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https://resources.wolframcloud.com/FunctionRepository/resources/TimeShiftedDistribution/
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Function Repository Resource:
# TimeShiftedDistribution
A derived distribution useful in actuarial science
Contributed by: Seth J. Chandler
ResourceFunction["TimeShiftedDistribution"][xmin,dist] represents a distribution dist that has been truncated to lie above xmin and then translated so that the lowest outcome is 0. ResourceFunction["TimeShiftedDistribution"][{xmin,ymin,…},dist] represents a multivariate truncation of the distribution dist, translated so the lowest outcome from the distribution is a vector of 0s.
## Details and Options
The translation works by subtracting xmin from the outputs so that the lowest outcome from the distribution is now 0.
A time-shifted distribution can be used to model the longevity of a person who had a certain longevity distribution at birth but who has survived to some attained age.
## Examples
### Basic Examples (2)
A BinomialDistribution, time-shifted so that its value must lie above 3 and be translated to the left by 3:
In[1]:=
Out[1]=
The probability density (mass) function of that distribution:
In[2]:=
Out[3]=
The distribution of the remaining life of a person aged 61, where longevity at birth is distributed according to a GompertzMakehamDistribution:
In[4]:=
Out[4]=
The survival function of that time-shifted distribution:
In[5]:=
Out[5]=
### Scope (2)
The mean of a time-shifted distribution of a ProductDistribution of two symbolic BetaDistributions:
In[6]:=
Out[6]=
A random variable drawn from a time-shifted MultinormalDistribution:
In[7]:=
Out[7]=
The numeric probability of a draw from a time-shifted CopulaDistribution falling into a particular set of values:
In[8]:=
Out[8]=
### Applications (2)
Show survival functions of a BetaDistribution for different time-shift values:
In[9]:=
Out[9]=
Show survival functions of a GompertzMakehamDistribution for different time-shift values:
In[10]:=
Out[10]=
### Properties and Relations (1)
Some distributions, such as the ExponentialDistribution, remain unchanged after time shifting :
In[11]:=
Out[11]=
### Neat Examples (3)
Consider a life insurance product in which the time a person dies and the time a person lets their policy lapse are given by the ProductDistribution of a GompertzMakehamDistribution (for longevity) and an ExponentialDistribution (for lapse). The insured has neither died nor lapsed before age 61. Compute the probability that an insurer will have to pay a death benefit between k and k+1 years thereafter and that the insurance policy will not have lapsed by that time:
In[12]:=
In[13]:=
Out[13]=
In[14]:=
Out[14]=
Compute the actuarial present value of a death benefit of \$1 for such a person, assuming the policy pays nothing if the policy lapses before death and assuming a 5% effective rate of interest:
In[15]:=
Out[15]=
Compute the maximum amount the insurer would be willing to pay a 75-year-old who holds such a policy to let it lapse, again assuming a 5% effective rate of interest:
In[16]:=
Out[16]=
Seth J. Chandler
## Version History
• 1.0.0 – 07 August 2019
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# If Rajendra obtains 10% marks more then Surendra then how many % less marks are obtained by Surendra than Rajendra?
245 views
If Rajendra obtains 10% marks more then Surendra then how many % less marks are obtained by Surendra than Rajendra?
posted Dec 31, 2015
Let Surendra obtain 100 marks.
Then Rajendra marks =100+10%of100=110
so surendra is 10 marks less than Rajendra's 110 marks.
% less =10/110*100=9.09%
Ans is 9.09%
answer Dec 31, 2015
Thanks sir
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This is an archived post. You won't be able to vote or comment.
you are viewing a single comment's thread.
[–] 3 points4 points (17 children)
Is this sarcasm? I can't tell.
[–] 9 points10 points (16 children)
Nope, im not even shitting you. Look at exponential growth, if a man starts out with a penny And he doubles the amount of money he has everyday, in 28 days he'll have over a million dollars, but during the first 10 days he'll barely have more than a couple dollars, once he gets to a million dollars the growth is immense.
[–] 1 point2 points (0 children)
It takes money to make money,
[–] 0 points1 point (0 children)
and he doubles the amount of money he has everyday
Oh I didn't realize it was that easy. Let me just go do that then.
[–] 0 points1 point (0 children)
He has a lot of skill, and a nation that has done everything in its power to fight deflation for the past two decades is behind him.
[–] -3 points-2 points (12 children)
True but investments dont really grow exponentially
[–] 9 points10 points (10 children)
Yes they do.. The potential gain is proportional to the current value, so of course it's exponential.
[–] -1 points0 points (0 children)
The compound interest formula is exponential. M(t) = M0 * (1 + r) ^ t. In this r=interest rate, M0 = the initial amount of money you invest, t = time in number of years (assuming yearly compounding).
This assumes that the interest rate is constant, while in reality (at least for the more financially interesting values of r) it's more likely that r = f(t), with f being an unknown function.
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1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question
# In the figure shown below, a horizontal force F is applied on 6 kg block towards left. If the coefficients of friction between the surfaces are 0.3 and 0.4 as shown in the figure. then the value of tension in the rope and force required just to slide the 6 kg block under the 3 kg block is (Take g=10 m/s2)
A
T=3 N,F=40 N
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B
T=6 N,F=45 N
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C
T=9 N,F=45 N
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D
T=9 N,F=40 N
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Open in App
Solution
## The correct option is C T=9 N,F=45 NAs per the question, the block of mass 6 kg will just slide under block of mass 3 kg. Then, the condition of friction will be limiting. The FBDs of the blocks are as shown At equilibrium, we have N1=3g=30 N f1=T In limiting case, f1=μ1N1=0.3×30=9 N ⇒ T=9 N Again, from the FBD of 6 kg block, we have N2=N1+6g=30+60=90 N F=f1+f2 In limiting case, f2=μ2N2=0.4×90=36 N ⇒ F=9+36=45 N
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https://www.teacherspayteachers.com/Product/5th-Grade-Daily-Math-Packet-1393918
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# Main Categories
Total:
\$0.00
Whoops! Something went wrong.
# 5th Grade Daily Math Packet
Product Rating
4.0
3 ratings
File Type
PDF (Acrobat) Document File
16 MB|7 pages
Product Description
I created this resource to incorporate math into the daily routine, as per Guided Math suggestion.
When students are younger, they often do Calendar Math. This is a spin off that. The teacher does need to do some prepping-planning out the daily number and parts of the worksheet. I plan to do a packet a week; it is 4 pages so the student should complete a page M-Th with math talk and then going over the process/answers on Friday, hopefully done as a morning activity before we get our day started!
We will start the school year slowly, maybe not even finishing the packet each week, as I need to review, and sometime TEACH!, the concepts. However, these are all important topics I feel my 5th graders need to know as they head into 6th grade and state tests. I have tried to incorporate various vocabulary, or places to use math talk, throughout the resource.
If you have any questions or suggestions feel free to contact me!
drazsclass@gmail.com
Topics covered:
Day 1-See Preview
Odd or Even
Factors
Prime/Composite
Factor Tree
Plotting on a number line
Ways to make
Day 2-
Write an equation (vocabulary)
Write a word problem
Geometry
School Day Fraction, simplify and decimal
Day 3-
Big number
Round, Place Value, Exapanded and word form
Data Set
Mean, Median, Mode, Range and Outlier
Day 4-
Big Multiplication
Big Division
T chart/table and graphing an equation
Total Pages
7 pages
N/A
Teaching Duration
30 minutes
• Product Q & A
\$3.00
\$3.00
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# `int_3^6 1/10x(6x^2-15)dx `
lemjay | Certified Educator
`int_3^6 1/10x(6x^2-15)dx`
First, factor out 1/10.
`=1/10int_3^6x(6x^2-15)dx`
Then, distribute x to (6x^2-15)dx.
`=1/10int_3^6(6x^3-15x)dx`
Then, integrate each term inside the parenthesis uisng the integral formula `intu^n du=u^(n+1)/(n+1)+C` .
`=1/10 ((6x^4)/4-(15x^2)/2)` `|_3^6`
`=1/10((3x^4)/2-(15x^2)/2) |_3^6`
`=1/10[((3*6^4)/2-(15*6^2)/2) -((3*3^4)/2-(15*3^2)/2)]`
`=1/10[(1944-270)-(243/2-135/2)]`
`=1/10(1674-54)`
`=1/10*1620`
`=162`
Hence, `int_3^6 1/10x(6x^2-15)dx=162` .
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# Help a new newbe Please!!!
#### Sand_in_my_Boots
##### New Member
I need some help figuring out what formula, would best suite my needs and how to make it meet several criteria’s: Age group, Repetitions, Gender for example if a 21 year old male soldier does 42 sit ups then his raw score for push up would be 60 percent, and a female same age, would score 100 percent.
Thus when I enter the age, M/F (gender), Repetitions the excel sheet will auto configure the Raw scores for each event, then if the soldier scores less than 60 percent
In any event it will enter a Fail for that soldier, and a pass if the soldier scores greater than 60 percent.
This message was edited by Sand_in_my_Boots on 2002-10-14 03:32
This message was edited by Sand_in_my_Boots on 2002-10-14 03:33
### Excel Facts
Does the VLOOKUP table have to be sorted?
No! when you are using an exact match, the VLOOKUP table can be in any order. Best-selling items at the top is actually the best.
On 2002-10-14 03:29, Sand_in_my_Boots wrote:
I need some help figuring out what formula, would best suite my needs and how to make it meet several criteria’s: Age group, Repetitions, Gender for example if a 21 year old male soldier does 42 sit ups then his raw score for push up would be 60 percent, and a female same age, would score 100 percent.
Thus when I enter the age, M/F (gender), Repetitions the excel sheet will auto configure the Raw scores for each event, then if the soldier scores less than 60 percent
In any event it will enter a Fail for that soldier, and a pass if the soldier scores greater than 60 percent.
This message was edited by Sand_in_my_Boots on 2002-10-14 03:32
This message was edited by Sand_in_my_Boots on 2002-10-14 03:33
Welcome to the board,
If you actually supply the relationships between these factors, how do you work out that Male 21 with 42 sit ups = 60%
and Female 21 with 42 sit ups = 100%
If you post up the drivers for how you score soldiers then several people on this board will be able to put these into a formula.
If you actually supply the relationships between these factors, how do you work out that Male 21 with 42 sit ups = 60%
and Female 21 with 42 sit ups = 100%
If you post up the drivers for how you score soldiers then several people on this board will be able to put these into a formula.
The relationship between male and female soldiers for the U.S. ARMY Physical Fitness Test i.e. APFT. Here is the Relationship http://www.benning.army.mil/usapfs/Training/APFT/
Underscoring standards, click on the Following hyperlinks: Push-Ups, Sit-Ups, and 2-Mile Run. Where I am at right now.
I have copied each standard to it’s own sheet in a workbook.
I have downloaded HtmlMaker2.32; I have failed to get it to work. I keep getting an error not located on this machine.
Purchased Excel for Dummies, the famous yellow know how books, my mistake! My son found it to be a very colorful by using the pages as a coloring book. Now covered in spaghetti and meat sauce With CHEESE! The book is useless to me.
Wow what a lot of data...
First, for Excel the table needs to be sorted the other way from lowest upwards.
Therefore I have entered the details of the 1st 5 rows (1-5) repetitions and have included 2 lookup tables that enables the correct column to be references.
You will just need to enter in the other columns.
Since this is big I hope it fits into the MrExcel limit for posts...
Training.xls
ABCDEFGH
1Age22AgeColumn
2Male/femalefemale152
3Repitions4224
4Column5276
5328
6Score203710
74212
84714
95216
10
11MaleFemaleMaleFemaleMaleFemaleMale
12Repetitions17-2117-2122-2622-2627-3127-3132-36
1312425253
142484104106
Sheet1
This message was edited by s-o-s on 2002-10-14 06:53
I see the post did not work correctly (edit: 1st time a smaller area has now been posted) but the detail you need is in there...Just click the formulas in the column and score fields to be able to put into your own spreadsheet.
If you have real problems drop me an email and I will send through the file for you.
_________________
Hope This Helps.
Sean.<A HREF= "http://website.lineone.net/~s-o-s/Index.html">
Win98, XL2000
This message was edited by s-o-s on 2002-10-14 06:54
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# HP Forums
You're currently viewing a stripped down version of our content. View the full version with proper formatting.
Pages: 1 2
(05-09-2017 04:03 PM)pier4r Wrote: [ -> ]Since I cannot send you a pm, remember that the wiki is open . Everyone can contribute.
Added the new result and code to the page of the challenge.
Thanks. I would swear that private messaging was enabled by default because I have a few private messages in my mailbox ... Strange ... Re-enabled ...
From Calculator Benchmark
0.130 FX-CG50 MicroPython / OS 3.20
Code:
``` MicroPython ------------- def nqueens(): for i in range(100): a = [0] * 9 r = 8 s = 0 x = 0 y = 0 t = 0 while True: x += 1 a[x] = r while True: s += 1 y = x while y>1: y -= 1 t = a[x]-a[y] if t==0 or x-y==abs(t): y=0 a[x] -= 1 while a[x]==0: x -= 1 a[x] -= 1 if y==1: break; if x==r: break; print(s)```
This seems a remarkably fast result.
Does it really outperform the prime D ?
0.158 HP-Prime Formula / List / G2 / HW.D
Code:
`EXPORT NQUEENS() BEGIN R:=8; L1:=MAKELIST(0,X,1,R,1); S:=0; X:=0; REPEAT X:=X+1; L1(X):=R; REPEAT S:=S+1; Y:=X; WHILE Y>1 DO Y:=Y-1; T:=L1(X)-L1(Y); IF T==0 OR X-Y==ABS(T) THEN Y:=0; L1(X):=L1(X)-1; WHILE L1(X)==0 DO X:=X-1; L1(X):=L1(X)-1; END; END; END; UNTIL Y==1 END; UNTIL X==R END; S; END;`
A simple explanation could be a more efficient bytecode interpreter. If I'm not mistaken, the CPU frequency of the CG50 is 59 MHz only, while executing MicroPython.
(12-28-2018 02:37 PM)xerxes Wrote: [ -> ]A simple explanation could be a more efficient bytecode interpreter. If I'm not mistaken, the CPU frequency of the CG50 is 59 MHz only, while executing MicroPython.
It could be. Another datapoint is the benchmark results given here:
Here the Prime G2 is an order of magnitude faster than CG50 MicroPython. Thats why I questioned the faster than G2 result here, esp given the low clockspeed of 59MHz.
Yes, the Prime is one of the fastest for many tasks, but an integer benchmark like n-queens is more focused on the efficiency of the
language itself and not on the mathematical or graphical functions. The HP-200LX with Turbo C for example is faster than the Prime
with n-queens but much slower at mathematical functions.
(12-28-2018 03:46 PM)xerxes Wrote: [ -> ]Yes, the Prime is one of the fastest for many tasks, but an integer benchmark like n-queens is more focused on the efficiency of the
language itself and not on the mathematical or graphical functions. The HP-200LX with Turbo C for example is faster than the Prime
with n-queens but much slower at mathematical functions.
Yes I can see how C could speed up a slow calc. Im just very surprised that taking into account clock differences, the language efficiency of the G2 is only about 10% of the CG50 running Micropython. My initial thought was the decimal point may have slipped to the left as mistakes can and do sometimes happen! Thanks for the explanation.
(12-28-2018 02:58 PM)Stevetuc Wrote: [ -> ]Here the Prime G2 is an order of magnitude faster than CG50 MicroPython.
The astronomer (and the rebellion) in me always comes to a hiccup when this particular term comes up.
Does anyone actually know what "one order of magnitude" really means?
One order of magnitude is the 5th root of 100 or ca 2,511886431509580111085032067799328 (according to my DM42)
Astronomically, stars are odered by brightness in magnitudes.
A magnitude 6 star is the faintest one (normally) can see with the naked eye in a lightunpolluted sky. A magnitude 1 star is 100 times brighter. Each magnitude is determined by ~2,512^x where x starts with 0 at mag. 6 (2,512^0=1) and 5 (2,512^5=~100) at mag. 1. So, a mag. 6 star is "1" bright and a mag 1 star is "100" bright.
Star magnitudes
(Post should be taken as )
Yes, and a continous merry christmas and a coming prosperous new year to all.
(12-28-2018 08:20 PM)DA74254 Wrote: [ -> ]Does anyone actually know what "one order of magnitude" really means?
Wikipedia is quite clear.
https://en.wikipedia.org/wiki/Order_of_magnitude
A sentence can be used in different contexts.
(12-28-2018 08:31 PM)pier4r Wrote: [ -> ]
(12-28-2018 08:20 PM)DA74254 Wrote: [ -> ]Does anyone actually know what "one order of magnitude" really means?
Wikipedia is quite clear.
https://en.wikipedia.org/wiki/Order_of_magnitude
A sentence can be used in different contexts.
Which means that "an order of magnitude" can be slower. For instance, let us set a calculation that my Prime performs in 0,432 secs. Then my "other" calc does the same in "an order of 10 times the magnitude". So, in 0,432 secs it has performed 0,000226379693794 of the work that my Prime has done.. (0,432^10=~2,26*10^-4)
(12-28-2018 09:09 PM)DA74254 Wrote: [ -> ]Which means that "an order of magnitude" can be slower. For instance, let us set a calculation that my Prime performs in 0,432 secs. Then my "other" calc does the same in "an order of 10 times the magnitude". So, in 0,432 secs it has performed 0,000226379693794 of the work that my Prime has done.. (0,432^10=~2,26*10^-4)
Not quite.
An order of magnitude. Or two orders of magnitude and so on.
When you say "an order of 10 times the magnitude" maybe you mean "ten orders of magnitude".
In this case it would be: 0.432 e10 or 0.432 x 10^10 .
(12-28-2018 09:18 PM)pier4r Wrote: [ -> ]
(12-28-2018 09:09 PM)DA74254 Wrote: [ -> ]Which means that "an order of magnitude" can be slower. For instance, let us set a calculation that my Prime performs in 0,432 secs. Then my "other" calc does the same in "an order of 10 times the magnitude". So, in 0,432 secs it has performed 0,000226379693794 of the work that my Prime has done.. (0,432^10=~2,26*10^-4)
Not quite.
An order of magnitude. Or two orders of magnitude and so on.
When you say "an order of 10 times the magnitude" maybe you mean "ten orders of magnitude".
In this case it would be: 0.432 e10 or 0.432 x 10^10 .
Yes, that was my first approach. But I wanted it to be "orders of ten" smaller, thus my approach. Anyway, "orders of magnitude" is an arbitrary and very unaccurate measure since one does not know which direction it goes and/or which "magnitude" one chooses. Stars are x^~2,512 decimal are x^10 and binary are x^2 and so on. Then again "zero point.." will always be less in any orders of magnitude unless the magnitude is negative.
(12-28-2018 08:20 PM)DA74254 Wrote: [ -> ]
(12-28-2018 02:58 PM)Stevetuc Wrote: [ -> ]Here the Prime G2 is an order of magnitude faster than CG50 MicroPython.
The astronomer (and the rebellion) in me always comes to a hiccup when this particular term comes up.
Does anyone actually know what "one order of magnitude" really means?
One order of magnitude is the 5th root of 100 or ca 2,511886431509580111085032067799328 (according to my DM42)
Astronomically, stars are odered by brightness in magnitudes.
A magnitude 6 star is the faintest one (normally) can see with the naked eye in a lightunpolluted sky. A magnitude 1 star is 100 times brighter. Each magnitude is determined by ~2,512^x where x starts with 0 at mag. 6 (2,512^0=1) and 5 (2,512^5=~100) at mag. 1. So, a mag. 6 star is "1" bright and a mag 1 star is "100" bright.
Star magnitudes
(Post should be taken as )
Yes, and a continous merry christmas and a coming prosperous new year to all.
You got me! I did forget to mention that I meant the commonly used definition of10^n and not the astronomical definition
The Cambridge dictionary says:
Meaning of “order of magnitude” in the English Dictionary
A level in a system used for measuring something in which each level is ten times larger than the one before:
Eg. These processor speeds have recently increased by two ordersof magnitude (= by a hundred times).
Just saying..
I tried those benchmarks on both my HP-35s and my dm42.
What's going on?
4:17 HP-35S Keystroke / RPN
My HP-35s can do this in well under 20 seconds. How do they report a time of 4 minutes and 17 seconds?
4.18 DM-42 Keystroke / RPN
My dm-42 gets an answer in under 1 second on battery and even faster when plugged in.
Both of my calculators seem to be much faster than the reported times.
Also, they both get an answer of 8 in the Y-reg and an answer of 36 in the X-reg.
I just copied the corresponding program into the calculator that was listed on the site.
I'm referring to:
(12-29-2018 11:08 PM)gomefun2 Wrote: [ -> ]I tried those benchmarks on both my HP-35s and my dm42.
... snip ...
4.18 DM-42 Keystroke / RPN
My dm-42 gets an answer in under 1 second on battery and even faster when plugged in.
Both of my calculators seem to be much faster than the reported times.
Also, they both get an answer of 8 in the Y-reg and an answer of 36 in the X-reg.
I just copied the corresponding program into the calculator that was listed on the site.
I'm referring to:
Using the program listed for the DM42 on the linked page, my DM42 gets about the same time as listed there (between 4-5 secs) on battery power, and about 2 secs with USB plugged in.
The correct results should be 876 (not 36) so there is probably an error (maybe a skipped line) in the code you entered.
what is this key:
"x<>y?"
I can't find it on my HP35s, DM42 or wp34s...
All three of my calculators give the same incorrect answer. I think it has to do with this key.
I can find the x<>y key. That just swaps the x-reg and y-reg.
EDIT: I assume it means !=
(does not equal)
But I still don't get the correct answer.
EDIT: Hmm actually I got it to work on the wp-34s, I'll go back and check the other two.
(12-30-2018 11:48 AM)gomefun2 Wrote: [ -> ]EDIT: I assume it means !=
(does not equal)
Yes, that's correct. The program as listed is for older machines, so some of the instructions may not look exactly the same as on the 35S or 42S, though except for the one you noted, they all seem clear to me; ask about them if they don't seem clear.
When programs are listed the way they are on that page (i.e. not using 1 instruction per line) it is easy to miss an instruction, particularly when many are similar as is the case for this program.
(12-30-2018 11:48 AM)gomefun2 Wrote: [ -> ]what is this key:
"x<>y?"
I can't find it on my HP35s, DM42 or wp34s...
That's the x≠y? test.
Since "≠" is not an ASCII character this is written as "<>" in various programming languages and similar environments.
(12-30-2018 11:48 AM)gomefun2 Wrote: [ -> ]All three of my calculators give the same incorrect answer. I think it has to do with this key.
...
EDIT: I assume it means !=
(does not equal)
But I still don't get the correct answer.
Most probably you got an incorrect program with one or more errors. Since the 35s program on the mentioned benchmark page is merely 40 lines, what about posting your (!) 35s program here? The errors should be easy to spot.
For example: STO (I) and STO (J) are different from STO I and STO J.
The same is true for DSE (I) and DSE I, all these are different commands.
Just to be sure: on the 35s please do a manual CLVARX 000 before you run the program.
The 35s code has a CLVARS command, but this does not clear the extended registers where the program stores its data!
Dieter
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Reference URL's
• HP Forums: https://www.hpmuseum.org/forum/index.php
• :
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How does an outlier affect the mean of a data set?
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An outlier can affect the mean of a data set by skewing the results so that the mean is no longer representative of the data set. There are solutions to this problem.
Explanation:
As we have seen in data collections that are used to draw graphs or find means, modes and medians the data arrives in relatively closed order. In other words, each element of the data is closely related to the majority of the other data. If not, the data set may have information that is too scattered to be useful in any analysis.
In some data sets there may be a point or two that can be out of context with the bulk of the data. These are referred to as outliers, which are out of line with the normal data set. The outlier can push the mean of the data out of its usual position.
For example, the data set $3 , 4 , 5 , 6 , 7$ has a mean of $5$, found by dividing the sum of the data by the number of data elements:
$m e a n = \frac{3 + 4 + 5 + 6 + 7}{5} = \frac{25}{5} = 5$
If the $4$ was mistakenly recorded as a $14$, the $14$ would be unusual for the data set and it would be an outlier.
Then: $m e a n = \frac{3 + 14 + 5 + 6 + 7}{5} = \frac{35}{5} = 7$
And we can see the outlier has moved the mean of the data set.
To solve this problem the unusual data element can either be re-investigated and corrected, or removed from the data set with an explanation.
The former solution may bring back our original $4$ after error checking is completed. The latter will return our mean closer to a representative evaluation of the data.
$m e a n \left(n o \cancel{14}\right) = \frac{3 + \cancel{14} + 5 + 6 + 7}{\left(\cancel{5}\right) 4} = \frac{21}{4} = 5.25$
There are pictures and graphs here:
https://www.mathsisfun.com/data/outliers.html
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The mean will move towards the outlier.
Explanation:
The mean is non-resistant. That means, it's affected by outliers. More specifically, the mean will want to move towards the outlier.
Think about it this way:
Let's say we have some data. $1 , 2 , 3$. The mean of this is $2$. But if we add an outlier of $94$ to the data set, the mean will become $25$. As you can see, the mean moved towards the outlier.
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# Maxwell–Boltzmann statistics
(Redirected from Boltzmann statistics)
Not to be confused with Maxwell–Boltzmann distribution.
Maxwell–Boltzmann statistics can be used to derive the Maxwell–Boltzmann distribution of particle speeds in an ideal gas. Shown: distribution of particle speed for 106 oxygen particles at
−100, 20 and 600 degrees Celsius.
In statistical mechanics, Maxwell–Boltzmann statistics describes the average distribution of non-interacting material particles over various energy states in thermal equilibrium, and is applicable when the temperature is high enough or the particle density is low enough to render quantum effects negligible.
The expected number of particles with energy $\varepsilon_i$ for Maxwell–Boltzmann statistics is $\langle N_i \rangle$ where:
$\langle N_i \rangle = \frac {g_i} {e^{(\varepsilon_i-\mu)/kT}} = \frac{N}{Z}\,g_i e^{-\varepsilon_i/kT}$
where:
• $\varepsilon_i$ is the i-th energy level
• $\langle N_i \rangle$ is the number of particles in the set of states with energy $\varepsilon_i$
• $g_i$ is the degeneracy of energy level i. i.e. no. of single-particle states with energy $\varepsilon_i$
• μ is the chemical potential
• k is Boltzmann's constant
• T is absolute temperature
• N is the total number of particles
$N=\sum_i N_i\,$
$Z=\sum_i g_i e^{-\varepsilon_i/kT}$
Equivalently, the particle number is sometimes expressed as
$\langle N_i \rangle = \frac {1} {e^{(\varepsilon_i-\mu)/kT}} = \frac{N}{Z}\,e^{-\varepsilon_i/kT}$
where the index i now specifies a particular state rather than the set of all states with energy $\varepsilon_i$, and $Z=\sum_i e^{-\varepsilon_i/kT}$
## Applications
Maxwell–Boltzmann statistics may be used to derive the Maxwell–Boltzmann distribution (for an ideal gas of classical particles in a three-dimensional box). However, they apply to other situations as well. Maxwell–Boltzmann statistics can be used to extend that distribution to particles with a different energy–momentum relation, such as relativistic particles (Maxwell–Jüttner distribution). In addition, hypothetical situations can be considered, such as particles in a box with different numbers of dimensions (four-dimensional, two-dimensional, etc.).
## Limits of applicability
Maxwell–Boltzmann statistics are often described as the statistics of "distinguishable" classical particles. In other words the configuration of particle A in state 1 and particle B in state 2 is different from the case where particle B is in state 1 and particle A is in state 2. This assumption leads to the proper (Boltzmann) statistics of particles in the energy states, but yields non-physical results for the entropy, as embodied in the Gibbs paradox.
Technically speaking, however, there are no real particles which have the characteristics required by Maxwell–Boltzmann statistics. Indeed, the Gibbs paradox is resolved if we treat all particles of a certain type (e.g., electrons, protons, etc.) as indistinguishable, and this assumption can be justified in the context of quantum mechanics. Once this assumption is made, however, the particle statistics change. Quantum particles are either bosons (following instead Bose–Einstein statistics) or fermions (subject to the Pauli exclusion principle, following instead Fermi–Dirac statistics).
Both of these quantum statistics approach the Maxwell–Boltzmann statistics in the limit of high temperature and low particle density, without the need for any ad hoc assumptions. The Fermi–Dirac and Bose–Einstein statistics give the energy level occupation as:
$\langle N_i \rangle = \frac{g_i}{e^{(\varepsilon_i-\mu)/kT}\mp 1}.$
It can be seen that the condition under which the Maxwell–Boltzmann statistics are valid is when
$e^{(\varepsilon_{\rm min}-\mu)/kT} \gg 1,$
where $\varepsilon_{\rm min}$ is the lowest (minimum) value of $\varepsilon_i$.
Maxwell–Boltzmann statistics are particularly useful for studying gases, whereas Fermi–Dirac statistics are most often used for the study of electrons in solids. Bose–Einstein statistics are important for blackbody radiation. Note however that none of these statistics are general, as they all assume that the particles are non-interacting (they all assume a static ladder of energy states).
## Derivations of Maxwell–Boltzmann statistics
Maxwell–Boltzmann statistics can be derived in various statistical mechanical thermodynamic ensembles:[1]
In each case it is necessary to assume that the particles are non-interacting, and that multiple particles can occupy the same state and do so independently.
### Derivation from microcanonical ensemble
Suppose we have a container with a huge number of very small particles all with identical physical characteristics (mass, charge, etc.). Let's refer to this as the system. Assume that though the particles have identical properties, they are distinguishable. For example, we might identify each particle by continually observing their trajectories, or by placing a marking on each one, e.g., drawing a different number on each one as is done with lottery balls.
The particles are moving inside that container in all directions with great speed. Because the particles are speeding around, they possess some energy. The Maxwell–Boltzmann distribution is a mathematical function that speaks about how many particles in the container have a certain energy.
In general, there may be many particles with the same amount of energy $\varepsilon$. Let the number of particles with the same energy $\varepsilon_1$ be $N_1$, the number of particles possessing another energy $\varepsilon_2$ be $N_2$, and so forth for all the possible energies {$\varepsilon_i$ | i=1,2,3,...}. To describe this situation, we say that $N_i$ is the occupation number of the energy level $i.$ If we know all the occupation numbers {$N_i$ | i=1,2,3,...}, then we know the total energy of the system. However, because we can distinguish between which particles are occupying each energy level, the set of occupation numbers {$N_i$ | i=1,2,3,...} does not completely describe the state of the system. To completely describe the state of the system, or the microstate, we must specify exactly which particles are in each energy level. Thus when we count the number of possible states of the system, we must count each and every microstate, and not just the possible sets of occupation numbers.
To begin with, let's ignore the degeneracy problem: assume that there is only one way to put $N_i$ particles into the energy level $i$ . What follows next is a bit of combinatorial thinking which has little to do in accurately describing the reservoir of particles.
The number of different ways of performing an ordered selection of one single object from N objects is obviously N. The number of different ways of selecting two objects from N objects, in a particular order, is thus N(N − 1) and that of selecting n objects in a particular order is seen to be N!/(N − n)!. It is divided by the number of permutations, n!, if order does not matter. The binomial coefficient, N!/(n!(N − n)!), is, thus, the number of ways to pick n objects from $N$. If we now have a set of boxes labelled a, b, c, d, e, ..., k, then the number of ways of selecting Na objects from a total of N objects and placing them in box a, then selecting Nb objects from the remaining N − Na objects and placing them in box b, then selecting Nc objects from the remaining N − Na − Nb objects and placing them in box c, and continuing until no object is left outside is
\begin{align} W & = \frac{N!}{N_a!(N-N_a)!} \times \frac{(N-N_a)!}{N_b!(N-N_a-N_b)!} ~ \times \frac{(N-N_a-N_b)!}{N_c!(N-N_a-N_b-N_c)!} \times \ldots \times \frac{(N-\ldots-N_l)!}{N_k!(N-\ldots-N_l-N_k)!} = \\ \\ & = \frac{N!}{N_a!N_b!N_c!\ldots N_k!(N-\ldots-N_l-N_k)!} \end{align}
and because not even a single object is to be left outside the boxes, implies that the sum made of the terms Na, Nb, Nc, Nd, Ne, ..., Nk must equal N, thus the term (N - Na - Nb - Nc - ... - Nl - Nk)! in the relation above evaluates to 0!. (0!=1) which makes possible to write down that relation as
\begin{align} W & = N!\prod_{i=a,b,c,...}^k \frac{1}{N_i!} \end{align}
Now going back to the degeneracy problem which characterize the reservoir of particles. If the i-th box has a "degeneracy" of $g_i$, that is, it has $g_i$ "sub-boxes", such that any way of filling the i-th box where the number in the sub-boxes is changed is a distinct way of filling the box, then the number of ways of filling the i-th box must be increased by the number of ways of distributing the $N_i$ objects in the $g_i$ "sub-boxes". The number of ways of placing $N_i$ distinguishable objects in $g_i$ "sub-boxes" is $g_i^{N_i}$ (the first object can go into any of the $g_i$ boxes, the second object can also go into any of the $g_i$ boxes, and so on). Thus the number of ways $W$ that a total of $N$ particles can be classified into energy levels according to their energies, while each level $i$ having $g_i$ distinct states such that the i-th level accommodates $N_i$ particles is:
$W=N!\prod \frac{g_i^{N_i}}{N_i!}$
This is the form for W first derived by Boltzmann. Boltzmann's fundamental equation $S=k\,\ln W$ relates the thermodynamic entropy S to the number of microstates W, where k is the Boltzmann constant. It was pointed out by Gibbs however, that the above expression for W does not yield an extensive entropy, and is therefore faulty. This problem is known as the Gibbs paradox. The problem is that the particles considered by the above equation are not indistinguishable. In other words, for two particles (A and B) in two energy sublevels the population represented by [A,B] is considered distinct from the population [B,A] while for indistinguishable particles, they are not. If we carry out the argument for indistinguishable particles, we are led to the Bose–Einstein expression for W:
$W=\prod_i \frac{(N_i+g_i-1)!}{N_i!(g_i-1)!}$
The Maxwell–Boltzmann distribution follows from this Bose–Einstein distribution for temperatures well above absolute zero, implying that $g_i\gg 1$. The Maxwell–Boltzmann distribution also requires low density, implying that $g_i\gg N_i$. Under these conditions, we may use Stirling's approximation for the factorial:
$N! \approx N^N e^{-N},$
to write:
$W\approx\prod_i \frac{(N_i+g_i)^{N_i+g_i}}{N_i^{N_i}g_i^{g_i}}\approx\prod_i \frac{g_i^{N_i}(1+N_i/g_i)^{g_i}}{N_i^{N_i}}$
Using the fact that $(1+N_i/g_i)^{g_i}\approx e^{N_i}$ for $g_i\gg N_i$ we can again use Stirlings approximation to write:
$W\approx\prod_i \frac{g_i^{N_i}}{N_i!}$
This is essentially a division by N! of Boltzmann's original expression for W, and this correction is referred to as correct Boltzmann counting.
We wish to find the $N_i$ for which the function $W$ is maximized, while considering the constraint that there is a fixed number of particles $\left(N=\textstyle\sum N_i\right)$ and a fixed energy $\left(E=\textstyle\sum N_i \varepsilon_i\right)$ in the container. The maxima of $W$ and $\ln(W)$ are achieved by the same values of $N_i$ and, since it is easier to accomplish mathematically, we will maximize the latter function instead. We constrain our solution using Lagrange multipliers forming the function:
$f(N_1,N_2,\ldots,N_n)=\ln(W)+\alpha(N-\sum N_i)+\beta(E-\sum N_i \varepsilon_i)$
$\ln W=\ln\left[\prod\limits_{i=1}^{n}\frac{g_i^{N_i}}{N_i!}\right] \approx \sum\limits_{i=1}^n\left(N_i\ln g_i-N_i\ln N_i + N_i\right)$
Finally
$f(N_1,N_2,\ldots,N_n)=\alpha N +\beta E + \sum\limits_{i=1}^n\left(N_i\ln g_i-N_i\ln N_i + N_i-(\alpha+\beta\varepsilon_i) N_i\right)$
In order to maximize the expression above we apply Fermat's theorem (stationary points), according to which local extrema, if exist, must be at critical points (partial derivatives vanish):
$\frac{\partial f}{\partial N_i}=\ln g_i-\ln N_i -(\alpha+\beta\varepsilon_i) = 0$
By solving the equations above ($i=1\ldots n$) we arrive to an expression for $N_i$:
$N_i = \frac{g_i}{e^{\alpha+\beta \varepsilon_i}}$
Substituting this expression for $N_i$ into the equation for $\ln W$ and assuming that $N\gg 1$ yields:
$\ln W = \alpha N+\beta E\,$
or, differentiating and rearranging:
$dE=\frac{1}{\beta}d\ln W-\frac{\alpha}{\beta}dN$
Boltzmann realized that this is just an expression of the second law of thermodynamics. Identifying dE as the internal energy, the second law of thermodynamics states that for variation only in entropy (S) and particle number (N):
$dE=T\,dS+\mu\,dN$
where T is the temperature and μ is the chemical potential. Boltzmann's famous equation $S=k\,\ln W$ is the realization that the entropy is proportional to $\ln W$ with the constant of proportionality being Boltzmann's constant. It follows immediately that $\beta=1/kT$ and $\alpha=-\mu/kT$ so that the populations may now be written:
$N_i = \frac{g_i}{e^{(\varepsilon_i-\mu)/kT}}$
Note that the above formula is sometimes written:
$N_i = \frac{g_i}{e^{\varepsilon_i/kT}/z}$
where $z=\exp(\mu/kT)$ is the absolute activity.
Alternatively, we may use the fact that
$\sum_i N_i=N\,$
to obtain the population numbers as
$N_i = N\frac{g_i e^{-\varepsilon_i/kT}}{Z}$
where Z is the partition function defined by:
$Z = \sum_i g_i e^{-\varepsilon_i/kT}$
### Derivation from canonical ensemble
In the above discussion, the Boltzmann distribution function was obtained via directly analysing the multiplicities of a system. Alternatively, one can make use of the canonical ensemble. In a canonical ensemble, a system is in thermal contact with a reservoir. While energy is free to flow between the system and the reservoir, the reservoir is thought to have infinitely large heat capacity as to maintain constant temperature, T, for the combined system.
In the present context, our system is assumed to have the energy levels $\varepsilon _i$ with degeneracies $g_i$. As before, we would like to calculate the probability that our system has energy $\varepsilon_i$.
If our system is in state $\; s_1$, then there would be a corresponding number of microstates available to the reservoir. Call this number $\; \Omega _ R (s_1)$. By assumption, the combined system (of the system we are interested in and the reservoir) is isolated, so all microstates are equally probable. Therefore, for instance, if $\; \Omega _ R (s_1) = 2 \; \Omega _ R (s_2)$, we can conclude that our system is twice as likely to be in state $\; s_1$ than $\; s_2$. In general, if $\; P(s_i)$ is the probability that our system is in state $\; s_i$,
$\frac{P(s_1)}{P(s_2)} = \frac{\Omega _ R (s_1)}{\Omega _ R (s_2)}.$
Since the entropy of the reservoir $\; S_R = k \ln \Omega _R$, the above becomes
$\frac{P(s_1)}{P(s_2)} = \frac{ e^{S_R(s_1)/k} }{ e^{S_R(s_2)/k} } = e^{(S_R (s_1) - S_R (s_2))/k}.$
Next we recall the thermodynamic identity (from the first law of thermodynamics):
$d S_R = \frac{1}{T} (d U_R + P \, d V_R - \mu \, d N_R).$
In a canonical ensemble, there is no exchange of particles, so the $d N_R$ term is zero. Similarly, $d V_R = 0.$ This gives
$S_R (s_1) - S_R (s_2) = \frac{1}{T} (U_R (s_1) - U_R (s_2)) = - \frac{1}{T} (E(s_1) - E(s_2)),$
where $\; U_R (s_i)$ and $\; E(s_i)$ denote the energies of the reservoir and the system at $s_i$, respectively. For the second equality we have used the conservation of energy. Substituting into the first equation relating $P(s_1), \; P(s_2)$:
$\frac{P(s_1)}{P(s_2)} = \frac{ e^{ - E(s_1) / kT } }{ e^{ - E(s_2) / kT} },$
which implies, for any state s of the system
$P(s) = \frac{1}{Z} e^{- E(s) / kT},$
where Z is an appropriately chosen "constant" to make total probability 1. (Z is constant provided that the temperature T is invariant.) It is obvious that
$\; Z = \sum _s e^{- E(s) / kT},$
where the index s runs through all microstates of the system. Z is sometimes called the Boltzmann sum over states (or "Zustandsumme" in the original German). If we index the summation via the energy eigenvalues instead of all possible states, degeneracy must be taken into account. The probability of our system having energy $\varepsilon _i$ is simply the sum of the probabilities of all corresponding microstates:
$P (\varepsilon _i) = \frac{1}{Z} g_i e^{- \varepsilon_i / kT}$
where, with obvious modification,
$Z = \sum _j g_j e^{- \varepsilon _j / kT},$
this is the same result as before.
Comments on this derivation:
• Notice that in this formulation, the initial assumption "... suppose the system has total N particles..." is dispensed with. Indeed, the number of particles possessed by the system plays no role in arriving at the distribution. Rather, how many particles would occupy states with energy $\varepsilon _i$ follows as an easy consequence.
• What has been presented above is essentially a derivation of the canonical partition function. As one can see by comparing the definitions, the Boltzmann sum over states is equal to the canonical partition function.
• Exactly the same approach can be used to derive Fermi–Dirac and Bose–Einstein statistics. However, there one would replace the canonical ensemble with the grand canonical ensemble, since there is exchange of particles between the system and the reservoir. Also, the system one considers in those cases is a single particle state, not a particle. (In the above discussion, we could have assumed our system to be a single atom.)
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Purchase Solution
# Optimal Solution for Clothing Retailer
Not what you're looking for?
A company makes three types of shirts: Athletic, Varsity, and Surfer. The shirts are made from different combinations of cotton and rayon. The cost per yard of cotton is \$5 and of rayon is \$7. The company can receive up to 4000 yards of cotton and 3000 yards of rayon per week. The table below shows relevant manufacturing information:
Shirt Fabric Requirement/Minimum Weekly Contracts Maximum Weekly Demand Selling Price
Athletic 1.00/At least 60% cotton 500 600 \$30
Varsity 1.20/No more than 30% rayon 650 850 \$40
Surfer 0.90 Up to 80% cotton 300 700 \$36
##### Solution Summary
Solution builds a linear programming model in the given case and determines the optimal output level of Athletic, Varsity and Surfer shirts. Solution is determined with the help of "Solver" in MS Excel.
##### Solution Preview
Please refer attached file for complete solution.
Let
the total number of Athletic shirts produced be A
the total number of Varsity shirts produced be V
the total number of Surfer shirts produced be S
Cotton cloth used in Athletic shirts be Xca
Cotton cloth used in Varsity shirts be Xcv
Cotton cloth used in Surfer shirts be Xcs
Rayon cloth used in Athletic shirts be Xra
Rayon cloth used ...
Solution provided by:
###### Education
• BEng (Hons) , Birla Institute of Technology and Science, India
• MSc (Hons) , Birla Institute of Technology and Science, India
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##### MS Word 2010-Tricky Features
These questions are based on features of the previous word versions that were easy to figure out, but now seem more hidden to me.
##### Marketing Research and Forecasting
The following quiz will assess your ability to identify steps in the marketing research process. Understanding this information will provide fundamental knowledge related to marketing research.
##### Production and cost theory
Understanding production and cost phenomena will permit firms to make wise decisions concerning output volume.
