${{\mathit K}^{-}}$ ${{\mathit p}}$ $\rightarrow$ ( ${{\mathit K}}{{\mathit \pi}}$) ${}^{-}$ ${{\mathit p}}$
$46$ $\pm5$ 1700 1, 3
 1964
HBC - 1.7 ${{\mathit K}^{-}}$ ${{\mathit p}}$ $\rightarrow$ ${{\overline{\mathit K}}^{0}}{{\mathit \pi}^{-}}{{\mathit p}}$
• • • We do not use the following data for averages, fits, limits, etc. • • •
$46.7$ $\pm0.2$ ${}^{+0.1}_{-0.2}$ 183k
 2019 AQ
BES $\pm{}$ ${{\mathit J / \psi}}$ $\rightarrow$ ${{\mathit K}^{+}}{{\mathit K}^{-}}{{\mathit \pi}^{0}}$
$43.6$ $\pm1.3$ 4K 4
 2017 C
BABR ${{\mathit J / \psi}}$ $\rightarrow$ ${{\mathit K}_S^0}$ ${{\mathit K}^{\pm}}{{\mathit \pi}^{\mp}}$
$47.2$ $\pm0.3$ $\pm2.3$ 190k 5
 2016 N
LHCB ${{\mathit D}^{0}}$ $\rightarrow$ ${{\mathit K}_S^0}$ ${{\mathit K}^{\pm}}{{\mathit \pi}^{\mp}}$
$54.8$ $\pm1.7$ 27k 6
 1999 D
CBAR $\pm{}$ 0.0 ${{\overline{\mathit p}}}$ ${{\mathit p}}$ $\rightarrow$ ${{\mathit K}^{+}}{{\mathit K}^{-}}{{\mathit \pi}^{0}}$
$45.2$ $\pm1$ $\pm2$ 80k 7
 1989
LASS - 11 ${{\mathit K}^{-}}$ ${{\mathit p}}$ $\rightarrow$ ${{\overline{\mathit K}}^{0}}{{\mathit \pi}^{-}}{{\mathit p}}$
$42.8$ $\pm7.1$ 3700
 1983
HBC + 70 ${{\mathit K}^{+}}$ ${{\mathit p}}$ $\rightarrow$ ${{\mathit K}^{0}}{{\mathit \pi}^{+}}$ X
$64.0$ $\pm9.2$ 800 1, 3
 1982
SPEC + 30 ${{\mathit K}^{+}}$ ${{\mathit p}}$ $\rightarrow$ ${{\mathit K}_S^0}$ ${{\mathit \pi}^{+}}{{\mathit p}}$
$62.0$ $\pm4.4$ 3200 1, 3
 1982
SPEC + 50 ${{\mathit K}^{+}}$ ${{\mathit p}}$ $\rightarrow$ ${{\mathit K}_S^0}$ ${{\mathit \pi}^{+}}{{\mathit p}}$
$55$ $\pm4$ 3600 1, 3
 1982
SPEC - 50 ${{\mathit K}^{+}}$ ${{\mathit p}}$ $\rightarrow$ ${{\mathit K}_S^0}$ ${{\mathit \pi}^{-}}{{\mathit p}}$
$62.6$ $\pm3.8$ 380
 1981
SPEC + 50 ${{\mathit K}^{\pm}}$ ${{\mathit p}}$ $\rightarrow$ ${{\mathit K}^{\pm}}{{\mathit \pi}^{0}}{{\mathit p}}$
$50.5$ $\pm3.9$ 187
 1981
SPEC - 50 ${{\mathit K}^{\pm}}$ ${{\mathit p}}$ $\rightarrow$ ${{\mathit K}^{\pm}}{{\mathit \pi}^{0}}{{\mathit p}}$
1  Width errors enlarged by us to 4${\times }\Gamma /\sqrt {\mathit N }$; see note.
2  Inclusive reaction. Complicated background and phase-space effects.
3  Number of events in peak reevaluated by us.
4  From a Dalitz plot analysis in an isobar model with charged and neutral ${{\mathit K}^{*}{(892)}}$ masses and widths floating.
5  Average of fit results with different parametrizations for the ${{\mathit K}}{{\mathit \pi}}$ ${\mathit S}{\mathrm -wave}$.
6  K-matrix pole.
7  From a partial wave amplitude analysis.
References:
 ABLIKIM 2019AQ
PR D100 032004 Partial-wave analysis of $J/\psi\to K^+K^-\pi^0$
 LEES 2017C
PR D95 072007 Dalitz Plot Analyses of ${{\mathit J / \psi}}$ $\rightarrow$ ${{\mathit \pi}^{+}}{{\mathit \pi}^{-}}{{\mathit \pi}^{0}}$ , ${{\mathit J / \psi}}$ $\rightarrow$ ${{\mathit K}^{+}}{{\mathit K}^{-}}{{\mathit \pi}^{0}}$ , and ${{\mathit J / \psi}}$ $\rightarrow$ ${{\mathit K}_S^0}$ ${{\mathit K}^{\pm}}{{\mathit \pi}^{\mp}}$ Produced via ${{\mathit e}^{+}}{{\mathit e}^{-}}$ Annihilation with Initial-State Radiation
 AAIJ 2016N
