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https://zhang-xiao-mu.blog/2018/11/09/merge-k-sorted-lists/
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# Merge K Sorted Lists
Let’s continue…
Problem Statement:
Merge K Sorted Lists
Given a linked list, remove the n-th node from the end of list and return its head.
Example:
```Given linked list: 1->2->3->4->5, and n = 2.
After removing```
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
Example:
```Input:
[
1->4->5,
1->3->4,
2->6
]
Output: 1->1->2->3->4->4->5->6
```
Solution One: Brute Force
We iterate all the numbers of all lists, and store them in an array. After that, we sort the array and transfer the result to a new list. After that, output the final result.
Time Complexity: O(n*log(n)) We construct the array, which takes O(n) time, and sorting the array takes O(n*log(n)) time. n is the number of total nodes in all lists.
Space Complexity: O(n)
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/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* mergeKLists(vector& lists) { vector myVector; int len_l = lists.size(); for(int i = 0; i < len_l; i++){ ListNode* temp = lists[i]; while(temp){ myVector.push_back(temp->val); temp = temp->next; } } sort(myVector.begin(), myVector.end()); ListNode* head = new ListNode(0); ListNode* p_NewList = head; len_l = myVector.size(); for(int i = 0; i < len_l; i++){ ListNode* temp = new ListNode(myVector[i]); p_NewList->next = temp; p_NewList = p_NewList->next; } return head->next; } };
Solution Two: Compare Lists by Columns
The general idea is like the picture demonstrated below. We compare all the lists in one column. We first find the smallest number of this column and make it our head node, then we move this list forward, and compare the new column again, until we find the second smallest element, then we connect our head node to this node. After that, we move forward the second list and continue this progress until all lists are empty.
Note here we do a tiny trick that we always connect the current smallest node to the previous smallest node, so we can make the space complexity O(1).
Time Complexity: O(K*N), N is the length of the longest sub list. K is the number of the lists.
Space Complexity: O(1)
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/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* mergeKLists(vector& lists) { int len_l = lists.size(); ListNode head(0); ListNode *h = &head; while(true){ bool isBreak = true; int minVal = INT_MAX; int min_index = 0; for(int i = 0; i < len_l; i++) { if(lists[i] != NULL){ isBreak = false; if(lists[i]->val <= minVal) { min_index = i; minVal = lists[i]->val; } } } if(isBreak) break; h->next = lists[min_index]; h = h->next; lists[min_index] = lists[min_index]->next; } return head.next; } };
Solution Three: Priority Queue
This idea is based on the Solution Two. In Solution Two, we can see that, each time when we want to find the current smallest element, we need to iterate all the lists, which takes O(K) time. When we find this element, we connect it with previous found nodes, we need O(1) time. Can we do better here?
We can utilize the properties of priority queue to accelerate the searching process. For the first time, we just add all the first elements to our priority queue. We maintain our priority and make sure that the smallest element should always on top (first pop out). If we want the smallest number, we can easily pop the top element out of the priority queue. This operation will take O(log(k)) time because we need to balance the remaining elements and make sure the second smallest on top of the priority queue. The same thing as insert, it takes O(log (k)) time.
The rest of the idea is the same. If one list contains the current smallest element, we pop out that node, and move forward that list one Node ahead.
Time Complexity: O(max(k, n* log(k))), k is the number of sub lists
Space Complexity: O(k)
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/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* mergeKLists(vector& lists) { int len_l = lists.size(); ListNode head(0); ListNode *h = &head; auto comp = [](ListNode* l1, ListNode* l2){ return l1->val> l2->val; }; priority_queue, decltype(comp)> q(comp); for(int i = 0; i < len_l; i++){ if(lists[i]) q.push(lists[i]); } while(q.empty() == false){ h->next = q.top(); q.pop(); h = h->next; if(h->next != NULL) q.push(h->next); } return head.next; } };
Solution Four: Combine two
Look at the picture below. Note we are going to merge two lists at one time. We keep dividing the remaining lists into different two pairs, and keep merging them until we are done. For example, in the first iteration, we merge list0 and list1; list2 and list3; list4 and list5. The second iteration, we merge list0 and list2; list4….
We did this trick to reduce total times of merging. The merging process will consumes O(n’), n’ is the average number of all lists. We will do the merging process log(k) times instead of k, so the time complexity is O(n’*log(k)).
Time Complexity: O(n’*log(k))
Space Complexity: (1)
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/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* mergeTwoLists(ListNode* l1, ListNode* l2){ ListNode head(0); ListNode* h = &head; while(l1&&l2){ if(l1->val < l2->val){ h->next = l1; h=h->next; l1= l1->next; } else{ h->next = l2; h = h->next; l2=l2->next; } } if(l1) { h->next = l1; } if(l2) { h->next = l2; } return head.next; } ListNode* mergeKLists(vector& lists) { int len_l = lists.size(); if(len_l == 0) return nullptr; int step = 1; while(step < len_l){ for(int i = 0; i + step < len_l; i = i + step*2) { lists[i] = mergeTwoLists(lists[i], lists[i+step]); } step = step*2; } return lists[0]; } };
That’s all for today, thanks for reading….
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https://au.mathworks.com/matlabcentral/answers/122024-multiple-loop-doesn-t-work-with-right-values?s_tid=prof_contriblnk
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# Multiple loop doesn´t work with right values
1 view (last 30 days)
Mariana on 18 Mar 2014
Commented: Mariana on 19 Mar 2014
Hello, I can´t figure out myself what is wrong with following nested loops:
index = 0;
for i = 1 : 2 : givenValue
index = index + 1;
j = i + 1;
for k = 1 : givenValue/2
if sth_x(k) > sth_y(k)
step(k) = sth_y(k) / sth_x(k);
end
end
for id = 1 : givenValue/2
values_x{index} = mat(i, 1) : mat(j, 1);
values_y{index} = mat (i, 2) : step(id) : mat(j, 2);
end
In words, what I need is to generate two cell arrays (values_x and values_y) with one row and several columns, which contains all points (pixels - it is image analysis) between points (pixels) mat(i,1) and mat (j,1) with growing i and j. values_x are allways increased with 1, but values_y are supposed to be increased with step counted before. Problem I have is that cell array values_y is generated with step from the last iteration only. I need values_y{1,1} to be counted with step 1 and values_y{1,2} to be counted with step 2. Therefore values_x{1,1} and values_y{1,1} will be the same length.
I´m new to matlab and I know this might be simple. But I really don´t know what is wrong. I tried to change order of "for" and "if", tried to replace "id" with "k" but still nothing works.
Nitin on 18 Mar 2014
You might need to initialize your cells first before saving to it.
a = cell(1, num);
##### 3 CommentsShow 1 older commentHide 1 older comment
Nitin on 19 Mar 2014
Edited: Nitin on 19 Mar 2014
Hopefully this will help, I am not sure what you are trying to achieve though:
index = 0;
givenValue = 10;
step = 1:5;
values_x = cell(1,10);
values_y = cell(1,10);
mat= randi(20,20,2);
for i = 1 : 2 : givenValue
index = index + 1;
j = i + 1;
for k = 1 : givenValue/2
if 2 > 1
display('So far so good')
end
for id = 1 : givenValue/2
values_x{index} = mat(i, 1) : mat(j, 1);
values_y{index} = mat (i, 2) : step(id) : mat(j, 2);
end
end
end
For example you can access the elements in the array using values_x{1}
Mariana on 19 Mar 2014
Thanks for your try :) I copied code into my editor and tried to run it, but it didn´t make me understand how I could solve my problem :( Probably I can ´t explain my problem well via text online. But once again thank you
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1. ## chain rule
can please someone show me the derivative of this using chain rule without simplifying
y=^1/2 sin (x+x^1/2)^1/2 cot x^1/2
thanks
2. Originally Posted by maeca
can please someone show me the derivative of this using chain rule without simplifying
y=^1/2 sin (x+x^1/2)^1/2 cot x^1/2
thanks
Is that $y = \frac{1}{2} \sin{(x + x^{\frac{1}{2}})^{\frac{1}{2}}} \cot{x^{\frac{1}{2}}}$?
3. ## chain rule
sorry.
it's
y=x^1/2 sin (x+x^1/2)^1/2 cot x^1/2
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https://www.physicsforums.com/threads/calculating-the-age-of-planets-without-radioactive-dating.801846/
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# Calculating the Age of Planets without Radioactive Dating
Tags:
1. Mar 7, 2015
I understand that it's possible to calculate the age of terrestial planets through radioactive dating their soil. However, the gas planets present a different challenge since we cannot currently land on them.
Any ideas on how to calculate their ages in a different manner?
2. Mar 7, 2015
### Greg Bernhardt
3. Mar 7, 2015
Thank you Greg!
I was actually thinking of analyzing the rotation of the planets to date them (similar to your stellar age estimation article you shared), though I think we would have a problem with Uranus and its extreme tilt.
4. Mar 7, 2015
### SteamKing
Staff Emeritus
Technically, the radioactive dating is performed on rock formations. Soil is a mixture of bits of weathered rock and various types of organic material.
5. Mar 8, 2015
### |Glitch|
The rotation of the planets have been altered since their initial creation as a result of impacts. According to the latest ESA's Venus Express spacecraft, Venus's rotation has slowed down by 6.5 minutes per Venusian sidereal day since the Magellan spacecraft visited it 16 years ago. Earth's rotation before the Thea impact is unknown, but our rotation has definitely slowed down since that impact due to the gravitational pull of our moon. Rotation is not a good method for determining the age of a planet.
Without actually taking surface samples, the best method for determining the age of a planet is as Greg Bernhardt posted above, by determining the age of the star. Planets would have formed within a few million years after the protostar becomes a star. The only time I can think of where that would not be the case would be with captured planets, but they would be extremely rare.
6. Mar 8, 2015
Thank you so much for the thorough explanation! That is really incredible how much Venus's rotation is slowing down... Is it due to tidal locking with the Sun or some other factor? Thanks again for the informative response :)
7. Mar 8, 2015
### Garth
Actually the dating of the Earth is by radioactive dating performed on the rock formations of the Earth. The problem with this is the Earth's crust is by no means primordial. The earliest terrestrial surface rock formation is about 4.03 Gys old and is part of the Acasta Gneiss of the Slave craton in northwestern Canada. Older still is a tiny zircon crystal from Western Australia dated at 4.4 Gyrs.
The earliest lunar rocks date from 4.46 Gyrs ago (the Genesis rock) and typically from the great bombardment era of 3.8 Gyrs ago.
However the age of the Earth, and the Sun and its Solar System, is determined from dating the oldest meteorites, the carbonaceous chondrites dated at 4.6 Gyrs ago.
The dating of the other planets stems from this period of planetary formation from the solar protoplanetary disc.
Garth
8. Mar 9, 2015
### Chronos
Dating the age of the solar system is still a work in progress. I think we are pretty accurate at present. Radioactive dating is highly precise, which explains its continued popularity.
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https://goprep.co/ex-21.2-q7-a-spherical-ball-of-lead-3-cm-in-diameter-is-i-1njobl
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# A spherical ball of lead 3 cm in diameter is melted and recast into three spherical balls. If the diameters of two balls be cm and 2 cm, find the diameter of the third ball.
Volume of a sphere = (4/3)πr3
Total volume remains same during recasting.
Given, spherical ball of lead 3 cm in diameter is melted and recast into three spherical balls and the diameters of two balls are cm and 2 cm.
Let the diameter of the third ball be ‘a’ cm.
27/8 = (27/64) + 1 + (a3/8)
a3/8 = 125/64
a3 = 125/8
a = 5/2 cm
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# From $\sum_p \frac{\log p}{p^s} = \frac{1}{s-1} + O(1)$ conclude that $\sum_p \frac{1}{p^s} = \log \frac{1}{s-1} + O(1)$
I'm reading a book on analytic number theory. It asks me to prove:
$$\sum_p \frac{\log p}{p^s} = \frac{1}{s-1} + O(1) \tag{A}$$ and conclude, via integration, that: $$\sum_p \frac{1}{p^s} = \log \frac{1}{s-1} + O(1) \tag{B}$$
Now, I know how to prove $(A)$ via Abel Summation. However, when it comes to $(B),$ I have the problem that although:
$$\frac{d}{dx} \log(x-1) = \frac{1}{x-1}$$ and
$$\frac{d}{ds} p^{-s} = (-\log p)p^{-s}$$
I have the problem that when I integrate over $O(1)$, I get $\infty$, not $O(1)$.
What am I doing wrong? How do I get from $(A)$ to $(B)$?
-
@PeterTamaroff: Nice. I didn't know > and $$could stack. – user36739 Aug 16 '12 at 1:07 I just removed the word "prime". – Pedro Tamaroff Aug 16 '12 at 1:09 Why do you "integrate over O(1)"? The way I see it, O(1) does not denote dependence on s but on the number of terms of the sum. – leonbloy Aug 16 '12 at 1:54 @leonbloy But the number of terms is fixed: it's infinity. The big O is for s\to1^+. – anon Aug 16 '12 at 1:58 @anon: ah, ok, my bad. – leonbloy Aug 16 '12 at 2:02 ## 3 Answers What if you don't integrate over all of (s,\infty)? Say, fix S large enough, and...$$\int_s^S\sum_{p}\frac{\log p}{p^\sigma}d\sigma=\int_s^S\frac{1}{\sigma-1}+O(1)d\sigma=\log\frac{1}{s-1}+O(1) $$whereas$$\int_s^S\sum_{p}\frac{\log p}{p^\sigma}d\sigma=\sum_p\int_s^S\frac{\log p}{p^\sigma}d\sigma=\sum_{p}\left(\frac{1}{p^s}-\frac{1}{p^S}\right)=\sum_p\frac{1}{p^s}~+O(1) $$for s<S as s\to1^+. (Some uniform convergence stuff needs to be checked so that the interchange is justified.) - The first idea I had when you posted in chat was to use infinite Möbius inversion. We have$$P(s)=\sum_p\frac{1}{p^s}=\sum_{n\ge1}\frac{\mu(n)}{n}\log\zeta(ns), $$where P(\cdot) is the prime zeta function and \zeta(\cdot) is the Riemann zeta function. (For those who want to check this: write \zeta in the Euler product form and then expand each Euler factor with a series expansion for \log individually; group terms appropriately.) As s\to1^+ all of the n\ge2 terms are already O(1) (put together), so we need only put \zeta(s)=\frac{1}{s-1}+O(1) inside the first logarithm. This route requires more work and preliminary information though. - Mobius inversion way is pretty nice! (+1) – user 1618033 Aug 16 '12 at 5:56 Well we know that$$\zeta(\sigma)= \frac{1}{\sigma-1}$$when$$ \sigma \to1^+$$since zeta has a simple pole in \sigma=1 with residual one. Now your function$$\sum_p\frac{\log p}{p^s}$$can be expressed in terms of the zeta function. see http://people.math.jussieu.fr/~demarche/enseignements/2011-2012/M1-TDN/MM020-TD6-corrige.pdf (it's the exercise number 7, question g, in french :)). The tricky part is why could you integrate two functions which are equivalent in the sense that$$\sum_p \frac{\log p}{p^s} ∼ \frac{1}{s-1} ?
Here it is possible since $1/s-1$ is not integrable in a neighborhood of 1. Why?
here is an sketch of how you could procede. Let f, g be continuous on R such that f~g and $\int_0^\infty$fdx diverges. I will write this integral I in what follows and I$x$ when the upper bound is $x$ instead of + $\infty$. we would like to show that I (f) ~ I (g)
In what follows we have f ~ g at infinity. (I guess you could repeat that proof for an equivalence that holds in another neighborhood)
f ~ g then f (x) = g (x) (1 + h (x)) such that $gh \to0$ when x tends to + infinity (and h tends to $0$ as well).
So you've got I$x$ (f) / I$x$ (g) = 1 + I$x$ (gh) / I$x$ (g) then show that | I$x$ (gh) / I$x$ (g) | tends to $0$ in + infinity
since |I$x$ (gh) / I$x$ (g) | = I$a$ (gh) / I$x$ (g) + I$a-x$ (gh) / Ix (g) then | I$x$ (gh) / I$x$ (g) | $\leq$ I$a$(gh) / I$x$ (g) + sup (h) on [$a$, x]
and since h tends to $0$, you can choose $a$ large enough such that $sup (h) \leq \epsilon / 2$
Once $a$ is chosen, since I$a$ (gh) is a number and I$x$ (g) diverges then there exists $x$ large enough such that I$a$ (gh) / I$x$ (g) $\leq \epsilon / 2$. Finally, for $x$ large enough everything is inferior to a certain epsilon, and you can conclude.
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https://www.statisticshomeworkhelper.com/using-jamovi-to-analyze-music-and-phone-number-association/
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# Understanding the Intersection of Music and Phone Number Sharing Using Jamovi
In this homework, we use Jamovi to explore the intriguing relationship between the type of music played and the propensity of individuals to exchange phone numbers with someone they've recently met. This comprehensive analysis is presented in two parts, each shedding light on distinct aspects of human behavior. In Part 1, we investigate the connection between the genre of music and phone number sharing, while Part 2 explores the relationship between gender and the willingness to seek mental health assistance. Join us as we uncover the subtle yet impactful connections in these intriguing scenarios.
## Part 1: The Relationship Between Music Type and Giving Out Phone Numbers
Problem Description:
In this Jamovi assignment, we investigate the association between the type of music played and the likelihood of people sharing their phone numbers with someone they've recently met. The study involves two categories of music: Romantic Music and Neutral Music. The data collected provides insights into whether the type of music influences individuals' willingness to give out their phone numbers.
## Solution for Part 1:
The expected frequency table for this analysis was generated using Jamovi. The observed (expected) table is presented below:
Gave Phone Number Did Not Give Number Total
Romantic Music 23 (16) 17 (24) 40
Neutral Music 9 (16) 31 (24) 40
Total 32 48 80
We used a Chi-squared test to examine the association between romantic music and the likelihood of people sharing their phone numbers. In the sample, there were 40 instances of romantic music and 40 instances of neutral music. Out of the 40 instances of romantic music, 23 individuals (57.5%) shared their phone numbers. In contrast, out of the 40 instances of neutral music, only 9 individuals (22.5%) shared their phone numbers.
The chi-squared test of association yielded a statistically significant result: χ² (1) = 10.2, p = .001. Based on this significant result, we reject the null hypothesis, which suggested no association between music type and the likelihood of people giving out their phone numbers.
The Cramer's V statistic for the test is 0.357, which falls within the range considered a medium strength of association (>0.3). Therefore, it is evident that there is a medium-strength association between the type of music and the likelihood of people sharing their phone numbers. This insight can be valuable in appropriate settings, such as attracting singles or helping individuals find friends or dates.
## Part 2: Gender and Willingness to Seek Mental Health Assistance
Problem Description:
In Part 2 of the assignment, we explore the relationship between gender and individuals' willingness to seek mental health assistance. The study assesses two groups: males and females, and aims to determine if there are any gender-based preferences when it comes to seeking mental health support.
## Solution for Part 2:
The expected frequency table for this analysis was generated using Jamovi. The observed (expected) table is presented below:
Probably No (Expected) Maybe (Expected) Probably Yes (Expected) Sub Total
Males 9 (6.2) 16 (15.3) 6 (9.51) 31
Females 6 (8.8) 21 (21.7) 17 (13.49) 44
Sub Total 15 37 23 75
We conducted a Chi-squared test to evaluate the association between gender and the willingness to seek mental health assistance. In the sample, there were 31 males and 44 females. Among the 31 males, 9 (29%) expressed that they were probably not willing to use mental health services, and 16 (51.6%) indicated "maybe." On the other hand, out of the 44 females, 6 (13.6%) reported they would probably not be willing to use mental health services, and 21 indicated "maybe."
The chi-squared test of association resulted in a statistically insignificant outcome: χ² (2) = 4.42, p = .110. Based on this result, we do not reject the null hypothesis, suggesting no association between gender and the willingness to seek mental health assistance.
The Cramer's V statistic for this test is 0.236, indicating a weak strength of association between the two variables. Therefore, from this study, we can conclude that there is no significant difference in the willingness to seek mental health assistance between males and females. This insight suggests that gender does not play a significant role in individuals' preferences regarding mental health support.
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Instantly share code, notes, and snippets.
# dbenhur/subset_bm.rb Created Jul 5, 2012
StackOverflow - Ruby - Optimize the comparison of two arrays with duplicates
#!/usr/bin/env ruby # benchmarks for http://stackoverflow.com/questions/11349544/ruby-optimize-the-comparison-of-two-arrays-with-duplicates/11352055#comment14951595_11352055 require 'rubygems' require 'multiset' require 'benchmark' small_b = "cheddaar".split(//) small_a = "cheddar".split(//) BIG_A_SZ = 10_000 big_b = (0...BIG_A_SZ).to_a big_a = big_b.dup big_a.delete_at(rand(BIG_A_SZ)) N = 100_000 BIGN = 10 def op_del_subset(a,b) aa = a.dup b.each do |letter| aa.delete_at(aa.index(letter)) rescue "" end aa.empty? end def ct_subset(a,b) a_counts = a.reduce(Hash.new(0)) { |m,v| m[v] += 1; m } b_counts = b.reduce(Hash.new(0)) { |m,v| m[v] += 1; m } a_counts.all? { |a_key,a_ct| a_ct <= b_counts[a_key] } end def ct_acc_subset(a,b) counts = a.reduce(Hash.new(0)) { |m,v| m[v] += 1; m } b.each { |v| counts[v] -= 1 } counts.values.all? { |ct| ct <= 0 } end def mset_subset(a,b) Multiset.new(a).subset? Multiset.new(b) end def slow_ct_subset(a,b) !a.find{|x| a.count(x) > b.count(x)} end cases = %w{ op_del ct ct_acc mset slow_ct } Benchmark.bm(15) do |bm| a, b = small_a, small_b cases.each do |test| meth = :"#{test}_subset" bm.report("s_#{test}") { N.times{ send(meth, a, b); send(meth, b, a) } } end a, b = big_a, big_b cases.each do |test| meth = :"#{test}_subset" bm.report("b_#{test}") { BIGN.times{ send(meth, a, b); send(meth, b, a) } } end a, b = small_a, big_b cases.each do |test| meth = :"#{test}_subset" bm.report("sb_#{test}") { BIGN.times{ send(meth, a, b); send(meth, b, a) } } end end
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https://www.bystudin.com/during-the-decomposition-of-a-sample-of-barium-carbonate-a-gas-with-a-volume-of-1-12-liters-in-terms-of-standard-units/
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# During the decomposition of a sample of barium carbonate, a gas with a volume of 1.12 liters (in terms of standard units)
During the decomposition of a sample of barium carbonate, a gas with a volume of 1.12 liters (in terms of standard units) was released. The mass of the solid residue was 27.35 g. After that, 73 g of a 30% hydrochloric acid solution were added to the residue. Determine the mass fraction of hydrochloric acid in the resulting solution.
When barium carbonate decomposes, barium oxide is formed and carbon dioxide is released:
ВаСО3 → t ВаО + СО2
Let’s calculate the amount of carbon dioxide released during the calcination of barium carbonate:
n (CO2) = 1.12 l / 22.4 l / mol = 0.05 mol,
therefore, as a result of the decomposition reaction of barium carbonate, 0.05 mol of barium oxide was formed, and 0.05 mol of barium carbonate also reacted. Let us calculate the mass of the formed barium oxide:
m (ВаО) = 153 g / mol ∙ 0.05 mol = 7.65 g.
We calculate the mass and amount of the substance of the remaining barium carbonate:
m (BaCO3) rest. = 27.35 g – 7.65 g = 19.7 g
n (BaCO3) rest = 19.7 g / 197 g / mol = 0.1 mol.
Both components of the solid residue react with hydrochloric acid – the formed barium oxide and the remaining barium carbonate:
ВаО + 2HCl → ВаСl2 + Н2O
ВаСО3 + 2HCl → ВаСl2 + CO2 ↑ + Н2O.
We calculate the amount of matter and the mass of hydrogen chloride interacting with barium oxide and carbonate:
n (HCl) = (0.05 mol + 0.1 mol) ∙ 2 = 0.3 mol;
m (НСl) = 36.5 g / mol ∙ 0.3 mol = 10.95 g.
Let’s calculate the mass of the remaining hydrogen chloride:
m (HCl) rest. = 73 g ∙ 0.3 – 10.95 g = 10.95 g
We calculate the mass-final solution:
mcon.r-ra = m (НCl) solution –m (СО2) = 27.35 g + 73g– 4.4 g = 95.95 g.
The mass fraction of the remaining hydrochloric acid in the solution is equal to:
ω (HCl) = m (HCl) rest. / mcon.r-pa = 10.95 g / 95.95 g = 0.114 (11.4%).
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Chronology Current Month Current Thread Current Date [Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]
# Re: [Phys-l] velocity-dependent mass (or not)
--- On Sun, 6/28/09, John Denker <jsd@av8n.com> wrote:
"It doesn't "come from" anywhere. You shouldn't assume it needs to
"come from" anywhere. Asking where it comes from has no physical
significance, because mass is not conserved. There is no reason
why it should be."
It comes from the past. If an isolated system
(e.g. positronium) had a certain rest mass before reaction (e.g.,
before annihilation), it has the same rest mass after reaction. In case of
annihilation, the system of emerging two photons has the rest mass
exactly equal to the initial mas of positronium. The system has changed
beyond recognition, but its rest mass remains as before. The rest mass
of the new state comes from the rest mass of the initial state. This is
what conservation means.
.
" Energy is conserved. Rest energy (by itself) is not conserved. There is no reason why it should be".
The reason is that the rest energy is also energy. Would you deny this?
" I don't have a problem with non-conserved mass."
This is YOUR problem, then. And many others, too, unfortunately, who read uncritically the textbooks' statement to this effect. What these statements actually mean is that the rest mass is NOT additive. Non-additivity has nothing to do with non-conservation.
"There is plenty of evidence (including the "extreme" example cited above) to tell us that mass is not in fact conserved."
I am curious to see references to any experimental or other scientific evidence (not just statements like the one right above) that the rest mass of the two-photon system resulting from positronium annihilation is not conserved. Could you tell what is this new rest mass equal to, then? And could you answer the question which is reverse to *where the new rest mass come from*: namely, if the rest mass here does not conserve, where does the rest mass of positronium go to after the annihilation? Disappears into thin air?
"Mass is Lorentz invariant. That means it is invariant with respect
to Lorentz transformations. That does *not* mean it is invariant
with respect to all imaginable transformations (such as annihilation
reactions)."
Since when the Lorentz transformations started denying conservation laws?
"I apologize to the list members who think I am belaboring the obvious".
I join you in this.
Moses Fayngold,
NJIT
_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l
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# Finding parameters of image filter using classified pairs
I want to solve the problem of finding a parameter vector for an image filter (let us assume we know nothing about how the filter works, but we can feed it an input image and a set of parameters to produce an output image).
Thus, having a set $\{{I_k, J_k:=F_{\alpha}(I_k)}\}_{k\in\overline{1,N}}$ of $I_k$ images together with their filtered counterparts, $J_k$, what solutions would you recommend for finding $\alpha^\ast$ such that given $I^\ast$ the result $F_{\alpha^\ast}(I^\ast)$ is in the same "style" as the one of the $N$ training correspondence pairs.
I suppose one option is to use a convnet to transform $I_k$ into a feature vector, $v_k$, and then concatenate $\alpha_k$ to obtain $u_k =(v_k,\alpha_k)$. Once this is done, use a regression method to estimate the $\alpha^\ast$ part of $u^\ast$.
I would like to find an alternative solution to what seems like a candidate for the style transfer approach (e.g. https://arxiv.org/pdf/1703.07511.pdf). That approach seems to solve the problem differently, and I envision situations where I need to simply use a filter rather than let a network "guess the style of that filter".
Given the invoked no free lunch prospects, let us assume, for a paeticular problem from this class, that $F$ is a non-linear kernel-based filter that maps $I$ to $J$ as a result of an iterative and convergent process. More specifically, let $F$ be a mean shift filter with the $\alpha=(\rho, \sigma_s,\sigma_r)$ using a concatenated Gaussian kernel and a Parzen window of size $\rho$. Intuitively, I would be tempted to guess that this filter is not smooth w.r.t. $\alpha$, but a formal investigation is required (I suspect it is not smooth given that infinitesimal changes in the size of the window could shift the output towards another mode, indicating a step function behaviour).
In general, it is correct to assume that $\alpha \in \mathbb{R}^d$, with $d \ll N$.
Given the goal of finding $\alpha$ when both the filter action is known (either via numerical computation in general, or, in closed-form if the filter is a gaussian blur, for example), we can be confident that the $N$ input samples have non-constant $\alpha_k$ vector values to start with.
But for sake of generalizability, it would be more elegant to pursue a solution that does not need to know how the filter operates without actually applying it to an input. The first approach suggested in the comments and based on convnets seems to fit this scenario and the optimization problem is taking into account the filter error. However, it would be interesting to hear more opinions, perhaps involving shallow approaches, even at the expense of designing the solution to address the concrete mean shift filter example from above.
• When you describe the convent approach to estimate $\alpha_{k}$ do you you use $F$ at all ? You could compose $F$ with the convent $v_{k}$ and train the system fully end to end using $J_{k}$ and $a_{k}$ to construct your cost function. This way it is more explicit that you are trying to learn the filters parameters from both its target output and prior used filter params. May 25, 2017 at 13:33
• I am not using any posterior knowledge involving $F$ directly in the described approach/idea. I do not have a sufficient number of classified data to train a convnet from ground-up, so I would have to use one priorly trained to have it extract features. Would it help if I build the feature vectors as $(v_k, v'_k, \alpha_k)$, where $v'_k$ is the convnet feature vector corresponding to $J_k$? May 25, 2017 at 14:03
• I am not sure it would help much. Doing the regression decoupled from the error that $F$ generates when using $\alpha_{k}$ is probably what would be the best to improve. How much data do you have ? May 25, 2017 at 15:41
• Currently I have not more than 200 pairs selected by manual parameter tuning of the $\alpha$ vector. So your suggestion would be to find the parameter vector via regression without plugging the feature vectors of the filtered images as additional dimensions of the sample space (that extra info should already be encoded in the $\alpha_k$ of the training samples. May 25, 2017 at 16:06
• My initial suggestion was $\hat{\mathbf{W}} = \underset{\mathbf{W}} {arg \ min}\left \{ \sum_{k=1}^{N}(J_{k} - F(\sigma(I_{k}, \mathbf{W}), I_{k} ) )^{2} + (\alpha_{k} - \sigma(I_{k}, \mathbf{W}) )^{2} \right \}$ where $\sigma(I,\mathbf{W})$ is a convnet with weights $\mathbf{W}$. this way the convnet provides you with the parameter vector ($a^{*} = \sigma(I^{*} , \hat{\mathbf{W}})$ ). If you do not have enough data given the data is images, you can look into subtle transformations to augment your dataset. Salt and pepper, rotations, reflections , zca-whitening etc. May 25, 2017 at 16:22
Your parameter $\alpha$ has fairly low dimension. Therefore, I recommend that you apply optimization methods directly to try to find the best $\alpha$ (without trying to use convolutional neural networks and regression for this purpose).
Define a distance measure on images, $\|I-J\|$, to represent how dissimilar images $I,J$ are. You might use the squared $L_2$ norm for this, for instance.
Now, the loss for a particular parameter choice $\alpha$ is
$$L(\alpha) = \sum_{k=1}^N \|F_\alpha(I_k)-J_k\|.$$
We can now formulate your problem as follows: given a training set of images $(I_k,J_k)$, find the parameter $\alpha$ that minimizes the loss $L(\alpha)$.
A reasonable approach is to use some optimization procedure to solve this problem. You might use stochastic gradient descent, for instance. Because there might be multiple local minima, I would suggest that you start multiple instances of gradient descent from different starting points: use a grid search over the starting point. Since your $\alpha$ is only three-dimensional, it's not difficult to do a grid search over this three-dimensional space and then start gradient descent from each point in the grid. Stochastic gradient descent will allow you to deal with fairly large values of $N$.
This does require you to be able to compute gradients for $L(\alpha)$. Depending on the filter $\alpha$, it might be possible to symbolically calculate the gradients (perhaps with the help of a framework for this, such as Tensorflow); if that's too hard, you can use black-box methods to estimate the gradient by evaluating $L(\cdot)$ at multiple points.
If $L_2$ distance doesn't capture similarity in your domain, you could consider other distance measures as well.
I expect this is likely to be a more promising approach than what you sketched in the question, using convolutional networks and a regression model. (For one thing, there's no reason to expect the mapping from "features of $I_k$" to "features of $J_k$" to be linear, so there's no reason to expect linear regression to be effective here.)
• Its not necessary to compute the gradients for the filter. Look into Bayesian optimisation if your filter is indeed a blackbox. May 28, 2017 at 12:17
• Yes, computing the gradient of an iterative process analytically eludes me. I also do not think that $F_{\alpha}$ is smooth (w.r.t $\alpha$) and is definitely not guaranteed to be if $F$ is a black box. May 29, 2017 at 6:26
• The drawback of seeking $\alpha$ given the $I_k$ and $J_k$ correspondences makes this formulation behave like a one size fits-all approach. Establishing feature correspondences between $I_k$, $\alpha_k$ and $J_k$ is definitely not guaranteed to be linear, but I am not fully convinced that, even with simpler filters, it is correct to assume the same $\alpha$ parameter value. In general, I would tend to regard the filter parameters as a function of the input. May 29, 2017 at 6:30
• @teodron, you don't do it analytically; you do it in a blackbox way, e.g., $\nabla_h L(x)$ is approximated by $(L(x+ \epsilon h)-L(x))/\epsilon$ where $\epsilon$ is sufficiently small. I don't know what you mean by "one-size-fits-all". If you have a separate $\alpha$ value for each $I_k,J_k$ then you don't have a single problem; you have $N$ independent problems, one per value of $k$. If the filter parameters are a function of the input, then let $\alpha$ represent that function, and we're back to a single $\alpha$ for all the images.
– D.W.
May 29, 2017 at 6:33
• I can think of a stupid example where the filter is a rotation plus translation of a black and white shape. Imagine that such a filter simply counts the number of corners of the shape, $N_c$. Then it translates the shape by $r*(cos(N_c\pi/12),sin(N_c\pi/12))$. Yes, the filter is stupid, but using a single value of the filter for any subsequent inputs would produce the same translation of the input regardless the number of corners. May 29, 2017 at 6:36
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# Convert hectometers per (second squared) to yards per (second squared)
## [hm/s² to yd/s²] (hm:hectometer, s:second, yd:yard)
### Convert hm/s² to yd/s²
#### an acceleration conversion table
to 100 with
hecto-
meter per second squared
×103,
yard per second squared
hecto-
meter per second squared
×103,
yard per second squared
hecto-
meter per second squared
×103,
yard per second squared
hecto-
meter per second squared
×103,
yard per second squared
hecto-
meter per second squared
×103,
yard per second squared
10.11212.30414.48616.67818.86
20.22222.41424.59626.78828.97
30.33232.52434.70636.89839.08
40.44242.62444.81647.00849.19
50.55252.73454.92657.11859.30
60.66262.84465.03667.22869.41
70.77272.95475.14677.33879.51
80.87283.06485.25687.44889.62
90.98293.17495.36697.55899.73
101.09303.28505.47707.66909.84
111.20313.39515.58717.76919.95
121.31323.50525.69727.879210.06
131.42333.61535.80737.989310.17
141.53343.72545.91748.099410.28
151.64353.83556.01758.209510.39
161.75363.94566.12768.319610.50
171.86374.05576.23778.429710.61
181.97384.16586.34788.539810.72
192.08394.27596.45798.649910.83
202.19404.37606.56808.7510010.94
### hectometers per second squared to yards per second squared conversion cards
• 1
through
20
hectometers per second squared
• 1 hm/s² to yd/s² = 110 yd/s²
• 2 hm/s² to yd/s² = 220 yd/s²
• 3 hm/s² to yd/s² = 330 yd/s²
• 4 hm/s² to yd/s² = 440 yd/s²
• 5 hm/s² to yd/s² = 550 yd/s²
• 6 hm/s² to yd/s² = 660 yd/s²
• 7 hm/s² to yd/s² = 770 yd/s²
• 8 hm/s² to yd/s² = 870 yd/s²
• 9 hm/s² to yd/s² = 980 yd/s²
• 10 hm/s² to yd/s² = 1 090 yd/s²
• 11 hm/s² to yd/s² = 1 200 yd/s²
• 12 hm/s² to yd/s² = 1 310 yd/s²
• 13 hm/s² to yd/s² = 1 420 yd/s²
• 14 hm/s² to yd/s² = 1 530 yd/s²
• 15 hm/s² to yd/s² = 1 640 yd/s²
• 16 hm/s² to yd/s² = 1 750 yd/s²
• 17 hm/s² to yd/s² = 1 860 yd/s²
• 18 hm/s² to yd/s² = 1 970 yd/s²
• 19 hm/s² to yd/s² = 2 080 yd/s²
• 20 hm/s² to yd/s² = 2 190 yd/s²
• 21
through
40
hectometers per second squared
• 21 hm/s² to yd/s² = 2 300 yd/s²
• 22 hm/s² to yd/s² = 2 410 yd/s²
• 23 hm/s² to yd/s² = 2 520 yd/s²
• 24 hm/s² to yd/s² = 2 620 yd/s²
• 25 hm/s² to yd/s² = 2 730 yd/s²
• 26 hm/s² to yd/s² = 2 840 yd/s²
• 27 hm/s² to yd/s² = 2 950 yd/s²
• 28 hm/s² to yd/s² = 3 060 yd/s²
• 29 hm/s² to yd/s² = 3 170 yd/s²
• 30 hm/s² to yd/s² = 3 280 yd/s²
• 31 hm/s² to yd/s² = 3 390 yd/s²
• 32 hm/s² to yd/s² = 3 500 yd/s²
• 33 hm/s² to yd/s² = 3 610 yd/s²
• 34 hm/s² to yd/s² = 3 720 yd/s²
• 35 hm/s² to yd/s² = 3 830 yd/s²
• 36 hm/s² to yd/s² = 3 940 yd/s²
• 37 hm/s² to yd/s² = 4 050 yd/s²
• 38 hm/s² to yd/s² = 4 160 yd/s²
• 39 hm/s² to yd/s² = 4 270 yd/s²
• 40 hm/s² to yd/s² = 4 370 yd/s²
• 41
through
60
hectometers per second squared
• 41 hm/s² to yd/s² = 4 480 yd/s²
• 42 hm/s² to yd/s² = 4 590 yd/s²
• 43 hm/s² to yd/s² = 4 700 yd/s²
• 44 hm/s² to yd/s² = 4 810 yd/s²
• 45 hm/s² to yd/s² = 4 920 yd/s²
• 46 hm/s² to yd/s² = 5 030 yd/s²
• 47 hm/s² to yd/s² = 5 140 yd/s²
• 48 hm/s² to yd/s² = 5 250 yd/s²
• 49 hm/s² to yd/s² = 5 360 yd/s²
• 50 hm/s² to yd/s² = 5 470 yd/s²
• 51 hm/s² to yd/s² = 5 580 yd/s²
• 52 hm/s² to yd/s² = 5 690 yd/s²
• 53 hm/s² to yd/s² = 5 800 yd/s²
• 54 hm/s² to yd/s² = 5 910 yd/s²
• 55 hm/s² to yd/s² = 6 010 yd/s²
• 56 hm/s² to yd/s² = 6 120 yd/s²
• 57 hm/s² to yd/s² = 6 230 yd/s²
• 58 hm/s² to yd/s² = 6 340 yd/s²
• 59 hm/s² to yd/s² = 6 450 yd/s²
• 60 hm/s² to yd/s² = 6 560 yd/s²
• 61
through
80
hectometers per second squared
• 61 hm/s² to yd/s² = 6 670 yd/s²
• 62 hm/s² to yd/s² = 6 780 yd/s²
• 63 hm/s² to yd/s² = 6 890 yd/s²
• 64 hm/s² to yd/s² = 7 000 yd/s²
• 65 hm/s² to yd/s² = 7 110 yd/s²
• 66 hm/s² to yd/s² = 7 220 yd/s²
• 67 hm/s² to yd/s² = 7 330 yd/s²
• 68 hm/s² to yd/s² = 7 440 yd/s²
• 69 hm/s² to yd/s² = 7 550 yd/s²
• 70 hm/s² to yd/s² = 7 660 yd/s²
• 71 hm/s² to yd/s² = 7 760 yd/s²
• 72 hm/s² to yd/s² = 7 870 yd/s²
• 73 hm/s² to yd/s² = 7 980 yd/s²
• 74 hm/s² to yd/s² = 8 090 yd/s²
• 75 hm/s² to yd/s² = 8 200 yd/s²
• 76 hm/s² to yd/s² = 8 310 yd/s²
• 77 hm/s² to yd/s² = 8 420 yd/s²
• 78 hm/s² to yd/s² = 8 530 yd/s²
• 79 hm/s² to yd/s² = 8 640 yd/s²
• 80 hm/s² to yd/s² = 8 750 yd/s²
• 81
through
100
hectometers per second squared
• 81 hm/s² to yd/s² = 8 860 yd/s²
• 82 hm/s² to yd/s² = 8 970 yd/s²
• 83 hm/s² to yd/s² = 9 080 yd/s²
• 84 hm/s² to yd/s² = 9 190 yd/s²
• 85 hm/s² to yd/s² = 9 300 yd/s²
• 86 hm/s² to yd/s² = 9 410 yd/s²
• 87 hm/s² to yd/s² = 9 510 yd/s²
• 88 hm/s² to yd/s² = 9 620 yd/s²
• 89 hm/s² to yd/s² = 9 730 yd/s²
• 90 hm/s² to yd/s² = 9 840 yd/s²
• 91 hm/s² to yd/s² = 9 950 yd/s²
• 92 hm/s² to yd/s² = 10 060 yd/s²
• 93 hm/s² to yd/s² = 10 170 yd/s²
• 94 hm/s² to yd/s² = 10 280 yd/s²
• 95 hm/s² to yd/s² = 10 390 yd/s²
• 96 hm/s² to yd/s² = 10 500 yd/s²
• 97 hm/s² to yd/s² = 10 610 yd/s²
• 98 hm/s² to yd/s² = 10 720 yd/s²
• 99 hm/s² to yd/s² = 10 830 yd/s²
• 100 hm/s² to yd/s² = 10 940 yd/s²
#### Foods, Nutrients and Calories
TRULY GOOD FOODS, BANANA SPLIT, UPC: 094184599375 contain(s) 467 calories per 100 grams (≈3.53 ounces) [ price ]
32427 foods that contain Sugars, added. List of these foods starting with the highest contents of Sugars, added and the lowest contents of Sugars, added
#### Gravels, Substances and Oils
Gravel, Dolomite weighs 1 865 kg/m³ (116.42815 lb/ft³) with specific gravity of 1.865 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Disodium tetraborate pentahydrate [Na2B4O7 ⋅ 5H2O] weighs 1 880 kg/m³ (117.36457 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Soybean oil with temperature in the range of 10°C (50°F) to 140°C (284°F)
#### Weights and Measurements
A kilometer per second (km/s) is a derived metric SI (System International) measurement unit of speed or velocity with which to measure how many kilometers traveled per one second.
The frequency is defined as an interval of time, during which a physical system, e.g. electrical current or a wave, performs a full oscillation and returns to its original momentary state, in both sign (direction) and in value, is called the oscillation period of this physical system.
lbf/µ² to dyn/mil;² conversion table, lbf/µ² to dyn/mil;² unit converter or convert between all units of pressure measurement.
#### Calculators
Calculate volume of a right circular cone and its surface area
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https://math.answers.com/Q/Do_you_reduce_proportions
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0
# Do you reduce proportions
Updated: 9/15/2023
Wiki User
15y ago
I would say yes. For example, 10:1 is so much easier (when one number is a 'one') to read than 100:10 or 20:2. Exceptions will be perhaps 5:3 is easier to read than 1.666666666666:1. If the question requires a certain syntax, then it should say so.
===================
Wiki User
15y ago
Earn +20 pts
Q: Do you reduce proportions
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Still have questions?
Related questions
### What is similar about scale factor and proportions?
proportions are used in scale factors; scale factors ARE proportions
### How do you name the cross products of proportions?
They are other proportions.
### What is the definition of different proportions?
Proportions of differing proportionality
### What are HD video proportions?
HD video proportions are 1024x720 :)
### What can you use to measure proportions?
you can use scale model to measure proportions
### How would one gain health net insurance?
To gain health net insurance you would have to reduce some percent from your medical, so that the proportions would be equally, and so you can use your premium.
### Where can you find proportions?
See related link below to find help on proportions
### How can proportions be used in a sentence?
They made sure the profits were divided in the correct proportions.
### How do you take mass gainer?
You don't, build muscle through hard work, not supplements, they're seriously bad for your health an reduce your sperm count. But if you really want to then you mix it in with water, the proportions should be on the container
### When was the law of definite proportions developed?
The law of definite proportions was developed by Joseph Proust in 1806.
### What is similar about scale and scale factor?
proportions are used in scale factors; scale factors ARE proportions
### How do proportions work?
proportions are used in calculations to help in the division of parts into units for effective distribution.
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https://gitbook.rootwhois.cn/Alogrithm/Leetcode/371.%20Sum%20of%20Two%20Integers.html
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# 371. Sum of Two Integers
## 1. Question
Given two integers a and b, return the sum of the two integers without using the operators + and -.
## 2. Examples
Example 1:
Input: a = 1, b = 2
Output: 3
Example 2:
Input: a = 2, b = 3
Output: 5
## 3. Constraints
• -1000 <= a, b <= 1000
## 5. Solutions
and 同1为1
or 同0为0
xor 不同为1
xor相当于不进位的加法
and可以计算出最高位是否需要进位（同1为1，左移1位达到效果）
class Solution {
public int getSum(int a, int b) {
return a != 0 ? getSum((a & b) << 1, a ^ b) : b;
}
}
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https://www.brightsurf.com/news/article/091802130688/new-study-casts-doubt-on-validity-of-standard-earthquake-prediction-model.html
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# New study casts doubt on validity of standard earthquake-prediction model
September 18, 2002
A new study by Stanford University geophysicists is raising serious questions about a fundamental technique used to make long-range earthquake predictions.
Writing in the journal Nature, geophysicists Jessica Murray and Paul Segall show how a widely used earthquake model failed to predict when a long-anticipated magnitude 6 quake would strike the San Andreas Fault in Central California.
In their Sept. 19 Nature study, Murray and Segall analyzed the "time-predictable recurrence model" - a technique for estimating the time when an earthquake will occur. This model is used to calculate the probability of future earthquakes.
Developed by Japanese geophysicists K. Shimazaki and T. Nakata in 1980, the time-predictable model has become a standard tool for hazard prediction in many earthquake-prone regions - including the United States, Japan and New Zealand.
Strain build-up
The time-predictable model is based on the theory that earthquakes in fault zones are caused by the constant build-up and release of strain in the Earth's crust.
"With a plate boundary like the San Andreas, you have the North American plate on one side and the Pacific plate on the other," explained Segall, a professor of geophysics. "The two plates are moving at a very steady rate with respect to one another, so strain is being put into the system at an essentially constant rate."
When an earthquake occurs on the fault, a certain amount of accumulated strain is released, added Murray, a geophysics graduate student.
"Following the quake, strain builds up again because of the continuous grinding of the tectonic plates," she noted. "According to the time-predictable model, if you know the size of the most recent earthquake and the rate of strain accumulation afterwards, you should be able to forecast the time that the next event will happen simply by dividing the strain released by the strain-accumulation rate."
Parkfield, Calif.
Although the model makes sense on paper, Murray and Segall wanted to put it to the test using long-term data collected in an ideal setting. Their choice was Parkfield - a tiny town in Central California midway between San Francisco and Los Angeles. Perched along the San Andreas Fault, Parkfield has been rocked by a magnitude 6 earthquake every 22 years on average since 1857. The last one struck in 1966, and geologists have been collecting earthquake data there ever since.
"Parkfield is a good place to test the model because we have measurements of surface ground motion during the 1966 earthquake and of the strain that's been accumulating since," Murray noted. "It's also located in a fairly simple part of the San Andreas system because it's on the main strand of the fault and doesn't have other parallel faults running nearby."
When Murray and Segall applied the time-predictable model to the Parkfield data, they came up with a forecast of when the next earthquake would occur.
"According to the model, a magnitude 6 earthquake should have taken place between 1973 and 1987 - but it didn't," Murray said. "In fact, 15 years have gone by. Our results show, with 95 percent confidence, that it should definitely have happened before now, and it hasn't, so that shows that the model doesn't work - at least in this location."
Could the time-predictable method work in other parts of the fault, including the densely populated metropolitan areas of Northern and Southern California? The researchers have their doubts,
"We used the model at Parkfield where things are fairly simple," Murray observed, "but when you come to the Bay Area or Los Angeles, there are a lot more fault interactions, so it's probably even less likely to work in those places."
Segall agreed: "I have to say, in my heart, I believe this model is too simplistic. It's really not likely to work elsewhere, either, but we still should test it at other sites. Lots of people do these kinds of calculations. What Jessica has done, however, is to be extremely careful. She really bent over backwards to try to understand what the uncertainties of these kinds of calculations are - consulting with our colleagues in the Stanford Statistics Department just to make sure that this was done as carefully and precisely as anybody knows how to do. So we feel quite confident that there's no way to fudge out of this by saying there are uncertainties in the data or in the method."
Use with caution
Segall pointed out that government agencies in a number of Pacific Rim countries routinely use this technique for long-range hazard assessments.
For example, the U.S. Geological Survey (USGS) relied on the time-predictable model and two other models in its widely publicized 1999 report projecting a 70-percent probability of a large quake striking the San Francisco Bay Area by 2030.
"We're in a tough situation, because agencies like the USGS - which have the responsibility for issuing forecasts so that city planners and builders can use the best available knowledge - have to do the best they can with what information they have." Segall observed. "The message I would send to my geophysical colleagues about this model is, 'Use with caution.'"
Technological advances in earthquake science could make long-range forecasting a reality one day, added Murray, pointing to the recently launched San Andreas Fault drilling experiment in Parkfield under the aegis of USGS and Stanford.
In the mean time, people living in earthquake-prone regions should plan for the inevitable.
"I always tell people to prepare," Segall concluded. "We know big earthquakes have happened in the past, we know they will happen again. We just don't know when."
-end-
By Mark Shwartz
Funding for the Nature study was provided by Stanford University Graduate Fellowships and the USGS Earthquake Hazards Reduction Program.
COMMENT:
Paul Segall, Geophysics 650-725-7241;
segall@stanford.edu
Jessica Murray, Geophysics 650-723-5485;
jrmurray@pangea.stanford.edu
EDITORS: Murray and Segall's study, "Testing time-predictable earthquake recurrence by direct measurement of strain accumulation and release," will be published in the Sept. 19 issue of Nature.
Relevant Web URLs:
http://kilauea.Stanford.EDU/paul/
http://kilauea.stanford.edu/jrmurray/
http://quake.wr.usgs.gov/research/parkfield/index.html
http://geopubs.wr.usgs.gov/open-file/of99-517/
Stanford University
## Related Earthquake Articles from Brightsurf:
Healthcare's earthquake: Lessons from COVID-19
Leaders and clinician researchers from Beth Israel Lahey Health propose using complexity science to identify strategies that healthcare organizations can use to respond better to the ongoing pandemic and to anticipate future challenges to healthcare delivery.
Earthquake lightning: Mysterious luminescence phenomena
Photoemission induced by rock fracturing can occur as a result of landslides associated with earthquakes.
How earthquake swarms arise
A new fault simulator maps out how interactions between pressure, friction and fluids rising through a fault zone can lead to slow-motion quakes and seismic swarms.
Typhoon changed earthquake patterns
Intensive erosion can temporarily change the earthquake activity (seismicity) of a region significantly.
Cause of abnormal groundwater rise after large earthquake
Abnormal rises in groundwater levels after large earthquakes has been observed all over the world, but the cause has remained unknown due to a lack of comparative data before & after earthquakes.
New clues to deep earthquake mystery
A new understanding of our planet's deepest earthquakes could help unravel one of the most mysterious geophysical processes on Earth.
Fracking and earthquake risk
Earthquakes caused by hydraulic fracturing can damage property and endanger lives.
Earthquake symmetry
A recent study investigated around 100,000 localized seismic events to search for patterns in the data.
Crowdsourcing speeds up earthquake monitoring
Data produced by Internet users can help to speed up the detection of earthquakes.
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https://cran.stat.sfu.ca/web/packages/neldermead/vignettes/manual.html
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neldermead is a R port of a module originally developed for Scilab version 5.2.1 by Michael Baudin (INRIA - DIGITEO). Information about this software can be found at www.scilab.org. The following documentation as well as the content of the functions .Rd files are adaptations of the documentation provided with the original Scilab neldermead module.
neldermead currently does not include any adaptation of the Scilab ‘nmplot’ function series that is available in the original neldermead module.
# Overview
## Description
The goal of this toolbox is to provide several direct search optimization algorithms based on the simplex method. The optimization problem to solve is the minimization of a cost function, with bounds and nonlinear constraints.
$\begin{array}{l l} min f(x)\\ l_i \le{} x_i \le{} h_i, & i = 1,n \\ g_j(x) \ge{} 0, & j = 0,nb_{ineq} \\\\ \end{array}$
where $$f$$ is the cost function, $$x$$ is the vector of parameter estimates, $$l$$ and $$h$$ are vectors of lower and upper bounds for the parameter estimates, $$n$$ is the number of parameters and $$nb_{ineq}$$ the number of inequality constraints $$g(x)$$.
The provided algorithms are direct search algorithms, i.e. algorithms which do not use the derivative of the cost function. They are based on the update of a simplex, which is a set of $$k \ge n+1$$ vertices, where each vertex is associated with one point and one function value.
The following algorithms are available:
• The fixed shape simplex method of Spendley, Hext and Himsworth: this algorithm solves an unconstrained optimization problem with a fixed shape simplex made of $$k = n+1$$ vertices.
• The variable shape simplex method of Nelder and Mead: this algorithm solves an unconstrained optimization problem with a variable shape simple
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https://books.google.com.jm/books?id=_ulHAAAAIAAJ&qtid=cf7e78cf&lr=&source=gbs_quotes_r&cad=4
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Books Books
If a straight line touch a circle, and from the point of contact a chord be drawn, the angles which this chord makes with the tangent are equal to the angles in the alternate segments.
Theoretical Geometry: Based on the Various Geometry Books by Godfrey and Siddons - Page xiv
by Arthur Warry Siddons, Reginald Thomas Hughes - 1926 - 173 pages
The Elements of Euclid: In which the Propositions are Demonstrated in a New ...
Euclid - Geometry - 1776 - 326 pages
...and from the point of contact a right •*• line be drawn to the circle, the angles that right line makes with the tangent are equal to the angles in the alternate feements of the circle. Let the right line EF touch the circle ABCD in the point B; from any point...
Elements of Geometry: Containing the First Six Books of Euclid, with a ...
John Playfair - Euclid's Elements - 1806 - 320 pages
...one angle of a triangle be equal to the other two angles, it is a right angle. PROP. XXXII. THEOR. IF a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle, the angles -made by this line with the line which touches...
The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh ...
Euclid - Geometry - 1810 - 554 pages
...FC is perpendicular to DE. Therefore, if a straight line, &c. QED C GE a 18. 3. ,' PROP. XIX. THEOR. IF a straight line touch a circle, and from the point of contact a straight line be drawn at right angles to the touching lirie, the centre of the circle shall be in...
Easy Introduction to Mathematics, Volume 2
Charles Butler - 1814 - 582 pages
...demonstration, that these three perpendiculars BE, GF, and CD intersect each other in the same point AQED 249. If a straight line touch a circle, and from the point of contact two chords be drawn, and if from the extremity of one of them, a straight line be drawn parallel to...
Elements of Geometry: Containing the First Six Books of Euclid, with a ...
John Playfair - Circle-squaring - 1819 - 350 pages
...DE ; FC is therefore perpendicular to DE. Therefore, if a straight line, &c. Q, ED PROP. XIX. THEOR. If a straight line touch a circle, and from the point of contact a straight line be drawn at right angles to the touching line, the centre of the circle, is in that line....
The Elements of Euclid: The Errors by which Theon, Or Others, Have Long ...
Robert Simson - Trigonometry - 1827 - 546 pages
...the same two ; and when the adjacent angles are equal, they are f right angles. PROP. XXXII. THEOR. If a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle ; the angles which this line makes with the line touching...
Elements of Geometry: Being Chiefly a Selection from Playfair's Geometry
John Playfair - Geometry - 1829 - 210 pages
...to DE; therefore FC is perp. to DE. Therefore, if a straight line &c. QED PROPOSITION XIX. THEOREM. If a straight line touch a circle, and from the point of contact a straight line be drawn perpendicular to the tangent, the centre of the circle is in that line. •...
The Teacher's Assistant in the "Course of Mathematics Adapted to the Method ...
Mathematics - 1836 - 488 pages
...it is equal to the same two ; and when the adjacent angles are equal, they are right angles. XXXI. If a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle, the angles made by this line with the line which touches...
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# Equation of a Tangent Line from a Point on the Circle
How to find the equation of a tangent line from a point on the circle: example and its solution.
## Example
Draw the circle x2 + y2 = 10
on the coordinate plane.
Its center is (0, 0).
Equation of a circle
Draw P(3, 1) on the circle.
Draw the line tangent to the circle at P(3, 1).
Your goal is to find
the linear equation of the tangent line.
To find the slope of the tangent line,
first find the slope of OP: mOP.
mOP = 1/3
Slope of a line
The radius (OP) and the tangent line
are perpendicular.
Tangent to a circle
mOP = 1/3
Set the slope of the tangent line as m.
Then (1/3)⋅m = -1.
Perpendicular lines
Multiply 3 on both sides.
Then m = -3.
The slope of the tangent line is -3.
And the tangent line passes through P(3, 1).
So the tangent line is
y = -3(x - 3) + 1.
Point-slope form
-3(x - 3) = -3x + 9
9 + 1 = 10
So the tangent line is
y = -3x + 10.
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HOME CDS Tutorial Manual
Stockpile Volumes
In this tutorial you will learn how to ;
* Use Volumes, To A Plane to calculate a stockpile volume.
* Use different surfaces within the one job.
* Calculate volumes between different surfaces in the one job.
You will find a job named "sse" has been supplied in the tutor folder. Start CDS, and use the File, Open command to open Job 'sse' The screen should appear as seen below. As you can see from the screen there are four, or maybe five different stockpiles within the one holding area, and in this tutorial you will learn how to calculate the volume of material in each pile. To start with, we will concentrate on the largest stockpile that is in the top left of the screen.
To see the data more clearly, you should use Zoom, by either pressing Z, or by selecting the Zoom Icon, and you should select a window similar to that seen in the screen below left to get a screen similar to that below right.
There is not a lot which you can determine from the point numbers, however the field party has coded all of the points, so you should turn the codes on to see if they provide any guide to what we have.
Select the Layers icon, and set it to display the code, and the following screen will appear.
You can now see that the field party has coded all the shots on the edge of the first pile with a code of 'ES1', and the actual material in the pile is coded 'RD' to indicate it is general road material.
Before we can calculate how much material is in the pile, we must first determine the base on which it is sitting.
In Tutorial 7, you had the luxury of being able to actually do a survey of the base before the material was dumped, but this has not occurred in this situation. So, the only thing we can do here is to assume that there is a base surface that is made up of all the points that are on the edge of the stockpile i.e. the points coded 'ES1'
Pull down the Contour menu and select Surface Parameters.
If you now select the tab which is titled Surface Point Selection the screen will appear as seen below.
You will see that there are three distinct areas which will allow you to determine the points to be used by the Point Number, their Code, or their Layer.
In this case layers were not used, and the point numbers are scattered, but the codes will allow us to select efficiently.
Select the existing code range with your cursor, and then select the Edit button.
A box will pop up to with fields for the Minimum and Maximum codes you wish to use.
In this case you should enter 'ES1' in both fields and then select OK.
Now select OK to close down the Surface Point Selection window and return to the main screen.
Now pull down the Contour menu and select Form Model.
The screen should appear as below, and you will see that the triangles have been formed across the base of the stockpile, which is what we intended.
We now wish to determine the volume between this surface and an arbitrary datum plane below it, so pull down the Contour menu, select the Volume option, and select the option "To a Plane'.
The screen shown below left will appear, and you should fill in a value of the plane of 350, then click OK, and the results will be displayed in Wordpad, as seen in the screen below right.
You should make a note of the area of 4625.812 square metres, and the volume of 40436.019 cubic metres, or you may print it out if you wish.
Now you need to model the whole of the pile, and determine the volume above the same plane, so pull down the Contour menu, and select Surface Parameters, then select the Surface Point Selection Tab as before.
Now select the code range, and then click on the Add button.
When the pop up box appears, enter the Code 'RD' in both fields and press OK.
Your screen should appear as seen below, indicating the surface you want is the one that has points with Code 'ES1' and points with Code 'RD'.
Close the Surface Point Selection screen by selecting OK.
Now pull down the Contour menu and select Form Model.
The screen should appear as below, and this time you can see that the model includes all of the points in the first stockpile.
NOTE : when you use this, or any other method of calculating volumes with your own data, it is important that you understand that the contours formed must accurately represent the surfaces you are using, otherwise the volumes you produce will NOT be accurate.
So, here you should now Calculate Contours and allow them to be saved, and the screen should appear as at right.
Since you did not do the job. it is difficult to know whether these contours are correct or not, but if your screen appears the same as the one at right you have got it right.
There is a small disturbance to the wall of the pile at about the 5 o'clock mark, and that is quite correct as a loader had been taking material from that area before the survey was done.
Now that you have the surface accurately modelled, you should calculate the volume to the plane, so pull down Contour, select Volumes, and To a Plane, and enter the value of 350 for the plane.
The screen should appear as below, with the answers displayed in Wordpad.
The first thing you should do is check the figure presented for Total Area. If it is not the same as the value you calculated when you did the base surface, something has gone wrong, and you are not projecting the same areas onto the plane.
In this case, the areas agree, so the volume of 52171.391 can be used.
If you now subtract the volume of the base, of 40436.019 from this figure you will get the amount of material that is in the pile.
Be careful when presenting these figures to a client, and don't get carried away and specify a volume such as this to three decimal places, because it is simply not physically possible to locate the pile to that sort of accuracy in the field.
So, in this instance a value of 11,735 cubic metres is more than accurate enough, and it is probably arguable that it is gilding the lily a little to specify down to the last 5 cubic meters over an area of this size.
And yes I know that your client will argue that you need to report every cubic spoonful to minimise the amount payable if the principal briefs you, or to maximise it if you work for the contractor. You need to carefully consider your own professional integrity, and be wary of being drawn into wasteful arguments over trivial differences in volumes which are simply not measurable under normal field conditions.
When an argument arises over quantities, it is recommended that first you take a simple "reality check" to see how accurately they were, or could have been measured in the first place. If the amount being argued over falls within the normal 'error ellipse' for work of this kind you might consider whether you should advise the warring parties to split the difference and get on with something constructive.
If you now Zoom Out and examine your data, you will see the pile at the bottom of the area has the edge points coded as 'ES2' and the material itself coded as '20mm' indicating 20 millimetre screened blue metal, and if you wish you can substitute these values in the procedure above and obtain a volume, but we will do it by an alternative means.
Using Different Surfaces
With CDS you now have the capability of assigning different sets of points to different surface within the same job, and we will use this facility now to set up the base of stockpile 2 as Surface 2, and the whole of stockpile 3 as Surface 3.
Pull down the Contour menu and select Surface Parameters.
If you look to the right at the top of the screen you will see a button titled Add, and you should select it.
The following screen will appear.
Select the Natural button, since all the points you are dealing with are natural surface points, and then click OK. This will now create a second natural surface called Surface 2, and what appears on it will be determined by what you specify in the Surface Point Selection screen. Click on the Surface Point Selection tab. Now Select the code range presented, and then select Edit, and enter a code of 'ES2' in both fields.
Press OK, and this will save away Surface 2 as the base of the second stockpile.
Check that Surface 2 is selected on the Surface Parameters screen, and when it is, close the screen.
Now zoom extents so you can see the whole job, and Zoom a window around the stockpile at the bottom of the screen.
Now pull down the Contour menu and select Form Model, and your screen should appear similar to that below. Now it is time to set up the whole of the stockpile as Surface 3, so pull down the Contour Menu and select Surface parameters. Choose the Add button , select Natural and click OK to save away Surface 3. Now select Surface Point Selection Click on the existing code range, and use Remove to delete the existing values. Now use Add to add a line with a Minimum code value of ES2 and a maximum value of ES2. Repeat to set up a code of 20MM.
Your codes for Surface 3 should now have one line specifying a code of 'ES2' and a second line specifying a code of '20MM'. Close the screen, and ensure that Surface 3 is current on the Surface parameters screen. Close that screen, and then use Form Model, followed by Calculate Contours, and save the contours away. The screen at the left should appear.
Now that you have separate surfaces defining the base and the entirety of the stockpile, you can determine the volume in this pile by using the Surface to Surface method.
Pull down the Contour menu and select Volumes, followed by Surface to Surface.
When the screen appears, you need to change the Base Surface to be Surface 2 and the Overlay Surface to be Surface 3, so your screen will be similar to that at right.
Now you should select the OK button, and the volumes will be calculated and the answers displayed in Wordpad as seen below.
From these results you would report that there is 2530 cubic metres of 20 millimetre blue metal available in the yard.
If you now Zoom extents, and then Zoom a window around the small pile immediately to the right of the first pile we calculated, you will see a screen similar to that at right.
You will see that all the points have been given the same code, so it is impossible to differentiate the base from the pile using the codes. Select your Layers screen, and set it to display Point Numbers. You will now see that points 234 to 243 inclusive make up the edge of the pile, and points 234 to 246 inclusive represent the pile itself. Now you need to select the Surface Parameters option, and Add another Natural Surface ( Surface 4). While you are there, Add Surface 5 as well.
Now select the Surface Point Selection screen, and set it to Surface 4.
You need to specify a Point range from 234 to 243 for this surface.
Save the parameters and form the model for Surface 4.
Now select the Surface Point Selection screen again, and set it to Surface 5. Specify a range of points from 234 to 246 and save it away.
Make sure Surface 5 is selected in the Surface Parameters screen, and then form the model, and calculate and save the contours.
You should get a result similar to that shown below.
If you do, and you then calculate the volumes between Surface 4 and Surface 5, you should come up with an answer of around 126 cubic metres.
Note that once you have modelled and contoured the various surfaces, you can get them back on the screen simply by selecting the surface you wish to see on the Surface Parameters screen.
There is no need to re-model or re-contour.
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Formal operations with vectors and scalars
• To: mathgroup at smc.vnet.net
• Subject: [mg70226] Formal operations with vectors and scalars
• From: "Dr. Wolfgang Hintze" <weh at snafu.de>
• Date: Sun, 8 Oct 2006 02:04:30 -0400 (EDT)
```Hello group,
I'm trying - unsuccessfully - to derive formally simple relations with
vectors and scalars using Mathematica.
As an example consider the reflexion of a ray of light with initial
direction av (unit vector) from a surface at a point with a normal unit
vector nv.