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# Aptitude Questions and answers
Topics
51). Taps A, B and C are connected to a water tank and the rate of flow of water is 42 litres/hr., 56 litres/hr, and 48 litres/hr, respectively. Taps A and B fill the tank while tap C empties the tank. If all the three taps are opened simultaneously, the tank gets completely filled up in 16 hours. What is the capacity of the tank ?
A). 960 litres B). 2346 litres C). 1600 litres D). 800 litres
52). Four of the following five parts numbered (1), (2), (3), (4) and (5) are equal. That one part which is not equal to other four parts is the answer.
A). $$\Large (2x + 3y)^{2} =$$ B). $$\Large (2y - x)^{2} + y (16x + 5y) =$$ C). $$\Large 4x (x + 2y) + y (4x + 9y) =$$ D). $$\Large (2x + 2y)^{2} + y (4x + 5y) =$$
53). A boat running down stream covers a distance of 16 km in 2 hours while for covering the same distance upstream it takes 4 hours. What is the speed of the boat in still water ?
A). 4 kmph B). 6 kmph C). 8 kmph D). Data inadequate
54). If 2x + 3y + z = 55; x + z - y = 4 and y - x + z = 12 then what is the value of y ?
A). 7 B). 8 C). 12 D). None of these
55). If 3y + 2x = 47 and 11x = 7y then what is value of y - x ?
A). 4 B). 6 C). 7 D). 5
56). Four of the following five parts numbered (1), (2), (3), (4) and (5) are exactly equal. The number of the part which is not equal to the other four is the answer
A). $$\Large 136 \times 12 \div 9 \times 3 =$$ B). $$\Large 17 \times 64 \div 6 \times 3 =$$ C). $$\Large 36 \times 17 \div 9 \times 8 =$$ D). $$\Large 76 \times 6 \div 19 \times 17$$
57). For which of the following values of x the inequality x (x + 3) < 10 is satisfied ?
A). x > 2 , x < - 5 B). = -5 < x < 2 C). = -2 < x < 5 D). x < - 2 , x > 5
58). A boat takes 6 hours to travel from place M to N downstream and back from N to M upstream. If the speed of the boat in still water is 4 km./hr., what is the distance between the two places?
A). 8 kms. B). 12 kms. C). 6 kms. D). Data inadequate
59). The speed of a boat when travelling downstream is 32 kmph whereas when travelling upstream it is 28 kmph. What is the speed of the boat in still water ?
A). 27 Kmph B). 29 Kmph C). 31 Kmph D). None of these
60). A boat covers a distance of 2.75 km upstream in 11 minutes. The ratio between speed of current and that of boat downstream is 1 : 7 respectively. The boat covers distance between A and B downstream in 52 minutes. What is the distance between point A and point B ?
A). 19.2 km. B). 17.2 Km. C). 18.2 km. D). 16.5 km.
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1. ## Counting Strings
Let $\displaystyle a_n$ be the number of strings of length n in which every $\displaystyle 0$ is immediately followed by three consecutive $\displaystyle 1's$. So for example, the string $\displaystyle 101111$ is allowed but $\displaystyle 01110$ is not.
Find a recurrence relation and initial conditions for $\displaystyle a_n$.
I know the first initial condition has to have 4 places because the $\displaystyle 0$ need three $\displaystyle 1's$ after it. From there I am lost.
2. Originally Posted by minkyboodle
Let $\displaystyle a_n$ be the number of strings of length n in which every $\displaystyle 0$ is immediately followed by three consecutive $\displaystyle 1's$. So for example, the string $\displaystyle 101111$ is allowed but $\displaystyle 01110$ is not.
Find a recurrence relation and initial conditions for $\displaystyle a_n$.
I know the first initial condition has to have 4 places because the $\displaystyle 0$ need three $\displaystyle 1's$ after it. From there I am lost.
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# Example of second order system
### Generate continuous second-order systems MATLAB ord2
Examples First-Order Systems MIT OpenCourseWare. Example: Simulation of second-order system The flrst program calls Matlab’s ode solver ode23. The command [T,X]=ode23(@sys2,[t_ini t_final],x_ini), Application of Second Order Differential Equations in Mechanical Engineering Analysis - Damped mass-spring system Review solution method of second order,.
### Generate continuous second-order systems MATLAB ord2
Second order system pole locations example Bodetechnics. Example calculations showing derivation of a second order control system pole locations in the Laplace domain. Rise time and peak time are also covered., 1 Page Control Laboratory Second Order and Higher Order Systems 1. Second Order System In this section, we shall obtain the response of a typical second-order.
Second Order Systems Second Order Equations 2 2 +2 +1 = s s K G s Example 5.5 • Heated tank + controller = 2nd order system (a) When feed rate changes from 0.4 Example: Step response of first order system (2) The graph below shows the effect of ζ on the unit step response of a second order system,
Time Response of Second Order Systems the natural frequency dimensionl ess damping ratio 2 ( 2 ) ( / ) / ( / ) ( ) Consider t he first term only: Examples of First-Order Differential Equations Phenomena in many disciplines are modeled by first-order differential equations. Some examples Mechanical Systems.
The Bode plot of second order system. Second order systems The Bode plot of second order system. The transfer function of a second order system If the system contained high losses, for example if the spring–mass experiment were conducted in a The effect of varying damping ratio on a second-order system.
First order LTI systems are characterized by the differential equation and ω = 2π f is the frequency in radians per second. Example solution 1 Page Control Laboratory Second Order and Higher Order Systems 1. Second Order System In this section, we shall obtain the response of a typical second-order
Using ode45 to solve a system of second order... Learn more about ode45, system, second order, differential equations, vector, matrix First order LTI systems are characterized by the differential equation and ω = 2π f is the frequency in radians per second. Example solution
Introduction to First-Order Systems Introduction There are two methods to analyze functioning of a control system that are time domain analysis and control domain Example 1 The motion of a spring-mass system from Section 3.8 was described by the equation This second order equation can be converted into a system of
Review of First- and Second-Order System system parameter that establishes the time scale of system responses in a flrst-order system. For example a resistor ECE 421 Second-Order System Example #3. This example considers the design of a second-order system that will satisfy certain time-domain specifications.
Transient Response of a Second-Order System ECEN 2830 Spring 2012 1. Introduction In connection with this experiment, you are selecting the gains in your feedback loop to What can be a simple practical example for a third order uncertain system which could be described by a family of tr functions?
Develop an understanding of dynamic behavior and take a look Consider now a second-order system with numerator tough challenges for process control systems. Response of 1st Order Systems. A first order system is described by In this model, x represents the measured and controlled output variable and f(t) the input function.
PHYSICS6770 Laboratory 1 University of Utah Department of Physics First and second order systems Week 1, First order systems slide 4 First order system: Example 1, RC Example 1 The motion of a spring-mass system from Section 3.8 was described by the equation This second order equation can be converted into a system of
What can be a simple practical example for a third order uncertain system which could be described by a family of tr functions? Examples: First-Order Systems In the above examples, first-order state equations are sufficient to describe energetic transactions within the system;
Mathematical methods for economic theory: one example, which shows how the start of this section to convert the two-equation system to a single second-order Examples such as the following can be used to illustrate the 2.6 Systems of two First Order second order linear differential equation is expected.
Control Systems Response of the First Order System Analysis, Response of the First Order, Second example, the response of the system for an impulse Control Systems Response of the First Order System Analysis, Response of the First Order, Second example, the response of the system for an impulse
JOURN.IL OF MATHEMATICAL ANALYSIS AND APPLICATIONS 8, 423-444 (1964) A Study of Second Order Nonlinear Systems Y. s. LIM Bell Telephone Laboratories Time Response of Second Order Systems the natural frequency dimensionl ess damping ratio 2 ( 2 ) ( / ) / ( / ) ( ) Consider t he first term only:
What can be a simple practical example for a third order uncertain system which could be described by a family of tr functions? A mass attached to a spring and a damper. The F in the diagram denotes an external force, which this example does not include.
Phenomena in many disciplines are modeled by first-order differential equations. Some examples the spring-mass system. an example of a linear second-order The Bode plot of second order system. Second order systems The Bode plot of second order system. The transfer function of a second order system
Examples of First-Order Differential Equations Phenomena in many disciplines are modeled by first-order differential equations. Some examples Mechanical Systems. Control Systems Response of Second Order System - Learn Control Systems in simple and easy steps starting from Introduction, Feedback, Mathematical Models, Modelling
Response of 1st Order Systems. A first order system is described by In this model, x represents the measured and controlled output variable and f(t) the input function. Second order system response. Unstable Re(s) Im(s) Overdamped or Critically damped Undamped Underdamped Underdamped
Examples of First-Order Differential Equations Phenomena in many disciplines are modeled by first-order differential equations. Some examples Mechanical Systems. Beat the System with Family Systems Therapy. sustainable than second-order changes. Example: order change can occur for a whole system and/or for an
Control Systems Response of Second Order System - Learn Control Systems in simple and easy steps starting from Introduction, Feedback, Mathematical Models, Modelling Second Order Systems - 7 minimum values of +/-1. The following figure gives one example of the step response for z = 0.10. The two envelopes are shown as dashed lines.
### SIMULATION OF DYNAMICAL SYSTEMS Г…bo Akademi
Response of 1st Order Systems FacStaff Home Page for CBU. If the system contained high losses, for example if the spring–mass experiment were conducted in a The effect of varying damping ratio on a second-order system., Barthes and the Second-Order Semiological System Saleem Suzah 02 / 09 / 2015 There are three important components in the sign system, signifier, signified, and.
A Study of Second Order Nonlinear Systems Deep Blue. Response of 1st Order Systems. A first order system is described by In this model, x represents the measured and controlled output variable and f(t) the input function., In this section we will look at some of the basics of systems of differential equations. an example of a system of first order, equation in the second.
### Examples First-Order Systems MIT OpenCourseWare
Second-Order Behavior Process Control Understanding. Second order differential equations come up frequently in oscillations What are practical applications of second-order ODEs? What is second order system? Review of First- and Second-Order System system parameter that establishes the time scale of system responses in a flrst-order system. For example a resistor.
Second order system response. Unstable Re(s) Im(s) Overdamped or Critically damped Undamped Underdamped Underdamped As a second example, consider the systems shown below in which we have swapped the components from the fires example. Unit Step Response of a Second Order System.
Example 1 The motion of a spring-mass system from Section 3.8 was described by the equation This second order equation can be converted into a system of Introduction to Second Order Systems Introduction As we discussed earlier we have two methods of analyzing the working and functioning of a control system named
Second Order Systems. From ControlTheoryPro.com. which this example does not include. natural frequency of the system . The second parameter, , JOURN.IL OF MATHEMATICAL ANALYSIS AND APPLICATIONS 8, 423-444 (1964) A Study of Second Order Nonlinear Systems Y. s. LIM Bell Telephone Laboratories
Standard form of second order system is given by: to “Introduction to Second Order Systems Introduction to First-Order Systems. Second order system response. Unstable Re(s) Im(s) Overdamped or Critically damped Undamped Underdamped Underdamped
Second order system response. Unstable Re(s) Im(s) Overdamped or Critically damped Undamped Underdamped Underdamped What can be a simple practical example for a third order uncertain system which could be described by a family of tr functions?
Response of 1st Order Systems. A first order system is described by In this model, x represents the measured and controlled output variable and f(t) the input function. Control Systems Response of Second Order System - Learn Control Systems in simple and easy steps starting from Introduction, Feedback, Mathematical Models, Modelling
Second-Order Systems Unit 3: Time Response, Part 2: Second-Order Responses Second-order systems For example, the natural Example 1 The motion of a spring-mass system from Section 3.8 was described by the equation This second order equation can be converted into a system of
And hence this time response of second-order control system is referred as critically damped. Table Method Examples History of Laplace Transform. Review of First- and Second-Order System system parameter that establishes the time scale of system responses in a flrst-order system. For example a resistor
Examples: First-Order Systems In the above examples, first-order state equations are sufficient to describe energetic transactions within the system; Second-Order Systems Unit 3: Time Response, Part 2: Second-Order Responses Second-order systems For example, the natural
Introduction to Second Order Systems Introduction As we discussed earlier we have two methods of analyzing the working and functioning of a control system named Response of First and Second Order Systems 19.11 SECTION 19 Christiansen-Sec.19 For a second example consider an electric RLC circuit with i(t)
The Bode plot of second order system. Second order systems The Bode plot of second order system. The transfer function of a second order system In this chapter we will start looking at second order differential equations. The 3-D Coordinate System; Note as well that while we example mechanical
## Introduction to Second Order Systems IDC-Online
Introduction to Second Order Systems. The Bode plot of second order system. Second order systems The Bode plot of second order system. The transfer function of a second order system, Second order differential equations come up frequently in oscillations What are practical applications of second-order ODEs? What is second order system?.
### The Bode plot of second order system Harvey Mudd College
SECOND ORDER MEASUREMENT SYSTEMS. This MATLAB function generates the state-space description (A,B,C,D) of the second-order system, Examples of First-Order Differential Equations Phenomena in many disciplines are modeled by first-order differential equations. Some examples Mechanical Systems..
PHYSICS6770 Laboratory 1 University of Utah Department of Physics First and second order systems Week 1, First order systems slide 4 First order system: Example 1, RC Second Order Systems Second Order Equations 2 2 +2 +1 = s s K G s Example 5.5 • Heated tank + controller = 2nd order system (a) When feed rate changes from 0.4
Phenomena in many disciplines are modeled by first-order differential equations. Some examples the spring-mass system. an example of a linear second-order Second order differential equations come up frequently in oscillations What are practical applications of second-order ODEs? What is second order system?
JOURN.IL OF MATHEMATICAL ANALYSIS AND APPLICATIONS 8, 423-444 (1964) A Study of Second Order Nonlinear Systems Y. s. LIM Bell Telephone Laboratories In this section we will look at some of the basics of systems of differential equations. an example of a system of first order, equation in the second
Response of First and Second Order Systems 19.11 SECTION 19 Christiansen-Sec.19 For a second example consider an electric RLC circuit with i(t) Application of Second Order Differential Equations in Mechanical Engineering Analysis - Damped mass-spring system Review solution method of second order,
Second Order Systems - 7 minimum values of +/-1. The following figure gives one example of the step response for z = 0.10. The two envelopes are shown as dashed lines. Example 1 The motion of a spring-mass system from Section 3.8 was described by the equation This second order equation can be converted into a system of
Example: Simulation of second-order system The flrst program calls Matlab’s ode solver ode23. The command [T,X]=ode23(@sys2,[t_ini t_final],x_ini) Example: Step response of first order system (2) The graph below shows the effect of ζ on the unit step response of a second order system,
In this chapter we will start looking at second order differential equations. The 3-D Coordinate System; Note as well that while we example mechanical The Bode plot of second order system. Second order systems The Bode plot of second order system. The transfer function of a second order system
Mathematical methods for economic theory: one example, which shows how the start of this section to convert the two-equation system to a single second-order In logic and mathematics second-order logic is an For example, the second-order Henkin proved that the standard deductive system for first-order logic
ECE 421 Second-Order System Example #3. This example considers the design of a second-order system that will satisfy certain time-domain specifications. Second order system response. Unstable Re(s) Im(s) Overdamped or Critically damped Undamped Underdamped Underdamped
And hence this time response of second-order control system is referred as critically damped. Table Method Examples History of Laplace Transform. Time Response of Second Order Systems the natural frequency dimensionl ess damping ratio 2 ( 2 ) ( / ) / ( / ) ( ) Consider t he first term only:
Mathematical methods for economic theory: one example, which shows how the start of this section to convert the two-equation system to a single second-order Second Order Systems. From ControlTheoryPro.com. which this example does not include. natural frequency of the system . The second parameter, ,
Mathematical methods for economic theory: one example, which shows how the start of this section to convert the two-equation system to a single second-order Second order system response. Unstable Re(s) Im(s) Overdamped or Critically damped Undamped Underdamped Underdamped
Second Order Systems Second Order Equations 2 2 +2 +1 = s s K G s Example 5.5 • Heated tank + controller = 2nd order system (a) When feed rate changes from 0.4 Second-Order LTI Systems In Fig. 1 we see an example of a second order response, y(t) In homework two you solved a second-order system with ode45
Second Order Systems SecondOrderSystems.docx 10/3/2008 11:39 AM Page 1 Background Here is an example of a second order system from EAS 206. F The Bode plot of second order system. Second order systems The Bode plot of second order system. The transfer function of a second order system
We have two types of system, first order system and second order system, Our main topic which we discuss here is First Order Control System. In theory, Second order differential equations come up frequently in oscillations What are practical applications of second-order ODEs? What is second order system?
This MATLAB function generates the state-space description (A,B,C,D) of the second-order system What can be a simple practical example for a third order uncertain system which could be described by a family of tr functions?
In logic and mathematics second-order logic is an For example, the second-order Henkin proved that the standard deductive system for first-order logic Examples: First-Order Systems In the above examples, first-order state equations are sufficient to describe energetic transactions within the system;
Standard form of second order system is given by: to “Introduction to Second Order Systems Introduction to First-Order Systems. Introduction to First-Order Systems Introduction There are two methods to analyze functioning of a control system that are time domain analysis and control domain
Mathematical methods for economic theory: one example, which shows how the start of this section to convert the two-equation system to a single second-order Applications of Second‐Order Equations. Skydiving. The spring‐block oscillator is an idealized example of a frictionless system. In real life,
Second-Order LTI Systems In Fig. 1 we see an example of a second order response, y(t) In homework two you solved a second-order system with ode45 First Order Control System. First Order System is the First and Second Order System Lyapunov Stability Analysis with Solved Examples Electrical Academia.
### PHYSICS6770 Laboratory 1 Week 1 First order systems First
Introduction to First-Order Systems IDC-Online. Example: Simulation of second-order system The flrst program calls Matlab’s ode solver ode23. The command [T,X]=ode23(@sys2,[t_ini t_final],x_ini), If the system contained high losses, for example if the spring–mass experiment were conducted in a The effect of varying damping ratio on a second-order system..
What can be a simple practical example for a third order. Second order system response. Unstable Re(s) Im(s) Overdamped or Critically damped Undamped Underdamped Underdamped, First order LTI systems are characterized by the differential equation and ω = 2π f is the frequency in radians per second. Example solution.
### Second Order Systems Educating Global Leaders
Barthes and the Second-Order Semiological System. First and second order change in systems. and the concepts of first and second order change can help us not be one of the system. For example, In this chapter we will start looking at second order differential equations. The 3-D Coordinate System; Note as well that while we example mechanical.
In logic and mathematics second-order logic is an For example, the second-order Henkin proved that the standard deductive system for first-order logic JOURN.IL OF MATHEMATICAL ANALYSIS AND APPLICATIONS 8, 423-444 (1964) A Study of Second Order Nonlinear Systems Y. s. LIM Bell Telephone Laboratories
We will classify second order systems from This implies that the second order system can be Step response and poles for an example of an underdamped system PHYSICS6770 Laboratory 1 University of Utah Department of Physics First and second order systems Week 1, First order systems slide 4 First order system: Example 1, RC
Develop an understanding of dynamic behavior and take a look Consider now a second-order system with numerator tough challenges for process control systems. First Order Control System. First Order System is the First and Second Order System Lyapunov Stability Analysis with Solved Examples Electrical Academia.
Examples such as the following can be used to illustrate the 2.6 Systems of two First Order second order linear differential equation is expected. Transient Response of a Second-Order System ECEN 2830 Spring 2012 1. Introduction In connection with this experiment, you are selecting the gains in your feedback loop to
Beat the System with Family Systems Therapy. sustainable than second-order changes. Example: order change can occur for a whole system and/or for an Review of First- and Second-Order System system parameter that establishes the time scale of system responses in a flrst-order system. For example a resistor
ECE 421 Second-Order System Example #4. This example considers the relationship between the locations of the closed-loop poles for the standard second-order system Example calculations showing derivation of a second order control system pole locations in the Laplace domain. Rise time and peak time are also covered.
First Order Control System. First Order System is the First and Second Order System Lyapunov Stability Analysis with Solved Examples Electrical Academia. As a second example, consider the systems shown below in which we have swapped the components from the fires example. Unit Step Response of a Second Order System.
In this section we will look at some of the basics of systems of differential equations. an example of a system of first order, equation in the second Example calculations showing derivation of a second order control system pole locations in the Laplace domain. Rise time and peak time are also covered.
As a second example, consider the systems shown below in which we have swapped the components from the fires example. Unit Step Response of a Second Order System. Example: Simulation of second-order system The flrst program calls Matlab’s ode solver ode23. The command [T,X]=ode23(@sys2,[t_ini t_final],x_ini)
Examples such as the following can be used to illustrate the 2.6 Systems of two First Order second order linear differential equation is expected. Example calculations showing derivation of a second order control system pole locations in the Laplace domain. Rise time and peak time are also covered.
Time Response of Second-Order Control System Published on 24/2/2012 and last updated on Friday 13th of July 2018 at 07:22:10 PM First Order Control System. First Order System is the First and Second Order System Lyapunov Stability Analysis with Solved Examples Electrical Academia.
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deffunc H1( addLoopStr ) -> Element of bool [:[: the carrier of \$1, the carrier of \$1:], the carrier of \$1:] = the addF of \$1;
let G be RealLinearSpace-Sequence; :: thesis:
reconsider GS = RLSStruct(# (product (carr G)),(),[:():],[:():] #) as non empty RLSStruct ;
dom G = Seg (len G) by FINSEQ_1:def 3;
then dom G = Seg (len (carr G)) by Def4;
then A1: dom G = dom (carr G) by FINSEQ_1:def 3;
now :: thesis: for a1, b1 being Real
for v, w being VECTOR of GS holds
( a1 * (v + w) = (a1 * v) + (a1 * w) & (a1 + b1) * v = (a1 * v) + (b1 * v) & (a1 * b1) * v = a1 * (b1 * v) & 1 * v = v )
let a1, b1 be Real; :: thesis: for v, w being VECTOR of GS holds
( a1 * (v + w) = (a1 * v) + (a1 * w) & (a1 + b1) * v = (a1 * v) + (b1 * v) & (a1 * b1) * v = a1 * (b1 * v) & 1 * v = v )
reconsider a = a1, b = b1 as Element of REAL by XREAL_0:def 1;
let v, w be VECTOR of GS; :: thesis: ( a1 * (v + w) = (a1 * v) + (a1 * w) & (a1 + b1) * v = (a1 * v) + (b1 * v) & (a1 * b1) * v = a1 * (b1 * v) & 1 * v = v )
reconsider v1 = v, w1 = w as Element of product (carr G) ;
A2: now :: thesis: for x being object st x in dom (carr G) holds
([:():] . (jj,v1)) . x = v1 . x
let x be object ; :: thesis: ( x in dom (carr G) implies ([:():] . (jj,v1)) . x = v1 . x )
assume x in dom (carr G) ; :: thesis: ([:():] . (jj,v1)) . x = v1 . x
then reconsider i = x as Element of dom (carr G) ;
reconsider vi = v1 . i as VECTOR of (G . i) by ;
([:():] . (jj,v1)) . x = 1 * vi by Lm4;
hence ([:():] . (jj,v1)) . x = v1 . x by RLVECT_1:def 8; :: thesis: verum
end;
A3: now :: thesis: for x being object st x in dom (carr G) holds
([:():] . ((a + b),v1)) . x = ([:():] . (([:():] . (a,v1)),([:():] . (b,v1)))) . x
let x be object ; :: thesis: ( x in dom (carr G) implies ([:():] . ((a + b),v1)) . x = ([:():] . (([:():] . (a,v1)),([:():] . (b,v1)))) . x )
assume x in dom (carr G) ; :: thesis: ([:():] . ((a + b),v1)) . x = ([:():] . (([:():] . (a,v1)),([:():] . (b,v1)))) . x
then reconsider i = x as Element of dom (carr G) ;
reconsider vi = v1 . i as VECTOR of (G . i) by ;
([:():] . ((a + b),v1)) . i = (a + b) * vi by Lm4
.= (a * vi) + (b * vi) by RLVECT_1:def 6
.= H1(G . i) . ((([:():] . (a,v1)) . i),(b * vi)) by Lm4
.= H1(G . i) . ((([:():] . (a,v1)) . i),(([:():] . (b,v1)) . i)) by Lm4 ;
hence ([:():] . ((a + b),v1)) . x = ([:():] . (([:():] . (a,v1)),([:():] . (b,v1)))) . x by Lm3; :: thesis: verum
end;
A4: now :: thesis: for x being object st x in dom (carr G) holds
([:():] . (a,([:():] . (v1,w1)))) . x = ([:():] . (([:():] . (a,v1)),([:():] . (a,w1)))) . x
let x be object ; :: thesis: ( x in dom (carr G) implies ([:():] . (a,([:():] . (v1,w1)))) . x = ([:():] . (([:():] . (a,v1)),([:():] . (a,w1)))) . x )
assume x in dom (carr G) ; :: thesis: ([:():] . (a,([:():] . (v1,w1)))) . x = ([:():] . (([:():] . (a,v1)),([:():] . (a,w1)))) . x
then reconsider i = x as Element of dom (carr G) ;
reconsider vi = v1 . i, wi = w1 . i as VECTOR of (G . i) by ;
([:():] . (a,([:():] . (v1,w1)))) . i = the Mult of (G . i) . (a,(([:():] . (v1,w1)) . i)) by Lm4
.= a * (vi + wi) by Lm3
.= (a * vi) + (a * wi) by RLVECT_1:def 5
.= H1(G . i) . ((([:():] . (a,v1)) . i),(a * wi)) by Lm4
.= H1(G . i) . ((([:():] . (a,v1)) . i),(([:():] . (a,w1)) . i)) by Lm4 ;
hence ([:():] . (a,([:():] . (v1,w1)))) . x = ([:():] . (([:():] . (a,v1)),([:():] . (a,w1)))) . x by Lm3; :: thesis: verum
end;
( dom ([:():] . (a,([:():] . (v1,w1)))) = dom (carr G) & dom ([:():] . (([:():] . (a,v1)),([:():] . (a,w1)))) = dom (carr G) ) by CARD_3:9;
hence a1 * (v + w) = (a1 * v) + (a1 * w) by ; :: thesis: ( (a1 + b1) * v = (a1 * v) + (b1 * v) & (a1 * b1) * v = a1 * (b1 * v) & 1 * v = v )
A5: now :: thesis: for x being object st x in dom (carr G) holds
([:():] . ((a * b),v1)) . x = ([:():] . (a,([:():] . (b,v1)))) . x
let x be object ; :: thesis: ( x in dom (carr G) implies ([:():] . ((a * b),v1)) . x = ([:():] . (a,([:():] . (b,v1)))) . x )
assume x in dom (carr G) ; :: thesis: ([:():] . ((a * b),v1)) . x = ([:():] . (a,([:():] . (b,v1)))) . x
then reconsider i = x as Element of dom (carr G) ;
reconsider vi = v1 . i as VECTOR of (G . i) by ;
([:():] . ((a * b),v1)) . i = (a * b) * vi by Lm4
.= a * (b * vi) by RLVECT_1:def 7
.= the Mult of (G . i) . (a,(([:():] . (b,v1)) . i)) by Lm4 ;
hence ([:():] . ((a * b),v1)) . x = ([:():] . (a,([:():] . (b,v1)))) . x by Lm4; :: thesis: verum
end;
( dom ([:():] . ((a + b),v1)) = dom (carr G) & dom ([:():] . (([:():] . (a,v1)),([:():] . (b,v1)))) = dom (carr G) ) by CARD_3:9;
hence (a1 + b1) * v = (a1 * v) + (b1 * v) by ; :: thesis: ( (a1 * b1) * v = a1 * (b1 * v) & 1 * v = v )
( dom ([:():] . ((a * b),v1)) = dom (carr G) & dom ([:():] . (a,([:():] . (b,v1)))) = dom (carr G) ) by CARD_3:9;
hence (a1 * b1) * v = a1 * (b1 * v) by ; :: thesis: 1 * v = v
( dom ([:():] . (jj,v1)) = dom (carr G) & dom v1 = dom (carr G) ) by CARD_3:9;
hence 1 * v = v by ; :: thesis: verum
end;
hence ( product G is vector-distributive & product G is scalar-distributive & product G is scalar-associative & product G is scalar-unital ) ; :: thesis: verum
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# Free Practical Geometry 02 Practice Test - 7th grade
If the distance between any two lines is always constant then the lines:
A.
are parallel to each other
B.
may not be parallel to each other
C.
may be parallel to each other
D.
are perpendicular to each other
#### SOLUTION
Solution : B and C
If the distance between any two lines is always constant then the lines may or may not be parallel. If the lines lie in the same plane, then they are parallel and if they are not, they are not parallel.
Pavan tried to draw a line parallel to a given line, 'l'. Few steps are shown below. What will be the next step?
Step 1: Mark a point A, not on the line 'l'.
Step 2: Mark point B on line 'l'.
Step 3: Draw line segment joining points A and B.
Step 4: Draw an arc with B as the centre, such that it intersects line 'l' at D and AB at E.
A.
Draw another arc with the same radius and A as the centre, such that it intersects AB at F
B.
Draw another arc with B as the centre and distance DE as the radius
C.
Draw another arc with some other radius and A as the centre
D.
Draw another arc with B as the centre and distance AB as the radius
#### SOLUTION
Solution : A
Following are the steps to draw a line parallel to a given line:
Step 1: Mark a point A, not on the line 'l'.
Step 2: Mark point B on line 'l'.
Step 3: Draw line segment joining points A and B.
Step 4: Draw an arc with B as the centre, such that it intersects line 'l' at D and AB at E.
Step 5: Draw another arc with the same radius and A as the centre, such that it intersects AB at F.
Step 6: Draw another arc with F as the centre and distance DE as the radius.
Step 7: Mark the point of intersections of this arc and the previous arc as G.
Step 8: Draw line 'm' passing through points A and G.
Line 'm' is the required parallel line.
In the figure shown, the two lines are parallel to each other. Which of the following options is incorrect?
A.
1 = 8
B.
6 = 8
C.
2 = 8
D.
4 = 8
#### SOLUTION
Solution : A
When two lines intersect each other, four angles are formed. The pair of angles which lie on the opposite sides of the point of intersection are called vertically opposite angle. If two parallel lines are cut by a transversal:
(i) Alternate interior angles are equal i.e., 2 = 8 and 3 = 5.
(ii) Alternate exterior angles are equal i.e., 1 = 7 and 4 = 6.
(iii) Corresponding angles are equal i.e., 1 = 5, 4 = 8, 2 = 6 and 3 = 7.
(iv) Co-interior angles are supplementary i.e., 2 + 5 = 180 and 3 + 8 =180
1 = 8 is incorrect.
Ram drew a line segment AB of length 3cm and another line segment CB of length 4cm which is perpendicular to line segment AB. What is the length of the third side of the triangle?
A. 5cm
B.
6cm
C.
2cm
D. 1cm
#### SOLUTION
Solution : A
The triangle will be a right angled triangle as CB is perpendicular to AB .So, the third side can be easily calculated by using Pythagoras theorem.
Let the length of the third side be zcm.
z2=32+42
z=25
z=5 cm
If line segment AB is perpendicular to CD then DCA = ? (in degrees)
__
#### SOLUTION
Solution :
When a line segment is perpendicular to another line segment then angle between them is 90.
In the shown figure, line segment BD is angle bisector of ABC. Line segment BE is angle bisector of ABD. Then ABC= __ ×ABE
#### SOLUTION
Solution : As stated in question line segment BD is angle bisector so, ABD=DBC.
So, ABC=4ABE
If the sides of a triangle are 3 cm, 4 cm, and 5 cm. Then, the triangle formed is a/an _____ triangle.
A.
right-angled
B.
obtuse-angled
C.
acute-angled
D.
equilateral
#### SOLUTION
Solution : A
Let us construct the required triangle as follows:
Step 1:
Draw the base line XY of the length 3 cm.
Step 2:
Use your compass to make an arc with centre at point Y and the radius 4 cm.
Step 3:
Use your compass to make a circle with centre at point X and the radius 5 cm.
Step 4:
The point of intersection between the arcs is named as X.
The triangle is completed by joining all these three points.
Now, from above construction we see that the triangle formed is a right angle triangle.
Pavan followed the steps given below to construct a triangle
Right-angled triangle LMN, with hypotenuse LN=8cm and side MN=5cm.
Step 1: Draw line 'l'.
Step 2: Mark a point on 'l' and name it M.
Step 3: Draw a line segment MN=5cm on 'l' .
Step 4: Construct a right angle XMN at M.
Step 5: Draw an arc with N as the centre and radius equal to 8cm, such that it intersects MX.
Step 6: Mark the point of intersection as L.
Step 7: Join points L and N.
Are the steps followed by him correct?
If it is correct then enter "Yes” if no then enter "No”.
__
#### SOLUTION
Solution :
Following the above steps, we can construct a right angled triangle with the given measurements. So, the given steps are correct.
The lines shown below never meet at any point. So, these lines are not parallel. Say true or false.
A.
True
B.
False
#### SOLUTION
Solution : B
Parallel lines never intersect. Parallel lines are lines in a plane which do not meet, that is, two lines in a plane that do not intersect or touch each other at any point are said to be parallel.
Raman had to draw an isosceles triangle with base angles as 65 each and base length of 5cm. Raman said the given data is insufficient. Is he correct?
A.
True
B.
False
#### SOLUTION
Solution : B
We are given two angles and one side. So, data is sufficient and the triangle can be drawn.
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Aids to Understanding
The Meaning of Capacitance
Whenever we are asked to explain what a capacitance of 2000 m F means we first look at the units of F. Thus 2000 m F means 2000 mCV-1. This means little on its own so we need to elaborate a little further. If we recall the activity that first led us to the concept of capacitance it should be clear that 2000 mCV-1 means that for every 2000 mC placed on the capacitor, its pd rises by 1 V.
The Coulombmeter
The Coulombmeter consists of a capacitor. Whenever we use it to measure the charge on a capacitor's plate we connect the Coulombmeter in parallel with the capacitor and the charge is transferred from the capacitor to the Coulombmeter. However unless the Coulombmeter's capacitor (2000 mF) is much larger than the capacitance of the the capacitor whose charge it is we wish to measure then the capacitors will share the charge and an inaccurate reading will occur. In the first activity we used a capacitor of 1.0 m F
Capacitors in parallel
For resistors in parallel we use the fact that the pd across components in parallel are equal. Use this fact to find an expression for capacitors inparallel. Compare this with the expression for resistors in parallel. If we have a number of resistances in parallel then the largest resistance will have the smallest current flowing in it. This is a fact. Write an equivalent expression for capacitances in parallel. Remember C = Q/V.
Energy and Capacitors
In unit 2.1 you learned that the work done in pushing a charge, Q, through a pd of V, is equal to QV and we say W = QV. The work done in charging a capacitor, by an external source, is exactly 1/2 QV, however. Thus although the source is placing charges on the capacitor against the field the field it is working against is only at a maximum at the end of the charging process. Given the field is zero inititally we can think of the source working against an average field and thus the energy stored in a capacitor with a charge Q on one of its plates and a pd of V between the plates, is equal to E = 1/2QV. In this argument it was assumed that the E field between the plates is proportional to the pd between the plates. This is entirely justified and it can be shown that E = V/d where d is the distance between the plates.
Capacitors in dc circuits
The characterist curves described in the main text should be memorised but also understood. To help us here we need to recall our circuit rules for dc circuits, ie the current in a dc circuit is the same at all points. This means there is only one characteristic graph for current building up in a CR circuit. The fact that the current falls to zero can be explained by thinking of the capacitor gradually becoming charged up and therefore opposing more charge from joining it. When the pd across the capacitor equals the pd of the cell the current is zero.
The voltage across components in series always sums to the voltage of the cell therefore as the pd rises across the capacitor it must fall across the resistor. whenever a capacitor discharges through a resistor the voltage across each must be the same across both components so the characteristic voltage graphs are identical.
Read this paragraph in conjuction with the graphs in the main text on the right.
Problem Solving
The following points must be borne in mind.
1. The charge stored on a capacitor is the area under its I-t graph.
2. The initial charging current is independent of the value of the capacitor and depends only on Voltage of cell and resistance of resistor, ie I = v/R.
3. The energy stored on a capacitor is the area under its Q-V graph.
4. For a CR circuit the voltage across each component always sum to the voltage of the cell at ALL times during charging or discharging. This list is not comprehensive but represents some points that students occasionally miss or forget.
ac supplies and Capacitors
The result that current increases with frequency in ac circuits contaning capacitors might surprise you but it ties in with our previous work. In a dc circuit the capacitor gradually charges up and prevents further charge from joining it. with an ac supply, the charge never quites get time to charge up the capacitor before the signal changes. As the frequency increases this effect is further enhanced therefore the opposition to charge decreases with increasing frequency and the current increases. We call the opposition to ac current reactance and clearly the reactance of a capacitor decreases with increasing currenty. This result should be compared with resistors in ac circuits. The current in an ac circuit with a resistance does not change with changing frequency.
## 1) Charge and pd relationship for two parallel conducting plates
If we place two conducting plates close together and deposit a charge on one then an equal but opposite charge appears on the inside face of the other. If we then investigate the pd across the plates as for different magnitudes of charge then we find both quantities are in direct proportion. This is illustrated below. This phenomenon introduces us to the concept of capacitance and we say the parallel plates have a capacity for storing charge. This will become clearer when we consider capacitors as components in circuits, ie circuit elements.
## 2) Investigating pd and charge for capacitors in circuits.
The following circuit is used to investigate the pd applied to a capacitor and the charge stored on one of its plates. A table and graph is shown. The pd is varied using the potentiometer in 0.2 V steps. when the switch is in position 1 the capacitor charges, in position 2 the capacitor discharges through the Coulombmeter and the reading is displayed.
## 3) Capacitance defined.
From the above activity we can see the gradient of the line is equal to Q/V = 1.0 mF. This means for every p.d of 1.0 V applied to a 1.0 m F capacitor, the capacitor will store 1.0 mC of charge and we write C = Q/V. The units of capacitance are CV-1 and we say 1 CV-1 = 1 F (remember in this statement C is for Coulombs). Circuit capacitors have capacitances much less than 1 F and tend to be stated in mF or mF. From this discussion it should be clear that the capacitance of a capacitor is equal to the ratio of charge to p.d.
## 4) Problems using C = Q/V.
The following problem is typical of the question you might get using C = Q/V.
A circuit capacitor has a capacitance of 2000 mF and a maximum working voltage of 15 V. This means if the voltage applied to the capacitor exceeds 15 V then it is likely the capacitor will be damaged. The following questions could now be asked
i) Calulate the maximum charge this capacitor can hold if it is not to be overloaded. Answer: Q = C V = 2000 mF x 15 V = 30 000 mC.
ii) If the capacitor is connected into an ac supply, then what is the maximum Vrms that can be used if the capacitor is not to be overloaded?
Answer: The peak voltage of the ac supply must not exceed 15 V, therfore Vrms = Vpk ⁄ √ 2 = 15 V ⁄ √ 2 = 10.6 V.
## 5) Energy matters and capacitance
Whenever two negative charges are pushed together their natural repulsion for each other must be overcome. Thus the force pushing them together does work and the work done is equal to the gain in electric potential energy of the two charges. This is precisely what is happening when charge builds up on a capacitor.
## 6) Work done and Q-V graphs
When a source of energy (cell or manual effort) deposits charge on a capacitor work needs to be done (energy expended) and this energy is stored between the capacitor's plate as electrical potential energy of the charges on its plates. Since the pd (energy per coulomb) between the plates rises in direct proportion to the charge deposited, we can draw the following graph. It should be clear that the area under this graph is the total energy needed to charge the capacitor and therefore the total energy stored between its plates, by virtue of the charge deposited there,
## 7) Investigating current and voltage in dc circuits containing capacitors
Now we have established some properties of capacitors it is appropriate to investigate current and voltage in RC circuits. An RC circuit is a circuit containing a capacitor and resistor in series. Firstly we will investigate RC circuits with dc supplies.
1. Investigating current in an RC circuit.
A capacitor consists of two conducting plates that are insulated from each other. The insulation can either be air or some other material. Since no charges can flow through a capacitor we might expect the current in circuits containing capacitors might also be zero. We find this is not the case as the investigations show.
The image above is colour coded. Red represents the circuit, data and graph for matters involving charging the capacitor. Blue represents the circuit, data and graph for matters involving discharging the capacitor. Note the initial current in both cases is 3 mA, which is V/R nd the area under each graph is equal to the charge transferred to the capacitor during charging and the charge that leaves the capacitor during discharging.
2. Investigating voltage in an RC circuit (i) Charging.
Whenever a current flows in an RC circuit, charges build up on the capacitor thus raising the pd across its plates Vc, additionally the current flowing in the resistor means a pd exists across the ends of the resistor, VR. since we have a series circuit we can say;
Vcell = Vc + VR. This means if we know the current characteristic (graph) for a charging and discharging capacitor we can draw the voltage characteristics.
Note that the right hand columns of each table add up to 3.0 V. This is to be expected since Vcell = Vc + VR.
2. Investigating voltage in an RC circuit (ii) Discharging.
If we throw the switch to position 2, the capacitor will discharge through the resistor and the pd between its plates will fall. Note that during discharging the capacitor and resistor are in parallel. This means the pd across them will be equal and thus their pd discharging graphs are identical.
## 8) Problems on CR circuits when the supply is dc
These problems require you to understand the steps involved in finding the current and pd characteristic graphs for charging and discharging. The labelling of the graphs are required as well. Study the graphs above and try the following problem.
Example
The pc terminals above monitor the pd across the capacitor and resistor when the switch is closed. If the capacitor takes 100.0s to completely charge...
i) Draw what will be seen on the pc monitors. Numerical values on axes are required for a full answer.
ii) Draw the current-time graph for the circuit. Numerical values on axes are required for a full answer.
iii) How much charge is stored on the capacitor when fully charged.
iv) How much energy is stored on the capacitor when it is fully charged.
## 9) Capacitors in ac circuits
Up to now we have discussed capacitors in dc circuits. If we connect a capacitor into an ac source the natural thing to investigate is current in the circuit and the frequency of the source. The circuit below enables us to do this and the result of the activity is shown on the graph.
The fact that current increases with frequency is a useful fact to engineers when they are designing electronic circuits.
## 10) Uses of capacitors
Modern electronics would not be possible without capacitors. You are invited to find out the following.
1. Why are capacitors are needed in flash lights for cameras.
2. What property of capacitors allow it to block dc signals but pass ac signals
3. Modern electronic circuits work with dc signals, but they are connected in the ac mains supply. Find out what rectification means and then find out what smoothing means. Find out how capacitors smooth a rectified signal.
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mersenneforum.org Sieving with powers of small primes in the Small Prime variation of the Quadratic Sieve
Register FAQ Search Today's Posts Mark Forums Read
2016-05-04, 14:39 #1 mickfrancis Apr 2014 Marlow, UK 23×7 Posts Sieving with powers of small primes in the Small Prime variation of the Quadratic Sieve As I understand it, in the Small Prime Variation of the Quadratic Sieve, primes less than a threshold (Pmin, say) are not used for sieving, as the cost of sieving is disproportionately high given the contribution of these primes. Sieving with powers of primes that are themselves used in sieving is also less beneficial than sieving with other primes of size comparable to these powers (e.g. the contribution for p^2 is the same as for p). However, it seems to me that for primes in the factor base less than Pmin it ought still to make sense to sieve with the lowest powers that are Pmin or above, as the contribution is just as high as that of similarly-sized primes (why would I sieve with the prime 257 but not 256 (28), for example)? In the implementations I've looked at, this does not appear to be done, and I was wondering why; is the benefit outweighed by the added complexity?
2016-05-06, 02:19 #2 jasonp Tribal Bullet Oct 2004 23·32·72 Posts Once the problem size gets above a fairly small threshold, the runtime needed for sieving small primes becomes insignificant compared to sieving everything else; there are only a few such, and they cache nicely, so they're done in a flash compared to the thousands of larger primes you must also sieve with.