PR D93 052018 Studies of the Resonance Structure in ${{\mathit D}^{0}}$ $\rightarrow$ ${{\mathit K}_S^0}$ ${{\mathit K}^{\pm}}{{\mathit \pi}^{\mp}}$ Decays
 ABELE 1999D
PL B468 178 Antiproton Proton Annihilation at Rest into ${{\mathit K}_L^0}{{\mathit K}^{+}}{{\mathit K}^{-}}{{\mathit \pi}^{0}}$
 BIRD 1989
SLAC-332 Cerenkov Ring Imaging and Spectroscopy of Charged ${{\mathit K}^{*}}$ Interactions at 11 ${\mathrm {GeV/}}\mathit c$
 BAUBILLIER 1984B
ZPHY C26 37 A Study of the ${{\overline{\mathit K}}}{{\mathit \pi}}$ System in the Reaction ${{\mathit K}^{-}}$ ${{\mathit p}}$ $\rightarrow$ ${{\overline{\mathit K}}^{0}}{{\mathit \pi}^{-}}{{\mathit p}}$ at 8.25 ${\mathrm {GeV/}}\mathit c$
 NAPIER 1984
PL 149B 514 ${{\mathit K}^{*}{(892)}^{\pm}}$ Production in ${{\mathit \pi}^{-}}{{\mathit N}}$ Interactions at 200 ${\mathrm {GeV/}}\mathit c$
 BARTH 1983
NP B223 296 Inclusive Resonance Production in ${{\mathit K}^{+}}{{\mathit p}}$ Interactions at 70 ${\mathrm {GeV/}}\mathit c$
 CLELAND 1982
NP B208 189 Study of the Reactions ${{\mathit K}^{\pm}}$ ${{\mathit p}}$ $\rightarrow$ ${{\mathit K}_S^0}$ ${{\mathit \pi}^{\pm}}{{\mathit p}}$ at 30 and 50 ${\mathrm {GeV/}}\mathit c$: Description of the Apparatus and Amplitude Analysis of the ${{\mathit K}_S^0}$ ${{\mathit \pi}}$ System
 DELFOSSE 1981
NP B183 349 Production of ${{\mathit \rho}^{\pm}}$, ${{\mathit g}{(1700)}^{\pm}}$, ${{\mathit A}_{{2}}^{\pm}}$, ${{\mathit K}^{*}{(890)}^{\pm}}$ and ${{\mathit K}^{*}{(1420)}^{\pm}}$ at 50 ${\mathrm {GeV/}}\mathit c$
 TOAFF 1981
PR D23 1500 Reaction ${{\mathit K}^{-}}$ ${{\mathit p}}$ $\rightarrow$ ${{\overline{\mathit K}}^{0}}{{\mathit p}}{{\mathit \pi}^{-}}$ at 6.5 ${\mathrm {GeV/}}\mathit c$
 AJINENKO 1980
ZPHY C5 177 Inclusive ${{\mathit K}^{*}{(892)}^{+}}$ and ${{\mathit K}^{*}{(892)}^{0}}$ Production in ${{\mathit K}^{+}}{{\mathit p}}$ Interaction at 32 ${\mathrm {GeV/}}\mathit c$
 AGUILAR-BENITEZ 1978B
NP B141 101 A New Experimental Determination of the ${{\mathit K}^{*}{(892)}}$ Electromagnetic Mass Difference
 COOPER 1978
NP B136 365 Inclusive Production of ${{\mathit K}^{*}{(892)}^{\pm}}$ and ${{\mathit K}_S^0}$ in ${{\overline{\mathit p}}}{{\mathit p}}$ Annihilations at 0.76 ${\mathrm {GeV/}}\mathit c$
 PALER 1975
NP B96 1 Inclusive ${{\mathit K}^{*}{(890)}^{-}}$ and ${{\overline{\mathit K}}^{*}{(890)}^{0}}$ Production in 14.3 ${\mathrm {GeV/}}\mathit c$ ${{\mathit K}^{-}}{{\mathit p}}$ Interactions
 CLARK 1973
NP B54 432 The Reactions ${{\mathit K}^{-}}$ ${{\mathit p}}$ $\rightarrow$ ${{\mathit K}^{*-}}$ nucleon at 3.13 and 3.3 ${\mathrm {GeV/}}\mathit c$
 AGUILAR-BENITEZ 1971B
PR D4 2583 Properties of the ${{\mathit K}^{*}{(890)}}$ and ${{\mathit K}^{*}{(1420)}}$ in ${{\mathit K}^{-}}{{\mathit p}}$ Interactions at 3.9 and 4.6 ${\mathrm {GeV/}}\mathit c$
 WOJCICKI 1964
PR 135 B484 ${{\mathit K}^{-}}$ ${{\mathit p}}$ $\rightarrow$ ${{\overline{\mathit K}}^{0}}{{\mathit \pi}^{-}}{{\mathit p}}$ Reaction from 1.0 to 1.7 ${\mathrm {GeV/}}\mathit c$
see fit info