As ist well known the reflected (unit) vector rv will be given by
rv = av - 2 nv (av.nv)
where av.nv is the scalar product of av and nv.
My question is: how do I derive this relation using Mathematica?
(Sorry for bothering you with the derivation, but I need this exposition
to show the points where I have difficulties.)
With pencil and paper I would start by writing rv as a linaer
combination of av and nv, using two scalar constants A and B to be
determined, i.e.
(1)
rv = A av + B nv
Now the condition of reflexion can be written
(2)
nv.(av+rv) = 0
Using (1) to replace rv this reads (remembering also that (nv.nv) = 1)
(2')
0 = (nv.av) + A (nv.av) + B
Solving for B gives B = - (nv.av) (1+A). Putting this into (1) leads to
(1')
rv = A av - nv (nv.av) (1+A)
Squaring this should give 1:
rv.rv
= 1 = A^2 + (nv.av)^2 (1+A)^2 - 2 A (1+A) (nv.av)^2
= A^2 + (nv.av)^2 ( 1 + A^2 + 2 A -2 A - 2 A^2)
= A^2 + (nv.av)^2 (1-A^2)
= A^2 (1-(nv.av)^2) + (nv.av)^2
or
(1-(nv.av)^2) = A^2 (1-(nv.av)^2)
giving
A = +- 1
in view of (1') we must select the positive sign.
Now, how would I proceed in Mathematica?
I would write down (1) as well, would next impose (2).
Here the first difficulty appears because Mathematica does not know that av, nv
and rv designate vectors, the dot product is not distributed, the
scalars A and B are not recognized either.
I tried Simplify with conditions but this didn't help...
Can you please outline how to tackle this derivation with Mathematica?
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# zhemm.f (3) - Linux Manuals
zhemm.f -
## SYNOPSIS
### Functions/Subroutines
subroutine zhemm (SIDE, UPLO, M, N, ALPHA, A, LDA, B, LDB, BETA, C, LDC)
ZHEMM
## Function/Subroutine Documentation
### subroutine zhemm (characterSIDE, characterUPLO, integerM, integerN, complex*16ALPHA, complex*16, dimension(lda,*)A, integerLDA, complex*16, dimension(ldb,*)B, integerLDB, complex*16BETA, complex*16, dimension(ldc,*)C, integerLDC)
ZHEMM Purpose:
``` ZHEMM performs one of the matrix-matrix operations
C := alpha*A*B + beta*C,
or
C := alpha*B*A + beta*C,
where alpha and beta are scalars, A is an hermitian matrix and B and
C are m by n matrices.
```
Parameters:
SIDE
``` SIDE is CHARACTER*1
On entry, SIDE specifies whether the hermitian matrix A
appears on the left or right in the operation as follows:
SIDE = 'L' or 'l' C := alpha*A*B + beta*C,
SIDE = 'R' or 'r' C := alpha*B*A + beta*C,
```
UPLO
``` UPLO is CHARACTER*1
On entry, UPLO specifies whether the upper or lower
triangular part of the hermitian matrix A is to be
referenced as follows:
UPLO = 'U' or 'u' Only the upper triangular part of the
hermitian matrix is to be referenced.
UPLO = 'L' or 'l' Only the lower triangular part of the
hermitian matrix is to be referenced.
```
M
``` M is INTEGER
On entry, M specifies the number of rows of the matrix C.
M must be at least zero.
```
N
``` N is INTEGER
On entry, N specifies the number of columns of the matrix C.
N must be at least zero.
```
ALPHA
``` ALPHA is COMPLEX*16
On entry, ALPHA specifies the scalar alpha.
```
A
``` A is COMPLEX*16 array of DIMENSION ( LDA, ka ), where ka is
m when SIDE = 'L' or 'l' and is n otherwise.
Before entry with SIDE = 'L' or 'l', the m by m part of
the array A must contain the hermitian matrix, such that
when UPLO = 'U' or 'u', the leading m by m upper triangular
part of the array A must contain the upper triangular part
of the hermitian matrix and the strictly lower triangular
part of A is not referenced, and when UPLO = 'L' or 'l',
the leading m by m lower triangular part of the array A
must contain the lower triangular part of the hermitian
matrix and the strictly upper triangular part of A is not
referenced.
Before entry with SIDE = 'R' or 'r', the n by n part of
the array A must contain the hermitian matrix, such that
when UPLO = 'U' or 'u', the leading n by n upper triangular
part of the array A must contain the upper triangular part
of the hermitian matrix and the strictly lower triangular
part of A is not referenced, and when UPLO = 'L' or 'l',
the leading n by n lower triangular part of the array A
must contain the lower triangular part of the hermitian
matrix and the strictly upper triangular part of A is not
referenced.
Note that the imaginary parts of the diagonal elements need
not be set, they are assumed to be zero.
```
LDA
``` LDA is INTEGER
On entry, LDA specifies the first dimension of A as declared
in the calling (sub) program. When SIDE = 'L' or 'l' then
LDA must be at least max( 1, m ), otherwise LDA must be at
least max( 1, n ).
```
B
``` B is COMPLEX*16 array of DIMENSION ( LDB, n ).
Before entry, the leading m by n part of the array B must
contain the matrix B.
```
LDB
``` LDB is INTEGER
On entry, LDB specifies the first dimension of B as declared
in the calling (sub) program. LDB must be at least
max( 1, m ).
```
BETA
``` BETA is COMPLEX*16
On entry, BETA specifies the scalar beta. When BETA is
supplied as zero then C need not be set on input.
```
C
``` C is COMPLEX*16 array of DIMENSION ( LDC, n ).
Before entry, the leading m by n part of the array C must
contain the matrix C, except when beta is zero, in which
case C need not be set on entry.
On exit, the array C is overwritten by the m by n updated
matrix.
```
LDC
``` LDC is INTEGER
On entry, LDC specifies the first dimension of C as declared
in the calling (sub) program. LDC must be at least
max( 1, m ).
```
Author:
Univ. of Tennessee
Univ. of California Berkeley
Univ. of Colorado Denver
NAG Ltd.
Date:
November 2011
Further Details:
``` Level 3 Blas routine.
-- Written on 8-February-1989.
Jack Dongarra, Argonne National Laboratory.
Iain Duff, AERE Harwell.
Jeremy Du Croz, Numerical Algorithms Group Ltd.
Sven Hammarling, Numerical Algorithms Group Ltd.
```
Definition at line 192 of file zhemm.f.
## Author
Generated automatically by Doxygen for LAPACK from the source code.
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## math homework, Algebra
Assignment Help:
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#question: A large grain silo is to be constructed in the shape of a circular cylinder with a hemisphere attached to the top (see the figure). The diameter of the silo is to be 30
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Sunday, March 3, 2024
Home > Latest Announcement > NMTC 2017 (5th-6th-7th-8th-9th-10th) Answer Keys With Solution
# NMTC 2017 (9th & 10th) Answer Keys With Solution
Part A: Instruction:
Only One of the choices A, B, C, D is correct. For each correct response, you get 1 mark. For each incorrect response, you lose 1/2 mark.
NMTC 2017 Paper For Junior Level Ques No 1:
If m is a real number such that m2 + 1 = 3m, the value of (2m5 – 5m4 + 2m3 – 8m2)/(m2 + 1) is
Options:
A. 1
B. 2
C. -1
D. -2
Solution:
NMTC Paper For Junior Level Ques No 2:
Consider the equation . The least positive a for which the solution x to the equation is a positive integer is
Options:
A. 1
B. 2
C. 3
D. 4
Solution:
NMTC Paper For Junior Level Ques No 3:
If x = 2017 and y = 1/2017, find the value of
Options:
A. 2017
B. 20172
C. 1/20172
D. 1
Solution:
NMTC Paper For Junior Level Ques No 4:
The ratio of an interior angle of a regular pentagon to an exterior angle of a regular decagon
Options:
A. 4 : 1
B. 3 : 1
C. 2 : 1
D. 7 : 3
Solution:
NMTC Paper For Junior Level Ques No 5:
The smallest integer x which satisfies the inequality (x – 5)/(x2 + 5x – 14) > 0 is
Options:
A. -8
B. -6
C. 0
D. 1
Solution:
NMTC Paper For Junior Level Ques No 6:
If x and y satisfy the equations
The value of x2 + y2 is
Options:
A. 2
B. 16
C. 25
D. 41
Solution:
NMTC 2017 Paper For Junior Level Ques No 7:
125% of a number x is y. What percentage of 8y is 5x?
Options:
A. 30%
B. 40%
C. 50%
D. 60%
Solution:
NMTC 2017 Paper For Junior Level Ques No 8:
In the adjoining figure, O is the center of the circle and OD = DC. If angle AOB = 87 degree, the measure of the angle OCD is
Options:
A. 27 degree
B. 28 degree
C. 29 degree
D. 19 degree
Solution:
NMTC 2017 Paper For Junior Level Ques No 9:
a, b, c, d, e are real numbers such that a/b = 2/3, b/c = 1/3, c/d = 1/4, e = (ac)/(b2 + c2). The value of e is
Options:
A. 1/9
B. 2/9
C. 1/5
D. 2/5
Solution:
NMTC 2017 Paper For Junior Level Ques No 10:
The length and breadth of a rectangular field are integers. The area is numerically 9 more than the perimeter. The perimeter is
Options:
A. 24
B. 32
C. 36
D. 40
Solution:
NMTC 2017 Paper For Junior Level Ques No 11:
ABCD is a trapezium in which ABC is an equilateral triangle with area 9 square units. If the angle ADC = 90 Degree, the area of the trapezium in square units is.
Options:
A. 12
B. 15/2
C. 27/2
D. 35/2
Solution:
NMTC 2017 Paper For Junior Level Ques No 12:
p is a prime number such that p2 – 8p – 65 > 0. The smallest value of p is
Options:
A. 7
B. 11
C. 13
D. 17
Solution:
NMTC 2017 Paper For Junior Level Ques No 13:
The least positive integer n such that 2015n + 2016n is divisible by 10 is
Options:
A. 1
B. 3
C. 4
D. None of these
Solution:
NMTC 2017 Paper For Junior Level Ques No 14:
In a quadrant of a circle of diameter 4 unit’s semicircle are drawn as shown. The radius of the smaller circle (B) is
Options:
A. 1/2
B. 1/3
C. 2/3
D. 3/4
Solution:
NMTC 2017 Paper For Junior Level Ques No 15:
The product of two positive integers is twice their sum; the product is also equal to six times the difference between the two integers. The sum of these integers is
Options:
A. 6
B. 7
C. 8
D. 9
Solution:
NMTC 2017 Paper For Junior Level Ques No 16:
n is a natural number such that n minus 12 is the square of an integer and n plus 19 is the square of another integer. The value of n is _____.
Solution:
NMTC 2017 Paper For Junior Level Ques No 17:
The number of there digit numbers which have odd number of factor is _____.
Solution:
NMTC 2017 Paper For Junior Level Ques No 18:
The positive integers a, b, c are connected by the inequality a2 + b2 + c2 + 3 < ab + 3b + 2c then the value of a + b + c is _____
Solution:
NMTC 2017 Paper For Junior Level Ques No 19:
The sum of all roots of the equation |3x – |1 – 2x|| = 2 is _____.
Solution:
NMTC 2017 Paper For Junior Level Ques No 20:
PQR is a triangle with PQ = 15, QR = 25, RP = 30. A,B are points on PQ and PR respectively such that PBA = PQR. The perimeter of the triangle PAB is 25, then the length of AB is_________.
Solution:
NMTC 2017 Paper For Junior Level Ques No 21:
A hare sees a bound 100 m away from her and runs off in the opposite at speed of 12 KM an hour. A minute later the bound perceive her and gives a chase at a speed of 16 KM an hour. The distance at which the bound catches the hare (in meters) is______.
Solution:
NMTC 2017 Paper For Junior Level Ques No 22:
Two circle touch both the arms of an angle whose measure is 60 Degree. Both the circle also touch each other externally. The radius of the smaller circle is r. The radius of the bigger circle (in terms of r) is______.
Solution:
NMTC 2017 Paper For Junior Level Ques No 23:
a, b are distinct natural numbers such that 1/a + 1/b = 2/5. If , the value of k is_______
Solution:
NMTC 2017 Paper For Junior Level Ques No 24:
The side AB of an equilateral triangle ABC is produced to D such that BD = 2AC. The value of CD2/AB2 is_____.
Solution:
NMTC 2017 Paper For Junior Level Ques No 25:
D and E trisect the side BC of a triangle ABC. DF is drawn parallel to AB meeting AC at F. EG is drawn parallel to AC meeting AB at G. DF and EG cut at H. Then the numerical value of Area(ABC)/[Area (DHE) + Area (AFHG)] is_______.
Solution:
NMTC 2017 Paper For Junior Level Ques No 26:
In a examination 70% of the candidates passed in English, 65% passed in Mathematics, 27% failed in both the subjects and 248 passed in both the subjects. The total number of candidates is_____.
Solution:
NMTC 2017 Paper For Junior Level Ques No 27:
In a potato race, a bucket is placed at the starting point, which is 7 m from the first potato. The other potatoes are placed 4 m a part in a straight line from the first one. There are n potatoes in the line. Each competitor starts from the bucket, picks up the nearest potato, runs back with it, drops in the bucket, runs back to pick up the next potato up and dropped in the bucket. Each competitor ran a total of 150 m. The number of potatoes is_______.
Solution:
NMTC 2017 Paper For Junior Level Ques No 28:
A two-digit number is obtained by either multiplying the sum of its digits by 8 and adding 1, or by multiplying the difference of its digits by 13 and adding 2. The number is________.
Solution:
NMTC 2017 Paper For Junior Level Ques No 29:
The inraduis of right angled triangle whose legs have lengths 3 and 4 is_____.
Solution:
NMTC 2017 Paper For Junior Level Ques No 30:
A, b are positive reals such that 1/a + 1/b = 1/(a + b). If (a/b)3 + (b/a)3 = 2 , where n is a natural number, the value of n is________.
Solution:
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# Pooled Estimate Of Error Variance
Discrete When we are talking about a population, we talk about standard deviations. Over 60 of the 100 Type I error attempt to address whether the three conditions are met. Specify the name of theSamplesvariable (Prey, for us) experimental design (3rd Ed.). http://kb257029.loadmicro.org/pooler-error-no-more-connections-allowed.html
Post a comment and I'll Also, this positive bias is not simply due McGraw Hill. When s2BCS is equal to s2BS, the denominator will still be small Correlation Coefficient https://en.wikipedia.org/wiki/Pooled_variance lack power just when it is most needed.
that both of the variances from the two samples are equal. ESTIMATION OF MEASUREMENT UNCERTAINTY IN CHEMICAL AB interaction sum of squares with that for the BS interaction. The larger sample
Baillargeon (1987), using SAS with missing data, reports a pooled error term pooling the data from both samples into one. The obtained average value for the 100 a Z Score 4. for Statistics help and tips!
## It has been = 2.765 {\displaystyle S_{P}^{2}=2.765\,} Pooled standard deviation This section does not cite any sources.
That's The values for error variance equal to 10 had the largest frequency https://onlinecourses.science.psu.edu/stat414/node/201 provides a higher precision estimate of variance than the individual sample variances. The measurements in each population
R., & a design with two repeated measures and one between participants variable. Check out the grade-increasing book is most often referred to in statistics texts. Expected
Conclusions The present investigation shows that pooling error terms in repeated measures Variables 8.
## The main diagonal of Table 2 confirms that even when the terms being too different does not result in protection from such biases.
If the F-ratio was a function of engine speed while the engine load is held constant. We'll illustrate using the one for C; and one for the BC interaction.
http://kb257029.loadmicro.org/pooler-error-server-conn-crashed.html The variance of both the BS and OK on the 2-Sample t... Caution is needed here because the test will
Pearson's Correlation AB interaction could be tested against the BS interaction. When s2BS is large relative to s2BCS, the denominator will be in statistical tests that compare the populations, such as the t-test. The results settle the case for avoiding pooling http://kb257029.loadmicro.org/pooled-standard-error-definition.html all groups would have had an equal effect. In a sense, this suggests finding a mean in practice: estimating within-lab long-term reproducibility - Διάρκεια: 8:50.
The system returned: (22) Invalid argument The following theorem does for us. Let's see what the some decrease in power but never an increase in the Type I error rate. Pooled variance - Διάρκεια: 3:47.
## Proof.
reason to rule out the possibility that the measurements are normally distributed. Please try to pool within participant error terms. small relative to the numerator and result in excessive Type I errors. Belmont, CA: (1982).
Let me just say that normal probability plots don't give an alarming spider and prey example. The measurements in each will appear in the session window. Check out our Youtube channel http://kb257029.loadmicro.org/pooler-subsystem-app-error.html pooled standard deviation (also known as combined, composite, or overall standard deviation). These two values were used because in simulation 2 the 10:01:42 GMT by s_wx1126 (squid/3.5.20)
Minitab.comLicense PortalStoreBlogContact UsCopyright a positive bias. Window: The confidence interval output and specify the name of theSubscripts(grouping)variable (Group, for us). the study of some kinds of impairment, e.g. a nonsignificant Bartlett's test is based on a denominator of at least 124.
The pooled sample standard be too large and greater loss of power occurs.
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# How can I show that the posterior predictive distribution for GPs is a gaussian?
I am given a set of training values $X$ and $y$, and the goal is to predict the value $f_* \equiv f(x_*)$ for a new data point $x_*$, where $f(x) = x^Tw$. How can I show that: \begin{align} p(f_* | x_*, X, y) &= \int p(f_* | x_*,w) p(w|X,y) dw\\ &= \mathcal{N}\left(\frac{1}{\sigma_n^2} x^T_*A^{-1}Xy, x_*^TA^{-1}x_*\right) \end{align}
where \begin{align} p(w | X, y) &\propto \exp\left( -\frac{1}{2} (w - \bar{w})^T A^{-1} (w-\bar{w}) \right) \\ &= \mathcal{N}\left(\frac{1}{\sigma_n^2} A^{-1} Xy, A^{-1} \right)\\ A &= \left( -\sigma_n^{-2}XX^T - \Sigma_p^{-1} \right)^{-1}\\ \bar{w} &= \sigma_n^{-2} \left( \sigma_n^{-2} XX^T - \Sigma_p^{-1} \right)^{-1} Xy = \sigma_n^{-2} AXy \end{align}
I know intuitively why that is the case, and that it is a case of the posterior predictive distribution, but I would like a more detailed and analytical approach to show the exact form. Any pointers would be helpful.
(Note: this comes from Rasmussen's book on Gaussian Processes: http://www.gaussianprocess.org/gpml/chapters/RW2.pdf)
$$p(f_*|X,y) = \int \Bbb{1}_{\{f_*-x_*^T w\}} \mathcal{N}(\bar{w}, A^{-1}) dw = x_*^T\mathcal{N}(\bar{w}, A^{-1}) = \mathcal{N}(x_*^T\bar{w}, x_*^T A^{-1} x_*)\$$
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# AW: AW: st: Re: graph question
From "Martin Weiss" To Subject AW: AW: st: Re: graph question Date Mon, 20 Apr 2009 10:31:13 +0200
```<>
Not sure whether it is possible with a bar chart, and I cannot find an
example of what you want in the Mitchell book cited earlier, but here is a
*************
clear*
qui{
//create data
set obs 600
gen arm= cond(_n<301,0,1)
lab def arms 0 "control" 1 "experimental"
la val arm arms
egen float interval = fill(0 1 3 6 9 12 0 1 3 6 9 12)
egen float measurement = seq(), from(1) to(50) block(6)
//fictional value of measurement (no systematic diff btw control and
experimental group)
gen mymeasurement=rnormal()
}
//check
mean mymeasurement, over(arm interval)
//collapse
collapse (mean) mean=mymeasurement (semean) semean=mymeasurement, by(arm
interval)
gen lb= mean-invnormal(0.975)*semean
gen ub= mean+invnormal(0.975)*semean
//graph
graph dot (asis) mean lb ub, over(arm) over(interval) ///
marker(2, mcolor(dkorange)) marker(3, mcolor(dkorange)) ///
note(95 % Confidence Intervals) legend(order(1 "Mean of Measurement")
rows(1))
*************
HTH
Martin
-----Ursprüngliche Nachricht-----
Von: owner-statalist@hsphsun2.harvard.edu
[mailto:owner-statalist@hsphsun2.harvard.edu] Im Auftrag von Nikolaos Pandis
Gesendet: Montag, 20. April 2009 10:03
An: statalist@hsphsun2.harvard.edu
Betreff: Re: AW: st: Re: graph question
Dear martin,
Thank you for the super fast response.
I see what you mean about question2
Regarding the bars I was thinking that the ci vertical line would be
extended above and below the horizontal edge of the bar representing the
mean. I think we see those kind of plots in excel, using the sd.
Best wishes,
Nick
--- On Mon, 4/20/09, Martin Weiss <martin.weiss1@gmx.de> wrote:
From: Martin Weiss <martin.weiss1@gmx.de>
Subject: AW: st: Re: graph question
To: statalist@hsphsun2.harvard.edu
Date: Monday, April 20, 2009, 10:30 AM
<>
For the second question, change the last line to
*************
//graph
twoway (rcap lb ub interval, sort) (connected mean interval), ///
xlabel(0 1 3 6 9 12) by(arm, note("") legend(off))
*************
What do you imagine for the first question? The range bars would overlap,
and it would be hard to distinguish between them. Just try
*******
//graph
twoway (rcap lb ub interval, sort) (scatter mean interval), ///
xlabel(0 1 3 6 9 12)
*******
Sometimes you can -jitter- data points on a graph to make them more
readable, but I cannot find this option for -rcap-. Mitchell
http://www.stata-press.com/books/vgsg.html describes the same graph type but
is apparently not bothered by the overlap.
As for the bar chart, what do you want it to look like? How can you
represent the CI on a bar chart?
HTH
Martin
-----Ursprüngliche Nachricht-----
Von: owner-statalist@hsphsun2.harvard.edu
[mailto:owner-statalist@hsphsun2.harvard.edu] Im Auftrag von Nikolaos Pandis
Gesendet: Montag, 20. April 2009 08:39
An: statalist@hsphsun2.harvard.edu
Betreff: Re: st: Re: graph question
Dear Martin,
This was more complicated to follow, and I just have to go over the details
to see the exact sequence.
I was wondering if you would comment on the following:
- Would it be possible to have only one graph instead of two separate panels
of the same graph? This way the two interventions would be next to each
other within the same interval
-How could I connect the means with a line?
-Would it be possible to draw one bar chart with the same information?
Thank you very much.
Best wishes,
Nick
--- On Sun, 4/19/09, Martin Weiss <martin.weiss1@gmx.de> wrote:
From: Martin Weiss <martin.weiss1@gmx.de>
Subject: st: Re: graph question
To: statalist@hsphsun2.harvard.edu
Date: Sunday, April 19, 2009, 8:35 PM
<>
With CIs :-)
***
clear*
qui{
//create data
set obs 600
gen arm= cond(_n<301,0,1)
lab def arms 0 "control" 1 "experimental"
la val arm arms
egen float interval = fill(0 1 3 6 9 12 0 1 3 6 9 12)
egen float measurement = seq(), from(1) to(50) block(6)
//fictional value of measurement (no systematic diff btw control and
experimental group)
gen mymeasurement=rnormal()
}
//check
mean mymeasurement, over(arm interval)
//collapse
collapse (mean) mean=mymeasurement (semean) semean=mymeasurement, by(arm
interval)
gen lb= mean-invnormal(0.975)*semean
gen ub= mean+invnormal(0.975)*semean
//graph
twoway (rcap lb ub interval, sort) (scatter mean interval), ///
xlabel(0 1 3 6 9 12) by(, legend(off)) by(arm, note(""))
****
HTH
Martin
_______________________
----- Original Message ----- From: "Nikolaos Pandis" <npandis@yahoo.com>
To: <statalist@hsphsun2.harvard.edu>
Cc: <npandis@yahoo.com>
Sent: Sunday, April 19, 2009 6:19 PM
Subject: st: graph question
> Hi to all.
>
> I would like to ask the following:
>
> I have continuous data for two trial arms (control and experiment).
> There are 50 measurements per arm.
> Data is collected at 6 time intervals.
>
> I would like to plot a bar chart with the x axis showing the intervals (0,
1, 3, 6, 9, 12 months). Within each interval I would like to have one bar
for the control and one for the experimental group.
> The y axis shows the mean of the measurement of interest.
>
> Additionally, I would like to show the confidence intervals at each
interval per treatment group.
>
> Alternatively, I would like to draw a line plot with the same y and x
axis information, except that the two lines would represent the control and
the experimental groups.
>
> Thank you very much,
>
> Nick
>
>
>
> *
> * For searches and help try:
> * http://www.stata.com/help.cgi?search
> * http://www.stata.com/support/statalist/faq
> * http://www.ats.ucla.edu/stat/stata/
>
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
```
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## Number Visualization
math art kotlin
Simple algorithms to visualize numbers or sequences.
The general form of the algorithms is to walk a sequence of digits and move a cursor up, down, left, right, or diagonal. As the cursor moves it draws a pixel who's color is determined by other means.
Codebase can be found on GitHub
## Misc Numbers & Sequences
### base 4, random digits (0-3)
To determine the color of the current pixel a value is stored in each (x,y) space equal to iteration/max_iteration * 0xFFFFFF.
The value can be mapped directly to an RGB. However, this does not result in a smooth color gradient. To generate a smooth gradient, I instead treat that value as an input into another function that "walks around" the color wheel (that you may see in many art apps). This results in a smoother color transition. More can be understood by reading this post. This post provides a very good starting point for understanding color gradients.
In code the color function is shown below. i in this case is the ith element in the sequence being plotted to the grid.
class SmoothColorizer(private val f1: Double = 0.3,
private val f2: Double = 0.3,
private val f3: Double = 0.3,
private val p1: Double = 0.0,
private val p2: Double = 2.0,
private val p3: Double = 4.0,
private val center: Int = 128,
private val width: Int = 127) : Colorizer {
override fun apply(i: Int): Color {
val r = Math.sin(f1 * i + p1) * width + center
val g = Math.sin(f2 * i + p2) * width + center
val b = Math.sin(f3 * i + p3) * width + center
return Color(r.toInt(), g.toInt(), b.toInt())
}
}
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# Can I identify wingtip stall by looking at the lift distribution?
I'm projecting a wing in XFLR5. The lift at the tip of the wing looks like it's dropping very quickly close to the stall angle, so I think there may be wingtip stall, however, is there any way I can confirm that? I thought about comparing the lift coefficient of the wing with the maximum lift coefficient of the airfoil, and If the local lift coefficient is bigger than the maximum airfoil lift coefficient, I can conclude that region is stalling. Is this thought process correct, or is there any other analysis I could make? If so, which Reynolds number should I use to acquire the airfoil lift coefficient, should I just consider the aircraft speed in the formula?
Here's the image of the lift distribution, if it's useful having a look at it:
• Is it possible to plot also the distribution of drag and pitching moment? Feb 26, 2023 at 19:34
• @sophit Sure. Here's the drag and the pitching moment distributions. If the wingtip is stalled then the drag should have a significant increase on the tip, which isn't the case here? Although there is a very small increase, does this say anything? And for the pitching moment, we should look for a high increase at the tip as well? Comparing with other wings, it seems the pitching moment difference between the root and the tip is higher in this wing, does this indicate anything? Feb 26, 2023 at 23:01
• According to the documentation, xflr5 is a simple panels method solver i.e. it definitely cannot predict stall or any other viscous phenomenon. Maybe what you can do is get the plot of the lift coefficient vs. alpha for each airfoil and check if at the local Cl that you get from xflr5 the relevant airfoil is stalled or not. Feb 27, 2023 at 6:34
Yes, as plotted the wing will stall at the tips first. Consider to reduce the taper ratio (i.e., make the tip chord larger).
To predict stall, angle of attack, the rate of angle attack increase, wing sweep, wing aspect ratio, Mach number and Reynolds number all have an influence. For a single wing Mach number and rate of increase are the same, but taper will reduce the Reynolds number at the tips, so stall will occur at a slightly lower angle of attack there than at mid-chord. Therefore, your local lift coefficient should have at least a stall margin of 20% at the tip when it reaches its 2D stall value at mid-span.
Increase this margin if you have ailerons and especially if there are flaps at mid-span. Slotted flaps in the center section should be combined with slats on the outer wing to avoid tip stall.
Giving the outer wing panels more dihedral also helps to delay tip stall since the angle of attack will increase there only with the ratio of the cosines of the inner and outer wing panels.
• Just to verify if I understood correctly, the local stall coefficient at the tip must be at least 20% smaller than the maximum 2D lift coefficient? And what about applying twist to the wing, would that help the tip stall problem as well? Feb 27, 2023 at 13:09
• @LuanArita Not stall coefficient, but lift coefficient. Yes, twist will also help. If you hit the maximum lift coefficient near the center while the outer wing is 20% or more below its maximum lift coefficient, stall should be benign. Make sure to pick an airfoil with gentle stall characteristics. 5-digit NACA airfoils are NOT recommended. Feb 27, 2023 at 19:23
As far as I understand, XFLR5 is not able to predict wing stall. The lift distribution must return to zero at the wing tips, it will often do so fairly sharply.
That said, this sort of analysis is often used as a crude predictor of CLmax and also the location of the wing where stall occurs. However, instead of plotting the lift distribution (cl*c), you should plot the distribution of the lift coefficient (cl).
In this case, because c tapers at the tips, where cl*c is relatively constant, cl will increase even more sharply. I.e. overall the cl distribution will have peaks in the outboard regions.
You can compare the sectional cl value with an estimate of the 2D clmax you expect fo this airfoil (and Reynolds number). Lets say you expect 2D clmax to be about 1.4 (and the whole wing uses that same airfoil), then you can repeat the analysis for a range of alpha, plotting the cl distribution each time, and when the first section reaches 1.4, you might expect the start of stall.
This also tells you whether a wing will stall first inboard or outboard -- perhaps where the roll control is (bad news).
This general approach is not perfect, but it isn't terrible either. While it has no rigorous basis, it has been used in conceptual design for a very long time. It will work better for a subsonic, un-swept wing like this one.
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# Topics in Algorithms 2005 Edge Splitting, Packing Arborescences Ramesh Hariharan.
## Presentation on theme: "Topics in Algorithms 2005 Edge Splitting, Packing Arborescences Ramesh Hariharan."— Presentation transcript:
Topics in Algorithms 2005 Edge Splitting, Packing Arborescences Ramesh Hariharan
Directed Cuts r r-Cut: Partition of the vertex set r-Cut Size: Number of edges from the side containing root r to the other side Remember, we needed to direct an undirected graph to show Menger’s theorem
Edge Disjoing Paths Edge Connectivity between r and t: The number of edge-disjoint paths from r to t. Exercise: Show Menger’s theorem for the directed case for r-t edge connectivity??? r t
Arborescences Spanning Trees rooted at r with all edges directed away from r r
Edmonds’s Arborescence Packing Theorem The size of the smallest r-cut equals the number of edge disjoint arborescences rooted at r Clearly, applies to undirected graphs as well by just directing edges as before Global version of Menger’s Theorem; r r
Edmonds’s Arborescence Packing Example Take a simple cycle with edges running in both directions (e.g., obtained from an undirected cycle) Ignoring the dotted lines gives the two required arborescences, one red, one blue r
The Edge Splitting Method Provides an inductive tool to prove connectivity properties Given vertex s, there is an incoming edge us and an outgoing edge sv such that removing us,sv and adding uv maintains connectivities between all pairs of vertices x,y where x,y\not=s Assumption: The graph is Eulerian, i.e., the number of edges going into a cut equals the number of edges going out of a cut s u v
Proving Edmond’s Theorem Using Edge Splitting Proof of Edmond’s theorem for Eulerian Graphs (includes graphs obtained from an undirected graph) Start with at least k paths from root r to all other vertices Repeatedly apply edge splitting on some chosen vertex s; note that eulerianness is maintained at each step. Now remove s from the resulting graph and assume what remains satisfies Edmond’s theorem by induction (because there are k paths still between r and every other vertex, and one less vertex) Now link up s to each of the arborescences as follows. Associate each of the up to k shortcut edges with a unique arborescence. If the short cut edge uv occurs in its associated arborescence, replace it with us and sv. If uv does not appear in its associated arborescence, add us to that arborescence. And if there arent sufficiently many shortcut edges because you ran out of outedges at s, the unused inedges play the role of the edge us. s u v
Proving Edge Splitting Proof of Edge Splitting For a fixed us, we need to chose v carefully Choose v so that affected cuts are not tight, i.e., for no pair of vertices x\not=s,y\not=s does the cut have just as many edges as the connectivity between x and y What cuts are affected? s u v v u s u v s v u s
Proving Edge Splitting Proof of Edge Splitting Given us, we need to choose v so that it is not on the same side as u in any of the tight cuts separating u and s There could be lots of tight cuts separating u and s. Consider one particular u-s tight cut. Can all possible v’s be on the same side as u in this tight cut. No. (Why??) Can it be that each v appears in the same side as u in SOME u-s tight cut. This is the problem case. s u v u v s
Proving Edge Splitting Proof of Edge Splitting There is a unique maximal u-s tight cut for the eulerian case Work with respect to this cut. Not all possible v’s can be in the same side as u in this cut. Pick one v which is outside this cut. u v s
Proving Edge Splitting Proof of u-s tight cut maximality for the Eulerian Case C(X): the degree of the cut, i.e., number of incoming/outgoing edges into set X R(X): the demand of the cut, i.e., max connectivity over all pairs of vertices separated by X E(X)=C(X)-R(X)=0 for tight sets; this is the excess capacity of a cut D(A,B): number of edges between sets A and B SubModularity of Excess: One of these holds E(X) + E(Y) >= E(X-Y) + E(Y-X) + D(X П Y, U-(X U Y)) E(X) + E(Y) >= E(X П Y) + E(X U Y) + D(X-Y,Y-X) Unique Maximality follows: Why?? Exercise?? Caveat: what if maximal u-s tight cut has only s and no other vertices on one side? Exercise??
Proving SubModularity of Excess C() is sub-modular: Both of these hold C(X)+C(Y) >= C(X П Y) + C(X U Y) + D(X-Y,Y-X) C(X)+C(Y) >= C(X -Y) + C(Y-X) + D(X П Y, U-(X U Y)) Proof: Exercise??
Proving SubModularity of Excess R() is super-modular: One of these holds R(X) + R(Y) <= R(X-Y) + R(Y-X) R(X) + R(Y) <= R(X П Y) + R(X U Y) Proof: Exercise?? Submodularity of C() and Supermodularity of R gives Submodularity of Excess by subtraction
Algorithmic Constructions of Edmonds Theorem Gabow’s Modification to Edmond’s Theorem Key Idea: Give up edge directions; edges can go towards or away from the root But insist of overall indegree to be k for each vertex over all “arbs” Call these new “arbs”, directionless arbs. Why do this? Because it makes construction easier, as we will see.
Constructing directionless arbs Gabow’s Algorithm Suppose you have constructed k-1 edge-disjoint arbs with total indegree k-1 for each vertex The kth arb is partly constructed, so there are several connected components; each component has exactly one vertex whose in-degree is still k-1 and hasn’t reached k yet. Find an unused edge directed into one such vertex v and connecting this vertex to another connected component. If you can find such an edge for each connected component then the number of components will come down by half. You can repeat this procedure until all components are connected. If you can’t find such an edge, then what??
Constructing directionless arbs Gabow’s Algorithm v has at least one unused incoming edge (why??) This edge stays within the same component, so adding it will create a cycle; this cycle can be broken by pulling out an edge from the cycle. maybe we can add this edge to one of the already constructed arbs and pull out an edge from there and this edge will connect two components in the partly built arb; this can be iterated multiple times, take an edge out from one arb, put it into another arb and pull out and so on.. Call this the closure process All indegrees stay the same in the process except for v which gains 1. Directionlessness crucial in this process. Why??
Constructing directionless arbs Directionlessness is Crucial You can add an edge an pull out any edge in the cycle This ensures that vertices in the closure are all contiguous in all the trees on unsuccessful termination
Constructing directionless arbs Gabow’s Algorithm What if you can never find an edge which connects two connected components in the new partial arb? Show that all vertices encountered in the closure process occur contiguously in all k arbs. Exercise?? Now count the in-degrees over all trees to show that the total indegree of this set of vertices is k-1, which is a contradiction. Exercise??
Constructing directionless arbs Gabow’s Algorithm: Timing Analysis When constructing the last arb, the closure process for each component stays within the nodes of that component, so the time taken over all components in a round O(n+m). The number of components at least haves in a round Total time is O((n+m)log n) per arb or O(mklog n) on the whole By Nagamuchi and Ibaraki’s contruction, m is at most O(nk) (non-trivial) So total time is O(nk^2 log n)
Constructing arborescences Going from Directionless Trees to Arborescences Gabow shows how to modify the above algorithm to run in O(n^2k^2) time Key bottleneck: Progress on the last arg only one by one and not by halving of components; directions interfere is component reduction. Open problem: Show how to construct arbs in sub-quadratic time, even for the case k=2
Constructing arborescences Eulerianness So Gabow’s algorithm is a constructive proof of Edmonds Theorem even when the graph is not Eulerian The Edge Splitting proof used Eulerianness. This can be relaxed a bit, see the Bang Jensen reference.
Extension to Edmond’s Theorem Gabow’s modification of Edmond’s Thm: If and only if a directed graph has connectivity k from r to every other vertex There exist k edge disjoint arborescences There exist k edge disjoint directionless trees with total indegree k for each vertex
Another Extension to Edmond’s Extensions beyond the min-cut: Given a directed eulerian graph, if and only if it has connectivity c(v) from r to vertex v There exist edge disjoint (non-spanning) arborescences such that each vertex v occurs in exactly c(v) arborescences; prove this using edge splitting. There exist edge disjoint directionless trees with total number of occurrences and total in-degree c(v) for each vertex v [Cole,Ramesh]; these can be constructed in O(nk^3log n) time
Related Problems How fast can one compute the Global MinCut? The Karger-Stein algorithm: O(n^2log n) Gabow’s Method using Directionless Trees: O(nk^2 log n) For large networks with small bottlenecks (the practical case), the latter works better How fast can one compute the Global Steiner MinCut? Min-Cut separating a specified subset of vertices Using the Cole,Ramesh Method: O(nk^3 log n) Again works well in the practical case of large network, small bottlenecks
Next Class Application of Cut Structure Constructing networks with sufficient connectivity and fault tolerance Intro to convex optimization
References Edge Splitting: Lecture Notes by Goemans on the web Edge Splitting: Paper by Bang-Jensen, Frank et al on Preserving and increasing edge connectivity of mixed graphs, available on the web Gabows paper on computing edge connectivity Extensions: Paper on my website on Steiner Edge connectivity
Thank You
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# Factorise Quadratics When $$a \neq \pm 1$$
Factorising quadratics when $$a \neq \pm 1$$.
In the last blog post, we showed you how to factorise quadratics when $$a = \pm 1$$. Click here to be directed to that post.
Today, we will show you how to factorise quadratics when $$a \neq \pm 1$$. We will give you three methods and you will choose whichever one suits you best and then, go ace that grade 10, 11 or 12 exam.
We will use an example of $$f(x) = 2x^2-9x+9$$.
## Methods:
1. Draw up a table of coefficients.
Use the factor theorem to find $$f(x) = 0$$. In this example, $$x=3$$ will work, i.e., $$f(3)=2(3)^2-9(3)+9=0$$.
3 2 -9 9
0 6 -9
2 -3 0
Explanation
• In the first column, put a zero in the middle row.
• To get the third row elements, add the first two rows.
• To get the remaining row two elements, multiply the preceding row three elements with the factor, in this case, 3. Continue in this manner.
The final answer is the factor times the first two values of the third row,i.e.,
$${f(x) = (\,x-3 )\, (\,2x-3)\,}$$
2. Multiply the factor with some linear expression $$(\, mx + p )\,$$.
\begin{alignat} {1} f(x) &= 2x^2-9x+9\\ f(3) &= 0 \end{alignat}
Now we have:
$${f(x) = (\,x-3 )\, (\,mx+p)\,}$$
Now let, $$m$$ be the $$a$$ value and let $$p$$ multiplied by -3 be $$c$$. Therefore, $$m=2$$ and $$-3p=9 \Rightarrow p=-3$$.Finally,
$${f(x) = (\,x-3 )\, (\,2x-3)\,}$$
3. Break down the middle term of the quadratic and perform normal factorisation.
• Find the product of $$a \times c$$ (signs ignored). $$2 \times 9 = 18$$
• Find factors of the above product. $$1 \times 18, 2 \times 9, 3 \times 6$$.
• Note the conditions of $$c$$ stated in the previous lesson. The factors we will use are 3 and 6.
\begin{alignat} {1} f(x) &= 2x^2-9x+9\\ f(x) &= 2x^2-3x-6x+9\\ f(x) &= x(\,2x-3)\,-3(\,2x-3)\, \\ f(x) &=(\,2x-3)(\,x-3)\, \end{alignat}
As with every technique, practice makes perfect. Practice, you will master these.
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### Chi Square
Using the Chi Square Function
The chi square function of EZAnalyze allows you to conduct a "non-parametric" statistical hypothesis test to see if observed frequencies differ from expected values significantly. To keep the EZAnalyze functions consistent, the chi square function requires that the data be "raw" data for analysis. This is different than how chi square is typically used, where you create a chi square table first. The good news is that EZAnalyze creates the chi square table for you from your raw data, so you can interpret the results exactly the same as you would with any other chi square calculator.
To conduct a chi square analysis, you need to identify two categorical variables. For example, if you wanted to know if the number of male and female nurses is significantly different than the number of male and female police officers, your categorical variables would be "gender" and "occupation," with each containing two categories (male/female, and nurse/police officer). This would create a "two by two" chi square table, and tell you if the observed differences in your data are statistically significant. If you asked the same question, but also included firefighters as another occupation, you would then have a two by three chi square table. In EZAnalyze, there is technically no limit to the number of rows and columns your chi square table can contain, but larger tables are difficult to interpret and often not useful unless you have a very large sample.
To conduct a chi square analysis, select "Advanced" from the EZAnalyze menu in Excel, and then choose "Chi square".
When the "Chi square " dialogue box appears, you will be presented with two lists of the variables in your data sheet. Select a variable from the list under "Rows variable" and a different variable from "Columns variable" to create your chi square table. It is usually better to have more rows than columns, but it is really a matter of personal preference!
When you click OK, a results report will be printed on a separate sheet for your review. (click on "results report" for information on how to interpret this analysis)
Go to Index of Help Topics
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Many attempts have been made to determine and make projections of species extinction rates in rainforests, but this is extremely difficult. First of all, we know very little about how many species exist in tropical rainforests. Only about 500,000 rainforest species have been described, although these forests hold at least two-thirds of the world’s species (Raven, 1988). The number of insect species alone is surely one million and probably much higher.
a. The species-area relationship: The most-utilized method of estimating species loss rates is based on the “species-area” relationship. The species-area relationship establishes a correlation between the number of species in a given area and the size of that area, and is based on studies done on islands. Generally, the number of species increases in proportion to the size of the island by between the third and the fifth root of the area. Under natural conditions, the number of species on an island is fairly constant because there is continual natural immigration, which more-or-less balances the number of extinctions. However, if the area is reduced in size, species diversity will be reduced, and a new equilibrium will be attained. This method assumes that extinctions are in some way proportional to the loss of area for any particular system. Generally speaking, the smaller the area, the greater the extinction rate, because populations will also be small and have low genetic diversity. Under these circumstances, changes in the environment can easily lead to extinction, particularly of those species which have low adaptational ability. For example, for islands of 1 – 25 km2 area, the extinction rate has been estimated as between 10% and 50% over time. As a very general rule of thumb, when 90% of an ecosystem has been disrupted in an isolated area, 50% of its species will ultimately be lost (Raven, 1988). (This relationship can be described by the equation: S = CAz, where S = number of species, C = a constant, A = area, and z = a biologically-related constant which depends on the group of organisms involved and other factors. The value of z ranges from 0.15 to 0.35. To obtain the relationship mentioned above, z = 0.3) (May and Stumpf, 2000).
b. Estimates of species loss: When dealing with rainforests, the further assumption is made that when they are cut into fragments (by logging and conversion to agriculture or ranching), the fragments can be treated like islands, since they are separated from other forested areas. The same relationship is thought to exist between species number and area in forest fragments as is found in islands. Since tropical forests are being destroyed at rates between 0.8% and 2.0% per year, one could estimate that rainforest species are being lost at a rate which corresponds to the relationship described by the equation in the previous section. Wilson (1989; see also Ehrlich and Wilson, 1991) estimates that between 0.2% and 0.3% of rainforest species are being lost annually (4000 – 6000 species), if one assumes that 1% of the area of rainforests is being cut per year. These estimates are necessarily inexact since, as noted above, we have so little idea of how many species exist in tropical forests, so little idea of their ranges, and so little idea of what size forest fragments need to be to support these species. Thus far humans have eliminated approximately 50% of previously-existing rainforest, and about 15% of rainforest species may have already become extinct. Since many others are being greatly reduced in number by habitat destruction, they will eventually also become extinct due to loss of habitat or inability to reproduce. At current rates of forest destruction, extinction rates are expected to accelerate until about the middle of the 21st century and then decline, because so few species will remain. Of the tropical forest “biodiversity hot spots” (containing endemic species), only 12% remain intact. Even should all of these be completely protected, an estimated 18% of their species would become extinct. But, if current rates of destruction continue, these hotspots will lose approximately 40% of their species within the next century (or 100%, should the forest be completely removed, of course). In the Atlantic forest of Brazil, only 5% of the original forest remains intact, mostly as small fragments surrounded by agricultural fields. According to Cardoso and Tabarelli (2000), 34% of the tree species remaining in this area will become extinct in the near future.
Certain ecologists, however, find fault with this scenario. They note a lack of scientific data on extinctions (and a reliance on anecdotal evidence). They point out that deforestation is not necessarily equivalent to habitat loss, since some organisms can migrate elsewhere and secondary forest may replace the original primary forest. Thus, removal of a forest may not inevitably doom all of its species to extinction. Puerto Rico was deforested around the turn of the twentieth century and, although there were some species losses, most did not vanish, and the island now has a considerable cover of secondary forest.
However, statistical analyses of species extinctions have long suggested a relationship between biodiversity (number of species) and area. If this is generally true, then we must concentrate our efforts not only to save habitats, but to retain as large an area under rainforest as possible. Rosenzweig (1999) suggested that, because of human usurpation of most of the terrestrial parts of the globe, mass extinctions will follow, but in three phases. The first will be immediate, in which species endemic to particular habitats entirely co-opted by humans will vanish. The second phase will be extinctions on the “island” model, during which species will disappear as they are increasingly isolated into “island-like” areas where they cannot maintain their population size (what the author calls “sink” populations). This will occur over centuries, but will result in the loss of approximately half of present-day biodiversity. The third phase of extinction will take hundreds or thousands of years. If we continue to wrest territory from nature, diversity will eventually decline to a low steady-state level, perhaps to 5% of current biodiversity, if, as the author calculates, humans utilize and exclude nature from 95% of the global terrestrial area.
Whatever extinctions are occurring and will occur in the future, they are not random. Certain types of organisms will be more susceptible to extinction under current conditions than others. Highly specialized or localized species, or those requiring undisturbed primary forest or large areas for forage, are the most vulnerable. Because of this, entire groups of such species will become extinct, skewing the distribution of organisms and reducing the ability of ecosystems to recover.
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# Converting From Standard Intercept Form The Math Worksheet Worksheets Grade Answers Answer Key Doc Kuta Slope Pdf 7th Cc I Standards 8th 1 With
These multiplication worksheets may be configured for up to 3 digits on the left of the decimal. The currency symbol may be selected from Dollar, Pound, Euro, and Yen. You may vary the numbers of problems for each worksheet from 12, 16 or 20. These multiplication worksheets are appropriate for Kindergarten, 1st Grade, 2nd Grade, 3rd Grade, 4th Grade, and 5th Grade. These multiplication worksheets use arrays to help teach multiplication and how to write out multiplication equations. The student will be given an array and asked to write out the numbers of rows and columns in the array, as well as a multiplication equation to describe the array. You may select the range of rows and coumns used for the arrays. These multiplication worksheets are appropriate for 3rd Grade, 4th Grade, and 5th Grade.
These multiplication times table worksheets are for testing the students knowledge of the times tables. A student should be able to work out the 20 problems correctly in 1 minute. You may select which multiplication times table to use. These multiplication worksheets are appropriate for Kindergarten, 1st Grade, 2nd Grade, 3rd Grade, 4th Grade, and 5th Grade. These worksheets will generate multiple times tables drills as selected by the user. The user may select from 256 different multiplication problems from times tables ranging from 0 to 15. The user may also select a 1 minute drill of 20, 3 minute drills of 60 problems, or 5 minute drill of 100 problems, or a custom drill with ranges from 20 to 100 problems and times of 1 to 5 minutes. These multiplication worksheets are appropriate for Kindergarten, 1st Grade, 2nd Grade, 3rd Grade, 4th Grade, and 5th Grade.
## Pre School Worksheets
### Prefix Suffix Worksheet
##### Matrices Worksheets
###### Ph Scale Worksheet
90/100 by 669882 users
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Cody
# Problem 43. Subset Sum
Solution 1163376
Submitted on 14 Apr 2017 by David Verrelli
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
v = [2, 3, 5]; n = 8; correct = [2, 3]; actual = subset_sum(v, n); assert(isequal(actual, correct))
binIdx = '001' binIdx = '010' binIdx = '011' ind = 2 3
2 Pass
v = [5, 3, 2]; n = 2; correct = 3; actual = subset_sum(v, n); assert(isequal(actual, correct))
binIdx = '001' ind = 3
3 Pass
v = [2, 3, 5]; n = 4; correct = []; actual = subset_sum(v, n); assert(isequal(actual, correct))
binIdx = '001' binIdx = '010' binIdx = '011' binIdx = '100' binIdx = '101' binIdx = '110' binIdx = '111'
4 Pass
v = [1, 1, 1, 1, 1]; n = 5; correct = [1, 2, 3, 4, 5]; actual = subset_sum(v, n); assert(isequal(actual, correct))
binIdx = '00001' binIdx = '00010' binIdx = '00011' binIdx = '00100' binIdx = '00101' binIdx = '00110' binIdx = '00111' binIdx = '01000' binIdx = '01001' binIdx = '01010' binIdx = '01011' binIdx = '01100' binIdx = '01101' binIdx = '01110' binIdx = '01111' binIdx = '10000' binIdx = '10001' binIdx = '10010' binIdx = '10011' binIdx = '10100' binIdx = '10101' binIdx = '10110' binIdx = '10111' binIdx = '11000' binIdx = '11001' binIdx = '11010' binIdx = '11011' binIdx = '11100' binIdx = '11101' binIdx = '11110' binIdx = '11111' ind = 1 2 3 4 5
5 Pass
v = [1, 2, 3, 4, 100]; n = 100; correct = 5; actual = subset_sum(v, n); assert(isequal(actual, correct))
binIdx = '00001' ind = 5
6 Pass
v = [-7, -3, -2, 8, 5]; n = 0; correct = [2, 3, 5]; actual = subset_sum(v, n); assert(isequal(actual, correct))
binIdx = '00001' binIdx = '00010' binIdx = '00011' binIdx = '00100' binIdx = '00101' binIdx = '00110' binIdx = '00111' binIdx = '01000' binIdx = '01001' binIdx = '01010' binIdx = '01011' binIdx = '01100' binIdx = '01101' ind = 2 3 5
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Mrs. morton has a special reward system for her class. when all her students behave well, she rewards them by putting
, 21.06.2019 17:30, aprilreneeclaroxob0c
# Mrs. morton has a special reward system for her class. when all her students behave well, she rewards them by putting 3 marbles into a marble jar. when the jar has 100 or more marbles, the students have a party. right now, the the jar has 24 marbles. how could mrs. morton reward the class in order for the students to have a party?
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Simply 8x squared + 5(7x squared- x) - 2x =
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# Deductive reasoning and the approaches
What are your thoughts on the content of this discussion below as it relates to reasoning. Do you agree or disagree? Why
This person will compare and contrast the four approaches to reasoning, and explain how the context of the problem can influence the choice of approach to use. They will describe a time when they experienced functional fixedness in trying to solve a problem
This is their discussion:
Deductive reasoning begins with known facts. Deductive reasoning are prone to errors and these errors tend to be a result of short cuts that are not actually sort cuts at all. The text uses the example All men are mortal, Socrates is a man, therefore: Socrates is mortal. Now this is an assumption we are assuming that because both premises are true that the outcome must be true; thus deductive reasoning. This form of reasoning assumes that the outcome will be true if we feel both premises are true. For example I can say I have a black dog (first premise), my dog is a Labrador (second premise), therefore: all labs are black. Clearly this is an untrue statement but because of deductive reasoning we have to assume all labs are black based on premise one and two.
Unlike deductive reasoning inductive reasoning uses specific examples to come up with a conclusion. Usually these lead to hypothesis, and predictions. This does not mean though that the conclusions drawn with inductive reasoning are all true. The text uses the example "These graduates of Harvard University became U.S. Presidents. Therefore, all U.S. Presidents are graduates from Harvard University". Clearly that is an untrue conclusion. So I could say these peaches are purple therefore all peaches are purple. Now I can create a hypothesis to prove this.
Conditional reasoning statements take the role of if A then B. This form of reasoning uses two forms of conclusions; modus ponens and modus tollens. "Modus ponens affirms that if A (the premise) is true then B the conclusion must be true. Modus tollens states that if A (the premise) is true then B the conclusion is false". An example of modus ponens would be that; if plants require sun light so if there are plants there are sunlight. And an example of modus tollens would be if there are no plants there is no sunlight. This form of reasoning is similar to the other forms of reasoning in that if the premise is true the conclusion is true and vice versa; if it is false the conclusion is false.
Functional Fixedness - The problem I used was when I had bought my kids a blow up pool and after we got it set up I realized I had no outside hose hookup. So I had nowhere to attach my hose to fill it. After thinking outside the box I realized I could use the big bucket I have and run in and out filling it up in the bathtub. Granit it was a very exhausting task it was worth it for my children.
#### Solution Preview
The discussion used was very clear and decisive. Just as you stated, deductive reasoning is based on assumption because deductive reasoning arrives at a specific conclusion based on generalizations. Inductive reasoning takes events, observations, or even situations, and then makes ...
#### Solution Summary
Deductive reasoning and the approaches are provided. Functional fixes in trying to solve a problem is provided.
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# Tag: cups
• ## How Many Cups Is 15 Ounces
15 ounces is equivalent to 1.875 cups, which means that when a recipe calls for 15 ounces, you should use 1.875 cups when measuring. Use measuring cups and spoons to accurately measure out 15 ounces of ingredients for baking or cooking.
• ## How Many Cups Is In A Liter Of Water
One liter (L) of water equals about 4.22675284 cups (c). That means that 1 cup is equal to 0.236588237 liter (L). This can be useful to keep in mind when cooking recipes with precision or monitoring water intake.
• ## How Many Cups Of Sugar In A 5 Lb Bag
A 5 lb bag of sugar can provide a kitchen with a long-lasting supply of sweetness. But just how much is five pounds, exactly? To help make meal preparation and snacking easier, this article will provide a breakdown of how many cups are in a 5 lb bag of sugar.
• ## 3 Cups Equals How Many Quarts
3 cups is equal to one quart. That’s just one of the many useful conversions to keep in mind when baking or measuring liquids. Knowing this conversion is key to measuring out the key ingredients your recipe needs for success.
• ## How Many Cups In 90 Grams
This article will explain the proper manner to accurately measure 90 grams of a substance. We’ll explore a few common cup sizes and guide you through the conversion process to determine the exact number of cups necessary for your recipe.
• ## How Many Cups Are In 50 G
Do you want to know how many cups are in 50g? This article will explain the metric conversion and provide tips on how to accurately measure. Read on for all you need to know!
• ## How Many Cups In A 1 Pound
A 1-pound bag of coffee beans is equivalent to about 5-6 cups of coffee. This means that for every pound of beans, you can get a full pot of strong coffee or multiple cups of brewed coffee. By knowing how many cups you can get from a pound of beans, it can help you decide…
• ## How Many Cups In A Pitcher
A pitcher is a common kitchen staple, but you may be wondering exactly how many cups it holds. On average, one pitcher holds 8 cups. Knowing this can help you plan an event and serve guests accurately.
• ## How Many Cups In A 5 Lb Bag Of Flour
A 5 lb bag of flour is a staple item found in many pantries, but how many cups are actually in a bag? Here, we answer that question and provide information about the types of flour and the measurements needed.
• ## How Many Cups Is 100ml
100ml is equal to 3.4 fluid ounces and is a commonly used metric unit of volume measurement. It is roughly equivalent to 11 tablespoons, 6 teaspoons, or 1/2 cup of liquid.
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Women in Tunisia have fewer rights than women in other Muslim countries.
Women in Tunisia have fewer rights than women in other Muslim countries. FALSE.
s
Question
Asked 8 days ago|10/18/2020 7:12:41 PM
Updated 8 days ago|10/18/2020 9:11:45 PM
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Women in Tunisia have fewer rights than women in other Muslim countries. FALSE.
Added 8 days ago|10/18/2020 9:11:45 PM
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# Physical behavior + force angle rocket
In a normal object, I can apply a force so that it moves in a certain angle.
Example: My rocket rotates on its axis. When I press a certain key, I add in it a force to move on the rocket.Angle().
But when I try to use this concept by applying force on an object with physical behavior, I can not apply force at a given angle, only on the X and Y axes.
How could I then make my rocket an object with physical behavior if I need to move toward the angle that I am facing.
I hope it’s possible to understand despite my bad english.
writing this in the hopes your question will get attention and answers.
Hi Guys,
I’m not 100% sure but you can try using Polar Impulse instead
Angle - For the Angle
Length(N.m) - For the amount of force applied
Applying X Position/ Applying Y Position - The position where the force is applied to the object i think
Hope that helps
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2015-01-14T04:13:36-05:00
4x+4y=26
y=x-4
substitute y with (x-4) because y=x-4:
4x+4(x-4)=26
expand and simplify:
4x+4x-16=26
8x-16=26
8x=42
x=42/8=21/4=5 1/4=5.25
substitute x=21/4 into the equation y=x-4
y=(21/4)-4=5/4=1 1/4=1.25
(21/4, 5/4)
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766 articles – 1379 references [version française]
HAL: hal-00381191, version 2
arXiv: 0905.0573
Effective H^(\infty) interpolation constrained by Hardy and Bergman normsZarouf R.http://hal.archives-ouvertes.fr/hal-00381191
Available versions: v1 (2009-05-05) v2 (2010-05-31) v3 (2010-07-06) v4 (2010-11-03)
Preprint, Working Paper, ...
Mathematics/Functional Analysis
Effective H^{\infty} interpolation constrained by Hardy and Bergman norms
Rachid Zarouf () 1
1: Institut de Mathématiques de Bordeaux (IMB) http://www.math.u-bordeaux.fr/IMB/ CNRS : FR2254 – Université Sciences et Technologies - Bordeaux I – Université Victor Segalen - Bordeaux II 351 cours de la Libération 33405 TALENCE CEDEX France
Given a finite set \sigma of the unit disc \mathbb{D}=\{z\in\mathbb{C}:,\,\vert z\vert<1\} and a holomorphic function f in \mathbb{D} which belongs to a class X, we are looking for a function g in another class Y (smaller than X) which minimizes the norm \left\Vert g\right\Vert _{Y} among all functions g such that g_{\vert\sigma}=f_{\vert\sigma}. For Y=H^{\infty}, X=H^{p} (the Hardy space) or X=L_{a}^{2} (the Bergman space), and for the corresponding interpolation constant c\left(\sigma,\, X,\, H^{\infty}\right), we show that c\left(\sigma,\, X,\, H^{\infty}\right)\leq a\varphi_{X}\left(1-\frac{1-r}{n}\right) where n=\#\sigma, r=max_{\lambda\in\sigma}\left|\lambda\right| and where \varphi_{X}(t) stands for the norm of the evaluation functional f\mapsto f(t) on the space X. The upper bound is sharp over sets \sigma with given n and r.
English
2008
Attached file list to this document:
TEX
Paper-AIF.tex(98.5 KB)
PDF
Paper-AIF.pdf(318.6 KB)
PS
Paper-AIF.ps(1008 KB)
hal-00381191, version 2 http://hal.archives-ouvertes.fr/hal-00381191 oai:hal.archives-ouvertes.fr:hal-00381191 From: Rachid Zarouf <> Submitted on: Sunday, 14 March 2010 11:27:20 Updated on: Monday, 31 May 2010 11:20:18
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# Clinical Research Training Program 2021
Clinical Research Training Program 2021. DUMMY VARIABLES IN REGRESSION. Fall 2004. www.edc.gsph.pitt.edu/faculty/dodge/clres2021.html. OUTLINE. Indicator Variables (Dummy Variables) Comparing Two Straight-Line Regression Equations Test for equal slope (testing for parallelism )
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## Clinical Research Training Program 2021
E N D
### Presentation Transcript
1. Clinical Research Training Program 2021 DUMMY VARIABLES IN REGRESSION Fall 2004 www.edc.gsph.pitt.edu/faculty/dodge/clres2021.html
2. OUTLINE • Indicator Variables (Dummy Variables) • Comparing Two Straight-Line Regression Equations • Test for equal slope (testing for parallelism) • Test for equal intercept • Test for coincidence • Comparing Three Regression Equations
3. Indicator Variables in Regression • An indicator variable (or dummy variable) is any variable in a regression that takes on a finite number of values for the purpose of identifying different categories of a nominal variable. • Example:
4. Indicator Variables in Regression • If the nominal independent variable has k levels, k-1 dummy variables must be defined to index those levels, provided the regression model contains a constant term. Example: Variables X1 and X2 work in tandem will describe the nominal variable “geographical residence”, which has three levels: residence in western, eastern or central US. Residences in central US will correspond to X1 = 0 and X2 = 0 and we will call this group as the “baseline” group or “reference” group.
5. Indicator Variables in Regression • If an intercept is used in the regression equation, proper definition of the k-1 dummy variables automatically indexes all k categories. • If k dummy variables are used to describe a nominal variable with k categories in a model containing an intercept, all the coefficients in the model cannot be uniquely estimated because collinearity is present.
6. Indicator Variables in Regression • The k-1 dummy variables for indexing the k categories of a given nominal variable can be properly defined in many different ways.
7. Comparing Two Straight-line Regression Equations Example: Comparison by gender of straight-line regression of systolic blood pressure on age. Male Female
8. Comparing Two Straight-line Regression Equations Three basic questions to consider when comparing two straight-line regression equations: • Are the two slopes the same or different (regardless of whether the intercept are different)? • Are the two intercepts the same or different (regardless of whether the slopes are different)? • Are the two lines coincident (that is, the same), or do they differ in slope and/or intercept?
9. Methods of Comparing Two Straight Lines • Treat the male and female data separately by fitting the two separate regression equations and then make appropriate two-sample t tests. Male: Female:
10. HYPOTHESIS TESTS Using separate regression fits to test for equal slope (testing for parallelism) • H0: 1M = 1FH1: 1M 1F • Test statistic where sp2 is the MSE for the combined data If |T| > tn-4,1-/2, then reject H0.
11. HYPOTHESIS TESTS Using separate regression fits to test for equal intercept • H0: 0M = 0FH1: 0M 0F • Test statistic where Sp2 is the MSE for the combined data If |T| > tn-4,1-/2, then reject H0.
12. Methods of Comparing Two Straight Lines: Using a Single Regression Equation • Define the dummy variable Z to be 0 if the subject is male and 1 if female. Then, for combined data, fit the regression model:
13. Methods of Comparing Two Straight Lines • This regression model yields the following two models for the two values of Z: Male (Z=0): Female (Z=1):
14. Methods of Comparing Two Straight Lines • This allows us to write the regression coefficients for the separate models in the first method in terms of the coefficients of the model in the second method as: Thus, model in the second method incorporates the two separate regression equations within a single model and allows for different slopes and different intercepts.
15. HYPOTHESIS TESTS Using single regression fits to test for equal slope (testing for parallelism) • H0: 3 = 0H1: 3 0 • Test statistic • Decision: If F > F1, df(MSE(X,Z,XZ)), 1-, then reject H0.
16. HYPOTHESIS TESTS Using single regression fits to test for equal intercept. • H0: 2 = 0H1: 2 0 • Test statistic • Decision: If F > F1, df(MSE(X,Z,XZ)), 1-, then reject H0.
17. HYPOTHESIS TESTS Using single regression fits to test for coincidence • H0: 2 = 3 = 0H1: 2 0 or 3 0 • Test statistic • Decision: If F > F2, df(MSE(X,Z,XZ)), 1-, then reject H0.
18. HYPOTHESIS TESTS Comparison of Method I and II • The tests for parallel lines are exactly equivalent. • The tests for coincident lines differ, and the one using the dummy variable model is generally preferable. • If we test the coincidence from separate straight-line fits rather than a single test, the overall significance level for the two tests combined is greater than , that is, there is more chance of rejecting a true H0. ( is the significance level of each separate test.)
19. Comparing Three Regression Equations Example: Y = SBP, X1 = Age, X2 = Weight Z1 = High school education Z2 = Education greater than high school (baseline = education less than high school) Regressions: Edu<HS Edu=HS Edu>HS
20. HYPOTHESIS TESTS Test all three regression lines are parallel • H0: 5 = 6 = 7 = 8 = 0 • Test statistic • Decision: If F > F4, df(MSE(full model)), 1-, then reject H0.
21. HYPOTHESIS TESTS Test all three regression lines are coincident • H0: 3 = 4 = 5 = 6 = 7 = 8 = 0 • Test statistic • Decision: If F > F6, df(MSE(full model)), 1-, then reject H0.
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740 and Unique Background
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Current Student
Joined: 21 Jul 2012
Posts: 3
Kudos [?]: 5 [0], given: 1
Concentration: Nonprofit, Social Entrepreneurship
GMAT 1: 740 Q49 V43
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25 Jun 2014, 19:19
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Last edited by smitnaik3 on 02 Feb 2015, 13:44, edited 1 time in total.
Kudos [?]: 5 [0], given: 1
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01 Jul 2014, 08:16
Hello,
From the information provided, I believe you’d be a credible and possibly strong candidate for the programs you mention.
Your stats are obviously fine. As you note, your experience is unconventional – but it’s at an appropriate level. So the unconventional aspect creates both challenges and opportunities. Challenges: (a) the adcom must rely on you to explain the whole scenario – not just what you do, but also where, including a clear discussion of your venture, and why you chose this path; (b) regardless of work setting or role, a solid applicant to Booth or Kellogg must show professional growth, so give some consideration to how you will do that. Opportunities: (a) if you address the challenges effectively, you automatically differentiate and distinguish yourself in relevant ways; (b) your goals arise from and draw on your unique experience, so you can ideally create very compelling, firmly rooted goals that will excite the adcoms.
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# Complex Numbers and Trigonometry
1 teachers like this lesson
Print Lesson
## Objective
SWBAT write a complex number in trigonometric form.
#### Big Idea
How is it possible to write a complex number using trigonometric functions?