2016-05-06, 08:13 #3
mickfrancis
Apr 2014
Marlow, UK
5610 Posts
Quote:
Originally Posted by jasonp Once the problem size gets above a fairly small threshold, the runtime needed for sieving small primes becomes insignificant compared to sieving everything else; there are only a few such, and they cache nicely, so they're done in a flash compared to the thousands of larger primes you must also sieve with.
Makes sense - thanks Jason.
Similar Threads Thread Thread Starter Forum Replies Last Post carpetpool Prime Gap Searches 45 2017-09-30 20:51 mickfrancis Factoring 5 2016-03-31 06:21 ThiloHarich Factoring 13 2009-01-04 18:19 Housemouse Math 2 2008-06-04 05:23 R1zZ1 Factoring 36 2007-11-02 15:59
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# Rodrigues’ rotation formula
Rodrigues’ rotation formula is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
Rotate a point about some axis by some angle using Rodrigues' rotation formula.
Reference
```with Ada.Text_Io;
procedure Rodrigues is
type Vector is record
X, Y, Z : Float;
end record;
function Image (V : in Vector) return String is
('[' & V.X'Image & ',' & V.Y'Image & ',' & V.Z'Image & ']');
-- Basic operations
function "+" (V1, V2 : in Vector) return Vector is ((V1.X + V2.X,
V1.Y + V2.Y,
V1.Z + V2.Z));
function "*" (V : in Vector; A : in Float) return Vector is ((V.X*A, V.Y*A, V.Z*A));
function "/" (V : in Vector; A : in Float) return Vector is (V*(1.0/A));
function "*" (V1, V2 : in Vector) return Float is (-- dot-product
(V1.X*V2.X + V1.Y*V2.Y + V1.Z*V2.Z));
function Norm(V : in Vector) return Float is (Sqrt(V*V));
function Normalize(V : in Vector) return Vector is (V /Norm(V));
function Cross_Product (V1, V2 : in Vector) return Vector is (-- cross-product
(V1.Y*V2.Z - V1.Z*V2.Y,
V1.Z*V2.X - V1.X*V2.Z,
V1.X*V2.Y - V1.Y*V2.X));
function Angle (V1, V2 : in Vector) return Float is (Arccos((V1*V2) / (Norm(V1)*Norm(V2))));
-- Rodrigues' rotation formula
function Rotate (V, Axis : in Vector;
Theta : in Float) return Vector is
K : constant Vector := Normalize(Axis);
begin
return V*Cos(Theta) + Cross_Product(K,V)*Sin(Theta) + K*(K*V)*(1.0-Cos(Theta));
end Rotate;
--
-- Rotate vector Source on Target
Source : constant Vector := ( 0.0, 2.0, 1.0);
Target : constant Vector := (-1.0, 2.0, 0.4);
begin
Put_Line ("Vector " & Image(Source));
Put_Line ("rotated on " & Image(Target));
Put_Line (" = " & Image(Rotate(V => Source,
Axis => Cross_Product(Source, Target),
Theta => Angle(Source, Target))));
end Rodrigues;
```
Output:
```Vector [ 0.00000E+00, 2.00000E+00, 1.00000E+00]
rotated on [-1.00000E+00, 2.00000E+00, 4.00000E-01]
= [-9.84374E-01, 1.96875E+00, 3.93750E-01]
```
## ALGOL 68
Translation of: JavaScript
```BEGIN # Rodrigues' Rotation Formula #
MODE VECTOR = [ 3 ]REAL;
MODE MATRIX = [ 3 ]VECTOR;
PROC norm = ( VECTOR v )REAL: sqrt( ( v[1] * v[1] ) + ( v[2] * v[2] ) + ( v[3] * v[3] ) );
PROC normalize = ( VECTOR v )VECTOR:
BEGIN
REAL length = norm( v );
( v[1] / length, v[2] / length, v[3] / length )
END # normalize # ;
PROC dot product = ( VECTOR v1, v2 )REAL: ( v1[1] * v2[1] ) + ( v1[2] * v2[2] ) + ( v1[3] * v2[3] );
PROC cross product = ( VECTOR v1, v2 )VECTOR: ( ( v1[2] * v2[3] ) - ( v1[3] * v2[2] )
, ( v1[3] * v2[1] ) - ( v1[1] * v2[3] )
, ( v1[1] * v2[2] ) - ( v1[2] * v2[1] )
);
PROC get angle = ( VECTOR v1, v2 )REAL: acos( dot product( v1, v2 ) / ( norm( v1 ) * norm( v2 ) ) );
PROC matrix multiply = ( MATRIX m, VECTOR v )VECTOR: ( dot product( m[1], v )
, dot product( m[2], v )
, dot product( m[3], v )
);
PROC a rotate = ( VECTOR p, v, REAL a )VECTOR:
BEGIN
REAL ca = cos( a ), sa = sin( a ), t = 1 - ca, x = v[1], y = v[2], z = v[3];
MATRIX r = ( ( ca + ( x*x*t ), ( x*y*t ) - ( z*sa ), ( x*z*t ) + ( y*sa ) )
, ( ( x*y*t ) + ( z*sa ), ca + ( y*y*t ), ( y*z*t ) - ( x*sa ) )
, ( ( z*x*t ) - ( y*sa ), ( z*y*t ) + ( x*sa ), ca + ( z*z*t ) )
);
matrix multiply( r, p )
END # a rotate # ;
VECTOR v1 = ( 5, -6, 4 );
VECTOR v2 = ( 8, 5, -30 );
REAL a = get angle( v1, v2 );
VECTOR cp = cross product( v1, v2 );
VECTOR ncp = normalize( cp );
VECTOR np = a rotate( v1, ncp, a );
print( ( "( ", fixed( np[ 1 ], -10, 6 )
, ", ", fixed( np[ 2 ], -10, 6 )
, ", ", fixed( np[ 3 ], -10, 6 )
, " )", newline
)
)
END```
Output:
```( 2.232221, 1.395138, -8.370829 )
```
## AutoHotkey
Translation of: JavaScript
```v1 := [5,-6,4]
v2 := [8,5,-30]
a := getAngle(v1, v2)
cp := crossProduct(v1, v2)
ncp := normalize(cp)
np := aRotate(v1, ncp, a)
for i, v in np
result .= v ", "
MsgBox % result := "[" Trim(result, ", ") "]"
return
norm(v) {
return Sqrt(v[1]*v[1] + v[2]*v[2] + v[3]*v[3])
}
normalize(v) {
length := norm(v)
return [v[1]/length, v[2]/length, v[3]/length]
}
dotProduct(v1, v2) {
return v1[1]*v2[1] + v1[2]*v2[2] + v1[3]*v2[3]
}
crossProduct(v1, v2) {
return [v1[2]*v2[3] - v1[3]*v2[2], v1[3]*v2[1] - v1[1]*v2[3], v1[1]*v2[2] - v1[2]*v2[1]]
}
getAngle(v1, v2) {
return ACos(dotProduct(v1, v2) / (norm(v1)*norm(v2)))
}
matrixMultiply(matrix, v) {
return [dotProduct(matrix[1], v), dotProduct(matrix[2], v), dotProduct(matrix[3], v)]
}
aRotate(p, v, a) {
ca:=Cos(a), sa:=Sin(a), t:=1-ca, x:=v[1], y:=v[2], z:=v[3]
r := [[ca + x*x*t, x*y*t - z*sa, x*z*t + y*sa]
, [x*y*t + z*sa, ca + y*y*t, y*z*t - x*sa]
, [z*x*t - y*sa, z*y*t + x*sa, ca + z*z*t]]
return matrixMultiply(r, p)
}
```
Output:
`[2.232221, 1.395138, -8.370829]`
## C
Translation of: JavaScript
```#include <stdio.h>
#include <math.h>
typedef struct {
double x, y, z;
} vector;
typedef struct {
vector i, j, k;
} matrix;
double norm(vector v) {
return sqrt(v.x*v.x + v.y*v.y + v.z*v.z);
}
vector normalize(vector v){
double length = norm(v);
vector n = {v.x / length, v.y / length, v.z / length};
return n;
}
double dotProduct(vector v1, vector v2) {
return v1.x*v2.x + v1.y*v2.y + v1.z*v2.z;
}
vector crossProduct(vector v1, vector v2) {
vector cp = {v1.y*v2.z - v1.z*v2.y, v1.z*v2.x - v1.x*v2.z, v1.x*v2.y - v1.y*v2.x};
return cp;
}
double getAngle(vector v1, vector v2) {
return acos(dotProduct(v1, v2) / (norm(v1)*norm(v2)));
}
vector matrixMultiply(matrix m ,vector v) {
vector mm = {dotProduct(m.i, v), dotProduct(m.j, v), dotProduct(m.k, v)};
return mm;
}
vector aRotate(vector p, vector v, double a) {
double ca = cos(a), sa = sin(a);
double t = 1.0 - ca;
double x = v.x, y = v.y, z = v.z;
matrix r = {
{ca + x*x*t, x*y*t - z*sa, x*z*t + y*sa},
{x*y*t + z*sa, ca + y*y*t, y*z*t - x*sa},
{z*x*t - y*sa, z*y*t + x*sa, ca + z*z*t}
};
return matrixMultiply(r, p);
}
int main() {
vector v1 = {5, -6, 4}, v2 = {8, 5, -30};
double a = getAngle(v1, v2);
vector cp = crossProduct(v1, v2);
vector ncp = normalize(cp);
vector np = aRotate(v1, ncp, a);
printf("[%.13f, %.13f, %.13f]\n", np.x, np.y, np.z);
return 0;
}
```
Output:
```[2.2322210733082, 1.3951381708176, -8.3708290249059]
```
## Factor
Note the following words already exist in Factor, which I have elected not to redefine:
Word Vocabulary Equivalent function in JavaScript (ES5) entry
normalize math.vectors normalize()
cross math.vectors crossProduct()
angle-between math.vectors getAngle()
mdotv math.matrices matrixMultiply()
Translation of: JavaScript
Works with: Factor version 0.99 2021-06-02
```USING: grouping kernel math math.functions math.matrices
math.vectors prettyprint sequences sequences.generalizations ;
:: a-rotate ( p v a -- seq )
a cos a sin :> ( ca sa )
ca 1 - v first3 :> ( t x y z )
x x t * * ca + x y t * * z sa * - x z t * * y sa * +
x y t * * z sa * + ca y y t * * + y z t * * x sa * -
z x t * * y sa * - z y t * * x sa * + ca z z t * * +
9 narray 3 group p mdotv ;
{ 5 -6 4 } { 8 5 -30 }
dupd [ cross normalize ] [ angle-between ] 2bi a-rotate .
```
Output:
```{ 2.232221073308229 1.395138170817642 -8.370829024905852 }
```
## FreeBASIC
This example rotates the vector [-1, 2, -0.4] around the axis [-1, 2, 1] in increments of 18 degrees.
```#define PI 3.14159265358979323
type vector
'define a 3 dimensional vector data type
x as double
y as double
z as double
end type
operator + ( a as vector, b as vector) as vector
dim as vector r
r.x = a.x + b.x
r.y = a.y + b.y
r.z = a.z + b.z
return r
end operator
operator * ( a as vector, b as vector ) as double
'dot product
return a.x*b.x + a.y*b.y + a.z*b.z
end operator
operator * ( c as double, a as vector ) as vector
'multiplication of a scalar by a vector
dim as vector r
r.x = c*a.x
r.y = c*a.y
r.z = c*a.z
return r
end operator
function hat( a as vector ) as vector
'returns a unit vector in the direction of a
return (1.0/sqr(a*a))*a
end function
operator ^ ( a as vector, b as vector ) as vector
'cross product
dim as vector r
r.x = a.y*b.z - a.z*b.y
r.y = a.z*b.x - a.x*b.z
r.z = a.x*b.y - a.y*b.x
return r
end operator
function rodrigues( v as vector, byval k as vector, theta as double ) as vector
k = hat(k)
return cos(theta)*v + sin(theta)*(k^v) + (1-cos(theta))*(k*v)*k
end function
dim as vector k, v, r
dim as double theta
k.x = 0 : k.y = 2 : k.z = 1
v.x = -1 : v.y = 2 : v.z = 0.4
print "Theta rotated vector"
print "-----------------------------"
for theta = 0 to 2*PI step PI/10
r = rodrigues( v, k, theta )
print using "##.### [##.### ##.### ##.###]"; theta; r.x; r.y; r.z
next theta```
Output:
```
Theta rotated vector
-----------------------------
0.000 [-1.000 2.000 0.400]
0.314 [-1.146 1.915 0.424]
0.628 [-1.269 1.818 0.495]
0.942 [-1.355 1.719 0.606]
1.257 [-1.397 1.629 0.745]
1.571 [-1.390 1.555 0.900]
1.885 [-1.335 1.505 1.055]
2.199 [-1.238 1.484 1.194]
2.513 [-1.107 1.494 1.305]
2.827 [-0.956 1.534 1.376]
3.142 [-0.800 1.600 1.400]
3.456 [-0.654 1.685 1.376]
3.770 [-0.531 1.782 1.305]
4.084 [-0.445 1.881 1.194]
4.398 [-0.403 1.971 1.055]
4.712 [-0.410 2.045 0.900]
5.027 [-0.465 2.095 0.745]
5.341 [-0.562 2.116 0.606]
5.655 [-0.693 2.106 0.495]
5.969 [-0.844 2.066 0.424]
6.283 [-1.000 2.000 0.400]
```
## Go
Translation of: JavaScript
```package main
import (
"fmt"
"math"
)
type vector [3]float64
type matrix [3]vector
func norm(v vector) float64 {
return math.Sqrt(v[0]*v[0] + v[1]*v[1] + v[2]*v[2])
}
func normalize(v vector) vector {
length := norm(v)
return vector{v[0] / length, v[1] / length, v[2] / length}
}
func dotProduct(v1, v2 vector) float64 {
return v1[0]*v2[0] + v1[1]*v2[1] + v1[2]*v2[2]
}
func crossProduct(v1, v2 vector) vector {
return vector{v1[1]*v2[2] - v1[2]*v2[1], v1[2]*v2[0] - v1[0]*v2[2], v1[0]*v2[1] - v1[1]*v2[0]}
}
func getAngle(v1, v2 vector) float64 {
return math.Acos(dotProduct(v1, v2) / (norm(v1) * norm(v2)))
}
func matrixMultiply(m matrix, v vector) vector {
return vector{dotProduct(m[0], v), dotProduct(m[1], v), dotProduct(m[2], v)}
}
func aRotate(p, v vector, a float64) vector {
ca, sa := math.Cos(a), math.Sin(a)
t := 1 - ca
x, y, z := v[0], v[1], v[2]
r := matrix{
{ca + x*x*t, x*y*t - z*sa, x*z*t + y*sa},
{x*y*t + z*sa, ca + y*y*t, y*z*t - x*sa},
{z*x*t - y*sa, z*y*t + x*sa, ca + z*z*t},
}
return matrixMultiply(r, p)
}
func main() {
v1 := vector{5, -6, 4}
v2 := vector{8, 5, -30}
a := getAngle(v1, v2)
cp := crossProduct(v1, v2)
ncp := normalize(cp)
np := aRotate(v1, ncp, a)
fmt.Println(np)
}
```
Output:
```[2.2322210733082275 1.3951381708176436 -8.370829024905852]
```
## JavaScript
### JavaScript: ES5
```function norm(v) {
return Math.sqrt(v[0]*v[0] + v[1]*v[1] + v[2]*v[2]);
}
function normalize(v) {
var length = norm(v);
return [v[0]/length, v[1]/length, v[2]/length];
}
function dotProduct(v1, v2) {
return v1[0]*v2[0] + v1[1]*v2[1] + v1[2]*v2[2];
}
function crossProduct(v1, v2) {
return [v1[1]*v2[2] - v1[2]*v2[1], v1[2]*v2[0] - v1[0]*v2[2], v1[0]*v2[1] - v1[1]*v2[0]];
}
function getAngle(v1, v2) {
return Math.acos(dotProduct(v1, v2) / (norm(v1)*norm(v2)));
}
function matrixMultiply(matrix, v) {
return [dotProduct(matrix[0], v), dotProduct(matrix[1], v), dotProduct(matrix[2], v)];
}
function aRotate(p, v, a) {
var ca = Math.cos(a), sa = Math.sin(a), t=1-ca, x=v[0], y=v[1], z=v[2];
var r = [
[ca + x*x*t, x*y*t - z*sa, x*z*t + y*sa],
[x*y*t + z*sa, ca + y*y*t, y*z*t - x*sa],
[z*x*t - y*sa, z*y*t + x*sa, ca + z*z*t]
];
return matrixMultiply(r, p);
}
var v1 = [5,-6,4];
var v2 = [8,5,-30];
var a = getAngle(v1, v2);
var cp = crossProduct(v1, v2);
var ncp = normalize(cp);
var np = aRotate(v1, ncp, a);
console.log(np);
```
### JavaScript: ES6
(Returning a value directly and avoiding console.log, which is often defined by browser libraries,
but is not part of JavaScript's ECMAScript standards themselves, and is not available to all JavaScript interpreters)
```(() => {
"use strict";
// --------------- RODRIGUES ROTATION ----------------
const rodrigues = v1 =>
v2 => aRotate(v1)(
normalize(
crossProduct(v1)(v2)
)
)(
angle(v1)(v2)
);
// ---------------------- TEST -----------------------
const main = () =>
rodrigues([5, -6, 4])([8, 5, -30]);
// ---------------- VECTOR FUNCTIONS -----------------
const aRotate = p =>
v => a => {
const
cosa = Math.cos(a),
sina = Math.sin(a),
t = 1 - cosa,
[x, y, z] = v;
return matrixMultiply([
[
cosa + ((x ** 2) * t),
(x * y * t) - (z * sina),
(x * z * t) + (y * sina)
],
[
(x * y * t) + (z * sina),
cosa + ((y ** 2) * t),
(y * z * t) - (x * sina)
],
[
(z * x * t) - (y * sina),
(z * y * t) + (x * sina),
cosa + (z * z * t)
]
])(p);
};
const angle = v1 =>
v2 => Math.acos(
dotProduct(v1)(v2) / (
norm(v1) * norm(v2)
)
);
const crossProduct = xs =>
// Cross product of two 3D vectors.
ys => {
const [x1, x2, x3] = xs;
const [y1, y2, y3] = ys;
return [
(x2 * y3) - (x3 * y2),
(x3 * y1) - (x1 * y3),
(x1 * y2) - (x2 * y1)
];
};
const dotProduct = xs =>
compose(
sum,
zipWith(a => b => a * b)(xs)
);
const matrixMultiply = matrix =>
compose(
flip(map)(matrix),
dotProduct
);
const norm = v =>
Math.sqrt(
v.reduce((a, x) => a + (x ** 2), 0)
);
const normalize = v => {
const len = norm(v);
return v.map(x => x / len);
};
// --------------------- GENERIC ---------------------
// compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
const compose = (...fs) =>
// A function defined by the right-to-left
// composition of all the functions in fs.
fs.reduce(
(f, g) => x => f(g(x)),
x => x
);
// flip :: (a -> b -> c) -> b -> a -> c
const flip = op =>
// The binary function op with
// its arguments reversed.
x => y => op(y)(x);
// map :: (a -> b) -> [a] -> [b]
const map = f =>
// The list obtained by applying f
// to each element of xs.
// (The image of xs under f).
xs => [...xs].map(f);
// sum :: [Num] -> Num
const sum = xs =>
// The numeric sum of all values in xs.
xs.reduce((a, x) => a + x, 0);
// zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
const zipWith = f =>
// A list constructed by zipping with a
// custom function, rather than with the
// default tuple constructor.
xs => ys => xs.map(
(x, i) => f(x)(ys[i])
).slice(
0, Math.min(xs.length, ys.length)
);
return JSON.stringify(
main(),
null, 2
);
})();
```
Output:
```[
2.2322210733082275,
1.3951381708176431,
-8.370829024905852
]```
## jq
Works with: jq
Works with gojq, the Go implementation of jq
In the comments, the term "vector" is used to mean a (JSON) array of numbers. Some of the functions have been generalized to work with vectors of arbitrary length.
```# v1 and v2 should be vectors of the same length.
def dotProduct(v1; v2): [v1, v2] | transpose | map(.[0] * .[1]) | add;
# Input: a vector
def norm: dotProduct(.; .) | sqrt;
# Input: a vector
def normalize: norm as \$n | map(./\$n);
# v1 and v2 should be 3-vectors
def crossProduct(v1; v2):
[v1[1]*v2[2] - v1[2]*v2[1], v1[2]*v2[0] - v1[0]*v2[2], v1[0]*v2[1] - v1[1]*v2[0]];
# v1 and v2 should be of equal length.
def getAngle(v1; v2):
(dotProduct(v1; v2) / ((v1|norm) * (v2|norm)))|acos ;
# Input: a matrix (i.e. an array of same-length vectors)
# \$v should be the same length as the vectors in the matrix
def matrixMultiply(\$v):
map(dotProduct(.; \$v)) ;
# \$p - the point vector
# \$v - the axis
# \$a - the angle in radians
def aRotate(\$p; \$v; \$a):
{ca: (\$a|cos),
sa: (\$a|sin)}
| .t = (1 - .ca)
| .x = \$v[0]
| .y = \$v[1]
| .z = \$v[2]
| [
[.ca + .x*.x*.t, .x*.y*.t - .z*.sa, .x*.z*.t + .y*.sa],
[.x*.y*.t + .z*.sa, .ca + .y*.y*.t, .y*.z*.t - .x*.sa],
[.z*.x*.t - .y*.sa, .z*.y*.t + .x*.sa, .ca + .z*.z*.t]
]
| matrixMultiply(\$p) ;
def example:
[5, -6, 4] as \$v1
| [8, 5,-30] as \$v2
| getAngle(\$v1; \$v2) as \$a
| (crossProduct(\$v1; \$v2) | normalize) as \$ncp
| aRotate(\$v1; \$ncp; \$a)
;
example```
Output:
```[2.2322210733082275,1.3951381708176436,-8.370829024905852]
```
## Julia
Translation of: Perl
```using LinearAlgebra # use builtin library for normalize, cross, dot
using JSON3
getangleradians(v1, v2) = acos(dot(v1, v2) / (norm(v1) * norm(v2)))
t = 1 - ca
x, y, z = rotationvector
return [[ca + x * x * t, x * y * t - z * sa, x * z * t + y * sa]';
[x * y * t + z * sa, ca + y * y * t, y * z * t - x * sa]';
[z * x * t - y * sa, z * y * t + x * sa, ca + z * z * t]'] * pointvector
end
v1 = [5, -6, 4]
v2 = [8, 5, -30]
cp = cross(v1, v2)
ncp = normalize(cp)
np = rodrotate(v1, ncp, a)
JSON3.write(np) # "[2.2322210733082284,1.3951381708176411,-8.370829024905854]"
```
## Nim
Translation of: Wren
Only changed most function names.
```import math
type
Vector = tuple[x, y, z: float]
Matrix = array[3, Vector]
func norm(v: Vector): float =
sqrt(v.x * v.x + v.y * v.y + v.z * v.z)
func normalized(v: Vector): Vector =
let length = v.norm()
result = (v.x / length, v.y / length, v.z / length)
func scalarProduct(v1, v2: Vector): float =
v1.x * v2.x + v1.y * v2.y + v1.z * v2.z
func vectorProduct(v1, v2: Vector): Vector =
(v1.y * v2.z - v1.z * v2.y, v1.z * v2.x - v1.x * v2.z, v1.x * v2.y - v1.y * v2.x)
func angle(v1, v2: Vector): float =
arccos(scalarProduct(v1, v2) / (norm(v1) * norm(v2)))
func `*`(m: Matrix; v: Vector): Vector =
(scalarProduct(m[0], v), scalarProduct(m[1], v), scalarProduct(m[2], v))
func rotate(p, v: Vector; a: float): Vector =
let ca = cos(a)
let sa = sin(a)
let t = 1 - ca
let r = [(ca + v.x * v.x * t, v.x * v.y * t - v.z * sa, v.x * v.z * t + v.y * sa),
(v.x * v.y * t + v.z * sa, ca + v.y * v.y * t, v.y * v.z * t - v.x * sa),
(v.z * v.x * t - v.y * sa, v.z * v.y * t + v.x * sa, ca + v.z * v.z * t)]
result = r * p
let
v1 = (5.0, -6.0, 4.0)
v2 = (8.0, 5.0, -30.0)
a = angle(v1, v2)
vp = vectorProduct(v1, v2)
nvp = normalized(vp)
np = v1.rotate(nvp, a)
echo np
```
Output:
`(x: 2.232221073308228, y: 1.395138170817643, z: -8.370829024905852)`
## Perl
```#!perl -w
use strict;
use Math::Trig; # acos
use JSON;
use constant PI => 3.14159265358979;
# Rodrigues' formula for vector rotation - see https://stackoverflow.com/questions/42358356/rodrigues-formula-for-vector-rotation
sub norm {
my(\$v)=@_;
return (\$v->[0]*\$v->[0] + \$v->[1]*\$v->[1] + \$v->[2]*\$v->[2]) ** 0.5;
}
sub normalize {
my(\$v)=@_;
my \$length = &norm(\$v);
return [\$v->[0]/\$length, \$v->[1]/\$length, \$v->[2]/\$length];
}
sub dotProduct {
my(\$v1, \$v2)=@_;
return \$v1->[0]*\$v2->[0] + \$v1->[1]*\$v2->[1] + \$v1->[2]*\$v2->[2];
}
sub crossProduct {
my(\$v1, \$v2)=@_;
return [\$v1->[1]*\$v2->[2] - \$v1->[2]*\$v2->[1], \$v1->[2]*\$v2->[0] - \$v1->[0]*\$v2->[2], \$v1->[0]*\$v2->[1] - \$v1->[1]*\$v2->[0]];
}
sub getAngle {
my(\$v1, \$v2)=@_;
return acos(&dotProduct(\$v1, \$v2) / (&norm(\$v1)*&norm(\$v2)))*180/PI; # remove *180/PI to go back to radians
}
sub matrixMultiply {
my(\$matrix, \$v)=@_;
return [&dotProduct(\$matrix->[0], \$v), &dotProduct(\$matrix->[1], \$v), &dotProduct(\$matrix->[2], \$v)];
}
sub aRotate {
my(\$p, \$v, \$a)=@_; # point-to-rotate, vector-to-rotate-about, angle(degrees)
my \$ca = cos(\$a/180*PI); # remove /180*PI to go back to radians
my \$sa = sin(\$a/180*PI);
my \$t=1-\$ca;
my(\$x,\$y,\$z)=(\$v->[0], \$v->[1], \$v->[2]);
my \$r = [
[\$ca + \$x*\$x*\$t, \$x*\$y*\$t - \$z*\$sa, \$x*\$z*\$t + \$y*\$sa],
[\$x*\$y*\$t + \$z*\$sa, \$ca + \$y*\$y*\$t, \$y*\$z*\$t - \$x*\$sa],
[\$z*\$x*\$t - \$y*\$sa, \$z*\$y*\$t + \$x*\$sa, \$ca + \$z*\$z*\$t]
];
return &matrixMultiply(\$r, \$p);
}
my \$v1 = [5,-6,4];
my \$v2 = [8,5,-30];
my \$a = &getAngle(\$v1, \$v2);
my \$cp = &crossProduct(\$v1, \$v2);
my \$ncp = &normalize(\$cp);
my \$np = &aRotate(\$v1, \$ncp, \$a);
my \$json=JSON->new->canonical;
print \$json->encode(\$np) . "\n";
```
Output:
`[2.23222107330823,1.39513817081764,-8.37082902490585]`
### Generalized
```use strict;
use warnings;
use feature <say signatures>;
no warnings 'experimental::signatures';
use Math::Trig;
use List::Util 'sum';
use constant PI => 2 * atan2(1, 0);
sub norm (\$v) { sqrt sum map { \$_*\$_ } @\$v }
sub normalize (\$v) { [ map { \$_ / norm \$v } @\$v ] }
sub dotProduct (\$v1, \$v2) { sum map { \$v1->[\$_] * \$v2->[\$_] } 0..\$#\$v1 }
sub getAngle (\$v1, \$v2) { 180/PI * acos dotProduct(\$v1, \$v2) / (norm(\$v1)*norm(\$v2)) }
sub mvMultiply (\$m, \$v) { [ map { dotProduct(\$_, \$v) } @\$m ] }
sub crossProduct (\$v1, \$v2) {
[\$v1->[1]*\$v2->[2] - \$v1->[2]*\$v2->[1], \$v1->[2]*\$v2->[0] - \$v1->[0]*\$v2->[2], \$v1->[0]*\$v2->[1] - \$v1->[1]*\$v2->[0]]
}
sub aRotate ( \$p, \$v, \$a ) {
my \$ca = cos \$a/180*PI;
my \$sa = sin \$a/180*PI;
my \$t = 1 - \$ca;
my(\$x,\$y,\$z) = @\$v;
my \$r = [
[ \$ca + \$x*\$x*\$t, \$x*\$y*\$t - \$z*\$sa, \$x*\$z*\$t + \$y*\$sa],
[\$x*\$y*\$t + \$z*\$sa, \$ca + \$y*\$y*\$t, \$y*\$z*\$t - \$x*\$sa],
[\$z*\$x*\$t - \$y*\$sa, \$z*\$y*\$t + \$x*\$sa, \$ca + \$z*\$z*\$t]
];
mvMultiply(\$r, \$p)
}
my(\$v1,\$v2) = ([5, -6, 4], [8, 5, -30]);
say join ' ', @{aRotate \$v1, normalize(crossProduct \$v1, \$v2), getAngle \$v1, \$v2};
```
Output:
`2.23222107330823 1.39513817081764 -8.37082902490585`
## Phix
Translation of: JavaScript
```with javascript_semantics
function norm(sequence v)
return sqrt(sum(sq_power(v,2)))
end function
function normalize(sequence v)
return sq_div(v,norm(v))
end function
function dotProduct(sequence v1, v2)
return sum(sq_mul(v1,v2))
end function
function crossProduct(sequence v1, v2)
atom {v11,v12,v13} = v1,
{v21,v22,v23} = v2
return {v12*v23-v13*v22, v13*v21-v11*v23, v11*v22-v12*v21}
end function
function getAngle(sequence v1, v2)
return arccos(dotProduct(v1, v2) / (norm(v1)*norm(v2)))
end function
function matrixMultiply(sequence matrix, v)
return {dotProduct(matrix[1], v), dotProduct(matrix[2], v), dotProduct(matrix[3], v)}
end function
function aRotate(sequence p, v, atom a)
atom ca = cos(a), sa = sin(a), t=1-ca, {x,y,z} =v
sequence r = {{ca + x*x*t, x*y*t - z*sa, x*z*t + y*sa},
{x*y*t + z*sa, ca + y*y*t, y*z*t - x*sa},
{z*x*t - y*sa, z*y*t + x*sa, ca + z*z*t}}
return matrixMultiply(r, p)
end function
sequence v1 = {5,-6,4},
v2 = {8,5,-30};
atom a = getAngle(v1, v2)
sequence cp = crossProduct(v1, v2),
ncp = normalize(cp),
np = aRotate(v1, ncp, a);
?np
```
Output:
```{2.232221073,1.395138171,-8.370829025}
```
## Processing
Translation of: C
```//Aamrun, 30th June 2022
class Vector{
private double x, y, z;
public Vector(double x1,double y1,double z1){
x = x1;
y = y1;
z = z1;
}
void printVector(int x,int y){
text("( " + this.x + " ) \u00ee + ( " + this.y + " ) + \u0135 ( " + this.z + ") \u006b\u0302",x,y);
}
public double norm() {
return Math.sqrt(this.x*this.x + this.y*this.y + this.z*this.z);
}
public Vector normalize(){
double length = this.norm();
return new Vector(this.x / length, this.y / length, this.z / length);
}
public double dotProduct(Vector v2) {
return this.x*v2.x + this.y*v2.y + this.z*v2.z;
}
public Vector crossProduct(Vector v2) {
return new Vector(this.y*v2.z - this.z*v2.y, this.z*v2.x - this.x*v2.z, this.x*v2.y - this.y*v2.x);
}
public double getAngle(Vector v2) {
return Math.acos(this.dotProduct(v2) / (this.norm()*v2.norm()));
}
public Vector aRotate(Vector v, double a) {
double ca = Math.cos(a), sa = Math.sin(a);
double t = 1.0 - ca;
double x = v.x, y = v.y, z = v.z;
Vector[] r = {
new Vector(ca + x*x*t, x*y*t - z*sa, x*z*t + y*sa),
new Vector(x*y*t + z*sa, ca + y*y*t, y*z*t - x*sa),
new Vector(z*x*t - y*sa, z*y*t + x*sa, ca + z*z*t)
};
return new Vector(this.dotProduct(r[0]), this.dotProduct(r[1]), this.dotProduct(r[2]));
}
}
void setup(){
Vector v1 = new Vector(5d, -6d, 4d),v2 = new Vector(8d, 5d, -30d);
double a = v1.getAngle(v2);
Vector cp = v1.crossProduct(v2);
Vector normCP = cp.normalize();
Vector np = v1.aRotate(normCP,a);
size(1200,600);
fill(#000000);
textSize(30);
text("v1 = ",10,100);
v1.printVector(60,100);
text("v2 = ",10,150);
v2.printVector(60,150);
text("rV = ",10,200);
np.printVector(60,200);
}
```
## Raku
```sub infix:<⋅> { [+] @^a »×« @^b }
sub norm (@v) { sqrt @v⋅@v }
sub normalize (@v) { @v X/ @v.&norm }
sub getAngle (@v1,@v2) { 180/π × acos (@v1⋅@v2) / (@v1.&norm × @v2.&norm) }
sub crossProduct ( @v1, @v2 ) {
my \a = <1 2 0>; my \b = <2 0 1>;
(@v1[a] »×« @v2[b]) »-« (@v1[b] »×« @v2[a])
}
sub aRotate ( @p, @v, \$a ) {
my \ca = cos \$a/180×π;
my \sa = sin \$a/180×π;
my \t = 1 - ca;
my (\x,\y,\z) = @v;
map { @p⋅\$_ },
[ ca + x×x×t, x×y×t - z×sa, x×z×t + y×sa],
[x×y×t + z×sa, ca + y×y×t, y×z×t - x×sa],
[z×x×t - y×sa, z×y×t + x×sa, ca + z×z×t]
}
my @v1 = [5,-6, 4];
my @v2 = [8, 5,-30];
say join ' ', aRotate @v1, normalize(crossProduct @v1, @v2), getAngle @v1, @v2;
```
Output:
`2.232221073308229 1.3951381708176411 -8.370829024905852`
Alternately, differing mostly in style:
```sub infix:<•> { sum @^v1 Z× @^v2 } # dot product
sub infix:<❌> (@v1, @v2) { # cross product
my \a = <1 2 0>; my \b = <2 0 1>;
@v1[a] »×« @v2[b] »-« @v1[b] »×« @v2[a]
}
sub norm (*@v) { sqrt @v • @v }
sub normal (*@v) { @v X/ @v.&norm }
sub angle-between (@v1, @v2) { acos( (@v1 • @v2) / (@v1.&norm × @v2.&norm) ) }
sub infix:<> is equiv(&infix:<×>) { \$^a × \$^b } # invisible times
sub postfix:<°> (\d) { d × τ / 360 } # degrees to radians
sub rodrigues-rotate( @point, @axis, \$θ ) {
my ( \cos𝜃, \sin𝜃 ) = cis(\$θ).reals;
my ( \𝑥, \𝑦, \𝑧 ) = @axis;
my \𝑡 = 1 - cos𝜃;
map @point • *, [
[ 𝑥²𝑡 + cos𝜃, 𝑦𝑥𝑡 - 𝑧sin𝜃, 𝑧𝑥𝑡 + 𝑦sin𝜃 ],
[ 𝑥𝑦𝑡 + 𝑧sin𝜃, 𝑦²𝑡 + cos𝜃, 𝑧𝑦𝑡 - 𝑥sin𝜃 ],
[ 𝑥𝑧𝑡 - 𝑦sin𝜃, 𝑦𝑧𝑡 + 𝑥sin𝜃, 𝑧²𝑡 + cos𝜃 ]
]
}
sub point-vector (@point, @vector) {
rodrigues-rotate @point, normal(@point ❌ @vector), angle-between @point, @vector
}
put qq:to/TESTING/;
Task example - Point and composite axis / angle:
{ point-vector [5, -6, 4], [8, 5, -30] }
Perhaps more useful, (when calculating a range of motion for a robot appendage,
for example), feeding a point, axis of rotation and rotation angle separately;
since theoretically, the point vector and axis of rotation should be constant:
{
(0, 10, 20 ... 180).map( { # in degrees
sprintf "Rotated %3d°: %.13f, %.13f, %.13f", \$_,
rodrigues-rotate [5, -6, 4], ([5, -6, 4] ❌ [8, 5, -30]).&normal, .°
}).join: "\n"
}
TESTING
```
Output:
```Task example - Point and composite axis / angle:
2.232221073308228 1.3951381708176427 -8.370829024905852
Perhaps more useful, (when calculating a range of motion for a robot appendage,
for example), feeding a point, axis of rotation and rotation angle directly;
since theoretically, the point vector and axis of rotation should be constant:
Rotated 0°: 5.0000000000000, -6.0000000000000, 4.0000000000000
Rotated 10°: 5.7240554466526, -6.0975296976939, 2.6561853906284
Rotated 20°: 6.2741883650704, -6.0097890410223, 1.2316639322573
Rotated 30°: 6.6336832449081, -5.7394439854392, -0.2302810114435
Rotated 40°: 6.7916170161550, -5.2947088286573, -1.6852289831393
Rotated 50°: 6.7431909410900, -4.6890966233686, -3.0889721249495
Rotated 60°: 6.4898764214992, -3.9410085899762, -4.3988584118384
Rotated 70°: 6.0393702908772, -3.0731750048240, -5.5750876118134
Rotated 80°: 5.4053609500356, -2.1119645522518, -6.5819205958338
Rotated 90°: 4.6071124519719, -1.0865831254651, -7.3887652531624
Rotated 100°: 3.6688791733663, -0.0281864202486, -7.9711060171693
Rotated 110°: 2.6191688576205, 1.0310667150840, -8.3112487584187
Rotated 120°: 1.4898764214993, 2.0589914100238, -8.3988584118384
Rotated 130°: 0.3153148442246, 3.0243546928699, -8.2312730024418
Rotated 140°: -0.8688274150348, 3.8978244887705, -7.8135845280911
Rotated 150°: -2.0265707929363, 4.6528608599741, -7.1584842417190
Rotated 160°: -3.1227378427887, 5.2665224084086, -6.2858770340300
Rotated 170°: -4.1240220834695, 5.7201633384526, -5.2222766334692
Rotated 180°: -5.0000000000000, 6.0000000000000, -4.0000000000000```
## RPL
This is a direct transcription from Wikipedia's formula.
```≪ DEG SWAP DUP ABS / @ set degrees mode and normalize k
→ v θ k
≪ v θ COS *
k v CROSS θ SIN * +
k DUP v DOT * 1 θ COS - * +
→NUM @ can be removed if using HP-28/48 ROM versions
≫ ≫ 'ROTV' STO @ ( vector axis-vector angle → rotated-vector )
```
```[-1 2 .4] [0 2 1] 18 ROTV
```
Output:
```[-1.11689243765 1.85005696279 .699886074428]
```
## Wren
Translation of: JavaScript
```var norm = Fn.new { |v| (v[0]*v[0] + v[1]*v[1] + v[2]*v[2]).sqrt }
var normalize = Fn.new { |v|
var length = norm.call(v)
return [v[0]/length, v[1]/length, v[2]/length]
}
var dotProduct = Fn.new { |v1, v2| v1[0]*v2[0] + v1[1]*v2[1] + v1[2]*v2[2] }
var crossProduct = Fn.new { |v1, v2|
return [v1[1]*v2[2] - v1[2]*v2[1], v1[2]*v2[0] - v1[0]*v2[2], v1[0]*v2[1] - v1[1]*v2[0]]
}
var getAngle = Fn.new { |v1, v2| (dotProduct.call(v1, v2) / (norm.call(v1) * norm.call(v2))).acos }
var matrixMultiply = Fn.new { |matrix, v|
return [dotProduct.call(matrix[0], v), dotProduct.call(matrix[1], v), dotProduct.call(matrix[2], v)]
}
var aRotate = Fn.new { |p, v, a|
var ca = a.cos
var sa = a.sin
var t = 1 - ca
var x = v[0]
var y = v[1]
var z = v[2]
var r = [
[ca + x*x*t, x*y*t - z*sa, x*z*t + y*sa],
[x*y*t + z*sa, ca + y*y*t, y*z*t - x*sa],
[z*x*t - y*sa, z*y*t + x*sa, ca + z*z*t]
]
return matrixMultiply.call(r, p)
}
var v1 = [5, -6, 4]
var v2 = [8, 5,-30]
var a = getAngle.call(v1, v2)
var cp = crossProduct.call(v1, v2)
var ncp = normalize.call(cp)
var np = aRotate.call(v1, ncp, a)
System.print(np)
```
Output:
```[2.2322210733082, 1.3951381708176, -8.3708290249059]
```
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# Thread: Calculating the week in the month (PHP)
1. ## Calculating the week in the month (PHP)
So there is a function for the day of the week (date('d')), the total days in a month (date('t')) and the week in the year, but how can I get the week number (1-4/5) in the month?
Any help would be appreciated.
2. Why don't you write your own function. You must put on your thinking cap though and fail a couple of dozen times depending on your experience.
Happy coding...
3. Thats what I was trying to do, but the maths involved have bested me. I can't get a reliable result.
4. mail me what you have done so far so that I can have a look see if I can help in any way.
5. PHP Code:
``` function getWeekNoByDay(\$year = 2007,\$month = 5,\$day = 5) { return ceil((\$day + date("w",mktime(0,0,0,\$month,1,\$year)))/7); } ```
and if you want the the week in the currentmonth devide result by month eg 5 here. Hope that helps you
6. Originally Posted by Kloon
PHP Code:
``` function getWeekNoByDay(\$year = 2007,\$month = 5,\$day = 5) { return ceil((\$day + date("w",mktime(0,0,0,\$month,1,\$year)))/7); } ```
and if you want the the week in the currentmonth devide result by month eg 5 here. Hope that helps you
Does not work for me. If I give it the current year, month and the 15th, it gives me the week of the month as 2. Did the divide by month as well. Shouldn't the result be 3?
7. 2 would be correct, it is only the 12th. This means your in the second week of feb, or so i think, if you think it should be 3 try rounding the value up
8. Sorry mate didnt check the code properly, no need to devide by month it already gives you the correct answer. If i run it with todays date i get 3
9. I'll have to speak to the guy that wrote the spec, because it seems that he only caters for 4 weeks in a month, which there aren't.
Thanks for all the help. It is appreciated.
10. Nope there can be 5 weeks in a month too
11. Originally Posted by Kloon
Nope there can be 5 weeks in a month too
I'm aware of that, yes.
12. Why over-complicate things? I googled this very thing (it's how I found your question) and it seems anyone who has "solved" this is generating like 2000 lines of code to do something so simple.
Here's my idea on how this should be done...
Code:
```// Code by RobbieF.com - determines current week of the month
// March 5, 2007
\$i=0;
\$week=0;
if (date("N", mktime(0, 0, 0, date(n), 1, date(Y))) <= "6") \$week++;
while (\$i <= date(j)) {
if (date("N", mktime(0, 0, 0, date(n), \$i, date(Y))) == "7") \$week++;
\$i++;
}
echo "Current week of the month: " . \$week;```
All that does is simply count how many Sundays have passed (the "7") since the beginning of the month, and also checks if the first day of the month falls on or before Saturday (which would cause it to be week 1... this solves the "week 5" thing). That makes sense to me to do what you're wanting. I dunno why we'd wanna over-complicate something like that.