text:
Centre of Polyhedra

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Quote:
 Original post by DonDickieDHey Erin,thanks for the reply. This was the kind of formula I was looking for. I have very bad days finding the formulas I need at the moment :-) E.g.What is the rotational energy of a rigid body. I remember that it is:T = 0.5 * I * w^2For the cases of a 3D rigid bodies w is a vector so could I calculate it like this?T = 0.5 * (I * w) * w

The above will work provided w is in the local frame. If you already have the momentum and velocity handy (in world space):

  flt getTotalKineticEnergyRB(RigidBody & rb) {    flt linE = rb.linM.dot(rb.linVel);    flt rotE = rb.rotM.dot(rb.rotVel);    return .5f*(linE + rotE);  } // getTotalKineticEnergyRB

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Some definitions:

(i) MOMENTS: Nth degree moment generating function of an arbitrary distribution D(X) of i samples x1,x2...x(i), around point c, is defined as: SUM(i=1,...,m)[D(x(i))*(x(i)-c)^n]
If the distribution is continuous, the formula turns into
Intergral[of x][D(x)*(x-c)^n][dx]

(ii) CENTROID: Centroid of a volume is the origin of coordinate axes for which first moments of the volume are zero. It is considered center of a volume. For a homogeneous body in a parallel gravity field, mass center and center of gravity coincide with the centroid.

(iii) FIRST MOMENT OF A FINITE DISTRIBUTION D(X(i)) AROUND POINT 'c':
SUM(i=1,...,m)[D(X(i))*(X(i)-c)]

(iv) THUS FOR C TO BE THE CENTROID OF VOLUME:
SUM(i=1,...,m)[D(X(i))*(X(i)-c)] = 0 <=>
D(x1)*(x1-c) + D(x2)*(x2-c) + ... + D(x(i))*(x(i)-c) = 0 <=>
(X1-C) + (X2-C) + ... + (X(i)-C) = 0 <=>
X1+X2+...+Xi = i*c <=>
C = (X1+X2+...+X(i))/i (points' average!)

where D(Xk)=D(Xj) for all k,j in 1,...,i since the body is homogenous

Quote:
 @someusernameI am looking for the center of the volume. If you consider this volume as the infently points masses, the integral over all these point masses will give you the center of gravity.(my comments: For a homogeneous body in a parallel gravity field, mass center and center of gravity coincide with the centroid of volume) You only consider 4 vertices from a solid body - what would be wrong...

You don't want to find the volume; you want the centroid. Even if you manage to define a valid continuous distribution of mass for this volume and solve the integral it would give you the very same result, because:
the convex closure of those 8 points in R^3 is not missing any of the volume if you found a symbolic parametrization for it. Nor is it adding any.
[ edit: I'm missing something here! The original 4vertices polygon needs to be convex for this to hold. ]
(And I din't imply 4 vertices, I implied all 8.)