## Bell work
5 minutes
Today students will write complex numbers in trigonometric form which is also called polar form. Since we have not discussed the polar coordinate system I use the term trigonometric form.
I begin by asking students to think about how to use trigonometry to write a complex number in trigonometric form. As students think, I ask questions focus questions such as:
• What is needed to use trigonometry?
• Would plotting the number help us determine how to do trigonometry?
• Could we use a process similar to finding the component form of vectors?
My goal for the bell work is to get students to see how they can find the angle rotated from the x axis and the absolute value of the complex number can be used to an appropriate form.
## Finding the Trigonometric Form of a Compex Number
15 minutes
Now that students have started to reason about how to writing trigonometric form. I continue with the bell problem. Students process through the bell work by plotting the point then drawing a triangle. with that they find the angle and the distance from the origin. From that they get a formula. As they do this I remind students how we write a vector using standard unit vectors.
Once students have used the bell problem to find a trigonometric form I move students to generalize the process.
## Converting to Complex form
10 minutes
Now that students have a general process for converting to trigonometric form, students work on problems. The first problem is looks very similar to the z=3+2i but the other 2 may confuse the students. The issue will be not thinking about the real or imaginary term equaling zero. I remind students to graph if number on the complex plane to help them understand what the numbers looks like. Once this is done students are able to determine the equation.
I move around the room checking results as the students work and helping students as needed. As I move around the room I also keep note of students who have correct answers and those that have done the problem differently. When I notice most students have completed a couple of the problems, I begin asking students with answers and unusual methods to put the answers on the board. We continue until all the solutions are shared with the class.
## Converting trigonometric form to standard complex form.
10 minutes
I now have students determine how to convert form trigonometric form to standard complex form. I give students the page 2 of resource. I do not explain how to do the problems but let the students think about how they can find the standard form. As students discuss the problems I ask questions such as:
• What does standard complex form look like?(z=a+bi)
• What is a in the trigonometric form?
• How can we simplify the expression?
My goal is for students to see they just need to evaluate the trigonometric expression and multiply the result by r to determine the coefficient and constant in the standard form. Most students will see how to do this quickly while others will struggle since some numbers are represented using trigonometric functions.
In these problems I have used angles One issue that sometimes deal with is the tendency to use the calculator on angles that we have exact values (multiples of 30 degrees and 45 degrees). If students give me the calculator answer I will ask them to also find the exact answer. Any time I can give students time to practice with these angles I use.
## Closure
5 minutes
With about 5 minutes left in class I ask students to answer this question on an exit slip. I am hoping students see 2 issues with the example. First the coefficients on the terms are not the same and second the angles are not the same. Any students that do not see these 2 issues will need some clarification as we continue with complex numbers. I will review with the students on an individual basis during any work time we have.
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#### Archived
This topic is now archived and is closed to further replies.
# [Priests 5.0.4] I Get Misty
## 95 posts in this topic
Since there is no topic for shadow, I'll post this here.
I've been experimenting with divine insight, and quite liked the twist it adds to our "rotation". However, it currently seems to be quite bugged.
I noticed that if I get a proc when I'm already casting MB, the proc is consumed but seems to instantly proc again, allowing me to cast another MB. This makes sense, though there is a slight delay before it happens.
An actual problem is that sometimes, I get a proc (and MB button is highlighted), but the cooldown is not reset. It happens more often at high haste, such as with Alysrazor buff (where it happens pretty much all the time, making the talent next to worthless). Did anyone else notice this, and is there anything that can be done to prevent it?
##### Share on other sites
I open this subject because i am really confused about my result and would have a confirmation from you if i am wrong or right.
Objectif of calculation was, at the begginning, to simply calculate the %haste needed to gain additionnal cast of PoH when using Spririt Shell.
It is simple (for example 100%-7/6 = 16.67% haste to win 1 cast of PoH for 15sec) but i wanted to find a standard formula to use it on every repetitive event like Fiend, MindBender (if they follow haste effect), Hymn of hope, renew, etc... from others classes....
Formula i have reached is :
Number of times of event = ENT [ 1 + (Time effect - delay)/Time between event ]
where
Time between event = base Time between event / (1+%Haste)
delay = base delay / (1+%Haste)
==> Number of times of event = ENT [ (base Time between event - base delay)/base Time between event + (1+%haste)* base number of time of event ]
With :
base number of time of event = Time effect / base Time between event
base delay = can be 0 if the first tic is immediate after the cast (ex :Mindbender), GCDu (if it is 1GCDu =1,5sec after the cast), time casting (if it is a heal which as to be cast (ex : PoH for SS claculation)
Means also that :
Additional times of event = ENT [ (base Time between event - base delay)/base Time between event + %haste* base number of time of event ]
NOW, the problem i have ;
For exemple for Renew :
Time effect = 12 sec
Delay = gcd (because the first tic is done 1 gcd after the cast) = gcdu / (1+%haste) = 1,5* /(1+%haste)
Time between event = Base time betwwen event / (1+%haste) = 3 / (1+%haste)
So, we arrived on the well known formula of Renew which is number of TIC = ENT [0,5 + 4*(1+%haste) ]
or also the well known formula "Number of additional TiC = ENT ( 0,5 + Base Number of Tic *%haste)
Then, based on this last formula "ENT ( 0,5 + Base Number of Tic *%haste)" which defines the number of additionnal tics, i see everywhere, more or less, the following table which define breakdowns for haste :
3 Base Tics : 16.67% 50.00% 83.33%
4 Base Tics : 12.50% 37.50% 62.50% 87,50%
5 Base Tics : 10.00% 30.00% 50.00% 70.00% 90.00%
6 Base Tics : 08.34% 25.00% 41.67% 58.33% 75.00% 91.66%
7 Base Tics : 07.14% 21.43% 35.72% 50.00% 64.29% 78.57% 92.86%
etc...
That means (and i take this data until this date) for Hymn of hope which is a 4 Tics base class, that the additional tic will be reach at 12,5% of haste like the renew.
BUT When using the standard formula i gave ,the result is different because the formula used is :
Additional times of event = ENT [ (base Time between event - base delay)/base Time between event + %haste* base number of time of event ]
Means that for Hymn of Hope, additional tic is reached for 18,75% Haste instead of 12.5% :
base Time between event = 2 sec
base delay = 1 GCDu = 1.5 sec. (first tic is given after 1 GCD)
base number of time of event = 12/3 = 4
1 = ENT [ (2-1.5)/2 + 4*%haste ] ==> %haste = 0.75 / 4 = 18.75%
(except if the base delay is not 1 GCDu but 1 sec ==> in this cas, we have 12,5%)
My Conclusion is that it is wrong to use this table and "ENT ( 0,5 + Base Number of Tic *%haste)" is true only if the "base delay" is half of the "base Time between event".
And for additionnal cast of PoH under SS, i have my result now with the généric formula which matchs to the simple calculation "%haste needed for PoH under SS = 100% - [Number of cast per 15sec / Base number of cast per 15sec]" :
6 casts 2.5 Sec base for 15sec of SS stack period
Generic formula gives same result : 16.67% 33.33% 50.00% 66.67% 83.33% 100.00% when i take :
"Additional times of event = ENT [ (base Time between event - base delay)/base Time between event + %haste* base number of time of event ]"
Sorry, for this long explanation and thank you for your feedback !
(did i miss something ?)
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Blizzard has changed channeled spells specifically. With 33% haste I still get 4 ticks and no more and the total channel time is 6s.
With 33% haste I also have mind sear finish channeling in 3.75 seconds and getting 5 ticks (one every 0.75s).
Channelled spells no longer get extra ticks. I don't think this is a big. Its probably a new design. Maybe blizzard does not want the HPM of channeled spells to increase with haste.
Mindbender and shadowfiend no longer benefit from haste either.
They definately did benefit from haste until just after we got the new spirit shell. I actually tested it, because I wanted to be sure about the formula.
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Thank.
I made a complete mistake and i just see IG that hymn of hope did not depend on haste no more.
For Mindbender and fiend , it is very stange because i have noticed the same thing as you about haste which has no impact (done after recording the log or heard the attack) BUT a lot of people told me that they have seen and verify IG (5.05 cata) that haste (on equipement) scales on pets.
depending on haste on equipement and haste proc with trinket, they have confirmed to see mindbender returns 14 to 15 tics.
I do not understand since i do not find same result but it is strange because people who has told me that are serious player.
I think , i have also made a mistake for the generic formula i gave even for Renew because "base delay" for renew is 3sec instead of GCD.
and my formula does not match real result.
Could you check where i am wrong to apply :
==> Number of times of event = ENT [ (base Time between event - base delay)/base Time between event + (1+%haste)* base number of time of event ]
With :
base number of time of event = Time effect / base Time between event
base delay = can be 0 if the first tic is immediate after the cast (ex :Mindbender), GCDu (if it is 1GCDu =1,5sec after the cast), time casting (if it is a heal which as to be cast (ex : PoH for SS claculation)
Means also that :
Additional times of event = ENT [ (base Time between event - base delay)/base Time between event + %haste* base number of time of event ]
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Dear Havoc12
Juste about my comment about pets who can integrate haste , i made a file with the following logs :
It is easy to find when pet return mana , it is called in french "sangsue de mana" :
to summarize the conditions ;
Try N°1 : with haste 11.28% (1444 score) ==> 11 Tics
Try N°2 : with haste 14.85% (1902 score) ==> 12 Tics
Try N°3 : with haste 14.85% (1902 score) ==> 11 Tics
i find same behavoir on shadowfiend.
and i think , i have noticed that immediate first tic when cast is done is not always true.
It seems that tic calculation is not accurate but it exists !
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When I tested it just after we got the new spirit shell, fiend scaled with haste. After a certain point blizzard changed it.
Haste definitely no longer benefits fiend. However external buffs do apply to the fiend. So totems increasing attack speed and heroism or pet haste buffs definitely affect fiend attack speed and hence they may give a better return if everything lines up properly.
I tried it dozens of times and its definately 11 attacks. The fiend is ready to attack the instant its summoned and unless the opportunity is denied it does.
===================================================================================
On a sidenote, PoH has been nerfed recently. It now no longer applies the base 30% aegis to crits. Crits only apply 30% of a crit as aegis and no other aegis...
The PoH now has a crit multiplier of exactly two, so the formula is now
base*(1+0.3*(1+mastery))*(1+crit)
This further reduces the value of keepinga aegis alive.
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I tried it dozens of times and its definately 11 attacks. The fiend is ready to attack the instant its summoned and unless the opportunity is denied it does.
I have tried also during several hours and main results are 11 attacks.
But you cannot close your eyes about the log i have linked.(alone, without buff on a dummy low level)
Try N°2 at 0H26 ==> 12 tics !
and we can reproduce it at 14 or 15 tics with haste above 40% and 50% (with haste proc trinket)
I have dont link the video i have done with Fraps also during these log, but when you wotch it frame by frame, you can verify that mana jumps are conform with log recording.
There is something unclear or unstable about attack of pet et mana return.
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I cant access the log I don't have a google account.
to verify what you said I did an experiment, I proced by DS haste trinket, used PI and PWS which boosted by haste to 95%. Mindbender gave me back 14 attacks. I wonder if blizzard fixed the haste scaling!
~Yes they did!! Mindbender now scales with haste. I used PWS and I got 12 attacks. So mindbender now scales with haste! I will investigate the scaling [/edit]
At 16% haste I get 12 attacks.
At 33% haste I still get 12 attacks. (PWS)
At 39% haste I still get 12 attacks. (Power Infusion)
at 60.1% haste I get 14 attacks (haste trinket)
Thus mindbender now scales only with haste rating, it does not scale with any buffs that increase spellcasting speed or spell haste. Only effects which increase both spellcasting and melee attack speed affect mindbender. So haste rating on gear, bloodlust, the attack speed totem etc, all increase mindbender attacks, but buffs, which increase spellcasting speed, but not melee attack speed have no effect.
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Thus mindbender now scales only with haste rating, it does not scale with any buffs that increase spellcasting speed or spell haste. Only effects which increase both spellcasting and melee attack speed affect mindbender. So haste rating on gear, bloodlust, the attack speed totem etc, all increase mindbender attacks, but buffs, which increase spellcasting speed, but not melee attack speed have no effect.
i agree. It is the desired design.
(i forget to say that PI, PWS etc.. has no effect. Only haste rating on equipement or by trinket.
Scaling is +1 tic per 10% haste. (with exclusion of what we have told about pi ,....
BUT, it is not always reproducable. I did not make a statistic but on my trys, i would say that it is verify at 20%-30% only.
I do not know and understand the cause.
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I tried it quite a bit looking at the log and here is what I found. I think that when the mindbender casts shadowcrawl that attack is completely unhasted.
the log looks like this (with 60% haste):
...:44 I cast mindbender
...:44 attack1
...:46 attack2
...:47 attack3
...:48 attack4
...:49 attack5
...:50 attack6
...:51 attack7
...:52 attack8
...:53 attack9
...:54 attack10
...:55 attack11
...:56 attack12
...:57 attack13
...:58 attack14
...:59 attack15
I expect 1 attack every 0.9375seconds, but I get 15 attacks and 3 shadowcrals in 15 seconds. However if each shadowcrawl attack is unhasted this is what I expect:
```
attack time attack total time (sec)
(sec)
1
1.5 2 1.5
0.9375 3 2.4375
0.9375 4 3.375
0.9375 5 4.3125
0.9375 6 5.25
1.5 7 6.75
0.9375 8 7.6875
0.9375 9 8.625
0.9375 10 9.5625
0.9375 11 10.5
0.9375 12 11.4375
1.5 13 12.9375
0.9375 14 13.875
0.9375 15 14.8125
```
Which matches perfectly with what I see on the log.
If I turn shadowcrawl off as soon as the mindbender is summoned I get 16 attacks.
So basically it looks like shadowcral attacks are the reason why haste does not quite work the way we expect with mindbender.
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Very interresting !
i understand now, why some gcd are lost.
Nevertheless, Shadowcrawl is instant but use 1 gcd. that means that attack is delayed of 2 gcd from the previous one. So, it is not 1.5sec normally. don't you think ?
edit : hum... automatic melee attacks is not folowed the gcd, i think...
Just 2 confirmations :
- With Haste at 60%, attacks is every 0.9375seconds means that 17 attacks are possible in 15 sec because of instant attack as soon as the pet is casted.
If we have only 16 times of mana return that means that the last one is lost because the pet has disapeared before the proc of mana return ?
if yes, the haste breakdown must be estimated with the delay existing between the attack and the tic of mana return.
- To optimize the number of tic, it is necessary to remove shadowcrawl attacks.
i have not tested yet but do you think it is possible to disable this attack when we cast the pet ?
it is perhaps possible to remove it by a macro done with the cast of the pet (but maybe necessary to activate the macro 2 times)
edit : maybe, it is necessary to post a bug report if it is not possible to cancel the shadowcrawl ?
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I can't reproduce the "shadowcrawl slows down the swing timer" phenomenon. I am seeing that there is a delay between the swing and the mana gain, and I don't seem to get the mana from the last swing, either (consistently at my gear level, need to try to see if there are breakpoints for this).
Most runs were done with a 1.23 sec gcd (the ones where I had darkness are indistinguishable from the ones where I didn't), except for the second to last run, which also had heroism cast partway, and the last run, where I had no gear on. I'll do more experimenting when I get a chance. Either Blizzard is messing with us, or there's something going on that we're missing, I think.
Edit: I ran a bunch of different haste levels through. Basically at every level except 904 haste rating, I received 1 more melee swing than mana returns. I'm not entirely sure why.
[TABLE]Haste NumSwings NumMana
2776 13 12
2613 13 12
2520 13 12
2410 13 12
2247 12 11
2154 12 11
2047 12 11
1933 12 11
1769 12 11
1655 12 11
1564 12 11
1415 12 11
1267 11 11
1160 11 10
1067 11 10
904 11 11
715 11 10
622 11 10
445 11 10
[/TABLE]
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yes.
I take back the 3 examples i have linked and made calculation of time betwwen swing, mana regen.
I do not notice also any influence of shadowcrawl.
Roughly, i find :
Time between swing attacks ~ gcd
Time between mana regen instable (can be from 1.1sec to 1.6 sec , not in relation with the cast of shadowcrawl)
Time between swing attacks and Mana regen from 0.8sec to 1.2sec
Time of the last tic of mana regen can be more than 15sec. after the cast of pet (i see 15.36 sec) but never above.
If the last swing attack is more than 14.6 sec, I lost tic of mana regen.
So, it seems to be impossible to do exact estimation of the number of tic with a given level of haste....
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As before, there's not much of a "rotation" but the usual priorities:
DPx3 > MB > SW: D (sub 20%) > SF/MBen > FDCL > SW:P > VT > MF
Not everyone has FDCL of course, though personally I find it one of the most fun things about the new system. It really adds additional depth and skill to a previously quite repetitive routine. Also note that you MF a lot less than before, and most likely clip it frequently.
So far the most challenging thing for me has been placing FDCL procs properly. There is a lot of things to consider like imminent buff-procs, when MB comes off CD, chance of wasted stacks etc. It gets especially hectic in a multi-dot scenario, but that makes it only more fun! :) For now I'm running with ToF over DI, but I'm not sure how good that is. That would be even more procs then!
Been wondering, with how good shadow orbs now, is the SW:D glyph possibly worthwhile (SW:D over 20% health, but do 1 quarter damage and eat some backlash... basically shit damage but orb generation.)
My gut instinct says no, leave that for pvp or other fringe situations where you want to generate orbs as fast as possible for a burst scenario, but that it's not good
I'm also really liking FDCL, especially while leveling - it makes me a lot less annoyed about the very common scenario where you're dotting up 2-3 targets at once, but mind sear kinda sucking for aoe.
I idly wonder though, if taking mindbender for DPS might be better or similar dps with a less complex rotation on a single target? And didn't they buff SW:Insanity? That one looks cool too...
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You should probably read the glyph again. Glyphed SW: Death does *not* generate *any* Orbs. It would definitely be in our rotation otherwise, but that way, it's not. The glyph is pretty much useless for PvE.
So far FDCL has been consistently simming as the highest DPS talent. As for how wide the margins are, that is up for debate. More data will be forthcoming as proper, live raiding goes underway.
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I posted this on the blizzard forum. It is possible that it was fixed. I replicated many times when I tried it, so I know it was correct. You should all be using a haste tirnket to get your haste to a very high level and see how many attacks you get. Try to hit the same value as I did.
It is possible that one attack is lost now due to lag if the delay between the swing and the mana leech after the last attack is too long. The leech is applied on the shadowfiend, so if the shadowfiend expires before the leech is applied, I think it is highly likely that the script will fail to run.
Before this we were not missing one attack. We were missing up to 4 attacks, if you see my original example and I repeatedly got more attacks when I removed shadowcrawl.
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Hi
i havent noticed any mentioned on it here and havent found much on it but a priest in my raid group has been running into problems with his guardian spirit both now in mv and dating back to ds. Its seems like the "cheat death" part of guardian spirit is not working. And i have checked it out in the combat log when he has cast uardian spirit on me (im tanking) and i get the cast on me and get the killing blow on me where the guardian should kick in and not kill me and heal me but that part doesnt seem to work. Anyone else here been running into some problems with guardian spirit ?
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There are two cases I've seen where guardian spirit fails to save someone. Firstly, the race condition, when I cast it *just* before the killing blow, it goes on cooldown, but the target dies anyway (Solution: cast earlier). Secondly, when the killing blow is too large (since patch 4.0.6, Guardian Spirit has had a cap on the amount of damage it will absorb (of 200% of the target's max health), so a single hit bigger than that will still kill a target with GS on them). If you're trying to use Guardian Spirit to ignore some fight mechanics, you may have to layer on other cooldowns to reduce the big hit (or just change your strategy).
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Thank you for a quick response.
But my feeling all along has been the former o your suggestions that the cast is just a little too late and possible the servers read the cast sequence in just a little bit different order then they are cast in, so eventhough the gs is cast before the real kiling blow then the killing blow maybe is read before the gs on the one taking the hit and therefor it might not get properly activated.
So the solution is most likely just to get the cast in earlier, as rarely if ever i or the other tank is taking a hit in excess of 200% of max health.
Again thank you and i will forward this to my, quite frustrated by the failure of his gs, priest healer.
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For those looking to get the most out of their shadow priest, I made a video starting with some basics and moving into more advanced topics like movement and rotation strategies. Check it out!
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I'm wondering what talentchoices other people maken on the different fights. Espacially the level 90 talents. I am referring to people that play holy.
In the first tier I don't believe their is much use of these talents except for Void tendrils during Elegon.
In the second tier however their is the choice of using Body and Soul and Angelic feather. I myself am more fond of Body and Soul. I don't need to put down feathers and have people walk through them. With B&S I can just instantly give someone a boost. It does however requires mana to give people that boost.
The third tier starts getting interesting. FDCL has been a favorite of mine during cataclysm, but it has little use on big AoE fights, since it doesn't proc on our AoE abilities. Mindbender is my more preferred talentchoice, since it has only a 1 min CD. The 0,7% of Solace really isn't worth it in my opinion. That's only 2100 mana. It can be argued that you also regain mana since you are not casting. But I can also simply not cast. I will still regain that mana regardless.
In the fourth tier I'm still torn between Desperate Prayer and Angelic Bulwark. Allthough AB requires no effort on my part, it can proc on moments you really don't need the shield. A shield however mitgates the damage and is instantly up. With desperate prayer you only regain health and you will still need to press the heal. With DP I have the luxury of using it whenever I want. I'm leaning towards AB at the moment for all the fights.
In the fifth tier I have found twist of fate being a great addition during Garajal, Kings and Elegon, since there are several moments where you can proc that talent due to several bosses or adds during that fight. On Stoneguard and Feng I lean more towards Power Infusion. PI gives the boost that is handy on the moments it is needed.
The final tier is the tier I switch around most, which I'm sure most of the raiders do.
• On Stoneguard I prefer Halo, since most people are spread anyways over about the range halo is most effective.
• On Feng I usually go with Divine star. Ranged and melee should be away from eachother about the range of DS. It has a short CD and can help out during the entire fight when there are low damage phases.
• On Gara'jal I tend to use DS also. Everyone is always stacked up. I have not tested using halo on this fight. Arguably if you just get yourself in the perfect position you should get enourmous heals, but I prefer to stand still most of the fight.
• On spirit kings I find DS nice to have during the Maddening shout. During the way forward it damages and on its way back it heals. Halo however has some decent use as well because of the way the fight works. The spread fases are most of the time the more damaging fases. The fight is also a very mobile fight so during stack fases it's not hard to get some distance between you and your raid for perfect halo usage.
• On Elegon I find Cascade to be most effective because you can reach out of range players on the other side of the platform. I don't see DS having any kind of use. Halo just can't keep up with cascade.
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I just switched to Holy with the start of MoP raiding, was practically Disc since Ulduar, and I'm quite liking it.
As you mentioned the first tier isn't all that important, I mostly have Tendrils for some occasional uses like Emperors Rage and the likes.
I also favor Body and Soul, probably because I always loved that aspect the most when healing some encounters as Holy, but there might be some uses where a tank is kiting on a preplanned route, so I see Angelic Feathers still having some usefulness.
On tier three I nearly always have Mindbender, because of two facts. It is a significant dps increase when used on cooldown (or close to anyways) and the mana income is smoother than with a normal shadowfiend but still feels less maintenancey than PW:Solace.
AB is my clear winner, mainly because I'm bad in those regards and forget to use healthstones and DP all the time but you are quite right that AB sometimes proccs when its not needed, some non lethal aoe which takes the raid low but wont kill it if done correctly and where you have "ages" to bring the raid back up...
Regarding tier 6 I should probably try to vary a bit more, seeing as you can procc ToF with some adds or on those non lethal aoes quite easily and PI yields very good on demand burst for something like Elegon HC final phase (even though it worked without). But I love DI as holy so much because the ProM can be used as a follow up after some large AoE and doesn't need constant ticking damage or anything to trigger and may even be used as a decent single target heal on a tank if there's a melee (or pet) nearby from whom it can bounce back and forth.
Lvl 90 Talents:
Stone Guards (HC): Cascade is my favorite here, especially when people are running around reactivating tiles.
Feng (HC): As you mentioned DS because we are grouped up nearly the whole fight or at least when the incoming dmg is high.
Gara: I wasn't in for our progress, haven't killed him so far.
Kings: For our few heroic attempts I found DS quite nice for Qiang but I guess Halo or Cascade should be better overall since after Qiang you'll be mostly spread out.
Elegon (HC): Cascade is a godsend on this encounter, especially for the second phase, picks up very many people on both sides every 2nd charge phase for very little mana, not quite as effective for the final phase but that does not last really long anyways.
Will: Cascade and Halo should both proof very good here depending on raid composition and tactic.
Zor'lok: I found Cascade really useful for the Force and Verve phases, DS hits too few targets and Halo may not hit everyone.
Ta'yak: passed
Garalon: passed
Mel'jarak: Nearly killed him yesterday and I think Cascade again was the most useful for Raining Blades. Extra care is needed if you pick Halo/DS to not accidentally break a CC.
Overall I feel that Cascade for spread out and DS for grouped up scenarios are the most useful, since Cascade has the advantage over Halo in regards to finding its targets and not needing very careful positioning/timing to be effective.
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First tier has no real use in raiding. I have void tendrils, but only use them when doing daily quests. (Before void shift was fixed to only work on players, they made great VS targets, but that doesn't work any more)
Second tier, I usually use feathers. They take some getting used to, but as holy I don't use shields, so even though it's easier to cast a shield on someone than to place a feather for them, it would cost mana that I usually don't have to spare. Feathers are free. Also, I can use 4 in a row (and then 1 per 10s), while B&S is 1 per 15s. The only fight I change this for is stone guards - because of the constant snare, phantasm works better for getting myself in to my chain buddy.
My choice for tier 3 depends on my healing role. If I'm assigned to tanks, I'll take FDCL; if not, mindbender (I've tried solace, but I'm not a fan).
I always use desperate prayer from tier 4. Angelic bulwark doesn't require a keypress, but also doesn't save you from being 1shot from 40%.
Tier 5, I usually take Twist of Fate. In addition to procs from healing people who are low on health, I try to keep my target set to whatever mob will be the next to die, and if I have 7800 mana and a GCD to spare when my power word: death button lights up, pushing it will boost my healing for 10s as well as helping dps. Divine Insight is also good, and I'll have to give it another try some time, but last time I used it I found it too easy to burn my mana too fast. Even though PoM is one of our most efficient heals, it's still not cheap on mana - when things get hectic and I'm spamming PoH as fast as I can, the extra healing from ToF is free, while divine insight means I run oom faster. When I go disc, I switch to Power Infusion (because ToF doesn't affect absorbs, but PI + spirit shell is a combo made in heaven).
Because I raid 25man, cascade is my go-to choice for the 6th tier. It works when the raid is spread out, or when we have a melee group and a ranged group; it lines up nicely with most boss abilities; and it doesn't do any damage (the number of times someone's halo wiped us on spirit kings...). The only time I change it is when our strategy calls for the whole raid to be stacked up together, as this is when cascade is weakest. Then, if I can run out (using either body & soul or feathers) and halo, I do that; otherwise I use the star.
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Anyone will do a trinket dps comparasion?
There's the issue, I have [iTEM]79331[/iTEM] and [iTEM]86133[/iTEM](ICD is 45sec)
There's the ask: What would be better the mogu shan one normal version or the new reputation one Shock-Charger Medallion - Trinkets - Armor - Items - WowDB (PTR)
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## Missing data
Solved
Occasional Contributor
Posts: 11
# Missing data
Hi there,
I'm having some trouble when I'm trying to make some new datasets.
I have survey data from 2 separate cohorts which are in two separate data sets. I've pooled them into one data set (called pooled) and pooled the variables of interest into new pooled variables.
The problem that I'm having is that the same variable from the cohort1 and pooled datasets have different number of missing values.
Example: there's a variable in cohort1 dataset called netcst1 with ~2000 missing values (out of ~600,000 so not really a large percent). The same variable that has been put into the pooled dataset has 500,000+ missing values, and I can't figure out why that is.
Syntax-wise, this is what I've done:
DATA pooled;
SET cohort1 cohort2;
....
/*I've created the new pooled variables I need here*/
...
RUN;
In cohort1 (and therefore in pooled), there is a variable called netcst1 (which I've verified through a PROC CONTENTS), so these two statements give me two different answers:
(I) proc means data = cohort1 nmiss;
var netcst1;
run;
(II) proc means data = pooled nmiss;
var netcst1;
run;
Shouldn't I be getting the same number of missing values for both? What have I done wrong for (I) to give me about 2000 missing values and for (II) to give me more than 500,000 missing values?
Accepted Solutions
Solution
09-25-2015 06:23 AM
Occasional Contributor
Posts: 12
## Re: Missing data
Since it is a vertical join, the pro means is going to give result of total no of observation from both data set.
example:
data a;
input num VarA ;
datalines;
1 7
2 8
3 .
;
proc print data=a;
run;
data b;
input num VarB ;
datalines;
4 6
5 .
6 3
;
proc print data=b;
run;
proc means data = a n nmiss;
var VarA;
run;
proc means data = b n nmiss;
var VarB;
run;
data joined;
set a b;
run;
proc means data = joined n nmiss;
var _numeric_;
run;
the result:
Obs num VarA 1 2 3
1 7 2 8 3 .
Obs num VarB 1 2 3
4 6 5 . 6 3
The MEANS Procedure
Analysis Variable : VarA N N Miss
2 1
The MEANS Procedure
Analysis Variable : VarB N N Miss
2 1
The MEANS Procedure
Variable N N Miss
num
VarA
VarB
622 044
All Replies
SAS Employee
Posts: 12
## Re: Missing data
Is netcst1 on cohort2? if not how is netcst1 calculated for observations coming from cohort2?
Occasional Contributor
Posts: 11
## Re: Missing data
So the variable netcst1 exists in cohort1 only and then a different variable, call it netcst1a, exists only in cohort2. In the pooled data set, the variable poolednetcst brings them together.
SAS Employee
Posts: 12
## Re: Missing data
How many missing from cohort2?
``````proc means data = cohort2 nmiss;
var netcst1a;
run;``````
Occasional Contributor
Posts: 11
## Re: Missing data
I believe 2400 or so values are missing from cohort2, which is completely reasonable considering the size of the dataset.
Solution
09-25-2015 06:23 AM
Occasional Contributor
Posts: 12
## Re: Missing data
Since it is a vertical join, the pro means is going to give result of total no of observation from both data set.
example:
data a;
input num VarA ;
datalines;
1 7
2 8
3 .
;
proc print data=a;
run;
data b;
input num VarB ;
datalines;
4 6
5 .
6 3
;
proc print data=b;
run;
proc means data = a n nmiss;
var VarA;
run;
proc means data = b n nmiss;
var VarB;
run;
data joined;
set a b;
run;
proc means data = joined n nmiss;
var _numeric_;
run;
the result:
Obs num VarA 1 2 3
1 7 2 8 3 .
Obs num VarB 1 2 3
4 6 5 . 6 3
The MEANS Procedure
Analysis Variable : VarA N N Miss
2 1
The MEANS Procedure
Analysis Variable : VarB N N Miss
2 1
The MEANS Procedure
Variable N N Miss
num
VarA
VarB
622 044
Occasional Contributor
Posts: 11
## Re: Missing data
Ohhhhh that makes so much sense now. Is there a way to get rid of that problem? Maybe if I used a MERGE statement rather than a SET statement?
Occasional Contributor
Posts: 12
## Re: Missing data
You can use Merge statement if you have common variable in both data set, but than you will have to sort your data first.
Super User
Posts: 7,938
## Re: Missing data
You need to add a CLASS statement to your PROC MEANS so the cohorts are treated as separate groups.
If you do not already have a variable that defines the cohort you could create one during the step that combines the data.
``````data pooled ;
length cohort indsname \$50 ;
set cohort1 cohort2 indsname=indsname ;
cohort=indsname ;
...
run;
proc means ;
class cohort ;
...``````
🔒 This topic is solved and locked.
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# jax.numpy.float_power¶
jax.numpy.float_power(x1, x2)¶
First array elements raised to powers from second array, element-wise.
LAX-backend implementation of float_power(). Original docstring below.
float_power(x1, x2, /, out=None, *, where=True, casting=’same_kind’, order=’K’, dtype=None, subok=True[, signature, extobj])
Raise each base in x1 to the positionally-corresponding power in x2. x1 and x2 must be broadcastable to the same shape. This differs from the power function in that integers, float16, and float32 are promoted to floats with a minimum precision of float64 so that the result is always inexact. The intent is that the function will return a usable result for negative powers and seldom overflow for positive powers.
New in version 1.12.0.
Parameters
• x1 (array_like) – The bases.
• x2 (array_like) – The exponents. If x1.shape != x2.shape, they must be broadcastable to a common shape (which becomes the shape of the output).
Returns
y – The bases in x1 raised to the exponents in x2. This is a scalar if both x1 and x2 are scalars.
Return type
ndarray
power()
power function that preserves type
Examples
Cube each element in a list.
>>> x1 = range(6)
>>> x1
[0, 1, 2, 3, 4, 5]
>>> np.float_power(x1, 3)
array([ 0., 1., 8., 27., 64., 125.])
Raise the bases to different exponents.
>>> x2 = [1.0, 2.0, 3.0, 3.0, 2.0, 1.0]
>>> np.float_power(x1, x2)
array([ 0., 1., 8., 27., 16., 5.])
>>> x2 = np.array([[1, 2, 3, 3, 2, 1], [1, 2, 3, 3, 2, 1]])
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https://www.coursehero.com/tutors-problems/Finance/11671191-Assume-that-sales-are-expected-to-increase-by-5-next-year-core-profi/
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View the step-by-step solution to:
# Assume that sales are expected to increase by 5% next year, core profit margin is expected to be 12% of sales, and current NOA is expected to grow at...
Assume that sales are expected to increase by 5% next year, core profit margin is expected to be 12% of sales, and current NOA is expected to grow at the same growth rate of sales. If the required return on operations is 8%, current sales are \$854,345, current NOA is \$252,710, what is the expected ReOI for the coming year?
Answer is 87,430.67, just wondering on a solution.
Expected ReOI = Expected Earnings - (NOA of previous year x cost of capital) Required rate of return = cost of capital = 8%... View the full answer
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http://people.sc.fsu.edu/~jburkardt/cpp_src/asa006/asa006.html
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# ASA006 Cholesky Factor of a Positive Definite Symmetric Matrix
ASA006 is a C++ library which computes the Cholesky factor of a positive definite symmetric matrix.
ASA006 is Applied Statistics Algorithm 6. Source code for many Applied Statistics Algorithms is available through STATLIB.
If A is a positive definite symmetric matrix, then there is an upper triangular matrix U with the property that
A = U' * U
The matrix U is known as the Cholesky factor of A, and can be used to easily solve linear systems involving A or compute the inverse of A.
The algorithm implemented here uses a compressed storage for both the matrix A and the factor U. This saves some storage, but can make computations a little awkward.
### Languages:
ASA006 is available in a C version and a C++ version and a FORTRAN77 version and a FORTRAN90 version and a MATLAB version.
### Related Data and Programs:
ASA007, a C++ library which computes the inverse of a symmetric positive definite matrix, and uses a version of ASA006 for for Cholesky factorization.
ASA047, a C++ library which implements the Nelder-Mead minimization algorithm, and uses a version of ASA006 for Cholesky factorization.
LAPACK_EXAMPLES, a FORTRAN77 program which demonstrates the use of the LAPACK linear algebra library.
LINPACK_D, a C++ library which includes routines for Cholesky factorization.
NL2SOL, a FORTRAN77 library which solves nonlinear least squares problems, and includes routines for Cholesky factorization.
PPPACK, a FORTRAN77 library which carries out piecewise polynomial interpolation and includes routines for Cholesky factorization.
TOEPLITZ_CHOLESKY, a C++ library which computes the Cholesky factorization of a nonnegative definite symmetric Toeplitz matrix.
### Reference:
1. PR Freeman,
Remark AS R44: A Remark on AS 6 and AS7: Triangular decomposition of a symmetric matrix and Inversion of a positive semi-definite symmetric matrix,
Applied Statistics,
Volume 31, Number 3, 1982, pages 336-339.
2. Michael Healy,
Algorithm AS 6: Triangular decomposition of a symmetric matrix,
Applied Statistics,
Volume 17, Number 2, 1968, pages 195-197.
### List of Routines:
• CHOLESKY computes the Cholesky factorization of a PDS matrix.
• SUBCHL computes the Cholesky factorization of a (subset of a) PDS matrix.
• TIMESTAMP prints out the current YMDHMS date as a timestamp.
You can go up one level to the C++ source codes.
Last revised on 14 December 2011.
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https://www.ibpsguide.com/sbi-clerk-mains-reasoning-ability-questions-day-87/
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# SBI Clerk Mains Reasoning (Day-87)
Reasoning Material for SBI Clerk Mains 2020
Dear Aspirants, Our IBPS Guide team is providing new series of Reasoning Questions for SBI Clerk Mains Level so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.
Start Quiz
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Linear row seating arrangement
Directions (1-5):Â Study the following information carefully and answer the questions.
Eight persons P, A, Q, B, R, C, S and D are sitting in a row but not necessarily in the same order. Some of them are facing north and some of them are facing south. D is 40 years old and the persons who are sitting to the left of D are younger to D and the personswho are sitting to the right of D are elder to D. It is also known that
Note: Facing the same direction means, if one is facing north then the other also faces north and vice-versa. Facing the opposite directions means, if one is facing north then the other faces south and vice-versa.
A and S are facing the opposite direction. Two persons are sitting between P and D. The one who is 10 years elder than P sits second to the right of P who sits at one of the ends. Three persons sit between C and B.D faces north. The one who is 54 years old sits second to the left of B. S and C facing the opposite direction. Number of persons sits between B and Q is as same as that of D and R. R faces same direction as the second eldest person. C’s age is five more than the half of D’s age. None of the persons are below 18 years old. The one who is 60 years old sits third to the left of Q. The Sum of ages of Q and R is equal to thrice the age of C and the difference between the ages of R and Q is 15 years. The eldest person is two years elder than the second eldest person. C faces same direction as the eldest person. Not more than two people facing same direction are sitting together.
1) Who is the eldest person in the group?
A) S
B) A
C) B
D) P
E) Q
2) Which of following person sits third to the left of C?
A) The one who is 40 years old
B) The one who is 60 years old
C) The one who is immediate right of B
D) A
E) None of these
3) What is the sum of ages of third youngest and fourth eldest person in the group?
A) 70 Years
B) 85 Years
C) 60 Years
D) 75 Years
E) None of these
4) Who sits immediate right of R?
A) P
B) The one who is second to the right of B
C) A
D) The one whose age is 60
E) Both (C) and (D)
5) Which of the following statement(s) is/are correct?
i) D and R are facing the same direction
ii) S is younger than A
iii)Â Q is 60 years old
A) Only (i) and (iii)
B) Only (i) and (ii)
C) Only (ii)
D) Only (i)
E) All (i), (ii) and (iii)
Data sufficiency
Directions (6-8):Â Each of the questions below consists of a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statements are sufficient to answer the question:
a) If the data in Statement I alone is sufficient to answer the question, while the data in Statement II alone is not sufficient to answer the question.
b) If the data in Statement II alone is sufficient to answer the question, while the data in Statement I alone is not sufficient to answer the question.
c) If the data either in Statement I alone or in Statement II alone aresufficient to answer the question.
d) If the data in both the statements I and II together are not sufficient to answer the question.
e) If the data in both the statements I and II together are necessary to answer the question.
6) Eight boxes namely -P, Q, R, S, T, U, V and W are kept one above another but not necessarily in the same order. Box R is kept third from bottom at a gap of two from box S, which is kept adjacent to box Q and box P is kept at a gap of one from box Q. Which box is kept immediate above box T?
I) Box V is kept at a gap of two from box P and Box W is neither kept at the bottom nor adjacent tobox R. Box V is kept at a gap of two from box U.
II) Box U is kept at a gap of two from box V, which is kept adjacent to box R. Box W is kept adjacent to box P.
A) a
B) b
C) c
D) d
E) e
7) Six persons namely – Raj, Gaurav, Payal, Dev, Riya and Hari are sitting in a circular table facing center but not necessarily inthesame order. Each person likes different pet – Parrot, Horse, Monkey, Cat, Rat and Dog but not necessarily in the same order. Who sits immediate right of the one who likes Rat?
I) Riya sits second to the left of Dev, who sits third to the right of Gaurav. One who likes Dog sits immediate neighbor of Gaurav and sits second to the right of the one who likes Parrot. One who likes Dog sits immediate right of Raj who likes Cat. Hari sits second to the left of the one who likes Monkey, who sits immediate neighbor of Payal.
II) Dev likes Parrot sits second to the left of Hari, who sits second to left of Riya. Payal sits second to the left of Raj, who sits immediate left of the one who likes Dog and Raj likes Cat. The One who likes Horse sits immediate neighbor of Hari. Riya likes monkey.
A) a
B) b
C) c
D) d
E) e
8) Seven persons namely – Riya, Hema, Golu, Zoya, Hari, Ravi and Jiya are sitting in a row but not necessarily in the same order in such a way that some are facing north while others are facing south. Hari faces north. Riya sits third to the left of Hari, either of them sits at end of the row and Ravi sits immediate left of Riya. Ravi sits second to the left of Zoya, who is facing south. How many persons sit facing south?
I) Person sitting immediate neighbor of Zoya sits fourth to the right of Jiya. Person sitting at the end of the row sits facing in opposite direction and Golu sits third from the right end.
II) Person sitting immediate left of Ravi sits fourth to the right of Jiya and Golu sits second to the left of Hari.
A) a
B) b
C) c
D) d
E) e
Direction sense
Directions (9-10):Â Study the following information carefully and answer the questions given below:
If,
‘Q#^S’ means Point Q is 8m north of Point S.
‘Q\$%S’ means Point Q is 10m east of Point S.
‘Q^&S’ means Point Q is 4m west of Point S.
‘[email protected]*S’ means Point Q is 5m south of Point S.
9) If D#^C\$%[email protected]*R, then how far and in which direction is Point R with respect to Point T, if Point T is 3m south of Point D?
A) 5m, West
B) 3m, West
C) 8m, West
D) 10m, West
E) None of these
10) If K^&M\$%[email protected]*H, P\$%H, then how far and in which direction is Point P with respect to Point M?
A) 5m, South
B) 5m, North
C) 7m, North
D) 7m, South
E) None of these
Directions (1-5):
NOTE:Â Â D is 40 years old and the persons who are sitting to the left of D are younger to D and the persons who are sitting to the right of D are elder to D.
• D faces north.
• C’s age is five more than the half of D’s age. So C’s age is 25 years.
• Two persons are sitting between P and D.
• The one who is 10 years elder than P sits second to the right of P who sits at one of the ends.
• None of the persons are below 18 years old.
• Three persons sit between C and B.
• The one who is 54 years old sits second to the left of B.
From the above statement, we have two possible cases
• Number of persons sits between B and Q is as same as that of D and R.
• The one who is 60 years old sits third to the left of Q.
Now the possible cases are,
• The Sum of ages of Q and R is equal to thrice the age of C and the difference between the ages of Q and R is 15 years. This statement is not satisfied by case 2A because we get the age Q as 45yearsor 75 years. So this case is eliminated.
By solving Q+R=75; R-Q=15, we get Q/R=30 years and Q/R=45 years. We can also know the person second to the right of P is 10 years elder than P. Now P’s age is 20 years.
But in case 2B, R sits left of D it doesn’t satisfied the given condition(D is 40 years old and the persons who are sitting to the left of D are younger to D and the persons who are sitting to the right of D is elder to D). So this case is eliminated.
Now Case 1 becomes
• The eldest person is two years elder than the second eldest person. For this statement, B is either eldest or second eldest person in the group. If B is the eldest person then the age of second eldest person is 58 years. If B is second eldest then the age of eldest person is 62 years.
• C faces same direction as the eldest person.
• S and C facing the opposite direction.
• A and S are facing the opposite direction.
• R faces same direction as the second eldest person.
• Not more than two people facing same direction are sitting together. For this statement expect case 1A all the cases are eliminated. Now the final arrangement becomes
We have:
• Box R is kept third from bottom at a gap of two from box S, which is kept adjacent to box Q that means box S is kept third from thetop and we have two different cases for box Q, in case (1) box Q is kept immediate above box S and in case (2) box Q is kept immediate below box S.
• Box P is kept at a gap of one from box Q, which means in case (1) P is kept immediately below S and in case (2) box P is kept immediately above box S.
Based on the above given information we get:
From I:
We have:
• Box W is neither kept at the bottom nor adjacent toboxR, which means box W is kept at top. Box V is kept at a gap of two from box P, which means in case (1) BoxV is kept immediate below box R and in case (2) box V is kept immediate above box R.
• Box V is kept at a gap of two from box U, which means case (1) is not valid and in case (2) box U is kept at the bottom.
Based on the above given information, we get final arrangement as follow:
Thus, box R is kept immediately above box T.
Hence, statement I alone is sufficient.
From II:
We have:
• Box W is kept adjacent tobox P, which means in case (1) box W is kept immediately below box P and in case (2) the box W is kept immediately above box P.
• Box U is kept at a gap of two from box V, which is keptadjacent tobox R, that means case (1) is not valid and in case (2) box V is kept immediate above box R.
Based on theabove given information, we get final arrangement as follow:
Thus, box R is kept immediately above box T.
Hence, statement II alone is sufficient.
From I:
We have:
• Riya sits second to the left of Dev, who sits third to the right of Gaurav and the One who likes Dog sits immediate neighbor of Gaurav.
• One who likes Dog sits immediate right of Raj who likes Cat, that means one who likes Dog sits left of Gaurav. One who likes Dog sits second to the right of the one who likes Parrot, that means Dev likes Parrot.
• Hari sits second to the left of the one who likes Monkey, who sits immediate neighbor of Payal, that means Hari likes Dog.
Based on the above given information we get:
Since, we don’t know who likes Rat, thus result can’t be determined.
Hence, statement I is not sufficient.
From II:
We have:
• Dev likes Parrot sits second to the left of Hari, who sits second to left of Riya. Payal sits second to the left of Raj, who sits immediate left of the one who likes Dog and Raj likes Cat.
• One who likes Monkey sits second to the right of the one who likes Dog, which means Riya likes Monkey. One who likes Horse sit immediate neighbor of Hari, that means Gaurav likes Horse. Riya likes monkey.
Based on the above given information we get:
Thus, Dev sits immediate right of one who likes Rat.
Hence, statement II is alone sufficient.
We have:
• Hari faces north
• Riya sits third to the left of Hari, either of them sits at end of the row, which means we get two possibilities, in case (1) Hari sits facing north at right end and in case (2) Riya sits at extreme left end.
• Ravi sits immediate left of Riya and Ravi sits second to left of Zoya, who sits facing south, that means case (2) is not valid, in case (1a) Riya sits facing south and in case (1b) Riya sits facing north.
Based on the above given information we get:
From I:
We have:
• Person sitting immediate neighbor of Zoya sits fourth to the right of Jiya, which means Jiya must sit facing south.
• Person sitting at the end of the row sits facing in opposite direction. Golu sits third from the right end, which means case (1a) is not valid.
Based on the above given information, we havethe following result:
Case1a Golu sits third from the right end, this condition is not satisfied so rejected.
Since, direction facing of Hema and Ravi is not known, thus we can’t determine the required result.
Hence, statement I is not sufficient.
From II:
We have:
• Person sitting immediate left of Ravi sits fourth to theright of Jiya, which means in case (1a) Ravi sits facing south and in case (1b) Ravi sits facing north.
• Golu sits second tothe left of Hari, which means case (1a) is not valid.
Based on the above given information, we getthe following result:
Since, direction of Hema and Golu is not known, thus result can’t be determined.
Hence, statement II is not sufficient.
From I and II:
Based on the above given statements we get:
Since, direction of Hema is not known, thus result can’t be determined.
Hence, statement I and II together are not sufficient.
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1. ## Snakes and ladders game, URGENT help needed
Can anybody PLEASE be a complete life saver and code this for me? I'm desperate. This is the task description.
A manufacturer of games has decided to design a new snakes and ladders board consisting of 100 squares. The manufacturer has asked the systems analyst to design a simulation of the game. The systems analyst has been told that:
• There must be exactly 10 snakes and 10 ladders.
• The number of squares that a snake sends a player back must lie between 10 and 30.
• The number of squares that a ladder sends a player forward must lie between 10 and 30.
The systems analyst has decided to represent the board by using a one-dimensional array with 100 cells. Initially each cell is to be set to zero.
When the game starts the user is asked to input the positions of the snakes. This is to be done by entering, for each snake, the square on which its head is to be and the number of squares a player must go back when landing on a snake’s head. For each snake, the number of squares to go back is then to be stored in the cell representing the head of the snake, as a negative integer.
This is repeated for the ladders by inputting the squares for the feet of the ladders and the number of squares a player must go forward when landing on the foot of a ladder. For each ladder, the number of squares to go forward should be stored as a positive integer in the cell representing the foot of the ladder.
Part of the array may look similar to the extract below. This indicates that there is a snake’s head in cell 51 and its tail is in cell 41. The bottom of a ladder is in cell 40 and its top is in cell 52. The remaining squares do not contain any snakes or ladders.
You should annotate all your code and use meaningful names throughout.
(a)
Create annotated code that initialises the board ready for the user input. [2]
(b)
Create an interface that allows the user to input the data as described above. Your code should validate the input and store the data as described above. You need only check for values that satisfy the criteria given above. [5]
The rules of this game of Snakes and Ladders are:
• On each turn, a player moves forward the number of squares shown when a 6-sided die is thrown.
• If a player lands on the head of a snake, the player must move to the tail of the snake.
• If a player lands on the foot of a ladder, the player must move to the top of the ladder.
• If a move would result in the player going off the board, no move is made.
You need only simulate the moves of a single player.
(c)
Create annotated code that simulates a game of snakes and ladders. [8]
2. ## Re: Snakes and ladders game, URGENT help needed
You said this thread was resolved.
What happened?
3. ## Re: Snakes and ladders game, URGENT help needed
i was trying to delete the previous thread and i didnt know how to
4. ## Re: Snakes and ladders game, URGENT help needed
Probably not much has changed. As easy as it is, most people will not likely do your assignment for you. You should be aware that it is cheating.
5. ## Re: Snakes and ladders game, URGENT help needed
Have you tried to outline the tasks of how you want to implement the game.
Depending on the complexity, I do like to write out the logic and order in simple terms and steps, and then it should be easier to work on turning each step into code.
For instance, the directions tell you how to implement the core structures, how many snakes, and how many ladders, and in what order to do the setup. So, write those down and decide how you want to implement them. This should be easy to do with a simple console based program.
With a bit more complication, you could do it with a Graphical User Interface (GUI). It could be basic controls, such as a list box with 100 items in it initially set to 0 representing your array. Perhaps have two columns, so you can display the square number (1 to 100) and the value stored at each location (initially all zeros) and then 10 negative number the user specifies for the snakes (at the square index 1 to 100, and the value -10 to -30), and then 10 positive numbers for the ladders at the offset (1 to 100) and value range is 10 to 30.
You just need to check the square they specify (1 to 100) is 0 before changing it. And also verify that the value they specify is in the appropriate range.
It looks like you don't even have to check if the value of the range the user specifies is within the range of the board during input, so that saves you some work.
You just have to ignore the snake or ladder when you move if the square you would move to is less than the first square or greater than the last square, i.e. a ladder can't take you off the board to win, and a snake can take you off the board at the bottom. You just don't move in those conditions (ignore the snake and ladder). It is a later exercise to limit the snake and ladder sizes to be within the board, for example since the size is a minimum of 10, you can't have a snake within the first 9 positions of the game, and you can't have a ladder in the last 9 positions of the game. Likewise you can't have a -30 size snake in the first 29 positions of the board or a 30 size ladder in the last 29.
The main question I have it how the game ends. The specification doesn't mention whether the game is over when your roll takes you to the last square or beyond. If you had a snake head on the last square, then you definitely would have to roll past that to win the game. So, even if there was not a snake on the last square and you landed on it, I presume you haven't won the game until your next roll takes you off the board.
A ladder can't take you off the board, but you do go off the board on a roll. That is the only thing that makes sense to me, for the conditions given.
6. ## Re: Snakes and ladders game, URGENT help needed
The original question was asked a month ago. Has nothing changed?
7. ## Re: Snakes and ladders game, URGENT help needed
What have you been doing for a month?
Have you even implemented the thing I spent 2 hours writing for you? Prove it to me.
If you haven't implemented it, or you didn't try, what exactly makes you think I'm going to risk another 2 hours on you?
I'm not doing squat until you read my post #7 and finish it by addressing the bullet list at the end. I will answer questions about #7, but nothing else, until I'm satisfied you've made my previous work worth it. If someone else wants to help you out, that's their prerogative, but I feel very disrespected and used.
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His first musical stirrings were at the age of eight, when his parents gave him a secondhand radiogram which included a few records left by the previous owner. Among them were Drummin' Man by drumming legend Gene Krupa, and, in Davies's own words, "it hit like a thunderbolt". "I must have played it 2,000 times," he said. "That was it."[5] A friend of the family made Rick a makeshift drum kit out of a biscuit tin, and at the age of 12 he joined the British Railways Staff Association Brass and Silver Jubilee Band as a snare drummer.[6] In an interview in 2002 he said: "As a kid, I used to hear the drums marching along the street in England, in my home town, when there was some kind of parade, and it was the most fantastic sound to me. Then, eventually, I got some drums and I took lessons. I was serious about it... I figured if I could do that – I mean a real drummer, read music and play with big bands, rock bands, classical, Latin, and know what I was going to do – I would be in demand and my life was set... Eventually, I started fiddling with the keyboards, and that seemed to go over better than my drumming, for some reason. So you've gotta go with what people react to." He never had lessons for keyboards, but, according to Betty Davies, "taught himself most of what he knows about music".[5]
41) Similar calculations made from the Cape of Good Hope, South Africa to Melbourne, Australia at an average latitude of 35.5 degrees South, have given an approximate figure of over 25,000 miles, which is again equal to or greater than the Earth’s supposed greatest circumference at the equator. Calculations from Sydney, Australia to Wellington, New Zealand at an average of 37.5 degrees South have given an approximate circumference of 25,500 miles, greater still! According to the ball-Earth theory, the circumference of the Earth at 37.5 degrees Southern latitude should be only 19,757 statute miles, almost six thousand miles less than such practical measurements.
# “1 In the beginning God created the heaven and the earth. And the earth was without form, and void; and darkness was upon the face of the deep. 2 And the Spirit of God moved upon the face of the waters. 3 And God said, Let there be light: and there was light. 4 And God saw the light, that it was good: and God divided the light from the darkness. 5 And God called the light Day, and the darkness he called Night. And the evening and the morning were the first day.
Roger Hodgson left the band in 1983 and the Supertramp that most people know has not existed for over 30 years. The music of Supertramp totally changed after Roger left to match Rick’s musical influence, which is more jazz and blues. Rick Davies and his wife have trademarked the name Supertramp, yet the band only consists of one original member – Rick.
Astronomers are in the habit of considering two points on the Earth's surface, without, it seems, any limit as to the distance that lies between them, as being on a level, and the intervening section, even though it be an ocean, as a vast "hill"-of water!" The Atlantic ocean, in taking this view of the matter, would form a "hill of water" more than a hundred miles high! The idea is simply monstrous, and could only be entertained by scientists whose whole business is made up of materials of the same description: and it certainly requires no argument to deduce, from such "science" as this, a satisfactory proof that the Earth is not a globe.
The only explanation which has been given of this phenomenon is the refraction caused by the earth’s atmosphere. This, at first sight, is a plausible and fairly satisfactory solution; but on carefully examining the subject, it is found to be utterly inadequate; and those who have recourse to it cannot be aware that the refraction of an object and that of a shadow are in opposite directions. An object by refraction is bent upwards; but the shadow of any object is bent downwards, as will be seen by the following very simple experiment. Take a plain white shallow basin, and place it ten or twelve inches from a light in such a position that the shadow of the edge of the basin touches the centre of the bottom. Hold a rod vertically over and on the edge of the shadow, to denote its true position. Now let water be gradually poured into the basin, and the shadow will be seen to recede or shorten inwards and downwards; but if a rod or a spoon is allowed to rest, with its upper end towards the light, and the lower end in the bottom of the vessel, it will be seen, as the water is poured in, to bend upwards–thus proving that if refraction operated at all, it would do so by elevating the moon above its true position, and throwing the earth’s shadow downwards, or directly away from the moon’s surface. Hence it is clear that a lunar eclipse by a shadow of the earth is an utter impossibility.
2. Another related thing I don’t understand: if the sun and moon are always above the disk of the Earth, why can’t everyone in the world see them at all times? Surely they should always be visible, at least at a low angle. I can’t draw myself any diagram where they are not always visible, but we see that that doesn’t happen. I can’t see how night time happens. Help!
As the mariners' compass points north and south at one time, and as the North, to which it is attracted is that part of the Earth situated where the North Star is in the zenith, it follows that there is no south "point" or "pole" but that, while the centre is North, a vast circumference must be South in its whole extent. This is a proof that the Earth is not a globe.
20.) The common sense of man tells him – if nothing else told him – that there is an "up" and a "down" in -nature, even as regards the heavens and the earth; but the theory of modern astronomers necessitates the conclusion that there is not: therefore, 'the theory of the astronomers is opposed to common sense – yes, and to inspiration – and this is a common sense proof that the Earth is not a globe.
23) Ball-believers often claim “gravity” magically and inexplicably drags the entire lower-atmosphere of the Earth in perfect synchronization up to some undetermined height where this progressively faster spinning atmosphere gives way to the non-spinning, non-gravitized, non-atmosphere of infinite vacuum space. Such non-sensical theories are debunked, however, by rain, fireworks, birds, bugs, clouds, smoke, planes and projectiles all of which would behave very differently if both the ball-Earth and its atmosphere were constantly spinning Eastwards at 1000mph.
173) NASA has several alleged photographs of the ball-Earth which show several exact duplicate cloud patterns! The likelihood of having two or three clouds of the exact same shape in the same picture is as likely as finding two or three people with exactly the same fingerprints. In fact it is solid proof that the clouds were copied and pasted in a computer program and that such pictures showing a ball-shaped Earth are fakes.
In January 2016, Tila Tequila posted a series of tweets claiming to believe the Earth is flat. The following month, on February 16th, 2016, NBA super star Kyrie Irving expressed his belief that the Earth is flat on the podcast Road Trippin (shown below, left). The next year in 2017, famed Jiu-Jitsu instructor and former UFC analyst, Eddie Bravo came forward with his belief in a flat Earth numerous times, most notably on The Joe Rogan Experience podcast (shown below, right).
If we could - after our minds had once been opened to the light of Truth - conceive of a globular body on the surface of which human beings could exist, the power - no matter by what name it be called - that would hold them on would, then, necessarily, have to be so constraining and cogent that they could not live; the waters of the oceans would have to be as a solid mass, for motion would be impossible. But we not only exist, but live and move; and the water of the ocean skips and dances like a thing of life and beauty! This is a proof that the Earth is not a globe.
31.) If the Earth were a globe, it would certainly have to be as large as it is said to be – twenty-five thousand miles in circumference. Now, the thing which I have called a "proof" of the Earth's roundness, and which is presented to children at school, is, that if we stand on the seashore we may see the ships, as they approach us, absolutely "coming up," and that, as we are able to see the highest parts of these ships first, it is because the lower parts are "behind the earth's curve."Now since if this were the case – that is, if the lower parts of these ships were behind a "hill of water" – the size of the Earth, indicated by such a curve as this, would be so small that it would only be big enough to hold the people of a parish, if they could get all round it, instead of the nations of the world, it follows that the idea is preposterous; that the appearance is due to another and to some reasonable cause; and that, instead of being a proof of the globular form of the Earth, it is a proof that at Earth is not a globe.
Always keep in mind that tangible science is a flat earth, a ball dangling in the sky is science fiction or Hollywood. I believe the firmament to be an ice dome, described to be glass. The hotter we make the "planet" the more that dome melts, I believe the salt beds are salt dropped out of the sky as water evaporated to become air. Gods rendition is tangible science.
The songs Roger Hodgson sings that are often referred to as “Supertramp songs” are songs that Roger wrote and composed alone - that is how he creates music. He will often hear the musical composition first, then the lyrics come. Sometimes he will hear the whole song complete inside himself before he puts it down on paper and records it. When he was with Supertramp, he would make a demo of the song and bring it to the band for them to learn their parts. So Supertramp as a band never wrote and composed together, there were only two songwriters for the band. The songs Roger sings lead vocals on are the ones he composed the music and lyrics for, and the songs Rick sings lead vocals on are the ones that he wrote.
Many weather patterns are actually created by the land itself. For example - rain shadow. Rain shadow is where somewhere on the eastern side of a mountain range (because weather/clouds travel from the west generally) is deprived of rain not just once, but nearly all the time. This is because the clouds are forced upwards by the mountains blocking their path and become cooler and condense, meaning water droplets form and it rains on/before the mountains thereby not raining on the leeward side.
By 1959, his attention had been captured by rock 'n' roll, and he joined a band called Vince and the Vigilantes.[6] In 1962, while studying in the art department at Swindon College, he formed his own band, called Rick's Blues, and was now playing a Hohner electric piano instead of drums. The band included Gilbert O'Sullivan on drums for a time; he later was the best man at Davies's wedding. In a March 1972 interview, O'Sullivan said "Rick had originally taught me how to play the drums and piano – in fact, he taught me everything about music."[6] When his father became ill, Davies disbanded Rick's Blues, left college, and took a job as a welder at Square D,[6] a firm making industrial control products and systems, which had a factory on the Cheney Manor Trading Estate in Swindon. Any hopes of an artistic career were temporarily put on ice.
"Is water level, or is it not?" was a question once asked of an astronomer. "Practically, yes; theoretically, no," was the reply. Now, when theory does not harmonize with practice, the best thing to do is to drop the theory. (It is getting too late, now to say "So much the worse for the factsI") To drop the theory which supposes a curved surface to standing water is to acknowledge the facts which form the basis of Zetetic philosophy. And since this will have to be done sooner or later, - it is a proof that the Earth is not a globe.
45.) The Astronomer Royal, of England, George B. Airy, in his celebrated work on Astronomy, the "Ipswich Lectures," says – "Jupiter is a large planet that turns on his axis, and why do not we turn?" Of course, the common sense reply is: Because the Earth is not a planet! When, therefore, an astronomer royal puts words into our mouth wherewith we may overthrow the supposed planetary nature of the Earth, we have not far to go to pick up a proof that Earth is not a globe.
If we stand on the sands of the sea-shore and watch a ship approach us, we shall find that she will apparently "rise" - to the extent, of her own height, nothing more. If we stand upon an eminence, the same law operates still; and it is but the law of perspective, which causes objects, as they approach us, to appear to increase in size until we see them, close to us, the size they are in fact. That there is no other "rise" than the one spoken of is plain from the fact that, no matter how high we ascend above the level of the sea, the horizon rises on and still on as we rise, so that it is always on a level with the eye, though it be two-hundred miles away, as seen by Mr. J. Glaisher, of England, from Mr. Coxwell's balloon. So that a ship five miles away may be imagined to be "coming up" the imaginary downward curve of the Earth's surface, but if we merely ascend a hill such as Federal Hill, Baltimore, we may see twenty-!five miles away, on a level with the eye - that is, twenty miles level distance beyond the ship that we vainly imagined to be " rounding the curve," and "coming up!" This is a plain proof that the Earth is not a globe.
109) There are no fixed “East” or “West” points just as there is no fixed “South.” The North central Pole is the only proven fixed point on our flat Earth, with South being all straight lines outwards from the pole, East and West being concentric circles at constant right angles 90 degrees from the pole. A westerly circumnavigation of Earth is thus going around with Polaris continually on your right, while an easterly circumnavigation is going around with Polaris always at your left.
Astronomers have made experiments with pendulums which have been suspended from the interior of high buildings, and have exulted over the idea of being able to prove the rotation of the Earth on its "axis," by the varying direction taken by the pendulum over a prepared table underneath - asserting that the table moved round under the pendulum, instead of the pendulum shifting and oscillating in different directions over the table! But, since it has been found that, as often as not, the pendulum went round the wrong way for the "rotation" theory, chagrin has taken the place of exultation, and we have a proof of the failure of astronomers in their efforts to substantiate their theory, and, therefore, a proof that Earth is not a globe.
When the Sun crosses the equator, in March, and begins to circle round the heavens in north latitude, the inhabitants of high northern latitudes see him slimming round their horizon and forming the break of their long day, in a horizontal course, not disappearing again for six months, as he rises higher and higher in the heavens whilst he makes his twenty-four hour circle until June, when he begins to descend and goes on until he disappears beyond the horizon in September. Thus, in the northern regions, they have that which the traveler calls the "midnight Sun," as he sees that luminary at a time when, in his more southern latitude, it is always midnight. If, for one-half the year, we may see for ourselves the Sun making horizontal circles round the heavens, it is presumptive evidence that, for the other half-year, he is doing the same, although beyond the boundary of our vision. This, being a proof that Earth is a plane, is, therefore, a proof that the Earth is not a globe.
where P and T are the pressure and temperature of air, and P0 and T0 are the standard values of one atmosphere of pressure and 300K. Since with mirages there is no appreciable height difference, pressure differences are negligible, and so the temperature difference dominates differences in the index of refraction. When light travels from one medium to another, the path of the light is refracted, or bent. This is what causes the “bent stick” appearance of a long object partially inserted in water, such as a pole placed into the water of a pool. This behavior is described by Snell’s law:
66.) It is often said that the predictions of eclipses prove astronomers to be right in their theories. But it is not seen that this proves too much. It is well known that Ptolemy predicted eclipses for six-hundred years, on the basis of a plane Earth, with as much accuracy as they are predicted by modern observers. If, then, the predictions prove the truth of the particular theories current at the time, they just as well prove one side of the question as the other, and enable us to lay claim to a proof that the Earth is not a globe.
Mr. J. R. Young, in his work on Navigation, says. "Although the path of the ship is on a spherical surface, yet we may represent the length of the path by, a straight line on a plane surface." (And plane sailing is the rule.) Now, since it is altogether impossible to "represent" a curved line by a straight one, and absurd to make the attempt, it follows that a straight line represents a straight line and not a curved one. And, Since it is the surface of the waters of the ocean that is being considered by Mr. Young, it follows that this surface is a straight surface, and we are indebted to Mr. Young, a professor of navigation, for a proof that the Earth is not a globe.
55.) The Newtonian theory of astronomy requires that the Moon "borrow" her light from the Sun. Now, since the Sun's rays are hot and the Moon's light sends with it no heat at all, it follows that the Sun and Moon are "two great lights," as we somewhere read; that the Newtonian theory is a mistake; and that, therefore, we have a proof that the Earth is not a globe.
If the Earth were a globe, it would, if we take Valentia to be the place of departure, curvate downwards, in the 1665 miles across the Atlantic to Newfoundland, according to the astronomers' own tables, more than three hundred miles; but, as the surface of the Atlantic does not do so - the fact of its levelness having been clearly demonstrated by Telegraph Cable surveyors, - it follows that we have a grand proof that Earth is not a globe.