I haven't really tested this, and it's just a quick mockup - but it should at least give you a good starting point... but it should work.
Hope it helps!!
www.RobbieF.com
13. Actually there's an easier way.
The date function can return week number from a timestamp. Use mktime() to format your date as a UNIX timestamp.
eg:
date ("W",mktime(0, 0, 0, 3, 22, 2007));
where 3 is month, 22 is day and 2007 is year.
Hope this helps.
14. This line of code will return the week number. Note: PHP's "W" parameter in the Date function assumes that Monday is the first day of the week, so if you need to work from Sunday as the first day of the week, you'd need to build that logic in yourself.
Code:
`\$weekNum = date("W") - date("W",strtotime(date("Y-m-01"))) + 1;`
We add the 1 at the end because the "W" parameter returns a zero based result, so if you leave off the + 1, then you must remember the first week will be returned as 0.
15. function getWeekOfTheMonth(\$YY,\$MM,\$DD){
\$weekNum = date("W",mktime(0,0,0,\$MM,\$DD,\$YY)) - date("W",mktime(0,0,0,\$MM,01,\$YY)) + 1;
return \$weekNum;
}
#### Posting Permissions
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# Introduction
## What is Cryptography?
Cryptography is the study of how to securely communicate a message even when a third party can intercept all transmissions. Plaintext is the original message to be sent. Encryption converts plaintext into ciphertext, which hides the original message. Finally, decryption converts ciphertext back into plaintext to reveal the initial message.
An important issue arises when this process happens on the Internet: how can we communicate securely when we must assume that eavesdroppers are able to intercept all communications, including the method of encryption? For example, it may be relatively difficult to crack a substitution cipher (a cipher in which each character is substituted for another) in a vacuum, but if the eavesdropper was able to intercept everything being sent, then they would also be able to intercept the cipher!
This problem is actually impossible to solve the way it is currently stated. If an eavesdropper has access to both the ciphertext and the method of encryption, they can simply encrypt every possible plaintext message until it matches the ciphertext. The goal is to make it computationally infeasible for the message to be intercepted. For example, trying trillions of possible messages is infeasible. However, smarter techniques can reverse engineer the plaintext input from the ciphertext, or at least narrow down the possibilities until trying each one becomes feasible.
This brings us to the concept of a trapdoor function. A trapdoor function is easy to calculate but difficult to find the inverse of, without some private secret. This is useful in cryptography: the recipient can generate a trapdoor function along with the private key needed to solve it, and only send out the details of the function. The sender can then use the function to encode plaintext into ciphertext, and the recipient can use their private key to invert it back into plaintext. The eavesdropper, however, does not know the private key, and so must manually find the function’s inverse, which takes far too long.
## Why Not RSA?
RSA encryption is a popular method used today that uses the above principle to safely communicate messages. If you don’t know anything about it, there is another article, "Modular arithmetic and public-key cryptography," dedicated to explaining how it works. That article also introduces mods, so if even if you’re not interested in RSA, it will still be helpful to read the first two sections.
RSA still has not been cracked, but many advancements have made it more feasible to solve. The algorithm’s safety relies on the difficulty of tasks such as factoring large numbers, which is something many mathematicians have made important progress in solving. As a result, the keys used in RSA encryption must get progressively larger to deter attacks. While this maintains security, it makes encryption and decryption less convenient. Elliptic curves, however, don’t have algorithms that are as effective in solving them, thus allowing for smaller keys with the same level of protection and faster processing.
# Elliptic Curves
So what exactly is an elliptic curve, and why is it so useful in cryptography?
An elliptic curve is the graph of the equation $$y^2=x^3+ax+b$$. A couple examples are displayed below:
We can define a group using the points on the elliptic curve. A group is a set of objects which are connected by some binary operation. In this case, we’ll define what it means to add points together. This operation must have certain properties:
• Closure: Performing the operation on any two elements of the set must yield another element of the set.
• Commutativity and Associativity: For a particular expression, no matter how you order the terms or perform the operations, the result must be the same.
• Identity: There must exist some identity 0 such that for all elements $$a$$ in the group, $$a+0=a$$.
• Inverse: For every element in the set $$a$$, there exists another element in in the group $$-a$$ such that $$a + -a = 0$$.
Let’s define this function. The identity is represented with $$0$$, which doesn’t represent a specific point on the curve but will still be important. For every point $$(x, y)$$, the inverse will be the point $$(x, -y)$$, or the point that lies opposite the x-axis on the curve. Finally, three points $$P$$, $$Q$$, and $$R$$ on the curve are collinear if and only if $$P+Q+R=0$$.
This equation allows us to define the sum of two points: $$P+Q=-R$$. This can be represented with the following process: draw the line $$PQ$$ on the curve, then find the point $$R$$ which is the third point where the line intersects the curve, and then flip it across the x-axis to get the point $$-R$$, which is the sum of the two points $$P$$ and $$Q$$!
## Algebraic Addition
The above definition works fine with a pen and paper, but how can a computer perform this addition? Firstly, let’s define the coordinates of the points $$P=(x_P, y_P)$$ and $$Q=(x_Q, y_Q)$$. We then define the slope $$m = \frac{y_P-y_Q}{x_P-x_Q}$$, and let $$k$$ be the $$y$$-intercept of the line passing through $$P$$ and $$Q$$ (which we don’t need to calculate for now). Then the line through $$P$$ and $$Q$$ has equation $$y = mx+k$$, and we want the third point which lies on this line and the elliptic curve. We can solve for it using the following process: $y^2=x^3+ax+b$ $(mx+k)^2=x^3+ax+b$ $x^3+ax+b-(mx+k)^2=0$ We can now use Vieta’s formulas to find the third root $$x_R$$. In a monic cubic, the sum of the roots is equal to $$-b$$, where $$b$$ is the coefficient of the $$x^2$$ term. Therefore, $x_P+x_Q+x_R=-(-m^2)$ $x_R=m^2-x_P-x_Q$ And once we find $$x_R$$, we can use our knowledge that this point lies on the line through $$P$$ and $$Q$$ to solve for $$y_R$$: $m=\frac{y_R-y_P}{x_R-x_P}$ $m(x_R-x_P)=y_R-y_P$ $y_R=y_P+m(x_R-x_P)$ Then, once we get the point $$R$$ that’s collinear with $$P$$ and $$Q$$, the answer to $$P+Q$$ is $$-R$$, or the point $$(x_R, -y_R)$$. These equations will work for almost every case. There is one exception, however: what if $$P=Q$$? First of all, we’re going to interpret the line through $$PQ$$ as the line tangent to the curve at point $$P$$ in this instance. And since the slope is now going to be the slope of the tangent line rather than the slope between $$P$$ and $$Q$$, we need a new formula for this case. $m=\frac{3x_P^2+a}{2y_P}$ For those who are familiar with calculus, you may recognize that this formula for $$m$$ corresponds with the first derivative of the function $$y=\pm \sqrt{x_P^3+ax_P+b}$$.
## Scalar Multiplication
If we can successfully define an addition operation, then we can also define a scalar multiplication operation: for an integer $$n$$ and a point $$P$$ on an elliptic curve, the expression $$nP$$ is the result of adding $$P$$ to itself $$n$$ times.
How can we do this operation on a computer? One technique is to simply add $$P$$ to itself $$n$$ times. But as $$n$$ gets extremely large, this technique no longer becomes feasible. For example, $$n$$ can get as large as $$2^{512}$$, but adding a point to itself that number of times is far too difficult for any computer to do.
We can introduce a new algorithm to quickly perform scalar multiplication. The procedure looks something like this: for a point $$P$$, first calculate $$2P$$, then $$4P$$, $$8P$$, and so on, adding the result to itself at each each step to get the next result in the sequence. Then, for each power of 2, if that power of 2 shows up in the bit representation of the desired coefficient $$n$$, then we add it to a running total. By doing this, we ensure that the amount of time we take is proportional to the number of powers of 2 we must cycle through in order to get the result. To be more precise, the time complexity of this algorithm is $$O(\log_2n)$$.
# Finite Fields
We can now apply a new concept to add another layer of complexity to the groups we have just defined. This particular section will use modular arithmetic to show how an elliptic curve can be limited to a finite field.
First, a finite field $$\mathbb{F}_p$$ is a group of the nonnegative integers less than $$p$$ that are connected by addition, subtraction, and multiplication in modular arithmetic. So for example, if we were dealing with $$\mathbb{F}_5$$ we could say that $$4\cdot3 = 12 \equiv 2 \pmod{5}$$.
Now we can define the elliptic curve over a finite field: for two nonnegative integers $$x$$ and $$y$$ both less than $$p$$, we say that the point $$(x, y)$$ lies on the elliptic curve if $$y^2 \equiv x^3+ax+b \pmod{p}$$. Now can we define addition over the finite field as well: we’ll still say that three points $$P$$, $$Q$$, and $$R$$ are collinear if they lie on the same line, but this time the line will be defined as the set of points satisfying $$ax+by+c \equiv 0 \pmod{p}$$, the form of a line in a field. The algebraic equations for finding the sum $$P+Q$$ will be very similar: we once again look for the point $$R$$ on the curve that’s collinear with $$P$$ and $$Q$$ using the following formulas: $x_R = (m^2 - x_P - x_Q)\pmod{p}$ $y_R = y_P+m(x_R-x_P)\pmod{p}$ Where if $$P \neq Q$$, then $m = \frac{y_P-y_Q}{x_P-x_Q}\pmod{p}$ And if $$P=Q$$, then $m=\frac{3x_P^2+a}{2y_P}\pmod{p}$ Note that in a finite field, division is defined as multiplying by the modular inverse rather than actually dividing. This is done to ensure that we remain within the field of integers.
Once we calculate the coordinates $$R=(x_R, y_R)$$, we can now flip it to get $$P+Q=-R=(x_R, -y_R)$$, reaching our answer.
For an elliptic curve taken on a finite field (aka an elliptic curve group), we will define the order of the elliptic curve group as the number of distinct points on the curve. Schoof’s algorithm was designed to calculate this quantity fast enough to justify using it. While the algorithm is too complex to lie within the scope of this article, just know that it exists and is able to calculate this useful quantity.
## Cyclic Subgroups
In a finite field, the multiples of a given point $$P$$, called the base point, will eventually form an infinite loop. It’s important to note that the points that lie along that loop form a cyclic subgroup, because adding any two multiples of $$P$$ will yield another multiple of $$P$$ that also lies within the subgroup. We can define the order of that subgroup as the number of points in the subgroup, but we can also create an equivalent definition that may be easier to use: the order of a cyclic subgroup generated with the multiples of $$P$$ is the lowest number $$n$$ such that $$nP=0$$.
How do we find the order of a subgroup when that order is potentially very large? Well, we can make use of Lagrange’s theorem, which states that the order of a subgroup is always a divisor of the order of the group. If we know the order of the group, $$N$$ (which can be calculated with Schoof’s algorithm), then we can generate all the divisors of that number and individually test each possibility to find the smallest $$n$$ where $$nP=0$$.
Next, the goal is to find a base point that generates a cyclic subgroup with a large, prime order. With the right approach, we can actually pick the order we want, and then go about finding a base point that creates a subgroup with that order. We can make use of the cofactor of $$n$$, or the integer quantity $$\frac{N}{n}$$, which we’ll call $$h$$. For every point $$P$$, we know that $$NP=0$$, since the order of the cyclic subgroup generated by $$P$$ must divide $$N$$. We can express that statement as $$nhP=0$$, since $$h$$ is defined as $$\frac{N}{n}$$. Finally, this can be expressed as $$n(hP)=0$$, which can be true in two different ways: either $$hP=0$$ and the equation is trivially true, or $$hP$$ is a nonzero point, in which case we can use it as a base point to generate a cyclic subgroup of order $$n$$. The probability that $$hP=0$$ is sufficiently small enough that picking random points $$P$$ and checking if $$hP$$ is a valid base point for a cyclic subgroup of order $$n$$ will yield an answer very quickly.
## The Discrete Logarithm Problem
After establishing the properties of this elliptic field, we can finally create a trapdoor function that’s easy to create but very difficult to solve: Given two points $$P$$ and $$Q$$, what’s the integer $$k$$ such that $$kP=Q$$? At the current moment, no classical computing algorithm can solve this problem in a feasible amount of time. This is a discrete logarithm problem: discrete because it relates to finite fields and cyclic subgroups, and logarithmic because it’s similar to the original logarithm, which determines how many times to repeat the operation of multiplication to get the result. What separates elliptic curve cryptography from other discrete logarithm problems is that it’s uniquely difficult to break, which allows for smaller keys with the same level of security, as well as faster computation.
# Applications of Elliptic Curve Cryptography
## Elliptic-curve Diffie-Hellman
Elliptic-curve Diffie-Hellman is a key-agreement protocol that allows two parties to agree on a key without an eavesdropper knowing that key. Once the two parties can achieve this, they can go on to communicate via some other encryption algorithm, like AES encryption, which requires a secret key that’s unknown to any eavesdroppers. The process goes as follows:
• Alice and Bob publicly agree on an elliptic curve field to perform their calculations on, as well as a base point $$P$$ which generates a cyclic subgroup with a large, prime order.
• Alice and Bob then each privately pick numbers $$a$$ and $$b$$, respectively. Alice calculates and publicly sends the point $$aP$$, and Bob calculates and publicly sends the point $$bP$$.
• Alice then multiplies the point $$bP$$ by $$a$$ to get $$abP$$. Bob does the same, multiplying the point $$aP$$ by $$b$$ to get $$abP$$.
Note that the value of the end result, $$abP$$, is completely random and unintentional. But if both parties have the same secret information, that’s enough to use it for a different form of encryption with, for example, the $$x$$-coordinate being used as the key for that encryption method.
## Authenticating Signatures
Let’s say I want to transfer some bitcoin from my account to a friend’s account. I would create and send an instruction that initiates that transaction. But anybody else can also create these instructions, which would perform the transfer whether I want to or not. So we must introduce a signature, some confirmation that I am the person intending to make the transaction. If I have a private key that nobody else knows, then my signature would involve demonstrating my knowledge of the private key without revealing the true value of this key. This is a very tough problem, but we can solve it using elliptic curves.
Let’s first define a hash function $$hash(m, R)$$ that takes an integer $$m$$ and a point $$R$$ on an elliptic curve to produce a pseudorandom integer output. There are many necessary properties of good hash functions, but the main one that matters here is that trying to reverse engineer the input from the output of a hash function is computationally infeasible.
Let’s give each user a private key $$x$$ that only they know. We can also take a predetermined base point $$P$$ and calculate $$X = x \cdot P$$, which will be publicly available. We can publish $$X$$ without revealing $$x$$ because of the discrete logarithm problem having no known solution.
The user will first privately pick integers $$m$$ and $$r$$. Then they will calculate the point $$R = r \cdot P$$, and then calculate the integer $$s = hash(m, R) \cdot x + r$$. Notice that the following equalities all hold true: $hash(m, r \cdot P) \cdot x \cdot P + r \cdot P = (hash(m, r \cdot P) \cdot x+r) \cdot P$ $hash(m, R) \cdot X + R = (hash(m, R) \cdot x+r) \cdot P$ $hash(m, R) \cdot X + R = s \cdot P$ So the signature works in the following manner: if the user really does have access to the private key $$x$$, then they will easily be able to generate $$m$$, $$R$$, and $$s$$ which satisfy this final equation using the above process. Even though nobody else knows the value of $$x$$, they can use the public point $$X$$ to confirm that the generated solution does indeed work.
The security of this method has two parts to it: first, nobody else should be able to feasibly generate this working solution with only knowledge of $$X$$, and second, publicly releasing this solution $$m$$, $$R$$, and $$s$$ should not reveal anything about the original value of $$x$$.
The first part relies on the security of both the elliptic curve discrete logarithm and the hash function. This prevents a hacker from randomly picking $$m$$ and $$R$$, and trying to reverse engineer some value $$s$$ such that $$s \cdot P = hash(m, R) \cdot X+R$$. It also prevents them from randomly selecting $$s$$ and $$R$$, then trying to find an $$m$$ which satisfies $$hash(m, R) \cdot X+R=s \cdot P$$.
The second part uses the security of the discrete logarithm problem. Note that $$m$$ and $$R$$ are completely random, and so they reveal nothing about $$x$$. We have that $$s = hash(m, R) \cdot x+r$$, so $$x = \frac{s-r}{hash(m, R)})$$. In order to evaluate that expression, we must know the value of $$r$$, but that requires being able to solve the discrete logarithm problem for the given point $$R$$. In fact, Bitcoin and Ethereum both use this algorithm to facilitate secure transactions, in which a user requesting a transfer of cryptocurrency from them to another user can confirm that they are indeed the one requesting this transfer.
# Citations
1. Corbellini, Andrea. “Elliptic Curve Cryptography: a Gentle Introduction,” May 17, 2015. https://andrea.corbellini.name/2015/05/17/elliptic-curve-cryptography-a-gentle-introduction/.
2. Knutson, Hans. “What Is the Math behind Elliptic Curve Cryptography?” Hacker Noon, April 16, 2018. https://hackernoon.com/what-is-the-math-behind-elliptic-curve-cryptography-f61b25253da3.
3. Silverman, Joseph H. An Introduction to the Theory of Elliptic Curves. Laramie, Wyoming: University of Wyoming, 2006.
https://www.math.brown.edu/johsilve/Presentations/WyomingEllipticCurve.pdf.
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# 240 kg to lbs
Friends in today’s post we will learn how to convert 240 kg to lbs. To convert kilograms to pounds, you can use the following formula:
1 kilogram = 2.20462 pounds
To convert 240 kilograms to pounds, you can multiply 240 by 2.20462:
240 kilograms * 2.20462 pounds/kilogram = 529.109 pounds
Therefore, 240 kilograms is equal to approximately 529.109 pounds.
## What is the best conversion unit for 240 kg?
The choice of the best conversion unit depends on the context and the specific requirements of the situation. Here are a few common conversion units for 240 kilograms:
1. Pounds: 240 kilograms is approximately 529.109 pounds. This is a commonly used unit for weight in some countries like the United States.
2. Metric Tons: 240 kilograms is equal to 0.24 metric tons. Metric tons are often used for larger weights, especially in industrial and scientific contexts.
3. Grams: 240 kilograms is equal to 240,000 grams. Grams are commonly used for smaller weights, such as in cooking or laboratory measurements.
4. Stone: 240 kilograms is approximately 37.8 stone. The stone is a unit of weight commonly used in some countries, especially for measuring body weight.
The best conversion unit would depend on the specific application and the system of measurement used in your region or industry.
## Table for values around 240 kilograms
Here’s a table showing values around 240 kilograms converted into different units:
Please note that the values in the table are rounded to three decimal places for pounds and metric tons, and whole numbers for grams.
## FAQs on 240 kg to lbs
Q: How many pounds are in 240 kilograms?
A: To convert kilograms to pounds, you can multiply the weight in kilograms by the conversion factor 2.2046. Applying this formula, 240 kilograms is equal to approximately 529.11 pounds.
Q: Is the conversion from kilograms to pounds accurate?
A: The conversion factor from kilograms to pounds, which is 2.2046, is an approximation. It provides a close estimate when converting between the two units, but it is not an exact value. Therefore, the result obtained after conversion may have a slight rounding error.
Q: Can I convert 240 kilograms to pounds using an online converter?
A: Yes, you can use various online conversion tools to convert 240 kilograms to pounds accurately. These converters use the standard conversion factor of 2.2046 to perform the calculation.
Q: What is the formula to convert kilograms to pounds?
A: The formula to convert kilograms to pounds is: pounds = kilograms × 2.2046. Simply multiply the weight in kilograms by 2.2046 to obtain the weight in pounds.
Q: Can I use the conversion factor 2.2 instead of 2.2046 to convert kilograms to pounds?
A: While 2.2 is a commonly used rounded value for the conversion from kilograms to pounds, it is not as precise as 2.2046. If you require a more accurate conversion, it is recommended to use 2.2046 as the conversion factor.
Q: Why is the conversion factor for kilograms to pounds not a whole number?
A: The conversion factor for kilograms to pounds, 2.2046, is not a whole number because it is derived from the exact conversion factor, which is defined as 1 kilogram = 2.20462262185 pounds. The decimal value ensures a more precise conversion between the two units.
Q: What is the closest whole number for 240 kilograms in pounds?
A: Using the conversion factor 2.2046, 240 kilograms is approximately equal to 529.11 pounds. The closest whole number would be 529 pounds.
Q: Are kilograms or pounds more commonly used for weight measurements?
A: The use of kilograms and pounds varies depending on the region and context. Kilograms are the primary unit of mass measurement in the metric system and are widely used worldwide, including in scientific and medical fields. Pounds, on the other hand, are more commonly used in the United States and a few other countries that have not fully adopted the metric system.
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Specify Weekdays Range
I want to get a weekday of the previous month and cut it into 1 week and tie it in a range.
I’m thinking about range (Monday: Friday), but I’m worried about the first and last weeks.
Use the date manipulation for this.
``````1. Assign: currentDate = DateTime.Now
3. Assign: weekdayOfFirstDay = firstDayOfPreviousMonth.DayOfWeek
4. Assign: daysToMonday = (weekdayOfFirstDay - DayOfWeek.Monday + 7) % 7
6. Assign: endDate = startDate.AddDays(6) ' This gives you the end of the first week (Friday) in the previous month.
7. Do While: startDate.Month = firstDayOfPreviousMonth.Month
- Add your processing logic for each week (Monday to Friday) here.
- You can use a For Each activity with a range from startDate to endDate.
- Assign: startDate = startDate.AddDays(7) ' Move to the next week's Monday.
- Assign: endDate = startDate.AddDays(6) ' Calculate the end of the current week (Friday).
8. End While
``````
Hope it helps!!
1 Like
@wjdehdnr456
Get the First Day of the Previous Month:
Get the Last Day of the Previous Month:
Loop Through Weeks and Extract Weekdays:
1 Like
Hi @wjdehdnr456
Use the below syntaxes in assign activity:
``````currentYear= Now.Year
currentMonth= Now.Month
previousMonth= If(currentMonth > 1, currentMonth - 1, 12)
previousYear= If(currentMonth > 1, currentYear, currentYear - 1)
``````
Note: `currentYear`, `currentMonth`, `previousMonth`, `previousYear` is of DataType System.Int32
=> Use the below code in Invoke Code activity:
``````Dim firstDayOfMonth As New DateTime(previousYear, previousMonth, 1)
Dim lastDayOfPreviousMonth As DateTime = firstDayOfMonth.AddDays(-1)
Dim firstMonday As DateTime = firstDayOfMonth.AddDays(-CInt(firstDayOfMonth.DayOfWeek))
Dim lastMonday As DateTime = lastDayOfPreviousMonth.AddDays(-CInt(lastDayOfPreviousMonth.DayOfWeek))
Dim currentMonday As DateTime = firstMonday
weeklyRanges = New List(Of Tuple(Of DateTime, DateTime))
While currentMonday <= lastMonday
Dim currentFriday As DateTime = currentMonday.AddDays(4)
End While
weeklyRanges.ToArray()
``````
Below are invoked arguments:
=> Use For Each loop to iterate through arr_week and it is of DataType System.Collections.Generic.List(System.Tuple(System.DateTime,System.DateTime))
=> Use the below syntax in write line:
``````"Range: " & weeklyRange.Item1.ToString("yyyy-MM-dd") & " to " & weeklyRange.Item2.ToString("yyyy-MM-dd")
``````
Sequence.xaml (10.2 KB)
Hope it helps!!
Regards
1 Like
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https://gitlab.mn.tu-dresden.de/amdis/amdis-core/-/raw/5da436789053f4ebf8e96383333d2463606879f1/dune/amdis/assembler/FirstOrderTestPartialTrial.hpp
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#pragma once #include #include #include #include namespace AMDiS { /** * \addtogroup operators * @{ **/ namespace tag { struct test_partialtrial { std::size_t comp; }; } /// first-order operator \f$\langle\psi, c\,\partial_i\phi\rangle \f$ template class GridFunctionOperator : public GridFunctionOperatorBase { using Super = GridFunctionOperatorBase; static_assert( Category::Scalar, "Expression must be of scalar type." ); public: GridFunctionOperator(tag::test_partialtrial tag, GridFct const& expr, QuadCreator const& quadCreator) : Super(expr, quadCreator, 1) , comp_(tag.comp) {} template void calculateElementMatrix(Context const& context, QuadratureRule const& quad, ElementMatrix& elementMatrix, RowNode const& rowNode, ColNode const& colNode, std::integral_constant, std::integral_constant) { static_assert(RowNode::isLeaf && ColNode::isLeaf, "Operator can be applied to Leaf-Nodes only."); auto const& rowLocalFE = rowNode.finiteElement(); auto const& colLocalFE = colNode.finiteElement(); for (std::size_t iq = 0; iq < quad.size(); ++iq) { // Position of the current quadrature point in the reference element decltype(auto) local = context.position(quad[iq].position()); // The transposed inverse Jacobian of the map from the reference element to the element const auto jacobian = context.geometry.jacobianInverseTransposed(local); // The multiplicative factor in the integral transformation formula double factor = context.integrationElement(quad[iq].position()) * quad[iq].weight(); double c = Super::coefficient(local); rowLocalFE.localBasis().evaluateFunction(local, rowShapeValues_); // The gradients of the shape functions on the reference element std::vector > referenceGradients; colLocalFE.localBasis().evaluateJacobian(local, referenceGradients); // Compute the shape function gradients on the real element std::vector colPartial(referenceGradients.size()); for (std::size_t i = 0; i < colPartial.size(); ++i) { colPartial[i] = jacobian[comp_][0] * referenceGradients[i][0][0]; for (std::size_t j = 1; j < jacobian.M(); ++j) colPartial[i] += jacobian[comp_][j] * referenceGradients[i][0][j]; } for (std::size_t j = 0; j < colLocalFE.size(); ++j) { const int local_j = colNode.localIndex(j); double value = factor * (c * colPartial[j]); for (std::size_t i = 0; i < rowLocalFE.size(); ++i) { const int local_i = rowNode.localIndex(i); elementMatrix[local_i][local_j] += value * rowShapeValues_[i]; } } } } private: std::size_t comp_; std::vector> rowShapeValues_; }; /** @} **/ } // end namespace AMDiS
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https://tulipindicators.org/linreg
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# Linear Regression
## Function Prototype
/* Linear Regression */
/* Type: overlay */
/* Input arrays: 1 Options: 1 Output arrays: 1 */
/* Inputs: real */
/* Options: period */
/* Outputs: linreg */
int ti_linreg_start(TI_REAL const *options);
int ti_linreg(int size,
TI_REAL const *const *inputs,
TI_REAL const *options,
TI_REAL *const *outputs);
## Description
This documentation is still a work in progress. It has omissions, and it probably has errors too. If you see any issues, or have any general feedback, please get in touch.
The Linear Regression is a smoothing functions that works by preforming linear least squares regression over a moving window. It then uses the linear model to predict the value for the current bar.
It takes one parameter, the period n. Larger values for n will have a greater smoothing effect on the input data but will also create more lag.
It is calculated as:
$$\overline{x} = \frac{1}{n} \sum_{i=1}^{n} i$$
$$\overline{y}_{t} = \frac{1}{n} \sum_{i=0}^{n-1} in_{t-i}$$
$$\widehat{\beta}_{t} = \frac {\sum_{i=1}^{n}(i-\overline{x})(in_{(t-n+i)}-\overline{y}_{t})} {\sum_{i=1}^{n}(i-\overline{x})^{2}}$$
$$\widehat{\alpha}_{t} = \overline{y}_{t} - \widehat{\beta}_{t}\overline{x}$$
$$linreg_{t} = \widehat{\alpha}_{t} + \widehat{\beta}_{t} n$$
## Example Usage
### Calling From C
/* Example usage of Linear Regression */
/* Assuming that 'input' is a pre-loaded array of size 'in_size'. */
TI_REAL *inputs[] = {input};
TI_REAL options[] = {5}; /* period */
TI_REAL *outputs[1]; /* linreg */
/* Determine how large the output size is for our options. */
const int out_size = in_size - ti_linreg_start(options);
/* Allocate memory for output. */
outputs[0] = malloc(sizeof(TI_REAL) * out_size); assert(outputs[0] != 0); /* linreg */
/* Run the actual calculation. */
const int ret = ti_linreg(in_size, inputs, options, outputs);
assert(ret == TI_OKAY);
### Calling From Lua (with Tulip Chart bindings)
-- Example usage of Linear Regression
linreg = ti.linreg(input, 5)
## Example Calculation
period = 5
dateinputlinreg
2005-11-0181.59
2005-11-0281.06
2005-11-0382.87
2005-11-0483.00
2005-11-0783.6183.62
2005-11-0883.1583.72
2005-11-0982.8483.11
2005-11-1083.9983.56
2005-11-1184.5584.17
2005-11-1484.3684.60
2005-11-1585.5385.40
2005-11-1686.5486.21
2005-11-1786.8986.95
2005-11-1887.7787.85
2005-11-2187.2987.75
## Other Indicators
Previous indicator: Lag
Next indicator: Linear Regression Intercept
Random indicator: Maximum In Period
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https://www.ammacement.in/help-construction/how-to-calculate-initial-setting-time-of-cement.html
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# How To Calculate Initial Setting Time Of Cement?
Standard Specification of Setting Time for Different Type of Cement – Depending upon the admixtures added into the cement in the process of manufacture of cement, the setting time differs for different cement. Standard setting times are listed in the below table as per IS Codes. Table-1: Setting Time for Different Type of Cement
Cement Type Initial Setting Time, min (min.) Final Setting Time, min (max.) OPC(33) 30 600 OPC(43) 30 600 OPC(53) 30 600 SRC 30 600 PPC 30 600 RHPC 30 600 PSC 30 600 High alumina 30 600 Super sulphated 30 600 Low heat 60 600 Masonry cement 90 1440 IRS-T-40 60 600
### How do you find the initial setting time of cement?
Determination of Final Setting Time – Replace the needle (C) of the Vicat apparatus by the needle with an annular attachment (F). The cement shall be considered as finally set when, upon applying the needle gently to the surface of the test block, the needle makes an impression thereon, while the attachment fails to do so. The period elapsing between the time when water is added to the cement and the time at which the needle makes an impression on the surface of test block while the attachment fails to do so shall be the final setting time. In the event of a scum forming on the surface of the test block, use the underside of the block for the determination.
## What is the setting time of 43 grade cement?
Physical Requirement for 43 Grade OPC (Ordinary Portland Cement)-(IS-8112) | March 5, 2014 | | When a 43 grade ordinary Portland cement is tested for its physical requirements, then it must satisfy the following specifications given by IS 8112.
You might be interested: Why Cement Is Grey In Colour?
The fineness value of 43 grade OPC when tested by Blaine’s air permeability method shall not be less than 225 m 2 /kg. The expansion value (i.e. soundness) of unaerated cement when tested by Le-Chaelier method shall not be more than 10 mm and when tested by autoclave method shall not be more than 0.8%. Initial setting time of 43 grade OPC shall not be less than 30 minutes. Final setting time of 43 grade OPC shall not be more than 600 minutes. The average compressive strength of at least 3 mortar cubes, when tested as per IS-4031-Part 6, shall be as follows.
Sl.No Time Compressive strength (MPa) 1 72 ± 1 hour Not less than 23 Mpa 2 168 ± 2 hour Not less than 33 Mpa 3 672 ± 4 hour Not less than 43 Mpa
Geotechnical-Material Engineer. You can connect me on, To know more about me just visit : Physical Requirement for 43 Grade OPC (Ordinary Portland Cement)-(IS-8112)
## Which test is used to find setting time of cement?
Free Gujarat Engineering Service 2019 Official Paper (Civil Part 1) 150 Questions 150 Marks 90 Mins The setting time of cement is of two types: 1) Initial Setting Time: It is defined as the time elapsed between the moments when water is added to the cement, to the time when the cement paste starts losing its plasticity.2) Final Setting Time: It is defined as the time elapsed between the moments when water is added to the cement, to the time when the cement paste has completely lost its plasticity.
Apparatus Test Le Chatelier apparatus Soundness Briquette testing apparatus Tensile strength Vicat apparatus Consistency, Initial and final setting time Casagrande’s apparatus Liquid limit determination
The setting time of cement is determined using the Vicat apparatus. Latest GPSC Engineering Services Updates Last updated on Oct 1, 2022 The Gujarat Public Service Commission (GPSC) has released a new notification for the GPSC Engineering Services Recruitment 2022.
• The commission has released 28 vacancies for the recruitment process.
• Candidates can apply for the applications from 15th October 2022 to 1st November 2022 and their selection will be based on Prelims, Mains and Interview.
• Candidates with a Graduation degree as the basic GPSC Engineering Services Eligibility Criteria are eligible to appear for the recruitment process.
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The finally selected candidates will get a salary range between Rs.53100 to Rs.208700.
#### What is the initial setting time *?
FAQ – What is initial & final setting time of cement? Initial setting time of cement is the time at which the cement paste starts losing its plasticity. Final setting time of cement is the time when the cement paste totally loses its plasticity. What is the initial setting time of cement? The time when water is added to cement to the time when the cement starts losing its plasticity is called initial setting time.
Initial setting time is the time when water is added to cement until the moment when the square needle penetrates to a depth of 33-35 mm from the top of the mould. Why is the initial setting of cement 30 minutes? The initial setting time of cement should not be less than 30 minutes because the time is required for handling operations of cement like mixing, placing, finishing, etc.
The cement operations cannot be carried out once the initial setting of cement occurs as cement paste starts losing plasticity which will then affect the strength gaining of the paste. What is cement setting? Upon addition of water, the cement paste formed remains plastic for some time.
### What is initial setting time of OPC 43 grade cement?
Setting Time in Minutes
Description IS Requirement Parasakti Results
Initial Setting time Not less than 30 Minutes 130-160 Minutes
Final Setting time Not more than 600 Minutes 230-260 Minutes
## What is meant by initial setting time?
2 The Setting Time – The setting time of cement includes the initial setting time and the final setting time. The initial time refers to the time that cement turns into paste by mixing with water and begins to lose its plasticity. And the time that cement completely loses its plasticity by mixing with water and begins to have a certain structural strength is known as the final setting time.
The national standards prescribe that the initial setting time of Portland cement should not be earlier than 45 min and the final setting time should not be later than 6.5 h. All the products off-grade at the initial setting time are spoiled products and those unqualified at the final setting time are sub-quality products.
The setting time of cement is measured by time determinator. The sample is the standard cement paste of which the temperature is 20 °C ± 3 °C and humidity is more than 90%. Various mineral components of the cement clinker are different in the water consumption of their normal consistency.
1. The finer the cement is ground, the more water the normal consistency will need.
2. The normal consistency of Portland cement is within 24% ~ 30%.
3. The setting time of cement is very important in the construction projects.
4. The initial setting time should not be too fast in order to ensure that there is enough time to complete every process, such as casting, before the initial setting time; and the final setting time should not be too late in order to enable the cement to complete its setting and hardening as soon as possible after pouring and tamping to make the next process occur earlier.
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## 6. Find a value of ‘a’ such that :- a .x = 3, y = -2, is a solution of the equation 5x – 2ay = 5. Now, find one more solution of<
Question
6. Find a value of ‘a’ such that :-
a .x = 3, y = -2, is a solution of the equation 5x – 2ay = 5. Now, find one more solution of
this equation
b. x=-1, y = 4 is a solution of the equation 2ax – 3y = 8. Now, find one more solution of
in progress 0
2 months 2021-09-25T04:37:12+00:00 2 Answers 0 views 0
## Answers ( )
1. Step-by-step explanation:
Equation:5x-2ay=5
Now putting values in this equation,we get
5 (3)-2a (-2)=5
15+4a=5
4a=5-15= -10
a= -10/4
Equation:2ax-3y=8
2a (-1) – 3 (4)=8
-2a-12=8
-2a=8+12=20
-a=20/2=10
a= -10
2. Step-by-step explanation:
1. letx=5;y=6
so,5x-2ay=5
5(5)-2a(6)=5
25-12a=5
12a=5-25
12a=20
a=20/12
a=5/3
b..
letx=-2
y=8
2ax-3y=8
2a(-2)-3(8)=8
-4a-24=8
-4a=8+24
-4a=32
a=32/4
a=8
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ACL2 !>(defun g (x y a) (if (zp x) a (g (- x 1) y (+ y a)))) The admission of G is trivial, using the relation O< (which is known to be well-founded on the domain recognized by O-P) and the measure (ACL2-COUNT X). We observe that the type of G is described by the theorem (OR (ACL2-NUMBERP (G X Y A)) (EQUAL (G X Y A) A)). We used primitive type reasoning. Summary Form: ( DEFUN G ...) Rules: ((:FAKE-RUNE-FOR-TYPE-SET NIL)) Time: 0.00 seconds (prove: 0.00, print: 0.00, other: 0.00) G ACL2 !>(defthm generalized-g-lemma (implies (and (natp x) (natp y) (natp a)) (equal (g x y a) (+ a (* x y))))) By the simple :definition NATP we reduce the conjecture to Goal' (IMPLIES (AND (INTEGERP X) (<= 0 X) (INTEGERP Y) (<= 0 Y) (INTEGERP A) (<= 0 A)) (EQUAL (G X Y A) (+ A (* X Y)))). Name the formula above *1. Perhaps we can prove *1 by induction. One induction scheme is suggested by this conjecture. We will induct according to a scheme suggested by (G X Y A). This suggestion was produced using the :induction rule G. If we let (:P A X Y) denote *1 above then the induction scheme we'll use is (AND (IMPLIES (AND (NOT (ZP X)) (:P (+ Y A) (+ -1 X) Y)) (:P A X Y)) (IMPLIES (ZP X) (:P A X Y))). This induction is justified by the same argument used to admit G. Note, however, that the unmeasured variable A is being instantiated. When applied to the goal at hand the above induction scheme produces six nontautological subgoals. Subgoal *1/6 (IMPLIES (AND (NOT (ZP X)) (EQUAL (G (+ -1 X) Y (+ Y A)) (+ (+ Y A) (* (+ -1 X) Y))) (INTEGERP X) (<= 0 X) (INTEGERP Y) (<= 0 Y) (INTEGERP A) (<= 0 A)) (EQUAL (G X Y A) (+ A (* X Y)))). By the simple :rewrite rule ASSOCIATIVITY-OF-+ we reduce the conjecture to Subgoal *1/6' (IMPLIES (AND (NOT (ZP X)) (EQUAL (G (+ -1 X) Y (+ Y A)) (+ Y A (* (+ -1 X) Y))) (INTEGERP X) (<= 0 X) (INTEGERP Y) (<= 0 Y) (INTEGERP A) (<= 0 A)) (EQUAL (G X Y A) (+ A (* X Y)))). This simplifies, using the :compound-recognizer rule ZP-COMPOUND-RECOGNIZER, the :definition G and the :rewrite rules COMMUTATIVITY-OF-* and DISTRIBUTIVITY, to Subgoal *1/6'' (IMPLIES (AND (NOT (ZP X)) (EQUAL (G (+ -1 X) Y (+ Y A)) (+ Y A (* -1 Y) (* X Y))) (<= 0 X) (INTEGERP Y) (<= 0 Y) (INTEGERP A) (<= 0 A)) (EQUAL (G (+ -1 X) Y (+ Y A)) (+ A (* X Y)))). But simplification reduces this to T, using the :compound-recognizer rule ZP-COMPOUND-RECOGNIZER, linear arithmetic, primitive type reasoning and the :type-prescription rule G. Subgoal *1/5 (IMPLIES (AND (NOT (ZP X)) (< (+ Y A) 0) (INTEGERP X) (<= 0 X) (INTEGERP Y) (<= 0 Y) (INTEGERP A) (<= 0 A)) (EQUAL (G X Y A) (+ A (* X Y)))). But we reduce the conjecture to T, by primitive type reasoning. Subgoal *1/4 (IMPLIES (AND (NOT (ZP X)) (NOT (INTEGERP (+ Y A))) (INTEGERP X) (<= 0 X) (INTEGERP Y) (<= 0 Y) (INTEGERP A) (<= 0 A)) (EQUAL (G X Y A) (+ A (* X Y)))). But we reduce the conjecture to T, by primitive type reasoning. Subgoal *1/3 (IMPLIES (AND (NOT (ZP X)) (< (+ -1 X) 0) (INTEGERP X) (<= 0 X) (INTEGERP Y) (<= 0 Y) (INTEGERP A) (<= 0 A)) (EQUAL (G X Y A) (+ A (* X Y)))). But we reduce the conjecture to T, by the :compound-recognizer rule ZP-COMPOUND-RECOGNIZER and primitive type reasoning. Subgoal *1/2 (IMPLIES (AND (NOT (ZP X)) (NOT (INTEGERP (+ -1 X))) (INTEGERP X) (<= 0 X) (INTEGERP Y) (<= 0 Y) (INTEGERP A) (<= 0 A)) (EQUAL (G X Y A) (+ A (* X Y)))). But we reduce the conjecture to T, by the :compound-recognizer rule ZP-COMPOUND-RECOGNIZER and primitive type reasoning. Subgoal *1/1 (IMPLIES (AND (ZP X) (INTEGERP X) (<= 0 X) (INTEGERP Y) (<= 0 Y) (INTEGERP A) (<= 0 A)) (EQUAL (G X Y A) (+ A (* X Y)))). This simplifies, using the :compound-recognizer rule ZP-COMPOUND-RECOGNIZER, the :executable-counterparts of <, INTEGERP, NOT and ZP and linear arithmetic, to Subgoal *1/1' (IMPLIES (AND (INTEGERP Y) (<= 0 Y) (INTEGERP A) (<= 0 A)) (EQUAL (G 0 Y A) (+ A (* 0 Y)))). This simplifies, using the :definition G and the :executable-counterpart of ZP, to Subgoal *1/1'' (IMPLIES (AND (INTEGERP Y) (<= 0 Y) (INTEGERP A) (<= 0 A)) (EQUAL A (+ A (* 0 Y)))). But simplification reduces this to T, using linear arithmetic and primitive type reasoning. That completes the proof of *1. Q.E.D. Summary Form: ( DEFTHM GENERALIZED-G-LEMMA ...) Rules: ((:COMPOUND-RECOGNIZER ZP-COMPOUND-RECOGNIZER) (:DEFINITION G) (:DEFINITION NATP) (:DEFINITION NOT) (:EXECUTABLE-COUNTERPART <) (:EXECUTABLE-COUNTERPART INTEGERP) (:EXECUTABLE-COUNTERPART NOT) (:EXECUTABLE-COUNTERPART ZP) (:FAKE-RUNE-FOR-LINEAR NIL) (:FAKE-RUNE-FOR-TYPE-SET NIL) (:INDUCTION G) (:REWRITE ASSOCIATIVITY-OF-+) (:REWRITE COMMUTATIVITY-OF-*) (:REWRITE DISTRIBUTIVITY) (:TYPE-PRESCRIPTION G)) Time: 0.01 seconds (prove: 0.00, print: 0.01, other: 0.00) Prover steps counted: 735 GENERALIZED-G-LEMMA ACL2 !>(defthm g-theorem (implies (and (natp x) (natp y)) (equal (g x y 0) (* x y)))) By the simple :definition NATP we reduce the conjecture to Goal' (IMPLIES (AND (INTEGERP X) (<= 0 X) (INTEGERP Y) (<= 0 Y)) (EQUAL (G X Y 0) (* X Y))). But simplification reduces this to T, using the :compound-recognizer rule NATP-COMPOUND-RECOGNIZER, the :definition FIX, primitive type reasoning and the :rewrite rules GENERALIZED-G-LEMMA and UNICITY-OF-0. Q.E.D. Summary Form: ( DEFTHM G-THEOREM ...) Rules: ((:COMPOUND-RECOGNIZER NATP-COMPOUND-RECOGNIZER) (:DEFINITION FIX) (:DEFINITION NATP) (:FAKE-RUNE-FOR-TYPE-SET NIL) (:REWRITE GENERALIZED-G-LEMMA) (:REWRITE UNICITY-OF-0)) Time: 0.00 seconds (prove: 0.00, print: 0.00, other: 0.00) Prover steps counted: 106 G-THEOREM ACL2 !>
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# Physics: Gravitational Force
label Physics
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
Two objects attract each other with a gravitational force of magnitude 9.50 10-9 N when separated by 19.3 cm. If the total mass of the objects is 5.07 kg, what is the mass of each?