I am trying to make my point through math here. If my math is wrong then my point is wrong!
This is the last post from me on this subject

[Edited by - someusername on December 1, 2005 12:03:17 PM]

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Thanks for your help so far. Actually I made the same assumption as you did, but became a little supicious through the behaviour of my simulator. So I tried to verify the formula by an example:

v1 = (1,1), v2 = (3,1), v3 = (1,2), v4 = (3,3)

These four points define a trapez. If you now build the centroid using you formula:

c = 0.25 * (v1 + v2 + v3 + v4) = 0.25 * (8, 7)

This isn't the centroid of the trapez. You can verify this by simply drawing the shape defined be the four points. You will find a rectangle and a triangle. The centroid of the trapez will lie on the line between the centroid of the triangle and the rectangle. The above point doesn't. Actually the x-coordinate can't be 2.

HTH,

-Dirk

PS:
Maybe you need to make a difference. I am not looking for the centroid of the point cloud, but I am looking for the centroid of the polyhedra defined by these points.

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Quote:
 c = 0.25 * (v1 + v2 + v3 + v4) = 0.25 * (8, 7)This isn't the centroid of the trapez. You can verify this by simply drawing the shape defined be the four points. You will find a rectangle and a triangle. The centroid of the trapez will lie on the line between the centroid of the triangle and the rectangle. The above point doesn't. Actually the x-coordinate can't be 2.

?
You're totally wrong. A trapezoid is a special case of quadrilaterals for which the following theorems hold.

1) theorem #1
The centroid of a quadrilateral occurs at the point of intersection of its bimedians. (the segments connecting pairs of opposite midpoints)

In your case one of the bimedians is the segment connecting the point M13(1,1.5) (Midpoint of segment connecting vertices 1 & 3) and M24(3,2) (the midpoint of the segment connecting the vertices 2 & 4)
The other is the segment connecting M12(2,1) M34(2,2.5)
They obviously intersect at (2,1.75) == .25*(8,7)

2) Theorem #2
The centroid of a quadrilateral occurs at the midpoint of the segment that connects the midpoints of its diagonals.

In your case midpoint of diagonal D14 is (2,2) and that of D32 is (2,1.5)
The midpoint of D14 - D32 is again (2, 1.75) == .25*(8,7)

I don't know what you're trying to calculate here, but it's NOT the centroid!

[edit: typo]

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The centroid is (only x_coordinate):

xs = ( xs1 * A1 + xs2 * A2 ) / A_total

A1 = 2 (Area of the lower rectangle)
A2 = 1 (Area of the upper triangle)

xs1 = 2 (x-coordinate of the centroid of the lower rectangle)
xs2 = 7/3 (x-coordinate of the centroid of the upper triangle)

xs = (2 * 2 + 7/3 * 1) / 3 = (12/3 + 7/3) / 3 = 19 / 9 = 2.1111 != 2

Maybe you are right. English is not my native language and what I actually calculate here is the center of gravity. Sorry if this was wrong!

Regards,

-Dirk

PS:

"A trapezoid is a special case of quadrilaterals for which the following theorems hold."

Are you sure this is correct? I would be careful to use theorems of quadrilaterals for trapezoids. The center of gravity of a shape composed of two simple shapes, must lie on the connection line between the centers of the two simple shapes. Forget all this math stuff and theorems for a moment and try to balance the trapezoid on your finger. I am sure that you agree that you can't balance it if you put you finger at xs = 2

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http://www.geometrictools.com/Documentation/PolyhedralMassProperties.pdf

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OK, I'm forgeting the math and the equations and the theorems...
Indeed centroid's X == 2 seems weird, because x=2 is in the middle and there is definitely more "area" to the right side of the trapezium.

BUT THE CENTROID'S Y ISN'T IN THE MIDDLE TOO!!! IT'S NOT 2!!! ITS 1.75 WITCH IS LOWER THAN MIDDLE, WHERE THERE IS ALSO MORE AREA!

How else can I put it?

And why not use theorems of quadrilaterals on trapezia? They're just quadrilaterals with 2 parallel sides.

[Edited by - someusername on December 1, 2005 6:07:58 PM]

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At DonDickieD
I want to apologize for the tone of my previous reply. It was an act of impulse, honestly, I didn't think this through. I'm sorry I came off as a dick... Honestly.