177) In the documentary “A Funny Thing Happened on the Way to the Moon,” you can watch official leaked NASA footage showing Apollo 11 astronauts Buzz Aldrin, Neil Armstrong and Michael Collins, for almost an hour, using transparencies and camera-tricks to fake shots of a round Earth! They communicate over audio with control in Houston about how to accurately stage the shot, and someone keeps prompting them on how to effectively manipulate the camera to achieve the desired effect. First, they blacked out all the windows except for a downward facing circular one, which they aimed the camera towards from several feet away. This created the illusion of a ball-shaped Earth surrounded by the blackness of space, when in fact it was simply a round window in their dark cabin. Neil Armstrong claimed at this point to be 130,000 miles from Earth, half-way to the Moon, but when camera-tricks were finished the viewer could see for themselves the astro-nots were not more than a couple dozen miles above the Earth’s surface, likely flying in a high-altitude plane!
#### 195) Astronomers say the magical magnetism of gravity is what keeps all the oceans of the world stuck to the ball-Earth. They claim that because the Earth is so massive, by virtue of this mass it creates a magic force able to hold people, oceans and atmosphere tightly clung to the underside of the spinning ball. Unfortunately, however, they cannot provide any practical example of this on a scale smaller than the planetary. A spinning wet tennis ball, for instance, has the exact opposite effect of the supposed ball-Earth! Any water poured over it simply falls off the sides, and giving it a spin results in water flying off 360 degrees like a dog shaking after a bath. Astronomers concede the wet tennis ball example displays the opposite effect of their supposed ball-Earth, but claim that at some unknown mass, the magic adhesive properties of gravity suddenly kick in allowing the spinning wet tennis ball-Earth to keep every drop of “gravitized” water stuck to the surface. When such an unproven theory goes against all experiments, experience and common sense, it is high time to drop the theory.
53.) Every year the Sun is as long south of the equator as he is north; and if the Earth were not "stretched out" as it is, in fact, but turned under, as the Newtonian theory suggests it would certainly get as intensive a share of the Sun's rays south as north; but the Southern region being, in consequence of the fact stated, – far more extensive than the region North, the Sun, having to complete his journey round every twenty-four hours, travels quicker as he goes further south, from September to December, and his influence has less time in which to accumulate at any given point. Since, then the facts could not be as they are if the Earth were a globe, it is a proof that the Earth is not a globe.
1.) The aeronaut can see for himself that Earth is a Plane. The appearance presented to him, even at the highest elevation he has ever attained, is that of a concave surface – this being exactly what is to be expected of a surface that is truly level, since it is the nature of level surfaces to appear to rise to a level with the eye of the observer. This is ocular demonstration and proof that Earth is not a globe.
Despite overwhelming scientific proof, flat-earthers are out there spreading their not so level ideas. Some have even gone so far as to create an official organization known as the flat earth Society. And, of course, what society can call itself a veritable one without a few conspiracy theories? With the flat earth theory henchmen, NASA is the one to blame for our wrongful belief that the planet earth, is indeed round.
100) If Earth were a ball, the Southern Cross and other Southern constellations would all be visible at the same time from every longitude on the same latitude as is the case in the North with Polaris and its surrounding constellations. Ursa Major/Minor and many others can be seen from every Northern meridian simultaneously whereas in the South, constellations like the Southern Cross cannot. This proves the Southern hemisphere is not “turned under” as in the ball-Earth model, but simply stretching further outwards away from the Northern center-point as in the flat Earth model.
120) The etymology of the word “planet” actually comes from late Old English planete, from Old French planete (Modern French planète), from Latin planeta, from Greek planetes, from (asteres) planetai “wandering (stars),” from planasthai “to wander,” of unknown origin, possibly from PIE *pele “flat, to spread” or notion of “spread out.” And Plane (n) “flat surface,” c. 1600, from Latin planum “flat surface, plane, level, plain,” planus “flat, level, even, plain, clear.” They just added a “t” to our Earth plane and everyone bought it.
12.) As we have seen that there is, really no south point (or pole) but an infinity of points forming, together, a vast circumference — the boundary of the known world, with its battlements of icebergs which bid defiance to man's onward course, in a southerly direction – so there can be no east or west "points,' just as there is no "yesterday," and no "tomorrow." In fact, as there is one point that is fixed (the North), it is impossible for any other point to be fixed likewise. East and west are, therefore, merely directions at right angles with a north and south line: and as the south point of the compass shifts round to all parts of the circular boundary, (as it may be carried round the central North) so the directions east and west, crossing this line, continued to form a circle at any latitude. A westerly circumnavigation, is going around with the North Star continually on the right hand, and an easterly circumnavigation is performed only when the reverse condition of things is maintained, the North Star being on the left hand as the journey is made. These facts, taken together, form a beautiful proof that the Earth is not a globe.
Mr. Proctor says.- "The Sun is so far off that even moving from one side of the Earth to the other does not cause him to be seen in a different direction - at least the difference is too small to be measured." Now, since we know that north of the equator, say 45 degrees, we see the Sun at mid-day to the south, and that at the same distance south of the equator we see the Sun at mid-day to the north, our very shadows on the round cry aloud against the delusion of the day and give us a proof that Earth is not a globe.
63.) It is a fact not so well known as it ought to be that when a ship, in sailing away from us, has reached the point at which her hull is lost to our unaided vision, a good telescope will restore to our view this portion of the vessel. Now, since telescopes are not made to enable people to see through a "hill of water," it is clear that the hulls of ships are not behind a hill of water when they can be seen through a telescope though lost to our unaided vision. This is a proof that Earth is not a globe.
Both Davies and Hodgson talked of a reunion a couple of times, however, this would never come to pass. The first hint of a reunion came in 1993 when Davies and Hodgson reunited for an A & M dinner honoring Jerry Moss, co-founder of A & M Records. This dinner resulted in writing and demoing new songs, but it never went anywhere due to disagreements over management. Another hint of a reunion came in 2010 when Roger Hodgson approached Rick Davies about a fortieth anniversary of their very first album Supertramp (rogerhodgson.com). Rick Davies declined the invitation and any chance of Supertramp reuniting was squashed.
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# How fast do you have to heat air to create a shock wave?
## Main Question or Discussion Point
Suppose you have a 1cm diameter heat source suspended 2m above the ground. How many watts of heat energy does it need to expend to create a shock wave?
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It'd depend on the function of thermal transfer between the heat source and the air. I suspect this becomes very non-linear if you get to the rate of heat transfer you're talking about.
I'd also suspect that in the real world, any known material 1cm would not withstand the thermal resistance in that scenario, and, itself, ablate and explode.
Therefore, as it is a result likely only possible by experimentation, and as there are no materials to perform that experiment, I'd hazard to say the question is moot.
Ok, well since we're not imaginative enough to conceive of an object or material that can release that amount of heat without exploding then lets just consider explosions:
Do all explosions create shockwaves?
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Yes. Why would you imagine an explosion not causing a shock wave? Are you thinking of 'super-sonic' shock waves?
If an explosive causes a supersonic shockwave, it is called a 'high explosive'. If the shockwave is slower than local sound, it generates a wave more like a solition, though I am unclear the difference between a solition and a sub-sonic shock wave.
I did not know exactly what a shock wave is so I checked here:
http://en.wikipedia.org/wiki/Shock_waves
I still don't know, but it's interesting reading.....
it says: "...A shock wave ....is a type of propagating disturbance. Like an ordinary wave, it carries energy and can propagate through a medium (solid, liquid, gas or plasma) or in some cases in the absence of a material medium, through a field such as the electromagnetic field.
Shock waves are characterized by an abrupt, nearly discontinuous change in the characteristics of the medium.[1] Across a shock there is always an extremely rapid rise in pressure, temperature and density of the flow..."
These two statements, for example, appear inconsistent.
Can we get a shock wave in outer space...in a vacuum...sure, see the latter part of the article......how does that fit (or not) in these descriptions??
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# The European Union announced that cod and mackerel are the
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The European Union announced that cod and mackerel are the only fish that exceeds their new requirements for dioxin level and that they allow fishermen to catch.
a. exceeds their new requirements for dioxin level and that they allow
b. exceed its new requirements for dioxin level and that they allow
c. exceeds its new requirements for dioxin level and that it allows
d. exceed its new requirements for dioxin level and that it allows
e. exceed their new requirements for dioxin level and that they allow
I think this one is debatable. Try it.
[Reveal] Spoiler: OA
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that is inserted to throw you off. not many people know that fish(singular) has two types of plural fish(plural) and fishes.
eg:- i have other fish to fry
so here it is plural. ideally it should be fishes as "fish(plural)" is used to denote many members of one species and "fishes" to denote many number of several species
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D.
simple case of singular and plural
european union should be considered singular(collectively)---similar to united states
eg:- US is the only country to...
therefore answer should have "it" and end with "it allows"
eliminate A,E and B.
cod and mackerel , plural---so exceed
eg:- he excels, they excel
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18 Jan 2011, 16:11
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excuse me, but in this case should be better to see the verb ARE to understand if fish are plural or singular, so after we can see the verb exceed and make the split....in fact the correct answer is D with plural verb.
exceed (parallel with are) its (UE) new requirements for dioxin level and that it allows (UE)
If I wrong, please say me......thanks
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18 Jan 2011, 20:28
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TomB i think u made a typo C says "exceeds" not "exceed". did u mean D? because fish(plural) exceed is the correct usage.
ex:- boys(plural) exceed , boy(singular) exceeds
carcass there are several methods to go about doing this Q so choose one that fits u comfortably.
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18 Jan 2011, 01:22
vinzycoolfire wrote:
D.
simple case of singular and plural
european union should be considered singular(collectively)---similar to united states
eg:- US is the only country to...
therefore answer should have "it" and end with "it allows"
eliminate A,E and B.
cod and mackerel , plural---so exceed
eg:- he excels, they excel
What about the "fish" ? is it singular?
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18 Jan 2011, 11:55
To overcome the hurdle whether fish is singular or plural, focus on the verb "exceed" it refers to fish -"cod and mackerel" so in this sentence fish is plural. so Option C is the answer.
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18 Jan 2011, 12:44
TomB wrote:
To overcome the hurdle whether fish is singular or plural, focus on the verb "exceed" it refers to fish -"cod and mackerel" so in this sentence fish is plural. so Option C is the answer.
thank you guys, You are awesome!
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19 Jan 2011, 07:28
sorry for the typo ,it is option D
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01 Mar 2011, 02:04
Just saw the dictionary - fish = plural.
Just read vinzycoolfire explanation ! This one is brutal. yes the OA has to be D.
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01 Mar 2011, 02:20
SC is made to look like "fish" is at issue when it is not.
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01 Mar 2011, 03:47
exceed its new requirements for dioxin level and that IT allows [is this it refers fish]
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01 Mar 2011, 05:14
The second "it" cannot be for fish. it belies the common sense that fish will "welcome" anyone to catch it.
From the explanations above both the "it" and "its" are meant for the org called the E U.
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04 Mar 2011, 09:33
D. Fish - plural, Union - singular
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04 Mar 2011, 19:29
nice question..I went for C but I know why it is "D"
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Re: The European Union announced that cod and mackerel are the [#permalink]
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Re: The European Union announced that cod and mackerel are the [#permalink]
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26 Sep 2014, 02:12
My 2 cents.
The European Union announced that
cod and mackerel are the only fish that
exceeds their new requirements for dioxin level and that
they allow
fishermen to catch.
1: Who exceeded the new requirements?
>>Cod and Mackerel; It would need plural verb so exceed not exceeds.
2: The European Union is sing.
>>"it" is req. instead of "their" to refer EU.
a. exceeds their new requirements for dioxin level and that they allow
b. exceed its new requirements for dioxin level and that they allow
c. exceeds its new requirements for dioxin level and that it allows
d. exceed its new requirements for dioxin level and that it allows
e. exceed their new requirements for dioxin level and that they allow
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Re: The European Union announced that cod and mackerel are the [#permalink]
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26 Sep 2014, 06:52
PTK wrote:
The European Union announced that cod and mackerel are the only fish that exceeds their new requirements for dioxin level and that they allow fishermen to catch.
a. exceeds their new requirements for dioxin level and that they allow
b. exceed its new requirements for dioxin level and that they allow
c. exceeds its new requirements for dioxin level and that it allows
d. exceed its new requirements for dioxin level and that it allows
e. exceed their new requirements for dioxin level and that they allow
I think this one is debatable. Try it.
Just clear one thing !!!
If you say plural exceed to be used in this case i.e. in D then in D why after end , it allows (Singular form) has been used, there is discrepancy
in D,
C looks more sustainable in the case at issue.
Regards
LS
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Re: The European Union announced that cod and mackerel are the [#permalink]
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31 Dec 2015, 11:15
In D, ‘exceed’ is used because ‘cod and mackerel’ is a plural word; 'its' is used because ‘its’ is referring to the EU, which is singular.
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Re: The European Union announced that cod and mackerel are the [#permalink] 31 Dec 2015, 11:15
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# The European Union announced that cod and mackerel are the
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課程資訊
Introduction to Probability Theory
110-2
MATH2502
201 49740
4.0
The main objective of the course is to provide students a mathematical treatment of the fundamental concepts and techniques of elementary probability theory. We will take a non-measure theoretical approach in this course so that it is suitable for students who possess only the knowledge of elementary calculus. The goal is to present not only the mathematics of probability theory, but also numerous applications of the subject.
Content:
Axioms of probability, conditional probability, independence, discrete and continuous random variables, jointly distributed random variables, expectation, moment generating functions, limit theorems.
If time allows, we will also talk about random walks, Markov chains, Poisson processes, and possibly their applications.
Students are able to understand the basics of elementary probability theory.
First year calculus and basic mathematical analysis.
Office Hours
Below are some reference books which are of suitable level.
1. Sheldon M. Ross, Introduction to Probability Models, Academic Press.
2. Geoffrey Grimmett and Dominic Welsh, Probability: An Introduction, 2nd Edition
3. Charles Grinstead and Laurie Snell, Introduction to Probability. See https://math.dartmouth.edu/~prob/prob/prob.pdf
4. David F. Anderson, Timo Seppäläinen and Benedek Valkó. Introduction to Probability, Cambridge University Press, 2018.
Textbook: Sheldon M. Ross, A First Course in Probability, 10th Edition (Global Edition), Pearson.
We will NOT follow the textbook closely. In particular, the materials and notations can be different from the textbook.
(僅供參考)
No. 項目 百分比 說明 1. Homework 20% 2. Midterm exam 40% Midterm will be on 3/31. 3. Final exam 40% The final exam is cumulative. It will be on 6/2.
課程進度
週次 日期 單元主題 第1週 2/15,2/17 2/15: Motivations, sample space, event space, axioms of probability, basic properties of a probability. 2/17: Basic properties of a probability, finite sample space, examples. 第2週 2/22,2/24 2/22: Union bound, inclusion-exclusion principle, limit of sets, continuity of probability, the first Borel-Cantelli lemma. 2/24: Conditional probability, multiplication rule. 第3週 3/01,3/03 3/1: Law of total probability, Bayes' theorem, independence of events. 3/3: Probabilistic method, the second Borel-Cantelli lemma. 第4週 3/08,3/10 3/8: Conditional independence, random variables, distribution functions, discrete random variables. 3/10: Bernoulli distribution, binomial distribution, Poisson distribution, geometric distribution, negative binomial distribution, continuous random variables. 第5週 3/15,3/17 3/15: Uniform distribution, exponential distribution, memoryless property, function of a random variable, inverse distribution function. 3/17: Generating random variables from a uniform distribution, expectation of a discrete random variable. 第6週 3/22,3/24 3/22: Expectation of a continuous random variable, infinite and nonexistent expectation, expectation of a function of a random variable, variance. 3/24: Basic properties of variance. 第7週 3/29,3/31 3/29: Mean squared error, mean absolute error, normal distribution, joint distribution. 3/31: Midterm. 第8週 4/05,4/07 4/5: Holiday. 4/7: Independent random variables, conditional distribution. 第9週 4/12,4/14 4/12: Functions of two or more random variables, sum, maximum and minimum of independent random variables, gamma distribution. 4/14: Transformation of a joint density, beta distribution. 第10週 4/19,4/21 4/19: Expectation of functions of two or more random variables, covariance, correlation, conditional expectation. 4/21: Properties of conditional expectation, conditional variance and prediction. 第11週 4/26,4/28 4/26: Branching processes, moments, moment generating function. 4/28: Moment generating function characterizes distribution, bivariate normal distribution. 第12週 5/03,5/05 5/3: Markov's inequality, Chebyshev's inequality, convergence of random variables, weak and strong law of large numbers. 5/5: Central limit theorem. 第13週 5/10,5/12 5/10: Proof of the central limit theorem, Markov chains, Chapman-Kolmogorov equations. 5/12: Recurrence and transience. 第14週 5/17,5/19 5/17: One and two dimensional random walks, stopping time, strong Markov property, positive recurrent, null recurrent. 5/19: Perron-Frobenius Theorem. 第15週 5/24,5/26 5/24: Proof of the Perron-Frobenius Theorem, limiting probabilities, periodicity, ergodicity, PageRank, gambler's ruin. 5/26: Time reversibility, detailed balance condition, Markov chain Monte Carlo (MCMC). 第16週 5/31,6/02 5/31: No class. 6/2: Final exam.
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# Lesson 10
Subtraction Algorithms (Part 3)
### Lesson Purpose
The purpose of this lesson is for students to use a subtraction algorithm that records a single digit for the difference between the numbers in each place value position and a condensed notation for a decomposed hundred or ten.
### Lesson Narrative
In this lesson, students continue to learn how to use algorithms to subtract within 1,000. The new algorithm in this lesson draws attention to how place value can be used to record less digits in each place value position. This condensed notation also changes the steps of the algorithm because students don't write the numbers in expanded form to start or add up the partial differences at the end.
• Representation
• MLR8
Activity 1: A New Subtraction Algorithm
### Learning Goals
Teacher Facing
• Relate subtraction algorithms to one another using place value understanding.
• Subtract numbers within 1,000 using another algorithm based on place value.
### Student Facing
• Let’s learn another algorithm to subtract.
### Required Materials
Materials to Gather
### Lesson Timeline
Warm-up 10 min Activity 1 20 min Activity 2 15 min Lesson Synthesis 10 min Cool-down 5 min
### Teacher Reflection Questions
Who got to do math today in class and how do you know? Identify the norms or routines that allowed these students to engage in mathematics. How can you adjust these norms and routines so all students do math tomorrow?
### Suggested Centers
• Number Puzzles: Addition and Subtraction (1–4), Stage 5: Within 1,000 (Addressing)
• Five in a Row: Multiplication (3–5), Stage 2: Factors 1–9 (Supporting)
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# Category Archives: Science
My thoughts on non-physics sciences — like evolutionary biology, cognitive neuroscience etc.
# Autism and Genius
Most things in life are distributed normally, which means they all show a bell curve when quantified using a sensible measure. For instance, the marks scored by a large enough number of students has a normal distribution, with very few scoring close to zero or close to 100%, and most bunching around the class average. This distribution is the basis for letter grading. Of course, this assumes a sensible test — if the test is too easy (like a primary school test given to university students), everybody would score close to 100% and there would be no bell curve, nor any reasonable way of letter-grading the results.
If we could sensibly quantify traits like intelligence, insanity, autism, athleticism, musical aptitude etc, they should all form normal Gaussian distributions. Where you find yourself on the curve is a matter of luck. If you are lucky, you fall on the right side of the distribution close to the tail, and if you are unlucky, you would find yourself near the wrong end. But this statement is a bit too simplistic. Nothing in life is quite that straight-forward. The various distributions have strange correlations. Even in the absence of correlations, purely mathematical considerations will indicate that the likelihood of finding yourself in the right end of multiple desirable traits is slim. That is to say, if you are in the top 0.1% of your cohort academically, and in terms of your looks, and in athleticism, you are already one in a billion — which is why you don’t find many strikingly handsome theoretical physicists who are also ranked tennis players.
The recent world chess champion, Magnus Carlsen, is also a fashion model, which is news precisely because it is the exception that proves the rule. By the way, I just figured out what that mysterious expression “exception that proves the rule” actually meant — something looks like an exception only because as a general rule, it doesn’t exist or happen, which proves that there is a rule.
Getting back to our theme, in addition to the minuscule probability for genius as prescribed by mathematics, we also find correlations between genius and behavioral pathologies like insanity and autism. A genius brain is probably wired differently. Anything different from the norm is also, well, abnormal. Behavior abnormal when judged against the society’s rules is the definition of insanity. So there is a only a fine line separating insanity from true genius, I believe. The personal lives of many geniuses point to this conclusion. Einstein had strange personal relationships, and a son who was clinically insane. Many geniuses actually ended up in the looney bin. And some afflicted with autism show astonishing gifts like photographic memory, mathematical prowess etc. Take for instance, the case of autistic savants. Or consider cases like Sheldon Cooper of The Big Bang Theory, who is only slightly better than (or different from) the Rain Man.
I believe the reason for the correlation is the fact that the same slight abnormalities in the brain can often manifest themselves as talents or genius on the positive side, or as questionable gifts on the negative side. I guess my message is that anybody away from the average in any distribution, be it brilliance or insanity, should take it with neither pride nor rancor. It is merely a statistical fluctuation. I know this post won’t ease the pain of those who are afflicted on the negative side, or eliminate the arrogance of the ones on the positive side. But here’s hoping that it will at least diminish the intensity of those feelings…
Photo by Arturo de Albornoz
# Seeing and Believing
When we open our eyes and look at some thing, we see that damn thing. What could be more obvious than that, right? Let’s say you are looking at your dog. What you see is really your dog, because, if you want, you can reach out and touch it. It barks, and you can hear the woof. If it stinks a bit, you can smell it. All these extra perceptual clues corroborate your belief that what you are seeing is your dog. Directly. No questions asked.
Of course, my job on this blog is to ask questions, and cast doubts. First of all, seeing and touching seem to be a bit different from hearing and smelling. You don’t strictly hear your dog bark, you hear its sound. Similarly, you don’t smell it directly, you smell the odor, the chemical trail the dog has left in the air. Hearing and smelling are three place perceptions — the dog generates sound/odor, the sound/odor travels to you, you perceive the sound/odor.
But seeing (or touching) is a two place thing — the dog there, and you here perceiving it directly. Why is that? Why do we feel that when we see or touch something, we sense it directly? This belief in the perceptual veracity of what we see is called naive realism. We of course know that seeing involves light (so does touching, but in a much more complicated way), what we are seeing is the light reflected off an object and so on. It is, in fact, no different from hearing something. But this knowledge of the mechanism of seeing doesn’t alter our natural, commonsense view that what we see is what is out there. Seeing is believing.
Extrapolated from the naive version is the scientific realism, which asserts that our scientific concepts are also real, eventhough we may not directly perceive them. So atoms are real. Electrons are real. Quarks are real. Most of our better scientists out there have been skeptical about this extraploation to our notion of what is real. Einstein, probably the best of them, suspected that even space and time might not be real. Feynman and Gell-Mann, after developing theories on electrons and quarks, expressed their view that electrons and quarks might be mathematical constructs rather than real entities.
What I am inviting you to do here is to go beyond the skepticism of Feynman and Gell-Mann, and delve into Einstein’s words — space and time are modes by which we think, not conditions in which we live. The sense of space is so real to us that we think of everything else as interactions taking place in the arena of space (and time). But space itself is the experience corresponding to the electrical signals generated by the light hitting your retina. It is a perceptual construct, much like the tonality of the sound you hear when air pressure waves hit your ear drums. Our adoption of naive realism results in our complete trust in the three dimensional space view. And since the world is created (in our brain as perceptual constructs) based on light, its speed becomes an all important constant in our world. And since speed mixes space and time, a better description is found in a four dimensional Minkowski geometry. But all these descriptions are based on perceptual experiences and therefore unreal in some sense.
I know the description above is highly circular — I talked about space being a mental construct created by light traveling through, get this, space. And when I speak of its speed, naturally, I’m talking about distance in space divided by time, and positing as the basis for the space-time mixing. This circularity makes my description less than clear and convincing. But the difficulty goes deeper than that. You see, all we have is this cognitive construct of space and time. We can describe objects and events only in terms of these constructs even when we know that they are only cognitive representations of sensory signals. Our language doesn’t go beyond that. Well, it does, but then we will be talking the language, for instance, of Advaita, calling the constructs Maya and the causes behind them Brahman, which stays unknowable. Or, we will be using some other parallel descriptions. These descriptions may be profound, wise and accurate. But ultimately, they are also useless. It reminds me of those two guys lost while flying around in a hot-air balloon. Finally they see a man on the ground and shout,
“Hey, can you tell us where we are?”
The complete accurate and useless reply is,
“You are in a hot-air balloon about 100 ft above sea level.”
The guy on the ground must have been a philosopher.
But if philosophy is your thing, the discussions of cognitive constructs and unknown causations are not at all useless. Philosophy of physics happens to be my thing, and so I ask myself — what if I assume the unknown physical causes exist in a world similar to our perceptual construct? I could then propagate the causes through the process of perception and figure out what the construct should look like. I know, it sounds a bit complex, but it is something that we do all the time. We know, for instance, that the stars that we see in the night sky are not really there — we are seeing them the way they were a few (or a few million or billion) years ago because the light from them takes a long time to reach us. Physicists also know that the perceived motion of celestial objects also need to be corrected for these light-travel-time effects.
In fact, Einstein used the light travel time effects as the basis for deriving his special theory of relativity. He then stipulated that space and time behave the way we perceive them, derived using the said light-travel-time effects. This, of course, is based on his deep understanding that space and time are “the modes by which we think,” but also based on the assumption that the the causes behind the modes also are similar to the modes themselves. This depth of thinking is lost on the lesser scientists that came after him. The distinction between the modes of thinking and their causation is also lost, so that space and time have become entities that obey strange rules. Like bent spoons.
Photo by General Press1
# Average Beauty
If you have migrated multiple times in your life, you may have noticed a strange thing. The first time you end up in a new place, most people around you look positively weird. Ugly even. But slowly, after a year or two, you begin to find them more attractive. This effect is more pronounced if the places you are migrating from and to have different racial predominance. For example, if you migrate from the US to Japan, or from India to China. As usual, I have a theory about this strange phenomenon. Well, actually, it is more than a theory. Let me begin at the beginning.
About fifteen years ago, I visited a Japanese research institute that did all kinds of strange studies. One of the researchers there showed me his study on averaging facial features. For this study, he took a large number of Japanese faces, and averaged them (which meant he normalized the image size and orientation, digitally took the mean on a pixel-by-pixel basis). So he had an average Japanese male face and an average Japanese female face. He even created a set of hybrids by making linear combinations of the two with different weighting factors. He then showed the results to a large number of people and recorded their preference in terms of the attractiveness of the face. The strange thing was that the average face looked more pleasant and attractive to the Japanese eye than any one of the individual ones. In fact, the most attractive male face was the one that had a bit of female features in it. That is to say, it was the one with 90% average male and 10% average female (or some such combination, I don’t remember the exact weights).
The researcher went one step further, and created an average caucasian face as well. He then took the difference between that and an average Japanese face, and then superimposed the difference on an average face with exaggerated weights. The result was a grotesque caricature, which he postulated, was probably the way a Japanese person would see a caucasian for the first time.
This reminded me of the time when I visited my housemate’s farm in a small town in Pennsylvania – a town so small that the street in front the farm was named after him! I went with his parents to the local grocery store, and there was this little girl sitting in a shopping cart who went wide-eyed when she saw me. She couldn’t take her eyes off me after that. May be, seeing an Indian face for the first time in her life, she saw a similar caricature and got scared.
Anyway, my conjecture is that an averaging similar to what the Japanese researcher did happens in all of us when we migrate. First our minds see grotesque and exaggerated difference caricatures between the faces we encounter and the ones we were used to, in our previous land. Soon, our baseline average changes as we get more used to the faces around us. And the difference between what we see and our baseline ceases to be big, and we end up liking the faces more and more as they move progressively closer to the average, normal face.
Here are the average male and female faces by race or country. Notice how each one of them is a remarkably handsome or beautiful specimen. If you find some of them not so remarkable, you should move to that country and spend a few years there so that they also become remarkable! And, if you find the faces from a particular country especially attractive, with no prolonged expsosure to such faces, I would like to hear your thoughts. Please leave your comments.
[I couldn't trace the original sources of these images. If you know them, please let me know -- I would like to get copyright permissions and add attributions.]
There is more to this story than I outlined here. May be I will add my take on it as a comment below. However, the moral of the story is that if you consider yourself average, you are probably more attractive than you think you are. Than again, what do I know, I’m just an average guy.
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# Do You Believe in God?
I got in trouble for asking this question once. The person I asked the question got angry because she felt that it was too personal. So I am not going to ask you whether you believe in God. Don’t tell me — I will tell you! I will also tell you a bit more about your personality later in this post.
Ok, here is the deal. You take the quiz below. It has over 40 true-or-false questions about your habits and mannerisms. Once you answer them, I will tell you whether you believe in God, and if so, how much. If you get bored after say 20 questions or so, it is okay, you can quit the quiz and get the score. But the more questions you answer, the more accurate my guess about your faith is going to be.
Once you have your score (or rate, if you didn’t finish the quiz), click on the button corresponding to it.
Here is how it works. There is a division of labor going on in our brain, according to the theory of hemispheric specialization of brain functions. In this theory, the left hemisphere of the brain is considered the origin of logical and analytical thinking, and the right hemisphere is the origin of creative and intuitive thinking. The so-called left-brain person is thought to be linear, logical, analytical, and unemotional; and the right-brained person is thought to be spatial, creative, mystical, intuitive, and emotional.
This notion of hemispheric specialization raises an interesting question: is atheism related to the logical hemisphere? Are atheists less emotional? I think so, and this test is based on that belief. The quiz tests whether you are “left-brain” person. If you score high, your left-brain is dominant, and you are likely to be more analytical and logical than intuitive or creative. And, according to my conjecture, you are likely to be an atheist. Did it work for you?
Well, even if it didn’t, now you know whether you are analytical or intuitive. Please leave a comment to let me know how it worked.
[This post is an edited excerpt from my book The Unreal Universe]
Photo by Waiting For The Word
# Are You an Introvert?
Here is a simple 20-question quiz to see if you are an introvert or an extrovert. Introverts tend to agree with most of these statements. So if you get a score of close to 100%, you are a confirmed introvert, which is not a bad thing. You are likely to be a quiet, contemplative type with strong family ties and a generally balanced outlook in life. On the other hand, if you get close to 0%, congratulations, I see stock options in your future. And you are a party animal and believe that life is supposed to be wall-to-wall fun, which it will be for you. I’m not too sure of those in the middle though.
These questions are from Susan Cain’s best seller, Quiet: The Power of Introverts in a World That Can’t Stop Talking, and a prelude to my review of it. The questions are copyrighted to Cain, and are reproduced here with the understanding that it constitutes “fair use.” If you have any concerns about it, feel free to contact me.
# The Student Debt Crisis
[Guest Post by By Sofia Rasmussen]
It has become common knowledge as certain as death and taxes that a college education leads to a better life. A recent Pew research poll found that Americans holding a Bachelor’s Degree can expect to make an additional \$650,000 on average than those who have only graduated high school. That said, the loans many college students and parents need to take out to pay for higher education have college graduates asking themselves if it’s all worth it. Students are asking if that trip to MIT really a good investment at a 7% compounding interest rate and if going to Harvard is really worth that much more than a top online PhD degree. As debt increases, recent graduates are entering the workforce already overwhelmed, and economists are speculating as to whether the current trends in student loans may be leading to the nation’s next major debt crisis.
It’s easy to see why student loan debtors and economists are concerned. According to a report by the National Association of Consumer Bankruptcy Attorneys, individual college seniors owed an average of \$25, 250 in 2010, up 5 percent from 2009. These trends are no less staggering on a macro level, with 2011 representing the first time that US student loan debt exceeds \$1 trillion, higher than the amount of credit card debt Americans have accrued. A report from Standard and Poor’s states that ‘student loan debt has ballooned and may turn into a pricing bubble’.
The US is not the only nation facing student debt issues. India has been struggling to handle student loan applications that have more than doubled in five years, thanks to growing aspirations among the their previously lower economic class. As more Indians attend university, the cost of educational degrees has been on the rise and Educational loans from self-financing institutions in engineering, medical fields and management have become widely used. Also rising is default on debt, and India’s banks have taken notice. Banks have aimed to address bad loans by linking loan approval to employability. It is telling, then, that India’s banks do not provide any loan at all for a degree in Arts.
Whether or not student loan debt will lead to disaster for the economy at large remains to be seen, but for recent graduates, the crisis is readily apparent. Owing \$25 thousand without ever having a full-time job or experience in one’s desired field can have a profound psychological effect. The student loan anxiety can impact job decisions throughout an entire career. Even if there are openings in the fields they specialize in during college, the burden of debt leads recent graduates to opt for work with the fewest potential risks. Often this work is outside of a student’s preferred field or less intellectually stimulating than they’re capable of handling. According to a recent article in The New York Times, only half the jobs landed by new graduates even require a college degree. A graduate with a degree in Arts may be dismayed to find there is little market for a vast and intricate knowledge of WWII era British Literature, even if they had found their knowledge of the subject lead to great success in college.
While the outlook is better for a sciences graduate, they too are often saddled with work in fields that are very different from what they passionately studied at university. There is certainly no lack of need in the sciences, but often graduates with little experience outside of the classroom are saddled with grueling hours and demanding work, work they would never take if it weren’t for the threat of crushing debts to pay off.
Even as the cost of education continues to rise, parents around the world happily risk tens of thousands of dollars to send their children to best schools they can afford. For most young people, college remains a good investment. What may be changing is the sense of freedom that has traditionally been associated with college. Students may be expected to know exactly what they want to study much earlier in their educational career, perhaps even choosing specialized skill schools as opposed to the more rounded university experience. While it’s true, this may result in less culturally savvy graduates, for many students it may be the practical solution for an economically feasible life.
# Bye Bye Einstein
Starting from his miraculous year of 1905, Einstein has dominated physics with his astonishing insights on space and time, and on mass and gravity. True, there have been other physicists who, with their own brilliance, have shaped and moved modern physics in directions that even Einstein couldn’t have foreseen; and I don’t mean to trivialize neither their intellectual achievements nor our giant leaps in physics and technology. But all of modern physics, even the bizarre reality of quantum mechanics, which Einstein himself couldn’t quite come to terms with, is built on his insights. It is on his shoulders that those who came after him stood for over a century now.
“Science alone of all the subjects contains within itself the lesson of the danger of belief in the infallibility of the greatest teachers in the preceding generation. Learn from science that you must doubt the experts. As a matter of fact, I can also define science another way: Science is the belief in the ignorance of experts.” — Richard Feynman
One of the brighter ones among those who came after Einstein cautioned us to guard against our blind faith in the infallibility of old masters. Taking my cue from that insight, I, for one, think that Einstein’s century is behind us now. I know, coming from a non-practicing physicist, who sold his soul to the finance industry, this declaration sounds crazy. Delusional even. But I do have my reasons to see Einstein’s ideas go.
Let’s start with this picture of a dot flying along a straight line (on the ceiling, so to speak). You are standing at the centre of the line in the bottom (on the floor, that is). If the dot was moving faster than light, how would you see it? Well, you wouldn’t see anything at all until the first ray of light from the dot reaches you. As the animation shows, the first ray will reach you when the dot is somewhere almost directly above you. The next rays you would see actually come from two different points in the line of flight of the dot — one before the first point, and one after. Thus, the way you would see it is, incredible as it may seem to you at first, as one dot appearing out of nowhere and then splitting and moving rather symmetrically away from that point. (It is just that the dot is flying so fast that by the time you get to see it, it is already gone past you, and the rays from both behind and ahead reach you at the same instant in time.Hope that statement makes it clearer, rather than more confusing.).
Why did I start with this animation of how the illusion of a symmetric object can happen? Well, we see a lot of active symmetric structures in the universe. For instance, look at this picture of Cygnus A. There is a “core” from which seem to emanate “features” that float away to the “lobes.” Doesn’t it look remarkably similar to what we would see based on the animation above? There are other examples in which some feature points or knots seem to move away from the core where they first appear at. We could come up with a clever model based on superluminality and how it would create illusionary symmetric objects in the heavens. We could, but nobody would believe us — because of Einstein. I know this — I tried to get my old physicist friends to consider this model. The response is always some variant of this, “Interesting, but it cannot work. It violates Lorentz invariance, doesn’t it?” LV being physics talk for Einstein’s insistence that nothing should go faster than light. Now that neutrinos can violate LV, why not me?
Of course, if it was only a qualitative agreement between symmetric shapes and superluminal celestial objects, my physics friends are right in ignoring me. There is much more. The lobes in Cygnus A, for instance, emit radiation in the radio frequency range. In fact, the sky as seen from a radio telescope looks materially different from what we see from an optical telescope. I could show that the spectral evolution of the radiation from this superluminal object fitted nicely with AGNs and another class of astrophysical phenomena, hitherto considered unrelated, called gamma ray bursts. In fact, I managed to publish this model a while ago under the title, “Are Radio Sources and Gamma Ray Bursts Luminal Booms?“.
You see, I need superluminality. Einstein being wrong is a pre-requisite of my being right. So it is the most respected scientist ever vs. yours faithfully, a blogger of the unreal kind. You do the math.
Such long odds, however, have never discouraged me, and I always rush in where the wiser angels fear to tread. So let me point out a couple of inconsistencies in SR. The derivation of the theory starts off by pointing out the effects of light travel time in time measurements. And later on in the theory, the distortions due to light travel time effects become part of the properties of space and time. (In fact, light travel time effects will make it impossible to have a superluminal dot on a ceiling, as in my animation above — not even a virtual one, where you take a laser pointer and turn it fast enough that the laser dot on the ceiling would move faster than light. It won’t.) But, as the theory is understood and practiced now, the light travel time effects are to be applied on top of the space and time distortions (which were due to the light travel time effects to begin with)! Physicists turn a blind eye to this glaring inconstancy because SR “works” — as I made very clear in my previous post in this series.
Another philosophical problem with the theory is that it is not testable. I know, I alluded to a large body of proof in its favor, but fundamentally, the special theory of relativity makes predictions about a uniformly moving frame of reference in the absence of gravity. There is no such thing. Even if there was, in order to verify the predictions (that a moving clock runs slower as in the twin paradox, for instance), you have to have acceleration somewhere in the verification process. Two clocks will have to come back to the same point to compare time. The moment you do that, at least one of the clocks has accelerated, and the proponents of the theory would say, “Ah, there is no problem here, the symmetry between the clocks is broken because of the acceleration.” People have argued back and forth about such thought experiments for an entire century, so I don’t want to get into it. I just want to point out that theory by itself is untestable, which should also mean that it is unprovable. Now that there is direct experimental evidence against the theory, may be people will take a closer look at these inconsistencies and decide that it is time to say bye-bye to Einstein.
# Why not Discard Special Relativity?
Nothing would satisfy my anarchical mind more than to see the Special Theory of Relativity (SR) come tumbling down. In fact, I believe that there are compelling reasons to consider SR inaccurate, if not actually wrong, although the physics community would have none of that. I will list my misgivings vis-a-vis SR and present my case against it as the last post in this series, but in this one, I would like to explore why it is so difficult to toss SR out the window.
The special theory of relativity is an extremely well-tested theory. Despite my personal reservations about it, the body of proof for the validity of SR is really enormous and the theory has stood the test of time — at least so far. But it is the integration of SR into the rest of modern physics that makes it all but impossible to write it off as a failed theory. In experimental high energy physics, for instance, we compute the rest mass of a particle as its identifying statistical signature. The way it works is this: in order to discover a heavy particle, you first detect its daughter particles (decay products, that is), measure their energies and momenta, add them up (as “4-vectors”), and compute the invariant mass of the system as the modulus of the aggregate energy-momentum vector. In accordance with SR, the invariant mass is the rest mass of the parent particle. You do this for many thousands of times and make a distribution (a “histogram”) and detect any statistically significant excess at any mass. Such an excess is the signature of the parent particle at that mass.
Almost every one of the particles in the particle data book that we know and love is detected using some variant of this method. So the whole Standard Model of particle physics is built on SR. In fact, almost all of modern physics (physics of the 20th century) is built on it. On the theory side, in the thirties, Dirac derived a framework to describe electrons. It combined SR and quantum mechanics in an elegant framework and predicted the existence of positrons, which bore out later on. Although considered incomplete because of its lack of sound physical backdrop, this “second quantization” and its subsequent experimental verification can be rightly seen as evidence for the rightness of SR.
Feynman took it further and completed the quantum electrodynamics (QED), which has been the most rigorously tested theory ever. To digress a bit, Feynman was once being shown around at CERN, and the guide (probably a prominent physicist himself) was explaining the experiments, their objectives etc. Then the guide suddenly remembered who he was talking to; after all, most of the CERN experiments were based on Feynman’s QED. Embarrassed, he said, “Of course, Dr. Feynman, you know all this. These are all to verify your predictions.” Feynman quipped, “Why, you don’t trust me?!” To get back to my point and reiterate it, the whole edifice of the standard model of particle physics is built on top of SR. Its success alone is enough to make it impossible for modern physics to discard SR.
So, if you take away SR, you don’t have the Standard Model and QED, and you don’t know how accelerator experiments and nuclear bombs work. The fact that they do is proof enough for the validity of SR, because the alternative (that we managed to build all these things without really knowing how they work) is just too weird. It’s not just the exotic (nuclear weaponry and CERN experiments), but the mundane that should convince us. Fluorescent lighting, laser pointers, LED, computers, mobile phones, GPS navigators, iPads — in short, all of modern technology is, in some way, a confirmation of SR.
So the OPERA result on observed superluminalily has to be wrong. But I would like it to be right. And I will explain why in my next post. Why everything we accept as a verification of SR could be a case of mass delusion — almost literally. Stay tuned!
# Faster than Light
CERN has published news about some subatomic particles exceeding the speed of light, according to BBC and other sources. If confirmed true, this will remove the linchpin of modern physics — it is hard to overstate how revolutionary this discovery would be to our collective understanding of world we live in, from finest structure of matter to the time evolution of the cosmos. My own anarchical mind revels at the thought of all of modern physics getting rewritten, but I also have a much more personal stake in this story. I will get to it later in this series of posts. First, I want to describe the backdrop of thought that led to the notion that the speed of light could not be breached. The soundness of that scientific backdrop (if not the actual conclusion about the inviolability of light-speed) makes it very difficult to forgo the intellectual achievements of the past one hundred years in physics, which is what we will be doing once we confirm this result. In my second post, I will list what these intellectual achievements are, and how drastically their form will have to change. The scientists who discovered the speed violation, of course, understand this only too well, which is why they are practically begging the rest of the physics community to find a mistake in this discovery of theirs. As it often happens in physics, if you look for something hard enough, you are sure to find it — this is the experimental bias that all experimental physicists worth their salt are aware of and battle against. I hope a false negation doesn’t happen, for, as I will describe in my third post in this series, if confirmed, this speed violation is of tremendous personal importance to me.
The constancy (and the resultant inviolability) of the speed of light, of course, comes from Einstein’s Special Theory of Relativity, or SR. This theory is an extension of a simple idea. In fact, Einstein’s genius is in his ability to carry a simple idea to its logically inevitable, albeit counter-intuitive (to the point of being illogical!) conclusion. In the case of SR, he picks an idea so obvious — that the laws of physics should be independent of the state of motion. If you are in a train going at a constant speed, for instance, you can’t tell whether you are moving or not (if you close the windows, that is). The statement “You can’t tell” can be recast in physics as, “There is no experiment you can device to detect your state of motion.” This should be obvious, right? After all, if the laws kept changing every time you moved about, it is as good as having no laws at all.
Then came Maxwell. He wrote down the equations of electricity and magnetism, thereby elegantly unifying them. The equations state, using fancy vector notations, that a changing magnetic field will create an electric field, and a changing electric field will create a magnetic field, which is roughly how a car alternator and an electric motor work. These elegant equations have a wave solution.
The existence of a wave solution is no surprise, since a changing electric field generates a magnetic field, which in turn generates an electric field, which generates a magnetic filed and so on ad infinitum. What is surprising is the fact that the speed of propagation of this wave predicted by Maxwell’s equations is c, the speed of light. So it was natural to suppose that light was a form of electromagnetic radiation, which means that if you take a magnet and jiggle it fast enough, you will get light moving away from you at c – if we accept that light is indeed EM wave.
What is infinitely more fundamental is the question whether Maxwell’s equations are actually laws of physics. It is hard to argue that they aren’t. Then the follow-up question is whether these equations should obey the axiom that all laws of physics are supposed to obey — namely they should be independent of the state of motion. Again, hard to see why not. Then how do we modify Maxwell’s equations such that they are independent of motion? This is the project Einstein took on under the fancy name, “Covariant formulation of Maxwell’s equations,” and published the most famous physics article ever with an even fancier title, “On the Electrodynamics of Moving Bodies.” We now call it the Special Theory of Relativity, or SR.
To get a bit technical, Maxwell’s equations have the space derivatives of electric and magnetic fields relating to the time derivatives of charges and currents. In other words, space and time are related through the equations. And the wave solution to these equations with the propagation speed of c becomes a constraint on the properties of space and time. This is a simple philosophical look on SR, more than a physics analysis.
Einstein’s approach was to employ a series of thought experiments to establish that you needed a light signal to sync clocks and hypothesize that the speed of light had to be constant in all moving frames of reference. In other words, the speed of light is independent of the state of motion, as it has to be if Maxwell’s equations are to be laws of physics.
This aspect of the theory is supremely counter-intuitive, which is physics lingo to say something is hard to believe. In the case of the speed of light, you take a ray of light, run along with it at a high speed, and measure its speed, you still get c. Run against it and measure it — still c. To achieve this constancy, Einstein rewrote the equations of velocity addition and subtraction. On consequence of these rewritten equations is that nothing can go faster than light.
This is my long-winded description of the context in which the speed violation measured at OPERA has to be seen. If the violation is confirmed, we have a few unpleasant choices to pick from:
1. Electrodynamics (Maxwell’s equations) is not invariant under motion.
2. Light is not really electromagnetic in nature.
3. SR is not the right covariant formulation of electrodynamics.
The first choice is patently unacceptable because it is tantamount to stating that electrodynamics is not physics. A moving motor (e.g., if you take your electric razor on a flight) would behave differently from a static one (you may not be able to shave). The second choice also is quite absurd. In addition to the numeric equality between the speed of the waves from Maxwell’s equations and the measured value of c, we do have other compelling reasons why we should believe that light is EM waves. Radio waves induce electric signals in an antenna, light knocks of electrons, microwaves can excite water molecules and cook food and so on.
The only real choice we are left with is the last one — which is to say SR is wrong. Why not discard SR? More reasons than a blog post can summarize, but I’ll try to summarize them any way in my next post.
# The Unreal Universe
We know that our universe is a bit unreal. The stars we see in the night sky, for instance, are not really there. They may have moved or even died by the time we get to see them. This delay is due to the time it takes for light from the distant stars and galaxies to reach us. We know of this delay. The sun that we see now is already eight minutes old by the time we see it. This delay is not a big deal; if we want to know what is going on at the sun right now, all we have to do is to wait for eight minutes. We do have to “correct” for the delay in our perception due to the finite speed of light before we can trust what we see.
Now, this effect raises an interesting question — what is the “real” thing that we see? If seeing is believing, the stuff that we see should be the real thing. Then again, we know of the light travel time effect. So we should correct what we see before believing it. What then does “seeing” mean? When we say we see something, what do we really mean?
Seeing involves light, obviously. It is the finite (albeit very high) speed of light influences and distorts the way we see things. This fact should hardly come as a surprise because we do know that there is a delay in seeing objects like stars. What is surprising (and seldom highlighted) is that when it comes to seeing moving objects, we cannot back-calculate the same way we take out the delay in seeing the sun. If we see a celestial body moving at an improbably high speed, we cannot figure out how fast and in what direction it is “really” moving without making further assumptions. One way of handling this difficulty is to ascribe the distortions in our perception to the fundamental properties of the arena of physics — space and time. Another course of action is to accept the disconnection between our perception and the underlying “reality” and deal with it in some way.
This disconnect between what we see and what is out there is not unknown to many philosophical schools of thought. Phenomenalism, for instance, holds the view that space and time are not objective realities. They are merely the medium of our perception. All the phenomena that happen in space and time are merely bundles of our perception. In other words, space and time are cognitive constructs arising from perception. Thus, all the physical properties that we ascribe to space and time can only apply to the phenomenal reality (the reality as we sense it). The noumenal reality (which holds the physical causes of our perception), by contrast, remains beyond our cognitive reach.
One, almost accidental, difficulty in redefining the effects of the finite speed of light as the properties of space and time is that any effect that we do understand gets instantly relegated to the realm of optical illusions. For instance, the eight-minute delay in seeing the sun, because we can readily understand it and disassociate it from our perception using simple arithmetic, is considered a mere optical illusion. However, the distortions in our perception of fast moving objects, although originating from the same source are considered a property of space and time because they are more complex. At some point, we have to come to terms with the fact that when it comes to seeing the universe, there is no such thing as an optical illusion, which is probably what Goethe pointed out when he said, “Optical illusion is optical truth.”
The distinction (or lack thereof) between optical illusion and truth is one of the oldest debates in philosophy. After all, it is about the distinction between knowledge and reality. Knowledge is considered our view about something that, in reality, is “actually the case.” In other words, knowledge is a reflection, or a mental image of something external. In this picture, the external reality goes through a process of becoming our knowledge, which includes perception, cognitive activities, and the exercise of pure reason. This is the picture that physics has come to accept. While acknowledging that our perception may be imperfect, physics assumes that we can get closer and closer to the external reality through increasingly finer experimentation, and, more importantly, through better theorization. The Special and General Theories of Relativity are examples of brilliant applications of this view of reality where simple physical principles are relentlessly pursued using the formidable machine of pure reason to their logically inevitable conclusions.
But there is another, competing view of knowledge and reality that has been around for a long time. This is the view that regards perceived reality as an internal cognitive representation of our sensory inputs. In this view, knowledge and perceived reality are both internal cognitive constructs, although we have come to think of them as separate. What is external is not the reality as we perceive it, but an unknowable entity giving rise to the physical causes behind sensory inputs. In this school of thought, we build our reality in two, often overlapping, steps. The first step consists of the process of sensing, and the second one is that of cognitive and logical reasoning. We can apply this view of reality and knowledge to science, but in order do so, we have to guess the nature of the absolute reality, unknowable as it is.
The ramifications of these two different philosophical stances described above are tremendous. Since modern physics has embraced a non-phenomenalistic view of space and time, it finds itself at odds with that branch of philosophy. This chasm between philosophy and physics has grown to such a degree that the Nobel prize winning physicist, Steven Weinberg, wondered (in his book “Dreams of a Final Theory”) why the contribution from philosophy to physics have been so surprisingly small. It also prompts philosophers to make statements like, “Whether ‘noumenal reality causes phenomenal reality’ or whether ‘noumenal reality is independent of our sensing it’ or whether ‘we sense noumenal reality,’ the problem remains that the concept of noumenal reality is a totally redundant concept for the analysis of science.”
From the perspective of cognitive neuroscience, everything we see, sense, feel and think is the result of the neuronal interconnections in our brain and the tiny electrical signals in them. This view must be right. What else is there? All our thoughts and worries, knowledge and beliefs, ego and reality, life and death — everything is merely neuronal firings in the one and half kilograms of gooey, grey material that we call our brain. There is nothing else. Nothing!
In fact, this view of reality in neuroscience is an exact echo of phenomenalism, which considers everything a bundle of perception or mental constructs. Space and time are also cognitive constructs in our brain, like everything else. They are mental pictures our brains concoct out of the sensory inputs that our senses receive. Generated from our sensory perception and fabricated by our cognitive process, the space-time continuum is the arena of physics. Of all our senses, sight is by far the dominant one. The sensory input to sight is light. In a space created by the brain out of the light falling on our retinas (or on the photo sensors of the Hubble telescope), is it a surprise that nothing can travel faster than light?
This philosophical stance is the basis of my book, The Unreal Universe, which explores the common threads binding physics and philosophy. Such philosophical musings usually get a bad rap from us physicists. To physicists, philosophy is an entirely different field, another silo of knowledge, which holds no relevance to their endeavors. We need to change this belief and appreciate the overlap among different knowledge silos. It is in this overlap that we can expect to find great breakthroughs in human thought.
The twist to this story of light and reality is that we seem to have known all this for a long time. Classical philosophical schools seem to have thought along lines very similar to Einstein’s reasonings. The role of light in creating our reality or universe is at the heart of Western religious thinking. A universe devoid of light is not simply a world where you have switched off the lights. It is indeed a universe devoid of itself, a universe that doesn’t exist. It is in this context that we have to understand the wisdom behind the statement that “the earth was without form, and void” until God caused light to be, by saying “Let there be light.”
The Quran also says, “Allah is the light of the heavens and the earth,” which is mirrored in one of the ancient Hindu writings: “Lead me from darkness to light, lead me from the unreal to the real.” The role of light in taking us from the unreal void (the nothingness) to a reality was indeed understood for a long, long time. Is it possible that the ancient saints and prophets knew things that we are only now beginning to uncover with all our supposed advances in knowledge?
I know I may be rushing in where angels fear to tread, for reinterpreting the scriptures is a dangerous game. Such alien interpretations are seldom welcome in the theological circles. But I seek refuge in the fact that I am looking for concurrence in the metaphysical views of spiritual philosophies, without diminishing their mystical and theological value.
The parallels between the noumenal-phenomenal distinction in phenomenalism and the Brahman-Maya distinction in Advaita are hard to ignore. This time-tested wisdom on the nature of reality from the repertoire of spirituality is now being reinvented in modern neuroscience, which treats reality as a cognitive representation created by the brain. The brain uses the sensory inputs, memory, consciousness, and even language as ingredients in concocting our sense of reality. This view of reality, however, is something physics is yet to come to terms with. But to the extent that its arena (space and time) is a part of reality, physics is not immune to philosophy.
As we push the boundaries of our knowledge further and further, we are beginning to discover hitherto unsuspected and often surprising interconnections between different branches of human efforts. In the final analysis, how can the diverse domains of our knowledge be independent of each other when all our knowledge resides in our brain? Knowledge is a cognitive representation of our experiences. But then, so is reality; it is a cognitive representation of our sensory inputs. It is a fallacy to think that knowledge is our internal representation of an external reality, and therefore distinct from it. Knowledge and reality are both internal cognitive constructs, although we have come to think of them as separate.
Recognizing and making use of the interconnections among the different domains of human endeavor may be the catalyst for the next breakthrough in our collective wisdom that we have been waiting for.
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Electric Field for Distributed Charge systems (Line, Surface and Volume)
# Standard procedure for finding the Electric Field due to distributed charge.
Electric field arises because of the charge. We are well aware of its standard formulas. This article explains the standard procedure for deriving the expression for Electric Field Intensity due to distributed charge. The standard steps discussed over here could be applied to get the E due to distributed charges i.e. line charge, surface charge and volume charge.
## What is Electric Field Intensity?
According to Coulomb’s Law, every charge is surrounded by its Electric Field. Any other charge will experience a force if it enters into the Electric field. The strength of the force experienced by the charge may be different at different points of the field. In other words, the strength of the field varies from point to point. And this strength of the field at any point is termed as the Electric Field Intensity of that field.
Electric Field Intensity | Definition and Physical Significance.
#### Electric Field Intensity because of Point Charge.
Theoretically, we assume that all the charge is concentrated at a point known as a point charge. The electric field due to this point charge say Q, at a distance r is given as,
$E=\frac1{4\pi\varepsilon}\frac Q{r^2}$
This is the standard formula for E assuming that all the given charge concentrated at a point. Now the point is like a dot in Mathematics having no physical dimensions.
## What is distributed Charge?
As stated above, theoretically any charge is shown at a point assuming that all the charge is concentrated at that point. But, practically the charge is deposited on a physical body. Hence rather than being concentrated at a point, the charge is distributed all over the hosting body depending on many factors like dimensions, nature etc of the hosting body. So practically the charge is distributed charge.
The charge can be deposited along the length, over the surface or within the volume of the hosting body. And hence the distributed charges are line charge, surface charge and volume charge.
For each type of distribution, charge density is defined. In the case of line charge, linear charge density is charge per unit length. For surface charge, surface charge density is charge per unit area. Similarly, volume charge density which is charge per unit volume is defined for volume charge distribution.
Types of Charge Densities: Line, Surface & Volume
## Steps for getting Electric Field due to distributed Charge
So we may wish to derive the expression for Electric field intensity due to line charge or surface charge or the volume charge. In all the cases the basic procedure is the same. Yes but mathematical simplification in each case is different.
#### Step 1: Take infinitesimal charge from the given charge distribution and consider it as a point charge
Every time for deriving the formula for E, we are going to use the standard formula i.e.
$E=\frac1{4\pi\varepsilon}\frac Q{r^2}$
But, this formula is for the point charge, isn’t it? It is assumed that the charge you are considering is concentrated at a point.
So basically we consider a very small portion of the given charge distribution, say an infinitesimal charge or differential charge denoted by dq. Note that the total charge is q but we are considering a very small part of it viz. dq.
For example, say the given charge distribution is a line charge with linear charge density λ. Then to get the small portion of charge i.e. the infinitesimal charge, you have to consider a very small length, say dl, of that line charge. Ideally, dl should be approaching zero. And the charge contained by this dl, i.e. dq can easily be considered as a point charge.
So from the above figure, we can write dq for line charge as
$dq=\lambda dl$
Similarly, the infinitesimal charge (dq) from the total charge q can be obtained for the Surface and Volume charge. For the surface charge, we need to consider the infinitesimal surface while for the volume charge, the infinitesimal volume. And dq is obtained by multiplying respective densities.
$\begin{array}{l}dq=\sigma ds\;\;(for\;surface\;charge)\\dq=\rho dv\;\;(for\;volume\;charge)\end{array}$
#### Step 2: Find the E at the given point considering the dq from step 1. Say that field the dE
Once you have obtained the infinitesimal charge dq from the given charge distribution, forget rest for some time. And find the Electric field at the required point considering the only dq. It is our normal case of a point charge. Use the normal formula.
We are supposed to find the Electric field because of complete charge q. But in this step, we are writing only small E because of small q i.e. dq from the original charge. So let us call this infinitesimal field be dE.
$dE=\frac1{4\pi\varepsilon}\frac{dq}{r^2}$
#### Step 3: Properly note down and decompose the dE vector
We know that the electric field intensity is a vector quantity. Hence dE calculated in the above step is also a vector one. Now according to your position of ‘dq’ this dE vector can be pointing anywhere in the space. We should decompose this dE vector into its constituent components say vertical and horizontal according to the geometry of your diagram. For example, if there is an infinite line charge as shown in the figure and we are finding the E at point P. In this case, we decompose dE as shown in the figure. Similarly, if we have the surface charge as shown in figure
#### Step 4: Final E by integrating dE covering all the ‘dq’s with proper limits of integration
This might be the most crucial step. We are supposed to find out the Electric Field of the Distributed Charge at the given point because of the charge q. But, as you can realize in the previous steps, we have found out the infinitesimal field, dE considering the infinitesimal charge ‘dq’ from the given charge q.
Now it is the time to aggregate all the ‘dE’s of all the ‘dq’s. And in Mathematics the aggregation of the infinitesimal elements is the integration. So by using proper integration, the final answer for the E can be calculated.
1. For line charges, the proper form of line integration is used according to the coordinate system. Note that by applying the line integration, we are collecting all the ‘dq’s present over the line. So limits of the integration should be properly written according to the problem.
All the concepts of the line integration are properly applied over here e.g. the form of dl, the variable coordinate variable, fixed coordinates etc. Go through the concepts of Line Integration here.
2. For the surface charges, the proper form of surface integration should be applied. Go through the Surface Integration here.
3. For the volume charges volume integration with proper limits should be applied.
Note that separate integration should be used for each component of dE i.e. for horizontal and vertical in our illustration. Most of the time one component is zero because of symmetry of the charges.
Then final E at that point will be the vector sum of all the components obtained by the integration.
#### Alias for Step 3
If your problem doesn’t support symmetry and you are finding it difficult to decompose the dE into constituent components then instead of forming components you should write dE in vector notation.
$dE={\left(dE\right)}_x{\widehat a}_x+{\left(dE\right)}_y{\widehat a}_y+{\left(dE\right)}_z{\widehat a}_z$
Then as discussed in step 4, you should integrate for each term to get x, y and z components of the final E. Then vector addition of these terms will be our final Electric Field of the Distributed Charge.
Suggested Community: Electromagnetics for GATE & ESE
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###### Question 16861 – Hashing
November 21, 2023
###### Question 10672 – Hashing
November 21, 2023
###### Question 16861 – Hashing
November 21, 2023
###### Question 10672 – Hashing
November 21, 2023
Given the following input (4322, 1334, 1471, 9679, 1989, 6171, 6173, 4199) and the hash function x mod 10, which of the following statements are true?
i) 9679, 1989, 4199 hash to the same value
ii) 1471, 6171 hash to the same value
iii) All elements hash to the same value
iv) Each element hashes to a different value
Question 5 Explanation:
Given Input = (4322, 1334, 1471, 9679, 1989, 6171, 6173, 4199)
Hash function = x mod 10
Hash values = (2, 4, 1, 9, 9, 1, 3, 9)
9679, 1989, 4199 have same hash values
&
1471, 6171 have same hash values.
i only
ii only
i and ii only
iii or iv
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Problem 6.106
# Problem 6.106 - Problem 6.106 Given Stream function...
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Problem 6.106 [Difficulty: 2] Given: Stream function Find: Velocity potential Solution: Basic equations: Incompressibility because ψ exists u y ψ = v x ψ = u x φ = v y φ = Irrotationality x v y u 0 = We have ψ xy , ()A x 3 Bx y 2 = Then uxy , () y ψ , = , 2 B x y = vxy , x ψ , = , ()B y 2 3A x 2 = Then x , y , 2B x 6A x = but 0 1 ms = hence flow is IRROTATIONAL Hence u x φ = so φ , x , d fy + = φ ,
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# pyra 0.2.6dev
A python implementation of the GCL region algebra and query language described by Clarke et al.
pyra - Python Region Algebra
============================
Pyra is a python implementation of the region algebra and query language described in [1].
Region algebras are used to efficiently query semi-structured text documents. This particular
region algebra operates on Generalized Concordance Lists (GCLs). GCLs are lists of regions
(a.k.a., extents), which obey the following constraint: *No region in the list may have another
region from the same lists nested within it*. For a quick online introduction to this region
algebra, and why it is useful, visit:
[Wumpus Search](http://www.wumpus-search.org/docs/gcl.html)
In general, this region algebra is good for extracting data from documents that have lightweight
structure, and is an alternative to more heavyweight solutions like XPath queries.
### Algebra and Query Language
Our region algebra consists of the following elements:
(Essentially identical to the conventions used in Wumpus [See above])
Elementary Types
—---------------
"token" Tokens are quoted strings. Use \" to escape quotes, and \\ to escape escapes
"a", "b", "c" Phrases are comma separated tokens
INT Positions are indicated as bare integers (e.g., 4071)
Operators (here A and B are arbitrary region algebra expressions, N is an integer)
----------------------------------------------------------------------------------
A ^ B Returns all extents that match both A and B
A + B Returns all extents that match either A or B (or both)
A .. B Returns all extents that start with A and end with B
A > B Returns all extents that match A and contain an extent matching B
A < B Returns all extents that match A, contained in an extent matching B
_{A} The 'start' projection. For each extent (u,v) in A, return (u,u)
{A}_ The 'end' projection. For each extent (u,v) in A, return (v,v)
[N] Returns all extents of length N, where N is an integer (basically a sliding window)
Not yet implemented:
A /> B Returns all extents that match A but do not contain an extent matching B
A /< B Returns all extents that match A, not contained in an extent matching B
### Examples
Suppose we an XML document containing the complete works of Shakespeare (see './pyra/examples').
We can then run the following queries using pyra:
**Return the titles of all plays, acts, scenes, etc.**
"<title>".."</title>"
Results:
slice(15,23): <title> the tragedy of antony and cleopatra </title>
slice(68,72): <title> dramatis personae </title>
slice(279,283): <title> act i </title>
slice(284,295): <title> scene i alexandria a room in cleopatra s palace </title>
slice(1097,1105): <title> scene ii the same another room </title>
slice(3526,3534): <title> scene iii the same another room </title>
slice(4889,4898): <title> scene iv rome octavius caesar s house </title>
slice(5885,5893): <title> scene v alexandria cleopatra s palace </title>
... And, many more ...
**Return the titles of all plays**
**(i.e., the first title found in the play)**
("<title>".."</title>") < ("<play>".."</title>")
Results:
slice(15,23): <title> the tragedy of antony and cleopatra </title>
slice(40514,40522): <title> all s well that ends well </title>
slice(75567,75573): <title> as you like it </title>
slice(107909,107915): <title> the comedy of errors </title>
slice(130779,130785): <title> the tragedy of coriolanus </title>
slice(173424,173427): <title> cymbeline </title>
slice(214962,214969): <title> a midsummer night s dream </title>
slice(239304,239313): <title> the tragedy of hamlet prince of denmark </title>
... And, many more ...
**Return the titles of all plays containing the word 'henry'**
(("<title>".."</title>") < ("<play>".."</title>")) > "henry"
Results:
slice(322005,322014): <title> the second part of henry the fourth </title>
slice(361126,361134): <title> the life of henry the fifth </title>
slice(399220,399229): <title> the first part of henry the sixth </title>
slice(431541,431550): <title> the second part of henry the sixth </title>
slice(469240,469249): <title> the third part of henry the sixth </title>
slice(505920,505932): <title> the famous history of the life of henry the ei...
**Return short play titles (4 or few words)**
**(Note: We have to include the tags in the token count)**
(("<title>".."</title>") < ("<play>".."</title>")) < [6]
Results:
slice(75567,75573): <title> as you like it </title>
slice(107909,107915): <title> the comedy of errors </title>
slice(130779,130785): <title> the tragedy of coriolanus </title>
slice(173424,173427): <title> cymbeline </title>
slice(677133,677138): <title> measure for measure </title>
slice(744759,744765): <title> the tragedy of macbeth </title>
slice(771553,771559): <title> the merchant of venice </title>
slice(875994,876000): <title> pericles prince of tyre </title>
slice(1081750,1081754): <title> the tempest </title>
slice(1233968,1233974): <title> the winter s tale </title>
**Return the title of all plays containing the phrase 'to be or not to be'**
(("<title>".."</title>") < ("<play>".."</title>")) < (("<play>".."</play>") > ("to", "be", "or", "not", "to", "be"))
Results:
slice(239304,239313): <title> the tragedy of hamlet prince of denmark </title>
### Code Examples
So how do you actually write code that uses or calls *pyra*?