Nov 23rd, 2017
the mass of object a) is 3.57kg, whereas that of object b)is 4.03 kg
Jun 30th, 2015
...
Nov 23rd, 2017
...
Nov 23rd, 2017
Nov 24th, 2017
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Dr. Phil's Home
Lectures in PHYS-2070 (12/13) / PHYS-2150
Updated: 28 April 2009 Tuesday.
Week of April 20-24, 2009.
Monday 4/20: Office Hours.
Tuesday 4/21: Office Hours. PHYS-2070(12) Noon section Final Exam 2:45-4:45pm.
Wednesday 4/22: Office Hours. PHYS-2070(13) 2pm section Final Exam 2:45-4:45pm.
Thursday 4/23: Dr. Phil working at home.
Friday 4/24: Office Hours.
Monday 4/27: Office Hours.
Tuesday 4/28: Grades due at Noon.
Week of January 5-9, 2009.
Monday 1/5: Class begins. Introduction to Dr. Phil.
Tuesday 1/6: Static electricity. The Two-Fluid Model of Electricity. Franklin's One-Fluid Model of Electricity. Occaam's Razor. The simple hydrogen atom -- whatever charge is, the charge on the electron (-e) and the proton (+e) exactly cancel.
Wednesday 1/7: Real Electric Charges. Two charges: like charges repel, unlike (opposite) charges attract. 1 Coulomb of charge is an enormous amount of charge. "Action at a distance" -- Gravity and the Electric Force are not contact forces. The Electric Force between two point charges, Coulomb's Law looks like Newton's Law of Universal Gravity. Four Fundamental Forces in Nature: Gravity, E & M, Weak Nuclear Force, Strong Nuclear Force. Distribute syllabus.
Thursday 1/8: The Hydrogen Atom: Gravity loses to Electric Force by a factor of 200 million dectillion (!!!). The Helium Atom: Putting more than one proton in the nucleus produces enormous forces on the tiny protons -- Need the Neutron and the Strong Nuclear Force (!!!).
Friday 1/9: 1 Coulomb of charge is an enormous amount of charge. Two 1.00 C charges separated by 1.00 meters have a force of nine-billion Newtons acting on each other. Finding the net vector electric force FE for a system of point charges. Remember: In PHYS-2070, Looking at Symmetry and Zeroes (problems where the answer is zero) as a way of solving problems. Four pages of Topic 1 assignment handed out. (Full 27-page Handout as PDF File -- Searchable HTML Page ) Q1 in-class, for attendance purposes. (If you missed class today, click here to print out a form to bring to class next time.)
FYI: Handout: SI Prefixes and Dr. Phil's Simplified Significant Figures. List of Topics covered in PHYS-2050 when Dr. Phil taught it last.
Week of January 12-16, 2009.
Monday 1/12: Demo these class web pages. Review of vectors and vector forces. Review of vector notation for components and Standard Form. Right Triangles and Adding and subtracting vectors: Analytical method. (Check to make sure your calculator is set for Degrees mode. Try cos 45° = sin 45° = 0.7071) Why arctangent is a stupid function on your calculator. Solving a vector Electric Force problem when there isn't symmetry to render the problem zero. How does q1 know that q2 is there? -- "Action at a Distance" -- Gravity and the Electric Force are not contact forces. The mathematical construct of the Electric Field. E is not an observable quantity. (Side example: Methods of measuring speed v, do not directly measure speed v.)
Remember: PHYS-2080 Lab Begins This Week.
Web comic XKCD's take on Guide To Converting To Metric. (Both funny and true, but there is some bad language.)
Tuesday 1/13: Conductors (metals) versus non-conductors (insulators). Conduction electrons in metals -- free to move around. Semi-Conductors sit in the middle. Sometimes they conduct and sometimes they don't. This means they act like a switch or valve, and this is the basis for the entire electronics semi-conductor industry. Charging a conductor by induction. Electric Field is a vector. FE = q E. For a point charge, E = k q1 /r2. SI units for E-field: (N/C). E-field lines radiate away from a positive point charge; converge towards a negative point charge. If the universe is charge neutral, can have all E-field lines from + charges terminating on - charges.Why use E-fields, when you need the force F = q E anyway? Because it allows us to examine the environment without needing another charge. E-field lines allow us to qualitatively sketch what happens when two charges are near to each other. (1) +q and -q, (2) +q and +q. Very close to each point charge, the E-field lines are radial outward, evenly spaced. In the system, the E-field lines interact with each other -- but E-field lines can never cross. Long range, the system of point charges looks like a single net charge. The density of E-field lines in an area gives you an indication of the strength of the E-field line. The numbers of E-field lines attached to a point charge is proportional to the charge. E-field lines radiate away from positive charges and terminate on negative charges. Q2 Take-Home, due in class on Thursday 15 January 2009.
Wednesday 1/14: Q2 comments about parts (a) and (b): You may be making this too hard. Part (a) is just asking what the net or total charge of the system is. Part (b) wants to know how to make the charge in (a) by either adding or subtracting electrons to a neutral system. See lecture notes about Real Electric Charges. E-field lines allow us to qualitatively sketch what happens when two charges are near to each other. (3) +2q and -q. Why use E-fields, when you need the force F = q E anyway? Because it allows us to examine the environment without needing another charge. Strength of E-field is important, in part because in air there is a limit to how big E can get. At breakdown, the E-field is large enough to start ripping electrons from gas molecules and the positive ions go one way, the electrons go another, and then we get a "spark" and no longer are static. Emax in air is 3,000,000 N/C. Direct integration of Electric Force and Electric Field are similar. Charge distributions -- lamda (linear charge density, C/m), sigma (surface charge density, C/m²), rho (volume charge density, C/m³). Examples: Rod in-line with line from point P (1-dimensional integration).
FYI for Future Physics Teachers -- brochures for the PhysTEC NOYCE Scholarship Program. Application Deadline: Friday February 27, 2009. You can also contact Prof. Al Rosenthal in Physics for further information.
Thursday 1/15: Direct integration of Electric Field continued. Rod perpendicular to line from point P. Note that in all these cases, we can predict the long range behavior (E-field behaves as a single point net charge), and anticipate the close-in short range behavior. Check Serway's examples (that's your textbook) -- watch out that his notation may be different.
Friday 1/16: Review of 2-D and 3-D Integration. Rectangular (area, volume), Polar (circumference, area), Cylindrical (volume, surface area). 1st Sample Exams for Exam 1. (Click here and here for a copy.) Q3 In-Class.
NOTE: Monday 19 January 2009 -- MLK Day activities on WMU campus. No classes.
NOTE: Tuesday 20 January 2009 -- Physics Help Room starts in 0077 Rood.
NOTE: Tuesday 20 January 2009 -- Obama inauguration shown live in Miller Auditorium.
NOTE: Tuesday 20 January 2009 -- Dr. Phil is still sick. Tuesday classes canceled.
Week of January 19-23, 2009.
Monday 1/19: Dr. Martin Luther King, Jr. Memorial Observance -- No Classes.
Tuesday 1/20: Dr. Phil's sick day -- Class canceled.
Wednesday 1/21: Review of 2-D and 3-D Integration (cont.) -- Spherical Co-ordinates. Direct integration of Electric Field continued. Thin ring of charge to center point P. (Symmetry!) Thin ring of charge perpendicular to line from point P. Note that in all these cases, we can predict the long range behavior (E-field behaves as a single point net charge), and anticipate the close-in short range behavior. Check Serway's examples (that's your textbook) -- watch out that his notation may be different.
Thursday 1/22: Return Q2. Direct integration of Electric Field continued. Disk of charge to center point P. Harder to see 1/r² dependence at long range, but it is clear that E goes to zero. Electric Flux: Electric field times Area. Analogy of a bag or box around a light, captures all the light rays no matter the size or shape. Use known E-field of a point charge to evaluate what the Electric Flux must be equal to. Review of Dot Product. Gauss' Law for Electricity.
Friday 1/23: Q4 in-class quiz (mini-exam).
Week of January 26-30, 2009.
Tuesday 1/27: . P.E. is minus the Work. Potential V is similar, but the integral is done on E-field not Force. More importantly the Potential V is an observable quantity. Simplified equation V = E d. Example: Lightning. Conductor in equilibrium is an equipotential throughout. Equipotential lines, where V is constant, are always perpendicular to E-field lines. Q5 Take-Home, due Thursday 29 January 2009.
Wednesday 1/28: Return Q3. Find components of E by negative of the partial derivative of Electric Potential function V. Finding V by direct integration. Direct integration of V for a whole and a half of a circular line of charge -- V really is a scalar, not a vector. In electrostatic equilibrium, E = 0 inside a charged conductor, but V = constant, not V = 0 automatically. Why charge accumulates on the tips of "pointy things". NOTE: The book is effectively closed for Exam 1 topics now.
Thursday 1/29: Return Q4A and Q4B. Moving from Field Theory to Applications leading to Devices. Start of Capacitors and Capacitance. The Capacitor stores charge +Q on one plate and -Q on second plate, stores energy in the E-field between the plates. This is different from a battery, which has energy stored in its chemical reaction. Dr. Phil & the camera flash. Stories: US Navy seaman vs. the tank capacitor (Cap-2, Seaman-0). Review Exam 1.
Friday 1/30: Exam 1.
Week of February 2-6, 2009.
Monday 2/2: Capacitor Equation. SI unit for Capacitance is the Farad. 1F is a large capacitor. Usually deal with µF (microfarad = 1/1,000,000th of a Farad) and pF (picofarad = 1/1,000,000,000,000th of a Farad). Circuit diagrams represent the elements of a circuit. So far: battery, wires, capacitor. Apply Gauss' Law for Electricity to the constant E-field of the Parallel Plate Capacitor. We now have an "operational equation", true for all capacitors, and a "by geometry" equation for the special case of the parallel plate capacitor. Two devices connected together in a circuit can only be connected two ways: series or parallel. In Series, same charge, share voltage. Equivalent capacitor is always smaller. NOTE: Remember to take the last reciprocal!
Tuesday 2/3: Two devices connected together in a circuit can only be connected two ways: series or parallel. In Parallel, same voltage, share charge. Equivalent capacitor is always larger. In Series, same charge, share voltage. Equivalent capacitor is always smaller. NOTE: Remember to take the last reciprocal! Work to assemble charges on a capacitor = Energy stored in the capacitor = U = ½CV² . Capacitor Network Reduction problem. Use table with columns for Q = C V. By going back through the intermediate diagrams, it is possible to know every value of every capacitor in the network. Extend the example in class with a fourth column, U=½CV², and find the energy stored in the equivalent capacitor and the sum of the energy stored in all four of the real capacitors -- if they agree, then our analysis and calculations are correct -- the battery cannot tell the difference! Q6 Take-Home, due Thursday 5 February 2009.
Wednesday 2/4: Making a real capacitor. What if not filled with air? Filling with conductor, must have at least one gap, otherwise will short outthe plates. A conducting slab inside a parallel plate capacitor makes two capacitors in series. Charge neutral slab stays charge neutral, but +Q of top plate attracts -Q on top of slab, and -Q of bottom plate attracts +Q on bottom of slab. Dielectrics -- an insulator where the +/- charge pairs are free to rotate, even if they do not move. Dielectric constant (kappa) and Dielectric strength (E-max). Dielectic constant increases capacitance over air gap. Dielectric strength usually bigger than E-max in air. Both allow you to (a) make bigger capacitors (or smaller for the same values) and (b) make non-hollow, self-supporting components. Examples of the uses of capacitors and dielectrics.
Thursday 2/5: Capacitive studfinder, uses edge effects of E-field from a capacitor to "see" the dielectric material behind the wall. Electrostatics to Electrodynamics (moving charges). Current defined: i = delta-Q/delta-t = dq/dt. The Simplest Circuit: Battery, wires, load (resistor). Resistance vs. Conductance. Discussion of microscopic theory of charges in a conductor. Drift velocity is the very slow net movement of the electrons moving randomly in the wire. Ohm's Law: V=IR form. (Ohm's "3 Laws") We usually treat the wires in a circuit as having R=0, but they usually are not superconductors. Joule Heating, Power Law: P = IV (also 3 forms). Q7 Take-Home, due Tuesday 10 February 2009.
Friday 2/6: Resistance by geometry. Discussion of heat from resistors. Our electronic devices generate heat, but heat can also kill them. Story of early "hot" 486 laptops. Discussion of microscopic theory of charges in a conductor. Drift velocity is the very slow net movement of the electrons moving randomly in the wire. This microscopic theory becomes more important as we go to smaller and smaller circuit elements in our microchips. Moore's Law. Ohm's Law: V=IR form. (Ohm's "3 Laws"). If R=constant over operating range, then we say the material is "ohmic". If R is not constant, it is "non-ohmic". Example: Because of the temperature dependence of R, the filament of an incandescent light bulb has a very different R when lit or dark. Therefore measuring the resistance of a light bulb with an ohm meter is useless. Kammerleigh Onnes 1916 work on extending the R vs. T curve toward T = 0 Kelvin. Discovered Superconductivity, where R=0 identically. We usually treat the wires in a circuit as having R=0, but they usually are not superconductors.
Week of February 9-13, 2009.
Monday 2/9: Go over Series and Parallel rules for Capacitors. Finish discussion of high temperature superconductors. "High temperature" superconductors (liquid nitrogen temperature, not liquid helium). The "Woodstock of Physics" in 1987. Discuss power cords -- flexible but hot cords for hair dryers, why power cords get recalled. Continuing with Simple Circuits... Two devices connected together in a circuit can only be connected two ways: series or parallel. In Series, same current, share voltage. Equivalent resistance is always larger. In Parallel, same voltage, share current. Equivalent resistance is always smaller. Resistor Network Reduction. (Similar rules to Capacitor Network Reduction except "opposite".) In this example, Resistor R1 sees the largest current and dissipates the largest amount of energy per second (Power in Watts). This means it is also the most vulnerable. Fault tolerant design. (Story of radio "repair" call from 4,000,000,000 miles.)
Tuesday 2/10: Real batteries consist of a "perfect" battery (Electromotive force = emf) in series with a small internal resistance, r. As chemical reaction in battery runs down, the internal resistance increases. Don't cut open batteries. Comments on different types of disposable (carbon-zinc, alkaline, lithium) and rechargable (Rayovac Renewal alkaline, NiCad, NiMH, Li-ion) batteries. Tip for weak car battery on cold day: Run headlights for 30 to 90 seconds. High internal resistance will warm the battery and make it more efficient. First Sample Exam 2. (Click here and here for a copy.) Q8 Take-Home, due Thursday 12 February 2009.
Wednesday 2/11: Class canceled by Dr. Phil. Reading Assignment: Sections 28.3 and 28.4 in Serway.
Thursday 2/12: Not all circuits can be reduced by serial and parallel network analysis. Kirchhoff's Laws: (1) The sum of all currents in and out of any junction must be zero. (2) The sum of all voltage gains and voltage drops about any closed loop is zero. Practically speaking, if there are N junctions, then (1) will give you (N-1) unique equations, and if there are M loops that can be made in the circuit by going around the perimeter of each "puzzle piece", then (2) will give you sufficient unique equations. You will get the same number of equations as you unknown currents through the resistors. NOTE: EE students and those who have had ECT-2100 (?) may know a "better" way to solve Kirchhoff's problems. But the brute force algebra approach has the advantage of being based on the Physics, so has instructional value. Example in class had 3 equations in 3 unknowns -- finish algebra and find i1, i2 and i3 for tomorrow. NOTE: I am providing a PDF of the problem worked out -- the order of the algebra might not be the same as on the board, but I believe the answers to be correct. Other examples of systems which require Kirchhoff Laws. Sometimes a resistor has zero current, in which case it does not contribute to the circuit. Proper procedure for jump starting a car. (And why doing it wrong ranges from dangerous to deadly.) Q9 Take-Home, due Thursday 19 February 2009.
Friday 2/13: FRIDAY THE THIRTEENTH (Not a WMU holiday) Return X1. RC series circuit. Calculus derivation of q(t) for charging capacitor .
Week of February 16-20, 2009.
Monday 2/16: PRESIDENT'S DAY (Not a WMU holiday). Calculus derivation of q(t) for charging capacitor and discharing circuit. RC current i(t). Who knew that (ohms) × (farads) = (seconds)? By time t=3RC, a charging capacitor will reach 95% of its top charge, or a discharging capacitor will be down to 5% of its original charge. Either way the current will be down to 5% of its maximum value.
Tuesday 2/17: Discussion of Christmas tree lights and what happens when one or more burns out. Building an ammeter or voltmeter -- non-digital version with a needle. The Galvanometer is a generic meter. It has a resistance and the needle moves in response to a current through a tiny coil. Since meters must be connected to the circuit, technically they change the circuit. However, we will show that the design of an ammeter and a voltmeter minimizes these changes. Ammeters measure current by connecting in series to the circuit. Voltmeters measure potential difference by connecting in parallel to the circuit. The Galvanometer is a generic meter. It has a resistance and the needle moves in response to a current through a tiny coil. The full-scale deflection current, iFS , is the current needed to move the needle to the maximum value on the scale -- it is very small. Second set of Sample Exam 2's handed out: (Click here and here for copies.)
NOTE: Thursday's weather may be heavy rain and snow = slush. Kirchhoff Law Q9 due date shifted to Friday 20 February 2009, just in case.
Wednesday 2/18: Ammeters measure current by connecting in series to the circuit. Voltmeters measure potential difference by connecting in parallel to the circuit. The Galvanometer is a generic meter. It has a resistance and the needle moves in response to a current through a tiny coil. The full-scale deflection current, iFS , is the current needed to move the needle to the maximum value on the scale -- it is very small. Ammeter: a galvanometer and a very small shunt resistor in parallel, together connected in series with the circuit. Voltmeter: a galvanometer amd a very large resistor in series, together connected in parallel with the circuit. In both cases, the role of the second resistor is to limit the current to the galvanometer, no matter what the design criteria of the meter in question. Does putting a real ammeter and voltmeter in a circuit, whether the very act of measuring V and I changes their value? It can't by much, because the full-scale deflection current and the voltage drop across the galvanometer are so small, compared to the values we are measuring.
NOTE: The numbers we found for the ammeter and voltmeter resistors were: rs = 0.001262 ohms and Rv = 49,940 ohms. The galvanometer had a resistance RG = 63.1 ohms and a full-scale deflection current iFS = 1.00 ×10-4 A. The high resistance wire we used for the shunt resistor in the ammeter had an R/L = 0.000147 ohm/cm.
Thursday 2/19: "Magnetism is just like Electricity, only different." Most people are familiar with (1) magnets sticking to some metals, not others such as stainless steel and (2) if you have two magnets, they may attract or repel. North and south are analogous to plus and minus charges. Real Magnets are dipoles (North and South ends, linked). Break a magnet in half, and you either get two new magnets -- or nothing. So far, there is no evidence that there are Magnetic Monopoles (magnetic charges: qM , isolated North or South poles). Rules similar to Electric Charges: Unlike poles attract, like poles repel. Demos: Cow magnets -- powerful cylindrical, rounded end magnets which get dropped into a cow's first stomach, to collect nails, bits of barbed wire, etc. from continuing on to the cow's other stomachs. The horizontal compass needle rotates until its North end points North (or rather to the North Magnetic Pole, which is of course a South pole of the Earth's magnetic core); the vertical compass rotates so that it lines up with the B-field along the surface of the Earth at the point. At the Equator, the vertical magnetic should be parallel to the ground, at the magnetic poles, it should be perpendicular to the ground. Is the Earth's magnetic field going to flip some day? And what about Mars? Q10 Take-Home, due on Friday 20 February 2009 now due Monday 23 February 2009.
NOTE: Exam 2 moved from Thursday 26 February 2009 to Wednesday 25 February 2009, due to memorial service for the late President Diether Haenicke.
Friday 2/20: Magnetic Force on a Moving Electric Charge - The Cross Product and Right-Hand Rule (R.H.R.). The Cross Product (or Vector Product) is the exact opposite of the Dot Product (or Scalar Product). Multiplying two vectors together by a cross product gives us another vector (instead of a scalar). And the cross product is not commutative, vector-A × vector-B = - (vector-B × vector-A), so the order is paramount. Using Right Hand Rule to assign directions to x,y,z coordinates. Constant speed, perpendicular constant magnetic force --> Uniform Circular Motion. Cyclotron frequency -- no dependence on the radius (constant angular velocity). Velocity Selector - the Magnetic Force is speed dependent, the Electric Force is not. So we can use an E-field to create an Electric Force to cancel the Magnetic Force on a moving charged particle, such that at the speed v = E / B, the particle travels exactly straight with no net force -- any other speed and the particle is deflected into a barrier. Hence a velocity selector "selects" velocities... Velocity Selector. Mass Spectrometer - different semi-circular paths for ions of different mass but same velocity. Can determine chemicals, molecules, and separate isotopes (same element, different number of neutrons in nucleus, so different mass -- cannot be separated by chemical means). Mass Spectrometer as Calutron -- detecting or separating isotopes, something that cannot be done by chemical means.
Week of February 23-27, 2009.
Monday 2/23: A current carrying wire consists of moving electric charges, and so therefore would see a magnetic force from a magnetic field. Discussion of microscopic theory of charges in a conductor. Drift velocity is the very slow net movement of the electrons moving randomly in the wire. Magnetic Force on a Current Carrying Wire. Demo -- hey it works and even in the right direction! Technically current is not a vector, despite the fact we talk of direction of current. J = current density = current/cross-sectional area is the vector related to current. NOTE: J-vector = sigma × E-vector (current density = conductivity × E-field) is the vector version of Ohm's Law. This will NOT be on Exam 2. For a Closed Loop, the net Magnetic Force from a constant B-field is zero. Magnetic Torque on a Current Carrying Wire. Left as is, this system is an osciallator -- the torque goes to zero after 90° and then points the other way. But if we can reverse the direction of the current after the torque goes to zero, then the rotation can continue -- and we have a primitive DC electric motor. Q11 Take-Home quiz, due on Tuesday 24 February 2009.
Tuesday 2/24: Demo Day: Light bulb boxes for PARALLEL RESISTORS (unscrew a bulb, remainder stay lit at same brightness), SERIES RESISTORS (unscrew a bulb and they all go out; put a 100W and a 15W bulb in series and the smaller bulb lights up okay, but little/no glow from bigger bulb -- remember that light bulb filaments heat up and are non-ohmic resistors), and RC CIRCUITS (start with three caps in series, connected in series with a light bulb. large (bright) initial current, fades over time as delta-V on plates builds up. 1 cap or 3 caps in parallel, takes longer -- RC constant is larger.). Review for X2.
Wednesday 2/25: Exam 2.
Thursday 2/26: Classes canceled due to Pres. Diether Haenicke memorial service.
Friday 2/27: Spirit Day - No Classes
WMU SPRING BREAK
Week of March 9-13, 2009.
Monday 3/9: Hall Effect -- a device with no moving electrical parts -- proves that charge carriers in a current carrying wire are negative, not positive. "The 200 Year Hall Effect Keyboards", will last "forever", but made obsolete in two years when Windows 95 added three keys. Discussion of Mid-Term Grades. Reset the course after Spring Break.
NOTE: Physics Help Room is moved from 0077 Rood to Bradley Commons 2202 Everett Tower -- next to Dr. Phil's office -- THIS WEEK ONLY.
Tuesday 3/10: The Biot-Savart Law. B-field from a infinitely long straight current carrying wire by direct integration. (Serway has a similar example, but rather than do the integral in x, he does this theta substitution which Dr. Phil does not think is straight forward.) Circular loop of current carrying wire by integration for P at the center of the loop. (Serway's example allows for P to be on a line perpendicular to the loop.) B-field for a circular current carrying wire at the center -- or any part of a circle. Gauss' Law for Magnetism. Not as useful as Gauss' Law for Electricity, because it is always zero (no magnetic monopoles).Q12 Take-Home, due Thursday 12 March 2009.
Wednesday 3/11: Return X2. Magnetic Field loops from a Current Carrying Wire. RHR has "two modes". Mode 1 uses three mutually perpendicular directions for when you have three vectors (A × B = C is 1-2-3, x-y-z). Mode 2 uses the curling of the fingers to represent the circulation of a field or the motion of a current, etc., with the thumb representing the relevent vector or direction. Magnetic Force between Two Current Carrying Wires. Combining problems, we find that for two parallel current carrying wires, with the currents in the same direction, the magnetic field from wire 1 creates an attractive magnetic force on wire 2. And the magnetic field from wire 2 creates an attractive magnetic force on wire 1. (Two forces, equal and opposite, acting on each other -- this is exactly as it should be with Newton's 3rd Law.) Anti-parallel currents (wires parallel, but currents in opposite directions) repel. Crossed currents (wires perpendicular to each other) see no magnetic force on each other. Operational defnition of the ampere and the Coulomb.
• NOTE: Corrected solution to Problem 2(b) on PHYS-2070 Exam 2 Form-B (2pm) can be found here.
Thursday 3/12: Gauss' Law for Magnetism. Not as useful as Gauss' Law for Electricity, because it is always zero (no magnetic monopoles). However, there is something we can use in a similar way which involves involving a path integral along a B-field and the current(s) contained inside -- Ampere's Law. Use in a way similar to the way we used Gauss' Law for Electricity. Use symmetry and geometry to select your Amperean Loop to your advantage. B-field of a Torroid (torroidal coil; a torus is like a donut). B-field of a Solenoid. (NOTE: The integrals for the L and R sides of the Amperean Loop for Ampere's Law are zero because: (1) the B-field is zero outside the solenoid and (2) for that part of the path which is inside the solenoid, the B-field and the ds-vector are perpendicular, so the dot product is zero as well.) Coils can have right-hand or left-hand turns -- but it is the direction that the current wraps around the coil which determines which way the B-field points. Comments about making a real velocity selector -- trying to stuff a capacitor for the E-field and a solenoid for the B-field in the same space!
Friday 3/13: Demo: Magnet moving into a coil, causing current to flow through galvanometer. Faraday's Law of Induction. A changing magnetic flux induces a current, induces an e.m.f., in the circuit, substituting for the battery as the power source. Lenz's Law "of maintaining the status quo." The coil acts as if it opposes any change of the magnetic flux inside, by inducing a magnetic field to cancel and increasing flux or maintain a decreasing flux. It is Lenz's Law that gives us the minus sign in Faraday's Law of Induction. Turn a coil in a magnetic field and the flux changes, thereby inducing a B-field, emf and current. Has same 180° problem that a DC motor has. Hand-crank generators. Electric generators and electric motors differ in which way the arrow points toward or away from mechanical energy. Regenerative braking -- turn electric motors into generators. An Inductor is a coil in a circuit. Why an Inductor has Self-Inductance -- running a current through a coil creates a magnetic field and therefore changes the magnetic flux in the coil. The inductor has to respond to that change. Inductance can be a big deal. Even our Simplest Circuit (a resistor hooked up to a battery) forms a loop, and the loop must respond to the circuit being turned on. Why induction is a big deal in electronics, industrial motors and electrical power distribution. Where they got it right (and wrong) regarding electrical power in the movie Jurassic Park. Q13 Take-Home, due Tuesday 17 March 2009.
Week of March 16-20, 2009.
Monday 3/16: Demonstration Day -- Hand-crank generators. Electric generators and electric motors differ in which way the arrow points toward or away from mechanical energy. Lenz Law race between cow magnets dropped through (a) plastic pipe, (b) non-magnetic aluminum pipe and (c) non-magnetic copper pipe. Induced B-fields due to changing B-fields of falling magnets are created by induced currents and induced emf -- as the magnet enters and leaves a circular region of metal pipe, it is slowed by magnetic forces between its magnetic field and the induced B-field. It is Lenz's Law that gives us the minus sign in Faraday's Law of Induction. Place bundle of iron rods in an AC coil and light bulb dims -- analogous to lights dimming when large electric motors drive vacuum cleaners and compressors for refrigerator, air conditioning and dehumidifiers. Demo: "Jumping Rings", making the bulb light, by Eddy Currents and Induction. Note that the metal rings get HOT, because there is a large induced current and metal has a low resistance. Adding metal increases the mass, but provides more eddy currents and therefore more induced B-fields repelling the solenoid's B-field. A split metal ring (a) does not get hot and (b) does not jump, because there is no circuit enclosing the changing magnetic flux. Ford test electric vehicle with inductive charger -- no exposed metal contacts, everything covered in smooth plastic. Heating the bottom of a metal cooking pot by induction: New type of cooking range uses sealed induction heating elements instead of exposed hot resistors or open gas fed flames -- usable for metal pans only.
Tuesday 3/17: Practical uses for induction: (First, how regular fuses and circuit breakers work -- and why that isn't fast enough to prevent some types of accidents.) Ground Fault Interupt -- if the current doesn't return via the return wire, because it has found another conductive path, then the 2 wires (hot and return) total a net-enclosed-current for Ampere's Law, generating a B-field in a metal ring, detected by an induction coil wrapped around the ring and this sets off the relay which breaks the circuit. An Inductor is a coil in a circuit. Why an Inductor has Self-Inductance -- running a current through a coil creates a magnetic field and therefore changes the magnetic flux in the coil. The inductor has to respond to that change. Inductance can be a big deal. Even our Simplest Circuit (a resistor hooked up to a battery) forms a loop, and the loop must respond to the circuit being turned on. The Inductor (L). (SI units = Henry = H) Self-Inductance. Back emf, back current. Opposing the status quo. Equations for Inductance. Q14 Take-Home quiz, due Thursday 19 March 2009.
Wednesday 3/18: RL Circuit, similar to RC Circuit, except that energy is stored in the magnetic field at the maximum current. UL = ½ L I ². RL Circuit for energizing the coil. Equation for current i(t) is similar in appearance to the equation for q(t) for a charging capacitor. Now we will de-energize the coil. NOTE: In class I used my usual brute force approach, instead of Serway's. The Kirchhoff Loop for the de-energining coil is -iR -L(di/dt) = 0. Since the current is decreasing, di/dt is negative and the induced emf from the coil becomes a voltage gain. Solution for i(t) the same form as the current i(t) for charging/discharging capacitor. Mutual Inductance between two inductors. 2nd coil responds only to changes in magnetic flux coming from 1st coil. And vice versa.
Thursday 3/19: LC Oscillator circuit. Same 2nd order differential equation as the Simple Harmonic Oscillator (PHYS-2050), such as a mass on a spring. Solutions are sines and cosines. Energy is held constant for all t between the capacitor and the inductor. Can't really have a true LC oscillator, since normal wires and coils have a resistance which dissipates energy through Joule heating. LC Oscillator solution: q(t) = Q0 cos(omega t + phi). Energy is held constant for all t between the capacitor and the inductor. Can't really have a true LC oscillator, since normal wires and coils have a resistance which dissipates energy through Joule heating. Comments ONLY about the RLC Damped Harmonic Oscillator. Mechanical analogue is the mass-on-a-spring with shock absorbers. First Sample Exam 3. (Click here for a copy.)
Friday 3/20: A.C. Circuits. Voltage is a sine or cosine functions, as is the Current. Problem: Average voltage is ZERO. Need to define a new average, the Root-Mean-Square. It is the RMS Voltage and Current that are usually reported in A.C. circuits. Typical A.C. frequency in U.S. is 60 Hz. Need to specify what type of A.C. For sine wave, define RMS Voltage as 0.7071 Maximum Voltage. Similar for RMS Current. Why A.C. power? (1) Transformers allow voltage to be raised or lowered. D.C. voltage can only be lowered by the voltage drop of a resistor, or raised by adding power sources. The transformer consists of two coils connected magnetically by an iron core. V2 = V1N2/N1. (2) D.C. power lines have huge power losses due to Joule heating, very low efficiency. Actual Efficiency = Power Used ÷ Total Power Generated. Power lines run at higher voltages to minimize power losses due to Joule heating in the powerlines (P = I²R). Q15 Take-Home quiz, due Tuesday 24 March 2009.
Week of March 23-27, 2009.
Monday 3/23: For resistive only circuits, can still use Ohm's Law, V = I R. Current and Voltage are both sine waves. Real A.C. circuits may have a Resistive nature, a Capacitive nature and an Inductive nature. For A.C. circuits with a Resistor only: I and V stay in phase with each other. RL Circuits: I and V out of phase by 90°. Inductive Reactance. Phasor diagrams -- taking the y-component of a rotating vector gives the sine function. RC Circuits: I and V out of phase by -90°. Capacitive Reactance. Many A.C. circuits have features of all three components (R, L and C), so we have to deal with Impedance, Z. Phasor diagrams (see textbook for diagrams).
Tuesday 3/24: Minimum impedance is when purely Resistive or when the two Reactances cancel each other -- the latter is frequency dependent. Can run into problems if expecting f=60Hz but get f=50Hz or f=25Hz. Phase angle, phi, for resultant Vmax vector relative to the Imax vector, is phi = tan-1((XL-XC)/R). For impedance matching , where XL=XC, we get the same equation for the angular frequency omega = 1/SQRT(LC) as for the LC oscillator. This is why power companies have to worry about maintaining their frequency -- it affects the impedance of the circuits. For DC circuits, P = IV. For AC, it is a little more complicated. Paverage = Irms Vrms cos(phi) -- because of the phase angle between V and I. For a purely resistive circuit, or one which looks like a purely resistive circuit, phi = 0°, and so get Paverage = Irms Vrms .
Wednesday 3/25: The problem with Ampere's Law -- it doesn't work properly in the gap between the plates of a capacitor while it is charging. So James Clerk Maxwell fixed it with a "displacement current" term, involving the time derivitive of the Electric flux in the gap. Maxwell's Equations in integral form. Note that Maxwell didn't invent the four equations, only half of one, but he figured out what todo with them. E & M Waves. Turning Maxwell's Equations in E and B, into 2nd order differential equations (Wave Equation) in E or in B. A constantly changing B-field creates a changing E-field, and a constantly changing E-field creates a changing B-field. The Great 19th Century Debate: Is Light a Particle or a Wave? (Wave-Particle Duality did not seem obvious at the time.) Second set of Sample Exam 3. (Click here for a copy.) Q16 Take-Home, now due Friday 27 March 2009.
Thursday 3/26: Looking at the solution to Traveling E-M Wave, with v in x-direction, E in y-direction and B in z-direction. Angular frequency omega, wave number k. c = Emax / Bmax. Derivation of c = E / B. Similar to the relationship between the E-field and the B-field in the velocity selector, where v = E / B. Poynting Vector, S. Traveling E-M Wave, Poynting Vector and Intensity. Energy stored equally in E- and B-fields of the E-M wave. Momentum and Pressure of light waves absorbed or reflected on contact. (Complete absorption like totally inelastic collision; complete reflection like totally elastic collision).
Friday 3/27: Light pressure and momentum transfer, despite the fact that light as mass = zero. NASA used solar panels as solar sails on Mariner 10 near the planet Mercury. Maxwell's Equations and Hertz's radio wave LC oscillator -- the spark gap radio. AM (amplitude modulation) radio versus FM (frequency modulation) and digital radio. The Marconi wireless telegraph of the RMS Titanic, and the modern cellphone. Q17 Take-Home, due Tuesday 31 March 2009.
Week of March 30-April 3, 2009.
Monday 3/30: Light as a wave. v = frequency × wavelength = c = 2.998 × 108 m/s (in vacuum). Light as a particle. The energy of a single photon ("particle" of light) is E = h f, where h = 6.626 × 10 -34 J·s is Planck's constant, a fundamental constant involved in Modern Physics. (If there was only Classical Physics, then h = 0.) The Electromagnetic Spectrum. Visible light (ROYGBIV=red orange yellow green blue indigo violet). Visible light is 400nm to 750nm (4000 angstroms to 7500 angstroms). Cannot "see" atoms with visable light, because the atom is about 1 angstrom across (1.00E-10 meters). The visible light wave is too large to see something that small. Frequencies LOWER and wavelengths LONGER than visible light (IR infrared, Microwave, Radio waves, ELF extremely low frequency). Microwave ovens have metal screens in their windows -- the centimeter-range sized EM waves cannot see the "small" holes in the screen, so they bounce off the window as if it were just like the metal in the other five walls. Discussion of how microwave ovens "cook" food. Radio includes what we consider consumer radio (AM and FM), television (analog and digital), WiFi, Bluetooth, cellular/wireless services, etc.
Tuesday 3/31: The Electromagnetic Spectrum. Visible light (ROYGBIV=red orange yellow green blue indigo violet). Frequencies HIGHER and wavelengths SHORTER than visible light (UV ultraviolet, X-rays, Gamma rays). Visible light is 400nm to 750nm (4000 angstroms to 7500 angstroms). Cannot "see" atoms with visable light, because the atom is about 1 angstrom across (1.00E-10 meters). The visible light wave is too large to see something that small. Discussion of why Superman's X-ray vision cannot work: Normal human vision -- light either reflected off of objects or directly from light source. Few X-rays get reflected, so what is the source of the X-rays for Superman to see? (If he's projecting the X-rays, he's killing anyone he looks at.) Gamma rays and food irradiation. Q18 Take-Home, due Thursday 2 April 2009.
Wednesday 4/1: April Fool's Day (Not a WMU holiday). Review of Sig. Figs. Comments on Optics: Transmission, Reflection (Scattering), Absorption. Optics: Geometric Optics (ray tracing, light as stream of particles), Physical Optics (wave nature of light). Constructive and Destructive Interference.
Thursday 4/2: Review for X3.
Friday 4/3: Exam 3.
Week of April 6-10, 2009.
Monday 4/6: When a light ray reaches a new material, it can undergo (1) reflection, (2) absorption, (3) transmission. The Law of Reflection. Measure all angles from the normal line perpendicular to the surface. Rough surfaces -- scattering. The Optical Lever -- move a mirror by 10° and the reflected ray moves by 20°. (Dr. Phil's theory on the origin of "seven years of bad luck for breaking a mirror".) The Law of Refraction - Snell's Law. Light bent at the interface between two media, because the speed of light changes in the media. (Analogy: If you are driving along the road and your right tires go off onto the soft shoulder, they can't go as fast and the car turns towards the shoulder until all four wheels are driving off the road.)
Tuesday 4/7: If going from an high index of refraction media to a lower index media, have a chance for Total Internal Reflection (T.I.R.). This is a "perfect" reflection, better than a mirror. Used in high-end optical systems instead of mirrors. Also useful in fiber optics cables. Dispersion -- in vacuum all speeds of light are the same, but in a medium, there are slighltly different n's for each wavelength. As one goes from a high index of refraction to a low index, increasing the angle of incidence, white light will start breaking into the rainbox spectrum of colors. This is due to dispersion, a change in the speed of light for each wavelength. Two interfaces: air-to-glass and glass-back-to-air. Coming down the normal, 0°, no deviation. At any other angle from the normal in air, the ray is refracted towards the normal in glass and then back out at the original angle when back in air. However, the light ray is offset from the line the ray would've followed had the glass not been there. Discussion of trying to see forward (large angle from normal) through the windows of a bus, train or plane. Very hard and very expensive to make true parallel-plano surfaces. Thin Lenses. Simplest lens surfaces are spherical (convex = bows out, concave = bows in) and flat (plano). So some lenses might appear to be biconvex, plano-convex, biconcave, convex-concave. A biconvex lens is also called a positive or converging lens. Parallel light rays coming into such a lens will all pass through the focal point, a distance f from the center of the lens. By itself, could use as a magnifying lens. Concentrating sunlight: burning paper or popping ants? Ray tracing gets same results as doing Snell's Law on mulitple curved lens surfaces. Handy not to have to do all that refraction calculations!
Wednesday 4/8: A biconvex lens is also called a positive or converging lens. Parallel light rays coming into such a lens will all pass through the focal point, a distance f from the center of the lens. By itself, could use as a magnifying lens. Concentrating sunlight: burning paper or popping ants? Ray tracing gets same results as doing Snell's Law on mulitple curved lens surfaces. Handy not to have to do all that refraction calculations! A biconcave lens is also called a negative or diverging lens. Parallel light rays will diverge, coming together at the near focal point, not the far focal point -- you can only see the bright spot by looking through the lens. Ray tracing gets same results as doing Snell's Law on mulitple curved lens surfaces. Handy not to have to do all that refraction calculations! Real image formed by passing three rays through a positive thin lens. Physical Optics. Based on wave properties of light. Constructive and Destructive Interference. Anti-Reflection coating for lenses. Step 1 - A thin coating of thickness t applied to a glass surface, means that light rays coming in from the air make two reflections (air to coating, and coating to glass). If the roundtrip distance of the 2nd reflection is out of phase with the 1st reflection (off by ½ wavelength) then the two reflections can cancel each other. Thickness = half the round-trip distance, yields a "quarter-wave coating". Step 2 - But the roundtrip distance is done in the coating, so need to find the wavelength in the coating, not the air or glass. Step 3 - When a light ray goes from a low index of refraction to a higher index of refraction, the reflection gains a ½ wavelength phase shift. If both reflections are the same kind (low to high, or high to low), then we still get the same quarterwave anti-reflection solution. If both reflections are different, we get halfwave anti-reflection coatings. Back to Anti-Reflection and Max-Reflection coatings with 0, 1 or 2 half-wavelength shifts upon reflections. Q19-20 Double Take-Home Quiz, due Friday 10 April 2009.
Thursday 4/9: White light getting two reflections off a thin coating of oil on water -- wavelength associated with maximum reflection will be the strongest irridescent color. Modern Physics -- goes to size/time/length scales far outside our normal experience. Classical Relativity (two observers, two frames of reference), Special Relativity (speed constant), General Relativity (accelerations or gravity). Einstein's postulates: (1) All observers see the same Physics laws. (2) All observers measure the speed of light in vacuum as c. Beta, gamma, Length Contraction and Time Dilation. No preferred observer in Special Relativity. Two observers cannot agree on what they see, distance or time. One sees the proper length: a length measurement where both ends are measured at the same time. First set of Sample Final Exams handed out. (Click here and here for a copy.) FIRST DAY to turn in Topic 1 Book Reports.
Friday 4/10: Einstein's postulates: (1) All observers see the same Physics laws. (2) All observers measure the speed of light in vacuum as c. Beta, gamma, Length Contraction and Time Dilation. Alpha Centauri is 4.20 LY from Earth (proper length). Those on a starship see a different distance and experience a different time than the observer left on the Earth. But both think the other observer is moving at v < c. No preferred observer in Special Relativity. Two observers cannot agree on what they see, distance or time. One sees the proper length: a length measurement where both ends are measured at the same time. One sees the proper time: a time measurement where beginning and end are measured at the same place. Experimental confirmation of Special Relativity: put atomic clocks on aircraft, spacecraft. Difference in time with identical clocks left on the ground. Two observers cannot agree on the order of events, either. The concept of "simultaneity" is gone. SECOND DAY to turn in Topic 1 Book Reports.
Week of April 13-17, 2009.