And of course you're obviously right about the specific counter-example.

If you're still interested in any more input on this, from me, I'll do my best to be of any assistance.

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First: No problem - I didn't understand it like this!

As I mentioned in an earlier post I made exactly the same assumption like you, namely that the centroid is simply the average of the vertices and used your formula. Than I got skeptic beacuse of the result in my application and setup the small example and tried to verify if the result holds for this specific example. As we both know know - it didn't.

The first thing that I believe is wrong is that we are merging two different things. I think the formula is right, but can't be used here. While the average is of course the center of a point cloud, it isn't necessarily the center of the surface/volume defined by the points. The formula for the center of gravity is:

xs = Sum( xs_i * A_i ) / A_total

This is a weighted average. Of course for a point cloud there is no such thing as an area and the formula simply becomes the arithmetic mean.

Regarding your theorem #1 from a previous post:

I am not sure here since I am not sure if what you mean by "BiMedian" is the same as I think it is. So again a little example:

The centroid of a triangle is the intersection of the perpendicular lines in the middle of each edge. Ii it that what you mean by BiMedian? If yes, then think how this line will look like for the edge (1,2) to (3,3). So you see a reason why I think that you can't carry out the formula of the rectangle to the trapez.

Dirk

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ok, I just wanted to let you know I regret, that's all.

The formula you are presenting for the centroid's x-coord seems to be fine, since when viewed as vector equation for the centroid's coordinates, it merely states that for any two objects their centroid is on the line connecting their own centroids at a distance depending on their ares over their total sum. And it generalizes for 'n' objects. This makes perfect sense.

The two theorems I posted above are correct. (for that I'm sure... You can check the site I linked in a previous reply) And there is also another one -as far as I remember- especially for trapezoids, stating that its centroid lies on the bimedian (more on that below) which is parallel to its 2 parallel edges, also providing a formula (which eludes me right now) to calculate the other coordinate too, based entirely on the edges length and its height.

Your counter-example though, got me thinking to be honest. Maybe some quadrilaterals' theormes don't hold true for trapezoids? But how can that be? I mean, they apply to the square and rhombus which are even more special/constrained categories of quadrilaterals.

As for the bimedian (of a quadrilateral), I understand it is the segment that connects the midpoints of two non-adjacent edges. I.e. in a trapezoid, the bimedian of its two non-parallel edges, is parallel to its other two edges.

Quote:
 The centroid of a triangle is the intersection of the perpendicular lines in the middle of each edge. Ii it that what you mean by BiMedian?

No not all. Bimedian isn't even defined for 3-gons

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# Something is wrong with this proof, limits

Could someone please tell me what is wrong with this proof?

Show that $\lim\limits_{(x,y) \to (0,0)} \dfrac{xy^3}{x^4 + 3y^4}$ does not have a limit or show that it does and find the limit.

I know it is wrong because the limit doesn't exist, but this proof is contradicing me

Proof

Case 1

Assume for $x,y > 0$, then $x^4 + 3y^4 > x^4 > x > 0$

\begin{align*} 0 < x < x^4 + 3y^4 &\iff 0 < \dfrac{x}{x^4 + 3y^4} < 1 \\ & \iff 0 < \dfrac{x|y^3|}{x^4 + 3y^4} < |y^3| \\ & \iff \lim\limits_{(x,y) \to (0,0)} 0 < \lim\limits_{(x,y) \to (0,0)} \dfrac{x|y^3|}{x^4 + 3y^4} < \lim\limits_{(x,y) \to (0,0)}|y^3|\\ &\iff 0 < \lim\limits_{(x,y) \to (0,0)} \dfrac{x|y^3|}{x^4 + 3y^4} < 0 \end{align*}

Case 2. WLOG Assume $x,y <0$ and combine both cases.

What's wrong the pproof? I don't find the flaw

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Marvis has mentioned the main errors in his answer, but there’s one more: when you pass to the limits, you have to change $<$ to $\le$. As it stands, your last line says that $0<0$, which is clearly wrong. –  Brian M. Scott May 25 '12 at 5:19
Ahhhhh! okay thanks! –  sidht May 25 '12 at 5:26
When $x=y\ne 0$, the function is $1/4$, so estimates cannot be right. When $x=-y\ne 0$ the function is $-1/4$, so limit cannot exist. –  André Nicolas May 25 '12 at 5:31

The statement $x^4>x$ is clearly wrong especially when $x$ is in the neighborhood of $0$.