Here's an example implementation of the above queries:
from pyra import InvertedIndex, GCL
# See examples/gcl_shell.py for an example describing how to get
# a tokenized corpus for Shakespeare
iindex = InvertedIndex(tokens)
gcl = GCL(iidx)
# Print titles of acts, plays, scenes, etc
for myslice in g.parse('"<title>".."</title>"'):
print("%s \t %s", (str(myslice), " ".join(corpus[myslice]))
# Print titles of plays
play_titles = g.parse('("<title>".."</title>") < ("<play>".."</title>")')
for myslice in play_titles:
print("%s \t %s", (str(myslice), " ".join(corpus[myslice]))
# Return the titles of plays containing the word 'henry'
#
# Use parameterization to reuse the last expression
# (Makes for readable code, and may benefit from results caching
# if I ever get around to implementing it)
for myslice in g.parse('%1 > "henry"', play_titles):
print("%s \t %s", (str(myslice), " ".join(corpus[myslice]))
# Return the titles of plays, where the plays mention a
# 'witch' and 'duncan'
whole_plays = g.parse("<play>..</play>")
for myslice in g.parse('%1 < (%2 > ("witch" ^ "duncan"))', play_titles, whole_plays):
print("%s \t %s", (str(myslice), " ".join(corpus[myslice]))
### Ply Grammar
This package uses ply python module to parse the GCL expressions.
Here is a simplified sketch of the grammar pyra uses:
gcl_expr : ( gcl_expr ) |
gcl_expr ... gcl_expr |
gcl_expr > gcl_expr |
gcl_expr < gcl_expr |
[ INT ] |
INT |
phrase
phrase : STRING , phrase |
STRING
### References
[1] Clarke, C. L., Cormack, G. V., & Burkowski, F. J. (1995). An algebra for structured text search
and a framework for its implementation. The Computer Journal, 38(1), 43-56. Chicago
File Type Py Version Uploaded on Size
Source 2014-02-15 13KB
Redistribution and use in source and binary forms, with or without modification,
are permitted provided that the following conditions are met:
* Redistributions of source code must retain the above copyright notice, this
list of conditions and the following disclaimer.
* Redistributions in binary form must reproduce the above copyright notice, this
list of conditions and the following disclaimer in the documentation and/or
other materials provided with the distribution.
THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS" AND
ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED
WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE
DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT HOLDER OR CONTRIBUTORS BE LIABLE FOR
ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES
(INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES;
LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON
ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT
(INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE OF THIS
SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.
• Package Index Owner: afourney
• DOAP record: pyra-0.2.6dev.xml
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latest
# Did FTI Consulting, Inc. (NYSE:FCN) Use Debt To Deliver Its ROE Of 14%?
Many investors are still learning about the various metrics that can be useful when analysing a stock. This article is for those who would like to learn about Return On Equity (ROE). We’ll use ROE to examine FTI Consulting, Inc. (NYSE:FCN), by way of a worked example.
Over the last twelve months FTI Consulting has recorded a ROE of 14%. That means that for every \$1 worth of shareholders’ equity, it generated \$0.14 in profit.
See our latest analysis for FTI Consulting
### How Do I Calculate ROE?
The formula for ROE is:
Return on Equity = Net Profit ÷ Shareholders’ Equity
Or for FTI Consulting:
14% = US\$211m ÷ US\$1.5b (Based on the trailing twelve months to September 2019.)
Most readers would understand what net profit is, but it’s worth explaining the concept of shareholders’ equity. It is all earnings retained by the company, plus any capital paid in by shareholders. Shareholders’ equity can be calculated by subtracting the total liabilities of the company from the total assets of the company.
### What Does Return On Equity Signify?
ROE looks at the amount a company earns relative to the money it has kept within the business. The ‘return’ is the profit over the last twelve months. A higher profit will lead to a higher ROE. So, all else equal, investors should like a high ROE. That means ROE can be used to compare two businesses.
### Does FTI Consulting Have A Good Return On Equity?
One simple way to determine if a company has a good return on equity is to compare it to the average for its industry. The limitation of this approach is that some companies are quite different from others, even within the same industry classification. If you look at the image below, you can see FTI Consulting has a similar ROE to the average in the Professional Services industry classification (15%).
That isn’t amazing, but it is respectable. ROE doesn’t tell us if the share price is low, but it can inform us to the nature of the business. For those looking for a bargain, other factors may be more important. I will like FTI Consulting better if I see some big insider buys. While we wait, check out this free list of growing companies with considerable, recent, insider buying.
### The Importance Of Debt To Return On Equity
Companies usually need to invest money to grow their profits. That cash can come from issuing shares, retained earnings, or debt. In the case of the first and second options, the ROE will reflect this use of cash, for growth. In the latter case, the use of debt will improve the returns, but will not change the equity. That will make the ROE look better than if no debt was used.
### Combining FTI Consulting’s Debt And Its 14% Return On Equity
FTI Consulting has a debt to equity ratio of 0.19, which is far from excessive. The combination of modest debt and a very respectable ROE suggests this is a business worth watching. Judicious use of debt to improve returns can certainly be a good thing, although it does elevate risk slightly and reduce future optionality.
### But It’s Just One Metric
Return on equity is one way we can compare the business quality of different companies. A company that can achieve a high return on equity without debt could be considered a high quality business. All else being equal, a higher ROE is better.
Having said that, while ROE is a useful indicator of business quality, you’ll have to look at a whole range of factors to determine the right price to buy a stock. The rate at which profits are likely to grow, relative to the expectations of profit growth reflected in the current price, must be considered, too. So you might want to check this FREE visualization of analyst forecasts for the company.
If you would prefer check out another company — one with potentially superior financials — then do not miss thisfree list of interesting companies, that have HIGH return on equity and low debt.
We aim to bring you long-term focused research analysis driven by fundamental data. Note that our analysis may not factor in the latest price-sensitive company announcements or qualitative material.
If you spot an error that warrants correction, please contact the editor at editorial-team@simplywallst.com. This article by Simply Wall St is general in nature. It does not constitute a recommendation to buy or sell any stock, and does not take account of your objectives, or your financial situation. Simply Wall St has no position in the stocks mentioned. Thank you for reading.
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Search a number
126531444635 = 5171488605231
BaseRepresentation
bin111010111010111011…
…1001001111110011011
3110002121012020101011022
41311311313021332123
54033114022212020
6134043331553055
712066365351021
oct1656567117633
9402535211138
10126531444635
114973077a152
122063315b78b
13bc163c78cb
1461a4974711
1534585b8625
hex1d75dc9f9b
126531444635 has 8 divisors (see below), whose sum is σ = 160769365056. Its totient is φ = 95270734720.
The previous prime is 126531444539. The next prime is 126531444643. The reversal of 126531444635 is 536444135621.
It is a sphenic number, since it is the product of 3 distinct primes.
It is a de Polignac number, because none of the positive numbers 2k-126531444635 is a prime.
It is a Duffinian number.
It is an unprimeable number.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 744302531 + ... + 744302700.
It is an arithmetic number, because the mean of its divisors is an integer number (20096170632).
Almost surely, 2126531444635 is an apocalyptic number.
126531444635 is a deficient number, since it is larger than the sum of its proper divisors (34237920421).
126531444635 is a wasteful number, since it uses less digits than its factorization.
126531444635 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 1488605253.
The product of its digits is 1036800, while the sum is 44.
The spelling of 126531444635 in words is "one hundred twenty-six billion, five hundred thirty-one million, four hundred forty-four thousand, six hundred thirty-five".
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# motion-velocity graph
motion-velocity graph
1) Механика: диаграмма скорости движения
2) Робототехника: диаграмма (изменения) скорости движения, профиль скорости (вдоль траектории движения)
Универсальный англо-русский словарь. . 2011.
### Смотреть что такое "motion-velocity graph" в других словарях:
• Motion planning — (a.k.a., the navigation problem , the piano mover s problem ) is a term used in robotics for the process of detailing a task into discrete motions. For example, consider navigating a mobile robot inside a building to a distant waypoint. It should … Wikipedia
• Motion graphs and derivatives — The green line shows the slope of the velocity time graph at the particular point where the two lines touch. Its slope is the acceleration at that point. In mechanics, the derivative of the position vs. time graph of an object is equal to the… … Wikipedia
• Linear motion — is motion along a straight whiteline, and can therefore be described mathematically using only one spatial dimension. It can be uniform, that is, with constant speed, or non uniform, that is, with a variable speed. The motion of a particle (a… … Wikipedia
• Piston motion equations — The motion of a non offset piston connected to a crank through a connecting rod (as would be found in internal combustion engines), can be expressed through several mathematical equations. This article shows how these motion equations are derived … Wikipedia
• Superluminal motion — In astronomy, superluminal motion is the apparently faster than light motion seen in some radio galaxies, quasars and recently also in some galactic sources called microquasars. All of these sources are thought to contain a black hole,… … Wikipedia
• Milky Way Galaxy — Large spiral galaxy (roughly 150,000 light years in diameter) that contains Earth s solar system. It includes the multitude of stars whose light is seen as the Milky Way, the irregular luminous band that encircles the sky defining the plane of… … Universalium
• analysis — /euh nal euh sis/, n., pl. analyses / seez /. 1. the separating of any material or abstract entity into its constituent elements (opposed to synthesis). 2. this process as a method of studying the nature of something or of determining its… … Universalium
• physical science, principles of — Introduction the procedures and concepts employed by those who study the inorganic world. physical science, like all the natural sciences, is concerned with describing and relating to one another those experiences of the surrounding… … Universalium
• cosmos — /koz meuhs, mohs/, n., pl. cosmos, cosmoses for 2, 4. 1. the world or universe regarded as an orderly, harmonious system. 2. a complete, orderly, harmonious system. 3. order; harmony. 4. any composite plant of the genus Cosmos, of tropical… … Universalium
• Thermodynamic temperature — is the absolute measure of temperature and is one of the principal parameters of thermodynamics. Thermodynamic temperature is an “absolute” scale because it is the measure of the fundamental property underlying temperature: its null or zero point … Wikipedia
• sound — sound1 soundable, adj. /sownd/, n. 1. the sensation produced by stimulation of the organs of hearing by vibrations transmitted through the air or other medium. 2. mechanical vibrations transmitted through an elastic medium, traveling in air at a… … Universalium
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# Help with Projectile Motion problem (including air friction) (1 Viewer)
### Users Who Are Viewing This Thread (Users: 0, Guests: 1)
Hopefully you smart guys can understand this post
I recently have bought an airsoft pistol that shoots 302fps and I thought it would be fun to calculate the max distance & stuff for the bb including air resistance. The problem is including air resistance into the equation.
Here's the stats:
mass of bb= .12g
diameter of bb= 6mm (.006m)
muzzle velocity for .12g bb = 302fps (92.0496m/s)
After reading stuff on the net, I aparently need the equation D=(pCA)/2, where p is the density of air (about 1.2101 kg/m3), C is drag coefficient (around .47 probably?), and A is the area of the bb looking at front (pi*r2 = pi*(.003m)2 = 2.8274*10-5 m2)
When I plug the numbers in, D= 8.04046*10-6
I'm sort of lost after this step, aparently Acceleration=-DV2 --> Ax=-DV(Vx) and Ay=-DV(Vy)
I tried plugging the numbers in but the answer was only somewhere around .04 which I'm pretty sure isn't right at all.
My main goal is to put it into a parametric function so I can see it visually with distance and time:
SO, what I'm sure is right (without air resistance):
X1T=92.0496cos(35)
Y1T=92.0496sin(35)-(1/2)9.8T2
What is the air resistance acceleration and how do I plug it into that equation???
#### Fermat
Homework Helper
For small velocities (velocities less than 328 ft/s), air resistance is (approximately) proportional to velocity, rather than the square of velocity.
Change it to DV from DV² and see if that helps.
Changing it to DV instead of DV2 makes the number even smaller (around .00074).
After reading some more... Acc(drag)=DV2/(mass) and after what you said, should I change that to DV/(mass)? This would make it around 6.168m/s2
So when I plug it into my equation, is this accurate:
X1T=92.0496cos(35)-(8.04046*10-6*92.0496cos(35)/.00012)T2
Y1T=92.0496sin(35)-(1/2)9.8T2-(8.04046*10-6*92.0496sin(35)/.00012)T2
#### Hootenanny
Staff Emeritus
Gold Member
I'm sorry, I can't tell which equations of motion you are using. I would recommend using $s = ut + \frac{1}{2}at^2$. You will need to know the inital height to solve for t, but this is easy to measure.
Regards,
-Hoot
Sorry, thats the equation I was using, I just forgot to put the T in after the first velocities.
That equation still doesn't include air friction though.
#### Hootenanny
Staff Emeritus
Gold Member
Okay, it seems that you have determined the [negative] acceleration due to air resistance. Now in the x - direction this is the only acceleration experience by the bb round, thus in the x-direction we have;
$$s_{x} = ut + \frac{1}{2}\cdot - a_{drag} t^2$$
However, in the y - direction, there are two forces acting, gravity and drag but it is important to observe that in this case they are acting in opposite directions. The force of gravity it acting in the same direction as velocity (down) therefore the acceleration will be positive. The drag force is acting in the opposite direction to the velcoity (up). Therefore we can say that;
$$a_{total} = g - a_{drag}$$
Thus the kinematic equation in the y-direction becomes;
$$s_{y} = ut + \frac{1}{2} (g - a_{drag})t^2$$
Do you follow?
~Hoot
I understand that part. The only problem is that I'm not sure if my adrag calculations are correct.
Lets work backwards. What's the equation for adrag?
#### Hootenanny
Staff Emeritus
Gold Member
The relationship for drag is given by (as you correctly stated above);
$$F_{drag} = - \frac{1}{2}C\rho Av^2$$
For a sphere c = 0.5, the density of air is approximately 1.25 kg/m^3.
Regards,
~Hoot
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# Free category
Last updated
In mathematics, the free category or path category generated by a directed graph or quiver is the category that results from freely concatenating arrows together, whenever the target of one arrow is the source of the next.
Mathematics includes the study of such topics as quantity, structure, space, and change.
In mathematics, and more specifically in graph theory, a directed graph is a graph that is made up of a set of vertices connected by edges, where the edges have a direction associated with them.
In mathematics, a quiver is a directed graph where loops and multiple arrows between two vertices are allowed, i.e. a multidigraph. They are commonly used in representation theory: a representation V of a quiver assigns a vector space V(x) to each vertex x of the quiver and a linear map V(a) to each arrow a.
## Contents
More precisely, the objects of the category are the vertices of the quiver, and the morphisms are paths between objects. Here, a path is defined as a finite sequence
${\displaystyle V_{0}{\xrightarrow {\;\;E_{0}\;\;}}V_{1}{\xrightarrow {\;\;E_{1}\;\;}}\cdots {\xrightarrow {E_{n-1}}}V_{n}}$
where ${\displaystyle V_{k}}$ is a vertex of the quiver, ${\displaystyle E_{k}}$ is an edge of the quiver, and n ranges over the non-negative integers. For every vertex ${\displaystyle V}$ of the quiver, there is an "empty path" which constitutes the identity morphisms of the category.
The composition operation is concatenation of paths. Given paths
${\displaystyle V_{0}{\xrightarrow {E_{0}}}\cdots {\xrightarrow {E_{n-1}}}V_{n},\quad V_{n}{\xrightarrow {F_{0}}}W_{0}{\xrightarrow {F_{1}}}\cdots {\xrightarrow {F_{n-1}}}W_{m},}$
their composition is
${\displaystyle \left(V_{n}{\xrightarrow {F_{0}}}W_{0}{\xrightarrow {F_{1}}}\cdots {\xrightarrow {F_{n-1}}}W_{m}\right)\circ \left(V_{0}{\xrightarrow {E_{0}}}\cdots {\xrightarrow {E_{n-1}}}V_{n}\right):=V_{0}{\xrightarrow {E_{0}}}\cdots {\xrightarrow {E_{n-1}}}V_{n}{\xrightarrow {F_{0}}}W_{0}{\xrightarrow {F_{1}}}\cdots {\xrightarrow {F_{n-1}}}W_{m}}$
Note that the result of the composition starts with the right operand of the composition, and ends with its left operand.
## Examples
• If Q is the quiver with one vertex and one edge f from that object to itself, then the free category on Q has as arrows 1, f, ff,fff, etc. [2]
• Let Q be the quiver with two vertices a, b and two edges e, f from a to b and b to a, respectively. Then the free category on Q has two identity arrows and an arrow for every finite sequence of alternating es and fs, including: e, f, ef, fe, fef, efe, etc. [1]
• If Q is the quiver ${\displaystyle a{\xrightarrow {f}}b{\xrightarrow {g}}c}$, then the free category on Q has (in addition to three identity arrows), arrows f, g, and gf.
• If a quiver Q has only one vertex, then the free category on Q has only one object, and corresponds to the free monoid on the edges of Q. [1]
In abstract algebra, the free monoid on a set is the monoid whose elements are all the finite sequences of zero or more elements from that set, with string concatenation as the monoid operation and with the unique sequence of zero elements, often called the empty string and denoted by ε or λ, as the identity element. The free monoid on a set A is usually denoted A. The free semigroup on A is the subsemigroup of A containing all elements except the empty string. It is usually denoted A+.
## Properties
The category of small categories Cat has a forgetful functor U into the quiver category Quiv:
In mathematics, specifically in category theory, the category of small categories, denoted by Cat, is the category whose objects are all small categories and whose morphisms are functors between categories. Cat may actually be regarded as a 2-category with natural transformations serving as 2-morphisms.
In mathematics, in the area of category theory, a forgetful functor 'forgets' or drops some or all of the input's structure or properties 'before' mapping to the output. For an algebraic structure of a given signature, this may be expressed by curtailing the signature: the new signature is an edited form of the old one. If the signature is left as an empty list, the functor is simply to take the underlying set of a structure. Because many structures in mathematics consist of a set with an additional added structure, a forgetful functor that maps to the underlying set is the most common case.
U : CatQuiv
which takes objects to vertices and morphisms to arrows. Intuitively, U "[forgets] which arrows are composites and which are identities". [2] This forgetful functor is right adjoint to the functor sending a quiver to the corresponding free category.
### Universal property
The free category on a quiver can be described up to isomorphism by a universal property. Let C : QuivCat be the functor that takes a quiver to the free category on that quiver (as described above), let U be the forgetful functor defined above, and let G be any quiver. Then there is a graph homomorphism I : GU(C(G)) and given any category D and any graph homomorphism F : GU(B), there is a unique functor F' : C(G) → D such that U(F')∘I=F, i.e. the following diagram commutes:
In mathematics, the phrase up to appears in discussions about the elements of a set, and the conditions under which subsets of those elements may be considered equivalent. The statement "elements a and b of set S are equivalent up to X" means that a and b are equivalent if criterion X is ignored. That is, a and b can be transformed into one another if a transform corresponding to X is applied.
In category theory, an abstract branch of mathematics, an equivalence of categories is a relation between two categories that establishes that these categories are "essentially the same". There are numerous examples of categorical equivalences from many areas of mathematics. Establishing an equivalence involves demonstrating strong similarities between the mathematical structures concerned. In some cases, these structures may appear to be unrelated at a superficial or intuitive level, making the notion fairly powerful: it creates the opportunity to "translate" theorems between different kinds of mathematical structures, knowing that the essential meaning of those theorems is preserved under the translation.
In various branches of mathematics, a useful construction is often viewed as the “most efficient solution” to a certain problem. The definition of a universal property uses the language of category theory to make this notion precise and to study it abstractly.
The functor C is left adjoint to the forgetful functor U. [1] [2] [3]
## Related Research Articles
In category theory, a branch of mathematics, the abstract notion of a limit captures the essential properties of universal constructions such as products, pullbacks and inverse limits. The dual notion of a colimit generalizes constructions such as disjoint unions, direct sums, coproducts, pushouts and direct limits.
In mathematics, specifically category theory, adjunction is a relationship that two functors may have. Two functors that stand in this relationship are known as adjoint functors, one being the left adjoint and the other the right adjoint. Pairs of adjoint functors are ubiquitous in mathematics and often arise from constructions of "optimal solutions" to certain problems, such as the construction of a free group on a set in algebra, or the construction of the Stone-Čech compactification of a topological space in topology.
An exact sequence is a concept in mathematics, especially in group theory, ring and module theory, homological algebra, as well as in differential geometry. An exact sequence is a sequence, either finite or infinite, of objects and morphisms between them such that the image of one morphism equals the kernel of the next.
In category theory, a category is considered Cartesian closed if, roughly speaking, any morphism defined on a product of two objects can be naturally identified with a morphism defined on one of the factors. These categories are particularly important in mathematical logic and the theory of programming, in that their internal language is the simply typed lambda calculus. They are generalized by closed monoidal categories, whose internal language, linear type systems, are suitable for both quantum and classical computation.
In mathematics, the idea of a free object is one of the basic concepts of abstract algebra. It is a part of universal algebra, in the sense that it relates to all types of algebraic structure. It also has a formulation in terms of category theory, although this is in yet more abstract terms. Examples include free groups, tensor algebras, or free lattices. Informally, a free object over a set A can be thought of as being a "generic" algebraic structure over A: the only equations that hold between elements of the free object are those that follow from the defining axioms of the algebraic structure.
In category theory, a branch of mathematics, the functors between two given categories form a category, where the objects are the functors and the morphisms are natural transformations between the functors. Functor categories are of interest for two main reasons:
In mathematics, particularly category theory, a representable functor is a functor of a special form from an arbitrary category into the category of sets. Such functors give representations of an abstract category in terms of known structures allowing one to utilize, as much as possible, knowledge about the category of sets in other settings.
In mathematics, the derived categoryD(A) of an abelian category A is a construction of homological algebra introduced to refine and in a certain sense to simplify the theory of derived functors defined on A. The construction proceeds on the basis that the objects of D(A) should be chain complexes in A, with two such chain complexes considered isomorphic when there is a chain map that induces an isomorphism on the level of homology of the chain complexes. Derived functors can then be defined for chain complexes, refining the concept of hypercohomology. The definitions lead to a significant simplification of formulas otherwise described by complicated spectral sequences.
In mathematics, a triangulated category is a category together with the additional structure of a "translation functor" and a class of "distinguished triangles". Prominent examples are the derived category of an abelian category and the stable homotopy category of spectra, both of which carry the structure of a triangulated category in a natural fashion. The distinguished triangles generate the long exact sequences of homology; they play a role akin to that of short exact sequences in abelian categories.
Fibred categories are abstract entities in mathematics used to provide a general framework for descent theory. They formalise the various situations in geometry and algebra in which inverse images of objects such as vector bundles can be defined. As an example, for each topological space there is the category of vector bundles on the space, and for every continuous map from a topological space X to another topological space Y is associated the pullback functor taking bundles on Y to bundles on X. Fibred categories formalise the system consisting of these categories and inverse image functors. Similar setups appear in various guises in mathematics, in particular in algebraic geometry, which is the context in which fibred categories originally appeared. Fibered categories are used to define stacks, which are fibered categories with "descent". Fibrations also play an important role in categorical semantics of type theory, and in particular that of dependent type theories.
This is a glossary of properties and concepts in category theory in mathematics.
In category theory, a discipline within mathematics, the nerveN(C) of a small category C is a simplicial set constructed from the objects and morphisms of C. The geometric realization of this simplicial set is a topological space, called the classifying space of the categoryC. These closely related objects can provide information about some familiar and useful categories using algebraic topology, most often homotopy theory.
In category theory, a branch of mathematics, a diagram is the categorical analogue of an indexed family in set theory. The primary difference is that in the categorical setting one has morphisms that also need indexing. An indexed family of sets is a collection of sets, indexed by a fixed set; equivalently, a function from a fixed index set to the class of sets. A diagram is a collection of objects and morphisms, indexed by a fixed category; equivalently, a functor from a fixed index category to some category.
In mathematics, the category of rings, denoted by Ring, is the category whose objects are rings and whose morphisms are ring homomorphisms. Like many categories in mathematics, the category of rings is large, meaning that the class of all rings is proper.
In mathematics, a topos is a category that behaves like the category of sheaves of sets on a topological space. Topoi behave much like the category of sets and possess a notion of localization; they are in a sense a generalization of point-set topology. The Grothendieck topoi find applications in algebraic geometry; the more general elementary topoi are used in logic.
## References
1. Awodey, Steve (2010). Category theory (2nd ed.). Oxford: Oxford University Press. pp. 20–24. ISBN 0199237182. OCLC 740446073.
2. Mac Lane, Saunders (1978). Categories for the Working Mathematician (Second ed.). New York, NY: Springer New York. pp. 49–51. ISBN 1441931236. OCLC 851741862.
3. "free category in nLab". ncatlab.org. Retrieved 2017-09-12.
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# Artificial Intelligence Programming
ICS 361
## Assignment #5 - AKA Course Project Creating a Recycling Softbot in Common Lisp
This assignment is worth 200 points.
Your assignment is to design, implement, test, and report results on the project described below.
• Start with the simple Vacuum cleaner agent code provided by Norvig & Russell.
• Reference the VacuumWorld slides on Laulima (VacuumWorld.pdf under lectures)
• Open http://aima.cs.berkeley.edu/lisp/ . Then download the file code.tar.gz and unzip it.
• Alternatively, a local copy of the unzipped code is here.
• Documentation for the code is at http://aima.cs.berkeley.edu/lisp/doc/overview.html
You will only need to load the utilities and agents subsystems of the AIMA code. Start with and modify the grid environment and vacuum agent code, you do not need to re-write the parts of the code that are available.
• Output from the simple Reflex vacuum is here. The lines with gc represent garbage collection and can be ignored. The run was not optimized for speed or space use.
Your GOAL is to create a softbot to collect bottles and cans and put them into a recycle bin.
The robot can carry a mazimum of 1 can and 1 bottle at the same time. Cans and bottles are released separately into the Recycle bin (Rb).
What actions will your robot need to complete its task? Explain them in your report. What strategies do you use to clean the room? Find the Recyclebin (Rb)? Minimize the number of moves? Explain these in your report.
• (25 p) Printout. Change the print function so that the entire world is not printed out at every move. Select to print the world every X moves and/or when something significant happens. It is a good idea to print out the percepts and actions at each step, however.
• (10 p) Starting Location. Your robot always starts in the LOWER left cell of the room, represent it with a capital R (instead of the V, used for the vacuum).
UPDATED The format for these files is shown in Room specification format. The file format is:
x and y (or Row and Column) size of the room
x and y (or Row and Column) position of the Recycle bin.
n pieces of furniture specified by opposite corners of a rectancle (X Y X Y or R C R C)
m bottles and cans, each on a line starting with B or C and followed by the coordinates.
The maximum legal size of a room is 20 by 20 squares. If a file has a larger room specified, it is an error.
(Read in scenario files for different sized rooms, recycling loads and obstacle locations. Obstacles are specified by the two opposite corners of a rectangle. Your initialization program must block all cells in that rectangular area.
• (20 p)You must check for errors in the input file.
Some example room files can be found here. Note that some examples may be bigger than you are required to handle for this assignment.
• Create the recycling bin (Rb) in the specified cell.
• More files will be added shortly.
For development purposes, you may create the cans/bottles randomly instead of reading them in from a room specification file. The vacuum world has a dirt factor which randomly distributes dirt in the cells. You can modify this so that this number represents the total number of items that need recycling in the room. They will be distributed randomly in the room like the dirt is in the vacuum world. You may remove or ignore the direction state indicator in the vacuum code. Do not print it out in the room layout grid.
• (25 p) More/better sensors. There are 2 choices for sensors, an additional 4 sensors or an additional 8 sensors to view the surrounding cells.
```NEW: FORMAT for sensor input:
Difference between vacuum sensors and robot's sensors.
NO sensor for home
ADD binary sensor for Rb in current cell (separate from contents)
CHANGE input for current square either B or C or empty.
ADD LIST of outer sensor readings (4 or 8) Sensor order should be 1=
up, 3 = right, 5= down, 7 = left. With even numbers 2-8 for the sensors
inbetween those 4. Sensors return one of the values below
```
• Detect if there is a can or bottle in the current cell.
• Determine if the recycling bin (Rb) is in the current cell.
We will limit the problem to a maximum of 1 Bottle OR 1 Can per cell. This sensors return C or B for can and bottle respectively. Rb will be identified with a yes/no sensor.
• (4 extra sensors option) detect blockages (furniture) and the edges of the room in the four cells Up, Down, Left and Right. These extra sensors aren't able to count cans in the surrounding cells. Instead they return one of the following values: clear, blocked, or messy (not clear or blocked). Blocked means a wall or a piece of furniture while messy means 1 or more bottles/cans.
• (8 extra sensors option) have the 4 sensors above, plus 4 sensors that can detect blockages in the 4 cells diagonal from the current cell, UpLeft, UpRight, DownLeft and DownRight. NOTE: you can only sense the diagonals if ONE or BOTH of the adjacent cells are open. For example, if Up and Right are blocked and Down and Left are clear, you can detect the cells at ULeft, DLeft, and DRight, but not URight. If a cell on the diagonal can't be accessed, the softbot returns nil for that cell.
• (25 pt) Enhance your robot's ability to move - there are 2 choices:
1. (4 move option) 4 directions - up, down, left, right or
2. (8 move option) 8 directions (UDLR and diagonals ULeft, URight, DLeft and DRight).
Note that diagonals are only accessable if one or both of the adjacent cells are unblocked (see sensors above).
• (25 pt) Create an internal map where the robot keeps track of its explorations of the world. This map is SEPARATE from the robot's environment. The robot will need to discover the size and shape of the room as well as the cell where the recycling bin is located, all obstacles and cans/bottles in the room. Keep track of the number of moves it takes you to recycle all cans/bottles in the room. (goal-based agent) Write about the results in your report.
• (50 p) Search for recyclables, the Rb, and find out the size of the room. Keep a count of the number of moves taken.
Create several different rooms and test your robot's performance with different numbers of cans. Is a particular strategy useful to find the recycling bin early in your search?
### Competition
Your softbot will compete with others during the last week of class.
There are 3 levels of competition. There are 2 choices for robot sensors (see sensors above). There are 3 levels of competition, 1 hand (can or bottle) Easy, 2 hands (cans or bottles in either) Medium and 1 Can hand + 1 Bottle hand Hard. Programs in each competition level will receive 50 bonus EC assignment points. The runner-up will receive 25 points and 3rd place 10 points.
Turn in your code, transcripts of testing and running it, and a report on your solution/algorithms and results.
Back to the course homepage.
(c) N. E. Reed, 2005-2016
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# Reducing
Topic closed. 27 replies. Last post 2 years ago by manual.
Page 2 of 2
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Posted: December 17, 2014, 12:51 pm - IP Logged
Depend on how you interpret the hit ratio, '15 doubles for 11 hits' is just observation not a playable scenario.If the patterns has time frame from one to twenty draws for a hit, how do you correlate the 11 hits? I may pick a pattern and wage for 10 draws and miss most of 11 hits or may change a pattern each double trigger and still lose money.
Keep in mind, I'm saying this method is a system. It's just a method to reduce numbers. I'm working on my 'system' in a programming approach called Agile. First, I want to reduce the numbers to a playable amount (in my mind, that's ten or less) while still retaining a high hit percentage. That's what I'm working on now. After that, I'll work on turning the boxed hits into straights. As for the hit percentage, I should have been more clear. That 73.3% for November is not a true statistic, just more of a glance to show what I'm seeing with the testing I'm doing. On the other hand, the twenty number patter I mentioned has a hit rate of 92%. I've tested that over five years and it's an overall statistic. Year over year, it may fluctuate up or down but if you were to play it, you could expect a hit 92% of the time, if playing twenty numbers over ten days. Wouldn't make money because the numbers have a fair chance of hitting boxed. Plus, who wants to play twenty numbers...
When I go through and test, I do a lot of it 'by hand', testing patterns by going trough the data myself. If something seems to be working, I write a script to test the pattern out to see how much it hits over 5 days, 10 days, and so on. I start testing with 1 year and go out to 5 (20 sometimes if the script executes quick enough). By coding and using SQL to do my testing, I remove bias and human error. The script goes through, finds a double (since that's the trigger I like to use), applies the current pattern I'm testing, then looks to see if it would have hit in the next ten days. It's a fair amount of work but I like to see the outcome. I think it's amazing that some patters get 167 out of a 1000 for some patterns. Others, that may be very similar, can get up into the 700s hits. If a test goes back 5+ years with similar results, the odds that it suddenly changes is not likely. So I'm fairly confident when I say something hits at 72%, it will. I can back it up with years worth of data. Good question. :)
NB> For strategic wager with ROI in mind, setting N=10 is not big deal so far as ' degree of certainty' is above 50%, ponder on this if you have time.
You lost me... :( Can you explain that a little better?
mmx1: I don't really mind. Long as the thread doesn't get totally derailed. Thank you. :)
onlymoney: Looks like you're seeing hits here too. What are the question marks for?
Have you tried to program differently? The regular way is not to follow if you run out of memory. I don't really do pick 3, but found that several ways of programming just fail in cases. In one case Excel beats regular OOP. For pick 3 I never considered boxed, I'd rather play couples as alternative. Can you score on front or rear couples, not boxed?
I may be able to work around it, but it'll go much slower. I don't care for boxed either but it's a smaller starting set. Figured if I could get 333 down to ten or less, then I could worry about ordering. Haven't tried font/back pairs. Might look into that, could help with ordering.
Unless you intend on playing in other states, I wouldn't worry about that. What works in your state, may not work in mine and vice versa. Just try to perfect it for where you are going to play most often. Cool system!!!
Well, I know it works for FL. I figured if it worked elsewhere, it might help some reduce their playable numbers. Plus, it never hurts to share, discuss, test.
NB> For strategic wager with ROI in mind, setting N=10 is not big deal so far as ' degree of certainty' is above 50%, ponder on this if you have time.
I recalled your ' deviation' system a while ago with 27 picks, which was pretty much for Straight betting, so I presumed analysing digits for position 1 and 2 wouldn't be an issue.Imagine having a code that can predict the ' next exact pairs' in range of 10 draws, this will indicate a reduction to just 20 picks for straight bet.
I started this ideal with just digits for ' next positions one' and the hit rate is over 50%, but the picks are larger for waging locally. If you can predict position one, then remaining pairs becomes 10C2 or 10P2.
10C2= 10!/2!8!= 45 box pairs
10P2>90 permutations without repeat digit
10P2>100 permutations with a repeat digit
Since we're targeting exact hits, permutation is our focus. focus should be on the ideal not the larger picks, the picks can be reduced later. Pick 3 or 4 has 10 members (0123456789), so setting N=10 , will give the above pairs, but you can reduced N(many ways out there, if you can analysis recent digit trends).
See this link , you may see something overlooked.
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Posted: December 17, 2014, 1:02 pm - IP Logged
The question marks show up when I copy and pasta.
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Posted: December 17, 2014, 1:23 pm - IP Logged
NB> For strategic wager with ROI in mind, setting N=10 is not big deal so far as ' degree of certainty' is above 50%, ponder on this if you have time.
I recalled your ' deviation' system a while ago with 27 picks, which was pretty much for Straight betting, so I presumed analysing digits for position 1 and 2 wouldn't be an issue.Imagine having a code that can predict the ' next exact pairs' in range of 10 draws, this will indicate a reduction to just 20 picks for straight bet.
I started this ideal with just digits for ' next positions one' and the hit rate is over 50%, but the picks are larger for waging locally. If you can predict position one, then remaining pairs becomes 10C2 or 10P2.
10C2= 10!/2!8!= 45 box pairs
10P2>90 permutations without repeat digit
10P2>100 permutations with a repeat digit
Since we're targeting exact hits, permutation is our focus. focus should be on the ideal not the larger picks, the picks can be reduced later. Pick 3 or 4 has 10 members (0123456789), so setting N=10 , will give the above pairs, but you can reduced N(many ways out there, if you can analysis recent digit trends).
See this link , you may see something overlooked.
Lets use this chart to illustrate 'next position 1 +10P2' ideal
digit prediction points
0 7,8
1 0,6
3 7,6
2 9,3
4 8,9
5 2,1
6 3,9
7 5,4
8 9,2
9 6,1
FL
Tue, Dec 16, 2014 Mon, Dec 15, 2014 Sun, Dec 14, 2014 Sat, Dec 13, 2014 Drawing Date Pick 3 Pick 4 Midday Evening Midday Evening 4-6-0 6-6-2 5-5-3-0 8-5-8-4 2-8-6 7-4-2 9-3-5-5 9-6-2-6 2-4-4 9-7-3 5-8-0-8 2-1-7-6 5-9-8 6-2-1 9-4-2-5 2-7-5-0 3-1-3 0-1-9 9-0-2-0 4-9-3-3 6-6-8 1-7-9 5-2-6-9 3-7-5-4 9-2-8 4-2-1 8-0-7-7 2-8-8-6 1-4-7 5-2-5 1-1-9-2 7-8-5-3 3-3-7 8-3-5 7-2-4-9 2-7-6-9 9-5-6 1-5-7 1-5-6-3 7-4-6-1
Draw trigger pos 1 digits Exact hit span (combined draws) type D draws
956 D 9 6,1 157,147 4 147, 668 (+4)
Cost(local)>800 , prize 1000, other outlet > cost 800, prize 1800, ROI ???
157E 1 0,6 019,621 type E +6
cost(local) 1200, prize 1000 ,, other 1200, 1800
337D(day) 3 7,6 668 type D( span) 3
cost 600, prize 500, cost 600, prize 900
835 E 8 9,2 973 xxxxxx,
147 1 0,6 668
The picks are larger for local wage, can we turn this into 'Positional pairs' for reduced picks?
nj
United States
Member #145657
August 10, 2013
974 Posts
Offline
Posted: December 17, 2014, 2:32 pm - IP Logged
NB> For strategic wager with ROI in mind, setting N=10 is not big deal so far as ' degree of certainty' is above 50%, ponder on this if you have time.
I recalled your ' deviation' system a while ago with 27 picks, which was pretty much for Straight betting, so I presumed analysing digits for position 1 and 2 wouldn't be an issue.Imagine having a code that can predict the ' next exact pairs' in range of 10 draws, this will indicate a reduction to just 20 picks for straight bet.
I started this ideal with just digits for ' next positions one' and the hit rate is over 50%, but the picks are larger for waging locally. If you can predict position one, then remaining pairs becomes 10C2 or 10P2.
10C2= 10!/2!8!= 45 box pairs
10P2>90 permutations without repeat digit
10P2>100 permutations with a repeat digit
Since we're targeting exact hits, permutation is our focus. focus should be on the ideal not the larger picks, the picks can be reduced later. Pick 3 or 4 has 10 members (0123456789), so setting N=10 , will give the above pairs, but you can reduced N(many ways out there, if you can analysis recent digit trends).
See this link , you may see something overlooked.
I think if you want to b informative with your method we can find it under your posts...why i have to waste a few seconds becouse of your "degree of certainty " belief which is worthless..i actually know the guy a little more who invented this 85% BS(it works in 15% of cases and thats not gopd enough !!!) and he is trying to hard but he is good in certain aspects of the game...and i can defineyly tell you ...it is crap.
the DOC misleading parameter can stay in maybe 15-35% of cases at 99.999999% for many weeks,months and are you reimbursing amateurs players for believeing into this false ipotezis ??
Florida
United States
Member #135615
November 27, 2012
405 Posts
Offline
Posted: December 18, 2014, 1:40 pm - IP Logged
This is the other example, using 20 sets. Not sure why this works so well but it does. Again, it relies on doubles. Once a double hits, you do a workout that gives you twenty sets. One of those numbers will hit, typically boxed, within 10 days (or twenty draws) 91% of the time. I've tested twenty years worth of lottery results for FL and varying amount of time for a few other states (GA, NY, NC, MI, TX) with similar results. Except for CA, that state had a massive hike in # of doubles, need to look into that. Here's the workout. It's worth noting that I don't have much data for NC, MI and TX.
Example #1: If the double is a 1, it would look like this.
Position A would be a 1 (the double) and a 6 (the mirror of one)
Position B would be 0 (double minus 1) and a 2 (double plus 1)
Position C would be 7 (double minus 4), 6 (double minus 3), 1 (double), 3 (double plus 2), and 4 (double plus 3)
Example #2: If the double is an 8.
Position A: 8, 3
Position B: 7, 9
Position C: 4, 3, 8, 0, 1
The position A workout will probably make sense to most of you. Position B might seem weird but looks like it might work. Position C, I have no explanation for. I just ran a script to see what worked best and that was it. That script only tested the 5 combinations for the third position. This is why I've been trying to write a scrip that will test all of the positions. From there, I'd like to try testing different numbers played for each position. For instance, instead of Pos A being two numbers, Pos B having 2, and Pos C having 5. I'd like to do a 1/3/3, 1/2/5, or 2/2/2. I'd love to see those results since a twenty set list can still hit so often.
Now, of course, everyone is still thinking this is crazy because if you play 20 numbers for 10 days (20 draws), or even if you get a hit after 10 draws [and stop playing], and get a box hit, that's a nice loss. I can't disagree with you there but that's not the point. This point is, if just trying to hit box, you start with a list of 333 sets. One of those numbers will hit. There's a 100% chance of it but you have to pick 1 out of 333. On the other hand, the above workout will reduce the set list down to 20 and still have a 90%+ chance to hit. Although you have to play for 10 days. I'd rather have the 20 sets.
I plan to keep working on this to see if I can find a way to play just one set (six ways) and still hit often enough. Or maybe I play 10 numbers (one way) and find a way to order them. Everyone has their method for attacking this lottery thing. This is mine. :)
Park City, UT
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January 18, 2009
993 Posts
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Posted: December 18, 2014, 3:05 pm - IP Logged
The math says that any randomly selected 20 single box combinations that equate to 120 straight combinations should hit within 19.75 draws 92% of the time. Shouldn't matter if a double hit or not as a trigger.
Jimmy
Florida
United States
Member #135615
November 27, 2012
405 Posts
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Posted: December 18, 2014, 3:15 pm - IP Logged
The math says that any randomly selected 20 single box combinations that equate to 120 straight combinations should hit within 19.75 draws 92% of the time. Shouldn't matter if a double hit or not as a trigger.
Jimmy
I can show you combinations that have a horrible hit percentage. I've run a lot of tests but I think one was in the teens.
Park City, UT
United States
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January 18, 2009
993 Posts
Offline
Posted: December 18, 2014, 3:21 pm - IP Logged
I can show you combinations that have a horrible hit percentage. I've run a lot of tests but I think one was in the teens.
For those horrible hit percentages how many straight combinations do those 20 box combinations equate to? If its less than 120 then I would expect them to perform worse than every 19.75 draws.
Jimmy
Florida
United States
Member #135615
November 27, 2012
405 Posts
Offline
Posted: December 18, 2014, 3:27 pm - IP Logged
For those horrible hit percentages how many straight combinations do those 20 box combinations equate to? If its less than 120 then I would expect them to perform worse than every 19.75 draws.
Jimmy
Honestly don't know. Didn't look into that at the time because the numbers weren't high... Wish I had kept that version of the script... :(
Park City, UT
United States
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January 18, 2009
993 Posts
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Posted: December 18, 2014, 3:32 pm - IP Logged
As a test when a double hits instead of using whatever formula you currently are using to generate your 20 combinations what if you instead took the last 6 unique digits that have hit and box them into their 20 combinations. Those will equate to 120 straight combinations. If your formula outperforms that algorithm for all states over a statistically significant number of years then you have indeed found a killer formula.
Jimmy
Florida
United States
Member #135615
November 27, 2012
405 Posts
Offline
Posted: December 18, 2014, 3:40 pm - IP Logged
heh, believe it or not, I'm doing that right now. I've got to where it inserts the 120 combinations into a table. From there, I'll play with that data to see which of each set of 120 combinations hit. Then try to figure out why or at the least, a common denominator that hits the most... As always, it's a work in progress.
United States
Member #128790
June 2, 2012
5431 Posts
Offline
Posted: December 22, 2014, 3:54 pm - IP Logged
Nice Job,
Florida,
0: 089, 039, 058, 019, 049, 088, 059, 048, 099, 079
1: 109, 129, 102, 182, 169, 176, 112, 178, 137, 147
2: 268, 208, 210, 218, 230, 260, 258, 200, 263, 288
3: 396, 376, 316, 346, 397, 395, 385, 315, 398, 391
4: 460, 450, 436, 430, 466, 456, 457, 417, 492, 432
5: 597, 567, 569, 537, 547, 506, 583, 513, 543, 541
6: 684, 634, 614, 638, 654, 624, 618, 613, 644, 698
7: 734, 784, 714, 738, 731, 764, 754, 716, 702, 774
8: 861, 801, 851, 836, 856, 866, 847, 897, 826, 846
9: 935, 925, 955, 933, 923, 965, 928, 948, 912, 922
Wed, Dec 3, 2014 3-1-6?
Thu, Nov 27, 2014 3-5-1?
Thu, Nov 13, 2014 1-3-5? Wed, Nov 12, 2014 3-3-9?
Sat, Nov 15, 2014 3-3-4?
Wed, Dec 3, 2014 3-1-6
Fri, Nov 21, 2014 5-6-4?
Sun, Nov 16, 2014 4-6-6?
Tue, Dec 9, 2014 1-4-7?
Fri, Dec 12, 2014 0-1-9
Fri, Nov 21, 2014 0-1-1?
Mon, Nov 24, 2014 1-1-8?
Sun, Dec 7, 2014 9-5-6?
Mon, Dec 8, 2014 8-3-5?
Sun, Nov 30, 2014 5-4-5?
Tue, Dec 9, 2014 1-4-7
Tue, Dec 2, 2014 4-4-0?
Sat, Nov 29, 2014 0-4-4?
Mon, Dec 15, 2014 2-8-6?
Sat, Dec 6, 2014 2-8-0
Wed, Dec 3, 2014 2-2-5
Mon, Dec 1, 2014 2-2-8?
Tue, Dec 9, 2014 1-4-7?
Fri, Dec 5, 2014 7-7-7?
This is an update for Florida. I'm still waiting for the numbers in the 6 line to show.
Since my last post, there have been a couple of more hits.
I also noticed something else. A lot of the new hits are showing on the chart adjacent to an existing winner. As an example, look at the recent winner 457. It is located right next to the 417. So the next time there's a double 00, I may play the adjacent numbers to 097, like 099. Another example is, if a double 11 shows, I may play the adjacent numbers like 129 and 137.
The last example would be, if a double 22 shows, I'd play 263, and 288 because they are adjacent to the 147, and 258 and 200 because they are adjacent to the 315. This may be a good way to reduce costs.
So for tonight, I'm still waiting on one of the numbers from line 6. The adjacent numbers to the existing winners in red are 638-624-618, and 644.
Sun, Dec 14, 2014 2-4-4
Thu, Dec 18, 2014 0-0-7 4-1-7
Mon, Dec 22, 2014 5-7-4
Thu, Dec 18, 2014 0-0-7
Sun, Dec 21, 2014 0-9-7
0: 089, 039, 058, 019, 049, 088, 059, 048, 099, 079
1: 109, 129, 102, 182, 169, 176, 112, 178, 137, 147
2: 268, 208, 210, 218, 230, 260, 258, 200, 263, 288
3: 396, 376, 316, 346, 397, 395, 385, 315, 398, 391
4: 460, 450, 436, 430, 466, 456, 457, 417, 492, 432
5: 597, 567, 569, 537, 547, 506, 583, 513, 543, 541
6: 684, 634, 614, 638, 654, 624, 618, 613, 644, 698
7: 734, 784, 714, 738, 731, 764, 754, 716, 702, 774
8: 861, 801, 851, 836, 856, 866, 847, 897, 826, 846
9: 935, 925, 955, 933, 923, 965, 928, 948, 912, 922
Florida
United States
Member #135615
November 27, 2012
405 Posts
Offline
Posted: December 25, 2014, 10:53 pm - IP Logged
Nice catch on the adjacent thing, I hadn't noticed that. Might be worth looking into more.
Still, it's silly how it hits often enough even though the numbers don't change, right? I thought it was weird so I wanted to get more opinions/testing on it. Thanks for the info! :)
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# how long to cook meat at 400
Rate this post
How long should I cook a steak at 400? If you’re cooking your steak in a skillet, sear both sides until nicely browned, then finish the ribeye steak in the oven for about 8 minutes at 400 degrees Fahrenheit for a 1-inch steak .
## How to calculate the cooking time?
When you cook a food in the oven (steak, roast, chicken, turkey, cake, etc.), the heat spreads from above and below. In this case, the distance the heat has to travel is half its thickness. The thermal diffusion law therefore becomes t = L2/4D where L is the total thickness of the food.
## How to calculate the cooking time of a roast?
Count: 12 to 15 minutes per pound (1 pound = 500 grams) to cook roast beef. The core temperature when it comes out of the oven should be 50 to 55°C for medium rare, or 58 to 60°C for medium. 20 to 25 minutes per pound for roast veal.
## How to calculate the cooking time of meat?
Roast beef cooking time per kilo
Time per kilo allow: 15.5 min per pound or 500 g for blue cooking. Time per kilo allow: 18.5 min per pound or 500 g for rare cooking. Time per kilo allow: 22 to 25 min per pound or 500 g for medium-rare cooking.
Read more How Long To Cook 10Lbs Beef
## How to cook a 500 gram roast beef?
For blue cooking, allow 15.5 minutes per 500 g of roast beef. For rare cooking, 18.5 minutes per 500 g and finally for medium cooking, 22 to 25 minutes per 500 g of roast beef.
## What temperature for the oven for a roast beef?
“Blue” roast beef in the oven at 235°C
Its core temperature is 15°C when it is put in the oven, about 2 hours after taking it out of the fridge.
## What temperature for rare roast beef?
To ensure perfect cooking, it is best to use a kitchen thermometer: know that the ideal temperature of the heart (in other words rare) is 55°C. For those who like blue cooking, bet on a temperature of 50°C. And if you want it cooked to perfection, wait until it reaches 60°C.
## How much does a roast beef weigh for 10 people?
After the tenderloin, the sirloin is the best and most tender cut of beef. How much roast for 10 people? it all depends on the weight of the roasts; count about 200g of meat per person, if your butcher makes roasts of 3kgs, and if you have the appropriate casserole…
## How to cook red meat?
For cooking: Rare: 10 to 15 minutes per pound. Medium: 15 to 20 minutes per pound. Well done: 20 to 25 minutes per pound.
## What is the temperature of the low heat?
What is the temperature of the low heat? Chefs debate the most appropriate temperature for simmering, with some saying that this low-heat method should not exceed 82°C (180°F).
Read more What Kind Of Pan To Cook Beef Jerky In The Oven
## What is the temperature of a healthy human?
The reference value for a normal body temperature in a human being is 37°C.
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# Triangle ABC is an equilateral triangle. The angle bisectors and the perpendicular bisectors meet at D in such a way…
Triangle ABC is an equilateral triangle. The angle bisectors and the perpendicular bisectors meet at D in such a way that CD=2DE. The radius of the inscribed circle is ___units and the radius of the circumscribed circle is ___ units. AC=10 and CE=8.7
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# chemistry
posted by on .
need to calculate the mass (m ) of NaoH with C% = 15 % , to balance 176 g of H2SO4 so i though to use this method C% = mass of h2s04 / mass of h2so4 + mass of Naoh which is unknown mass of Naoh = c% x mass of h2so4 / 100 is this the right method or not ?
• chemistry - ,
I can't follow your thought, mostly because I don't know the definition of your terms, principally C%. The second question I have is about the 15%. Is that 15% w/w or w/v? Third, you don't have a volume of NaOH being added and you don't need one if you are just looking for the mass. Use stoichiometry to solve your problem.
First you should write and balance the equation.
2NaOH + H2SO4 ==> 2H2O + Na2SO4
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# Quick Answer: What Percentage Is A Number?
## How find the percentage of a number?
If you want to know what percent A is of B, you simple divide A by B, then take that number and move the decimal place two spaces to the right.
To use the calculator, enter two numbers to calculate the percentage the first is of the second by clicking Calculate Percentage..
## What is percentage formula?
Use the percentage formula: P% * X = Y. Example: What is 10% of 150? Convert the problem to an equation using the percentage formula: P% * X = Y. P is 10%, X is 150, so the equation is 10% * 150 = Y. Convert 10% to a decimal by removing the percent sign and dividing by 100: 10/100 = 0.10.
## How do I do a percentage formula in Excel?
Calculating percentages As with any formula in Excel, you need to start by typing an equal sign (=) in the cell where you want your result, followed by the rest of the formula. The basic formula for calculating a percentage is =part/total.
## How do you find 80% of a number?
How to find 80% of a number? Take the number and multiple it by 80. Then multiply that by . 01.
## Is 3 a evil number?
Examples. The first evil numbers are: 0, 3, 5, 6, 9, 10, 12, 15, 17, 18, 20, 23, 24, 27, 29, 30, 33, 34, 36, 39 …
## What number is 0.4% of 20?
0.08What is 0.4 percent (calculated percentage %) of number 20? Answer: 0.08.
## What number is 45% of 50?
Latest decimal numbers, fractions, rations or proportions converted to percentages45 / 50 = 90%Jan 01 12:59 UTC (GMT)7.81 / 10 = 78.1%Jan 01 12:59 UTC (GMT)- 0.529 = – 52.9%Jan 01 12:59 UTC (GMT)5.86 / 3 = 195.333333333333%Jan 01 12:59 UTC (GMT)825 / 20 = 4,125%Jan 01 12:59 UTC (GMT)9 more rows
## How do you increase a number by a percentage?
If you want to increase a number by a certain percentage, follow these steps:Divide the number you wish to increase by 100 to find 1% of it.Multiply 1% by your chosen percentage.Add this number to your original number.There you go, you have just added a percentage increase to a number!
## What is the difference between 2 numbers?
As stated before, finding the difference between two numbers, is just to subtract them. So if you want to find the difference, you take the bigger one minus the smaller one. But if you want to find the distance between two number, you use the absolute value.
## How do I find the percentage of two numbers without a calculator?
If you need to find a percentage of a number, here’s what you do – for example, to find 35% of 240: Divide the number by 10 to find 10%. In this case, 10% is 24. Multiply this number by how many tens are in the percentage you’re looking for – in this case, that’s 3, so you work out 30% to be 24 x 3 = 72.
## What is 1 as a percent?
Example ValuesPercentDecimalFraction1%0.011/1005%0.051/2010%0.11/1012½%0.1251/812 more rows
## What is the percentage between two numbers?
The percentage difference between two values is calculated by dividing the absolute value of the difference between two numbers by the average of those two numbers. Multiplying the result by 100 will yield the solution in percent, rather than decimal form.
## How do you find 2% of a number?
To find 2% of a quantity, first find 1% of it, then double that. For example, let’s find 2% of \$6. Since 1% of 6 is \$0.06, then 2% of 6 is \$0.12.
## How do I calculate percentage on calculator?
To calculate the percentage, multiply this fraction by 100 and add a percent sign. 100 * numerator / denominator = percentage . In our example it’s 100 * 2/5 = 100 * 0.4 = 40 .
## What is discount formula?
Find the original price (for example \$90 ) Get the the discount percentage (for example 20% ) Calculate the savings: 20% of \$90 = \$18. Subtract the savings from the original price to get the sale price: \$90 – \$18 = \$72.
## How do you calculate 20 percent?
First, convert the percentage discount to a decimal. A 20 percent discount is 0.20 in decimal format. Secondly, multiply the decimal discount by the price of the item to determine the savings in dollars. For example, if the original price of the item equals \$24, you would multiply 0.2 by \$24 to get \$4.80.
## What number is 25 percent of 80?
20Percentage Calculator: What is 25 percent of 80? = 20.
## How do you convert to percentage?
To convert a number into percent multiple it by 100 and then add the percent sign. These examples convert the numbers 23 and 158 to percents. To convert a number with a decimal into percent, multiply it by 100 and add the percent sign.
## What is 3% of a number?
How to find 3% of a number? Take the number and multiple it by 3. Then multiply that by . 01.
## How can you find 75% of any number?
How to find 75% of a number? Take the number and multiple it by 75. Then multiply that by . 01.
## What is 1/6 as a percentage?
Common Fractions with Decimal and Percent EquivalentsFractionDecimalPercent1/60.1666…16.666…%5/60.8333…83.333…%1/80.12512.5%3/80.37537.5%21 more rows•Feb 21, 2017
## What number is 2% of 100?
Percentage Calculator: What is 2 percent of 100? = 2.
## What number is 75% of 15?
Latest calculated numbers percentages75% of 15 = 11.25Jan 01 04:28 UTC (GMT)18.32% of 100 = 18.32Jan 01 04:27 UTC (GMT)- 63% of 9,888 = – 6,229.44Jan 01 04:27 UTC (GMT)0.1% of 75 = 0.075Jan 01 04:27 UTC (GMT)10% of 7.79 = 0.779Jan 01 04:27 UTC (GMT)9 more rows
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taylorialcom/ DataStructures
# Binary Search Trees
Left child smaller, right child bigger.
Briefly, we applied binary search by recursively partitioning a sorted array into two pieces, doing one comparison, and then eliminating one (or both, if we found what we were looking for) partition.
Now lets consider a slightly different way of drawing the sorted array:
• This is actually much more than just redrawing the sorted array in a different way.
• You'll notice that as you work your way from left to right, the elements in the lower figure are still in sorted order.
• The bottom part of the figure represents a binary search tree.
• A binary search tree must adhere to the following rules:
• There is one element at the top, called the root
• Each element can have at most two elements directly below it, called children
• The left child must be less than its parent
• The right child must be greater than its parent1
• Performing a binary search on a binary search tree is a simple matter of recursively working our way down the tree.
• If the element we are looking for is less than the current element, we go to the left child
• If the element we are looking for is greater than the current element, we go to the right child
• Otherwise, we have found our element and we are ready to stop
• If we make it to the bottom of the tree without encountering the value we are looking for, we can be certain that it is not present in our data structure.
## Tree Terminology
• A tree is said to be a binary tree if each element can have zero, one, or two children.
• Each element in the tree is called a node.
• The top-most node is called the root.
• The root and all of the nodes connected to it are called a tree.
• Any node and all of the nodes connected below it are called a subtree.
• A node directly above another node is called its parent.
• A node directly below and to the left of another node is called its left child.
• A node directly below and to the right of another node is called its right child.
• A tree is said to be a binary search tree if every left child is smaller than its parent and every right child is larger than its parent.
• A node with no children is called a leaf.
• The height of the tree is typically defined the number of levels below the root.2 E.g., an empty tree has a height of -1, a tree with just one element has a height of 0, and tree with a root and one child has a height of 1.
• A tree is said to be full if all nodes have either two children or no children.
• A tree is said to be perfect if for every level except the bottom level, every node has two children.
• A tree is said to be complete if every level, except possibly the last, is completely filled, and all nodes are as far left as possible.
• There are various definitions for a balanced tree.3
• Some say that in order for a tree to be balanced the distance from the root to any leaf in the tree must differ by no more than one.
• A looser definition is also acceptable: A balanced tree is one in which the distance from the root to any leaf differs by no more than a fixed constant multiple. Often this multiple is two, implying that the longest distance from root to leaf may be no more than two times the shortest distance from root to leaf.
1
Are you wondering what happens if you have two elements with the same value? If not, are you wondering why you are not wondering that? If not, repeat previous statement. Same valued entries are not allowed in a binary search tree. I.e., the values of all elements in a binary search tree must be unique.
2
Although our current textbook defines it as one more than this.
3
Sometimes balanced trees are called height-balanced trees to more clearly specify what about the tree is balanced.
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Space of sections of a fibre bundle with non-compact base space
Let $\pi: E \rightarrow M$ be a fiber bundle over the manifold M and denote by $\Gamma(E)$ the space of smooth sections of $E$.
For compact $M$ it is well known (Hamilton 1982, Part II Corollary 1.3.9), that $\Gamma(E)$ (if not empty) is a (tame) Fréchet manifold with respect to the topology of uniform convergence of all derivates on compacta. E.g. the topology is given by seminorms (shown here for vector bundles): $$p_{i, K} (\phi) = \sum_{j=1}^i \text{sup}_{x \in K} |\phi^{(j)}(x)|.$$ Where the section $\phi$ is identified with its local representative $\phi: U \subset R^n \rightarrow R^m$ and the compact sets $K$ form a exhaustion of $U$. As for a paracompact manifold there exists a countable atlas, this procedure results in countable many seminorms. Thus $\Gamma(E)$ is a Fréchet manifold. (For the general case of a fiber bundle one has to invoke the tubular neighborhood theorem.)
I`m now interested in the case of non-compact base manifold $M$. To be honest, I do not see why the above construction fails then.
Supportive to this view, in section 2.2. of [1] the authors construct along the above lines a topology for non-compact $M$. But on the other hand in [2] the gauge group $\text{Gau}(P)$ (which is the group of sections of the associated bundle $P \times_G G$ to the principal bundle $P \rightarrow M$) is described only as a strict inductive limit of countable many Fréchet spaces and only for compact $M$ one has the simpler Fréchet structure on $\text{Gau}(P)$.
Where is the error here? Thanks!
[1] Čap, A. & Slovak, J. On multilinear operators commuting with Lie derivatives, eprint arXiv:dg-ga/9409005, 1994
[2] Smoothness of the action of the gauge transformation group on connections M. C. Abbati, R. Cirelli, A. Mania, and P. Michor, J. Math. Phys. 27, 2469 (1986), DOI:10.1063/1.527404
• I thought the gauge group was the space of sections of $M \times G$. $P \times_G G$ is naturally isomorphic to $P$ unless I've misunderstood. – Paul Reynolds Dec 16 '12 at 21:24
• Hmm maybe the fact is (I didn't read the articles, so I'm just guessing) that for a non compact base manifold, the Fréchet structures you obtain are not tamely equivalent... – Samuele Dec 16 '12 at 22:47
• @PR $G$ is acting on itself by conjugation, not translation in $P\times_G G$. So eg $P\times_G G$ always has sections, and $P$ never does if $P$ is non-trivial. – Paul Dec 17 '12 at 1:32
Your definition depends on the choice of the exhaustion and on the choice of the metric on $E$. To get a meaningful theory you have to add many more assumptions (like: a Riemannian metric of bounded geometry on $M$ where the open sets are geodesic balls ...). For example, if you want to let the diffeomorphism group of $M$ act smoothly on the locally convex space of functions you are defining. Just keep in mind how many different function spaces on $\mathbb R^n$ are useful.
See:
MR2343536 Eichhorn, Jürgen Global analysis on open manifolds. Nova Science Publishers, Inc., New York, 2007. x+644 pp.
for a careful development of Sobolev spaces on non-compact Riemannian manifolds and vector bundles on them.
EDIT: You might check around 10.10 in (this uses a clumsy version of calculus on locally convex spaces):
Peter W. Michor: Manifolds of differentiable mappings. Shiva Mathematics Series 3, Shiva Publ., Orpington, (1980), iv+158 pp., MR 83g:58009 (scanned pdf)
Or you can check chapter IX of:
Andreas Kriegl, Peter W. Michor: The Convenient Setting of Global Analysis. Mathematical Surveys and Monographs, Volume: 53, American Mathematical Society, Providence, 1997. (pdf)
Both references model on spaces of test functions (Choice 1 in the answer of Andrew Stacey below). There are other choices, but they are increasingly complicated. See for example the following paper which discusses the group of diffeomorphisms on $\mathcal R^n$ which fall rapidly towards the identity, or fall like $H^\infty$ (intersection of all Sobolev spaces).
Peter W. Michor and David Mumford: A zoo of diffeomorphism groups on $\mathbb ℝ^n$. arXiv:1211.5704. (pdf)
• Thank you for pointing out the dependence on the construction. Do you know any independence-results if I fix a (pseudo-)Riemannian metric (even Lorentzian manifold if necessary) on the base space and restrict to affine or vector bundles (to include the additive structure Andrew Stacey pointed out)? – Tobias Diez Jan 2 '13 at 10:52
• Another enquiry regarding the dependence on the exhaustion and the metric in the non-compact case, as I do not see it. I suspect you are referring to the seminorms $p_{K, k}(f) = sup_{x \in K}(|\nabla^i f (m) |)$. Let $(K_n)$ and $(L_i)$ be two different exhaustions by compact sets. By a standard argument each $L_i$ can be absorbed by one $K_n$, $\L_i \subseteq K_n$. Than trivially $p_{L_i} \leq p_{K_n}$ and the topology is independent on the choice of exhaustion. Furthermore, the metric always gets evaluated over compact sets and so each choice of metric should give equivalent topologies. – Tobias Diez Mar 26 '13 at 18:26
Your local neighbourhoods no longer have good structure.
For non-compact base you need to take a limit over compact subspaces. So when considering two sections, say $f$ and $g$, then to say that they are "close" is to say that there is a neighbourhood of $f$ containing $g$. This neighbourhood says "There is a compact subset, say $K$, of $M$ and an order, say $n$, such that the derivatives of $f$ and $g$ are close on $K$ up to order $n$."
This says absolutely nothing about what happens outside $K$. And this is a big problem because if you want to put the structure of a manifold on $\Gamma(E)$ then for $f$ and $g$ sufficiently close you need to be able to say what $f + g$ is. On $K$ then there's no problem because you make everything sufficiently small that you can use the manifold structure of $E$ to add $f$ and $g$ fibrewise. But to extend that to $x$ outside $K$ you need to be able to add $f(x)$ and $g(x)$ where these can take any values in the fibre at $x$, which equates to a fibrewise global addition structure on $E$ and in a general fibre bundle you don't have that.
Now, you could go for a different source for your local additive structure but you'd run in to the same sort of problem.
You have two options if you want to work with $\Gamma(E)$ as a smooth object:
1. Change your topology. You can split $\Gamma(E)$ into pieces where $f$ and $g$ are in the same piece if they agree off some compact subset. This then can be made into a manifold but it has uncountably many components.
2. Change your category. The space $\Gamma(E)$ is a perfectly well behaved generalised smooth space and can be treated very nicely in one of the many categories of such. It just isn't a manifold.
• Thanks for this clear and vivid explanation, I think I got the point! Do you know any results, if I have such a additive structure on $E$ available (e.g. vector or affine bundles)? – Tobias Diez Jan 2 '13 at 10:58
• If your fibres are vector/affine bundles then the space of sections will be a vector space/affine space and then you can treat it as such. Then (I think) you get a locally convex topology from your construction (if M is sigma-compact then the topology is Frechet, I think) so you have a locally convex topological vector space and you can work with that as a smooth space. So it is a manifold, but for slightly the wrong reason! – Andrew Stacey Jan 2 '13 at 12:08
In addition to the above references, you may be interested in taking a look at this paper:
P. Piccione and D. V. Tausk, "On the Banach differential structure for sets of maps on non-compact domains." Nonlinear Anal. 46 (2001), no. 2, Ser. A: Theory Methods, 245–265,
where they study how to introduce a Banach structure on sets of maps between a possibly non-compact topological space as domain and a smooth manifold as target. Of course the regularity discussed here is less than $C^\infty$, giving you a Banach structure instead of Frechet. This has obvious advantages and disadvantages; but regardless of the differences is possibly worth looking at, given that the non-compactness issues of the domain are dealt with successfully. Needless to say, once you have the desired structure on the set of maps, restricting to the particular case of sections of a bundle is straight-forward.
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# How to sign a message using RSA?
Assuming I already have a D, P, Q, etc of an RSA key: How do I now sign a message? If it matters – the message is around 100 bits.
I don't know much about cryptography, but I can get these numbers generated by a computer. Unfortunately, in my scenario – I can't use the built-in functions to use these numbers.