Monday 4/13: Another confirmation of Special Relativity: Muons (a form of heavy electron) are created in the upper atmosphere -- they're unstable and will decay. Muons measured at mountaintop -- by sea level, nearly all should have already decayed. But you detect almost as many at sea level as on the mountaintop, because the muon lifetime is measured in the muon's rest frame not while we are watching it moving. The Correspondence Principle -- at some point our Classical Physics results need to match the Modern Physics results. So when do we need Special Relativity? For eyeball measurements, we have trouble distiguishing the size of things that are only off by 10%. That would correspond to a gamma = 1.10, and a beta = 0.417 c. Relativistic momentum, prel = gamma mv, and relativistic Kinetic Energy, KErel = (gamma - 1) mc². Total Energy, Etotal = gamma mc². The Einstein Relation, E = m c², and conversion between energy and matter (mass). Pair creation of an electron-positron pair from two high energy gamma rays. Once we used to talk of a "relativistic mass", to try to explain why an object of matter cannot be accelerated up to the speed of light in vacuum, c. You can just as easily use the Work-Energy theorem to show that it would take infinite work to get a matter object up to c. (British SF writer Charlie Stross talks about the difficulties of space travel, including traveling to another star here.) LAST DAY to turn in Topic 1 Papers. Second set of Sample Final Exams handed out. (Click here and here for a copy.) Q21-22 Double Take-Home, due Wednesday 15 April 2009.
Tuesday 4/14: All models of the atom are fundamentally wrong at some level. It's the nature of Quantum Mechanics, which operates at the land of the very small. But we've already used the "planetary" model of the atom back when we looked at the electron in orbit of a hydrogen atom, using U.C.M. and Coulomb's Law. Chemistry suggests a finite number of elements or starting blocks for all materials. Everything made of individual atoms, as opposed to say the Velveeta cheese model, where you can forever slice the Velveeta ever finer. Plum pudding model of the atom -- some hard bits in a matrix. Rutherford experiment -- firing particles at a very thin piece of gold foil (discussion of gold leaf), some of the particles rebounded at nearly 180°. "It was as if I had fired cannonballs at a piece of paper and some bounced back towards me." Better model of the atom: electrons on the outside (size of the atom is about 1 angstrom = 1 ×10-10 meters) and protons (and neutrons, not yet discovered) concentrated into a nucleus (about 1 femtometer = 1 ×10-15 meters). The Bohr atom is really quite a triumph of the Physics of PHYS-1130 and PHYS-1150: We showed early in the semester that the gravitational attraction between an electron and a proton doesn't matter in the hydrogen atom, so if we have an electron in a circular orbit about the proton (or better yet, the nucleus with a total proton charge Q = +Z e, so we can do elements other than hydrogen), then the Coulomb Force provides the centripetal force for Uniform Circular Motion (UCM). That allows us to find the speed v as a function of the radius r. The deBroglie wavelength -- Wave/Particle Duality for Matter. Planck's constant -- a very small number, but it is NOT zero ( h = 0 in Classical Physics). So the deBroglie wavelength only matters for very small objects, not Buicks. NOTE: This material is covered more fully in an Atomic & Nuclear Physics handout, along with the Periodic Table with equations sheet which will be handed out on Wednesday. (Click here for the handout (not given in class) and here for the Periodic Table.)
Wednesday 4/15: Return X3. Some comments about the Periodic Table of Elements. Niels Bohr postulated that the electron could only exist in certain orbits, so he proposed that the angular momentum (mvr) can only have integer values of h-bar. This introduces the principle quantum number n. This also means that the circumference of the orbital is some integer of the electron's deBroglie wavelength -- and we have a circular standing wave only for those orbits which are allowed. By the time we get the radius equation, there are only two variables (and both are integers!): the quantum state number n and the proton/element number Z. All the other items are fundamental constants. NOTE: This material is covered more fully in an Atomic & Nuclear Physics handout, along with the Periodic Table with equations sheet which will be handed out on Wednesday. (Click here for the handout (not given in class) and here for the Periodic Table which was handed out 4/15.)
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# Integration by Parts
In this tutorial we shall develop a technique that will help us to evaluate a wide variety of integrals that do not fit any of the basic integration formulas. The technique of integration that is quite useful here is integration by parts. It depends upon the formula for the derivative of a product.
$\frac{d}{{dx}}\left[ {f\left( x \right)\int {g\left( x \right)dx} } \right] = f’\left( x \right)\int {g\left( x \right)dx + f\left( x \right)\frac{d}{{dx}}\left[ {\int {g\left( x \right)dx} } \right]}$
$\frac{d}{{dx}}\left[ {f\left( x \right)\int {g\left( x \right)dx} } \right] = f’\left( x \right)\int {g\left( x \right)dx + f\left( x \right)g\left( x \right)}$
Integrating both sides, we have
$\begin{gathered} f\left( x \right)\int {g\left( x \right)dx = \int {\left[ {f’\left( x \right)\int {g\left( x \right)dx} } \right]dx + \int {f\left( x \right)g\left( x \right)dx} } } \\ \Rightarrow \int {f\left( x \right)g\left( x \right)dx} = f\left( x \right)\int {g\left( x \right)dx – \int {\left[ {f’\left( x \right)\int {g\left( x \right)dx} } \right]dx} } \,\,\,{\text{ – – – }}\left( {\text{i}} \right) \\ \end{gathered}$
Formula (i) is called the formula for integration by parts. With the help of this formula we integrate the product of two functions. The basic idea here is that we shall consider the two functions as the first and second functions.
The function whose integration can easily be found is considered the second function while the other is considered the first function. In formula (i), $f\left( x \right)$ and $g\left( x \right)$ are considered the first and second functions, respectively. Sometimes both functions are easily integrable; in this case we may choose either of them as the first function provided that the integrand does not become complicated.
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ENG RUS Timus Online Judge
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## 2109. Tourism on Mars
Time limit: 4.0 second
Memory limit: 300 MB
Few people know, but a long time ago a developed state existed on Mars. It consisted of n cities, numbered by integers from 1 to n, the capital had the number 1. Some pairs of cities were connected by a road. The residents of the state were very prudent, therefore, between any two cities, there was exactly one path (possibly consisting of several roads).
Due to the fact that the state was developed, its residents loved traveling. Tourist route on Mars was described by two numbers L and R. This meant that the tourist started the route in the city L, then went to the city L + 1 (without going into the cities, that did not lie on the path between L and L + 1), then went to the city L + 2, and so on. The last city on the route was the city R. A city that was the closest to the capital among all cities visited on this route (if to count a distance between the cities by the roads) was considered the main attraction of the route.
Knowing the map of the Martian state and all the tourist routes, find for each route its main attraction.
### Input
The first line contains an integer n that is the number of cities (1 ≤ n ≤ 2 · 105).
The following n − 1 lines describe the roads. Each road is described by two numbers of cities that are connected by it (1 ≤ viui ≤ n; vi ≠ ui).
The (n + 1)-th line contains an integer q that is the number of tourist routes (0 ≤ q ≤ 106).
Then q lines describe the routes themselves. Each route is described by a pair of integers Li, Ri (1 ≤ LiRin).
### Output
Output q integers, one per line — for each route the number of its main attraction. These numbers should be output in the same order in which the respective routes were described.
inputoutput
```7
1 2
1 3
2 4
2 5
2 6
3 7
3
4 6
3 4
5 7
```
```2
1
1
```
```7
1 3
3 5
5 6
7 5
1 4
2 4
3
4 5
5 6
6 7
```
```1
5
5
```
### Notes
This problem has a big input and output data size and a strict Time Limit. If you write your solution in C++ we recommend you to use Visual C++ 2013 compiler.
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Structural Analysis (Aero 103)
## Introduction
In the preceding chapter we learned to design beams for strength. In this chapter we will be concerned with another aspect in the design of beams, namely, the determination of deflection. Of particular interest is the determination of the maximum deflection of a beam under a given loading, since the design specifications of a beam will generally include a maximum allowable value for its deflection. Also of interest is that the knowledge of the deflections is required to analyze indeterminate beams. These are beams in which the number of reactions at the support exceeds the number of equilibrium equations available to determine these unknowns.
We saw before that a prismatic beam subjected to pure bending is bent into an arc of circle and that, within the elastic rang, the curvature of the neutral surface can be expressed as
Where M is the bending moment, E is the modulus of elasticity, and the moment of inertia of the cross section about its neutral axis.
When a beam is subjected to transverse loading, remains valid for any given transverse section, provided that Saint-Venant’s principle applies. However, both the bending moment and the curvature of the neutral surface will vary from section to section. Denoting by the distance of the section from the left end of the beam, we write
The knowledge of the curvature at various points of the beam will enable us to draw some general conclusions regarding the deformation of the beam under loading
To determine the slope and deflection of the beam at any given point, we first derive the following second-order linear differential equation, which governs the elastic curve characterizing the shape of the deformed beam
Where is the bending moment in the beam. We now recall from RELATIONS AMONG LOAD, SHEAR, AND BENDING MOMENT that when a beam supports a distributed load , we have and at any point of the beam. Differentiating both members of the last equation w.r.t. and assuming to be constant. We have therefore
And, differentiate again,
We conclude that, when a prismatic beam supports a distributed load , its elastic curve is governed by the fourth-order linear differential equation
Multiplying both members of the last equation by the constant and integrating four times, we write
This means that you have to integrate four times , sound like fun !
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## Physics demonstrations: invisibility on the cheap!
I spend a lot of time talking about invisibility on this blog, and it really has become a fascinating and vibrant area of optics, with lots of remarkable results. However, most of those results are theoretical, and the experimental results which do exist are very limited, and not typically done for visible light (with a few notable exceptions).
While we wait for our own personal invisibility cloaks, however, we can do some cute invisibility tricks at home to demonstrate some nice optics! In the video below, I show versions of the same disappearing act, each of which is quite inexpensive and can be done with very simple ingredients.
The first demonstration is the simplest: all you need are a glass of water and a set of clear water gems, the latter of which can be bought at craft stores.
Drop the gems into the glass of water and — presto! — they seem to disappear once they vanish below the surface. If you look very hard, you can spot the faint outlines of the gems in the water, but the illusion is very effective.
The second demonstration involves only a little more effort, requiring a glass of mineral oil, which can be found at a drug store, and one or more Pyrex (borosilicate) glass stirring rods. The rods will appear to disappear below the surface of the mineral oil, though again a faint outline of the rods can be seen if you look hard.
Why do these beads and rods seem to disappear in their respective liquids? The trick in both cases is that the objects are index-matched: they have a refractive index that is almost the same as the liquid they are immersed in.
The refractive index is one of the oldest quantities introduced and measured in optical science: it is a simple number that indicates, through Snell’s law, how strongly a ray of light gets bent when it passes a surface from one medium to another. Snell’s law is a trigonometric relationship between the angles of the incident and transmitted rays and the refractive indices of the media, given as
$n_1\sin\theta_1=n_2\sin\theta_2$.
This is illustrated below.
The refractive index of absolute vacuum is n=1. For visible light, the refractive index of water is roughly n = 1.33, and the refractive index of common glass is usually between n =1.5 and n=2.0.
There’s a lot we can say about refraction, and I have said a lot in an earlier post. It turns out that, when we consider the phenomenon of refraction from the perspective of the wave properties of light, the refractive index also indicates the degree by which the speed of light is slowed in the medium. Labeling the speed of light in vacuum as c, the speed of light in matter is therefore c/n. In common glass, then, the speed of light is c/1.5, or 2/3rds the vacuum speed of light.
When light passes through a material with a different refractive index, it therefore gets deflected from its original trajectory. Furthermore, some amount of light is always reflected at the interface between two different index media. Therefore, even though an object might be transparent, it is still visible due to distortions of light and reflections. It is for these reason that we can make glass doors and building facades without people constantly crashing into them (well, most of the time).
However, if we submerge one transparent object into a medium which has the same refractive index, or almost the same index, there will be no refraction or reflection! The object becomes almost completely invisible.
In the case of the water gems, this index matching works because the gems themselves are almost entirely made out of water. They are fashioned from a carbon-based polymer that can soak up 300 times their own weight. Therefore, even though they are solid, they have almost the same optical properties as water. If you’re interested in seeing how much water they can soak up, you can purchase dry “phantom crystals” at some science supply stores. These can be soaked overnight and swell to remarkable sizes.
The mineral oil-Pyrex invisibility relies on the remarkable coincidence that the oil and the glass have almost the same refractive index over the entire visible light spectrum. As noted in the video, the only challenge in this demo is buying a huge amount of mineral oil laxative at your local drug store without getting embarrassed.
Index matching forms the basis of one of the classic works of science fiction, H.G. Wells’ 1897 The Invisible Man!
The titular man describes the process to a colleague late in the book:
“If a sheet of glass is smashed, Kemp and beaten into a powder, it becomes much more visible while it is in the air; it becomes at last an opaque white powder. This is because the powdering multiplies the surfaces of the glass at which refraction and reflection occur. In the sheet of glass there are only two surfaces; in the powder the light is reflected or refracted by each grain it passes through, and very little gets right through the powder. But if the white powdered glass is put into water, it forthwith vanishes. The powdered glass and water have much the same refractive index; that is, the light undergoes very little refraction or reflection in passing from one to the other.
“You make the glass invisible by putting it into a liquid of nearly the same refractive index; a transparent thing becomes invisible if it is put in any medium of almost the same refractive index. And if you will consider only a second, you will see also that the powder of glass might be made to vanish in air, if its refractive index could be made the same as that of air; for then there would be no refraction or reflection as the light passed from glass to air.”
“Yes, yes,” said Kemp. “But a man’s not powdered glass!” “No,” said Griffin. “He’s more transparent!”
Both of the demonstrations suggested require the submersion of the object to be hidden in an appropriate liquid. In fact, many undersea creatures have evolved a remarkable degree of transparency as a defense against predators, as this article from Scientific American demonstrates (pdf). You might wonder: is it possible to find materials which are index matched to air and therefore can be hidden in plain sight, such as on one’s desktop? The problem is that air is a very tenuous medium, and the refractive index of our atmosphere is very, very close to that of vacuum. We would need a material that does not inhibit light at all, and in fact allows it to travel unimpeded at “Einstein’s speed limit,” a criterion that seems unlikely to ever be achieved.
The absolute speed limit of light in a vacuum turns out to be a limitation for more sophisticated cloaking devices, as well. Looking at the diagram of one of the original 2006 spherical cloaking designs shown below, we can see that rays passing close to the central hidden region must take a longer path than rays that would travel in straight lines outside the cloak.
However, if our cloak is trying to hide things in air, this means that light traveling near the center must travel faster than the speed of light in air in order to keep up with those rays traveling outside; otherwise, the total light field would be distorted and in principle detectable. I only note at this time that it is possible to “break” the vacuum speed of light at a single frequency, but it is not possible to break it for a large range of frequencies. Therefore the cloak pictured above will only work for a small number of colors. It might appear invisible to red, but not blue, for instance!
The theory of optical cloaking is still in its infancy, however, and it is possible that someone in the future come up with a clever and surprising new way to hide objects for all frequencies. Until that day arrives, however, at least we have a few simple clever tricks to demonstrate what such invisibility might look like if we lived underwater!
This entry was posted in Invisibility, Optics, Physics demos. Bookmark the permalink.
### 4 Responses to Physics demonstrations: invisibility on the cheap!
1. Laura says:
great! i have those water gems in my lab right now because we were doing some index matching for tomogrpahy projects….here’s my version of index matching:
• Nice! I have to try some more plastic shapes in a mineral oil bath.
2. Test QI says:
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Keep on posting!
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# f(x) = x^3 - 2x^2 - 5x - 12 / 4x^2 - 9?
asked Feb 3, 2013 in CALCULUS
# f(x) = x^3 - 2x^2 - 5x - 12 / 4x^2 - 9
Recall the quotient rule d/dx(u/v)= [ u'·v -u·v' ] / v2(x)
Here u = x3-2x2-5x-12;v = 4x2-9
u'(x) =3x2-4x-5;v'(x) =8x
f'(x)=(3x2-4x-5)(4x2-9)-(x3-2x2-5x-12)(8x)/(4x2-9)^2
=12x4-27x2-16x3-36x-20x2+45-8x4+16x3-40x2-96x/16x4-72x2+81
=4x4-87x2-132x+45/16x4-72x2+81
The derivative of the given function is 4x4-87x2-132x+45/16x4-72x2+81
Hope that helps u
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https://www.dummies.com/article/business-careers-money/business/accounting/general-accounting/how-profits-are-measured-168118/
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Managerial Accounting For Dummies
To earn money and succeed, a business — whether it’s a service company, a retailer, or a manufacturer — needs to create profits. To generate profits, companies must make sales to customers that exceed however much those sales cost. The three key components of profits are revenues, the cost of sales, and other types expenses. Companies use this information to compute net income.
Revenues, a component of profits
Revenues are inflows from customers. They’re the actual amounts that your customers give you or, in some cases, promise to give you. The term sales is a synonym for revenues.
For example, an airline’s revenues come from selling tickets (and charging baggage fees, handling fees, boarding fees, carry-on fees, and late-arrival fees). A clothing retailer’s revenues come from peddling clothing to customers. A bakery’s revenues come from getting tasty treats to hungry customers.
One caveat: A company obviously can’t record revenue until it knows the actual price and feels assured that the customer will pay. For example, if Michelangelo’s contract requires a team of evaluators to judge a completed work in order to set its price, he can’t record revenue until the actual price is set.
Revenues come only when you sell whatever you usually sell to your customers. If you sell a product you normally don’t sell (say, your delivery truck or the naming rights to your building), you record the inflow as a gain or loss rather than as a revenue.
The cost of sales as a component of profits
Cost of sales is the cost of buying or making the products that you sell. For example, suppose that a soft-drink bottler sells a can of cola for \$0.50. Consider what goes into that can of cola:
• Water
• Caramel color
• Carbonation
• Sweeteners
• Natural flavors
• Caffeine
• Various preservatives
• Aluminum
• Labor
Now, given all these (inexpensive) ingredients and costs, consider how much it probably costs to produce a can of cola. Would you say \$0.15 per can?
Because they don’t manufacture the goods themselves, when retailers measure cost of sales, they base it on the amount of money they paid to purchase the goods from a supplier.
Many manufacturers and retailers use the term cost of goods sold instead of cost of sales. This point often confuses students, so don’t be alarmed if your head is spinning a little. Just know that regardless of the name, these two concepts are largely the same thing.
Don’t confuse the accounting treatment for cost of goods sold with the cost of unsold goods. When buying or making goods, companies record the cost of unsold goods on the balance as inventory — an asset. Then, when the goods are sold, their cost is taken out of inventory and put into cost of sales, which is subtracted from revenue on the income statement.
Say you own a convenience store that sold 1,000 bags of chips last year. The chips cost you \$0.50 per bag, for a total cost of \$500, but you sold them for \$0.95 per bag, for total revenues of \$950. Use this information to construct the beginning of an income statement (where all revenue and expenses are reported).
Operating expenses are a component of profits
Sales don’t just materialize by themselves. To bring in sales, you must pay employees, advertisers, and other marketers, and that’s just the beginning. Typical expenses for a company include the following:
• Selling, general, and administrative expenses: Costs such as sales commissions and managerial salaries needed to generate sales and run the company
• Research and development expense: Costs of finding and advancing innovative technologies needed to create new products
• Interest expense: Costs of borrowing money, which companies must pay to their creditors
• Income tax expense (also known as provision for income taxes): The portion of profits companies are usually required to pay to government authorities
Although cost of sales is always an expense, not every expense qualifies as cost of sales. Cost of sales is the actual cost of producing or buying the item you sell. However, your company may need to pay additional expenses, such as sales commissions and advertising, to sell this item.
The distinction between cost of sales and other expenses is important for determining gross profit, the difference between revenues and cost of sales. When calculating gross profit, you deduct cost of sales but not operating or other expenses.
How net income is measured
Net income (also known as profit) is the difference between revenues and all expenses, including cost of sales. Investors and managers often refer to net income because it provides a single bottom-line number to measure a company’s performance. (Of course, that figure is only considered profit when it’s a positive number; a negative net income is called net loss.)
You can use this formula to describe net income:
Net Income = Revenues – Expenses
Suppose Al’s Used Cars earns \$1 million in revenue and has \$600,000 in cost of goods sold and \$300,000 in other expenses. To compute the net income, start with revenue and subtract all expenses, including cost of goods sold:
Net Income = Revenues – Expenses
Net Income = \$1,000,000 – \$600,000 – \$300,000
Net Income = \$100,000
This net income formula sets the structure for the income statement, where you start with revenues and then subtract from it cost of sales and all other expenses.
Having positive net income doesn’t necessarily guarantee positive cash flow. Even a profitable company may have to pay cash for additional costs not recorded as expenses, such as for new equipment or increased stores of inventory (assets, not expenses).
These payments require valuable cash flow but don’t affect net income. Similarly, companies don’t always collect sales immediately. Such a sale made on credit creates sales revenues, which help net income but not cash flows.
How return on sales is determined
Return on sales asks how much profit your company can squeeze out of its sales. Some companies, such as retailer Costco or the paper manufacturer Weyerhaeuser Company, benefit from a low-price, high-volume strategy. They keep profit margins very low so that they can charge customers the lowest prices possible. Because these companies usually yield relatively little profit from their sales, they usually have to make it up in volume to earn a healthy income.
Other companies, such as leather goods maker Coach or semiconductor manufacturer Intel, develop a high-price, low-volume strategy. These companies differentiate their products so that they can charge customers premium prices that yield rich profit margins and fairly high profits from their sales.
So how do you measure return on sales? Here’s the formula:
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# Re: Querying distances between two coordinates
From: Charles Hooper <hooperc2001_at_gmail.com>
Date: Mon, 13 Jun 2011 12:48:33 -0700 (PDT)
On Jun 11, 12:37 pm, Jeremy <jeremy0..._at_gmail.com> wrote:
> In article <ba7de41c-7ce8-465a-bc20-67c4e2484871
> > On Jun 10, 12:15 pm, Jeremy <jeremy0..._at_gmail.com> wrote:
> > > Hi this is a 10gR2 Standard Edition question.
>
> > > The requirement is to be able to calculate the distance between two co-
> > > oordinates expressed as latitude and longitude.
>
> > I vaguely remember teaching a class how to do this calculation of
> > distance around a spherical object in the early 1990s (with only a
> > calculator) - I could probably search and find the lesson plan which
> > is probably right next to the proof that 1=2. While I am searching
> > for that lesson plan, see if the following article is helpful:
> >http://www.movable-type.co.uk/scripts/latlong.html
>
> Thanks. Of course what I have failed to do is actually to ask the real
> question I wanted to answer - ultimately we need to be able to query a
> table to find the rows where the lat/long defined on those rows is
> 10gR2 SE that provide this capability efficiently?
>
> --
> jeremy
Have a look at the following article:
http://hoopercharles.wordpress.com/2011/06/13/calculate-the-distance-between-two-latitudelongitude-points-using-plain-sql/
An up-front warning - the method is CPU intensive if there are many rows in the database table.
Charles Hooper
Co-author of "Expert Oracle Practices: Oracle Database Administration from the Oak Table"
http://hoopercharles.wordpress.com/
IT Manager/Oracle DBA
K&M Machine-Fabricating, Inc. Received on Mon Jun 13 2011 - 14:48:33 CDT
Original text of this message
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# Interactive Real Analysis
Next | Previous | Glossary | Map
## 4.1. Series and Convergence
### Example 4.1.1: Zeno Paradox (Achilles and the Tortoise)
Achilles is racing against a tortoise. Achilles can run 10 meters per second, the tortoise only 5 meter per second. The track is 100 meters long. Achilles, being a fair sportsman, gives the tortoise 10 meter advantage. Who will win ?
Let us look at the difference between Achilles and the tortoise:
Time Difference t = 0 10 meters t = 1 5 = 10 / 2 meters t = 1 + 1/2 2.5 = 10 / 4 meters t = 1 + 1/2 + 1/4 1.25 = 10 / 8 meters t = 1 + 1/2 + 1/4 + 1/8 0.625 = 10 / 16 meters
and so on. In general we have:
Time Difference t = 1 + 1 / 2 + 1 / 2 2 + 1 / 2 3 + ... + 1 / 2 n 10 / 2 n meters
Now we want to take the limit as n goes to infinity to find out when the distance between Achilles and the tortoise is zero. But that involves adding infinitely many numbers in the above expression for the time, and we don't know how to do that. However, if we define
S n = 1 + 1 / 2 + 1 / 2 2 + 1 / 2 3 + ... + 1 / 2 n
then, dividing by 2 and subtracting the two expressions:
S n - 1/2 S n = 1 - 1 / 2 n+1
or equivalently, solving for S n:
S n = 2 ( 1 - 1 / 2 n+1)
But now S n is a simple sequence, for which we know how to take limits. In fact, from the last expression it is clear that
lim S n = 2
as n approaches infinity. Hence, we have - mathematically correct - computed that Achilles reaches the tortoise after exactly 2 seconds, and then, of course passes it and wins the race.
A much simpler calculation not involving infinitely many numbers gives the same result:
• Achilles runs 10 meters per second, so he covers 20 meters in 2 seconds
• The tortoise runs 5 meters per second, and has an advantage of 10 meters. Therefore, it also reaches the 20 meter mark after 2 seconds
• Therefore, both are even after 2 seconds
Of course, Achilles will finish the race after 10 seconds, while the tortoise needs 18 seconds to finish, and Achilles will clearly win.
The problem with Zeno's paradox is that Zeno was uncomfortable with adding infinitely many numbers. In fact, his basic argument was: if you add infinitely many numbers, then - no matter what those numbers are - you must get infinity. If that was true, it would take Achilles infinitely long to reach the tortoise, and he would loose the race. However, reducing the infinite addition to the limit of a sequence, we have seen that this argument is false.
Next | Previous | Glossary | Map
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Difficult Problems
1
Here, we show you a step-by-step solved example of limits of exponential functions. This solution was automatically generated by our smart calculator:
$\lim_{x\to0}\left(1+3sinx\right)^{\frac{1}{x}}$
2
Rewrite the limit using the identity: $a^x=e^{x\ln\left(a\right)}$
$\lim_{x\to0}\left(e^{\frac{1}{x}\ln\left(1+3\sin\left(x\right)\right)}\right)$
Multiplying the fraction by $\ln\left(1+3\sin\left(x\right)\right)$
$\lim_{x\to0}\left(e^{\frac{1\ln\left(1+3\sin\left(x\right)\right)}{x}}\right)$
Any expression multiplied by $1$ is equal to itself
$\lim_{x\to0}\left(e^{\frac{\ln\left(1+3\sin\left(x\right)\right)}{x}}\right)$
3
Multiplying the fraction by $\ln\left(1+3\sin\left(x\right)\right)$
$\lim_{x\to0}\left(e^{\frac{\ln\left(1+3\sin\left(x\right)\right)}{x}}\right)$
4
Apply the power rule of limits: $\displaystyle{\lim_{x\to a}f(x)^{g(x)} = \lim_{x\to a}f(x)^{\displaystyle\lim_{x\to a}g(x)}}$
${\left(\lim_{x\to0}\left(e\right)\right)}^{\lim_{x\to0}\left(\frac{\ln\left(1+3\sin\left(x\right)\right)}{x}\right)}$
5
The limit of a constant is just the constant
$e^{\lim_{x\to0}\left(\frac{\ln\left(1+3\sin\left(x\right)\right)}{x}\right)}$
Plug in the value $0$ into the limit
$\frac{\ln\left(1+3\sin\left(0\right)\right)}{0}$
The sine of $0$ equals $0$
$\frac{\ln\left(1+3\cdot 0\right)}{0}$
Multiply $3$ times $0$
$\frac{\ln\left(1+0\right)}{0}$
Add the values $1$ and $0$
$\frac{\ln\left(1\right)}{0}$
Calculating the natural logarithm of $1$
$\frac{0}{0}$
6
If we directly evaluate the limit $\lim_{x\to 0}\left(\frac{\ln\left(1+3\sin\left(x\right)\right)}{x}\right)$ as $x$ tends to $0$, we can see that it gives us an indeterminate form
$\frac{0}{0}$
7
We can solve this limit by applying L'Hôpital's rule, which consists of calculating the derivative of both the numerator and the denominator separately
$\lim_{x\to 0}\left(\frac{\frac{d}{dx}\left(\ln\left(1+3\sin\left(x\right)\right)\right)}{\frac{d}{dx}\left(x\right)}\right)$
Find the derivative of the numerator
$\frac{d}{dx}\left(\ln\left(1+3\sin\left(x\right)\right)\right)$
The derivative of the natural logarithm of a function is equal to the derivative of the function divided by that function. If $f(x)=ln\:a$ (where $a$ is a function of $x$), then $\displaystyle f'(x)=\frac{a'}{a}$
$\frac{1}{1+3\sin\left(x\right)}\frac{d}{dx}\left(1+3\sin\left(x\right)\right)$
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{1}{1+3\sin\left(x\right)}\frac{d}{dx}\left(3\sin\left(x\right)\right)$
The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function
$3\left(\frac{1}{1+3\sin\left(x\right)}\right)\frac{d}{dx}\left(\sin\left(x\right)\right)$
The derivative of the sine of a function is equal to the cosine of that function times the derivative of that function, in other words, if ${f(x) = \sin(x)}$, then ${f'(x) = \cos(x)\cdot D_x(x)}$
$3\left(\frac{1}{1+3\sin\left(x\right)}\right)\cos\left(x\right)$
Multiplying the fraction by $3\cos\left(x\right)$
$\frac{3\cos\left(x\right)}{1+3\sin\left(x\right)}$
Find the derivative of the denominator
$\frac{d}{dx}\left(x\right)$
The derivative of the linear function is equal to $1$
$1$
Any expression divided by one ($1$) is equal to that same expression
$e^{\lim_{x\to0}\left(\frac{3\cos\left(x\right)}{1+3\sin\left(x\right)}\right)}$
8
After deriving both the numerator and denominator, the limit results in
$e^{\lim_{x\to0}\left(\frac{3\cos\left(x\right)}{1+3\sin\left(x\right)}\right)}$
Evaluate the limit $\lim_{x\to0}\left(\frac{3\cos\left(x\right)}{1+3\sin\left(x\right)}\right)$ by replacing all occurrences of $x$ by $0$
$e^{\frac{3\cos\left(0\right)}{1+3\sin\left(0\right)}}$
The sine of $0$ equals $0$
$e^{\frac{3\cos\left(0\right)}{1+3\cdot 0}}$
Multiply $3$ times $0$
$e^{\frac{3\cos\left(0\right)}{1+0}}$
Add the values $1$ and $0$
$e^{\frac{3\cos\left(0\right)}{1}}$
The cosine of $0$ equals $1$
$e^{\frac{3\cdot 1}{1}}$
Multiply $3$ times $1$
$e^{\frac{3}{1}}$
Divide $3$ by $1$
$e^{3}$
9
Evaluate the limit $\lim_{x\to0}\left(\frac{3\cos\left(x\right)}{1+3\sin\left(x\right)}\right)$ by replacing all occurrences of $x$ by $0$
$e^{3}$
Final answer to the problem
$e^{3}$
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# 3.4 Perpendicular Lines Worksheet Answers
Friday, August 5th 2022. | Worksheet
3.4 Perpendicular Lines Worksheet Answers - There are a lot of affordable templates out there, but it can be easy to feel like a lot of the best cost a amount of money, require best special design template. Making the best template format choice is way to your template success. And if at this time you are looking for information and ideas regarding the 3.4 Perpendicular Lines Worksheet Answers then, you are in the perfect place. Get this 3.4 Perpendicular Lines Worksheet Answers for free here. We hope this post 3.4 Perpendicular Lines Worksheet Answers inspired you and help you what you are looking for.
3.4 Perpendicular Lines Worksheet Answers. Lines a and b b. Not enough information to know for sure.
Label points on the two creases, as shown. Yes, because it is perpendicular to l. Yes, it is parallel to l and l is perpendicular to n.
### Identify The Highlighted Pair Of Lines As Parallel Or Perpendicular In Each Object.
Proving lines are parallel interactive notebook page for geometry algebra worksheets teaching geometry geometry worksheets 140 1 70 2 3. Chapter 3 parallel and perpendicular lines worksheet answers. All spreadsheets have been carefully developed for all types of students that you can download in class lines 7 pdf cbse class and angles question bank.
### A) Parallel B) Perpendicular C) Intersecting A) Parallel B) Perpendicular C) Intersecting A) Parallel B) Perpendicular C) Intersecting A) Parallel B) Perpendicular C) Intersecting A) Parallel B) Perpendicular C) Intersecting A) Parallel B) Perpendicular C) Intersecting
Section 3.4 proofs with perpendicular lines 147 3.4 proofs with perpendicular lines writing conjectures work with a partner. Write a conjecture about ao — and ob —. Which lines or planes appear to fit the description.
### If A Transversal Is Perpendicular To One Of Two Parallel Lines It Is Perpendicular To The Other Line.
Write a conjecture about ao — and ob —. Two lines in the same plane are parallel. Parallel and perpendicular lines last modified by:
### Printable Worksheets @ Www.mathworksheets4Kids.com Name :
Geometry chapter 3 parallel and perpendicular lines worksheet answers. Based on this, tell how the lines listed below are related. Write and graph the equation that goes through given point a that is parallel to the original line.
### Perpendicular Lines Name Homework 6.
A sample problem is solved. Two parallel lines are coplanar. Find angle measures using the theorem.
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Super-logarithm on the imaginary line andydude Long Time Fellow Posts: 510 Threads: 44 Joined: Aug 2007 11/15/2007, 08:40 AM I just found a way to calculate slog on the imaginary axis! It depends very much on Jay's observation that slog is imaginary-periodic. Let $S(x) = \text{slog}_e(i x)$. S is periodic with period $2\pi$, because $S(x + 2\pi) = \text{slog}(i (x + 2\pi)) = \text{slog}(i x + 2i\pi) = \text{slog}(i x) = S(x)$. Since S is periodic, we can use Fourier series to represent it. Let $R(x) = S(-i\ln(x))$, then $R(e^{ix}) = S(x)$. The Taylor series coefficients of R will then be the Fourier series coefficients of S. In terms of the super-logarithm, $R(x) = S(-i\ln(x)) = \text{slog}(-ii\ln(x)) = \text{slog}(\ln(x)) = \text{slog}(x) - 1$. This means the Fourier series coefficients of $\text{slog}(ix)$ are the Taylor series coefficients of $\text{slog}(x) - 1$ which we already know. In other words, $\text{slog}(ix) = \text{slog}(e^{ix}) - 1$, so: $ \text{slog}_e(ix)_4 = -2 + \frac{12}{13}e^{ix} + \frac{16}{65}e^{2ix} - \frac{12}{65}e^{3ix} + \frac{1}{65}e^{4ix}$ The nice thing about this is that it seems to bypass the radius of convergence problem near z=i since its a Fourier series and not a Taylor series. Is this right? I've included a plot with multiple approximations, which seem to converge much faster than doing analytic continuation section-by-section. The top line is the imaginary part, and the bottom line is the real part, and the "y" axis is $\text{slog}(ix)$: PDF version Andrew Robbins jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 11/15/2007, 09:08 AM Actually, if you think about it, we're still limited by the same radius of convergence, because it's the same series. HOWEVER, because we're plugging e^z into the series, rather than z, we're limited by abs(e^z) < 1.37445, not abs(z)<1.37445. This essentially means that for all complex values with real part less than 0.31813 (real part of primary fixed point), the series will converge. So if you plug e^(iz) into the series, and use only real z, then (because the imaginary part is 0, and hence the real part of iz is 0) you will always be inside the radius of convergence. Effectively, you're calculating points on the unit circle with the original series. ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 11/15/2007, 09:25 AM By the way, now that I see it written down, I see that I made a simple but important mistake a few days ago, by forgetting the i in the exponents for the exponential form of the fourier series. I knew that we had real exponents for slog, but didn't quite link it up with the fact that we're periodic in the imaginary direction. ~ Jay Daniel Fox andydude Long Time Fellow Posts: 510 Threads: 44 Joined: Aug 2007 11/15/2007, 05:52 PM Right, since the real part of all points in the "backbone" of the slog are less than the logarithm of the radius of convergence, the exponential of them is within the radius of convergence (of the series expansion about z=0). Andrew Robbins « Next Oldest | Next Newest »
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Batool Alsamadi Finite Math - Randolph MWF 11:00 - 11:50 am Apr 13, 2007 due
I. Discuss the three ways of visually representing data? Describe them, tell when and how they’re used?
1) Bar Graph -Equally spaced, and each bar has the same width as the rest. -This method is usually used for comparisons. 2) Pie Graph -A circle that is divided into “slices of a pie” where each slice corresponds to a category of the data. The area of each slice is proportional to the percentage of items in that category. This is done by making the center angle of each slice equal to 360 degrees times the percentage associated with the segment. -This method is used to present data in percentiles or figures given in fractions. 3) Histogram -Just like a bar graph; except no spaces. -Can be automatically converted into a line graph by placing a dot in the center of each bar, and then connecting the dots. -Each bar is centered over its corresponding data. This is done by having the bottom of each bar extend a ½ hunt on each side of it’s data. For example, if a bar is representing the data of “12” then the base of the bar extends from 11 ½ to 12 ½. -This method is used for looking for trends. It’s usually used in data that consists a collectin of numbers. For example, the data could consist of ages, weights, test scores, etc.
II. How are medians and quartiles calculated for data?
Medians / Calculation: -The median is the center-most data in a set. The median data is the data that divides the bottom 50% of the rest of the data from the top 50%. The median is also known as the 2nd Quartile. -To find the median in a set, the data must be first arranged in increasing or decreasing numerical order. -The median is the “middle” data if the set has an odd amount of data -If the set has an even amount of data, the median then is the average of the two “middle” data’s. e.g. Let’s say the data we’re given is x: 0 1 2 3 4 5 6 7 8 y: 7 6 4 5 2 1 3 0 8 ….we place the data in order so that: { 0 1 2 3 4 5 6 7 8} *The median in this set is 4. However, in the case of an even amount of data in a set, we would end up with two center-most data’s as the median, so in this case we would find the mean of these two data’s: e.g. {1 2 3 4 5 6 7 8} *Since the medians are both 4 and 5, we find the mean of 4 and 5, so the median of this set is: 4 + 5 = 9 / 2 = 4 ½ The median is 4 ½. Quartiles/ Calculation: -The data must be first arranged in numerical order to compute the quartiles. -Each quartile divides the data into approx. 4 equal parts, each part roughly
consisting of 25%. -Each quartile is used for information about the dispersion of the data. -The Interquartile Range is the middle 50% of where the data lies, or Q3-Q1 (the data that lies in the regions of the first quartile to the third quartile):
-Another way to think about it, Q1 is 25%; Q2 is 50%; Q3 is 75%; and Q4 is 100%. e.g. If the data set below is given: {1 2 3 4 5 6 7 8} *Then, Q1=2; Q2=4; Q3=6; Q4=8.
e.g. If the data set given has a median that arrives at two numbers {the average of the middle two numbers}, the quartiles would be computed this way: {1 4 7 11 16 22 29 37 46 56} *The min=1; the max=56; the next number we should find is the median: 16 + 22 = 38 / 2 = 19. Thus, Q2 (median) = 19. -The to the left of the median are 1, 4, 7, 11, and 16, and the numbers to the right of the median are 22, 29, 37, 46, and 56. -Therefore, Q1=7; Q2=19; Q3=37; Q4=56. -For the interquartile range, these lists have medians 7 and 37, respectively. Therefore, the interquartile range is Q3 - Q1 = 37 - 7 = 30.
III. What is a frequency distribution? A relative frequency distribution? A probability distribution? How are they different? How are they similar? How are they used?
Frequency Distribution -Data collected from surveying and presented in a form where for each possible value of a statistical variable there’ll be a tabulated number of occurrences. -A frequency distribution lists out all the outcomes “x” of the experiment and the number “y” of times each occurred. -A frequency distribution is the most efficient form to analyze data that comes from survey data. -A frequency distribution is different than a relative frequency distribution and a probability distribution because it is used to display and summarize the survey data. Example:
“Table A”
sold Monthly sales
2 3 4 5 6
3 3 14 21 11 21 0 0 11 13
at Dierberg’ s
7 8
5 2
51 13
Relative Frequency Distribution -The relative frequency distribution is different than the frequency distribution and probability distribution because it is used for making comparisons. -The relative frequency distribution is just like frequency distribution, except it uses proportions instead of the actual number of occurrences. -The relative frequency distribution pairs each outcome “x” with its “y” relative frequency. It is calculated for each outcome X1 = Y1/n (“n” is the total number of occurrences); then the relative frequency of outcome X2 is = Y2/n, etc. -The frequency and the relative frequency distributions are both taken directly from the performance of a survey/experiment and the collection of data observed at each trial “x” of the survey/experiment. -The sum of the relative frequency distribution should always add up to 1. “Table B”
Proportion of
Sold Monthly sales
2 3 4 5 6 7 8 3/59 3/59 14/59 21/59 11/59 5/59 2/59 Total:
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God & Science Forum Message Forums: Atm · Astrophotography · Blackholes · Blackholes2 · CCD · Celestron · Domes · Education Eyepieces · Meade · Misc. · God and Science · SETI · Software · UFO · XEphem
Be the first pioneers to continue the Astronomy Discussions at our new Astronomy meeting place...The Space and Astronomy Agora Maybe The Water's Not So Deep! Forum List | Follow Ups | Post Message | Back to Thread Topics | In Response ToPosted by Richard D. Stafford, Ph.D. on March 31, 2003 22:02:28 UTC
Hi Tim,
There are some subtleties to physics not really presented in most lay publications. First, relativity was not invented by Einstein. The idea that physics (at least some aspects of physics) was not dependent upon the coordinate system used to display physical phenomena was first put forth by Galileo. Newton was the guy who really gave significance to the concept; however, at that time, it was presumed that the only reasonable coordinate system was the old Euclidean geometry. That is to say, that the "equations of relativity" (the relationships used to convert from one geometry to another) were a quite straight forward mechanical procedures. There was no "theory of relativity" as Euclidean relativity was held to be a fact and not a theory.
Newton's laws (as stated) were valid in an inertial coordinate system. If one were not in an inertial coordinate system, then one could use the Euclidean relativistic transformations to obtain those laws as they should be stated in a non inertial coordinate system. It is these transformations which give rise to what are often called pseudo forces. When free objects follow curved lines in your coordinate system, under Newton's laws they must be experiencing a force. There are two equivalent ways of viewing this circumstance. You can either say that your "true force" measurements must be made in the inertial frame or one can say that the use of a non-inertial frame produces a force.
In the second case (since the force is entirely due to the chosen coordinate system) that created force is generally referred to as a pseudo force. In many cases, it is actually more convenient to work in a non-inertial coordinate system. In particular, if most objects of interest remain at rest in that non-inertial frame (the earth, the interior of you car, whatever) descriptions of object behavior in the non-inertial frame become useful. Thus we have come to give names to some of these pseudo-forces. Centrifugal force is one and the coriolis force (very important in weather) is another.
Now pseudo-forces have one very important characteristic. Since they are entirely due to the coordinate system (if the coordinate system is not changing, the force vanishes) they can have absolutely nothing to do with any characteristics of the "free object" being observed. If the acceleration is entirely determined by the motion of the coordinate system (as the acceleration is actually a characteristic of the coordinate system being used all objects must display exactly the same acceleration), and the motion is to satisfy Newton's law, F=ma, the pseudo-force clearly must be directly proportional to mass.
Now two very important things came together around the turn of the ninetieth century. First, there were the problems created by the Michelson experiments. Maxwell's equations insisted that the speed of light was not a function of the coordinate system which was taken at the time to indicate that there must exist a unique coordinate system: i.e., that those equations only worked because our absolute velocity was slow as compared to the velocity of light (if we were moving, one would have to make the, very small, corrections required by the Euclidean relativity). Michelson was essentially intending to use those "required" small corrections to determine the exact speed of the earth through space.
We all know what Michelson discovered: the earth was standing absolutely still! Now this kind of blew a lot of physics away. Lorentz was able to come up with a set of relativistic transformations which would take care of the problem but could give no logical reason why they were necessary.