Also, you need to look at cases $x>0,y<0$ and $x<0,y>0$. But the main error is the statement $x^4 > x$.

EDIT

The answer is that the function is not continuous at the origin. This can be seen as follows. Remember that in two dimensions there are infinitely many different directions from which you can approach origin unlike in one dimension where you need to look only at two different cases $x>0$ and $x<0$. Hence, in two dimensions it is not sufficient/advantageous to split it into just four cases $x,y>0$, $x,y<0$, $x>0,y<0$, $x<0,y>0$ and analyze.

Approach the origin along the straight line $y=mx$. Then we have that $$\lim_{\overset{x \rightarrow 0}{y \rightarrow 0}} \dfrac{xy^3}{x^4 + 3y^4} = \lim_{\overset{x \rightarrow 0}{y = mx}} \dfrac{xy^3}{x^4 + 3y^4} = \lim_{x \rightarrow 0} \dfrac{x(mx)^3}{x^4 + 3(mx)^4} \\ = \lim_{x \rightarrow 0} \dfrac{m^3 x^4}{x^4 + 3m^4 x^4} = \lim_{x \rightarrow 0} \dfrac{m^3 x^4}{(1 + 3m^4) x^4} = \lim_{x \rightarrow 0} \dfrac{m^3}{(1 + 3m^4)} = \dfrac{m^3}{1+3m^4}$$ Hence, approaching the origin along different straight lines give different answers.

For instance, if we approach the origin along the line $y = x$, the limit is $\dfrac14$.

If we approach the origin along the line $y = -x$, the limit is $-\dfrac14$.

If we approach the origin along the line $y = 0$, the limit is $0$.

This proves that the function is not continuous at the origin.

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I realize the proof is wrong, but I want to understand the technique. I know it's kinda hard to do with this function since it has no limit anyways, but if I were to go with x,y>0, how do continue with my proof? –  sidht May 25 '12 at 5:30
@jak I have updated my answer and hope it clarifies your question. –  user17762 May 25 '12 at 5:38
@jak: The $x$, $y$ positive part doesn't really help, approach $(0,0)$ along $y=x$, or $y=2x$, get different answers. –  André Nicolas May 25 '12 at 5:39
@jak Remember that in two dimensions there are infinitely many different directions from which you can approach origin unlike in one dimension where you need to look only at two different cases $x>0$ and $x<0$. Hence, in two dimensions it is not sufficient/advantageous to split it into just four cases $x,y>0$, $x,y<0$, $x>0,y<0$, $x<0,y>0$ and analyze. –  user17762 May 25 '12 at 5:44
One cannot simply look at approaching the origin along a line. Approaching the origin along any line, $$\frac{x^2-y}{(x^2+y^2)^2+x^2-y}$$ limits to $1$. However, approaching along the curve $y=x^2$, the limit is $0$. –  robjohn May 25 '12 at 9:28
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The easiest way to see that a limit doesn't exist is to write $$\frac{xy^3}{x^4+y^4}=\frac{x/y}{(x/y)^4+1}\tag{1}$$ Using $(1)$, it is easy to see that, over any circle around the origin, $\frac{xy^3}{x^4+y^4}$ can take on any value that $\frac{t}{t^4+1}$ can; that is, any value in $\left[{-}\sqrt[4]{\frac{27}{256}}\;,\;\sqrt[4]{\frac{27}{256}}\right]$. Therefore, $$\lim_{x,y\to0}\frac{xy^3}{x^4+y^4}\tag{2}$$ does not exist.

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In the first line of your proof, $x^4$ is not greater than $x$ just because $x$ is positive. For numbers smaller than $1$, this is false.

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Try a different approach. Literally.

If the limit exists, then it will be the same regardless of how you approach $(0,0)$.

So, consider the limit first along the line $y=x$, and you will see that as long as $x\neq 0$, then $\frac{xy^3}{x^4 + 3y^4} = \frac{1}{4}$. Now try approaching along the line $y=2x$, and see what limit you get. If they are different, then this is proof that the limit as $(x,y) \to (0,0)$ doesn't exist.

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Just adding to the answers. When you have limits with several variables it is very useful to try to visualize it first.
When you see that different values are trying to "get closer and closer" to the same point, then you might expect that something fishy is going on.

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