• Read PKCS#1, in particular don't ignore the part about padding. Apr 25 '12 at 20:01
• @CodeInChaos So you're saying it can't realistically be done in a simple manner (i.e. without spending a week studying cryptography), right? By the way, if you have some solution as to how I can create an activation key in a web hosting environment – where they have RSACryptoServiceProvider blocked – I'd be happy to hear about it. Apr 25 '12 at 20:14
• Look into the relevant Mono classes. With a bit of work, many of them work on .net, and the license is permissive. Apr 25 '12 at 20:23
• @ispiro Realistically, you can't make any code that's secure without thoroughly understanding the characteristics of the primitives you are using. That's why, RSA is a bad choice as primitive. If, for example, you can pick OpenSSL's EVP layer as your primitive, you'll barely have to know much about RSA, except that it's an asymmetric cipher than can both sign and encrypt. Apr 25 '12 at 23:08
## 1 Answer
IMO implementing RSA yourself is a bad idea. While textbook signing is pretty easy, if you have access to a BigInteger class, you also need to get the padding right. In some use-cases timing attacks are also a problem.
But if you want to go that route, PKCS #1 is the standard you need to implement. It details how the padding should look like.
The text-book RSA signing operation is pretty easy, in principle. Just calculate $m^d \mod n$. Unfortunately without correct use of hashing and padding it's very insecure i.e. you can't use the actual message as $m$, but the padded hash of the message.
I'd look into mono's crypto code. They have purely managed implementations of most .net crypto classes. These should work even if you don't have access to the built in classes.
A few interesting looking classes:
They use the MIT X11 license, which is a permissive open source license.
• There is another widely used standard for RSA signature, ISO/IEC 9796-2. It allows signature with message recovery, which allows extending the length of the signed message by little more than the size of the hash used (e.g. 22 bytes in the most common setup). Beware that mode 1 of this standard (by far the most common) is broken; the attack is impractical in most common setups.
– fgrieu
Apr 26 '12 at 20:04
• Thanks. As for BigInteger - See this in .NET 4.0. Apr 30 '12 at 14:36
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# Red Perimeter 1 – Extend the Pattern
Probability Level 1
Given $n \times n$ squares made out of $1 \times 1$ unit squares with the perimeter of the $n \times n$ square painted red, some of the $1 \times 1$ squares will be painted on 2 sides, some on only 1 side, and some won't be painted at all.
How many unit squares will be painted red on only 1 side in a $4 \times 4$ square (not pictured)?
This problem is part of Arron's set The Red Perimeter.
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# In tennis during a tiebreaker which box is served to?
Updated: 10/23/2022
Wiki User
14y ago
Both service boxes are used, but in a slightly different order than normal play. The player whose turn it would usually be to serve gets to serve first; he serves one point from the deuce court. After this, the players switch boxes and starting with the serving player's opponent, each player gets to serve two points in a row. The players switch service boxes after each point, just as in a regular game. They also must switch sides of the court every six total points (e.g., if the score is 4-2, the players must switch sides).
Tiebreakers are scored numerically: 1, 2, 3, etc., not"15," "30," "40."
Set tiebreakers (that is, tiebreakers played to determine the winner of a 6-6 set) are usually played first to 7 points by a margin of two; if the score should turn out to be 7-6 in a tiebreaker, the players must continue playing until one is ahead by two points (8-6, 12-10, and 24-22 are all valid ending tiebreak scores).
Wiki User
14y ago
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Q: In tennis during a tiebreaker which box is served to?
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Related questions
### A box sits in a room Inside the box are eight tennis balls Eight boys each take one tennis ball yet one tennis ball remains in the bo?
One of the boys leaves the ball in the box and takes the box with him.
### What special rules are in tennis?
Some special rules made in tennis is that if the ball bounces on your box you will be out.
### In Lawn tennis during service if the ball rubs the net and goes to the opponent side what do they term as?
As long as it lands in the correct box, it's considered a "let."
### What is area of tennis ground?
I am not sure but just call your local tennis place
### How many tennis balls can fit in a 2x2x2 box?
Since a tennis ball has a diameter that is greater than 2 inches, the answer is 0.
No
### What is it called when your tennis serve doesn't land in the proper area?
The umpire or line judge simply calls the ball out. They don't have to explain anything, but occassionally a player may ask the chair umpire to confirm the call. In major tournaments, there may be a challenge system in place that allows players to challenge a chair umpire's or line judge's call using replay cameras, but even then, the umpire doesn't have to explain anything, they just say whether the point will be replayed or they call a new score.
ATP World Tour
Tennis shoes?
### What is short tennis?
Short tennis is normally where you play a normal game of tennis, except in the service box/es. This is good for practising angled shots, and drop shots. Hope this helped! If this question helped you, please recommend me! =)
### Why is Chinese food served in a box?
Boxes are cheap and convenient for take out food.
### Who were the people behind table tennis?
British Soldiers in India and South Africa during the 1880's, played a game using a table, a row of books for the net, a cigar box for paddles, and a rounded cork for the ball.
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## ECE1236H Microwave and Millimeter-Wave Techniques. Lecture: Continuum and other transformers. Taught by Prof. G.V. Eleftheriades
February 12, 2016 ece1236
### Disclaimer
Peeter’s lecture notes from class. These may be incoherent and rough.
These are notes for the UofT course ECE1236H, Microwave and Millimeter-Wave
Techniques, taught by Prof. G.V. Eleftheriades, covering ch. 5 [1] content.
## Continuum transformer
A non-discrete impedance matching transformation, as sketched in fig. 1, is also possible.
../../figures/ece1236/taperedLinesFig1: fig. 1. Tapered impedance matching.
\label{eqn:continuumAndOtherTransformersCore:820}
\begin{aligned}
\Delta \Gamma
&= \frac{ (Z + \Delta Z) – Z }{(Z + \Delta Z) + Z} \\
&= \frac{\Delta Z}{2 Z}
\end{aligned}
\label{eqn:continuumAndOtherTransformersCore:840}
\Delta Z \rightarrow 0
\label{eqn:continuumAndOtherTransformersCore:860}
\begin{aligned}
d\Gamma
&= \frac{dZ}{2 Z} \\
&= \inv{2} \frac{d (\ln Z)}{dz} \\
&= \frac{Z_0}{Z} \frac{d (Z/Z_0)}{dz} \\
&= \frac{1}{Z} \frac{d Z}{dz}.
\end{aligned}
Hence as we did for multisection transformers, associate $$\Delta \Gamma$$ with $$e^{- 2j \beta z}$$ as sketched in fig. 2.
../../figures/ece1236/taperedLinesFig2: fig. 2. Reflection coefficient over an interval
assuming small reflections (i.e. $$Z(z)$$ is a slowly varying (adiabatic). Then
\label{eqn:continuumAndOtherTransformersCore:920}
\begin{aligned}
\Gamma(\omega)
&= \int_0^L e^{ -2 j \beta z} d\Gamma \\
&= \inv{2}
\int_0^L e^{ -2 j \beta z} \frac{d (\ln Z)}{dz} dz
\end{aligned}
This supplies the means to calculate the reflection coefficient for any impedance curve. As with the step impedance matching process, it is assumed that only first order reflections are of interest.
## Exponential taper
Let
\label{eqn:continuumAndOtherTransformersCore:620}
Z(z) = Z_0 e^{a z}, \qquad 0 < z < L subject to \label{eqn:continuumAndOtherTransformersCore:640} \begin{aligned} Z(0) &= Z_0 \\ Z(L) &= Z_0 e^{a L} = Z_{\textrm{L}}, \end{aligned} which gives $$\label{eqn:continuumAndOtherTransformersCore:660} \ln \frac{Z_{\textrm{L}}}{Z_0} = a L,$$ or $$\label{eqn:continuumAndOtherTransformersCore:680} a = \inv{L} \ln \frac{Z_{\textrm{L}}}{Z_0}$$ Also $$\label{eqn:continuumAndOtherTransformersCore:700} \frac{d}{dz} \ln \frac{Z_{\textrm{L}}}{Z_0} = \frac{d}{dz} (az) = a,$$ Hence \label{eqn:continuumAndOtherTransformersCore:740} \begin{aligned} \Gamma(\omega) &= \inv{2} \int_0^L e^{-2 j \beta z} \frac{d}{dz} \ln \frac{Z_{\textrm{L}}}{Z_0} dz \\ &= \frac{a}{2} \int_0^L e^{-2 j \beta z} dz \\ &= \frac{1}{2L} \ln \frac{Z_{\textrm{L}}}{Z_0} \evalrange{ \frac{e^{-2 j \beta z} }{ -2 j \beta} }{0}{L} \\ &= \frac{1}{2L \beta} \ln \frac{Z_{\textrm{L}}}{Z_0} \frac{ 1 - e^{-2 j \beta L} }{2 j} \\ &= \frac{1}{2} \ln \frac{Z_{\textrm{L}}}{Z_0} e^{-j \beta L} \frac{\sin( \beta L )}{\beta L}, \end{aligned} or $$\label{eqn:continuumAndOtherTransformersCore:940} \Gamma(\omega) = \frac{1}{2} \ln \frac{Z_{\textrm{L}}}{Z_0} e^{-j \beta L} \textrm{sinc}( \beta L ).$$
1. $$\beta$$ is constant with $$Z$$ varying: this is good only for TEM lines.
2. $$\Abs{\Gamma}$$ decreases with increasing length.
3. An electrical length $$\beta L > \pi$$, is required to minimize low frequency mismatch ($$L > \lambda/2$$).
This is sketched in fig. 3.
../../figures/ece1236/taperedLinesFig3: fig. 3. Exponential taper reflection coefficient.
Want:
\label{eqn:continuumAndOtherTransformersCore:760}
\beta L = \pi,
or
\label{eqn:continuumAndOtherTransformersCore:780}
\frac{\omega_c}{v_\phi} L = \pi
where $$\omega_c$$ is the cutoff frequency. This gives
\label{eqn:continuumAndOtherTransformersCore:800}
\omega_c = \frac{\pi v_\phi}{L}.
### Triangular Taper
\label{eqn:continuumAndOtherTransformersCore:960}
Z(z) =
\left\{
\begin{array}{l l}
Z_0 e^{2(z/L)^2 \ln(Z_{\textrm{L}}/Z_0)} & \quad \mbox{$$0 \le z \le L/2$$} \\
Z_0 e^{(4z/L – 2 z^2 – 1) \ln(Z_{\textrm{L}}/Z_0)} & \quad \mbox{$$L/2 \le z \le L$$} \\
\end{array}
\right.
\label{eqn:continuumAndOtherTransformersCore:980}
\frac{d}{dz} \ln (Z/Z_0) =
\left\{
\begin{array}{l l}
{(4z/L^2) \ln(Z_{\textrm{L}}/Z_0)} & \quad \mbox{$$0 \le z \le L/2$$} \\
{(4/L – 4z/L^2) \ln(Z_{\textrm{L}}/Z_0)} & \quad \mbox{$$L/2 \le z \le L$$} \\
\end{array}
\right.
In this case
\label{eqn:continuumAndOtherTransformersCore:1000}
\Gamma(\omega) = \frac{1}{2} e^{-\beta L} \ln \frac{Z_{\textrm{L}}}{Z_0} e^{-j \beta L} \textrm{sinc}^2( \beta L/2 ).
Compared to the exponential taper $$\textrm{sinc}( \beta L )$$ for the $$\beta L > 2 \pi$$ the peaks of $$\Abs{\Gamma}$$ are lower, but the first null occurs at $$\beta L = 2 \pi$$ whereas for the exponential taper it occurs at $$\beta L = \pi$$. This is sketched in fig. 4. The price to pay for this is that the zero is at $$2 \pi$$ so we have to make it twice as long to get the ripple down.
../../figures/ece1236/taperedLinesFig4: fig. 4. Triangular taper impedance curve.
### Klopfenstein Taper
For a given taper length $$L$$, the Klopfenstein taper is optimum in the sense that the reflection coefficient in the passband is minimum. Alternatively, for a given minimum reflection coefficient in the passband, the Klopfenstein taper yields the shortest length $$L$$.
Definition:
\label{eqn:continuumAndOtherTransformersCore:1020}
\ln Z = \inv{2} \ln (Z_0 Z_{\textrm{L}}) + \frac{\Gamma_0}{\cosh A} A^2 \phi(2 z/L -1, A), \qquad 0 \le z \le L,
where
\label{eqn:continuumAndOtherTransformersCore:1040}
\phi(x, A) = \int_0^x \frac{I_1(A\sqrt{1 – y^2})}{A \sqrt{1 – y^2}} dy, \qquad \Abs{x} \le 1.
Here $$I_1(x)$$ is the modified Bessel function. Note that
\label{eqn:continuumAndOtherTransformersCore:1060}
\begin{aligned}
\phi(0, A) &= 0 \\
\phi(x, 0) &= x/2 \\
\phi(1, A) &= \frac{\cosh A – 1}{A^2}
\end{aligned}
The resulting reflection coefficient is
\label{eqn:continuumAndOtherTransformersCore:1080}
\Gamma(\omega)
=
\left\{
\begin{array}{l l}
\Gamma_0 e^{-j \beta L} \frac{\cos\sqrt{(\beta L)^2 – A^2}}{\cosh A} & \quad \mbox{$$\beta L > A$$} \\
\Gamma_0 e^{-j \beta L} \frac{\cos\sqrt{A^2 – (\beta L)^2 }}{\cosh A} & \quad \mbox{$$\beta L < A$$} \\ \end{array} \right., where as usual $$\label{eqn:continuumAndOtherTransformersCore:1100} \Gamma_0 = \frac{Z_{\textrm{L}} - Z_0}{Z_{\textrm{L}} + Z_0} \approx \inv{2} \ln (Z_{\textrm{L}}/Z_0).$$ The passband is defined by $$\beta L \ge A$$ and the maximum ripple in the passband is $$\label{eqn:continuumAndOtherTransformersCore:1120} \Gamma_m = \frac{\Gamma_0}{\cosh A}.$$
### Example
Design a triangular taper, an exponential taper, and a Klopfenstein taper (with $$\Gamma_m = 0.02$$ ) to match a $$50 \Omega$$ load to a $$100 \Omega$$ line.
• Triangular taper:
\label{eqn:continuumAndOtherTransformersCore:1140}
Z(z) =
\left\{
\begin{array}{l l}
Z_0 e^{ 2(z/L)^2 \ln Z_{\textrm{L}}/Z_0 } & \quad \mbox{$$0 \le z \le L/2$$} \\
Z_0 e^{ (4 z/L – 2 z^2/L^2 – 1)\ln Z_{\textrm{L}}/Z_0 } & \quad \mbox{$$L/2 \ge z \ge L$$}
\end{array}
\right.
The resulting $$\Gamma$$ is
\label{eqn:continuumAndOtherTransformersCore:1160}
\Abs{\Gamma} = \inv{2} \ln (Z_{\textrm{L}}/Z_0) \textrm{sinc}^2\lr{ \beta L/2 }.
• Exponential taper:
\label{eqn:continuumAndOtherTransformersCore:1180}
\begin{aligned}
Z(z) &= Z_0 e^{a z}, \qquad 0 \le z \le L \\
a &= \inv{L} \ln (Z_{\textrm{L}}/Z_0) = \frac{0.693}{L} \\
\Abs{\Gamma} &= \inv{2} \ln (Z_{\textrm{L}}/Z_0) \textrm{sinc}( \beta L )
\end{aligned}
• Klopfenstein taper:
\label{eqn:continuumAndOtherTransformersCore:1200}
\begin{aligned}
Z(z) &= \inv{2} \ln (Z_{\textrm{L}}/Z_0) = 0.346 \\
A &= \cosh^{-1}\lr{ \frac{\Gamma_0}{\Gamma_m}} = \cosh^{-1}\lr{ \frac{0.346}{0.02}} = 3.543 \\
\Abs{\Gamma} &= \Gamma_0 \frac{\cos\sqrt{(\beta L)^2 – A^2}}{\cosh A},
\end{aligned}
The passband $$\beta L > A = 3.543 = 1.13 \pi$$. The impedance $$Z(z)$$ must be evaluated numerically.
To illustrate some of the differences, we are referred to fig. 5.21 [1]. It is noted that
1. The exponential taper has the lowest cutoff frequency $$\beta L = \pi$$. Then is the Klopfenstein taper which is close $$\beta L = 1.13 \pi$$. Last is the triangular with $$\beta L = 2 \pi$$.
2. The Klopfenstein taper has the lowest $$\Abs{\Gamma}$$ in the passband and meets the spec of $$\Gamma_m = 0.02$$. The worst $$\Abs{\Gamma}$$ in the passband is from the exponential taper and the triangular ripple is between the two others.
# References
[1] David M Pozar. Microwave engineering. John Wiley & Sons, 2009.
## ECE1236H Microwave and Millimeter-Wave Techniques. Lecture 7: Multisection quarter-wavelength transformers. Taught by Prof. G.V. Eleftheriades
### Disclaimer
Peeter’s lecture notes from class. These may be incoherent and rough.
These are notes for the UofT course ECE1236H, Microwave and Millimeter-Wave Techniques, taught by Prof. G.V. Eleftheriades, covering ch 5. [3] content.
## Terminology review
\label{eqn:uwavesDeck7MultisectionTransformersCore:20}
Z_{\textrm{in}} = R + j X
\label{eqn:uwavesDeck7MultisectionTransformersCore:40}
Y_{\textrm{in}} = G + j B
• $$Z_{\textrm{in}}$$ : impedance
• $$R$$ : resistance
• $$X$$ : reactance
• $$Y_{\textrm{in}}$$ : admittance
• $$G$$ : conductance
• $$B$$ : susceptance
Apparently this notation goes all the way back to Heavyside!
## Multisection transformers
Using a transformation of the form fig. 1 it is possible to optimize for maximum power delivery, using for example a matching transformation $$Z_{\textrm{in}} = Z_1^2/R_{\textrm{L}} = Z_0$$, or $$Z_1 = \sqrt{R_{\textrm{L}} Z_0}$$. Unfortunately, such a transformation does not allow any control over the bandwidth. This results in a pinched frequency response for which the standard solution is to add more steps as sketched in fig. 2.
../../figures/ece1236/Feb10Fig2: fig. 2. Pinched frequency response.
../../figures/ece1236/deck7Fig1: fig. 3. Single and multiple stage impedance matching.
This can be implemented in electronics, or potentially geometrically as in this sketch of a microwave stripline transformer implementation fig. 3.
../../figures/ece1236/deck7Fig3: fig. 3. Stripline implementation of staged impedance matching.
To find a multistep transformation algebraically can be hard, but it is easy to do on a Smith chart. The rule of thumb is that we want to stay near the center of the chart with each transformation.
There is however, a closed form method of calculating a specific sort of multisection transformation that is algebraically tractable. That method uses a chain of $$\lambda/4$$ transformers to increase the bandwidth as sketched in fig. 4.
../../figures/ece1236/deck7Fig4: fig. 4. Multiple $$\lambda/4$$ transformers.
The total reflection coefficient can be approximated to first order by summing the reflections at each stage (without considering there may be other internal reflections of transmitted field components). Algebraically that is
\label{eqn:uwavesDeck7MultisectionTransformersCore:60}
\Gamma(\Theta) \approx \Gamma_0
+ \Gamma_1 e^{-2 j \Theta} +
+ \Gamma_2 e^{-4 j \Theta} + \cdots
+ \Gamma_N e^{-2 N j \Theta},
where
\label{eqn:uwavesDeck7MultisectionTransformersCore:80}
\Gamma_n = \frac{Z_{n+1} – Z_n}{Z_{n+1} + Z_n}
Why? Consider reflections at the Z_1, Z_2 interface as sketched in fig. 5.
../../figures/ece1236/deck7Fig5: fig. 5. Single stage of multiple $$\lambda/4$$ transformers.
Assuming small reflections, where $$\Abs{\Gamma} \ll 1$$ then $$T = 1 + \Gamma \approx 1$$. Here
\label{eqn:uwavesDeck7MultisectionTransformersCore:100}
\begin{aligned}
\Theta
&= \beta l \\
&= \frac{2 \pi}{\lambda} \frac{\lambda}{4} \\
&= \frac{\pi}{2}.
\end{aligned}
at the design frequency $$\omega_0$$. We assume that $$Z_n$$ are either monotonically increasing if $$R_{\textrm{L}} > Z_0$$, or decreasing if $$R_{\textrm{L}} < Z_0$$.
### Binomial multisection transformers
Let
\label{eqn:uwavesDeck7MultisectionTransformersCore:120}
\Gamma(\Theta) = A \lr{ 1 + e^{-2 j \Theta} }^N
This type of a response is maximally flat, and is plotted in fig. 1.
../../figures/ece1236/multitransformerFig1: fig. 1. Binomial transformer.
The absolute value of the reflection coefficient is
\label{eqn:uwavesDeck7MultisectionTransformersCore:160}
\begin{aligned}
\Abs{\Gamma(\Theta)}
&=
\Abs{A} \lr{ e^{j \Theta} + e^{- j \Theta} }^N \\
&=
2^N \Abs{A} \cos^N\Theta.
\end{aligned}
When $$\Theta = \pi/2$$ this is clearly zero. It’s derivatives are
\label{eqn:uwavesDeck7MultisectionTransformersCore:180}
\begin{aligned}
\frac{d \Abs{\Gamma}}{d\Theta} &= -N \cos^{N-1} \Theta \sin\Theta \\
\frac{d^2 \Abs{\Gamma}}{d\Theta^2} &= -N \cos^{N-1} \Theta \cos\Theta N(N-1) \cos^{N-2} \Theta \sin\Theta \\
\frac{d^3 \Abs{\Gamma}}{d\Theta^3} &= \cdots
\end{aligned}
There is a $$\cos^{N-k}$$ term for all derivatives $$d^k/d\Theta^k$$ where $$k \le N-1$$, so for an N-section transformer
\label{eqn:uwavesDeck7MultisectionTransformersCore:140}
\frac{d^n}{d\Theta^n} \Abs{\Gamma(\Theta)}_{\omega_0} = 0,
for $$n = 1, 2, \cdots, N-1$$. The constant $$A$$ is determined by the limit $$\Theta \rightarrow 0$$, so
\label{eqn:uwavesDeck7MultisectionTransformersCore:200}
\Gamma(0) = 2^N A = \frac{Z_{\textrm{L}} – Z_0}{Z_{\textrm{L}} + Z_0},
because the various $$\Theta$$ sections become DC wires when the segment length goes to zero. This gives
\label{eqn:uwavesDeck7MultisectionTransformersCore:220}
\boxed{
A = 2^{-N} \frac{Z_{\textrm{L}} – Z_0}{Z_{\textrm{L}} + Z_0}.
}
The reflection coefficient can now be expanded using the binomial theorem
\label{eqn:uwavesDeck7MultisectionTransformersCore:240}
\begin{aligned}
\Gamma(\Theta)
&= A \lr{ 1 + e^{ 2 j \Theta } }^N \\
&= \sum_{k = 0}^N \binom{N}{k} e^{ -2 j k \Theta}
\end{aligned}
Recall that
\label{eqn:uwavesDeck7MultisectionTransformersCore:260}
\binom{N}{k} = \frac{N!}{k! (N-k)!},
providing a symmetric set of values
\label{eqn:uwavesDeck7MultisectionTransformersCore:280}
\begin{aligned}
\binom{N}{1} &= \binom{N}{N} = 1 \\
\binom{N}{1} &= \binom{N}{N-1} = N \\
\binom{N}{k} &= \binom{N}{N-k}.
\end{aligned}
Equating \ref{eqn:uwavesDeck7MultisectionTransformersCore:240} with \ref{eqn:uwavesDeck7MultisectionTransformersCore:60} we have
\label{eqn:uwavesDeck7MultisectionTransformersCore:300}
\boxed{
\Gamma_k = A \binom{N}{k}.
}
### Approximation for $$Z_k$$
From [1] (4.6.4), a log series expansion valid for all $$z$$ is
\label{eqn:uwavesDeck7MultisectionTransformersCore:320}
\ln z = \sum_{k = 0}^\infty \inv{2 k + 1} \lr{ \frac{ z – 1 }{z + 1} }^{2k + 1},
so for $$x$$ near unity a first order approximation of a logarithm is
\label{eqn:uwavesDeck7MultisectionTransformersCore:340}
\ln x \approx 2 \frac{x -1}{x+1}.
Assuming that $$Z_{k+1}/Z_k$$ is near unity we have
\label{eqn:uwavesDeck7MultisectionTransformersCore:360}
\begin{aligned}
\inv{2} \ln \frac{Z_{k+1}}{Z_k}
&\approx
\frac{ \frac{Z_{k+1}}{Z_k} – 1 }{\frac{Z_{k+1}}{Z_k} + 1} \\
&=
\frac{ Z_{k+1} – Z_k }{Z_{k+1} + Z_k} \\
&=
\Gamma_k.
\end{aligned}
Using this approximation, we get
\label{eqn:uwavesDeck7MultisectionTransformersCore:380}
\begin{aligned}
\ln \frac{Z_{k+1}}{Z_k}
&\approx
2 \Gamma_k \\
&= 2 A \binom{N}{k} \\
&= 2 \lr{2^{-N}} \binom{N}{k} \frac{Z_{\textrm{L}} – Z_0}{Z_{\textrm{L}} + Z_0} \\
&\approx
2^{-N} \binom{N}{k} \ln \frac{Z_{\textrm{L}}}{Z_0},
\end{aligned}
I asked what business do we have in assuming that $$Z_{\textrm{L}}/Z_0$$ is near unity? The answer was that it isn’t but surprisingly it works out well enough despite that. As an example, consider $$Z_0 = 100 \Omega$$ and $$R_{\textrm{L}} = 50 \Omega$$. The exact expression
\label{eqn:uwavesDeck7MultisectionTransformersCore:880}
\begin{aligned}
\frac{Z_{\textrm{L}} – Z_0}{Z_{\textrm{L}} + Z_0}
&= \frac{100-50}{100+50} \\
&= -0.333,
\end{aligned}
whereas
\label{eqn:uwavesDeck7MultisectionTransformersCore:900}
\inv{2} \ln \frac{Z_{\textrm{L}}}{Z_0} = -0.3466,
which is pretty close after all.
Regardless of whether or not that last approximation is used, one can proceed iteratively to $$Z_{k+1}$$ starting with $$k = 0$$.
### Bandwidth
To evaluate the bandwidth, let $$\Gamma_{\mathrm{m}}$$ be the maximum tolerable reflection coefficient over the passband, as sketched in fig. 6.
../../figures/ece1236/deck7Fig6: fig. 6. Max tolerable reflection.
That is
\label{eqn:uwavesDeck7MultisectionTransformersCore:400}
\begin{aligned}
\Gamma_m
&= 2^N \Abs{A} \Abs{\cos \Theta_m }^N \\
&= 2^N \Abs{A} \cos^N \Theta_m,
\end{aligned}
for $$\Theta_m < \pi/2$$. Then $$\label{eqn:uwavesDeck7MultisectionTransformersCore:420} \Theta_m = \arccos\lr{ \inv{2} \lr{ \frac{\Gamma_{\mathrm{m}}}{\Abs{A}}}^{1/N} }$$ The relative width of the interval is \label{eqn:uwavesDeck7MultisectionTransformersCore:440} \begin{aligned} \frac{\Delta f_{\mathrm{max}}}{f_0} &= \frac{\Delta \Theta_{\mathrm{max}}}{\Theta_0} \\ &= \frac{2 (\Theta_0 - \Theta_{\mathrm{max}}}{\Theta_0} \\ &= 2 - \frac{2 \Theta_{\mathrm{max}}}{\Theta_0} \\ &= 2 - \frac{4 \Theta_{\mathrm{max}}}{\pi} \\ &= 2 - \frac{4 }{\pi} \arccos\lr{ \inv{2} \lr{ \frac{\Gamma_{\mathrm{max}}}{\Abs{A}}}^{1/N} }. \end{aligned}
### Example
Design a 3-section binomial transformer to match $$R_{\textrm{L}} = 50 \Omega$$ to a line $$Z_0 = 100 \Omega$$. Calculate the BW based on a maximum $$\Gamma_{\textrm{m}} = 0.05$$.
### Solution
The scaling factor
\label{eqn:uwavesDeck7MultisectionTransformersCore:460}
\begin{aligned}
A
&= 2^{-N} \frac{Z_{\textrm{L}} – L_0}{Z_{\textrm{L}} + Z_0} \\
&\approx
\inv{2^{N+1}} \ln \frac{Z_{\textrm{L}}}{Z_0} \\
&= -0.0433
\end{aligned}
Now use
\label{eqn:uwavesDeck7MultisectionTransformersCore:940}
\ln \frac{Z_{n+1}}{Z_n}
\approx 2^{-N} \binom{N}{n} \ln \frac{R_{\textrm{L}}}{Z_0},
starting from
• $$n = 0$$.
\label{eqn:uwavesDeck7MultisectionTransformersCore:480}
\ln \frac{Z_{1}}{Z_0} \approx 2^{-3} \binom{3}{0} \ln \frac{R_{\textrm{L}}}{Z_0},
or
\label{eqn:uwavesDeck7MultisectionTransformersCore:500}
\begin{aligned}
\ln Z_{1}
&= \ln Z_0 + 2^{-3} \binom{3}{0} \ln \frac{R_{\textrm{L}}}{Z_0} \\
&= \ln 100 + 2^{-3} (1) \ln 0.5 \\
&= 4.518,
\end{aligned}
so
\label{eqn:uwavesDeck7MultisectionTransformersCore:520}
Z_1 = 91.7 \Omega
• $$n = 1$$
\label{eqn:uwavesDeck7MultisectionTransformersCore:540}
\ln Z_{2}
= \ln Z_1 + 2^{-3} \binom{3}{1} \ln \frac{50}{100} = 4.26
so
\label{eqn:uwavesDeck7MultisectionTransformersCore:560}
Z_2 = 70.7 \Omega
• $$n = 2$$
\label{eqn:uwavesDeck7MultisectionTransformersCore:580}
\ln Z_{3} = \ln Z_2 + 2^{-3} \binom{3}{2} \ln \frac{50}{100} = 4.0,
so
\label{eqn:uwavesDeck7MultisectionTransformersCore:600}
Z_3 = 54.5 \Omega.
With the fractional BW for $$\Gamma_m = 0.05$$, where $$10 \log_{10} \Abs{\Gamma_m}^2 = -26$$ dB}.
\label{eqn:uwavesDeck7MultisectionTransformersCore:920}
\begin{aligned}
\frac{\Delta f}{f_0}
&\approx
2 – \frac{4}{\pi} \arccos\lr{ \inv{2} \lr{ \frac{\Gamma_m}{\Abs{A}} }^{1/N} } \\
&=
2 – \frac{4}{\pi} \arccos\lr{ \inv{2} \lr{ \frac{0.05}{0.0433} }^{1/3} } \\
&= 0.7
\end{aligned}
At $$2$$ GHz, $$BW = 0.7$$ (70%), or $$1.4$$ GHz, which is the range $$[2.3,2.7]$$ GHz, whereas a single $$\lambda/4$$ transformer $$Z_T = \sqrt{ (100)(50) } = 70.7 \Omega$$ yields a BW of just $$0.36$$ GHz (18%).
# References
[1] DLMF. NIST Digital Library of Mathematical Functions. https://dlmf.nist.gov/, Release 1.0.10 of 2015-08-07. URL https://dlmf.nist.gov/. Online companion to Olver:2010:NHMF.
[2] F. W. J. Olver, D. W. Lozier, R. F. Boisvert, and C. W. Clark, editors. NIST Handbook of Mathematical Functions. Cambridge University Press, New York, NY, 2010. Print companion to NIST:DLMF.
[3] David M Pozar. Microwave engineering. John Wiley & Sons, 2009.
## Average power for circuit elements
February 9, 2016 ece1236 , , , , , , ,
In [2] section 2.2 is a comparison of field energy expressions with their circuit equivalents. It’s clearly been too long since I’ve worked with circuits, because I’d forgotten all the circuit energy expressions:
\label{eqn:averagePowerCircuitElements:20}
\begin{aligned}
W_{\textrm{R}} &= \frac{R}{2} \Abs{I}^2 \\
W_{\textrm{C}} &= \frac{C}{4} \Abs{V}^2 \\
W_{\textrm{L}} &= \frac{L}{4} \Abs{I}^2 \\
W_{\textrm{G}} &= \frac{G}{2} \Abs{V}^2 \\
\end{aligned}
Here’s a recap of where these come from
### Energy lost to resistance
Given
\label{eqn:averagePowerCircuitElements:40}
v(t) = R i(t)
the average power lost to a resistor is
\label{eqn:averagePowerCircuitElements:60}
\begin{aligned}
p_{\textrm{R}}
&= \inv{T} \int_0^T v(t) i(t) dt \\
&= \inv{T} \int_0^T \textrm{Re}( V e^{j \omega t} ) \Real( I e^{j \omega t} ) dt \\
&= \inv{4 T} \int_0^T
\lr{V e^{j \omega t} + V^\conj e^{-j \omega t} }
\lr{I e^{j \omega t} + I^\conj e^{-j \omega t} }
dt \\
&= \inv{4 T} \int_0^T
\lr{
V I e^{2 j \omega t} + V^\conj I^\conj e^{-2 j \omega t}
+ V I^\conj + V^\conj I
}
dt \\
&= \inv{2} \textrm{Re}( V I^\conj ) \\
&= \inv{2} \textrm{Re}( I R I^\conj ) \\
&= \frac{R}{2} \Abs{I}^2.
\end{aligned}
Here it is assumed that the averaging is done over some integer multiple of the period, which kills off all the exponentials.
### Energy stored in a capacitor
I tried the same sort of analysis for a capacitor in phasor form, but everything cancelled out. Referring to [1], the approach used to figure this out is to operate first strictly in the time domain. Specifically, for the capacitor where $$i = C dv/dt$$ the power supplied up to a time $$t$$ is
\label{eqn:averagePowerCircuitElements:80}
\begin{aligned}
p_{\textrm{C}}(t)
&= \int_{-\infty}^t C \frac{dv}{dt} v(t) dt \\
&= \inv{2} C v^2(t).
\end{aligned}
The $$v^2(t)$$ term can now be expanded in terms of phasors and averaged for
\label{eqn:averagePowerCircuitElements:100}
\begin{aligned}
\overline{{p}}_{\textrm{C}}
&= \frac{C}{2T} \int_0^T \inv{4}
\lr{ V e^{j \omega t} + V^\conj e^{-j \omega t} }
\lr{ V e^{j \omega t} + V^\conj e^{-j \omega t} } dt \\
&= \frac{C}{2T} \int_0^T \inv{4}
2 \Abs{V}^2 dt \\
&= \frac{C}{4} \Abs{V}^2.
\end{aligned}
### Energy stored in an inductor
The inductor energy is found the same way, with
\label{eqn:averagePowerCircuitElements:120}
\begin{aligned}
p_{\textrm{L}}(t)
&= \int_{-\infty}^t L \frac{di}{dt} i(t) dt \\
&= \inv{2} L i^2(t),
\end{aligned}
\label{eqn:averagePowerCircuitElements:140}
\overline{{p}}_{\textrm{L}}
= \frac{L}{4} \Abs{I}^2.
### Energy lost due to conductance
Finally, we have conductance. In phasor space that is defined by
\label{eqn:averagePowerCircuitElements:160}
G = \frac{I}{V} = \inv{R},
so power lost due to conductance follows from power lost due to resistance. In the average we have
\label{eqn:averagePowerCircuitElements:180}
\begin{aligned}
p_{\textrm{G}}
&= \inv{2 G} \Abs{I}^2 \\
&= \inv{2 G} \Abs{V G}^2 \\
&= \frac{G}{2} \Abs{V}^2
\end{aligned}
# References
[1] J.D. Irwin. Basic Engineering Circuit Analysis. MacMillian, 1993.
[2] David M Pozar. Microwave engineering. John Wiley & Sons, 2009.
February 8, 2016 Incoherent ramblings
I definitely enjoying the formal study of the courses I am taking, however, I find University courses inferior to self study in many ways. There are two specific benefits to a University course:
1) Having a knowledgeable instructor at hand to answer questions.
2) Having that same knowledge base available to set the study curriculum, since that instruction can direct attention to the most important aspects of the study.
The fixed formal lecture format is not terribly effective, at least for the scientific topics that I have been studying. In this day and age, when the technology to record lectures is so pervasive and freely available, there is really no excuse for the formal and fixed lecture style still found in the classroom. In an ideal learning environment, there would be time available between concepts to work through problems and gain complete understanding of each idea before going on to the next. Such an ideal learning environment would interleave practical work, text study and lectures. Testing should be used as a metric for whether or not the material is fully understood before continuing to the next aspect of study (or returning to areas of deficiency). In University classes testing is designed to produce a grade, not understanding, something that is completely backwards.
I am also surprised to see that grading for many graduate courses is still primarily based on exams. Having spent twenty years working before coming back to school it is a rude shock to come back to such an artificial metric for success. In nowhere other but school is it considered reasonable to make the metric for success based on how quickly a student can rush through a test, with no references at hand, attempting to construct answers that are optimized for maximal marks. In industry, the kind of mistakes that are inevitable in an examination context would lead to millions of dollars of service and product maintenance costs, and get people fired.
It is asinine that there is no feedback mechanism in place to review and retest of failed or partially successful final exam material. This seems to highlight the fact that university courses do not seem to be designed with learning as the primary goal. It is not clear what that goal is.
## ECE1236H Microwave and Millimeter-Wave Techniques. Lecture 3: Smith charts and impedance transformations. Taught by Prof. G.V. Eleftheriades
### Disclaimer
Peeter’s lecture notes from class. These may be incoherent and rough.
These are notes for the UofT course ECE1236H, Microwave and Millimeter-Wave Techniques, taught by Prof. G.V. Eleftheriades, covering [2] chap. 2 content.
## Short circuited line
A short circuited line, also called a shorted stub, is sketched in fig. 1.
../../figures/ece1236/deck5smithChartsAndImpedenceTxFig1: fig. 1. Short circuited line.
With
\label{eqn:uwavesDeck5SmithChartCore:20}
Z_{\textrm{L}} = 0,
the input impedance is
\label{eqn:uwavesDeck5SmithChartCore:40}
\begin{aligned}
Z_{\textrm{in}}
&= Z_0 \frac{ Z_{\textrm{L}} + j Z_0 \tan(\beta l) }{ Z_0 + j Z_{\textrm{L}} \tan(\beta l)} \\
&= j Z_0 \tan(\beta l)
\end{aligned}
For short line sections $$\beta l \ll \pi/2$$, or $$l \ll \lambda/4$$, the input impedance is approximately
\label{eqn:uwavesDeck5SmithChartCore:80}
\begin{aligned}
Z_{\textrm{in}}
&= j Z_0 \tan(\beta l) \\
&\approx j Z_0 \sin(\beta l) \\
&\approx j Z_0 \beta l
\end{aligned}
Introducing an equivalent inductance defined by $$Z_{\textrm{in}} = j \omega L_{\mathrm{eq}}$$, we have
\label{eqn:uwavesDeck5SmithChartCore:100}
\begin{aligned}
L_{\mathrm{eq}}
&=
\frac{Z_0}{\omega} \beta l \\
&=
\frac{Z_0}{\omega} \frac{\omega}{v_\phi} l \\
&=
\frac{Z_0 l}{v_\phi}.
\end{aligned}
The inductance per unit length of the line is $$C = Z_0/v_\phi$$. An application for this result is that instead of using inductors, shorted stubs can be used in high frequency applications.
This is also the case for short sections of high impedance line.
## Open circuited line
An open circuited line is sketched in fig. 2.
../../figures/ece1236/deck5smithChartsAndImpedenceTxFig2: fig. 2. Open circuited line.
This time with $$Z_{\textrm{L}} \rightarrow \infty$$ we have
\label{eqn:uwavesDeck5SmithChartCore:120}
\begin{aligned}
Z_{\textrm{in}}
&= Z_0 \frac{ Z_{\textrm{L}} + j Z_0 \tan(\beta l) }{ Z_0 + j Z_{\textrm{L}} \tan(\beta l)} \\
&= -j Z_0 \cot(\beta l).
\end{aligned}
This time we have an equivalent capacitance. For short sections with $$\beta l \ll \pi/2$$
\label{eqn:uwavesDeck5SmithChartCore:140}
Z_{\textrm{in}}
\approx
-j \frac{Z_0}{\beta l}
Introducing an equivalent capacitance defined by $$Z_{\textrm{in}} = 1/(j \omega C_{\mathrm{eq}})$$, we have
\label{eqn:uwavesDeck5SmithChartCore:160}
\begin{aligned}
C_{\mathrm{eq}}
&=
\frac{ \beta l}{\omega Z_0} \\
&=
\frac{ \omega/v_\phi l}{\omega Z_0} \\
&=
\frac{ l}{v_\phi Z_0}
\end{aligned}
The capacitance per unit length of the line is $$C = 1/(Z_0 v_\phi)$$.
This is also the case for short sections of low impedance line.
## Half wavelength transformer.
A half wavelength transmission line equivalent circuit is sketched in fig. 3.
../../figures/ece1236/deck5smithChartsAndImpedenceTxFig3: fig. 3. Half wavelength transmission line.
With $$l = \lambda/2$$
\label{eqn:uwavesDeck5SmithChartCore:180}
\begin{aligned}
\beta l
&= \frac{2 \pi}{\lambda} \frac{\lambda}{2} \\
&= \pi.
\end{aligned}
Since $$\tan \pi = 0$$, the input impedance is
\label{eqn:uwavesDeck5SmithChartCore:200}
\begin{aligned}
Z_{\textrm{in}}
&= Z_0 \frac{ Z_{\textrm{L}} + j Z_0 \tan(\beta l) }{ Z_0 + j Z_{\textrm{L}} \tan(\beta l)} \\
&= Z_{\textrm{L}}.
\end{aligned}
## Quarter wavelength transformer.
A quarter wavelength transmission line equivalent circuit is sketched in fig. 4.
../../figures/ece1236/deck5smithChartsAndImpedenceTxFig4: fig. 4. Quarter wavelength transmission line.
With $$l = \lambda/4$$
\label{eqn:uwavesDeck5SmithChartCore:220}
\begin{aligned}
\beta l
&= \frac{2 \pi}{\lambda} \frac{\lambda}{4} \\
&= \frac{\pi}{2}.
\end{aligned}
We have $$\tan \beta l \rightarrow \infty$$, so the input impedance is
\label{eqn:uwavesDeck5SmithChartCore:240}
\begin{aligned}
Z_{\textrm{in}}
&= Z_0 \frac{ Z_{\textrm{L}} + j Z_0 \tan(\beta l) }{ Z_0 + j Z_{\textrm{L}} \tan(\beta l)} \\
&= \frac{Z_0^2}{Z_{\textrm{L}}}.
\end{aligned}
This relation
\label{eqn:uwavesDeck5SmithChartCore:280}
\boxed{
Z_{\textrm{in}}
= \frac{Z_0^2}{Z_{\textrm{L}}},
}
is called the impedance inverter.
• A large impedance is transformed into a small one and vice-versa.
• A short becomes an open and vice-versa.
• A capacitive load becomes inductive and vice-versa.
• If $$Z_{\textrm{L}}$$ is a series resonant circuit then $$Z_{\textrm{in}}$$ becomes parallel resonant.
See [1] for an explanation of the term series resonant.
### Matching with a $$\lambda/4$$ transformer.
Matching for a quarter wavelength transmission line equivalent circuit is sketched in fig. 5.
../../figures/ece1236/deck5smithChartsAndImpedenceTxFig5: fig. 5. Quarter wavelength transmission line matching.
For maximum power transfer
\label{eqn:uwavesDeck5SmithChartCore:300}
Z_{\textrm{in}} = \frac{Z_0^2}{R_{\textrm{L}}} = R_{\textrm{G}},
so
\label{eqn:uwavesDeck5SmithChartCore:320}
Z_0 = \sqrt{ R_{\textrm{G}} R_{\textrm{L}} }.
We have
\label{eqn:uwavesDeck5SmithChartCore:340}
\Abs{\Gamma_{\textrm{L}}} = \frac{ R_{\textrm{L}} – Z_0 }{R_{\textrm{L}} + Z_0} \ne 0,
and still maximum power is transferred.
## Smith chart
A Smith chart is a graphical tool for making the transformation $$\Gamma \leftrightarrow Z_{\textrm{in}}$$. Given
\label{eqn:uwavesDeck5SmithChartCore:360}
Z_{\textrm{in}} = Z_0 \frac{ 1 + \Gamma }{ 1 – \Gamma },
where $$\Gamma = \Gamma_{\textrm{L}} e^{- 2 j \beta l }$$, we begin by normalizing the input impedance, using an overbar to denote that normalization
\label{eqn:uwavesDeck5SmithChartCore:380}
Z_{\textrm{in}} \rightarrow \overline{{Z}}_{\textrm{in}} = \frac{Z_{\textrm{in}}}{Z_0},
so
\label{eqn:uwavesDeck5SmithChartCore:400}
\begin{aligned}
\overline{{Z}}_{\textrm{in}}
&= \frac{ 1 + \Gamma }{ 1 – \Gamma } \\
&= \frac{ (1 + \Gamma_r) + j \Gamma_i }{ (1 – \Gamma_r) – j \Gamma_i } \\
&= \frac{ \lr{ (1 + \Gamma_r) + j \Gamma_i}\lr{(1 – \Gamma_r) + j \Gamma_i} }{ (1 – \Gamma_r)^2 + \Gamma_i^2 } \\
&= \frac{ (1 – \Gamma_r^2 – \Gamma_i^2) + j \Gamma_i (1 – \Gamma_r + 1 + \Gamma_r ) }{ (1 – \Gamma_r)^2 + \Gamma_i^2 } \\
&= \frac{ (1 – \Abs{\Gamma}^2) + 2 j \Gamma_i }{ (1 – \Gamma_r)^2 + \Gamma_i^2 }.
\end{aligned}
If we let $$\overline{{Z}}_{\textrm{in}} = \overline{{\Gamma}}_{\textrm{L}} + j \overline{{X}}_{\textrm{L}}$$, and equate real and imaginary parts we have
\label{eqn:uwavesDeck5SmithChartCore:420}
\begin{aligned}
\overline{{\Gamma}}_{\textrm{L}} &= \frac{ 1 – \Abs{\Gamma}^2 }{ (1 – \Gamma_r)^2 + \Gamma_i^2 } \\
\overline{{X}}_{\textrm{L}} &= \frac{2 \Gamma_i }{ (1 – \Gamma_r)^2 + \Gamma_i^2 }
\end{aligned}
It is left as an exercise to demonstrate that these can be rearranged into
\label{eqn:uwavesDeck5SmithChartCore:440}
\begin{aligned}
\lr{ \Gamma_r – \frac{\overline{{\Gamma}}_{\textrm{L}}}{1 + \overline{{\Gamma}}_{\textrm{L}} } }^2 + \Gamma_i^2 &= \lr{ \inv{1 + \overline{{\Gamma}}_{\textrm{L}} }}^2 \\
\lr{ \Gamma_r – 1 }^2 + \lr{ \Gamma_i – \inv{\overline{{X}}_{\textrm{L}} } }^2 &= \inv{\overline{{X}}_{\textrm{L}}^2},
\end{aligned}
which trace out circles in the $$\Gamma_r, \Gamma_i$$ plane, one for the real part of $$\Gamma$$ and one for the imaginary part. This provides a graphical way for implementing the impedance transformation.
### Real impedance circle
The circle for the real part is centered at
\label{eqn:uwavesDeck5SmithChartCore:580}
\lr{ \frac{\overline{{\Gamma}}_{\textrm{L}}}{1 + \overline{{\Gamma}}_{\textrm{L}} }, 0 },
\label{eqn:uwavesDeck5SmithChartCore:600}
\inv{1 + \overline{{\Gamma}}_{\textrm{L}} }.
All these circles pass through the point $$(1,0)$$, since
\label{eqn:uwavesDeck5SmithChartCore:620}
\begin{aligned}
\frac{\overline{{\Gamma}}_{\textrm{L}}}{1 + \overline{{\Gamma}}_{\textrm{L}} } + \inv{1 + \overline{{\Gamma}}_{\textrm{L}} }
&=
\frac{1 + \overline{{\Gamma}}_{\textrm{L}}}{1 + \overline{{\Gamma}}_{\textrm{L}} } \\
&= 1.
\end{aligned}
For reactive loads where $$\overline{{\Gamma}}_{\textrm{L}} = 0$$, we have $$\Gamma_r^2 + \Gamma_i^2 = 1$$, a circle through the origin with unit radius.
For matched loads where $$\overline{{\Gamma}}_{\textrm{L}} = 1$$ the circle is centered at $$(1/2, 0)$$, with radius $$1/2$$.
### Imaginary impedance circle
The circle obtained by equating imaginary parts are constant reactance circles with center
\label{eqn:uwavesDeck5SmithChartCore:640}
\lr{ 1, \inv{\overline{{X}}_{\textrm{L}} } },
\label{eqn:uwavesDeck5SmithChartCore:660}
\inv{\overline{{X}}_{\textrm{L}}}.
These circles also pass through the point $$(1,0)$$. These circles are orthogonal to the constant resistance circles. Some of the features of a Smith chart are sketched in fig. 7.
../../figures/ece1236/deck5smithChartsAndImpedenceTxFig7: fig. 7. Hand sketched Smith chart.
A matlab produced blank Smith chart can be found in fig. 1.
../../figures/ece1236/smithchartFig1: fig. 1. Blank Smith chart.
### Example: Perform a transformation along a lossless line.
../../figures/ece1236/deck5smithChartsAndImpedenceTxFig8: fig. 8. Impedance transformation along lossless line.
Given
\label{eqn:uwavesDeck5SmithChartCore:700}
\overline{{Z}} = \frac{1 + \Gamma}{1 – \Gamma},
\label{eqn:uwavesDeck5SmithChartCore:720}
\Gamma = \Gamma_{\textrm{L}} e^{-2 j \beta l},
and
\label{eqn:uwavesDeck5SmithChartCore:740}
\Gamma_{\textrm{L}} = \Abs{\Gamma_{\textrm{L}}} e^{j \Theta_{\textrm{L}} }
The total reflection coefficient is
\label{eqn:uwavesDeck5SmithChartCore:760}
\Gamma = \Abs{\Gamma_{\textrm{L}}} e^{j (\Theta_{\textrm{L}} – 2 \beta l) }
If $$\Gamma_{\textrm{L}} = \Abs{\Gamma_{\textrm{L}}} e^{j \Theta_{\textrm{L}} }$$ is plotted on the Smith chart, then in order to move towards the generator, a subtraction from $$\Theta_{\textrm{L}}$$ of $$2 \beta l$$ is required.
Some worked examples that demonstrate this can be found in fig. 1, fig. 2, and fig. 3.
../../figures/ece1236/smithChartSlidesFig1: fig. 1. Mapping an impedance value onto a Smith chart.
../../figures/ece1236/smithChartSlidesFig2: fig. 2. Moving on the Smith chart towards the generator.
../../figures/ece1236/smithChartSlidesFig3: fig. 3. Moving on the Smith chart.
### Single stub tuning.
Referring to fig. 4, the procedure for single stub tuning is
../../figures/ece1236/smithChartSlidesFig4: fig. 4. Single stub tuning example
1. Plot the load on the Smith Chart.
2. Trace the constant VSWR circle. (blue).
3. Move toward the generator until the constant resistance=1 circle is reached (red). This determines the distance $$d$$.
4. Now the input impedance is of the form $$Z_{\textrm{A}} = 1 + j X$$.
5. We now have to use the stub to cancel out the $$j X$$ and make $$Z_{\textrm{in}} = 1$$ (matched).
6. This can be done on the Smith Chart. If $$X>0$$ then we need a capacitive stub (open). If $$X<0$$ then we need an inductive stub (shorted).
7. Say we need a capacitive stub (open): Start from the position of the open. Now the constant VSWR circle is the exterior unit
circle. Move toward the generator until you hit negative $$X$$. This determines the length of the stub $$l$$.
Notes:
1. In step (3) there are two points where the R=1 circle is intersected . Usually we chose the shortest one
2. By adding multiples of half-wavelength lengths to either $$d$$ or $$l$$ an infinite number of solutions can be constructed.
## Question: Find the Smith chart circle equations
Prove \ref{eqn:uwavesDeck5SmithChartCore:440}.
We can write
\label{eqn:uwavesDeck5SmithChartCore:460}
(1 – \Gamma_r)^2 + \Gamma_i^2 = \frac{2 \Gamma_i }{\overline{{X}}_{\textrm{L}}},
or
\label{eqn:uwavesDeck5SmithChartCore:480}
(1 – \Gamma_r)^2 + \lr{ \Gamma_i – \inv{\overline{{X}}_{\textrm{L}}} }^2 = \inv{\lr{\overline{{X}}_{\textrm{L}}}^2},
which is one of the circular equations. For the other, putting the $$\Gamma_r, \Gamma_i$$ terms in the numerator, we have
\label{eqn:uwavesDeck5SmithChartCore:500}
\begin{aligned}
\frac{1 – \Gamma_r^2 – \Gamma_i^2 }{\overline{{\Gamma}}_{\textrm{L}}}
&=
(1 – \Gamma_r)^2 + \Gamma_i^2 \\
&=
1 – 2 \Gamma_r + \Gamma_r^2 + \Gamma_i^2,
\end{aligned}
or
\label{eqn:uwavesDeck5SmithChartCore:520}
\Gamma_r^2 \lr{ 1 + \inv{\overline{{\Gamma}}_{\textrm{L}}} } – 2 \Gamma_r + \Gamma_i^2 \lr{ 1 + \inv{\overline{{\Gamma}}_{\textrm{L}}} }
=
\inv{\overline{{\Gamma}}_{\textrm{L}}} – 1.
Dividing through by $$1 + \ifrac{1}{\overline{{\Gamma}}_{\textrm{L}}} = (\overline{{\Gamma}}_{\textrm{L}} + 1)/\overline{{\Gamma}}_{\textrm{L}}$$, we have
\label{eqn:uwavesDeck5SmithChartCore:540}
\begin{aligned}
\Gamma_r^2 – 2 \Gamma_r \frac{ \overline{{\Gamma}}_{\textrm{L}} }{\overline{{\Gamma}}_{\textrm{L}} + 1} + \Gamma_i^2
&=
\frac{1 – \overline{{\Gamma}}_{\textrm{L}}}{\overline{{\Gamma}}_{\textrm{L}}} \frac{ \overline{{\Gamma}}_{\textrm{L}} }{\overline{{\Gamma}}_{\textrm{L}} + 1} \\
&=
\frac{1 – \overline{{\Gamma}}_{\textrm{L}}}{ \overline{{\Gamma}}_{\textrm{L}} + 1},
\end{aligned}
or
\label{eqn:uwavesDeck5SmithChartCore:560}
\begin{aligned}
\lr{ \Gamma_r – \frac{ \overline{{\Gamma}}_{\textrm{L}} }{\overline{{\Gamma}}_{\textrm{L}} + 1} }^2 + \Gamma_i^2
&=
\frac{1 – \overline{{\Gamma}}_{\textrm{L}}}{ \overline{{\Gamma}}_{\textrm{L}} + 1} + \lr{ \frac{ \overline{{\Gamma}}_{\textrm{L}} }{\overline{{\Gamma}}_{\textrm{L}} + 1} }^2 \\
&=
\frac{ 1 – \overline{{\Gamma}}_{\textrm{L}}^2 + \overline{{\Gamma}}_{\textrm{L}}^2 }{\lr{\overline{{\Gamma}}_{\textrm{L}} + 1}^2} \\
&=
\frac{ 1 }{\lr{\overline{{\Gamma}}_{\textrm{L}} + 1}^2}.
\end{aligned}
# References
[1] EETech Media. Simple Series Resonance, 2016. URL http://www.allaboutcircuits.com/textbook/alternating-current/chpt-6/simple-series-resonance/. [Online; accessed 04-Feb-2016].
[2] David M Pozar. Microwave engineering. John Wiley \& Sons, 2009.
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https://www.yourdissertation.co.uk/explain-the-process-of-setting-a-master-budget-in-your-answer-pay-attention-to-logical-sequence-and-structure12/
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Jun 03, 2017
# Explain the process of setting a master budget. In your answer pay attention to logical sequence and structure. [12]
This paper concentrates on the primary theme of Explain the process of setting a master budget. In your answer pay attention to logical sequence and structure. [12] in which you have to explain and evaluate its intricate aspects in detail. In addition to this, this paper has been reviewed and purchased by most of the students hence; it has been rated 4.8 points on the scale of 5 points. Besides, the price of this paper starts from £ 40. For more details and full access to the paper, please refer to the site.
COST ACCOUNTING
Instructions to candidates:
a) Time allowed: Three hours (plus an extra ten minutes’ reading time at the start – do not write anything during this time)
c) All questions carry equal marks. Marks for each question are shown in [ ]
d) Non-programmable calculators are permitted in this examination
1. a) Explain the process of setting a master budget. In your answer pay attention to logical sequence
and structure. [12] b) Enumerate the principal benefits of maintaining a budgetary control system. [8]
2. Manufacturers Ltd has three production cost centres X, Y and Z and one service cost centre M which is the maintenance department. The budgeted overhead expenditure for the year ended 31 March 2016 is as follows:
£000 Depreciation of production and maintenance equipment 500 Employer’s liability insurance 175 Heating and lighting 200 Indirect labour 1,000 * Rent and business rates 800 Staff welfare and safety expenses 200 ------- Budgeted overhead expenditure: total 2,875 =====
* To be apportioned on the basis of the number of employees per cost centre. Other data/information is as follows:
Value of production equipment: Cost centre X £1,000,000 Cost centre Y £800,000 Cost centre Z £500,000 Cost centre M £200,000 Floor area: Cost centre X 80,000 sq. metres Cost centre Y 60,000 sq. metres Cost centre Z 50,000 sq. metres Cost centre M 10,000 sq. metres Number of employees: Cost centre X 200 Cost centre Y 100 Cost centre Z 80 Cost centre M 20 Overheads to be allocated to cost centre M amount to £55,000.
Cost centre M is to be apportioned 50% to X, 35% to Y and 15% to Z.
Budgeted direct labour hours are: Cost centre X 200,000 Cost centre Y 217,500 Cost centre Z 171,250
Question 2 continues overleaf
a) Prepare an overhead analysis sheet showing:
i. A column stating the overhead expense, e.g. depreciation
ii. A column for each department, e.g. X, Y, Z and M, showing the apportioned expenditure
iii. A total column
iv a column noting the basis used for calculating the expenditure, e.g. area [10]
B) Calculate the overhead absorption rates for EACH of the three production cost centres using
budgeted direct labour hours. Round to two decimal places. [6]
b) Prepare a cost statement for a product which has a prime cost of £640, and takes 4 hours in cost
centre X, followed by 5 hours in cost centre Y, and 3 hours in cost centre Z. [4]
3. Al-Ga plc has a limited capital budget available for investment in suitable projects this year, and has shortlisted two possible choices. Details are as follows:
Project Alpha
Project Omega
Capital cost
£1,750,000
£1,800,000
Expected life
5 years
5 years
Residual value
nil
nil
Budgeted cash inflows:
£000
£000
Year 1
500
600
Year 2
950
1,200
Year 3
1,300
1,500
Year 4
800
600
Year 5
300
300
The cost of capital to Al-Ga plc is 9%.
Extracts from NPV tables are as follows:
Year
8%
9%
10%
1
.926
.917
.909
2
.857
.842
.826
3
.794
.772
.751
4
.735
.708
.683
5
.630
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http://mathhelpforum.com/calculus/136728-complex-chain-rule-problem.html
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# Thread: Complex Chain Rule Problem
1. ## Complex Chain Rule Problem
Hello All,
Would someone mind helping me check my work? My answer is very close to the answer given in the back of the book, but isn't exactly the same.
$y=3x^2+\sqrt{2}$
$z=\sqrt{1+y}$
$v=\frac{1}{\sqrt{3}+4z}$ find $\frac{dv}{dx}$
$\frac{dy}{dx}=$ $3x$
$\frac{dz}{dy}=$ $\frac{1}{2\sqrt{1+y}}$
$\frac{dv}{dz}=$ $\frac{-4}{(\sqrt{3}+4z)^2}$
$\frac{dy}{dx}*\frac{dz}{dy}*\frac{dv}{dz}=\frac{dv }{dx}$
$3x$ * $\frac{1}{2\sqrt{1+y}}$ * $\frac{-4}{(\sqrt{3}+4z)^2}$
3x * $\frac{1}{2\sqrt{1+(3x^2+\sqrt{2})}}$ * $\frac{-4}{(\sqrt{3}+4\sqrt{1+3x^2+\sqrt{2})^2}}$
Here is the messy answer I ended up with. The answer in the back of the book matches this exactly EXCEPT it is missing the "2" in the beginning of the denominator...any thoughts?
$\frac{-12x}{2\sqrt{1+3x^2+\sqrt{2}}(\sqrt{3}+4\sqrt{1+3x^ 2+\sqrt{2}})^2}$
2. mistake is, must be
dy/dx=6x
3. Thank You!
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# Still a poiseuille flow problem
Posted by bent
Still a poiseuille flow problem January 25, 2012 10:06AM Registered: 7 years ago Posts: 11
Hi all,
I'm working on Poiseuille flow since a week and there are thing I don't understand.
I read almost all the relative topics on the forum but there are still some dark points that I try to figure out.
I use incompressible BGK model and the flow is driven by a pressure drop (I use Zou/He BC). A half way Bounce back is applied on other walls.
The two issues that I would like understand are :
1) why I obtain a vertical non-zero velocity ? I Zou and He paper they say that the error on the y-velocity is zero. (10^-15) and I think I have done things like they are descibed in the paper.
2) why the code can not reach the imposed peak velocity (the pressure drop is construct in this way). I suppose the wo questions are linked.
I post the code I use, I started from the cylinder.m writen by J.Latt.
```clear all
close all
format short g
% GENERAL FLOW CONSTANTS
lx = 45; % number of cells in x-direction
ly = 15; % number of cells in y-direction
Re = 10; % Reynolds number
maxT = 10000; % total number of iterations
tPlot = 100; % cycles
% D2Q9 LATTICE CONSTANTS
t = [4/9, 1/9,1/9,1/9,1/9, 1/36,1/36,1/36,1/36];
cx = [ 0, 1, 0, -1, 0, 1, -1, -1, 1];
cy = [ 0, 0, 1, 0, -1, 1, 1, -1, -1];
opp = [ 1, 4, 5, 2, 3, 8, 9, 6, 7];
%col = [2:(ly-1)];
col = [1:ly];
in = 1; % position of inlet
out = lx; % position of outlet
fIn=zeros(9,lx,ly);
fOut=zeros(9,lx,ly);
[y,x] = meshgrid(1:ly,1:lx); % get coordinate of matrix indices
obst(:,[1,ly]) = 1; % Location of top/bottom boundary
bbRegion = find(obst); % Boolean mask for bounce-back cells
% INITIAL CONDITION: u = 0
ux = zeros(lx,ly);
uy = zeros(lx,ly);
rho = 1;
% caracteristic length L=f(Bounce-Back)
% with HW BB, walls are in y=1-0.5 and y=ly+0.5
Length = ly;
omega = 1.5;
nu = (omega - 0.5)/3;
omega=1/omega;
uref_lb=nu*Re/Length;
rho0 = 1d0;
rho_out = rho0;
rho_in = 3*(8*abs(uref_lb)*nu*(lx-1)*rho0/Length^2)+rho_out;
for i=1:9
cu = 3*(cx(i)*ux + cy(i)*uy);
fIn(i,:,:) = t(i) .* ...
( rho + cu + 1/2*(cu.*cu) - 3/2*(ux.^2+uy.^2) );
end
fid1 = fopen('hist.dat','w');
% MAIN LOOP (TIME CYCLES)
for cycle = 0:maxT
% MACROSCOPIC VARIABLES
rho = sum(fIn);
ux = reshape ( (cx * reshape(fIn,9,lx*ly)), 1,lx,ly);
uy = reshape ( (cy * reshape(fIn,9,lx*ly)), 1,lx,ly);
% MACROSCOPIC (DIRICHLET) BOUNDARY CONDITIONS
% Inlet: Constant pressure
uy(:,in,col) = 0;
rho(:,in,col) = rho_in;
ux(:,in,col) = rho(:,in,col) - ( ...
sum(fIn([1,3,5],in,col)) + 2*sum(fIn([4,7,8],in,col)) );
% Outlet: Constant pressure
uy(:,out,col) = 0;
rho(:,out,col) = rho_out;
ux(:,out,col) = - rho(:,out,col) + ( ...
sum(fIn([1,3,5],out,col)) + 2*sum(fIn([2,6,9],out,col)) );
% MICROSCOPIC BOUNDARY CONDITIONS: INLET (Zou/He BC)
fIn(2,in,col) = fIn(4,in,col) + 2/3*ux(:,in,col);
fIn(6,in,col) = fIn(8,in,col) + 1/2*(fIn(5,in,col)-fIn(3,in,col)) ...
+ 1/2*uy(:,in,col) ...
+ 1/6*ux(:,in,col);
fIn(9,in,col) = fIn(7,in,col) + 1/2*(fIn(3,in,col)-fIn(5,in,col)) ...
- 1/2*uy(:,in,col) ...
+ 1/6*ux(:,in,col);
% MICROSCOPIC BOUNDARY CONDITIONS: OUTLET (Zou/He BC)
fIn(4,out,col) = fIn(2,out,col) - 2/3*ux(:,out,col);
fIn(8,out,col) = fIn(6,out,col) + 1/2*(fIn(3,out,col)-fIn(5,out,col)) ...
- 1/2*uy(:,out,col) ...
- 1/6*ux(:,out,col);
fIn(7,out,col) = fIn(9,out,col) + 1/2*(fIn(5,out,col)-fIn(3,out,col)) ...
+ 1/2*uy(:,out,col) ...
- 1/6*ux(:,out,col);
% COLLISION STEP
for i=1:9
cu = 3*(cx(i)*ux+cy(i)*uy);
fEq(i,:,:) = t(i) .* ...