The other issue which had bothered some physicists very much was the fact that gravitational forces seemed to be directly proportional to mass. That is a very strong indicator that gravitational forces are actually pseudo-forces. The problem was, how could one change the coordinate system in order to remove the gravitational forces (i.e., get back to a real inertial system). In Newton's time, the only solution was to come up with a "real" force, but over the years, with the advances in non- Euclidean geometry, it was always hoped that someone might someday realize a transformation which would make gravity a pseudo-force.
At this point it should be evident to you why special relativity is called "special" relativity. Euclidean relativity was general (it told you how to make the relativistic transformations for absolutely any motion of your coordinate system). Einstein's theory of "special" relativity was an explanation of Lorenz's relativistic transformation equations based on the idea that the universe was not described by Euclidean geometry but was instead described by Minkowski geometry. On the other hand, the general transformations implied by Minkowski geometry just didn't work so there had to be an error in the idea somewhere. The solution eventually proposed by Einstein was a variation based on the "space-time" concept first suggested by Minkowski geometry.
I won't even talk about it here as "simple" just isn't an adjective generally applied to Einstein's general theory. In fact, very few physicists even attempt to express any common experiments in "General Relativisticly" correct terms. The actual achievement which makes Einstein's theory so overpowering is that it solved that old problem with gravity. He was able to construct a transformation which made gravity a pseudo-force.
Einstein's success led to the idea "that all physical phenomena must be independent of the coordinate system used to represent the phenomena". This is commonly expressed by the requirement that all phenomena should properly be expressed in what is called "covariant" form. What I was getting at with all this bull is that the idea of an "inertial" coordinate system is, very much, a Newtonian concept.
With regard to your second comment, yes, it is very clear that "time can be reasonably measured in inertial frames of reference". In that case, proper time (Einstein's path length in his geometry) and time (the coordinate in his geometry) are proportional and it makes no difference (mechanically) which you use. And, yes, "the measurement of time becomes more complex in accelerated frames"; however, the path length followed by your "clock" is a truly trivial thing to measure in that same accelerated frame. It is always exactly proportional to the reading on the clock. Doesn't that seem just a little strange to you?
Finally, I am willing to help anyone understand anything that I understand. Ask me a question and I will give you my best answer. Take a look at:
http://home.jam.rr.com/dicksfiles/flaw/Fatalfla.htm
Have fun -- Dick
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# 1279949 (number)
1,279,949 (one million two hundred seventy-nine thousand nine hundred forty-nine) is an odd seven-digits composite number following 1279948 and preceding 1279950. In scientific notation, it is written as 1.279949 × 106. The sum of its digits is 41. It has a total of 2 prime factors and 4 positive divisors. There are 1,163,580 positive integers (up to 1279949) that are relatively prime to 1279949.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 7
• Sum of Digits 41
• Digital Root 5
## Name
Short name 1 million 279 thousand 949 one million two hundred seventy-nine thousand nine hundred forty-nine
## Notation
Scientific notation 1.279949 × 106 1.279949 × 106
## Prime Factorization of 1279949
Prime Factorization 11 × 116359
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 1279949 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,279,949 is 11 × 116359. Since it has a total of 2 prime factors, 1,279,949 is a composite number.
## Divisors of 1279949
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 1.39632e+06 Sum of all the positive divisors of n s(n) 116371 Sum of the proper positive divisors of n A(n) 349080 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1131.35 Returns the nth root of the product of n divisors H(n) 3.66664 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,279,949 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 1,279,949) is 1,396,320, the average is 349,080.
## Other Arithmetic Functions (n = 1279949)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 1163580 Total number of positive integers not greater than n that are coprime to n λ(n) 581790 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 98371 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 1,163,580 positive integers (less than 1,279,949) that are coprime with 1,279,949. And there are approximately 98,371 prime numbers less than or equal to 1,279,949.
## Divisibility of 1279949
m n mod m 2 3 4 5 6 7 8 9 1 2 1 4 5 6 5 5
1,279,949 is not divisible by any number less than or equal to 9.
## Classification of 1279949
• Arithmetic
• Semiprime
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (1279949)
Base System Value
2 Binary 100111000011111001101
3 Ternary 2102000202112
4 Quaternary 10320133031
5 Quinary 311424244
6 Senary 43233405
8 Octal 4703715
10 Decimal 1279949
12 Duodecimal 518865
20 Vigesimal 7jjh9
36 Base36 rfm5
## Basic calculations (n = 1279949)
### Multiplication
n×y
n×2 2559898 3839847 5119796 6399745
### Division
n÷y
n÷2 639974 426650 319987 255990
### Exponentiation
ny
n2 1638269442601 2096901334787707349 2683926766560191233645201 3435289380931950209312941374749
### Nth Root
y√n
2√n 1131.35 108.575 33.6355 16.6509
## 1279949 as geometric shapes
### Circle
Diameter 2.5599e+06 8.04216e+06 5.14678e+12
### Sphere
Volume 8.78348e+18 2.05871e+13 8.04216e+06
### Square
Length = n
Perimeter 5.1198e+06 1.63827e+12 1.81012e+06
### Cube
Length = n
Surface area 9.82962e+12 2.0969e+18 2.21694e+06
### Equilateral Triangle
Length = n
Perimeter 3.83985e+06 7.09391e+11 1.10847e+06
### Triangular Pyramid
Length = n
Surface area 2.83757e+12 2.47122e+17 1.04507e+06
## Cryptographic Hash Functions
md5 ce9abec08fe41360483503d36daea5ba fe02c8c11e031c090e1f9f8c544fd8ab066e6f40 1b7dd7d6579a78b69eb8bf4a732dd821cceebde8f0d78caf110022eee3f97799 512c429cdb4a94f3c065a555b6c566293e48272210f0e9d4c56d209562bc5e94c2fe376eb1da033e1974c007815731c38b96fd4e7dcbb23e89c862a569866b09 a4252416ac0083431ab47b5ca884a514d7f9f07d
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## CBSE Class 12th NCERT Solution
Get Prepared for Engineering Entrance !
## NCERT Solutions for CBSE Class 12 Mathematics ‒ Chapter 3: Matrices (Part VI)
Jul 28, 2017
Find Class 12 Maths NCERT Solutions for Chapter 3 Matrices. In this article, you will find solutions of exercise 3.2 from question number 17 to question number 22. These questions are important for CBSE Class 12 Maths board exam & engineering entrance exams.
## NCERT Exemplar Solutions for CBSE Class 12 Physics ‒ Chapter 4: Moving Charges & Magnetism (Part II)
Jul 28, 2017
Class 12 Physics NCERT Exemplar Solution for Chapter 4 (Moving Charges & Magnetism) are available here. In this article, you will get solutions to multiple choice questions with multiple correct answers (i.e., question number 4.7 to 4.11). These questions are important for various competitive exams.
## NCERT Exemplar Solutions for CBSE Class 12 Physics ‒ Chapter 4: Moving Charges & Magnetism (Part I)
Jul 27, 2017
NCERT Exemplar Solutions for Class12 Physics – Chapter 4 (Moving Charges & Magnetism) are available here. In this article, you will get solutions of multiple choice questions with single correct answer (i.e., question number 4.1 to 4.6). These questions are important for various engineering and medical entrance exams.
## NCERT Solutions for CBSE Class 12 Mathematics ‒ Chapter 3: Matrices (Part V)
Jul 26, 2017
Get NCERT Solutions for Class 12 Mathematics, Chapter 3 (Matrices). In this article, you will find solutions of exercise 3.2 from question number 11 to question number 16. These questions are important for Class 12 Maths board exam & other competitive exams.
## NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 3 (Part V)
Jul 26, 2017
CBSE Class 12 Physics NCERT Exemplar Solutions for Current Electricity (Chapter 3) are available here. In this article, you will get solutions from question number 3.28 to question number 3.31. These are long answer type questions and their solutions are quite complicated. These questions & solutions are important for competitive exams.
## NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 3 (Part IV)
Jul 24, 2017
Class 12 Physics NCERT Exemplar Solutions for Current Electricity (Chapter 3) are available here. From this article, you will get solutions from question number 3.22 to question number 3.27. These are short answer type questions of the chapter. These questions are important for CBSE Class 12th Physics board exam and other competitive exams.
## NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 3 (Part III)
Jul 24, 2017
Find Physics NCERT Exemplar Solution for Class 12 Physics – Chapter 3 (Current Electricity). With this article, you will get solutions from question number 3.12 to question number 3.21. These are basically very short answer types questions of the chapter given in NCERT Exemplar. These questions can be asked in CBSE Class 12th Physics board exam and various engineering & medical entrance exams.
## NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 3 (Part II)
Jul 24, 2017
Class 12 Physics NCERT Exemplar solutions for Chapter 3 (Current Electricity) are available here. In this article, students will find solutions from question number 3.7 to question number 3.11. These questions are multiple choice questions with multiple correct answers. The solutions of this chapter are helpful for the preparation of CBSE Class 12 Physics board exam 2017 and other competitive exams.
## NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 3 (Part I)
Jul 21, 2017
NCERT Exemplar solutions for Class 12 Physics - Chapter 3 (Current Electricity) are available here. In this article, students will find solutions from question number 3.1 to question number 3.6. These questions are multiple choice questions with single correct answer. The solutions of this chapter are helpful for the preparation of CBSE Class 12 Physics board exam.
## NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 2 (Part III)
Jul 21, 2017
Get Class 12 Physics NCERT Exemplar Solutions for Chapter 2: Electrostatic Potential and Capacitance. Here, you will find solutions to very short answer type questions (i.e., question number 2.14 to question number 2.18). These questions are important for CBSE board exams and other competitive exams like NEET etc.
## NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 2 (Part V)
Jul 20, 2017
Find NCERT Exemplar Solutions for Class 12 Physics – Chapter 2 (Electrostatic potential and capacitance). Here, you will find solutions to long answer type questions (i.e. question number 2.24 to question number 2.33). These questions can be asked in CBSE Class 12 Physics board exams and various engineering & medical entrance exams.
## NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 2 (Part IV)
Jul 20, 2017
CBSE Class 12 Physics NCERT Exemplar Solutions for Chapter 2: Electrostatic Potential and Capacitance are available here. In this article, solutions to short answer type questions (i.e., question number 2.19 to question number 2.23) are available. These questions are important for the coming CBSE Class 12 board exams.
## NCERT Solutions for CBSE Class 12 Mathematics ‒ Chapter 3: Matrices (Part IV)
Jul 7, 2017
NCERT Solutions for Class 12 Mathematics, Chapter 3 (Matrices) are available here. In this article, you will find solutions of exercise 3.2 from question number 6 to question number 10. These questions are important for Class 12 Maths board exam and other competitive exams.
## NCERT Solutions for CBSE Class 12 Mathematics ‒ Chapter 3: Matrices (Part III)
Jul 7, 2017
Class 12 Maths NCERT Solutions for Chapter 3: Matrices are available here. With this article, you will get solutions of exercise 3.2 from question number 1 to question number 5. One can also download these solutions in PDF format. These questions are important for CBSE Class 12 Maths exams and other competitive exams.
## NCERT Solutions for CBSE Class 12 Mathematics ‒ Chapter 3: Matrices (Part II)
Jul 7, 2017
Get NCERT Solutions for Class 12 Maths, Chapter 3: Matrices. In this article, you will find solutions of exercise 3.1, from question number 6 to question number 10. These questions can be asked in CBSE Class 12 Maths board exam and other competitive exams.
## NCERT Solutions for CBSE Class 12 Mathematics ‒ Chapter 3: Matrices (Part I)
Jul 7, 2017
NCERT Solutions for Class 12 Maths, Chapter 3: Matrices are available here. In this article, you will find solutions of exercise 3.1 from question number 1 to question number 5. These questions can be asked in CBSE Class 12 board exam and other competitive exams.
## NCERT Solutions for CBSE Class 12 Mathematics ‒ Chapter 2: Inverse Trigonometric Functions (Part VI)
Jul 5, 2017
Class 12 Maths NCERT Solutions, Chapter 2: Inverse Trigonometric Functions are available here. Here, you will find solutions for exercise number 2.2 from question number 16 to question number 21. These solutions are important for CBSE Class 12 Maths board exam and other engineering entrance exams.
## NCERT Solutions for CBSE Class 12 Mathematics ‒ Chapter 2: Inverse Trigonometric Functions (Part V)
Jul 5, 2017
NCERT Solutions for Class 12 Maths, Chapter 2: Inverse Trigonometric Functions are available here. With this article, you will get solutions of exercise number 2.2 from question number 11 to question number 15. These questions and solutions are important for CBSE Class 12 Maths board exam and other competitive exams.
## NCERT Solutions for CBSE Class 12 Mathematics ‒ Chapter 2: Inverse Trigonometric Functions (Part IV)
Jun 29, 2017
NCERT Solutions for CBSE Class 12th Maths, Chapter 2: Inverse Trigonometric Functions are available here. From this article, you will get solutions of exercise number 2.2 from question number 5 to question number 10. These questions and solutions are important for CBSE Class 12 board exams.
## NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 2 (Part II)
Jun 29, 2017
Find NCERT Exemplar Solutions for Class 12 Physics, Chapter 2: Electrostatic Potential and Capacitance. In this article, you will find solutions of multiple choice questions with multiple correct answers (i.e., question number 2.7 to question number 2.13). These questions are important for CBSE board exams and other competitive exams.
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# Donut Delights Inc Has Determined That When X Donuts Are Made Daily The Profit P 1
Donut Delights, Inc. has determined that when x donuts are made daily, the profit P, in dollars, is given by P(x) = -0.002 x 2 + 4.1x – 1700
(a) What is the company’s profit if 900 donuts are made daily?
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Frequent Visitor
## Choose earliest value as true value
Hi,
I'm stuck on this annoying problem. I would like to indicate where the highest cost arises based on
- NAME
- ALIAS
- BRANCH
- MONTH
Now my following MAX(COST)-column makes the following calculation:
IF(CALCULATE(MAX('TABLE'[COST]),
ALLEXCEPT('TABLE', 'TABLE'[MONTH],
'TABLE'[NAME],
'TABLE'[ALIAS],
'TABLE'[BRANCH))='TABLE'[COST],'TABLE'[COST],BLANK())
And my INDICATOR-column then makes it as a "1" if the entry is the maximum.
I have following table
ID NAME ALIAS BRANCH MONTH YEAR COST MAX(COST) INDICATOR 1 VICTOR VLA CAR 1 2020 620 620 1 2 VICTOR VLA CAR 1 2020 453 3 VICTOR VLA CAR 1 2020 620 620 1 4 VICTOR VLA CAR 1 2020 249 5 VICTOR VLA TRUCK 2 2020 1219 1219 1 6 CARL CHO CAR 2 2020 310 310 1 7 CARL CHO TRUCK 2 2020 918 8 CARL CHO TRUCK 2 2020 1328 1328 1
But, I only want my INDICATOR to count the earliest MAX-value, i.e. the second "620" given the same filters should just be blank as following:
ID NAME ALIAS BRANCH MONTH YEAR COST MAX(COST) INDICATOR 1 VICTOR VLA CAR 1 2020 620 620 1 2 VICTOR VLA CAR 1 2020 453 3 VICTOR VLA CAR 1 2020 620 4 VICTOR VLA CAR 1 2020 249 5 VICTOR VLA TRUCK 2 2020 1219 1219 1 6 CARL CHO CAR 2 2020 310 310 1 7 CARL CHO TRUCK 2 2020 918 8 CARL CHO TRUCK 2 2020 1328 1328 1
As you can se from the table MAX(COST) is blank for ID='3'. How do I make sure my INDICATOR takes this into account?
4 REPLIES 4
Super User
@martin1212 here is the measure and the result
``````Max Cost =
VAR __table =
ALLEXCEPT (
'Cost',
'Cost'[MONTH],
'Cost'[NAME],
'Cost'[ALIAS],
'Cost'[BRANCH]
)
VAR __id =
CALCULATE (
MIN ( Cost[ID] ),
TOPN ( 1, __table, CALCULATE ( MAX ( Cost[COST] ) ), DESC )
)
VAR __return =
IF ( MIN ( Cost[ID] ) = __id , 1 )
RETURN __return``````
Check my latest blog post Compare Budgeted Scenarios vs. Actuals I would Kudos if my solution helped. 👉 If you can spend time posting the question, you can also make efforts to give Kudos to whoever helped to solve your problem. It is a token of appreciation!
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Frequent Visitor
When I try the code it just gives me the circular reference error. How can that be?
Community Support
Hi @martin1212 ,
You need to copy the code to measure instead of calculated column.
Best Regards,
Henry
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Frequent Visitor
What if I don't have an ID-column? How should I then approach it? I know it sounds silly, but making this code work without using the ID would be better in my case 🙂
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https://www.mathworks.com/matlabcentral/answers/1460469-compute-intersection-between-lines?s_tid=prof_contriblnk
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# Compute intersection between lines
2 views (last 30 days)
sparsh garg on 25 Sep 2021
Edited: Walter Roberson on 25 Sep 2021
Le't say I have 2 matrices that represent lines
Let's say you have A a 16x2 matrix that represents one set of lines shown below
And then there is B a 12x2 matrix representing another set of lines shown below
I know interx can be used to find intersection between lines,curves etc. but what i want to do is take the first line of A and find its intersection with the last line of B
similarly second line of A intersects with second last line of B
As of now this is what i came up with
for i=1:2:size(L_seg_proj,1)
L1=L_seg_proj(i:i+1,:);
for j=1:2:size(R_seg_proj,1)
L2=R_seg_proj(j:j+1,:);
P=InterX(L1,L2);
end
end
But the issue with this is that for every line in L_seg_proj(L1) it computes intersection between L1 and all the lines in R_seg_proj.
Any ideas on what I can do to rectify this?
sparsh garg on 25 Sep 2021
Edited: Walter Roberson on 25 Sep 2021
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# Math/Physics
215,876 results, page 42
2. ## Physics
1. Given t1/2 to be 0.0137 seconds, how long should it take to reach 87.5% of maximum charge?
3. ## physics
what is the heat capacity of a gold ring that has a mass of 5.00g?
4. ## Physics
calculate the force between two identical charges of 1.0 x 10 exp -9 C separated by 1.0 Cm
5. ## physics
What.is.the.weight.of.a.150.lb.person.in.newtons?.what.is.his.mass.in.kilograms?
6. ## physics
What is the apparent weight of a 55 kg person in an elevator accelerating downward at 2.2 m/s2?
7. ## PHYSICS
IS IT BETTER TO PUSH A GROCERY CART FROM THE BACK,OR TO PULL IT FROM THE FRONT.
8. ## physics
If the speed of an object is doubled its kinetic energy is multiplied by
9. ## Physics
What is the weight (in Newtons) of an object with a mass of 25 kg on the surface of the Earth?
10. ## physics
What is the weight of a 0.2kg toy on or near the surface of the Earth? (g = 9.8m/s2? in newton
11. ## physics
how i cld rise the base of a house wd the help f electric circuit
12. ## Physics
If a person runs the 100m dash in 10.0s, calculate that speed in m/s and km/h.
13. ## Physics
A ball thrown upward at 60.0 m/s has what speed after 5 seconds? What equation do I use?
14. ## Physics
What is the kinetic energy of a 1500 kilogram object moving at 25 m/s?
15. ## physics
How will the current in a wire change as the wavelength of light changes? I have no idea how to do this.
16. ## physics
How will the current in a wire change as the wavelength of light changes? I have no idea how to do this.
17. ## physics
Find the density of a 4.2- kg solid cylinder that is 13 cm tall with a radius of 5.0 cm.
18. ## physics
What,is,the,weight,of,150lb,person,in,newtons?,What,is,his,mass,in,kilograms?
19. ## physics
How much kinetic energy does a 0.03 kg bullet traveling at a speed of 300 m/s have?
20. ## Physics
For centripetal acceleration, if the radius increases, does that mean the angle is decreasing?
21. ## physics
Find the density of a 4.2- kg solid cylinder that is 13 cm tall with a radius of 5.0 cm.
22. ## Physics
Calculate AMA given μk = 0.019 and an angle of 11 degrees.
23. ## Physics
Does an object still possess any kinetic or potential energy at absolute zero?
24. ## physics
What is the acceleration of a 0.30 kg volleyball when a player uses a force of 93 N to spike the ball?
25. ## Physics
Which of the following are the correct units for ACCELERATION? Choose all that apply. m/s2 h mm mm/d2 pc mi NM yd/d km ft/s cm min
26. ## physics(URGENT!!!!!)
8.01xers Can anybody help with the falling ruler. I submission left(
27. ## Physics
Write 2.38x10 to the 6th power in standard notation
28. ## Physics
What is the frequency of an electromagnetic wave if it has a wavelength of 1.4 km? The speed of light is 3 × 10 8 m
29. ## physics
with what speed does a hammer dropped from a height of 23 m strike the ground?
30. ## physics
A fully loaded 737 aircraft takes off at 250 {\rm km/h}.
31. ## physics
when a car goes around an 8-km track in 20-minutes. what is the average speed and velocity.
32. ## physics
What is the period (in seconds) of a simple pendulum 62.0 cm. long, on the earth?
33. ## physics
As the wavelength of a wave in a uniform medium increases, its speed will ?
34. ## physics
if The magnitude of the nuclear spinangular momentum of a nuclei is 15 /2 units. Then what will be the value of I
35. ## Physics 102
What would be the path of the Moon if somehow all gravitational forces on it vanished to zero?
36. ## Physics
A 20 kg block is placed on a 35 degree ramp. What is its true metric weight?
37. ## physics
Calculate the gravity on a planet using a pendulum that has a period of 2.45 s having a length of 1.25 m.
38. ## physics
An automobile with a mass of 1400 kg has a speed of 30.0 m/s. What is its kinetic energy
39. ## Physics
I just have a question....How to know if the volt is -ve or not? and How to know if a charge has deficit or excess electrons ?
40. ## Physics
Part A - What is the magnitude of the electric force on charge in the figure?
41. ## physics
The rate at which distance changes in a time period is called acceleration
42. ## physics
What is the force on the charge located at x = +8.00 cm in Figure 17.40(a) given that q = 5.00 n C and a = 7.00? (The positive direction is to the right.) N
43. ## Physics
Given the following vectors: A = (28, 110°) B = (63, -30°) Determine A + B. Specify the sum as both Cartesian and polar.
44. ## physics
if The magnitude of the nuclear spinangular momentum of a nuclei is 15 /2 units. Then what will be the value of I
45. ## Physics
why the ray reflected by the prism is brighter than that by the plane mirror?
46. ## Physics
What is the application of simple harmonic motion to our daily lives?
47. ## PHYSICS
ADDITIONAL INO ADDED AT BOTTOM a spectator at a hockey game is sitting in a seat situated 10.4m above ground leve. if the spectator has a mass of 52.6 kg, calculate her gravitational potential energy relative to: a)ground level Eg=mgh =52.6X9.8X10.4
How do you determine the ymax? I know you have to use Q which I have values that were given to me as well as vo values but I am not sure what to do with these values. Physics - drwls, Saturday, September 15, 2012 at 1:29am Define your terms. I have no idea
49. ## bobpursley
ok im doing my physics homework and we have to submit the answers online. i keep getting these questions wrong and i only have one more try can someone please tell me how to work it? a 2.5 kg otter starts from rest at the top of a muddy incline 94.6 cm
50. ## math
need to make a venn diagram using 3 circles. A survey on subject being taken by 250 students at a certain college revealed the following info: 1. 90 were taking math 2. 145 were taking history 3. 88 were taking english 4. 25 were taking math and history 5.
51. ## Math
Could someone please help me with this math problem. I am lost on trying to solve this math problem. Add. Simplify if possible. s+r/sr^2 + 2s+r/s^2r Thanks.
52. ## Physics Newton law
A 10O0Kg elevator is hoisted upward from rest by a cable. In the first two seconds there is an 1100N tension force in the cable hoisting the elevator. What is the speed after two seconds and how far has it travelled ? Get the acceleration first. Tension=mg
53. ## Physics help
Are you sure that's how you do #4 because they have 2 different directions. I need physics help. I don't understand how to do these questions. Can someone show me what I should use? and explain. I'll do calculations by myself. A stationary curling stone is
54. ## physics multiple choice
A 3.00 lb. physics book rests on an ordinary scale that is placed on a horizontal table. The reaction force to the downward 3.00 lb force on the book is A. an upward 3.00 lb force on the scale due to the table. B.an upward 3.00 lb force on the book due to
55. ## physics multiple choice
A 3.00 lb. physics book rests on an ordinary scale that is placed on a horizontal table. The reaction force to the downward 3.00 lb force on the book is A. an upward 3.00 lb force on the scale due to the table. B.an upward 3.00 lb force on the book due to
56. ## Math
16.Eliot spent 24 minutes solving 8 math problems.At that rate, how many math problems can he solve in 54 minutes? a.12 B.16 C.18 D.24 A? Can you explain why it isnt a
57. ## Physics repost
Posted by Mary on Saturday, October 6, 2007 at 5:13pm. Suppose you wish to make a solenoid whose self-inductance is 1.2 mH. The inductor is to have a cross-sectional area of 1.2 10-3 m2 and a length of 0.048 m. How many turns of wire are needed? For
58. ## Physics
Two 0.09g pith balls are suspended from the same point by threads 40cm long. (Pith is a light insulating material once used to make helmets worn in tropical climates.) When the balls are given equal charges, they come to rest 19 cm apart What is the
59. ## Research and Statistics
I want to ensure I've done this problem correctly. Thanks! 3. In one elementary school, 200 students are tested on the subject of Math and English. The table below shows the mean and standard deviation for each subject. Mean SD Math 67 9.58 English 78
60. ## Re: PHYSICS
"Relative" is an important word. Block L of mass mL = 1.90 kg and block R of mass mR = 0.500 kg are held in place with a compressed spring between them. When the blocks are released, the spring sends them sliding across a frictionless floor. (The spring
61. ## Physics
A planet has two satellite moons. Moon X has an orbital period of 2.13 days. Moon Y has an orbital period of about 3.53 days. Both moons have nearly circular orbits. Use Kepler's third law to find the distance of each satellite from the planet's center.
62. ## phsyics
Posted by MELANA on Thursday, October 18, 2007 at 5:14pm. An object in equilibrium has three forces exerted on it. A 30 N force acts at 90° from the x-axis and a 44 N force acts at 60° from the x-axis. What are the magnitude and direction of the third
63. ## physics
An inquisitive physics student, wishing to combine pleasure with scientific inquiry, rides on a rollercoaster sitting on a bathroom scale. (Do not try this yourself on a rollercoaster that forbids loose heavy packages.) The bottom of the seat in the
64. ## Physics
A space station, in the form of a wheel 197 m in diameter, rotates to provide an “artificial gravity” of 2.5 m/s2 for persons who walk around on the inner wall of the outer rim. The rate of the wheel’s rotation in revolutions per minute that will
65. ## Math and Music
I have to do a presentation on how math and music relate. I have found some things and facts but everything is so complicated. If someone could help explain to me how I can narrow my topic down and make a presentation that is simple but effective. It can
66. ## 1st grade Math Fair
My school is planning a Math Fair. Each grade will compete with same grade team members. The rules are: TO make it fun and use Al. math strands . The problem is I don't know where or how to begin this project. Can anyone help?
67. ## physics
determine the pressure of a 20 lb force is applied to an area of 10 inch squared
68. ## Physics Class
The equation describing the motion of an object is y=0.8cos(4.2t+0.32). What is its period?
69. ## Physics
How long is Mayra, a ballerina, in the air when she leaps straight up with a speed of 1.23 m/s?
70. ## physics
What is the electric field intensity of a point 15cm from a charge 15microcoulomb
71. ## physics
how do you compare mass and weight on the earth, moon, and outer space?
72. ## physics
An object has a length of 5.0 cm, a height of 3.0 cm and a width of 15.0 cm. It has a mass of 24 grams. What is its density?
73. ## physics
a year ago she measured 98cm from her tree . now it 1.1m tall how much has it grown
74. ## physics
What is the weight of each of the following masses at Earths surface? A) 25g. B)102kg. C) 12mg
75. ## Physics
If I know the length of all 3 sides in a right angle triangle, how do I get the degrees of the other angles? Which formula do I use again?
76. ## physics
A certain car can go from 0 to 100 km/h in 10 s. If the engine delivered twice the power to the wheels, how many seconds would it take?
77. ## physics
the air pressure in the tyres of a car increases during driving.Explain.
78. ## physics
You throw a ball straight up at 25 m/s. How many seconds elapse before it is traveling downward at 13 m/s? 1 s
79. ## physics
A car traveling at 5.9 m/s accelerates at 2.12 m/s2 for 19.5 seconds. To the nearest meter how far does it travel.
80. ## physics
The time required for a car to reach 28.5 m/s from rest, accelerating at 3.70 m/s2 is __________ .
81. ## physics
how is a stone tied on an elastic string balanced when in air and water ?
82. ## Physics
Identify the correct neutrino (Ve, Vµ, Vτ or their anti-particles) a) π+ —> p° + e+ + v b) π+ —> p° + µ+ + v c) τ+ —> e- + v + v Please explain why too...I'm really lost. I'd appreciate any help.
83. ## physics
A rock is dropped from a 154-m-high cliff. How long does it take to fall (a) the first 77.0 m and (b) the second 77.0 m?
84. ## Physics
an elastic band is stretched a distance of 0.150m by a force of 100.0
85. ## Physics
If you are viewing a hologram and you close one eye, will you still perceive depth? Explain.
86. ## physics
a student driving a car travels 10.0 km in 30,0 min.What was her average speed?
87. ## physics
determine the pressure of a 20 lb force is applied to an area of 10 inch squared
88. ## physics
The energy of the em waves is of the order of 15 kev.to which part of the spectrum does it belong?
Which of the following is closer to a guess? hypothesis<<<<< law question theory
90. ## Physics/chemistry
Describe the partcle Theory of a HOt air Balloon?
91. ## physics
A lizard accelerates from 2m/s to 10m/s in 4 seconds. What is the lizards average acceleration?
92. ## Physics
Two point charges of magnitude 2.0 X 10^-7 C and 85 X 10^-8 C are placed 12 cm apart .What is the field intensity of each charge at the site of other ?
93. ## physics (motion)
the initial velocity of a particle is u (at t=0) and acceleration f is given by at.which relation is valid: (a)v=u+at^2 (b)v=u +at^2/2 (c)v=u+at (d)v=u
94. ## Physics
A stone weighs 34.0 N. What force must be applied to make it accelerate upward at 4.00 m/s2?
95. ## PHYSICS
A hypothetical planet has a mass 1.49 times that of Earth, but the same radius. What is g near its surface?
96. ## physics
where would the image of a 4.0 cm tall object that is 12 cm in front of a flat mirror be located?
97. ## physics
a 40kg girl climbs a vertical ladder at .5 m/s. what is her power output?
98. ## PHYSICS
WHAT IS THE MASS AND WEIGHT OF 1420LB OF SNOMOBILE AND 412KG OF HEAT PUMP?
99. ## physics
the rate at which position changes with time is called acceleration. true or false?
100. ## physics
Calculate the momentum of a 100 kg object moving with velocity 10m/s
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# Gold Board Game
## Awesome Gold Board Game, and Templates to Print Your Own
As a child, I had a favorite gold mining game that was so much fun to play. However, they no longer make it!! So when I started making my own board games with my own kids, I thought it would be fun to make a similar game.
You can see a picture of the original game board, and if you keep scrolling you'll find links to some free printable board game templates to make a similar game for yourself or your classroom.
## My Ravensburger Knock-off
Ravensburger's Gold Mining and Bandits game is great for math and strategy!
## Unofficial Gold Rules - The Rules as I remember Them (haven't found the original rules yet)
Setup: all hexagon game chips are placed face down on board spaces. The bandits, gole pieces and blank pieces should be mixed together and randomly placed on spaces with holes on them.
Train starts at the edge of the game board. Every time a player rolls a 1 (with the Locomotive) the train gets ome space closer to the town. When the train arrives in town the game is over.
Each player takes turns rolling the die. They move however many spaces they roll. If they land on a chip they flip it over to see what it is. If it's gold they keep it. If it's blank they discard. If they get a bandit they have to roll again. If they roll a number higher than the number on the bandit's head they get to keep the bandit as part of their treasure (and he is worth in gold the number on his head). If the player rolls a number lower than the bandit than all the gold that is not safe at home that the player collected is lost to the bandit and placed underneath the space where the player found the bandit. Other players can attempt to win everything underneath the bandit by landing on that space and rolling a number higher than the number on the bandit's head.
If a player goes back to his or her home, he can safely store his gold there (no bandit can take it once it's home).
Rafts: A player must have the exact number rolled to get on a raft, but they can park the raft on any shore bank of the lake they are in. The raft stays where they player leaves it, and someone else can ride it if they roll the exact number to get on it in is new location.
Once the train returns to town everyone counts the gold they have safe at their home. If you have gold in your hand that you haven't brought home, it does not count in your score. Whoever has the highest number of gold nuggets wins (one game token can be work five pieces of gold, and one bandit can also be worth as much as five pieces of gold).
Update: I did just find the original rules here.
## Eureka - Also Known as Gold - Great Game for Math Building Skills and Strategy
Strike it rich in Golden City!
Gold has been discovered in the foothills outside Golden City and every able-bodied person from miles around is off to grab a share.
But things can be tough for a prospector, who must be lucky enough to find the gold and to get it safely back to town - fending off bandits and claim-jumpers - before the train arrives,
Time is running out - can't you hear the train in the distance? May good fortune be yours!
Contents
1 Game Board
48 Gold Nugget chips
16 Bandit chips
10 Spare blank chips
6 Raft chips
6 Wooden Prospector figures
1 Wooden Locomotive
1 Die (with Locomotive 1
Strategy game for 3 - 6 players ages 8 - adult.
Playing time: 40 minutes
## Sometimes You Can Buy it From Amazon.com
Eureka
You can't buy it new, but you can sometimes snag a used copy. It's worth a shot anyway.
## Printable "Gold" Templates
When I couldn't buy the Gold game I decided to make my own. It's very much simplified compared to the original game, but it does work!
19
44
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## Have you ever played the Ravensburger Gold game?
0 of 8192 characters used
• Amy Trumpeter 3 years ago from Oxford
Looks great! I am the 'Game Night Contributor' on Squidoo, and have given your lens a 'BOOST'!
• Author
shauna1934 3 years ago
@shauna1934: Your template has been reposted.
• Author
shauna1934 3 years ago
@zardoz1970: I love your fantasy game idea. I may have to steal it. I'm currently working on a version based on Disney's Frozen for my kiddos. Thanks for writing.
• zardoz1970 4 years ago
could you repost the editable version of your game board. i printed the color version and my daughter and i made our own custom pieces an we have not been able to stop playing. we changed the theme and subbed a dragon instead of a train making it a more fantasy based game.
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Written By Dave Martin
• 6/7/2023
• Read Time : 3 min.
# Mathcad Community Challenge May 2023: Optimize Horizontal Distance
The Mathcad Community Challenge for May 2023 was as follows:
A ball with a mass of 1 kilogram is at the top of a frictionless ramp 10 meters above the ground. The ball rolls down the incline and launches from a height of 2 meters and an angle of 30 degrees above the ground.
The problems to be solved are:
1. Create a function that calculates the horizontal distance as a function of initial height, launch height, and launch angle.
2. Calculate the horizontal distance the ball will land from the end of the ramp.
3. Solve for the angle that will optimize the horizontal distance.
4. How will the horizontal distance change if this were performed on the Moon instead of on the Earth’s surface? Assume the acceleration due to gravity on the Moon’s surface is 1/6 that of Earth.
5. Use the Chart Component to depict how the horizontal landing distance changes as a function of angle.
6. Use a 3D Plot to show how the horizontal landing distance changes as a function of ramp height and launch angle. Assume the ball starts at a height of 10 meters.
## The Submissions
Frequent contributor PPal submitted the first and multiple worksheets. I like that they build on previous worksheets. I hadn’t intended people to take moment of inertia into account, but PPal did. Functionality in the solutions included user-defined functions, symbolic evaluation, the polyroots function, XY plots, and 3D plots. The worksheets incorporated multiple aspects to make the worksheet friendly to others, including text with different fonts, images, and highlighting.
Fred Kohlhepp, another loyal contributor, submitted an attractive Mathcad Express worksheet with numerous XY plots that greatly help to convey information to the reader. Fred also uses another technique I like: documenting the thinking behind your equations using the equality comparison operator (the thick equals sign). Fred also figured out the trick behind question 4. Since the acceleration due to gravity falls out of the equations, the horizontal distance is the same on the Earth, Moon, or any planetary body.
Jeff Henning’s submission wins the award for the most publication-ready worksheet. It could be used to teach physics to high school and college students. Jeff uses collapsible areas to organize the worksheet into assumptions, derivations, inputs, function definitions, and the answers to each question. Jeff gets extra points for being the first person since the challenge started to use the Chart Component. He plots the trajectory of the ball as well as the distance versus launch ramp angle. The 3D Plot for question 6 is also beautifully formatted. Another small thing that makes a huge difference: Jeff uses the document footer including page numbers. Just a beautiful work of art.
A collage of the different entrants' Mathcad Prime worksheets.
As usual, TTokoro submitted an inviting worksheet complemented with text, highlighting, XY Plots, 3D Plots, and headers and footers. Draft View was used to hide some calculations, which helps keep worksheets neat. The math made heavy use of symbolic evaluation. A nice, compact piece of work. If you want to take your XY Plots to the next level, I recommend checking out TTokoro’s challenge worksheets.
Alan Stevens posted an economical Mathcad Express worksheet. (Since the Express version lacks the Chart Component and 3D plots, he and Fred were not able to complete questions 5 and 6.) As I had suggested in a subsequent comment, Alan considered conservation of potential energy into kinetic energy, resulting in compact equations and functions. The optimum angle was solved by defining a custom function and applying the root function.
Jan Claeys’s worksheet started off with a table of contents consisting of hyperlinks to other parts of the worksheet. The calculations were documented well with text, highlighting, and use of the equality comparison operator (like Fred and others). The calculations included symbolic evaluation, the root function, an XY Plot, and even a Solve Block (my personal favorite Mathcad functionality). Jan completed questions 5 and 6 with a very nice Chart Component and 3D Plot, then concluded the worksheet with hyperlinks to external references. Like Jeff’s, this worksheet makes a very nice teaching tool.
## The Outcome
There were two sets of answers, depending on whether rotational inertia was taken into account. The people who submitted worksheets found those results. The more time I have worked in industry, the more I have realized the importance of documentation. It doesn’t matter how great your results are if you fail to communicate them effectively. We’re also not able to explain our results personally (especially years after the fact), so it’s important that the work stands on its own. Like previous months, we see a variety of solution tools, but also a variety of different tools to document that work. I recommend checking out the submissions to see which communication methods resonate with you to include in your own work. You can also check out all of the previous challenges and solutions that have covered a wide range of topics so far.
## Subscribe to the Mathcad Minute
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https://forums.wolfram.com/mathgroup/archive/2012/Dec/msg00223.html
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Re: Clean output of mathematica
• To: mathgroup at smc.vnet.net
• Subject: [mg129081] Re: Clean output of mathematica
• From: Bob Hanlon <hanlonr357 at gmail.com>
• Date: Thu, 13 Dec 2012 04:08:02 -0500 (EST)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• Delivered-to: l-mathgroup@wolfram.com
• Delivered-to: mathgroup-newout@smc.vnet.net
• Delivered-to: mathgroup-newsend@smc.vnet.net
• References: <20121212005617.818F1691F@smc.vnet.net>
```s = DSolve[
{y''[x] == -a^2 y[x], y[0] == 0},
y, x][[1]] /. C[2] -> 2;
t = {Factor[a /.
Solve[#, a][[1]] & /@
Simplify[
Reduce[{(y[L] /. s) == 0, L != 0}, a],
{Element[C[1], Integers],
L != 0}]] /.
(2 C[1] x_ || (2 C[1] + 1) x_) :> n*x}
{(n*Pi)/L}
Bob Hanlon
On Tue, Dec 11, 2012 at 7:56 PM, AZEEM MIR <docpasmon at gmail.com> wrote:
> Dear All
> Following code solves the 2nd order differential equation.
>
> s = DSolve[{y''[x] == - a^2 y[x]}, y, x]
> t = Reduce[y[0] == 0 && y[L] == 0 && C[1] == 0 && C[2] ==
= 2 /. s, a]
>
> These give output
>
> {{y->({x}\[Function]Subscript[c, 2] sin(a x)+Subscript[c, 1] cos(a x))}}
>
> Subscript[c, 3]\[Element]\[DoubleStruckCapitalZ]\[And]((Subscript[c, 1]\[LongEqual]0\[And]Subscript[c, 2]\[LongEqual]2\[And](L\[LongEqual]0\[Or](L!=0\[And]a\[LongEqual](2 \[Pi] Subscript[c, 3])/L)))\[Or](L!=0\[And]Subscript[c, 1]\[LongEqual]0\[And]Subscript[c, 2]\[LongEqual]2\[And]a\[LongEqual](\[Pi] (2 Subscript[c, 3]+1))/L))
>
> Now it is possible to extract the possible eigen values of a but that is not efficient automation. I wish to write a program that effectively plots eigenvalues solution without customizing the extract command each time.
>
> Is it possible to produce a clean output like all possible eigen-values enclosed by bracket defined by just{n Pi/L}. Sure it won't be so hard to overwrite some basic files.
>
```
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https://converterin.com/cooking/quart-uk-to-megaliter-ml.html
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# QUART [UK] TO MEGALITER CONVERTER
FROM
TO
The result of your conversion between quart [UK] and megaliter appears here
## QUART [UK] TO MEGALITER (qt TO ml) FORMULA
To convert between Quart [UK] and Megaliter you have to do the following:
First divide 0.00454609/4 / 1000 = 0.00000114
Then multiply the amount of Quart [UK] you want to convert to Megaliter, use the chart below to guide you.
## QUART [UK] TO MEGALITER (qt TO ml) CHART
• 1 quart [UK] in megaliter = 0.00000114 qt
• 10 quart [UK] in megaliter = 0.00001137 qt
• 50 quart [UK] in megaliter = 0.00005683 qt
• 100 quart [UK] in megaliter = 0.00011365 qt
• 250 quart [UK] in megaliter = 0.00028413 qt
• 500 quart [UK] in megaliter = 0.00056826 qt
• 1,000 quart [UK] in megaliter = 0.00113652 qt
• 10,000 quart [UK] in megaliter = 0.01136523 qt
Symbol: qt
No description
Symbol: ml
No description
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https://web2.0calc.com/questions/you-roll-a-pair-of-six-sided-number-cubes-numbered
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+0
# You roll a pair of six-sided number cubes, numbered 1 through 6. help
0
483
1
You roll a pair of six-sided number cubes, numbered 1 through 6. What is the probability of rolling two even numbers? (Enter your probability as a fraction.)
= _________
Oct 29, 2018
#1
+18846
+2
Each cube has 3/6 even numbers or 1/2 chance of being even when rolled
1/2 x 1/2 = 1/4 chance of BOTH being EVEN
Oct 29, 2018
#1
+18846
+2
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https://edurev.in/studytube/Water-Demand/e8b95dac-5297-4e88-bbbb-1b372ef2de85_t
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Water Demand
# Water Demand | Environmental Engineering - Civil Engineering (CE) PDF Download
### Fire Demand
Rate of fire demand is sometimes treated as a function of population and is worked out on the basis of empirical formulas:
1. As per GO Fire Demand
= 100(P)1/2
2. Kuichling’s Formula
Where Q = Amount of water required in liters/minute.
P = Population in thousand.
3. Freeman Formula
4. National Board of Fire Under Writers Formula
(i) For a central congested high valued city
(a) Where population < 200000
(b) where population > 200000
Q = 54600 lit/minute for first fire
and Q = 9100 to 36,400 lit/minute for a second fire.