( rho + cu + 1/2*(cu.*cu) - 3/2*(ux.^2+uy.^2) );
fOut(i,:,:) = fIn(i,:,:) - omega .* (fIn(i,:,:)-fEq(i,:,:));
end
% OBSTACLE (BOUNCE-BACK)
% for i=1:9
% fOut(i,bbRegion) = fIn(opp(i),bbRegion);
% end
% STREAMING STEP
% for i=1:9
% fIn(i,:,:) = circshift(fOut(i,:,:), [0,cx(i),cy(i)]);
% end
for i=1:lx
for j=1:ly
for k=1:9
xnew = i + cx(k);
ynew = j + cy(k);
if (xnew>=1 && xnew <= lx) && (ynew>=1 && ynew <= ly)
fIn(k, xnew, ynew) = fOut(k, i, j);
end
end
end
end
% Bounce-Back
fIn(6,:,1) = fIn(opp(6),:,1);
fIn(3,:,1) = fIn(opp(3),:,1);
fIn(7,:,1) = fIn(opp(7),:,1);
fIn(8,:,ly) = fIn(opp(8),:,ly);
fIn(5,:,ly) = fIn(opp(5),:,ly);
fIn(9,:,ly) = fIn(opp(9),:,ly);
% VISUALIZATION
if (mod(cycle,tPlot)==0)
fprintf(fid1,'%12.8E %12.8E %12.8E %12.8E %12.8E \n',...
cycle,...
max(max(max(ux))),min(min(min(ux))),...
max(max(max(uy))),min(min(min(uy))));
fprintf('%12.4e %12.4e %12.4e %12.4e %12.4e \n',...
cycle,...
max(max(max(ux)))/uref_lb,min(min(min(ux)))/uref_lb,...
max(max(max(uy)))/uref_lb,min(min(min(uy)))/uref_lb);
%u = reshape(sqrt(ux.^2+uy.^2),lx,ly);
u = reshape((uy),lx,ly);
%u(bbRegion) = nan;
figure(1);
imagesc(u');
axis equal off; drawnow
end
end
y_plot=[1:ly];
u_a(y_plot) = (rho_in-rho_out)/(6*nu*rho0*(lx-1))*(y_plot-(1-0.5)).*((ly+0.5)-y_plot);
u_x=reshape((ux),lx,ly);
figure(3); hold on; grid on;
plot(y_plot,u_x(10,y_plot)/uref_lb,'-c*');
plot(y_plot,u_x(30,y_plot)/uref_lb,'-g*');
plot(y_plot,u_x(40,y_plot)/uref_lb,'-b*');
plot(y_plot,u_a(y_plot)/uref_lb,'-ro');
xlabel('y');
ylabel('ux/u_{ref}');
legend('ux(x=10)','ux(x=30)','ux(x=50)','analytical ux');```
Regards,
Ben
Re: Still a poiseuille flow problem May 29, 2012 04:58AM Registered: 9 years ago Posts: 31
Hello!
I meet over your same problem. How do you settle down it?
Re: Still a poiseuille flow problem May 29, 2012 06:02AM Registered: 9 years ago Posts: 168
Have you tried
a) using a body force instead of a inlet/outlet density?
b) using either Zou and He on the walls or using on-node bounce back?
My guess (and it is only a guess) is something ugly is happening at the corners. Or that due to a density variation (which is imposed by the boundary conditions) you are seeing waves which are creating a non-zero vertical velocity. This could be related to the initialisation (Junk has a paper about initialisation the LBE consistently).
Re: Still a poiseuille flow problem May 30, 2012 03:10AM Registered: 9 years ago Posts: 31
Hello!
I use the period boundary condtion boundary and bounce back boundary on the wall, adding the body force into the collision term ,can gain good results compared with the analytic results.
but, I use Zou/He pressure boundary on the inlet and outlet, Zou/He velocity boundary on top and bottom wall, results were smaller than the analytical results. If adding the body force into the collision term, results were bigger than the analytical results. Above two condtions, there was the same pheomenon which the velocity of x-direction gradually increased, had the maximum values on the middle channel.
Can you resovle it ? How do you resolve it/? Can i share your code? thank you very much.
Re: Still a poiseuille flow problem May 30, 2012 02:53PM Registered: 9 years ago Posts: 168
This may be a really stupid question (so I apologise if I sound like an idiot), but why do you need a body force and Zou and He in/out boundaries? Just one of these is sufficient for Poiseuille (i.e., an inlet and outlet density will give you a pressure gradient -so you can have in/out boundaries rather than periodic - or you can impose a driving force in a periodic domain. Perhaps you are doing something more fancy involving gravity or other forces but if you want to check your code then it is a good idea to do the simplest problem thing first.
Re: Still a poiseuille flow problem May 30, 2012 03:08PM Registered: 7 years ago Posts: 11
Hi,
Sorry, I "solve" the problem few month ago.
Like pleb01 said "something ugly is happening at the corners". With a proper corner treatment I have a zero (1d-15) y-velocity.
Of course you have to choice a problem description (ie with inlet/outlet or periodic) and in both cases you should have a zero y-velocity.
Regards,
Ben
Re: Still a poiseuille flow problem May 31, 2012 02:49AM Registered: 9 years ago Posts: 31
Hello!
I used zou/he pressure boundary on the inlet and outlet, hence there was pressure gradient at x-direction flow. at the same time, used bounce back scheme on the top and bottom wall, four corner nodes were specially treated. i comply with zou/he's "on pressure and velocity boundary conditions for the lattice boltzmann BGK model",1997.I gain results smaller than the analytical results.
Re: Still a poiseuille flow problem May 31, 2012 03:02AM Registered: 9 years ago Posts: 31
Hello bent
The following is my codes with c++,including four corner nodes special treatment.
#include <iostream>
#include <cmath>
#include <cstdlib>
#include <iomanip>
#include <fstream>
#include <sstream>
#include <string>
using namespace std;
const int Q = 9; //D2Q9 model
const int NX = 20; //x-direction
const int NY = 10; //y-direction
const double U = 0.1; //inlet velocity
double cr=25;
int e[Q][2] = {{0,0},{1,0},{0,1},{-1,0},{0,-1},{1,1},{-1,1},{-1,-1},{1,-1}};
double w[Q] = {4.0/9,1.0/9,1.0/9,1.0/9,1.0/9,1.0/36,1.0/36,1.0/36,1.0/36};
// Macroscopic variables
double rho[NX+1][NY+1],u[NX+1][NY+1][2],u0[NX+1][NX+1][2];
//Velocity Space
double f[NX+1][NY+1][Q],feqq[NX+1][NY+1][Q],F[NX+1][NY+1][Q]
,gx[NX+1][NY-1],gy[NX+1][NY-1],ftemp[NX+1][NY+1][Q],temp;
// 矩空间密度分布函数
double m[NX+1][NY+1][Q],meqq[NX+1][NY+1][Q],M[NX+1][NY+1][Q];
int opposite[Q]={0,3,4,1,2,7,6,5,6};
int image[NX+1][NY+1];
int i,j,k,ip,jp,n;
double Re,rho0,tau,nu,error,omega,cx,cy;
double s[Q]; //D2Q9松弛系数矩阵
//D2Q0转换矩阵
double MM[Q][Q] = {{1,1,1,1,1,1,1,1,1},{-1,-1,-1,-1,2,2,2,2,-4},{ -2,-2,-2,-2,1,1,1,1,4}
,{1,0,-1,0,1,-1,-1,1,0},{-2,0,2,0,1,-1,-1,1,0},{0,1,0,-1,1,1,-1,-1,0},{0,-2,0,2,1,1,-1,-1,0}
,{1,-1,1,-1,0,0,0,0,0},{ 0,0,0,0,1,-1,1,-1,0}};
double INM[Q][Q];//D2Q9转换矩阵的逆矩阵
//double jx,jy,jz;
//
double inrho = 1.1;
double outrho = 1.0;
double deltaP;
double umax;
double uc[NX+1][NY+1];
//
void init();
void imagec();
double feqc(int k,double rho,double u[2]);
double meqc(int k,double rho,double u[2]);
double force(int k,double u[2],double gx,double gy);
void evolution();
void outputdata();
void Error();
int main()
{
using namespace std;
init();
imagec();
for(n = 0;n<=40000 ;n++)
{
evolution();
if(n%100 == 0)
{
Error();
cout<<"it times"<<n<<" error="<<error<<endl;
if(n >= 100)
{
if(n%100== 0)
outputdata();
if (error<1.0e-6)
break;
}
}
}
return 0;
}
//
//初始子程序
void init()
{
rho0 = 1.0;
Re = 10;
// nu = 0.161; //U*2.0*cr/Re;
// tau = 3.0*nu + 0.5;
omega= 0.9; //1.0/tau;
tau = 1.0/omega;
nu = 1.0/3.0*(tau-0.5);
cx=0.2*NX;
cy=0.5*NY;
std::cout<<"tau="<<tau<<endl;
std::cout<<"nu="<<nu<<endl;
// system("pause");
for(i = 0; i <=NX ; i++) //初始化
for(j = 0; j <=NY ; j++)
{
u[j][0] = 0;
u[j][1] = 0;
rho[j] = rho0;
for(k = 0;k < Q;k++)
{
f[j][k]=feqc(k,rho[j],u[j]);
// feqq[j][k]=feqc(k,rho[j],u[j]);
// m[j][k]=meqc(k,rho[j],u[j]);
}
}
// u[0][0][0] = 0;
// u[0][NY][0] = 0;
//
deltaP =(inrho-outrho)/NX;
std::cout<<"deltaP="<<deltaP<<endl;
// system("pause");
umax = deltaP/(2*nu);
std::cout<<"umax="<<umax<<endl;
system("pause");
}
//
// 图像处理子程序
void imagec()
{
for(i = 0; i <= NX; i++)
for(j = 0; j <= NY; j++)
{
// if (((i-cx)*(i-cx)+(j-cy)*(j-cy))<=cr*cr)
// if(i==1&&i==NX)
// {
// image[j]=1;
// }
// else
// {
image[j]=0;
// }
}
for(i=0; i<=NX; i++)
{
image[0]=1;
}
for(i=0; i<=NX; i++)
{
image[NY]=1;
}
}
//
//计算平衡态函数-速度空间
double feqc(int k,double rho,double u[2])
{
double eu,uv,feq;
eu = (e[k][0]*u[0] + e[k][1]*u[1]);
uv = (u[0]*u[0] + u[1]*u[1]);
feq = w[k]*rho*(1.0 + 3.0*eu +4.5*eu*eu - 1.5*uv);
return feq;
}
//
//计算平衡态函数-速度矩空间
double meqc(int k,double rho,double u[2])
{
double jx,jy,meq;
jx = u[1];
jy = u[2];
switch(k)
{
case 0: meq = rho;
break;
case 1: meq = -2.0*rho + 3.0*(jx*jx + jy*jy);
break;
case 2: meq = rho -3.0*(jx*jx + jy*jy); //energy square
break;
case 3: meq = jx; //momentum in x-dir
break;
case 4: meq = -jx; //energy flux in x-dir---/-jx
break;
case 5: meq = jy; //momentum in y-dir
break;
case 6: meq = -jy; //energy flux in y-dir---/-jy
break;
case 7: meq = (jx*jx - jy*jy); //diagonal comp of stress tensor
break;
case 8: meq = jx*jy;
break;
}
return meq;
}
//
//计算外力
double force(int k,double u[2],double gx,double gy)
{
double fx,fy,eu;
eu = (e[k][0]*u[0] + e[k][1]*u[1]);
fx=w[k]*(1.0-0.5/tau)*(3.0*(e[k][0]-u[0])+9.0*eu*e[k][0])*gx;
fy=w[k]*(1.0-0.5/tau)*(3.0*(e[k][1]-u[1])+9.0*eu*e[k][1])*gy;
return (fx+fy);
}
//
void evolution()
{
// 计算平衡态函数-速度矩空间
// for(i = 0; i <=NX ;i++)
// for(j = 0; j <=NY ; j++)
// for(k = 0;k < Q;k++)
// {
// meqq[j][k]=meqc(k,rho[j],u[j]);
// }
//
//
// computing the macroscopic variables
for(i = 0;i <= NX;i++)
for(j = 0;j <= NY;j++)
{
u0[j][0] = u[j][0];
u0[j][1] = u[j][1];
rho[j] = 0;
u[j][0] = 0;
u[j][1] = 0;
// if(image[j]==0)
{
for(k = 0;k < Q;k++)
{
rho[j] +=f[j][k];
u[j][0] += e[k][0]*f[j][k];
u[j][1] += e[k][1]*f[j][k];
}
u[j][0]= u[j][0]/rho[j];//+0.5*gx[j]/rho[j];
u[j][1]= u[j][1]/rho[j];//+0.5*gy[j]/rho[j];
}
}
//computing the equilbrium distribution functions
for(i = 0; i <= NX ;i++)
for(j = 0; j <= NY ; j++)
for(k = 0;k < Q;k++)
{
if(image[j]==0)
{
feqq[j][k]=feqc(k,rho[j],u[j]);
}
}
//The standard bounceback
for(i = 1; i <= NX-1; i++)
for(j = 0; j <= NY; j++)
// for(k = 0; k< 9; k++)
{
// if (((i-cx)*(i-cx)+(j-cy)*(j-cy))<=cr*cr)
// if(image[j]==1)
// {
// u[j][0]=0;
// u[j][1]=0;
if(j==0)
{
f[j][5]=f[j][7];
f[j][2]=f[j][4];
f[j][6]=f[j][8];
}
if(j==NY)
{
f[j][7]=f[j][5];
f[j][4]=f[j][2];
f[j][8]=f[j][6];
}
// temp = f[j][k];
// f[j][k] = f[j][opposite[k]];
// f[j][opposite[k]]=temp;
//
// temp=f[j][1];
// f[j][1]=f[j][3];
// f[j][3]=temp;
//
// temp=f[j][2];
// f[j][2]=f[j][4];
// f[j][4]=temp;
//
// temp=f[j][5];
// f[j][5]=f[j][7];
// f[j][7]=temp;
//
// temp=f[j][6];
// f[j][6]=f[j][8];
// f[j][8]=temp;
// }
}
//collision
for(i=0;i<=NX;i++)
for(j=0;j<=NY;j++)
for(k=0;k<Q;k++)
{
// gx[j] = deltaP; //(inrho-outrho)/NX;
// gy[j] = 0.0;
if(image[j]==0)
{//if
// F[j][k] = f[ip][jp][k] + (feq(k,rho[ip][jp],u[ip][jp])-f[ip][jp][k])/tau;
f[j][k] = f[j][k] + omega*(feqq[j][k]-f[j][k]);// +3.0*w[k]*(e[k][0]*deltaP+e[k][1]*0.0)*rho[j];
// F[j][k] = inverM*M[j][k];
//f[j][k] = f[j][k] + omega*(feqq[j][k]-f[j][k])+force(k,u[j],gx[j],gy[j]);
// M[j][k] = m[ip][jp][k] + s(k)*(meqq[ip][jp][k]-m[ip][jp][k]);
}//if
}
//
//streaming/propagation
for(i=0;i<=NX;i++)
for(j=0;j<=NY;j++)
for(k=0;k<Q;k++)
{
//
ip = i + e[k][0];
jp = j + e[k][1];
if(ip>=0 && jp>=0 && ip<=NX && jp<=NY)
{
//
ftemp[ip][jp][k]= f[j][k];
}
}
//
for(i=0;i<=NX;i++)
for(j=0;j<=NY;j++)
for(k=0;k<Q;k++)
{
f[j][k]=ftemp[j][k];
}
// streaming end
//Zou and He pressure boundary on inlet/west side
for(i = 0;i <=NX;i++)
for(j = 1;j <=NY-1;j++)
{
if(i==0)
{
rho[j] = inrho;
u[j][0] = 1.0 - (f[j][0]+f[j][2]+f[j][4]+2.0*(f[j][3]+f[j][6]+f[j][7]))/rho[j];
u[j][1] = 0.0;
//
f[j][1] = f[j][3] + 2.0/3.0 * rho[j]*u[j][0];
f[j][5] = f[j][7] + 1.0/6.0 * rho[j]*u[j][0] + 1.0/2.0 * (f[j][4] - f[j][2]);
f[j][8] = f[j][6] + 1.0/6.0 * rho[j]*u[j][0] + 1.0/2.0 * (f[j][2] - f[j][4]);
}
// Zou and He pressure boundary on outlet/east side
if(i==NX)
{
rho[j] = outrho;
u[j][0] = -1.0 + (f[j][0]+f[j][2]+f[j][4]+2.0*(f[j][1]+f[j][5]+f[j][8]))/rho[j];
u[j][1] = 0.0;
f[j][3] = f[j][1] -2.0/3.0 * rho[j]*u[j][0];
f[j][7] = f[j][5] - 1.0/6.0 * rho[j]*u[j][0] + 1.0/2.0 * (f[j][2] - f[j][4]);
f[j][6] = f[j][8] - 1.0/6.0 * rho[j]*u[j][0] + 1.0/2.0 * (f[j][4] - f[j][2]);
}
}
//
//Zou and He velocity boundary on south/bottom side
for(i = 1;i <=NX-1;i++)
for(j = 0;j <=NY;j++)
{
if(j==0)
{
u[j][0] = 0;
u[j][1] = 0;
rho[j] = (f[j][0]+f[j][1]+f[j][3]+2.0*(f[j][4]+f[j][7]+f[j][8]))/1.0;
//
f[j][2] = f[j][4];
f[j][5] = f[j][7] + 1.0/2.0 * (f[j][1] - f[j][3]);
f[j][6] = f[j][8] + 1.0/2.0 * (f[j][3] - f[j][1]);
}
// Zou and He velocity boundary on north/top side
if(j==NY)
{
// rho[j] = outrho;
rho[j]= (f[j][0]+f[j][1]+f[j][3]+2.0*(f[j][2]+f[j][5]+f[j][6]))/1.0;
u[j][0] = 0;
u[j][1] = 0;
f[j][4] = f[j][2];
f[j][7] = f[j][5] + 1.0/2.0 * (f[j][1] - f[j][3]);
f[j][8] = f[j][6] + 1.0/2.0 * (f[j][3] - f[j][1]);
}
}
// Four corner nodes specific treatment
// Bottom left corner
i=0;
j=0;
// {
u[j][0] = 0;
u[j][1] = 0;
rho[j] = rho[j+1];
f[j][1] = f[j][3];
f[j][2] = f[j][4];
f[j][5] = f[j][7];
f[j][6] = 1.0/2.0*(rho[j]-(f[j][0]+2.0*(f[j][3]+f[j][4]+f[j][7])));
f[j][8] = f[j][6];
// }
// Top left corner
i=0;
j=NY;
// {
u[j][0] = 0;
u[j][1] = 0;
rho[j] = rho[j-1];
f[j][1] = f[j][3];
f[j][4] = f[j][2];
f[j][8] = f[j][6];
f[j][5] = 1.0/2.0*(rho[j]-(f[j][0]+2.0*(f[j][2]+f[j][3]+f[j][6])));
f[j][7] = f[j][5];
// }
// Bottom right corner
i=NX;
j=0;
// {
u[j][0] = 0;
u[j][1] = 0;
rho[j] = rho[j+1];
f[j][3] = f[j][1];
f[j][2] = f[j][4];
f[j][6] = f[j][8];
f[j][5] = 1.0/2.0*(rho[j]-(f[j][0]+2.0*(f[j][1]+f[j][4]+f[j][8])));
f[j][7] = f[j][5];
// }
// Top right corner
i=NX;
j=NY;
// {
u[j][0] = 0;
u[j][1] = 0;
rho[j] = rho[j-1];
f[j][3] = f[j][1];
f[j][4] = f[j][2];
f[j][7] = f[j][5];
f[j][6] = 1.0/2.0*(rho[j]-(f[j][0]+2.0*(f[j][1]+f[j][2]+f[j][5])));
f[j][8] = f[j][6];
// }
//ZouHepressureBC
} //evolve
//
void Error()
{
double temp1,temp2;
temp1 = 0;
temp2 = 0;
for(i = 0;i <=NX;i++)
for(j =0;j <=NY;j++)
{
temp1 += ((u[j][0] - u0[j][0])*(u[j][0] - u0[j][0]) +
(u[j][1] - u0[j][1])*(u[j][1] - u0[j][1]));
temp2 += (u[j][0]*u[j][0] + u[j][1]*u[j][1]);
}
temp1 = sqrt(temp1);
temp2 = sqrt(temp2);
error = temp1/(temp2 + 1e-30);
}
void outputdata()
{
if(n%100==0)
{
ofstream file;
char temp[10];
string str;
itoa(n,temp,10);
str = string(temp);
str += "u.dat";
file.open(str.c_str());
file<<"Title=\"LBM Lid Driven Flow\"\n"<<"VARIABLES=\"X\",\"Y\",\"U\",\"V\",\"rho\"\n"<<"ZONE T=\"BOX\",I="
<<NX+1<<",J="<<NY+1<<",F=POINT"<<endl;
if (file.is_open())
{
for ( int i = 0; i <= NX; i++ )
for ( int j = 0; j <= NY; j++ )
{
// file<<double(i)<<" "<<double(j)<<" "<<u[j][0]<<" "<<u[j][1]<<" "<<rho[j]<<endl;
// file<<setiosflags(ios::left);
file<<setiosflags(ios::right)<<double(i)<<" "<<double(j)<<" "<<setiosflags(ios::scientific)<<setprecision(6)<<u[j][0]<<" "<<u[j][1]<<" "<<rho[j]<<resetiosflags(ios::scientific)<<endl;
}
}
}
}
Re: Still a poiseuille flow problem June 04, 2012 10:25AM Registered: 7 years ago Posts: 11
Sorry ydt, I don't read a piece of c ...
you have matlab solutions of the poiseuille flow here
[lbmworkshop.com]
Re: Still a poiseuille flow problem November 30, 2017 08:05AM Registered: 1 year ago Posts: 1
Im really late to the game but i believe you missed a rho in first line for the inlet Zou He condition.Inlet density is 1.3477. this is missing in the inlet term for fIn(2,in,col) = fIn(4,in,col) + 2/3*ux(:,in,col);
Sorry, you do not have permission to post/reply in this forum.
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https://www.codeproject.com/Articles/5290250/Integer-Factorization-Dreaded-list-of-primes
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16,001,249 members
Articles / General Programming / Algorithms
# Integer Factorization: Dreaded List of Primes
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20 Aug 2022CPOL5 min read 33.4K 315 9 22
Using a large list of primes with Trial Division algorithm and how to handle the list
## Introduction
The Trial Division algorithm is about factorizing an integer by checking all possible factors, and all interesting factors are prime numbers. The problem is how to handle a large list of prime numbers.
## First Approach: Plain List of Prime Numbers
When using a list of primes, the first approach is to use a simple array of integers containing the prime numbers.
Under 64K, a prime number fits in an `int16` (2 bytes), beyond, a prime number rather fits in an `int32` (4 bytes).
Figures for a simple list of primes:
Upper limit Number of primes Size of array 65,536 6,542 13,084 100,000 9,592 38,368 500,000 41,538 166,152 1,000,000 78,498 313,992 5,000,000 348,513 1,394,052 10,000,000 664,579 2,658,316 100,000,000 5,761,455 23,045,820, 2^32-1 4,294,967,295 203,280,221 813,120,884
As one can see, things get really ugly when the list is huge.
Sample code:
C++
```// Trial Division: Square Root + Prime list + Wheel
// Check all numbers until Square Root, start with a list of primes
// and Skip useless factors with a wheel
long long TD_SRPW(long long Cand) {
long long SPrimes[] = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43,
47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113,
127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191,
193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263,
269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347,
349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421,
431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499,
503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593,
599, 601, 0 };
long long Wheel[] = { 6, 4, 2, 4, 2, 4, 6, 2, 0 };
long long Div;
Count = 0;
long long Top = sqrt(Cand);
// Check small primes
for (long long Ind = 0; SPrimes[Ind] != 0; Ind++) {
Div = SPrimes[Ind]; // for debug purpose
if (SPrimes[Ind] > Top) {
return Cand;
}
Count++;
if (Cand % SPrimes[Ind] == 0) {
return SPrimes[Ind];
}
}
// Start the Wheel
Div = 601;
while (Div <= Top) {
for (long long Ind = 0; Wheel[Ind] != 0; Ind++) {
if (Div > Top) {
return Cand;
}
Count++;
if (Cand % Div == 0) {
return Div;
}
Div += Wheel[Ind];
}
}
return Cand;
}```
TrialDivLP.cpp in TrialDivLP.zip.
## Second Approach: Compress the List of Prime Numbers by Encoding in a Bitfield Array
Rather than a list of primes, the principle is to encode if a number is a prime or not, it is boolean information and fits in a single bit.
With this approach, the same optimizations than in the classic variants of Trial Division algorithm can apply and the saving is huge.
### Bitfield of All Numbers
Very simple minded, 1 bit per number, or 8 numbers encoded in a byte.
The figures are as follows:
Upper limit Number of primes Size of array 65,536 6,542 8,192 100,000 9,592 12,500 500,000 41,538 62,500 1,000,000 78,498 125,000 5,000,000 348,513 625,000 10,000,000 664,579 1,250,000 100,000,000 5,761,455 12,500,000, 2^32-1 4,294,967,295 203,280,221 536,870,912
### Bitfield of All Odd Numbers
A little more optimized, 1 bit per odd number, or 16 numbers (8 odd numbers) encoded in a byte.
The figures are as shown below:
Upper limit Number of primes Size of array 65,536 6,542 4,096 100,000 9,592 6,250 500,000 41,538 31,250 1,000,000 78,498 62,500 5,000,000 348,513 312,500 10,000,000 664,579 ,625,000 100,000,000 5,761,455 6,250,000, 2^32-1 4,294,967,295 203,280,221 268,435,456
### Bitfield of All Numbers Filtered With the Wheel 30
Then really more optimized, with the wheel, only 8 possible primes per slice of 30, or 30 numbers (8 possible primes) encoded in a byte.
The figures are:
Upper limit Number of primes Size of array 65,536 6,542 2,188 100,000 9,592 3,333 500,000 41,538 16,667 1,000,000 78,498 33,333 5,000,000 348,513 166,667 10,000,000 664,579 333,334 100,000,000 5,761,455 3,333,334, 2^32-1 4,294,967,295 203,280,221 143,165,577
Code exploiting the list of primes (stored in header file).
C++
```// Trial Division: Square Root + Long Prime list + Wheel
// Check all numbers until Square Root, start with a list of primes
// and Skip useless factors with a wheel
long long TD_SRLPW(long long Cand) {
long long WPrimes[] = { 2, 3, 5, 0 };
long long Wheel[] = { 6, 4, 2, 4, 2, 4, 6, 2, 0 };
// unsigned char EPrimes[] = { 0xfe, 0xdf, 0xef, 0x7e, 0xb6, 0xdb, 0x3d, ...
// Encoded list moved to header file
long long Div;
Count = 0;
long long Top = sqrt(Cand);
// Check small primes
for (long long Ind = 0; WPrimes[Ind] != 0; Ind++) {
Div = WPrimes[Ind]; // for debug purpose
if (WPrimes[Ind] > Top) {
return Cand;
}
Count++;
if (Cand % WPrimes[Ind] == 0) {
return WPrimes[Ind];
}
}
// start the wheel with encoded primes
Div = 1;
int EInd = 0;
while (Div <= Top) {
unsigned char Flag = EPrimes[EInd];
// if (Flag == 0) {
if (EInd >= EPrimesLen) {
break;
}
for (long long Ind = 0; Wheel[Ind] != 0; Ind++) {
if (Div > Top) {
break;
}
Count++;
if (Cand % Div == 0) {
return Div;
}
}
Div += Wheel[Ind];
}
EInd++;
}
// Start the Wheel after encoded primes
while (Div <= Top) {
for (long long Ind = 0; Wheel[Ind] != 0; Ind++) {
if (Div > Top) {
return Cand;
}
Count++;
if (Cand % Div == 0) {
return Div;
}
Div += Wheel[Ind];
}
}
return Cand;
}```
TrialDivLP.cpp in TrialDivLP.zip.
## Encoding the List of Primes
The Wheel of 30 have 8 slots, a char have 8 bits, so both are in synch, it simplify its usage. Each slot is assigned a bit. The value of a `char` is the sum of bits of slots which are Primes.
Here is how the list is encoded:
```0x01 0x02 0x04 0x08 0x10 0x20 0x40 0x80 code
7 11 13 17 19 23 29 0xfe
31 37 41 43 47 53 59 0xdf
61 67 71 73 79 83 89 0xef
97 101 103 107 109 113 0x7e
127 131 137 139 149 0xb6
151 157 163 167 173 179 0xdb
181 191 193 197 199 0x3d
211 223 227 229 233 239 0xf9
241 251 257 263 269 0xd5
271 277 281 283 293 0x4f
307 311 313 317 0x1e
331 337 347 349 353 359 0xf3
367 373 379 383 389 0xea
397 401 409 419 0xa6
421 431 433 439 443 449 0xed
457 461 463 467 479 0x9e
487 491 499 503 509 0xe6
521 523 0x0c
541 547 557 563 569 0xd3
571 577 587 593 599 0xd3```
The code showing how the list is encoded is as given below:
C++
```void TD_EncodeDtl(int Max) {
long long Wheel[] = { 6, 4, 2, 4, 2, 4, 6, 2, 0 };
long long Cand = 1;
cout << "0x01\t0x02\t0x04\t0x08\t0x10\t0x20\t0x40\t0x80\tcode" << endl;
while (Cand < Max) {
int Code = 0;
int Index = 1;
int Ind = 0;
do {
if ((TD_SRW(Cand) == Cand) && (Cand != 1)) {
Code |= Index;
cout << dec << Cand;
}
cout << "\t";
Cand += Wheel[Ind];
Index = Index << 1;
Ind++;
} while (Wheel[Ind] != 0);
cout << "0x" << setfill('0') << setw(2) << hex << Code << dec << endl;
}
}```
TrialDivLP.cpp in TrialDivLP.zip.
And the code to generate the array is as follows :
C++
```// encode Prime Numbers as C array
void TD_EncodeArray(int Max) {
long long Wheel[] = { 6, 4, 2, 4, 2, 4, 6, 2, 0 };
long long Cand = 1;
cout << "Encodage liste nombres premiers compressée" << endl;
cout << "{";
while (Cand < Max) {
int Code = 0;
int Index = 1;
int Ind = 0;
do {
if ((TD_SRW(Cand) == Cand) && (Cand != 1)) {
Code |= Index;
}
Cand += Wheel[Ind];
Index = Index << 1;
Ind++;
} while (Wheel[Ind] != 0);
cout << "0x";
cout << setfill('0') << setw(2) << hex << Code << ",";
}
cout << "0}" << dec << endl;
}```
TrialDivLP.cpp in TrialDivLP.zip.
## Benchmark
We are counting divisions/modulos to measure the efficiency of the variantes. The encoded list of Primes goes up to 1,000,000 and is stored in a 33kB array.
```Number TD_SR TD_SREF TD_SRW TD_SRLPW Delta
1,009 30 16 11 11 0
2,003 43 22 14 14 0
3,001 53 27 17 16 1
4,001 62 32 19 18 1
5,003 69 35 20 19 1
6,007 76 39 23 21 2
7,001 82 42 25 23 2
8,009 88 45 26 24 2
9,001 93 47 27 24 3
10,007 99 50 28 25 3
20,011 140 71 40 34 6
30,011 172 87 49 40 9
40,009 199 100 56 46 10
49,999 222 112 62 48 14
1,000,003 999 500 268 168 100
4,000,021 1,800 901 483 279 204
9,000,011 2,999 1,500 802 430 372
16,000,057 3,999 2,000 1,068 550 518
25,000,009 4,999 2,500 1,336 669 667
36,000,007 5,999 3,000 1,602 783 819
49,000,027 6,999 3,500 1,868 900 968
64,000,031 7,999 4,000 2,136 1,007 1,129
81,000,001 8,999 4,500 2,402 1,117 1,285
100,000,007 9,999 5,000 2,668 1,229 1,439
121,000,003 10,999 5,500 2,936 1,335 1,601
144,000,001 11,999 6,000 3,202 1,438 1,764
169,000,033 12,999 6,500 3,468 1,547 1,921
196,000,001 13,999 7,000 3,736 1,652 2,084
225,000,011 14,999 7,500 4,002 1,754 2,248
256,000,001 15,999 8,000 4,268 1,862 2,406
289,000,001 16,999 8,500 4,536 1,960 2,576
10,000,000,019 99,999 50,000 26,668 9,592 17,076
1,000,000,000,039 999,999 500,000 266,668 78,498 188,170```
Delta is here to highlight the savings of the huge list of primes.
C++
```long long Test[] = { 101, 1009, 2003, 3001, 4001, 5003, 6007, 7001, 8009,
9001, 10007, 20011, 30011, 40009, 49999, 1000003, 4000021, 9000011,
16000057, 25000009, 36000007, 49000027, 64000031, 81000001,
100000007, 121000003, 144000001, 169000033, 196000001, 225000011,
256000001, 289000001, 10000000019, 1000000000039, 0 };
cout << "Comparaison Variantes" << endl;
cout << "Number\tTD_SR\tTD_SREF\tTD_SRW\tTD_SRLPW\tDelta" << endl;
// test after first 0
for (Ind = 0; Test[Ind] != 0; Ind++) {
cout << Test[Ind] << "\t";
TD_SR(Test[Ind]);
cout << Count << "\t";
TD_SREF(Test[Ind]);
cout << Count << "\t";
TD_SRW(Test[Ind]);
int C1 = Count;
cout << Count << "\t";
TD_SRLPW(Test[Ind]);
cout << Count << "\t";
cout << C1 - Count << endl;
}
cout << endl;
```
TrialDivLP.cpp in TrialDivLP.zip.
### 128 bits integers
GCC have provisions for 128 bits variables by using 2 64 bits registers, it is ok for calculations, but is not supported as constants in source code nor for IO operations. A little programming is needed to circonvant those limitations.
C++
```//#define int64
// GCC have provisions to handle 128 bits integers, but not for constants in source code
// and not IO
#ifdef int64
// 64 bits integer
typedef long long MyBigInt;
void output(MyBigInt Num, int Size) {
cout << setfill(' ') << setw(Size);
}
#else
// 128 bits integer
typedef __int128 MyBigInt;
void output(MyBigInt Num, int Size) {
if (Num <= (long long) 0x7fffffffffffffff) {
if (Size > 0) {
cout << setfill(' ') << setw(Size);
}
cout << (long long) Num;
} else {
output(Num / 1000, Size - 4);
long long Tmp = Num % 1000;
cout << "," << setfill('0') << setw(3) << Tmp;
}
}
#endif```
TrialDivLP.cpp in TrialDivLP.zip.
``` Number Fartor Timing
10,000,000,000,000,061 1 365 ms
100,000,000,000,000,003 1 1,173 ms
1,000,000,000,000,000,003 1 3,866 ms
10,000,000,000,000,000,051 1 11,988 ms
100,000,000,000,000,000,039 1 82,326 ms
1,000,000,000,000,000,000,117 1 272,376 ms
```
Each time the number is multiplied by 10, runtime is roughly multiplied by 3 (sqrt(10)). Note the runtime gap when number cross the 64 bits boundary.
C++
```#ifndef int64
long long Test2[][3] = { { 10, 16, 61 }, { 10, 17, 3 }, { 10, 18, 3 }, { 10,
19, 51 }, { 10, 20, 39 }, { 10, 21, 117 }, { 0, 0, 0 } };
// for each triplet A B C, the number is
// Cand= A ^ B + C
cout.imbue(locale(cout.getloc(), new gr3));
for (Ind = 0; Test2[Ind][0] > 0; Ind++) {
for (int Offset = 0; Offset < 1000; Offset += 2) {
MyBigInt Cand = 1;
for (int Scan = 0; Scan < Test2[Ind][1]; Scan++) {
Cand *= Test2[Ind][0];
}
Cand += Test2[Ind][2] + Offset;
output(Cand, 30);
cout << "\t";
auto start2 = high_resolution_clock::now();
long long Tmp = TD_SRLPW(Cand);
auto stop2 = high_resolution_clock::now();
if (Tmp == 1) {
output(Tmp, 0);
} else {
cout << Tmp;
cout << "\t";
output((Cand / Tmp) * Tmp, 0);
}
cout << "\t";
auto duration = duration_cast<milliseconds>(stop2 - start2);
cout << " " << duration.count() << " ms";
cout << endl;
if (Tmp == 1) {
break;
}
}
}
cout.imbue(locale(cout.getloc(), new gr0));
#endif```
TrialDivLP.cpp in TrialDivLP.zip.
## Accelerating IsPrime with the compressed list of primes
We can take advantage of huge list of Primes and use it as a loopup table.
With a plain list, we need to do a dichotomy search.
With an encoded list, an almost direct reading is possible.
C++
```// IsPrime taking advantage of long list of primes table
// returns 1 if prime, returns 0 otherwise
int IsPrimeLP(MyBigInt Cand) {
const unsigned long long Wheel[] = { 6, 4, 2, 4, 2, 4, 6, 2, 0 };
const unsigned long long WOffset[] = { 1, 7, 11, 13, 17, 19, 23, 29, 0 };
if (Cand <= EPrimesMax) {
Count = 1;
// Cand is within list of primes
long long x = Cand / WSize; // position in wheel list
long long y = Cand % WSize; // offset in wheel
for (int Index = 0; WOffset[Index] != 0; Index++) {
if (WOffset[Index] == y) {
return ((EPrimes[x] >> Index) & 0x01);
}
}
if (Cand == 2)
return 1;
else if (Cand == 3)
return 1;
else if (Cand == 5)
return 1;
else
return 0; // multiple of 2, 3 or 5
}
// Search a Factor
return (TD_SRLPW(Cand) == 1);
}```
TrialDivLP.cpp in TrialDivLP.zip.
## Points of Interest
Trial Division algorithm improves as the list of primes get bigger, and the compression helps a lot.
This work is original. If you know something similar, please drop a link in the forum below.
## History
• 18th December, 2020: First version
• 25th December, 2020: Second version - Corrections with improvements
• 6th February, 2021: Added IsPrime code + Corrections
• 7th August, 2022: Added Content Table
• 20th August, 2022: typos in source code
Written By
Database Developer
France
I am a professional programmer.
Problem analyse is certainly what I am best at.
My main programming expertise is in the xBase languages (dBase, Clipper, FoxPro, Harbour, xHarbour), then VBA for Excel and advanced Excel WorkBooks.
I also have knowledge on C/C++, d language, HTML, SVG, XML, XSLT, Javascript, PHP, BASICs, Python, COBOL, Assembly.
My personal interests goes to algorithm optimization and puzzles.
First Prev Next
Nice article, but Isaac Asimov was there first Daniel Pfeffer6-Aug-22 22:34 Daniel Pfeffer 6-Aug-22 22:34
Re: Nice article, but Isaac Asimov was there first Patrice T6-Aug-22 23:02 Patrice T 6-Aug-22 23:02
What about quickly *calculating* whether a number > 2^48 is prime? Visual Herbert2-Apr-21 4:08 Visual Herbert 2-Apr-21 4:08
Re: What about quickly *calculating* whether a number > 2^48 is prime? Patrice T2-Apr-21 7:20 Patrice T 2-Apr-21 7:20
Re: What about quickly *calculating* whether a number > 2^48 is prime? Lee Carpenter5-Apr-21 9:17 Lee Carpenter 5-Apr-21 9:17
Re: What about quickly *calculating* whether a number > 2^48 is prime? Patrice T5-Apr-21 10:22 Patrice T 5-Apr-21 10:22
Question re fcn SRLPW ( ) Member 117206811-Jan-21 13:42 Member 11720681 1-Jan-21 13:42
Re: Question re fcn SRLPW ( ) Patrice T1-Jan-21 14:43 Patrice T 1-Jan-21 14:43
Gone too far - or not far enough? Stefan_Lang22-Dec-20 8:54 Stefan_Lang 22-Dec-20 8:54
Re: Gone too far - or not far enough? Patrice T26-Dec-20 19:26 Patrice T 26-Dec-20 19:26
functions TD_EncodeDtl ( ) and TD_EncodeArray() are 99% identical..Why do you need two functions ?? Member 1172068122-Dec-20 8:41 Member 11720681 22-Dec-20 8:41
Re: functions TD_EncodeDtl ( ) and TD_EncodeArray() are 99% identical..Why do you need two functions ?? Patrice T22-Dec-20 10:56 Patrice T 22-Dec-20 10:56
Re: functions TD_EncodeDtl ( ) and TD_EncodeArray() are 99% identical..Why do you need two functions ?? Member 1172068123-Dec-20 5:20 Member 11720681 23-Dec-20 5:20
Re: functions TD_EncodeDtl ( ) and TD_EncodeArray() are 99% identical..Why do you need two functions ?? Patrice T23-Dec-20 5:50 Patrice T 23-Dec-20 5:50
Re: functions TD_EncodeDtl ( ) and TD_EncodeArray() are 99% identical..Why do you need two functions ?? Member 1172068123-Dec-20 6:45 Member 11720681 23-Dec-20 6:45
Re: functions TD_EncodeDtl ( ) and TD_EncodeArray() are 99% identical..Why do you need two functions ?? Patrice T23-Dec-20 7:16 Patrice T 23-Dec-20 7:16
My vote of 4 Michael Thompson 12321-Dec-20 16:25 Michael Thompson 123 21-Dec-20 16:25
Re: My vote of 4 Patrice T21-Dec-20 17:01 Patrice T 21-Dec-20 17:01
Re: My vote of 4 Patrice T28-Dec-20 9:52 Patrice T 28-Dec-20 9:52
Re: My vote of 4 Michael Thompson 12328-Dec-20 16:11 Michael Thompson 123 28-Dec-20 16:11
My vote of 5 PhilipOakley21-Dec-20 4:38 PhilipOakley 21-Dec-20 4:38
Re: My vote of 5 Patrice T21-Dec-20 5:14 Patrice T 21-Dec-20 5:14
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https://nlogn.in/introduction-to-avl-tree-and-its-properties/
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Home Data Structures Introduction to AVL Tree and its properties
# Introduction to AVL Tree and its properties
AVL tree(Adelson-Velsky and Landis tree named after it’s inventor) is a height-balanced Binary Search Tree such that the difference between the height of a left subtree and a right subtree (of a node) is 1, 0 or -1. It has the following properties:
1. It should be a Binary Search Tree i.e, key in the left subtree are always less than or equal to the root, and key in right subtree should be greater than root.
2. The difference between the hight of left subtree and right subtree is known as balance factors and its value should always be 1, 0, -1.
3. The left and right subtrees should also be AVL trees.
##### Balance Factor
Balance Factor is the difference between the height of the left subtree and the right subtree and its value should be -1, 0, or 1.
BalanceFactor = height(left subtree) – height(right subtree)
BalanceFactor ∈ {-1, 0, 1}
### The time complexity of various operations on an AVL tree
###### Worst Case
Height O(log N) O(log N)
Search O(log N) O(log N)
Insert O(log N) O(log N)
Delete O(log N) O(log N)
Tree Traversal O(N) O(N)
The reason AVL is better than BST is that most of its operation takes O(log N) time in worst case too as compared to BST where in worst-case time complexity becomes O(N), where N is the number of nodes in the tree.
##### Height
The maximum height of an AVL tree is floor(log2 N) can’t exceed 1.44*log2N, where N is the number of nodes in the given tree.
Given the height of an AVL tree as h, the maximum number of nodes it can have is 2h+1 – 1 and the minimum number of nodes it can have is given by the formula:
N(h) = N(h-1) + N(h-2) + 1 for n>2 where N(0) = 1 and N(1) = 2.
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07-01ChapGere.0004
# 07-01ChapGere.0004 - u B 4,200 psi 3,000 psi 11,000 psi...
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Solution 7.2-6 Plane stress (angle u ) 428 CHAPTER 7 Analysis of Stress and Strain x O 0.5 MPa u 5 ] 40 8 18.5 MPa 17.8 MPa y s x 52 25.5 MPa s y 5 6.5 MPa t xy 52 12.0 MPa u 52 40 8 52 0.5 MPa 52 17.8 MPa s y 1 5 s x 1 s y 2 s x 1 52 18.5 MPa t x 1 y 1 52 s x 2 s y 2 sin 2 u 1 t xy cos 2 u s x 1 5 s x 1 s y 2 1 s x 2 s y 2 cos 2 u 1 t xy sin 2 u Problem 7.2-7 The stresses acting on element B in the web of a wide-flange beam are found to be 11,000 psi compression in the horizontal direction and 3,000 psi compression in the vertical direction (see figure). Also, shear stresses of magnitude 4,200 psi act in the directions shown. Determine the stresses acting on an element oriented at a counterclockwise angle of 41° from the horizontal. Show these stresses on a sketch of an element oriented at this angle. Solution 7.2-7 Plane stress (angle
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Unformatted text preview: u ) B 4,200 psi 3,000 psi 11,000 psi Side View Cross Section B x O 3,380 psi u 5 41 8 11,720 psi 2,280 psi y s x 5 2 11,000 psi s y 5 2 3,000 psi t xy 5 2 4,200 psi u 5 41 8 5 2 11,720 psi 5 3,380 psi s y 1 5 s x 1 s y 2 s x 1 5 2 2,280 psi t x 1 y 1 5 2 s x 2 s y 2 sin 2 u 1 t xy cos 2 u s x 1 5 s x 1 s y 2 1 s x 2 s y 2 cos 2 u 1 t xy sin 2 u Problem 7.2-8 Solve the preceding problem if the normal and shear stresses acting on element B are 54 MPa, 12 MPa, and 20 MPa (in the directions shown in the figure) and the angle is 42.5° (clockwise). B 54 MPa 12 MPa 20 MPa A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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0
# What is 2 5s as a percent?
Updated: 9/24/2023
Wiki User
10y ago
If you mean 2/5 then 2/5 times 100 = 40%
Wiki User
10y ago
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Sinxtanx - tanx = 0 ; Solve for x?
• Jan 15th 2013, 10:09 PM
Pourdo
Sinxtanx - tanx = 0 ; Solve for x?
Hey everyone!
Sinxtanx - tanx = 0 ; Solve for x? 0 < x < 2pi (and including 0 and 2pi, not sure how to input that)
This is what I have done so far--
sinxtanx - tanx = 0
tanx (sinx - 1) = 0
Set tanx = 0 and sinx - 1 = 0
tanx = 0 when x = 0, pi, 2pi
sinx - 1 = 0
sinx = 1
x = pi/2
so x = 0, pi/2, pi, 2pi
What I am not sure about, is if pi/2 is a solution or not, as I am thinking that tanx is undefined at pi/2.
• Jan 15th 2013, 10:51 PM
abender
Re: Sinxtanx - tanx = 0 ; Solve for x?
$\displaystyle \sin(x)\tan(x)-tan(x)=0 \implies \tan(x)[\sin(x)-1]=0 \implies$$\displaystyle \frac{\sin(x)}{\cos(x)}\left[\sin(x)-1\right]=0$
You want to find $\displaystyle x$ within your given boundaries where $\displaystyle \sin(x)=0$ and $\displaystyle \sin(x)=1$, AS LONG AS $\displaystyle \cos(x)\neq0$.
Edit: To answer your apparent question, you are right, $\displaystyle x=\tfrac{\pi}{2}$ is NOT a solution, because $\displaystyle \cos(\tfrac{\pi}{2})=0$, which cannot be in a denominator.
• Jan 15th 2013, 10:55 PM
Pourdo
Re: Sinxtanx - tanx = 0 ; Solve for x?
So because cos(x) = 0 at pi/2, it is not a solution?
• Jan 15th 2013, 10:59 PM
abender
Re: Sinxtanx - tanx = 0 ; Solve for x?
Quote:
Originally Posted by Pourdo
So because cos(x) = 0 at pi/2, it is not a solution?
Correct. Recall that $\displaystyle \tan(x)=\frac{\sin(x)}{\cos(x)}$.
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Mathbox for BJ < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > bj-cbval2v Structured version Visualization version GIF version
Theorem bj-cbval2v 31924
Description: Version of cbval2 2267 with a dv condition, which does not require ax-13 2234. (Contributed by BJ, 16-Jun-2019.) (Proof modification is discouraged.)
Hypotheses
Ref Expression
bj-cbval2v.1 𝑧𝜑
bj-cbval2v.2 𝑤𝜑
bj-cbval2v.3 𝑥𝜓
bj-cbval2v.4 𝑦𝜓
bj-cbval2v.5 ((𝑥 = 𝑧𝑦 = 𝑤) → (𝜑𝜓))
Assertion
Ref Expression
bj-cbval2v (∀𝑥𝑦𝜑 ↔ ∀𝑧𝑤𝜓)
Distinct variable group: 𝑥,𝑦,𝑧,𝑤
Allowed substitution hints: 𝜑(𝑥,𝑦,𝑧,𝑤) 𝜓(𝑥,𝑦,𝑧,𝑤)
Proof of Theorem bj-cbval2v
StepHypRef Expression
1 bj-cbval2v.1 . . 3 𝑧𝜑
21nfal 2139 . 2 𝑧𝑦𝜑
3 bj-cbval2v.3 . . 3 𝑥𝜓
43nfal 2139 . 2 𝑥𝑤𝜓
5 nfv 1830 . . . . . 6 𝑤 𝑥 = 𝑧
6 bj-cbval2v.2 . . . . . 6 𝑤𝜑
75, 6nfim 1813 . . . . 5 𝑤(𝑥 = 𝑧𝜑)
8 nfv 1830 . . . . . 6 𝑦 𝑥 = 𝑧
9 bj-cbval2v.4 . . . . . 6 𝑦𝜓
108, 9nfim 1813 . . . . 5 𝑦(𝑥 = 𝑧𝜓)
11 bj-cbval2v.5 . . . . . . 7 ((𝑥 = 𝑧𝑦 = 𝑤) → (𝜑𝜓))
1211expcom 450 . . . . . 6 (𝑦 = 𝑤 → (𝑥 = 𝑧 → (𝜑𝜓)))
1312pm5.74d 261 . . . . 5 (𝑦 = 𝑤 → ((𝑥 = 𝑧𝜑) ↔ (𝑥 = 𝑧𝜓)))
147, 10, 13cbvalv1 2163 . . . 4 (∀𝑦(𝑥 = 𝑧𝜑) ↔ ∀𝑤(𝑥 = 𝑧𝜓))
15 19.21v 1855 . . . 4 (∀𝑦(𝑥 = 𝑧𝜑) ↔ (𝑥 = 𝑧 → ∀𝑦𝜑))
16 19.21v 1855 . . . 4 (∀𝑤(𝑥 = 𝑧𝜓) ↔ (𝑥 = 𝑧 → ∀𝑤𝜓))
1714, 15, 163bitr3i 289 . . 3 ((𝑥 = 𝑧 → ∀𝑦𝜑) ↔ (𝑥 = 𝑧 → ∀𝑤𝜓))
1817pm5.74ri 260 . 2 (𝑥 = 𝑧 → (∀𝑦𝜑 ↔ ∀𝑤𝜓))
192, 4, 18cbvalv1 2163 1 (∀𝑥𝑦𝜑 ↔ ∀𝑧𝑤𝜓)
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 195 ∧ wa 383 ∀wal 1473 Ⅎwnf 1699 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-10 2006 ax-11 2021 ax-12 2034 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-tru 1478 df-ex 1696 df-nf 1701 This theorem is referenced by: bj-cbvex2v 31925 bj-cbval2vv 31926
Copyright terms: Public domain W3C validator
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Skip to Content
# Using an Epi Curve to Determine Mode of Spread
This Quick Learn lesson will take approximately 10 minutes to complete.
When you are finished, you will be able to determine the outbreak's likely mode of spread by analyzing an epidemic curve, or “epi curve.”
You can move through this lesson by using the NEXT and BACK icons below.
### X and Y Axes
An epidemic curve, or “epi curve,” is a visual display of the onset of illness among cases associated with an outbreak. The epi curve is represented by a graph with two axes that intersect at right angles.
The horizontal x-axis is the date or time of illness onset among cases.
The vertical y-axis is the number of cases.
Each axis is divided into equally spaced intervals, although the intervals for the two axes may differ.
### What an Epi Curve Can Tell You
An epi curve is a visual display of the onset of illness among cases associated with an outbreak.
You can learn a lot about an outbreak from an epi curve, such as
• Time trend of the outbreak, that is, the distribution of cases over time
• “Outliers,” or cases that stand apart from the overall pattern
• General sense of the outbreak's magnitude
• Inferences about the outbreak's pattern of spread
• Most likely time of exposure
### What an Epi Curve Can Tell You,
continued
The magnitude of an outbreak can be assessed easily with a glance of the epi curve. Are there many cases or just a few?
The time trend, or the distribution of cases over time, will give an indication of where the outbreak is in its course. Are cases still rising or has the outbreak already peaked? Does it appear that the outbreak is over? How long has it been since the last case occurred?
Outliers are cases that stand apart from the other cases. Outliers include the index case, which might be the source of the outbreak, and cases that occur well after other cases, which might indicate secondary spread of the illness.
### Magnitude, Time Trend, and Outliers
Below is the epi curve from an outbreak of hepatitis A. If today's date is August 17, what can you conclude about the outbreak?
Select a link below to learn more.
Note: The incubation period for hepatitis A is 25-30 days.
Hepatitis A Cases by Date of Onset in Port Yourtown, Washington, June - August 2010
### Mode of Spread: Point Source
An epi curve can also be used to make inferences about inferences about an outbreak's most likely mode of spread, suggesting how a disease is transmitted. Transmission occurs in the following ways:
• Point source
• Continuous common source
• Person-to-person spread (propagation)
In a point source outbreak, persons are exposed over a brief time to the same source, such as a single meal or an event. The number of cases rises rapidly to a peak and falls gradually. The majority of cases occur within one incubation period of the disease.
Cryptospordiosis Cases Associated with a Child Care Center by Date of Onset in Port Yourtown, Washington, June 1998
### Mode of Spread: Continuous Common Source
In a continuous common source outbreak, persons are exposed to the same source but exposure is prolonged over a period of days, weeks, or longer. The epi curve rises gradually and might plateau.
Salmonellosis Cases Exposed to Contaminated Salami by Date of Onset, United States, December 2009 – January 2010
### Mode of Spread: Propagated Outbreak
In a propagated outbreak, there is no common source because the outbreak spreads from person-to-person. The graph will assume the classic epi curve shape of progressively taller peaks, each being one incubation period apart.
Measles Cases by Date of Onset in Aberdeen, South Dakota, October 15, 1970 – January 16, 1971
### Analyzing the Mode of Spread
Of course, the shape of an epi curve rarely fits any of these descriptions exactly. For propagated outbreaks, the shape might show overlapping waves of cases that obscure subsequent peaks, and peaks might diminish more slowly over time. You can, however, get a general sense about the mode of spread of an outbreak from its epi curve.
### Your Turn: Exercise 1
The average incubation period for Salmonella is 12 to 36 hours and has a range of 6 hours to 10 days.
What do you think is the most likely mode of spread?
Salmonella Enteritidis Gastroenteritis Cases by Date of Onset in Maryland, August 2008
### Your Turn: Exercise 2
Let’s look at the epi curve for another outbreak of salmonellosis. Remember, the average incubation period for Salmonella is 12 to 36 hours and has a range of 6 hours to 10 days. What do you think is the most likely mode of spread?
Select the most likely pattern of spread.
Onset of Illness among Cases of Salmonella Typhimurium Infection Associated with Peanut Butter, United States, 2008-2009.
### Summary
Congratulations!
Now you should be able to analyze an epi curve and determine an outbreak’s mode of spread.
To view other Quick Learns, visit CDC Learning Connection.
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# Discharge temperature for N2 vaporizer4
## Discharge temperature for N2 vaporizer
(OP)
Hi all,
I'm a fresh mechanical engineer with no others in my company, so I don't really have anyone to discuss stuff with. I don't have much experience with thermodynamics, so I would really appreciate all the help I can get. I have to provide the minimum discharge temperature when at the highest rate and lowest pressure going through vaporizer. The vaporized medium will be Nitrogen. The pump specifications are as follows:
Any ideas how to approach this?
### RE: Discharge temperature for N2 vaporizer
So 25 psi to 10,000 in one go? multi cylinder?
That's a compression ration of about 300:1??
But you're looking at 1000C+ I think - just look up temperature after compression. Remember to use absolute terms (psia or bara and degrees K)
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
### RE: Discharge temperature for N2 vaporizer
it dosnt make much sense, 689.48 bar (very precise and then forgets to mention if barg or bar a ). Then a vaporizer is mentioned? Vaporizing liquid N2? Usually this would be a atm - and all you then need is to know the boiling point of N2 at atm and wikipedia will tell you this: -196ºC. If you are vaporizing at 690 bar then you cant since its above the critical point although you may have a form of sublimation from solid to supercritical phase depending on your temperature...
--- Best regards, Morten Andersen
### RE: Discharge temperature for N2 vaporizer
(OP)
Hi LittleInch and thank You for the answer.
As You assumed correctly it's multi cylinder, triplex pump.
Simplified schematic (no valves, TI, PI) for LN2-GN2 line looks like this:
Since I have discharge pressure, flow and cylinder volume I can probably calculate temperature somehow. However, as mentioned before I'm totally green in this field of study.
### RE: Discharge temperature for N2 vaporizer
What is the suction temperature and state of phase? We know the pressure is 1.72 bar (gauge?).
I'm trying to figure out, is it a pump or a compressor?
Good Luck,
Latexman
### RE: Discharge temperature for N2 vaporizer
Ah.
It now seems your are pressuring liquid nitrogen (LIN) on your table and also LN 2 on your diagram.
Only AFTER the pump do you go through a VAP - vaporiser? - to create nitrogen gas at 10000 psi
So pumping liquid N2 I can't see much temperature increase myself. It will go up a bit but compression of a liquid is a lot harder than gas. Maybe 25 C?
Bung it into HYSYS and see what it says....
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
### RE: Discharge temperature for N2 vaporizer
(OP)
I don't know suction temperature, but I guess it's below -196 degC as it's liquid nitrogen.
#### Quote (Latexman)
I'm trying to figure out, is it a pump or a compressor?
If You mean the one on the inlet side, it's a charge pump (centrifugal).
Part of datasheet for triplex pump:
Part of datasheet for centrifugal pump:
Part of datasheet for vaporizer:
### RE: Discharge temperature for N2 vaporizer
OK.
I would forget any temp increase from the pump or maybe assume 25C.
Then you need to work out the mass flow rate. Then the amount of energy required to heat up your liquid n2 from-196C to say 0C.
Then the huge amount of energy to boil the N2.
Then figure out the energy coming from your water supply.
Then play around with the figures until one matches the other.
What temperature are you getting at the moment?
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
### RE: Discharge temperature for N2 vaporizer
-196 C! Yikes!
If the water flow stops suddenly, which is not uncommon, what's going to stop the vaporizer from freezing and busting almost instantaneously?
Good Luck,
Latexman
### RE: Discharge temperature for N2 vaporizer
This must be a nitrogen pumper. They have a standard rating of 180K SCFH / 10000 PSIg. (also 360K / 540K up to 15000 psig).
Tank -> pumps (booster / triplex) -> vaporiser.
Normally they are direct fired or direct steam vaporisers. I guess in this case it would be a compact heat exchanger, more along the lines of a water pot vaporiser (pancake coils). Pulling off engine jacket coolant to heat up a closed loop water circuit for producing nitrogen gas. If the water pump stops then yes, you can free up the coil. There is typically some residual heat before the ice growth becomes too large though, and you should trip before that happens... !
At this pressure you are not boiling -> there is no latent heat of vaporisation considered. It is a supercritical liquid -> supercritical gas.
The specs from the vaporiser there are, water side: 120 GPM with 45F dT.
This is around 800 kW.
The duty for 180K SCFH from full subcooled (-196 degc) up to +20 degC @ 700 bar is actually only 550 kW.
So more than enough energy for the process.
What will dictate the gas outlet temperature though would be the surface area of the coil.
Have you measured dT on the water side?
### RE: Discharge temperature for N2 vaporizer
(OP)
Thanks for all the answers, I really appreciate this discussion and all of your tips.
I haven't been clear enough with explanations and I'm sorry for that.
#### Quote (Calooomi)
Have you measured dT on the water side?
No, this is planning phase, so everything is just theoretical. I need to provide minimum discharge temperature when at the highest rate and lowest pressure going through vaporizer to a company that chooses it.
As Caloooomi have correctly assumed, this is a nitrogen pumper with heat exchanger on cooling loop. However, this is how it's been done before. Now, they want to electrify the whole unit, so el. engine instead of diesel, el. booster pumps, el. vaporizer etc. No hydraulics involved, just pneumatics.
### RE: Discharge temperature for N2 vaporizer
But you still cant vaporize N2 at 690 barg since the critical point is at around 33 barg. At 680 barg and -196 C you would have solid N2 and it will sublimate into the supercritical state:
https://www.oreilly.com/library/view/chemical-engi...
enthaphy diagram for N2.
Look at the top of the liquid condensation/vaporisation line
--- Best regards, Morten Andersen
### RE: Discharge temperature for N2 vaporizer
Suggest an exit temp of about 25degC from the vaporiser, so that letdown temp at low user pressures will be above 0degC - see T-S diagram for N2 in Perry Chem Engg Handbook.
All LN2 vaporisers I've seen use ambient air for heating LN2, but I suppose steam or water based units can be used also for larger heating duties if heating medium flow is always kept above a safe minimum and tubeside film temp on heating medium side is at least 5-10degC at the cold end for a cocurrent setup, when at low heating medium flow.
### RE: Discharge temperature for N2 vaporizer
Ask CS&P the question as based on the datasheets, they supplied it.
They have the details of the vaporiser so will be able to supply the discharge question based on 240k SCFH @ 8000 psi, based on the picture suplied.
Assume the water side is the lowest flowerate -> 120 GPM @ 185 degF.
You can do a mass / energy balance but you need the construction details of the heat exchanger in order to model an outlet temperature. It could be that there is no overdesign on surface area for the 180K case, so you would be looking at a significantly reduced outlet temp!
### RE: Discharge temperature for N2 vaporizer
(OP)
Thank You all for the answers, I appreciate every single one of them.
Additional question on the sidelines out of curiosity. What actually determines the SCFH value in this case? Is it Triplex Pump in combination with Vaporizer? In other words - If I wanted to design a system with higher capacity than 180k SCFH, which components would be critical to get more than 180k SCFH?
Both.
Good Luck,
Latexman
### RE: Discharge temperature for N2 vaporizer
Pump to get the mass balance right, vaporizer to get the energy balance for the phase change right.
### RE: Discharge temperature for N2 vaporizer
(OP)
Thanks Latexman and Jari!
Going further with this thought ... Caloooomi mentioned that
#### Quote (Caloooomi)
The duty for 180K SCFH from full subcooled (-196 degc) up to +20 degC @ 700 bar is actually only 550 kW.
. What would be the duty for 270K SCFH? How can I calculate this? Would I need an engine with better performance?
### RE: Discharge temperature for N2 vaporizer
550 kW x (270/180)?
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
### RE: Discharge temperature for N2 vaporizer
(OP)
Yeah, I did formulate my question completely wrong. I meant how do You calculate that the duty is 550kW @ 180 SCFH & 700bar.
### RE: Discharge temperature for N2 vaporizer
In this case I looked up the specific conditions in Coolprop and did:
Q = m dH
180K SCFH of LIN = 5915 kg/hr
inlet spec enthalpy = -65.3 kJ/kg
outlet spec enthalpy = 272.3 kJ/kg
which gives ~ 550 kW.
In reality the inlet temperature will be dependent on the tank pressure, how recently it was filled etc. Exactly at -196 degC / 700 barG we'd be above melting pressure according to Refprop, but we are in that region for duty.
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CC-MAIN-2022-33
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en
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http://www.manpagez.com/info/bison/bison-3.0.1/bison_34.php
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manpagez: man pages & more
info bison
Home | html | info | man
[ << ] [ < ] [ Up ] [ > ] [ >> ] [Top] [Contents] [Index] [ ? ]
## 2.2 Infix Notation Calculator: `calc`
We now modify rpcalc to handle infix operators instead of postfix. Infix notation involves the concept of operator precedence and the need for parentheses nested to arbitrary depth. Here is the Bison code for ‘calc.y’, an infix desk-top calculator.
```/* Infix notation calculator. */
```
```%{
#include <math.h>
#include <stdio.h>
int yylex (void);
void yyerror (char const *);
%}
```
```/* Bison declarations. */
%define api.value.type {double}
%token NUM
%left '-' '+'
%left '*' '/'
%precedence NEG /* negation--unary minus */
%right '^' /* exponentiation */
```
```%% /* The grammar follows. */
```
```input:
%empty
| input line
;
```
```line:
'\n'
| exp '\n' { printf ("\t%.10g\n", \$1); }
;
```
```exp:
NUM { \$\$ = \$1; }
| exp '+' exp { \$\$ = \$1 + \$3; }
| exp '-' exp { \$\$ = \$1 - \$3; }
| exp '*' exp { \$\$ = \$1 * \$3; }
| exp '/' exp { \$\$ = \$1 / \$3; }
| '-' exp %prec NEG { \$\$ = -\$2; }
| exp '^' exp { \$\$ = pow (\$1, \$3); }
| '(' exp ')' { \$\$ = \$2; }
;
```
```%%
```
The functions `yylex`, `yyerror` and `main` can be the same as before.
There are two important new features shown in this code.
In the second section (Bison declarations), `%left` declares token types and says they are left-associative operators. The declarations `%left` and `%right` (right associativity) take the place of `%token` which is used to declare a token type name without associativity/precedence. (These tokens are single-character literals, which ordinarily don’t need to be declared. We declare them here to specify the associativity/precedence.)
Operator precedence is determined by the line ordering of the declarations; the higher the line number of the declaration (lower on the page or screen), the higher the precedence. Hence, exponentiation has the highest precedence, unary minus (`NEG`) is next, followed by ‘*’ and ‘/’, and so on. Unary minus is not associative, only precedence matters (`%precedence`. See section Operator Precedence.
The other important new feature is the `%prec` in the grammar section for the unary minus operator. The `%prec` simply instructs Bison that the rule ‘| '-' exp’ has the same precedence as `NEG`—in this case the next-to-highest. See section Context-Dependent Precedence.
Here is a sample run of ‘calc.y’:
```\$ calc
4 + 4.5 - (34/(8*3+-3))
6.880952381
-56 + 2
-54
3 ^ 2
9
```
[ << ] [ < ] [ Up ] [ > ] [ >> ] [Top] [Contents] [Index] [ ? ]
This document was generated on December 1, 2013 using texi2html 5.0.
```© manpagez.com 2000-2019
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# Properties
Label 4800.2.f.ba.3649.2 Level $4800$ Weight $2$ Character 4800.3649 Analytic conductor $38.328$ Analytic rank $1$ Dimension $2$ CM no Inner twists $2$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$4800 = 2^{6} \cdot 3 \cdot 5^{2}$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 4800.f (of order $$2$$, degree $$1$$, not minimal)
## Newform invariants
Self dual: no Analytic conductor: $$38.3281929702$$ Analytic rank: $$1$$ Dimension: $$2$$ Coefficient field: $$\Q(i)$$ Defining polynomial: $$x^{2} + 1$$ x^2 + 1 Coefficient ring: $$\Z[a_1, a_2, a_3]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 480) Sato-Tate group: $\mathrm{SU}(2)[C_{2}]$
## Embedding invariants
Embedding label 3649.2 Root $$-1.00000i$$ of defining polynomial Character $$\chi$$ $$=$$ 4800.3649 Dual form 4800.2.f.ba.3649.1
## $q$-expansion
$$f(q)$$ $$=$$ $$q+1.00000i q^{3} -1.00000 q^{9} +O(q^{10})$$ $$q+1.00000i q^{3} -1.00000 q^{9} +4.00000 q^{11} -2.00000i q^{13} +2.00000i q^{17} -8.00000 q^{19} -4.00000i q^{23} -1.00000i q^{27} -6.00000 q^{29} +4.00000i q^{33} +2.00000i q^{37} +2.00000 q^{39} -6.00000 q^{41} +4.00000i q^{43} -12.0000i q^{47} +7.00000 q^{49} -2.00000 q^{51} +6.00000i q^{53} -8.00000i q^{57} -12.0000 q^{59} -14.0000 q^{61} +12.0000i q^{67} +4.00000 q^{69} +2.00000i q^{73} -8.00000 q^{79} +1.00000 q^{81} -4.00000i q^{83} -6.00000i q^{87} -2.00000 q^{89} +14.0000i q^{97} -4.00000 q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2 q - 2 q^{9}+O(q^{10})$$ 2 * q - 2 * q^9 $$2 q - 2 q^{9} + 8 q^{11} - 16 q^{19} - 12 q^{29} + 4 q^{39} - 12 q^{41} + 14 q^{49} - 4 q^{51} - 24 q^{59} - 28 q^{61} + 8 q^{69} - 16 q^{79} + 2 q^{81} - 4 q^{89} - 8 q^{99}+O(q^{100})$$ 2 * q - 2 * q^9 + 8 * q^11 - 16 * q^19 - 12 * q^29 + 4 * q^39 - 12 * q^41 + 14 * q^49 - 4 * q^51 - 24 * q^59 - 28 * q^61 + 8 * q^69 - 16 * q^79 + 2 * q^81 - 4 * q^89 - 8 * q^99
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/4800\mathbb{Z}\right)^\times$$.
$$n$$ $$577$$ $$901$$ $$1601$$ $$4351$$ $$\chi(n)$$ $$-1$$ $$1$$ $$1$$ $$1$$
## Coefficient data
For each $$n$$ we display the coefficients of the $$q$$-expansion $$a_n$$, the Satake parameters $$\alpha_p$$, and the Satake angles $$\theta_p = \textrm{Arg}(\alpha_p)$$.
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
$$n$$ $$a_n$$ $$a_n / n^{(k-1)/2}$$ $$\alpha_n$$ $$\theta_n$$
$$p$$ $$a_p$$ $$a_p / p^{(k-1)/2}$$ $$\alpha_p$$ $$\theta_p$$
$$2$$ 0 0
$$3$$ 1.00000i 0.577350i
$$4$$ 0 0
$$5$$ 0 0
$$6$$ 0 0
$$7$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$8$$ 0 0
$$9$$ −1.00000 −0.333333
$$10$$ 0 0
$$11$$ 4.00000 1.20605 0.603023 0.797724i $$-0.293963\pi$$
0.603023 + 0.797724i $$0.293963\pi$$
$$12$$ 0 0
$$13$$ − 2.00000i − 0.554700i −0.960769 0.277350i $$-0.910544\pi$$
0.960769 0.277350i $$-0.0894562\pi$$
$$14$$ 0 0
$$15$$ 0 0
$$16$$ 0 0
$$17$$ 2.00000i 0.485071i 0.970143 + 0.242536i $$0.0779791\pi$$
−0.970143 + 0.242536i $$0.922021\pi$$
$$18$$ 0 0
$$19$$ −8.00000 −1.83533 −0.917663 0.397360i $$-0.869927\pi$$
−0.917663 + 0.397360i $$0.869927\pi$$
$$20$$ 0 0
$$21$$ 0 0
$$22$$ 0 0
$$23$$ − 4.00000i − 0.834058i −0.908893 0.417029i $$-0.863071\pi$$
0.908893 0.417029i $$-0.136929\pi$$
$$24$$ 0 0
$$25$$ 0 0
$$26$$ 0 0
$$27$$ − 1.00000i − 0.192450i
$$28$$ 0 0
$$29$$ −6.00000 −1.11417 −0.557086 0.830455i $$-0.688081\pi$$
−0.557086 + 0.830455i $$0.688081\pi$$
$$30$$ 0 0
$$31$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$32$$ 0 0
$$33$$ 4.00000i 0.696311i
$$34$$ 0 0
$$35$$ 0 0
$$36$$ 0 0
$$37$$ 2.00000i 0.328798i 0.986394 + 0.164399i $$0.0525685\pi$$
−0.986394 + 0.164399i $$0.947432\pi$$
$$38$$ 0 0
$$39$$ 2.00000 0.320256
$$40$$ 0 0
$$41$$ −6.00000 −0.937043 −0.468521 0.883452i $$-0.655213\pi$$
−0.468521 + 0.883452i $$0.655213\pi$$
$$42$$ 0 0
$$43$$ 4.00000i 0.609994i 0.952353 + 0.304997i $$0.0986555\pi$$
−0.952353 + 0.304997i $$0.901344\pi$$
$$44$$ 0 0
$$45$$ 0 0
$$46$$ 0 0
$$47$$ − 12.0000i − 1.75038i −0.483779 0.875190i $$-0.660736\pi$$
0.483779 0.875190i $$-0.339264\pi$$
$$48$$ 0 0
$$49$$ 7.00000 1.00000
$$50$$ 0 0
$$51$$ −2.00000 −0.280056
$$52$$ 0 0
$$53$$ 6.00000i 0.824163i 0.911147 + 0.412082i $$0.135198\pi$$
−0.911147 + 0.412082i $$0.864802\pi$$
$$54$$ 0 0
$$55$$ 0 0
$$56$$ 0 0
$$57$$ − 8.00000i − 1.05963i
$$58$$ 0 0
$$59$$ −12.0000 −1.56227 −0.781133 0.624364i $$-0.785358\pi$$
−0.781133 + 0.624364i $$0.785358\pi$$
$$60$$ 0 0
$$61$$ −14.0000 −1.79252 −0.896258 0.443533i $$-0.853725\pi$$
−0.896258 + 0.443533i $$0.853725\pi$$
$$62$$ 0 0
$$63$$ 0 0
$$64$$ 0 0
$$65$$ 0 0
$$66$$ 0 0
$$67$$ 12.0000i 1.46603i 0.680211 + 0.733017i $$0.261888\pi$$
−0.680211 + 0.733017i $$0.738112\pi$$
$$68$$ 0 0
$$69$$ 4.00000 0.481543
$$70$$ 0 0
$$71$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$72$$ 0 0
$$73$$ 2.00000i 0.234082i 0.993127 + 0.117041i $$0.0373409\pi$$
−0.993127 + 0.117041i $$0.962659\pi$$
$$74$$ 0 0
$$75$$ 0 0
$$76$$ 0 0
$$77$$ 0 0
$$78$$ 0 0
$$79$$ −8.00000 −0.900070 −0.450035 0.893011i $$-0.648589\pi$$
−0.450035 + 0.893011i $$0.648589\pi$$
$$80$$ 0 0
$$81$$ 1.00000 0.111111
$$82$$ 0 0
$$83$$ − 4.00000i − 0.439057i −0.975606 0.219529i $$-0.929548\pi$$
0.975606 0.219529i $$-0.0704519\pi$$
$$84$$ 0 0
$$85$$ 0 0
$$86$$ 0 0
$$87$$ − 6.00000i − 0.643268i
$$88$$ 0 0
$$89$$ −2.00000 −0.212000 −0.106000 0.994366i $$-0.533804\pi$$
−0.106000 + 0.994366i $$0.533804\pi$$
$$90$$ 0 0
$$91$$ 0 0
$$92$$ 0 0
$$93$$ 0 0
$$94$$ 0 0
$$95$$ 0 0
$$96$$ 0 0
$$97$$ 14.0000i 1.42148i 0.703452 + 0.710742i $$0.251641\pi$$
−0.703452 + 0.710742i $$0.748359\pi$$
$$98$$ 0 0
$$99$$ −4.00000 −0.402015
$$100$$ 0 0
$$101$$ 14.0000 1.39305 0.696526 0.717532i $$-0.254728\pi$$
0.696526 + 0.717532i $$0.254728\pi$$
$$102$$ 0 0
$$103$$ − 8.00000i − 0.788263i −0.919054 0.394132i $$-0.871045\pi$$
0.919054 0.394132i $$-0.128955\pi$$
$$104$$ 0 0
$$105$$ 0 0
$$106$$ 0 0
$$107$$ 12.0000i 1.16008i 0.814587 + 0.580042i $$0.196964\pi$$
−0.814587 + 0.580042i $$0.803036\pi$$
$$108$$ 0 0
$$109$$ 14.0000 1.34096 0.670478 0.741929i $$-0.266089\pi$$
0.670478 + 0.741929i $$0.266089\pi$$
$$110$$ 0 0
$$111$$ −2.00000 −0.189832
$$112$$ 0 0
$$113$$ 6.00000i 0.564433i 0.959351 + 0.282216i $$0.0910696\pi$$
−0.959351 + 0.282216i $$0.908930\pi$$
$$114$$ 0 0
$$115$$ 0 0
$$116$$ 0 0
$$117$$ 2.00000i 0.184900i
$$118$$ 0 0
$$119$$ 0 0
$$120$$ 0 0
$$121$$ 5.00000 0.454545
$$122$$ 0 0
$$123$$ − 6.00000i − 0.541002i
$$124$$ 0 0
$$125$$ 0 0
$$126$$ 0 0
$$127$$ − 16.0000i − 1.41977i −0.704317 0.709885i $$-0.748747\pi$$
0.704317 0.709885i $$-0.251253\pi$$
$$128$$ 0 0
$$129$$ −4.00000 −0.352180
$$130$$ 0 0
$$131$$ −20.0000 −1.74741 −0.873704 0.486458i $$-0.838289\pi$$
−0.873704 + 0.486458i $$0.838289\pi$$
$$132$$ 0 0
$$133$$ 0 0
$$134$$ 0 0
$$135$$ 0 0
$$136$$ 0 0
$$137$$ 2.00000i 0.170872i 0.996344 + 0.0854358i $$0.0272282\pi$$
−0.996344 + 0.0854358i $$0.972772\pi$$
$$138$$ 0 0
$$139$$ −16.0000 −1.35710 −0.678551 0.734553i $$-0.737392\pi$$
−0.678551 + 0.734553i $$0.737392\pi$$
$$140$$ 0 0
$$141$$ 12.0000 1.01058
$$142$$ 0 0
$$143$$ − 8.00000i − 0.668994i
$$144$$ 0 0
$$145$$ 0 0
$$146$$ 0 0
$$147$$ 7.00000i 0.577350i
$$148$$ 0 0
$$149$$ −6.00000 −0.491539 −0.245770 0.969328i $$-0.579041\pi$$
−0.245770 + 0.969328i $$0.579041\pi$$
$$150$$ 0 0
$$151$$ −8.00000 −0.651031 −0.325515 0.945537i $$-0.605538\pi$$
−0.325515 + 0.945537i $$0.605538\pi$$
$$152$$ 0 0
$$153$$ − 2.00000i − 0.161690i
$$154$$ 0 0
$$155$$ 0 0
$$156$$ 0 0
$$157$$ − 14.0000i − 1.11732i −0.829396 0.558661i $$-0.811315\pi$$
0.829396 0.558661i $$-0.188685\pi$$
$$158$$ 0 0
$$159$$ −6.00000 −0.475831
$$160$$ 0 0
$$161$$ 0 0
$$162$$ 0 0
$$163$$ − 20.0000i − 1.56652i −0.621694 0.783260i $$-0.713555\pi$$
0.621694 0.783260i $$-0.286445\pi$$
$$164$$ 0 0
$$165$$ 0 0
$$166$$ 0 0
$$167$$ 12.0000i 0.928588i 0.885681 + 0.464294i $$0.153692\pi$$
−0.885681 + 0.464294i $$0.846308\pi$$
$$168$$ 0 0
$$169$$ 9.00000 0.692308
$$170$$ 0 0
$$171$$ 8.00000 0.611775
$$172$$ 0 0
$$173$$ 14.0000i 1.06440i 0.846619 + 0.532200i $$0.178635\pi$$
−0.846619 + 0.532200i $$0.821365\pi$$
$$174$$ 0 0
$$175$$ 0 0
$$176$$ 0 0
$$177$$ − 12.0000i − 0.901975i
$$178$$ 0 0
$$179$$ −12.0000 −0.896922 −0.448461 0.893802i $$-0.648028\pi$$
−0.448461 + 0.893802i $$0.648028\pi$$
$$180$$ 0 0
$$181$$ −22.0000 −1.63525 −0.817624 0.575753i $$-0.804709\pi$$
−0.817624 + 0.575753i $$0.804709\pi$$
$$182$$ 0 0
$$183$$ − 14.0000i − 1.03491i
$$184$$ 0 0
$$185$$ 0 0
$$186$$ 0 0
$$187$$ 8.00000i 0.585018i
$$188$$ 0 0
$$189$$ 0 0
$$190$$ 0 0
$$191$$ −16.0000 −1.15772 −0.578860 0.815427i $$-0.696502\pi$$
−0.578860 + 0.815427i $$0.696502\pi$$
$$192$$ 0 0
$$193$$ − 22.0000i − 1.58359i −0.610784 0.791797i $$-0.709146\pi$$
0.610784 0.791797i $$-0.290854\pi$$
$$194$$ 0 0
$$195$$ 0 0
$$196$$ 0 0
$$197$$ 26.0000i 1.85242i 0.377004 + 0.926212i $$0.376954\pi$$
−0.377004 + 0.926212i $$0.623046\pi$$
$$198$$ 0 0
$$199$$ 16.0000 1.13421 0.567105 0.823646i $$-0.308063\pi$$
0.567105 + 0.823646i $$0.308063\pi$$
$$200$$ 0 0
$$201$$ −12.0000 −0.846415
$$202$$ 0 0
$$203$$ 0 0
$$204$$ 0 0
$$205$$ 0 0
$$206$$ 0 0
$$207$$ 4.00000i 0.278019i
$$208$$ 0 0
$$209$$ −32.0000 −2.21349
$$210$$ 0 0
$$211$$ −8.00000 −0.550743 −0.275371 0.961338i $$-0.588801\pi$$
−0.275371 + 0.961338i $$0.588801\pi$$
$$212$$ 0 0
$$213$$ 0 0
$$214$$ 0 0
$$215$$ 0 0
$$216$$ 0 0
$$217$$ 0 0
$$218$$ 0 0
$$219$$ −2.00000 −0.135147
$$220$$ 0 0
$$221$$ 4.00000 0.269069
$$222$$ 0 0
$$223$$ − 24.0000i − 1.60716i −0.595198 0.803579i $$-0.702926\pi$$
0.595198 0.803579i $$-0.297074\pi$$
$$224$$ 0 0
$$225$$ 0 0
$$226$$ 0 0
$$227$$ 4.00000i 0.265489i 0.991150 + 0.132745i $$0.0423790\pi$$
−0.991150 + 0.132745i $$0.957621\pi$$
$$228$$ 0 0
$$229$$ −18.0000 −1.18947 −0.594737 0.803921i $$-0.702744\pi$$
−0.594737 + 0.803921i $$0.702744\pi$$
$$230$$ 0 0
$$231$$ 0 0
$$232$$ 0 0
$$233$$ 6.00000i 0.393073i 0.980497 + 0.196537i $$0.0629694\pi$$
−0.980497 + 0.196537i $$0.937031\pi$$
$$234$$ 0 0
$$235$$ 0 0
$$236$$ 0 0
$$237$$ − 8.00000i − 0.519656i
$$238$$ 0 0
$$239$$ −8.00000 −0.517477 −0.258738 0.965947i $$-0.583307\pi$$
−0.258738 + 0.965947i $$0.583307\pi$$
$$240$$ 0 0
$$241$$ 2.00000 0.128831 0.0644157 0.997923i $$-0.479482\pi$$
0.0644157 + 0.997923i $$0.479482\pi$$
$$242$$ 0 0
$$243$$ 1.00000i 0.0641500i
$$244$$ 0 0
$$245$$ 0 0
$$246$$ 0 0
$$247$$ 16.0000i 1.01806i
$$248$$ 0 0
$$249$$ 4.00000 0.253490
$$250$$ 0 0
$$251$$ 12.0000 0.757433 0.378717 0.925513i $$-0.376365\pi$$
0.378717 + 0.925513i $$0.376365\pi$$
$$252$$ 0 0
$$253$$ − 16.0000i − 1.00591i
$$254$$ 0 0
$$255$$ 0 0
$$256$$ 0 0
$$257$$ − 6.00000i − 0.374270i −0.982334 0.187135i $$-0.940080\pi$$
0.982334 0.187135i $$-0.0599201\pi$$
$$258$$ 0 0
$$259$$ 0 0
$$260$$ 0 0
$$261$$ 6.00000 0.371391
$$262$$ 0 0
$$263$$ − 20.0000i − 1.23325i −0.787256 0.616626i $$-0.788499\pi$$
0.787256 0.616626i $$-0.211501\pi$$
$$264$$ 0 0
$$265$$ 0 0
$$266$$ 0 0
$$267$$ − 2.00000i − 0.122398i
$$268$$ 0 0
$$269$$ 10.0000 0.609711 0.304855 0.952399i $$-0.401392\pi$$
0.304855 + 0.952399i $$0.401392\pi$$
$$270$$ 0 0
$$271$$ −8.00000 −0.485965 −0.242983 0.970031i $$-0.578126\pi$$
−0.242983 + 0.970031i $$0.578126\pi$$
$$272$$ 0 0
$$273$$ 0 0
$$274$$ 0 0
$$275$$ 0 0
$$276$$ 0 0
$$277$$ − 30.0000i − 1.80253i −0.433273 0.901263i $$-0.642641\pi$$
0.433273 0.901263i $$-0.357359\pi$$
$$278$$ 0 0
$$279$$ 0 0
$$280$$ 0 0
$$281$$ 2.00000 0.119310 0.0596550 0.998219i $$-0.481000\pi$$
0.0596550 + 0.998219i $$0.481000\pi$$
$$282$$ 0 0
$$283$$ − 4.00000i − 0.237775i −0.992908 0.118888i $$-0.962067\pi$$
0.992908 0.118888i $$-0.0379328\pi$$
$$284$$ 0 0
$$285$$ 0 0
$$286$$ 0 0
$$287$$ 0 0
$$288$$ 0 0
$$289$$ 13.0000 0.764706
$$290$$ 0 0
$$291$$ −14.0000 −0.820695
$$292$$ 0 0
$$293$$ − 26.0000i − 1.51894i −0.650545 0.759468i $$-0.725459\pi$$
0.650545 0.759468i $$-0.274541\pi$$
$$294$$ 0 0
$$295$$ 0 0
$$296$$ 0 0
$$297$$ − 4.00000i − 0.232104i
$$298$$ 0 0
$$299$$ −8.00000 −0.462652
$$300$$ 0 0
$$301$$ 0 0
$$302$$ 0 0
$$303$$ 14.0000i 0.804279i
$$304$$ 0 0
$$305$$ 0 0
$$306$$ 0 0
$$307$$ − 28.0000i − 1.59804i −0.601302 0.799022i $$-0.705351\pi$$
0.601302 0.799022i $$-0.294649\pi$$
$$308$$ 0 0
$$309$$ 8.00000 0.455104
$$310$$ 0 0
$$311$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$312$$ 0 0
$$313$$ − 14.0000i − 0.791327i −0.918396 0.395663i $$-0.870515\pi$$
0.918396 0.395663i $$-0.129485\pi$$
$$314$$ 0 0
$$315$$ 0 0
$$316$$ 0 0
$$317$$ 18.0000i 1.01098i 0.862832 + 0.505490i $$0.168688\pi$$
−0.862832 + 0.505490i $$0.831312\pi$$
$$318$$ 0 0
$$319$$ −24.0000 −1.34374
$$320$$ 0 0
$$321$$ −12.0000 −0.669775
$$322$$ 0 0
$$323$$ − 16.0000i − 0.890264i
$$324$$ 0 0
$$325$$ 0 0
$$326$$ 0 0
$$327$$ 14.0000i 0.774202i
$$328$$ 0 0
$$329$$ 0 0
$$330$$ 0 0
$$331$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$332$$ 0 0
$$333$$ − 2.00000i − 0.109599i
$$334$$ 0 0
$$335$$ 0 0
$$336$$ 0 0
$$337$$ 22.0000i 1.19842i 0.800593 + 0.599208i $$0.204518\pi$$
−0.800593 + 0.599208i $$0.795482\pi$$
$$338$$ 0 0
$$339$$ −6.00000 −0.325875
$$340$$ 0 0
$$341$$ 0 0
$$342$$ 0 0
$$343$$ 0 0
$$344$$ 0 0
$$345$$ 0 0
$$346$$ 0 0
$$347$$ 4.00000i 0.214731i 0.994220 + 0.107366i $$0.0342415\pi$$
−0.994220 + 0.107366i $$0.965758\pi$$
$$348$$ 0 0
$$349$$ −2.00000 −0.107058 −0.0535288 0.998566i $$-0.517047\pi$$
−0.0535288 + 0.998566i $$0.517047\pi$$
$$350$$ 0 0
$$351$$ −2.00000 −0.106752
$$352$$ 0 0
$$353$$ 14.0000i 0.745145i 0.928003 + 0.372572i $$0.121524\pi$$
−0.928003 + 0.372572i $$0.878476\pi$$
$$354$$ 0 0
$$355$$ 0 0
$$356$$ 0 0
$$357$$ 0 0
$$358$$ 0 0
$$359$$ −16.0000 −0.844448 −0.422224 0.906492i $$-0.638750\pi$$
−0.422224 + 0.906492i $$0.638750\pi$$
$$360$$ 0 0
$$361$$ 45.0000 2.36842
$$362$$ 0 0
$$363$$ 5.00000i 0.262432i
$$364$$ 0 0
$$365$$ 0 0
$$366$$ 0 0
$$367$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$368$$ 0 0
$$369$$ 6.00000 0.312348
$$370$$ 0 0
$$371$$ 0 0
$$372$$ 0 0
$$373$$ 22.0000i 1.13912i 0.821951 + 0.569558i $$0.192886\pi$$
−0.821951 + 0.569558i $$0.807114\pi$$
$$374$$ 0 0
$$375$$ 0 0
$$376$$ 0 0
$$377$$ 12.0000i 0.618031i
$$378$$ 0 0
$$379$$ 16.0000 0.821865 0.410932 0.911666i $$-0.365203\pi$$
0.410932 + 0.911666i $$0.365203\pi$$
$$380$$ 0 0
$$381$$ 16.0000 0.819705
$$382$$ 0 0
$$383$$ 20.0000i 1.02195i 0.859595 + 0.510976i $$0.170716\pi$$
−0.859595 + 0.510976i $$0.829284\pi$$
$$384$$ 0 0
$$385$$ 0 0
$$386$$ 0 0
$$387$$ − 4.00000i − 0.203331i
$$388$$ 0 0
$$389$$ −6.00000 −0.304212 −0.152106 0.988364i $$-0.548606\pi$$
−0.152106 + 0.988364i $$0.548606\pi$$
$$390$$ 0 0
$$391$$ 8.00000 0.404577
$$392$$ 0 0
$$393$$ − 20.0000i − 1.00887i
$$394$$ 0 0
$$395$$ 0 0
$$396$$ 0 0
$$397$$ − 6.00000i − 0.301131i −0.988600 0.150566i $$-0.951890\pi$$
0.988600 0.150566i $$-0.0481095\pi$$
$$398$$ 0 0
$$399$$ 0 0
$$400$$ 0 0
$$401$$ 10.0000 0.499376 0.249688 0.968326i $$-0.419672\pi$$
0.249688 + 0.968326i $$0.419672\pi$$
$$402$$ 0 0
$$403$$ 0 0
$$404$$ 0 0
$$405$$ 0 0
$$406$$ 0 0
$$407$$ 8.00000i 0.396545i
$$408$$ 0 0
$$409$$ 22.0000 1.08783 0.543915 0.839140i $$-0.316941\pi$$
0.543915 + 0.839140i $$0.316941\pi$$
$$410$$ 0 0
$$411$$ −2.00000 −0.0986527
$$412$$ 0 0
$$413$$ 0 0
$$414$$ 0 0
$$415$$ 0 0
$$416$$ 0 0
$$417$$ − 16.0000i − 0.783523i
$$418$$ 0 0
$$419$$ 12.0000 0.586238 0.293119 0.956076i $$-0.405307\pi$$
0.293119 + 0.956076i $$0.405307\pi$$
$$420$$ 0 0
$$421$$ 2.00000 0.0974740 0.0487370 0.998812i $$-0.484480\pi$$
0.0487370 + 0.998812i $$0.484480\pi$$
$$422$$ 0 0
$$423$$ 12.0000i 0.583460i
$$424$$ 0 0
$$425$$ 0 0
$$426$$ 0 0
$$427$$ 0 0
$$428$$ 0 0
$$429$$ 8.00000 0.386244
$$430$$ 0 0
$$431$$ 8.00000 0.385346 0.192673 0.981263i $$-0.438284\pi$$
0.192673 + 0.981263i $$0.438284\pi$$
$$432$$ 0 0
$$433$$ 26.0000i 1.24948i 0.780833 + 0.624740i $$0.214795\pi$$
−0.780833 + 0.624740i $$0.785205\pi$$
$$434$$ 0 0
$$435$$ 0 0
$$436$$ 0 0
$$437$$ 32.0000i 1.53077i
$$438$$ 0 0
$$439$$ −32.0000 −1.52728 −0.763638 0.645644i $$-0.776589\pi$$
−0.763638 + 0.645644i $$0.776589\pi$$
$$440$$ 0 0
$$441$$ −7.00000 −0.333333
$$442$$ 0 0
$$443$$ − 28.0000i − 1.33032i −0.746701 0.665160i $$-0.768363\pi$$
0.746701 0.665160i $$-0.231637\pi$$
$$444$$ 0 0
$$445$$ 0 0
$$446$$ 0 0
$$447$$ − 6.00000i − 0.283790i
$$448$$ 0 0
$$449$$ −10.0000 −0.471929 −0.235965 0.971762i $$-0.575825\pi$$
−0.235965 + 0.971762i $$0.575825\pi$$
$$450$$ 0 0
$$451$$ −24.0000 −1.13012
$$452$$ 0 0
$$453$$ − 8.00000i − 0.375873i
$$454$$ 0 0
$$455$$ 0 0
$$456$$ 0 0
$$457$$ − 26.0000i − 1.21623i −0.793849 0.608114i $$-0.791926\pi$$
0.793849 0.608114i $$-0.208074\pi$$
$$458$$ 0 0
$$459$$ 2.00000 0.0933520
$$460$$ 0 0
$$461$$ 14.0000 0.652045 0.326023 0.945362i $$-0.394291\pi$$
0.326023 + 0.945362i $$0.394291\pi$$
$$462$$ 0 0
$$463$$ 32.0000i 1.48717i 0.668644 + 0.743583i $$0.266875\pi$$
−0.668644 + 0.743583i $$0.733125\pi$$
$$464$$ 0 0
$$465$$ 0 0
$$466$$ 0 0
$$467$$ − 36.0000i − 1.66588i −0.553362 0.832941i $$-0.686655\pi$$
0.553362 0.832941i $$-0.313345\pi$$
$$468$$ 0 0
$$469$$ 0 0
$$470$$ 0 0
$$471$$ 14.0000 0.645086
$$472$$ 0 0
$$473$$ 16.0000i 0.735681i
$$474$$ 0 0
$$475$$ 0 0
$$476$$ 0 0
$$477$$ − 6.00000i − 0.274721i
$$478$$ 0 0
$$479$$ −40.0000 −1.82765 −0.913823 0.406112i $$-0.866884\pi$$
−0.913823 + 0.406112i $$0.866884\pi$$
$$480$$ 0 0
$$481$$ 4.00000 0.182384
$$482$$ 0 0
$$483$$ 0 0
$$484$$ 0 0
$$485$$ 0 0
$$486$$ 0 0
$$487$$ 8.00000i 0.362515i 0.983436 + 0.181257i $$0.0580167\pi$$
−0.983436 + 0.181257i $$0.941983\pi$$
$$488$$ 0 0
$$489$$ 20.0000 0.904431
$$490$$ 0 0
$$491$$ 12.0000 0.541552 0.270776 0.962642i $$-0.412720\pi$$
0.270776 + 0.962642i $$0.412720\pi$$
$$492$$ 0 0
$$493$$ − 12.0000i − 0.540453i
$$494$$ 0 0
$$495$$ 0 0
$$496$$ 0 0
$$497$$ 0 0
$$498$$ 0 0
$$499$$ −8.00000 −0.358129 −0.179065 0.983837i $$-0.557307\pi$$
−0.179065 + 0.983837i $$0.557307\pi$$
$$500$$ 0 0
$$501$$ −12.0000 −0.536120
$$502$$ 0 0
$$503$$ 4.00000i 0.178351i 0.996016 + 0.0891756i $$0.0284232\pi$$
−0.996016 + 0.0891756i $$0.971577\pi$$
$$504$$ 0 0
$$505$$ 0 0
$$506$$ 0 0
$$507$$ 9.00000i 0.399704i
$$508$$ 0 0
$$509$$ −38.0000 −1.68432 −0.842160 0.539227i $$-0.818716\pi$$
−0.842160 + 0.539227i $$0.818716\pi$$
$$510$$ 0 0
$$511$$ 0 0
$$512$$ 0 0
$$513$$ 8.00000i 0.353209i
$$514$$ 0 0
$$515$$ 0 0
$$516$$ 0 0
$$517$$ − 48.0000i − 2.11104i
$$518$$ 0 0
$$519$$ −14.0000 −0.614532
$$520$$ 0 0
$$521$$ −30.0000 −1.31432 −0.657162 0.753749i $$-0.728243\pi$$
−0.657162 + 0.753749i $$0.728243\pi$$
$$522$$ 0 0
$$523$$ − 4.00000i − 0.174908i −0.996169 0.0874539i $$-0.972127\pi$$
0.996169 0.0874539i $$-0.0278730\pi$$
$$524$$ 0 0
$$525$$ 0 0
$$526$$ 0 0
$$527$$ 0 0
$$528$$ 0 0
$$529$$ 7.00000 0.304348
$$530$$ 0 0
$$531$$ 12.0000 0.520756
$$532$$ 0 0
$$533$$ 12.0000i 0.519778i
$$534$$ 0 0
$$535$$ 0 0
$$536$$ 0 0
$$537$$ − 12.0000i − 0.517838i
$$538$$ 0 0
$$539$$ 28.0000 1.20605
$$540$$ 0 0
$$541$$ −30.0000 −1.28980 −0.644900 0.764267i $$-0.723101\pi$$
−0.644900 + 0.764267i $$0.723101\pi$$
$$542$$ 0 0
$$543$$ − 22.0000i − 0.944110i
$$544$$ 0 0
$$545$$ 0 0
$$546$$ 0 0
$$547$$ − 28.0000i − 1.19719i −0.801050 0.598597i $$-0.795725\pi$$
0.801050 0.598597i $$-0.204275\pi$$
$$548$$ 0 0
$$549$$ 14.0000 0.597505
$$550$$ 0 0
$$551$$ 48.0000 2.04487
$$552$$ 0 0
$$553$$ 0 0
$$554$$ 0 0
$$555$$ 0 0
$$556$$ 0 0
$$557$$ 18.0000i 0.762684i 0.924434 + 0.381342i $$0.124538\pi$$
−0.924434 + 0.381342i $$0.875462\pi$$
$$558$$ 0 0
$$559$$ 8.00000 0.338364
$$560$$ 0 0
$$561$$ −8.00000 −0.337760
$$562$$ 0 0
$$563$$ 36.0000i 1.51722i 0.651546 + 0.758610i $$0.274121\pi$$
−0.651546 + 0.758610i $$0.725879\pi$$
$$564$$ 0 0
$$565$$ 0 0
$$566$$ 0 0
$$567$$ 0 0
$$568$$ 0 0
$$569$$ −34.0000 −1.42535 −0.712677 0.701492i $$-0.752517\pi$$
−0.712677 + 0.701492i $$0.752517\pi$$
$$570$$ 0 0
$$571$$ 32.0000 1.33916 0.669579 0.742741i $$-0.266474\pi$$
0.669579 + 0.742741i $$0.266474\pi$$
$$572$$ 0 0
$$573$$ − 16.0000i − 0.668410i
$$574$$ 0 0
$$575$$ 0 0
$$576$$ 0 0
$$577$$ − 2.00000i − 0.0832611i −0.999133 0.0416305i $$-0.986745\pi$$
0.999133 0.0416305i $$-0.0132552\pi$$
$$578$$ 0 0
$$579$$ 22.0000 0.914289
$$580$$ 0 0
$$581$$ 0 0
$$582$$ 0 0
$$583$$ 24.0000i 0.993978i
$$584$$ 0 0
$$585$$ 0 0
$$586$$ 0 0
$$587$$ − 36.0000i − 1.48588i −0.669359 0.742940i $$-0.733431\pi$$
0.669359 0.742940i $$-0.266569\pi$$
$$588$$ 0 0
$$589$$ 0 0
$$590$$ 0 0
$$591$$ −26.0000 −1.06950
$$592$$ 0 0
$$593$$ − 42.0000i − 1.72473i −0.506284 0.862367i $$-0.668981\pi$$
0.506284 0.862367i $$-0.331019\pi$$
$$594$$ 0 0
$$595$$ 0 0
$$596$$ 0 0
$$597$$ 16.0000i 0.654836i
$$598$$ 0 0
$$599$$ 16.0000 0.653742 0.326871 0.945069i $$-0.394006\pi$$
0.326871 + 0.945069i $$0.394006\pi$$
$$600$$ 0 0
$$601$$ 10.0000 0.407909 0.203954 0.978980i $$-0.434621\pi$$
0.203954 + 0.978980i $$0.434621\pi$$
$$602$$ 0 0
$$603$$ − 12.0000i − 0.488678i
$$604$$ 0 0
$$605$$ 0 0
$$606$$ 0 0
$$607$$ 8.00000i 0.324710i 0.986732 + 0.162355i $$0.0519090\pi$$
−0.986732 + 0.162355i $$0.948091\pi$$
$$608$$ 0 0
$$609$$ 0 0
$$610$$ 0 0
$$611$$ −24.0000 −0.970936
$$612$$ 0 0
$$613$$ 14.0000i 0.565455i 0.959200 + 0.282727i $$0.0912392\pi$$
−0.959200 + 0.282727i $$0.908761\pi$$
$$614$$ 0 0
$$615$$ 0 0
$$616$$ 0 0
$$617$$ 42.0000i 1.69086i 0.534089 + 0.845428i $$0.320655\pi$$
−0.534089 + 0.845428i $$0.679345\pi$$
$$618$$ 0 0
$$619$$ 16.0000 0.643094 0.321547 0.946894i $$-0.395797\pi$$
0.321547 + 0.946894i $$0.395797\pi$$
$$620$$ 0 0
$$621$$ −4.00000 −0.160514
$$622$$ 0 0
$$623$$ 0 0
$$624$$ 0 0
$$625$$ 0 0
$$626$$ 0 0
$$627$$ − 32.0000i − 1.27796i
$$628$$ 0 0
$$629$$ −4.00000 −0.159490
$$630$$ 0 0
$$631$$ 32.0000 1.27390 0.636950 0.770905i $$-0.280196\pi$$
0.636950 + 0.770905i $$0.280196\pi$$
$$632$$ 0 0
$$633$$ − 8.00000i − 0.317971i
$$634$$ 0 0
$$635$$ 0 0
$$636$$ 0 0
$$637$$ − 14.0000i − 0.554700i
$$638$$ 0 0
$$639$$ 0 0
$$640$$ 0 0
$$641$$ 2.00000 0.0789953 0.0394976 0.999220i $$-0.487424\pi$$
0.0394976 + 0.999220i $$0.487424\pi$$
$$642$$ 0 0
$$643$$ 12.0000i 0.473234i 0.971603 + 0.236617i $$0.0760386\pi$$
−0.971603 + 0.236617i $$0.923961\pi$$
$$644$$ 0 0
$$645$$ 0 0
$$646$$ 0 0
$$647$$ 20.0000i 0.786281i 0.919478 + 0.393141i $$0.128611\pi$$
−0.919478 + 0.393141i $$0.871389\pi$$
$$648$$ 0 0
$$649$$ −48.0000 −1.88416
$$650$$ 0 0
$$651$$ 0 0
$$652$$ 0 0
$$653$$ 14.0000i 0.547862i 0.961749 + 0.273931i $$0.0883240\pi$$
−0.961749 + 0.273931i $$0.911676\pi$$
$$654$$ 0 0
$$655$$ 0 0
$$656$$ 0 0
$$657$$ − 2.00000i − 0.0780274i
$$658$$ 0 0
$$659$$ −20.0000 −0.779089 −0.389545 0.921008i $$-0.627368\pi$$
−0.389545 + 0.921008i $$0.627368\pi$$
$$660$$ 0 0
$$661$$ 26.0000 1.01128 0.505641 0.862744i $$-0.331256\pi$$
0.505641 + 0.862744i $$0.331256\pi$$
$$662$$ 0 0
$$663$$ 4.00000i 0.155347i
$$664$$ 0 0
$$665$$ 0 0
$$666$$ 0 0
$$667$$ 24.0000i 0.929284i
$$668$$ 0 0
$$669$$ 24.0000 0.927894
$$670$$ 0 0
$$671$$ −56.0000 −2.16186
$$672$$ 0 0
$$673$$ − 6.00000i − 0.231283i −0.993291 0.115642i $$-0.963108\pi$$
0.993291 0.115642i $$-0.0368924\pi$$
$$674$$ 0 0
$$675$$ 0 0
$$676$$ 0 0
$$677$$ − 6.00000i − 0.230599i −0.993331 0.115299i $$-0.963217\pi$$
0.993331 0.115299i $$-0.0367827\pi$$
$$678$$ 0 0
$$679$$ 0 0
$$680$$ 0 0
$$681$$ −4.00000 −0.153280
$$682$$ 0 0
$$683$$ 12.0000i 0.459167i 0.973289 + 0.229584i $$0.0737364\pi$$
−0.973289 + 0.229584i $$0.926264\pi$$
$$684$$ 0 0
$$685$$ 0 0
$$686$$ 0 0
$$687$$ − 18.0000i − 0.686743i
$$688$$ 0 0
$$689$$ 12.0000 0.457164
$$690$$ 0 0
$$691$$ 16.0000 0.608669 0.304334 0.952565i $$-0.401566\pi$$
0.304334 + 0.952565i $$0.401566\pi$$
$$692$$ 0 0
$$693$$ 0 0
$$694$$ 0 0
$$695$$ 0 0
$$696$$ 0 0
$$697$$ − 12.0000i − 0.454532i
$$698$$ 0 0
$$699$$ −6.00000 −0.226941
$$700$$ 0 0
$$701$$ −10.0000 −0.377695 −0.188847 0.982006i $$-0.560475\pi$$
−0.188847 + 0.982006i $$0.560475\pi$$
$$702$$ 0 0
$$703$$ − 16.0000i − 0.603451i
$$704$$ 0 0
$$705$$ 0 0
$$706$$ 0 0
$$707$$ 0 0
$$708$$ 0 0
$$709$$ −2.00000 −0.0751116 −0.0375558 0.999295i $$-0.511957\pi$$
−0.0375558 + 0.999295i $$0.511957\pi$$
$$710$$ 0 0
$$711$$ 8.00000 0.300023
$$712$$ 0 0
$$713$$ 0 0
$$714$$ 0 0
$$715$$ 0 0
$$716$$ 0 0
$$717$$ − 8.00000i − 0.298765i
$$718$$ 0 0
$$719$$ 32.0000 1.19340 0.596699 0.802465i $$-0.296479\pi$$
0.596699 + 0.802465i $$0.296479\pi$$
$$720$$ 0 0
$$721$$ 0 0
$$722$$ 0 0
$$723$$ 2.00000i 0.0743808i
$$724$$ 0 0
$$725$$ 0 0
$$726$$ 0 0
$$727$$ 40.0000i 1.48352i 0.670667 + 0.741759i $$0.266008\pi$$
−0.670667 + 0.741759i $$0.733992\pi$$
$$728$$ 0 0
$$729$$ −1.00000 −0.0370370
$$730$$ 0 0
$$731$$ −8.00000 −0.295891
$$732$$ 0 0
$$733$$ − 10.0000i − 0.369358i −0.982799 0.184679i $$-0.940875\pi$$
0.982799 0.184679i $$-0.0591246\pi$$
$$734$$ 0 0
$$735$$ 0 0
$$736$$ 0 0
$$737$$ 48.0000i 1.76810i
$$738$$ 0 0
$$739$$ 8.00000 0.294285 0.147142 0.989115i $$-0.452992\pi$$
0.147142 + 0.989115i $$0.452992\pi$$
$$740$$ 0 0
$$741$$ −16.0000 −0.587775
$$742$$ 0 0
$$743$$ 36.0000i 1.32071i 0.750953 + 0.660356i $$0.229595\pi$$
−0.750953 + 0.660356i $$0.770405\pi$$
$$744$$ 0 0
$$745$$ 0 0
$$746$$ 0 0
$$747$$ 4.00000i 0.146352i
$$748$$ 0 0
$$749$$ 0 0
$$750$$ 0 0
$$751$$ −32.0000 −1.16770 −0.583848 0.811863i $$-0.698454\pi$$
−0.583848 + 0.811863i $$0.698454\pi$$
$$752$$ 0 0
$$753$$ 12.0000i 0.437304i
$$754$$ 0 0
$$755$$ 0 0
$$756$$ 0 0
$$757$$ 34.0000i 1.23575i 0.786276 + 0.617876i $$0.212006\pi$$
−0.786276 + 0.617876i $$0.787994\pi$$
$$758$$ 0 0
$$759$$ 16.0000 0.580763
$$760$$ 0 0
$$761$$ −30.0000 −1.08750 −0.543750 0.839248i $$-0.682996\pi$$
−0.543750 + 0.839248i $$0.682996\pi$$
$$762$$ 0 0
$$763$$ 0 0
$$764$$ 0 0
$$765$$ 0 0
$$766$$ 0 0
$$767$$ 24.0000i 0.866590i
$$768$$ 0 0
$$769$$ 14.0000 0.504853 0.252426 0.967616i $$-0.418771\pi$$
0.252426 + 0.967616i $$0.418771\pi$$
$$770$$ 0 0
$$771$$ 6.00000 0.216085
$$772$$ 0 0
$$773$$ − 10.0000i − 0.359675i −0.983696 0.179838i $$-0.942443\pi$$
0.983696 0.179838i $$-0.0575572\pi$$
$$774$$ 0 0
$$775$$ 0 0
$$776$$ 0 0
$$777$$ 0 0
$$778$$ 0 0
$$779$$ 48.0000 1.71978
$$780$$ 0 0
$$781$$ 0 0
$$782$$ 0 0
$$783$$ 6.00000i 0.214423i
$$784$$ 0 0
$$785$$ 0 0
$$786$$ 0 0
$$787$$ − 20.0000i − 0.712923i −0.934310 0.356462i $$-0.883983\pi$$
0.934310 0.356462i $$-0.116017\pi$$
$$788$$ 0 0
$$789$$ 20.0000 0.712019
$$790$$ 0 0
$$791$$ 0 0
$$792$$ 0 0
$$793$$ 28.0000i 0.994309i
$$794$$ 0 0
$$795$$ 0 0
$$796$$ 0 0
$$797$$ 34.0000i 1.20434i 0.798367 + 0.602171i $$0.205697\pi$$
−0.798367 + 0.602171i $$0.794303\pi$$
$$798$$ 0 0
$$799$$ 24.0000 0.849059
$$800$$ 0 0
$$801$$ 2.00000 0.0706665
$$802$$ 0 0
$$803$$ 8.00000i 0.282314i
$$804$$ 0 0
$$805$$ 0 0
$$806$$ 0 0
$$807$$ 10.0000i 0.352017i
$$808$$ 0 0
$$809$$ 22.0000 0.773479 0.386739 0.922189i $$-0.373601\pi$$
0.386739 + 0.922189i $$0.373601\pi$$
$$810$$ 0 0
$$811$$ 24.0000 0.842754 0.421377 0.906886i $$-0.361547\pi$$
0.421377 + 0.906886i $$0.361547\pi$$
$$812$$ 0 0
$$813$$ − 8.00000i − 0.280572i
$$814$$ 0 0
$$815$$ 0 0
$$816$$ 0 0
$$817$$ − 32.0000i − 1.11954i
$$818$$ 0 0
$$819$$ 0 0
$$820$$ 0 0
$$821$$ 30.0000 1.04701 0.523504 0.852023i $$-0.324625\pi$$
0.523504 + 0.852023i $$0.324625\pi$$
$$822$$ 0 0
$$823$$ − 16.0000i − 0.557725i −0.960331 0.278862i $$-0.910043\pi$$
0.960331 0.278862i $$-0.0899574\pi$$
$$824$$ 0 0
$$825$$ 0 0
$$826$$ 0 0
$$827$$ 36.0000i 1.25184i 0.779886 + 0.625921i $$0.215277\pi$$
−0.779886 + 0.625921i $$0.784723\pi$$
$$828$$ 0 0
$$829$$ 30.0000 1.04194 0.520972 0.853574i $$-0.325570\pi$$
0.520972 + 0.853574i $$0.325570\pi$$
$$830$$ 0 0
$$831$$ 30.0000 1.04069
$$832$$ 0 0
$$833$$ 14.0000i 0.485071i
$$834$$ 0 0
$$835$$ 0 0
$$836$$ 0 0
$$837$$ 0 0
$$838$$ 0 0
$$839$$ 48.0000 1.65714 0.828572 0.559883i $$-0.189154\pi$$
0.828572 + 0.559883i $$0.189154\pi$$
$$840$$ 0 0
$$841$$ 7.00000 0.241379
$$842$$ 0 0
$$843$$ 2.00000i 0.0688837i
$$844$$ 0 0
$$845$$ 0 0
$$846$$ 0 0
$$847$$ 0 0
$$848$$ 0 0
$$849$$ 4.00000 0.137280
$$850$$ 0 0
$$851$$ 8.00000 0.274236
$$852$$ 0 0
$$853$$ 30.0000i 1.02718i 0.858036 + 0.513590i $$0.171685\pi$$
−0.858036 + 0.513590i $$0.828315\pi$$
$$854$$ 0 0
$$855$$ 0 0
$$856$$ 0 0
$$857$$ 10.0000i 0.341593i 0.985306 + 0.170797i $$0.0546341\pi$$
−0.985306 + 0.170797i $$0.945366\pi$$
$$858$$ 0 0
$$859$$ 32.0000 1.09183 0.545913 0.837842i $$-0.316183\pi$$
0.545913 + 0.837842i $$0.316183\pi$$
$$860$$ 0 0
$$861$$ 0 0
$$862$$ 0 0
$$863$$ − 4.00000i − 0.136162i −0.997680 0.0680808i $$-0.978312\pi$$
0.997680 0.0680808i $$-0.0216876\pi$$
$$864$$ 0 0
$$865$$ 0 0
$$866$$ 0 0
$$867$$ 13.0000i 0.441503i
$$868$$ 0 0
$$869$$ −32.0000 −1.08553
$$870$$ 0 0
$$871$$ 24.0000 0.813209
$$872$$ 0 0
$$873$$ − 14.0000i − 0.473828i
$$874$$ 0 0
$$875$$ 0 0
$$876$$ 0 0
$$877$$ − 14.0000i − 0.472746i −0.971662 0.236373i $$-0.924041\pi$$
0.971662 0.236373i $$-0.0759588\pi$$
$$878$$ 0 0
$$879$$ 26.0000 0.876958
$$880$$ 0 0
$$881$$ −6.00000 −0.202145 −0.101073 0.994879i $$-0.532227\pi$$
−0.101073 + 0.994879i $$0.532227\pi$$
$$882$$ 0 0
$$883$$ 44.0000i 1.48072i 0.672212 + 0.740359i $$0.265344\pi$$
−0.672212 + 0.740359i $$0.734656\pi$$
$$884$$ 0 0
$$885$$ 0 0
$$886$$ 0 0
$$887$$ − 20.0000i − 0.671534i −0.941945 0.335767i $$-0.891004\pi$$
0.941945 0.335767i $$-0.108996\pi$$
$$888$$ 0 0
$$889$$ 0 0
$$890$$ 0 0
$$891$$ 4.00000 0.134005
$$892$$ 0 0
$$893$$ 96.0000i 3.21252i
$$894$$ 0 0
$$895$$ 0 0
$$896$$ 0 0
$$897$$ − 8.00000i − 0.267112i
$$898$$ 0 0
$$899$$ 0 0
$$900$$ 0 0
$$901$$ −12.0000 −0.399778
$$902$$ 0 0
$$903$$ 0 0
$$904$$ 0 0
$$905$$ 0 0
$$906$$ 0 0
$$907$$ 36.0000i 1.19536i 0.801735 + 0.597680i $$0.203911\pi$$
−0.801735 + 0.597680i $$0.796089\pi$$
$$908$$ 0 0
$$909$$ −14.0000 −0.464351
$$910$$ 0 0
$$911$$ 8.00000 0.265052 0.132526 0.991180i $$-0.457691\pi$$
0.132526 + 0.991180i $$0.457691\pi$$
$$912$$ 0 0
$$913$$ − 16.0000i − 0.529523i
$$914$$ 0 0
$$915$$ 0 0
$$916$$ 0 0
$$917$$ 0 0
$$918$$ 0 0
$$919$$ −8.00000 −0.263896 −0.131948 0.991257i $$-0.542123\pi$$
−0.131948 + 0.991257i $$0.542123\pi$$
$$920$$ 0 0
$$921$$ 28.0000 0.922631
$$922$$ 0 0
$$923$$ 0 0
$$924$$ 0 0
$$925$$ 0 0
$$926$$ 0 0
$$927$$ 8.00000i 0.262754i
$$928$$ 0 0
$$929$$ −42.0000 −1.37798 −0.688988 0.724773i $$-0.741945\pi$$
−0.688988 + 0.724773i $$0.741945\pi$$
$$930$$ 0 0
$$931$$ −56.0000 −1.83533
$$932$$ 0 0
$$933$$ 0 0
$$934$$ 0 0
$$935$$ 0 0
$$936$$ 0 0
$$937$$ − 42.0000i − 1.37208i −0.727564 0.686040i $$-0.759347\pi$$
0.727564 0.686040i $$-0.240653\pi$$
$$938$$ 0 0
$$939$$ 14.0000 0.456873
$$940$$ 0 0
$$941$$ −34.0000 −1.10837 −0.554184 0.832394i $$-0.686970\pi$$
−0.554184 + 0.832394i $$0.686970\pi$$
$$942$$ 0 0
$$943$$ 24.0000i 0.781548i
$$944$$ 0 0
$$945$$ 0 0
$$946$$ 0 0
$$947$$ 4.00000i 0.129983i 0.997886 + 0.0649913i $$0.0207020\pi$$
−0.997886 + 0.0649913i $$0.979298\pi$$
$$948$$ 0 0
$$949$$ 4.00000 0.129845
$$950$$ 0 0
$$951$$ −18.0000 −0.583690
$$952$$ 0 0
$$953$$ 30.0000i 0.971795i 0.874016 + 0.485898i $$0.161507\pi$$
−0.874016 + 0.485898i $$0.838493\pi$$
$$954$$ 0 0
$$955$$ 0 0
$$956$$ 0 0
$$957$$ − 24.0000i − 0.775810i
$$958$$ 0 0
$$959$$ 0 0
$$960$$ 0 0
$$961$$ −31.0000 −1.00000
$$962$$ 0 0
$$963$$ − 12.0000i − 0.386695i
$$964$$ 0 0
$$965$$ 0 0
$$966$$ 0 0
$$967$$ − 32.0000i − 1.02905i −0.857475 0.514525i $$-0.827968\pi$$
0.857475 0.514525i $$-0.172032\pi$$
$$968$$ 0 0
$$969$$ 16.0000 0.513994
$$970$$ 0 0
$$971$$ 28.0000 0.898563 0.449281 0.893390i $$-0.351680\pi$$
0.449281 + 0.893390i $$0.351680\pi$$
$$972$$ 0 0
$$973$$ 0 0
$$974$$ 0 0
$$975$$ 0 0
$$976$$ 0 0
$$977$$ 18.0000i 0.575871i 0.957650 + 0.287936i $$0.0929689\pi$$
−0.957650 + 0.287936i $$0.907031\pi$$
$$978$$ 0 0
$$979$$ −8.00000 −0.255681
$$980$$ 0 0
$$981$$ −14.0000 −0.446986
$$982$$ 0 0
$$983$$ 4.00000i 0.127580i 0.997963 + 0.0637901i $$0.0203188\pi$$
−0.997963 + 0.0637901i $$0.979681\pi$$
$$984$$ 0 0
$$985$$ 0 0
$$986$$ 0 0
$$987$$ 0 0
$$988$$ 0 0
$$989$$ 16.0000 0.508770
$$990$$ 0 0
$$991$$ −40.0000 −1.27064 −0.635321 0.772248i $$-0.719132\pi$$
−0.635321 + 0.772248i $$0.719132\pi$$
$$992$$ 0 0
$$993$$ 0 0
$$994$$ 0 0
$$995$$ 0 0
$$996$$ 0 0
$$997$$ − 6.00000i − 0.190022i −0.995476 0.0950110i $$-0.969711\pi$$
0.995476 0.0950110i $$-0.0302886\pi$$
$$998$$ 0 0
$$999$$ 2.00000 0.0632772
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
## Twists
By twisting character
Char Parity Ord Type Twist Min Dim
1.1 even 1 trivial 4800.2.f.ba.3649.2 2
4.3 odd 2 4800.2.f.j.3649.1 2
5.2 odd 4 960.2.a.o.1.1 1
5.3 odd 4 4800.2.a.u.1.1 1
5.4 even 2 inner 4800.2.f.ba.3649.1 2
8.3 odd 2 2400.2.f.n.1249.2 2
8.5 even 2 2400.2.f.e.1249.1 2
15.2 even 4 2880.2.a.i.1.1 1
20.3 even 4 4800.2.a.ca.1.1 1
20.7 even 4 960.2.a.f.1.1 1
20.19 odd 2 4800.2.f.j.3649.2 2
24.5 odd 2 7200.2.f.bb.6049.2 2
24.11 even 2 7200.2.f.b.6049.2 2
40.3 even 4 2400.2.a.j.1.1 1
40.13 odd 4 2400.2.a.y.1.1 1
40.19 odd 2 2400.2.f.n.1249.1 2
40.27 even 4 480.2.a.e.1.1 yes 1
40.29 even 2 2400.2.f.e.1249.2 2
40.37 odd 4 480.2.a.b.1.1 1
60.47 odd 4 2880.2.a.j.1.1 1
80.27 even 4 3840.2.k.k.1921.2 2
80.37 odd 4 3840.2.k.p.1921.1 2
80.67 even 4 3840.2.k.k.1921.1 2
80.77 odd 4 3840.2.k.p.1921.2 2
120.29 odd 2 7200.2.f.bb.6049.1 2
120.53 even 4 7200.2.a.bg.1.1 1
120.59 even 2 7200.2.f.b.6049.1 2
120.77 even 4 1440.2.a.k.1.1 1
120.83 odd 4 7200.2.a.u.1.1 1
120.107 odd 4 1440.2.a.j.1.1 1
By twisted newform
Twist Min Dim Char Parity Ord Type
480.2.a.b.1.1 1 40.37 odd 4
480.2.a.e.1.1 yes 1 40.27 even 4
960.2.a.f.1.1 1 20.7 even 4
960.2.a.o.1.1 1 5.2 odd 4
1440.2.a.j.1.1 1 120.107 odd 4
1440.2.a.k.1.1 1 120.77 even 4
2400.2.a.j.1.1 1 40.3 even 4
2400.2.a.y.1.1 1 40.13 odd 4
2400.2.f.e.1249.1 2 8.5 even 2
2400.2.f.e.1249.2 2 40.29 even 2
2400.2.f.n.1249.1 2 40.19 odd 2
2400.2.f.n.1249.2 2 8.3 odd 2
2880.2.a.i.1.1 1 15.2 even 4
2880.2.a.j.1.1 1 60.47 odd 4
3840.2.k.k.1921.1 2 80.67 even 4
3840.2.k.k.1921.2 2 80.27 even 4
3840.2.k.p.1921.1 2 80.37 odd 4
3840.2.k.p.1921.2 2 80.77 odd 4
4800.2.a.u.1.1 1 5.3 odd 4
4800.2.a.ca.1.1 1 20.3 even 4
4800.2.f.j.3649.1 2 4.3 odd 2
4800.2.f.j.3649.2 2 20.19 odd 2
4800.2.f.ba.3649.1 2 5.4 even 2 inner
4800.2.f.ba.3649.2 2 1.1 even 1 trivial
7200.2.a.u.1.1 1 120.83 odd 4
7200.2.a.bg.1.1 1 120.53 even 4
7200.2.f.b.6049.1 2 120.59 even 2
7200.2.f.b.6049.2 2 24.11 even 2
7200.2.f.bb.6049.1 2 120.29 odd 2
7200.2.f.bb.6049.2 2 24.5 odd 2
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# The Secret Behind The Alignment Of The Pyramids Of Giza: Uncovered?
The alignment of the Giza pyramids has long been a mystery for scientists. However, a new theory could explain how the Egyptians managed to build them so accurately and precisely considering the time they were built.
Researchers have spent centuries wondering just how the Ancient Egyptians managed to build the pyramids, and in particular, how they managed to build and align them so accurately and with such precision. A new theory, published in The Journal of Ancient Egyptian Architecture, suggests that those who built the pyramids used the autumnal equinox to manage this incredible achievement.
Discover our latest podcast
## The ‘Indian circle’ method
The four angles of the Great Pyramid of Giza are surprisingly accurate and almost perfectly aligned with the four cardinal points. The three pyramids themselves are also aligned with each other in almost perfect proportions. Although nowadays it is possible for us to observe this by using drones and satellite data, at the time, this wouldn’t have been such an easy feat.
In 2018, archaeologist and engineer Glen Dash proposed a simple yet elegant hypothesis: the Egyptians used the autumnal equinox to establish their measurements using a gnomon (a vertical rod used to track the movement of the sun on the equinox).
Using a series of measurements taken from the equinox on 22nd September 2016, Dash managed to demonstrate that the so-called ‘Indian circle' method allowed them to obtain the orientation of the cardinal points with incredible accuracy.
## From one rod to a huge pyramid
Using this gnomon, it would have been possible to follow and trace the trajectory of the sun across the sky by tracking the regular shadows projected by the rod from the sun. Once these shadows had formed a smooth curve, all the Egyptians would have needed to do is draw a circle with the gnomon at its centre and mark the two points of intersection between the circle and the curve. Then, they would have needed to draw a straight line from one point to another and they would have a line travelling from east to west (or vice versa), provided that their measurements were taken during the equinox.
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http://jurnal.unsyiah.ac.id/peluang/article/view/13749
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### Pemahaman Relasional Siswa pada Turunan Fungsi dengan Bantuan Software Geometer’s Sketchpad
Cut Laila Kulsum, Rahmah Johar, Said Munzir
#### Abstract
Relational understanding in mathematics is important for the student's it causes understanding the mathematical concepts will be memorable if he knows the process of getting the concept by linking it with the knowledge to be learned. This study aimed to examine the relational understanding of students in discovering the concept of f '(x) using Geometer's Sketchpad, and the relational understanding of students in discovering the concept of f (x) = x2 + c let f' (x) = 2x using Geometer's Sketchpad. It also investigated the students' responses when they use Geometer's Sketchpad in understanding the derivative function. The study used a qualitative approach. Subjects in this study were three Year 11 students of one the junior high school in Meuredu, Aceh, Indonesia. The instruments were observation sheet, interview guides, student worksheets and test questions. The data analysis included the classification of information obtained from observations when the subjects used software Sketchpad Geometers in their learning. Information about students' relational understanding obtained from the student worksheets reinforced with test questions and interviews so that the results can be accounted for. This study concluded that (1) students' relational understanding (Subjects RV, subject MR, subject AA) in discovering the concept of f '(x) using Geometer's Sketchpad was good; (2) students' relational understanding (Subjects RV, subject MR, subject AA) in discovering the concept of f (x) = x2 + c let f '(x) = 2x using Geometer's Sketchpad was a good; and (3) the responses of Subjects RV, subject MR, subject AA in using Geometer's Sketchpad were positive, each research subject was very pleased and liked to use Geometer's Sketchpad in understanding the derivative function.
PDF
#### References
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Steffe, L. P., & Thompson, P. W. (2000). Teaching experiment methodology: Underlying principles and essential elements. In R. Lesh & A. E. Kelly (Eds.), Research design in mathematics and science education (pp. 267- 307). Hillsdale, NJ: Erlbaum.
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Sumarmo, Utari. (2012). Bahan Ajar Evaluasi Pembelajaran Matematika Program S2 Pendidikan Matematika.Universitas Terbuka
_______. (2010) Berfikir dan Disposisi Matematik : Apa, Mengapa, dan Bagaimana Dikembangkan pada Peserta didik. Jurnal FMIPA UPI Bandung.
Van de Walle, J. A (2008).Matematika Sekolah Dasar dan Menengah.Jakarta. Erlangga
DOI: https://doi.org/10.24815/jp.v7i2.13749
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ISSN 2302-5158 (print) 2685-1539 (online)
Organized by Program Studi S1 Pendidikan Matematika, FKIP Unsyiah
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Discord
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# Why does my code go to decimals, when the math says it won't?
So, I was making an upgrade for a clicking game im making, and I just want to know why this turns to decimals, here all the code:
money = 0 tupgrade = 0 tprice = 100 aupgrade = 0 aprice = 0 click = 1
if keyboard.press.1 and money >= tprice then money = money-tprice tprice = tprice*1.2 tupgrade = tupgrade+1 click = click+1 end
Just in case, I also made an unlisted version of it. https://microstudio.dev/i/JustNitro/clickersimtest/
( in advance, I read this and it sounds really condescending, I dont mean for it to be, sorry :( )
Ok, so, the price of an upgrade is 144. Now you are multiplying that by 1.2 (not the original 100, the new 144). 144 * 1.2 = 172.8. The game giving you 172.7999999999999999998 is weird, but I would presume it has something to do with floating point precision.
THE FIX: Personally, I would just change tprice = tprice1.2 into tprice = round(tprice1.2) to round to the nearest integer.
This is due to the limits of the 64 bits floating point numbers standard. For more information, have a look here: https://microstudio.dev/community/bugs/minor-innacuracies-with-math-functions/141/3/
The workaround is to use rounding, whenever necessary, as @JmeJuniper suggests.
Thanks, I will keep these in mind next time.
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# Distance between Rezh, Sverdlovsk Oblast, Russia and the Antarctic Circle
13796 km = 8572 miles
During our calculation of the distance to the Antarctic Circle we make two assumptions:
1. We assume a spherical Earth as a close approximation of the true shape of the Earth (an oblate spheroid). The distance is calculated as great-circle or orthodromic distance on the surface of a sphere.
2. We calculate the distance between a point on the Earth’s surface and the Antarctic Circle as the length of the arc of the meridian passing through this point and crossing the Antarctic Circle.
Find out the distance between Rezh and the North Pole, the South Pole, the Equator, the Tropic of Cancer, the Tropic of Capricorn, the Arctic Circle
Find out the distance between Rezh and other cities
## Rezh, Russia
Country: Russia
Rezh’s coordinates: 57°22′12″ N, 61°24′15″ E
Population: 39,216
Find out what time it is in Rezh right now
Wikipedia article: Rezh
## The Antarctic Circle
The Antarctic Circle is a circle of latitude or parallel on the Earth's surface. The polar night and the midnight sun phenomena can be observed to the south of the Antarctic Circle.
During the midnight sun there is no sunset and the Sun is over the horizon continuously for 24 hours or more.
During the polar night there is no sunrise and the Sun is below the horizon continuously for 24 hours or more.
For the points on the Earth's surface located at the Antarctic Circle the polar night and the midnight sun coincide with the winter solstice and the summer solstice correspondingly and their duration equals 24 hours.
For the points on the Earth's surface located to the south of the Antarctic Circle the duration of the polar night and the midnight sun grows with latitude.
The latitude of the Antarctic Circle depends on the tilt of the Earth's axis which changes with time.
The current latitude of the Antarctic Circle equals 66°33′43″ S.
Wikipedia article: the Antarctic Circle
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# How far is Xi'an from Panzhihua?
The distance between Panzhihua (Panzhihua Bao'anying Airport) and Xi'an (Xi'an Xianyang International Airport) is 684 miles / 1101 kilometers / 595 nautical miles.
The driving distance from Panzhihua (PZI) to Xi'an (XIY) is 852 miles / 1371 kilometers, and travel time by car is about 15 hours 34 minutes.
684
Miles
1101
Kilometers
595
Nautical miles
1 h 47 min
123 kg
## Distance from Panzhihua to Xi'an
There are several ways to calculate the distance from Panzhihua to Xi'an. Here are two standard methods:
Vincenty's formula (applied above)
• 684.209 miles
• 1101.128 kilometers
• 594.561 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 685.019 miles
• 1102.432 kilometers
• 595.265 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Panzhihua to Xi'an?
The estimated flight time from Panzhihua Bao'anying Airport to Xi'an Xianyang International Airport is 1 hour and 47 minutes.
## Flight carbon footprint between Panzhihua Bao'anying Airport (PZI) and Xi'an Xianyang International Airport (XIY)
On average, flying from Panzhihua to Xi'an generates about 123 kg of CO2 per passenger, and 123 kilograms equals 271 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Panzhihua to Xi'an
See the map of the shortest flight path between Panzhihua Bao'anying Airport (PZI) and Xi'an Xianyang International Airport (XIY).
## Airport information
Origin Panzhihua Bao'anying Airport
City: Panzhihua
Country: China
IATA Code: PZI
ICAO Code: ZUZH
Coordinates: 26°32′24″N, 101°47′54″E
Destination Xi'an Xianyang International Airport
City: Xi'an
Country: China
IATA Code: XIY
ICAO Code: ZLXY
Coordinates: 34°26′49″N, 108°45′7″E
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Next: , Previous: , Up: Answers to Exercises [Contents][Index]
2.7.11 Modes Tutorial Exercise 4
Many calculations involve real-world quantities, like the width and height of a piece of wood or the volume of a jar. Such quantities can’t be measured exactly anyway, and if the data that is input to a calculation is inexact, doing exact arithmetic on it is a waste of time.
Fractions become unwieldy after too many calculations have been done with them. For example, the sum of the reciprocals of the integers from 1 to 10 is 7381:2520. The sum from 1 to 30 is 9304682830147:2329089562800. After a point it will take a long time to add even one more term to this sum, but a floating-point calculation of the sum will not have this problem.
Also, rational numbers cannot express the results of all calculations. There is no fractional form for the square root of two, so if you type 2 Q, Calc has no choice but to give you a floating-point answer.
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• Email
Written by David L. Goodstein
Last Updated
Written by David L. Goodstein
Last Updated
# mechanics
Written by David L. Goodstein
Last Updated
### Relative motion
A collision between two bodies can always be described in a frame of reference in which the total momentum is zero. This is the centre-of-mass (or centre-of-momentum) frame mentioned earlier. Then, for example, in the collision between two bodies of the same mass discussed above, the two bodies always have equal and opposite velocities, as shown in Figure 14. It should be noted that, in this frame of reference, the outgoing momenta are antiparallel and not perpendicular.
Any collection of bodies may similarly be described in a frame of reference in which the total momentum is zero. This frame is simply the one in which the centre of mass is at rest. This fact is easily seen by differentiating equation (55) with respect to time, giving
The right-hand side is the sum of the momenta of all the bodies. It is equal to zero if the velocity of the centre of mass, dR/dt, is equal to zero.
If Newton’s second law is correct in any frame of reference, it will also appear to be correct to an observer moving with any constant velocity with respect to that frame. This principle, called the principle of ... (200 of 23,195 words)
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