(ii) For a residential city.
(a) Small or low building,
Q = 2,200 lit/minutes.
(b) Larger or higher buildings,
Q = 4500 lit/minute.
(c) High value residences, apartments, tenements
Q = 7650 to 13,500 lit/minute.
(d) Three storeyed buildings in density built-up sections,
Q = 27000 lit/minute.
5. Buston’s Formula
The probability of occurrence of a fire, which, in turn, depends upon the type of the city served, has been taken into consideration in developing a above formula on the basis of actual water consumption in fire fighting for Jabalpur city of India. The formula is given as
Where,
R = Recurrence interval of fire i.e., period of occurrence of fire in years, which will be different for residential, commercial and industrial cities.
Per Capita Demand (q)
Assessment of Normal Variation
### Population forecasting Methods
1. Arithmetic increase method
Where,
Prospective or forecasted population after n decades from the present (i.e., last known census)
Population at present (i.e., last known census)
n = Number of decades between now & future.
Average (arithmetic mean) of population increases in the known decades.
2. Geometric Increase Method
where,
Po = Initial population.
Pn = Future population after ‘n’ decades.
r = Assumed growth rate (%).
where,
P2 = Final known population
P1 = Initial known population
t = Number of decades (period) between P1 and P2.
3. Incremental Increases Method
Where,
Average increase of population of known decades
Average of incremental increases of the known decades.
4. Decreasing rate of growth method
Since the rate of increase in the population goes on reducing, as the cities reach towards saturation, a method which makes use of the decrease in the percentage increase, in many times used, and gives quite rational results. In this method, the average decrease in the percentage increase is worked out, and is then subtracted from the latest percentage increase for each successive decade. This method is, however, applicable only in cases, where the rate of growth shows a downward trend.
5. Logistic Curve Method
(a)
Where,
Po = Population of the start point.
Ps = Saturation population
P = Population at any time t from the origin.
k = Constant.
The document Water Demand | Environmental Engineering - Civil Engineering (CE) is a part of the Civil Engineering (CE) Course Environmental Engineering.
All you need of Civil Engineering (CE) at this link: Civil Engineering (CE)
## Environmental Engineering
14 videos|74 docs|98 tests
## FAQs on Water Demand - Environmental Engineering - Civil Engineering (CE)
1. What is water demand in civil engineering?
Ans. Water demand in civil engineering refers to the quantity of water required for various activities such as domestic use, industrial processes, irrigation, and firefighting. It is an essential factor in designing water supply systems and ensuring the availability of sufficient water resources to meet the needs of a specific area or project.
2. How is water demand calculated in civil engineering projects?
Ans. Water demand in civil engineering projects is calculated by considering various factors such as population growth, per capita water consumption, and specific water requirements for different activities. The calculation involves estimating the water demand for each category and then summing them up to determine the total water demand for the project.
3. What are the factors influencing water demand in civil engineering?
Ans. Several factors influence water demand in civil engineering, including population growth, urbanization, climate conditions, economic development, lifestyle patterns, and technological advancements. These factors affect the overall water consumption and usage patterns, thereby impacting the estimation of water demand for a particular project or area.
4. How is water demand forecasted in civil engineering?
Ans. Water demand forecasting in civil engineering involves analyzing historical water consumption data, considering the population growth rate, and evaluating the impact of various factors like climate change and socio-economic conditions. Statistical models and simulation techniques are commonly used to forecast future water demand, helping in the planning and design of water supply systems.
5. What are some strategies to manage water demand in civil engineering projects?
Ans. To manage water demand in civil engineering projects, several strategies can be adopted, including: - Implementing water-efficient fixtures and appliances in buildings to reduce water consumption. - Promoting water conservation practices such as rainwater harvesting and greywater recycling. - Educating the community about the importance of water conservation and encouraging responsible water use. - Enhancing water infrastructure and distribution systems to minimize water losses and leakage. - Implementing water pricing mechanisms and incentives to encourage efficient water use.
## Environmental Engineering
14 videos|74 docs|98 tests
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# Calculate battery capacity from load
I have a 5000 VA UPS which is connected to eight 65 AH batteries. I am looking to replace these old batteries with new ones. I was wondering whether the UPS will run safely with less number of batteries or with less battery capacity.
I did some research and found out that the formula for calculating the battery capacity goes something like this:
Battery capacity = (Load / Voltage) * Back up time
I am assuming that the load is 5000 VA which will roughly convert to some 4000 Watts and its a 12 V battery. The back up time required is 30 minutes.
So battery capacity = (4000 / 12) * 0.5 = 166.
Is my formula wrong? I don't have an electrical background. Any help would be appreciated. Thanks.
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https://math.stackexchange.com/questions/8173/what-is-an-example-of-a-lambda-system-that-is-not-a-sigma-algebra?noredirect=1
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# What is an example of a lambda-system that is not a sigma algebra?
What is an example of a lambda-system that is not a sigma algebra?
Here is a somewhat more natural example.
Let $(\Omega, \mathcal{F})$ be a measurable space, and let $P,Q$ be two probability measures on $\mathcal{F}$. It is a good exercise to verify that $$\mathcal{L} := \{ A \in \mathcal{F} : P(A) = Q(A) \}$$ is a $\lambda$-system. (This is a common application of the $\pi$-$\lambda$ theorem : if one can show that $P$ and $Q$ agree on a $\pi$-system that generates $\mathcal{F}$, then $P$ and $Q$ must be the same.)
However, $\mathcal{L}$ need not be a $\sigma$-algebra. For instance, consider a sample space consisting of two coin flips: $$\Omega = \{ HH, HT, TH, TT \}, \quad \mathcal{F} = 2^\Omega.$$ Let $P$ be the probability measure under which the coins are independent and unbiased, and let $Q$ be the measure under which the first coin is unbiased but the second coin is stuck to the first so that they always come up the same. Explicitly, $$P(HH)=P(HT)=P(TH)=P(TT)=\frac{1}{4}$$ $$Q(HH)=Q(TT)=\frac{1}{2}; \quad Q(HT)=Q(TH)=0.$$ Then one can check that the events on which $P$ and $Q$ agree are those which only look at one of the coins (or none), so that $$\mathcal{L} = \{ \{HH,HT\}, \{HH,TH\}, \{HT,TT\}, \{TH,TT\}, \emptyset, \Omega\}.$$ This is not a $\sigma$-algebra since it is not closed under intersections.
• correct me if I'm wrong, but L doesn't appear to be closed under relative complement... Commented Oct 28, 2010 at 15:49
• I think you only need relative complement of included sets? I.e., $A \subseteq B \Rightarrow B \setminus A \in L$. Commented Oct 28, 2010 at 16:07
• @Neil G: Right, and note that there are no nontrivial inclusions among the sets of $\mathcal{L}$. That's also why the "increasing unions" axiom is satisfied. Commented Oct 28, 2010 at 16:35
• Even though both answers are correct, I'm going to mark this answer because it gives me a bit of intuition about what's going. The other answer would have been a better solution to an exam question. Commented Oct 28, 2010 at 17:56
For another example, let $$(\Omega, \mathcal{F}, P)$$ be a probability space and fix some event $$A \in \mathcal{F}$$. Let $$\mathcal{L}$$ be the collection of all events which are independent of $$A$$, i.e. $$\mathcal{L} = \{ B \in \mathcal{F} : P(A \cap B) = P(A) P(B)\}.$$ It is not hard to check that $$\mathcal{L}$$ is a $$\lambda$$-system. To see it need not be a $$\sigma$$-algebra, take as in my other answer a probability space $$\Omega = \{HH, HT, TH, TT\}$$, $$\mathcal{F} = 2^\Omega$$, $$P(A) = \frac{1}{4} |A|$$ consisting of two independent fair coin flips. Set $$A = \{HH, HT\}$$, the event that the first coin is heads. Then $$\{HH, TH\}, \{HH, TT\}$$ are in $$\mathcal{L}$$ but their union $$\{HH, TH, TT\}$$ is not.
Incidentally, this is really of the same form as my other answer if we take $$Q$$ to be the conditional probability measure $$Q(E) = P (E \mid A) = P(E \cap A)/P(A)$$. (Except when $$P(A)=0$$, but that case is trivial.)
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http://www.tutorsglobe.com/homework-help/physics/terrestrial-magnetism-75353.aspx
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#### Terrestrial Magnetism, Physics tutorial
Terrestrial Magnetism
Introduction:
A magnetic needle, if hanged freely in such a way that it can turn in a horizontal and a vertical plane, turns till it comes to rest in one definite direction. This proposes that the Earth should have a few of the properties of a magnet. The Earth's magnet consists of a magnetic field that acts on the suspended magnets. As compared by the field of a bar magnet the Earth's field is extremely weak. The magnetic needle turns only when it is hanged on a thread; the Earth's field is not strong adequate to make the magnet rotate if it is lying on the table, for the magnetic force is not adequate to overcome the friction.
The direction and magnitude of the Earth's field differs with position over the surface of Earth and it as well seems to be changing steadily with time. The prototype of field lines is identical to that which would be given when there was a strong bar magnet at the centre of Earth.
At present there is no commonly accepted theory of the Earth's magnetism however is might be caused through electric currents circulating in its core due to convection currents occurring from the radioactive heating within the earth.
Magnetic Field of the Earth:
The pattern of the magnetic field of earth's lines is identical to that which would be given when there was a strong bar magnet at the centre of the earth.
The points of Earth at which the magnetic field intensity is directed vertically are termed as magnetic poles. The Earth consists of two such poles; the north magnetic pole (that is, in the southern hemisphere) and the south magnetic pole (that is, in the northern hemisphere).
The straight line passing via the magnetic poles is termed as the Earth's magnetic axis. The circumference of the great circle in the plane perpendicular to the magnetic axis is termed as the magnetic equator. The magnetic field intensity at point on the magnetic equator is directly horizontally.
Definitions of some Relevant Terms:
The geographic meridian is the vertical plane in the direction, geographic N and S, that is, which passes via the geographic poles of the Earth.
The magnetic meridian is the vertical plane in which the magnet sets itself at a specific place.
The angle of variation or declination of the compass is the angle between the magnetic and geographic meridians.
The angle of dip or inclination is the angle between the horizontal and magnetic axis of the magnet free to swing in the magnetic meridian around a horizontal axis. This is the angle between the directions of the magnetic field of the earth and the horizontal.
The earth's magnetic field often termed as the net intensity, is resolved for expediency into a horizontal component and a vertical component.
The quantities, declination, dip, total intensity, horizontal component and vertical component are termed as the magnetic elements. The magnetic field of earth at a specific place might be specified through the declination and any two of the magnetic elements.
This is suitable to resolve the strength of earth's field into horizontal and vertical components, correspondingly. We then have:
BH = BR cos α
Bv = BR sin α
As well, tan α = Bv/BH
Here, 'α' is the angle of dip and 'θ' is the variation or declination.
Note: Instruments like compass needles whose motion is imprisoned to a horizontal plane are influenced by BH only.
Determination of Declination:
The determination of the declination at a place comprises determining two direction, geographic N and magnetic North.
The former can be set up precisely only through an astronomical process - observation of the sun and star. This can be found by means of fair accuracy from the fact that the shadow of a vertical slide cast through the sun at mid-day is due to N.
Magnetic North is determined by hanging a bar magnet freely on the vertical axis. As the magnetic axis of the magnet might not coincide by its geometric axis, the magnet should be turned over and the mean of the two directions found.
The vertical and horizontals components of the earth's magnetic flux density can be measured through the earth inductor. The instrument comprises of a coil of wire that can be rotated around an axis capable of being set in any direction through turning a movable frame. The principle of the instrument is that the coil is turned via a right angle between positions if it is threaded through maximum magnetic flux and zero magnetic flux and the quantity of electricity, induced is evaluated through a ballistic galvanometer joined in series by the coil.
The quality of electricity induced is represented by:
Q = - N Φ/R
Here, 'N' is the number of turns in the coil; Φ is the change in the magnetic flux threading it and 'R' is the net resistance of the coil and ballistic galvanometer circuit.
Assume that the coil is perpendicular to the field of magnetic flux density 'B' and is then turned via a right angle in such a way that no magnetic flux threads it.
Then, Φ = BA
Here, A is the area of the coil.
∴ B = - QR/NA
To find out the horizontal component of the earth's magnetic flux density, the frame is made up of vertical and the whole instrument set magnetic E and W. In this place, the coil is perpendicular to the magnetic field of earth and the maximum horizontal magnetic flux threads it. The coil is turned via 90o (about its vertical axis in such a way that its plane then lies in the magnetic meridian and none of the earth's magneic flux threads it.
To find out the vertical component of the magnetic field frame of the earth is turned in such a way that the coil can move around a horizontal axis in the magnetic meridian. If the coil is rotate via 90o, say from the vertical to a horizontal position, the quantity of electricity induced is proportional to the vertical component.
Determination of Dip:
The angle of dip 'D' can be computed from the values of horizontal and vertical components employing
tan α = Bv/BH
Tan D = Vertical component/horizontal component = θvh
Here θv and θh are the throws of the ballistic galvanometer in the earth's inductor experiment.
The Deflection Magnetometer:
The two magnetic fields can be compared by means of the deflection magnetometer that comprises of a small magnet, pivoted on the vertical axis and carrying a light pointer that can move over a circular scale.
Generally one of the fields is the earth's horizontal component and the other field is arranged to be at right angles to this. The pivoted magnet sets itself all along the resultant of the two fields at an angle θ to its direction if it is merely in the earth's field. If BH is the magnetic flux density of the earths horizontal component and B is the magnetic flux density of the other field then,
B = BH tanθ
The horizontal component of the earth's magnetic flux density can be computed by using a deflection magnetometer.
The magnetic flux density at the centre of the circular coil of acknowledged radius and identified number of turns, if a measured current is passing via the coil can be computed. This can be compared through means of a deflection magnetometer, having the horizontal component of the earth's magnetic flux density, letting the latter to be determined.
A suitable instrument for the aim is a tangent galvometer that comprises of a circular coil, at the centre of which there is a deflection magnetometer. If 'N' is the number of turns in the coil, 'a' its radius, 'I' the current and 'B' the magnetic flux density at the centre of the coil, then
B = μo NI/2a
Using the similar nomenclature as above
B = BH tanθ
BH = μo NI/2a tan θ
Variation of Dip over the Earth's Surface:
The angle of dip is 0o just about at the geomagnetic equator. It rises steadily northward or southward, till it becomes 90° at the magnetic pole. In the northern hemisphere, the 'N' pole dips, and in the southern hemisphere, the S pole dips.
Changes in the values of the Magnetic Elements:
The magnetic field of earth at any place is not constant however is subject to changes which might be categorized as follows:
1) Secular Change:
The magnetic elements experience a gradual cycle of changes that extent over a long interval after which they return to their original values. Such changes are relatively large and occur steadily.
2) Annual Change:
These changes are periodic and the value of an element differs steadily between a maximum value when the declination at a place attains a maximum value in February and a minimum value in the course of a year. As an illustration and the minimum in August every year, then it is the annual change.
3) Daily Change:
The periodic change extending over 24-hours in the value of an element is as well noticed. An element arrives at the maximum value at some hour of the day and the minimum value at some other hour, features of the element.
4) Magnetic Storm:
This has been found that throughout volcanic eruptions, display of Aurora Borealis, appearance of sunspots, and so on, sudden and violent changes take place in the indications of recording instruments computing the magnetic elements. Such changes are stated to be due to magnetic storms. They are not periodic.
Magnetic Maps:
The values of the magnetic elements at diverse places are not generally similar and magnetic maps have been drawn by joining such places on the geographical maps in which a magnetic element has equivalent values. In magnetic maps, we have the given lines.
1) Isogonic and Agonic Line:
Isogonic lines are lines connecting places on the map of the earth where the declination is similar. Agonic lines are such which pass via places having zero declination.
2) Isoclinic and Aclinic Lines:
Isoclinic lines are the lines connecting places on the map of the earth where the magnetic dip is similar. A line passing via places having no dip is termed as Aclinic line.
3) Isodynamic Lines:
Such lines join up places on the map of the earth where the value of horizontal intensity is similar.
The belt round the earth's surface passing via places of no dip is the magnetic equator. The part of the earth's surface comprised between the magnetic pole and the magnetic equator has been divided into 90 equivalent parts. Via each such point of division a circle has been drawn round the surface of earth parallel to the great circle of the magnetic equator. These circles are termed as the geomagnetic latitudes.
Tutorsglobe: A way to secure high grade in your curriculum (Online Tutoring)
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## TPTP Problem File: CSR119^3.p
View Solutions - Solve Problem
```%------------------------------------------------------------------------------
% File : CSR119^3 : TPTP v7.2.0. Released v5.3.0.
% Domain : Commonsense Reasoning
% Problem : Did someone like Bill in 2009?
% Version : Especial.
% English : During 2009 Mary liked Bill and Sue liked Bill. Is it the case
% that someone liked Bill during 2009?
% Refs : [BP10] Benzmueller & Pease (2010), Progress in Automating Hig
% : [Ben11] Benzmueller (2011), Email to Geoff Sutcliffe
% Source : [Ben11]
% Names :
% Status : Theorem
% Rating : 0.56 v7.2.0, 0.50 v7.1.0, 0.62 v7.0.0, 0.86 v6.4.0, 0.83 v6.3.0, 0.80 v6.2.0, 0.71 v6.1.0, 0.86 v5.5.0, 0.83 v5.4.0, 0.60 v5.3.0
% Syntax : Number of formulae : 5014 ( 0 unit;1435 type; 0 defn)
% Number of atoms : 20864 ( 406 equality;6086 variable)
% Maximal formula depth : 26 ( 4 average)
% Number of connectives : 16473 ( 0 ~; 77 |;1314 &;14029 @)
% ( 105 <=>; 948 =>; 0 <=; 0 <~>)
% ( 0 ~|; 0 ~&)
% Number of type conns : 1319 (1319 >; 0 *; 0 +; 0 <<)
% Number of symbols : 1441 (1435 :; 0 =)
% Number of variables : 2571 ( 0 sgn;2075 !; 494 ?; 2 ^)
% (2571 :; 0 !>; 0 ?*)
% ( 0 @-; 0 @+)
% SPC : TH0_THM_EQU_NAR
%------------------------------------------------------------------------------
%----Include SUMO axioms
include('Axioms/CSR005^0.ax').
%------------------------------------------------------------------------------
%----The extracted signature
thf(lBill_THFTYPE_i,type,(
lBill_THFTYPE_i: \$i )).
thf(lMary_THFTYPE_i,type,(
lMary_THFTYPE_i: \$i )).
thf(lSue_THFTYPE_i,type,(
lSue_THFTYPE_i: \$i )).
thf(likes_THFTYPE_IiioI,type,(
likes_THFTYPE_IiioI: \$i > \$i > \$o )).
thf(n2009_THFTYPE_i,type,(
n2009_THFTYPE_i: \$i )).
%----The translated axioms
thf(ax,axiom,
( holdsDuring_THFTYPE_IiooI @ ( lYearFn_THFTYPE_IiiI @ n2009_THFTYPE_i )
@ ( ( likes_THFTYPE_IiioI @ lMary_THFTYPE_i @ lBill_THFTYPE_i )
& ( likes_THFTYPE_IiioI @ lSue_THFTYPE_i @ lBill_THFTYPE_i ) ) )).
%----The translated conjecture
thf(con,conjecture,(
? [Y: \$i] :
( holdsDuring_THFTYPE_IiooI @ ( lYearFn_THFTYPE_IiiI @ n2009_THFTYPE_i ) @ ( likes_THFTYPE_IiioI @ Y @ lBill_THFTYPE_i ) ) )).
%------------------------------------------------------------------------------
```
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www.platodemusgo.com
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# Horizontal And Vertical Analysis Excel
Thereby, achieving a goal of the budgeting process to determine the firm’s game plan. This ratio is a measure of the ability of a firm to turn Inventory into Sales.
For example, the current period’s profits may appear excellent when only compared with those of the previous month, but are actually quite poor when compared to the results for the same month in the preceding year. When creating a Vertical Analysis for a balance sheet, total assets are used as basis for analyzing each asset account. Total liabilities and stockholder’s equity is used as the basis for each liability and stockholder account. Prepare a vertical analysis of the income statement data for SPENCER Corporation in columnar form for both years. First calculate dollar change from the base year and then translate it into percentage change. Horizontal analysis is valuable because analysts assess past performance along with the company’s current financial position or growth.
The horizontal analysis technique uses a base year and a comparison year to determine a company’s growth. The same process applied to ABC Company’s balance sheet would likely reveal further insights into how the company is structured and how that structure is changing over time. Applying common-size analysis to firm’s balance sheet gives us a clear understanding of its capital structure, which can be compared to other firms or some optimal capital structure for the industry. It also allows to estimate whether some of the company’s debts being too high. E.g. HGY Company’s income statement for the year ended 2016 is shown below along with the financial results for the year 2015. If the decrease in cost of goods sold is in proportion to decrease in sales, it indicates that the management is able to manage it effectively.
Whether you do a horizontal analysis quarterly or yearly, it’s worth the time and effort to perform this calculation regularly. Accounting Accounting software helps manage payable and receivable accounts, general ledgers, payroll and other accounting activities. John Freedman’s articles specialize in management and financial responsibility. He is a certified public accountant, graduated summa cum laude with a Bachelor of Arts in business administration and has been writing since 1998. His career includes public company auditing and work with the campus recruiting team for his alma mater. The amount shown in the horizontal analysis will be of 200%, since ”Year 2” \$ 10,000 of cash corresponds to 200% of the cash in ”Year 1”. The amount shown in the horizontal analysis will be of 100%, since ”Year 2” \$ 5,000 of cash corresponds to 100% of the cash in ”Year 1”.
## Steps To Perform A Horizontal Analysis
Though this format does take longer to create, it makes it much easier to spot trends and get a look at business performance compared to the previous year or previous quarter. With horizontal analysis, you look at changes line-by-line, between specific accounting periods – whether it be monthly, quarterly, or annually.
In a horizontal analysis the the changes in income statement and balance sheet items are computed and compared with the expected changes. For example, you start an advertising campaign and expect a 25% increase in sales. But if sales revenue increases by only 5%, then it needs to be investigated.
Vertical analysis is a method of financial statement analysis in which each line item is listed as a percentage of a base figure within the statement. Financial statement analysis is the process of analyzing a company’s financial statements for decision-making purposes. For example, the vertical analysis of an income statement results in every income statement amount being restated as a percent of net sales. If a company’s net sales were \$2 million, they will be presented as 100% (\$2 million divided by \$2 million).
A percentage or an absolute comparison may be used in horizontal analysis. Let us now look at the horizontal analysis of Colgate’s Income Statement. Here we have the YoY growth rates of Colgate’s Income statement from 2008 until 2015. We calculate the growth rate of each of the line items with respect to the previous year.
Hello, if the problem only request the horizontal analysis show Net Sales, Gross profit and operating income of a company, how would it all be calculated and or determined? Are the numbers given by looking at the income statement or are there any calculations needed? The dollar and percentage changes of the items of balance sheet, schedule of current assets, or the statement of retained earnings are computed in the similar way. In above analysis, 2007 is the base year and 2008 is the comparison year. All items on the balance sheet and income statement for the year 2008 have been compared with the items of balance sheet and income statement for the year 2007. The earliest period is usually used as the base period and the items on the statements for all later periods are compared with items on the statements of the base period. A common size income statement is an income statement in which each line item is expressed as a percentage of the value of sales, to make analysis easier.
Just like analysing the income statement, historical data comparison of the balance sheet can be done in whole or in part. Horizontal analysis of the balance sheet is also usually in a two-year format, such as the one shown below, with a variance showing the difference between the two years for each line item. An alternative format is to add as many years as will fit on the page, without showing a variance, so that you can see general changes by account over multiple years. A less-used format is to include a vertical analysis of each year in the report, so that each year shows each line item as a percentage of the total assets in that year.
Hi I just want to know how to calculate the % difference for horizontal analysis. to investigate unexpected increases or decreases in financial statement items. As stated before, this method is best used when comparing similar companies apples-to-apples. No two companies are the same, and this analysis shows only a very small piece of the overall pie when determining whether a company is a good buy, or not. A cash flow Statement contains information on how much cash a company generated and used during a given period. You don’t need any special financial skill to ascertain the difference between the previous year’s data and last year’s data. All you need is diligence, attention to details, and a logical mind to decipher why the change happens.
## Understanding Horizontal Analysis
It would require the arrangement and calculation of interlinked numbers and dates. Particularly, interlinks among the numbers make financial analysis tiresome and complex for a typical businessperson. A solution is to create Comparative Financial Statements, which depicts the results of Horizontal Analysis and show the trends relative to only one base year. The baseline acts as a peg for the other figures while calculating percentages. For example, in this illustration, the year 2012 is chosen as a representative year of the firm’s activity and is therefore chosen as the base. Without analysis, a business owner may make mistakes understanding the firm’s financial condition. For example, an Assets to Sales ratio is a measure of a firm’s productive use of Assets.
You use horizontal analysis to find and monitor trends over a period of time. Instead of creating an income statement or balance sheet for one period, you would also create a comparative balance sheet or income statement to cover quarterly or annual business activities.
Once you create a template, you can use it again and again as needed. This means that some organizations maneuver the growth and profitability trends reported in the analysis with a combination of methods to break down business segments. Even so, one-off events and accounting changes can be implemented to correct these anomalies to improve the accuracy of the analysis. The comparability constraint dictates that your statements and documents need to be evaluated against companies similar to yours within the same industry. Horizontal analysis improves and enhances the constraints during financial reporting.
• Next, changes in significant individual items, such as receivables and inventory, would be examined.
• For example, this analysis can be performed on revenues, cost of sales, expenses, assets, cash, equity and liabilities.
• Horizontal analysis, also called “trend analysis,” is used to discover trends in the earnings, assets and liabilities of a company over the course of several years.
• The percentage change in gross profit has been relatively higher than that of net sales due to a lower increase in the cost of goods sold.
• If the decrease in cost of goods sold is in proportion to decrease in sales, it indicates that the management is able to manage it effectively.
Frequently used tools of financial statement analysis consist of horizontal analysis, vertical analysis and ratio analysis. However, this was followed by a slight decrease of this ratio during the year 3. Notable is also an increasing trend of gross profit margin over the period of three years. From this common-size statement of profit and loss we also can notice a big percent of research expenses, which means the company is trying to be innovative and invests resources in the development.
## Horizontal Company Financial Statement Analysis
The length of each bar indicates the quantity that the bar represents. Organizing and listing information in a way which shows the relationships between the information is called a table. Plans, activities, and statements of progress can be presented in a table. Numbers such as survival rates of seedlings by species are also easily presented in a table. Tables which present only a few items of information are most effective.
Using actual dollar amounts would be ineffective when analyzing an entire industry, but the common-sized percentages of the vertical analysis solve that problem and make industry comparison possible. ABC Company’s income statement and vertical analysis demonstrate the value of using common-sized financial statements to better understand the composition of a financial statement. It also shows how a vertical analysis can be very effective in understanding key trends over time. To do that, we’ll create a “common size income statement” and perform a vertical analysis. For each account on the income statement, we divide the given number by the company’s sales for that year.
For example, to find the growth rate of Net Sales of 2015, the formula is (Net Sales 2015 – Net Sales 2014) / Net Sales 2014. Cost Of Goods SoldThe cost of goods sold is the cumulative total of direct costs incurred for the goods or services sold, including direct expenses like raw material, direct labour cost and other ledger account direct costs. However, it excludes all the indirect expenses incurred by the company. First, we need to take the previous year as the base year and last year as the comparison year. For example, let’s say we are comparing between 2015 and 2016; we will take 2015 as the base year and 2016 as the comparison year.
The company reported a net income of \$25 million and retained total earnings of \$67 million in the current year. Since we do not have any further information about the segments, normal balance we will project the future sales of Colgate on the basis of this available data. We will use the sales growth approach across segments to derive the forecasts.
These documents can also show a company’s emerging successes and potential weaknesses, based on metrics such as inventory turnover, profit margin, and return on equity. So, in vertical analysis, the figures are not only compared to the past year, but they are also represented as a percentage of the total cost or total assets/liabilities as may be the case. A notable problem with the horizontal analysis is that the compilation of financial information may vary over time. It means that elements of financial statements, such as liabilities, assets, or expenses, may change between different accounting periods, leading to variation when account balances for each accounting period are sequentially compared. Horizontal analysis, or trend analysis, is a method where financial statements are compared to reveal financial performance over a specific period of time. Horizontal Analysis doesn’t conclude with finding the change in sales over a period. To get a clear picture of the performance of our business, we need to do a horizontal analysis of each item in our income statement.
To make the best use of your financial data, you need a robust toolkit with plenty of options for slicing and dicing information in meaningful how to perform horizontal analysis ways. Further analysis via horizontal analysis will likely be required to unlock those insights, and make use of them in a strategic way.
Horizontal Analysis – analyzes the trend of the company’s financials over a period of time. Business owners can use company financial analysis both internally and externally. They can use them internally to examine issues such as employee performance, the efficiency of operations and credit policies.
In the same vein, a company’s emerging problems and strengths can be detected by looking at critical business performance, such as return on equity, inventory turnover, or profit margin. Trends or changes are measured by comparing the current year’s values against those of the base year. The goal is to determine any increase or decline in specific values that has taken place.
## How To Interpret The Vertical Analysis Of A Balance Sheet And Income Statement
However, these expenses don’t, at first glance, appear large enough to account for the decline in net income. For example, when a vertical analysis is done on an income statement, it will show the top-line sales number as 100%, and every other account will show as a percentage of the total sales number. Financial analysts use a broad range of techniques that are collectively known as ratio analysis. The general procedure involves calculating various financial ratios — such as profit margin, accounts receivable-to-sales, and inventory turnover ratios — and comparing them to other companies or general rules of thumb. There are hundreds of financial ratios employed and even different methods of calculating the same ratios. For this reason, ratio analysis is considered to be more of an art than a science. In horizontal analysis of balance sheet, you compute the increase or decrease of each balance sheet item in comparison to prior years.
Author: Roman Kepczyk
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Community Profile
# asif iqbal
### cetpa infotech pvt. ltd.
26 total contributions since 2013
MATLAB Expert at Cetpa Infotech Pvt. Ltd.Working in research field,did my project on various parts of Rayleigh,Rician and Nakagami fading channels as well as many signal processing projects including voice recognition system.Strong background in signal processing and Mathematics..Trained more than 1000 students,researchers and professionals across the country from renowned institute like IIT,DCE,schneider electrical,galgotia,GGSIPU,COER and many more.Delivered so many workshops and seminars across the country.
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A056813 Largest non-unitary prime factor of LCM(1,...,n); that is, the largest prime which occurs to power > 1 in prime factorization of LCM(1,..,n). 3
%I
%S 1,1,1,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,5,5,5,5,5,5,5,5,5,5,
%T 5,5,5,5,5,5,5,5,5,5,5,5,5,5,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,
%U 7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7
%N Largest non-unitary prime factor of LCM(1,...,n); that is, the largest prime which occurs to power > 1 in prime factorization of LCM(1,..,n).
%C For n>0, prime(n) appears {(prime(n+1))^2 - (prime(n))^2} times [from n=(prime(n))^2 to n=(prime(n+1))^2 - 1], that is, A000040(n) appears A069482(n) times (from n=A001248(n) to n=A084920(n+1)). - _Lekraj Beedassy_, Mar 31 2005
%C a(n) is the largest prime factor of A045948(n). [_Matthew Vandermast_, Oct 29 2008]
%C Alternative definition: a(n) = largest prime <= sqrt(n) (considering 1 as prime for this occasion, see A008578 for the 19th century definition of primes). - _Jean-Christophe Hervé_, Oct 29 2013
%H Jean-Christophe Hervé, <a href="/A056813/b056813.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = prime(w) if prime(w)^2 <= n < prime(w+1)^2.
%F To get the sequence, repeat 1 three times, and then for any k >= 1, repeat A000040(k) A069482(k) times; or equivalently, for any k >= 1, repeat A008578(k) a number of times equal to A008578(k+1)^2 - A008578(k)^2. - _Jean-Christophe Hervé_, Oct 29 2013
%e The j-th prime appears at the position of its square, at n = prime(j)^2.
%t Table[f = Transpose[FactorInteger[LCM @@ Range[n]]]; pos = Position[f[[2]], _?(# > 1 &)]; If[pos == {}, 1, f[[1, pos[[-1]]]][[1]]], {n, 100}] (* _T. D. Noe_, Oct 30 2013 *)
%Y Cf. A054041, A056170.
%Y Cf. A000040, A001248, A008578, A069482.
%K nonn,easy
%O 1,4
%A _Labos Elemer_, Aug 28 2000
%E Corrected offset by _Jean-Christophe Hervé_, Oct 29 2013
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Last modified June 22 20:06 EDT 2021. Contains 345388 sequences. (Running on oeis4.)
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| 3.359375
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CC-MAIN-2021-25
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latest
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www.goldsoonhis.com
| 1,368,983,240,000,000,000
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# perhaps higher mathematics can calculate
70 17 Critical Strike Rating = 0.773% crit chance, if your basic 20% crit, then the effect of this card is equivalent to an increase of 3% crit probability, if your base crit 10%, then this card representing an increase of 3.37% the effect of the explosion hit probability. This figure is derived, may be a slight error, but the total will not exceed 0.2% of the difference. The calculation idea is to calculate a hit rate of the burst to the burst rate for 20 consecutive hit and then averaged out, Also note that the burst rate for 20 consecutive hit is not a buff superposed burst rate, but appear to exclude the top 20 hit crit probability derived. The calculation below: 1 hit consecutive burst rate of 20% hit 1 for 2, the first strike crit, hit the second explosion hit a separate 20% probability, the probability of the second explosion hit consecutive hit * 20% = 20% 4% 2 The first strike is not critical strike, the second hit of the individual probability of 20.773% critical strike, the second consecutive hit probability of 80% crit * 20.773% = 16.62% probability of the second hit always crit = 4% +16.62% = 20.62% hit 1 for 3, hit crit the second and third hit probability of 20.62% crit * 20% = 4.124% 2, the first strike crit, hit not crit the second and third hit of the explosion hit probability = 20 % * (1-20.62%) * 20.773% = 3.3% 3, the first strike does not crit, hit not crit the second and third hit of the explosion hit probability = 80% * (1-20.62%) * 21.546% = 13.68% of total third-strike probability = 4.124% crit +3.3% +13.68% = 21.104% for four consecutive hit a third strike crit, crit fourth hit probability 21.104% * 20% = 4.22% 2. first strike crit, hit the second explosion hit, the third strike does not crit, crit fourth hit probability = 20% * 20% * (1-21.104%) * 20.773% = 0.66% 3, the first strike crit, the second hit does not crit,gold wow, the third strike does not crit, crit fourth hit probability = 20% * (1-20.62%) * (1-21.104%) 2.7% * 21.546% = 4, blow not crit, hit the second explosion hit, the third strike does not crit, crit fourth hit probability = 80% * 20.773% * (1-21.104%) * 20.773% = 2.72% 5, the first strike not crit,aion online gold, and the second hit does not crit, hit the third strike is not explosive, the fourth hit of the explosion hit probability = 80% * (1-20.62%) * (1-21.104%) * 22.319% = 11.19% the fourth The total probability of hit crit = 4.22% +0.66% +2.7% +2.72% +11.19% = 21.49% and then not in detail later in the projection, the projection of the steps needed too much, hit 20 consecutive 19-th power of 2 to calculate the possible and, perhaps higher mathematics can calculate, I die. But the overall trend of a crit can see, with the continuous increase in the number of fire, explosion blow down the curve to enhance the start, the first three times I used to replace the linear relationship between the decline in subsequent calculations, so the result will be slightly higher than the actual figure, but because the difference itself is relatively small, this result is a reference value. According to my projections, 20 consecutive crit hit probability at 24.4%, while the probability of a critical strike with a 1-20 overall averaged 23% probability of explosion hit about, this card is less than 3% raise crit probability. Also note that, if the underlying probability crit down,runescape money, and we think the same, to enhance the probability will increase the critical strike, but also very limited, I calculated the 10 percent under the basic,warhammer sale, enhanced also 3.37% only. Another 3% crit at 70 = 66 critical strike rating, choose to see for yourself, not very bt things, can not be considered waste, a matter of opinion.
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| 3.71875
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https://www.queryhome.com/puzzle/29282/what-probability-leave-none-them-wearing-either-their-shoes
| 1,725,857,984,000,000,000
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# What is the probability that when they leave that none of them is wearing either of their own shoes?
318 views
Three ladies get together for a bit of fun. When they arrive they all kick off their shoes. Unfortunately, when they leave they are not quite in the same frame of mind so they each grab a left shoe and a right shoe at random, and put them on.
What is the probability that when they leave that none of them is wearing either of their own shoes?
posted Sep 30, 2018
If the shoes are named
L1 R1 ==> Women 1 (W1)
L2 R2 ==> Women 2 (W2)
L3 R3 ==> Women 3 (W3)
Now all the possible favourable events for the given probability is
[L2 R2 (W1)]×[L3 R3 (W2)]×[L1 R1 (W3)] = 1/9 × 1 × 1/4 = 1/36
[L3 R3 (W1)]×[L1 R1 (W2)]×[L2 R2 (W3)] = 1/9 × 1/4 × 1 = 1/36
[L2 R3 (W1)]×[L3 R1 (W2)]×[L1 R2 (W3)] = 1/9 × 1/2 × 1/2 = 1/36
[L3 R2 (W1)]×[L1 R3 (W2)]×[L2 R1 (W3)] = 1/9 × 1/2 × 1/2 = 1/36
Required Probability = 1/36 + 1/36 + 1/36 + 1/36 = 1/9.
1/9
.
Let's look at the left foot first.
The chance that none has their own shoe is 2/6.
1/2/3 - at least 1 correct
1/3/2 - at least 1 correct
2/1/3 - at least 1 correct
2/3/1 - all wrong
3/1/2 - all wrong
3/2/1 - at least 1 correct
.
The same holds for the right shoe.
Thus the change that none has either shoe correct is 2/6 * 2/6 = 4/36 = 1/9
Similar Puzzles
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http://www.jiskha.com/display.cgi?id=1221985819
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Sunday
February 7, 2016
# Homework Help: Integral Calculus
Posted by Akitsuke on Sunday, September 21, 2008 at 4:30am.
Find the area of the surface generated by revolving the given curve about the y-axis.
8xy^2=2y^6+1 , 1<=y<=2.
thank you so much.. :)
• Integral Calculus - drwls, Sunday, September 21, 2008 at 8:01am
You need x as a function of y to do the integration.
In this case, x(y) = 1/(8y^2) + (y^4)/4
To get the surface area, evaluate:
Integral of 2 pi x(y) dy
(y = 1 to 2)
The indefinite integral is
2 pi [-1/(24y^3) + y^5/20]
= pi[(y^5)/10 - 1/(24y^3)]
Evaluate it at y=2 and y=1, and take the difference. Check my math.
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https://republicofsouthossetia.org/question/find-the-radius-of-a-circle-in-which-the-central-angle-a-intercepts-an-arc-of-the-given-length-s-15162755-15/
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## Find the radius of a circle in which the central angle, a, intercepts an arc of the given length s. a = 156°, s = 25 km
Question
Find the radius of a circle in which the central angle, a, intercepts an arc of the given length s. a = 156°, s = 25 km
in progress 0
3 weeks 2021-09-08T13:03:33+00:00 1 Answer 0
Radius of the circle is 9.178
Step-by-step explanation:
Given that,
Central angle ,a = 156°
Arc length,s =25 km
The formula to calculate the radius of a circle is;
Arc length= Θ / 360° x (2πr)
We are given that the Θ =20° and that arc length is 66 in. Substituting, we have;
L = Θ / 360° x (2πr)
25 = (156 / 360°) x 2 x 22/7 x r
25 = 6864r / 2520
r= (2520 x 25) / 6864
r= 63000 / 6864
r = 9.178
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tensorflow实现BP算法遇到了问题，求大神指点！！！
import tensorflow as tf
import numpy as np
#from tensorflow.examples.tutorials.mnist import input_data
#载入数据集
#每个批次的大小
#batch_size = 100 #?????????????????????????????????
#计算一共有多少个批次
#n_batch = mnist.train.num_examples // batch_size
#定义placeholder
x_data=np.mat([[0.4984,0.5102,0.5213,0.5340],
[0.5102,0.5213,0.5340,0.5407],
[0.5213,0.5340,0.5407,0.5428],
[0.5340,0.5407,0.5428,0.5530],
[0.5407,0.5428,0.5530,0.5632],
[0.5428,0.5530,0.5632,0.5739],
[0.5530,0.5632,0.5739,0.5821],
[0.5632,0.5739,0.5821,0.5920],
[0.5739,0.5821,0.5920,0.5987],
[0.5821,0.5920,0.5987,0.6043],
[0.5920,0.5987,0.6043,0.6095],
[0.5987,0.6043,0.6095,0.6161],
[0.6043,0.6095,0.6161,0.6251],
[0.6095,0.6161,0.6251,0.6318],
[0.6161,0.6251,0.6318,0.6387],
[0.6251,0.6318,0.6387,0.6462],
[0.6318,0.6387,0.6462,0.6518],
[0.6387,0.6462,0.6518,0.6589],
[0.6462,0.6518,0.6589,0.6674],
[0.6518,0.6589,0.6674,0.6786],
[0.6589,0.6674,0.6786,0.6892],
[0.6674,0.6786,0.6892,0.6988]])
y_data=np.mat([[0.5407],
[0.5428],
[0.5530],
[0.5632],
[0.5739],
[0.5821],
[0.5920],
[0.5987],
[0.6043],
[0.6095],
[0.6161],
[0.6251],
[0.6318],
[0.6387],
[0.6462],
[0.6518],
[0.6589],
[0.6674],
[0.6786],
[0.6892],
[0.6988],
[0.7072]])
xs = tf.placeholder(tf.float32,[None,4]) # 样本数未知，特征数为1，占位符最后要以字典形式在运行中填入
ys = tf.placeholder(tf.float32,[None,1])
#创建一个简单的神经网络
W1 = tf.Variable(tf.truncated_normal([4,10],stddev=0.1))
b1 = tf.Variable(tf.zeros([10])+0.1)
L1 = tf.nn.tanh(tf.matmul(x,W1)+b1)
W2 = tf.Variable(tf.truncated_normal([10,1],stddev=0.1))
b2 = tf.Variable(tf.zeros([1])+0.1)
L2 = tf.nn.softmax(tf.matmul(L1,W2)+b2)
#二次代价函数
#loss = tf.reduce_mean(tf.square(y-prediction))
#loss = tf.reduce_mean(tf.nn.softmax_cross_entropy_with_logits(labels=ys,logits=prediction))
loss = tf.reduce_mean(tf.reduce_sum(tf.square((ys-L2)),reduction_indices = [1]))#需要向相加索引号，redeuc执行跨纬度操作
#使用梯度下降法
#初始化变量
#init = tf.global_variables_initializer()
init = tf.initialize_all_variables()
#结果存放在一个布尔型列表中
#correct_prediction = tf.equal(tf.argmax(y,1),tf.argmax(prediction,1))
#求准确率
#accuracy = tf.reduce_mean(tf.cast(correct_prediction,tf.float32))
with tf.Session() as sess:
sess.run(init)
for epoch in range(21):
for i in range(22):
#batch_xs,batch_ys = mnist.train.next_batch(batch_size) #?????????????????????????
sess.run(train_step,feed_dict={xs:x_data,ys:y_data})
#test_acc = sess.run(accuracy,feed_dict={x:mnist.test.images,y:mnist.test.labels,keep_prob:1.0})
#train_acc = sess.run(accuracy,feed_dict={x:mnist.train.images,y:mnist.train.labels,keep_prob:1.0})
print (sess.run(prediction,feed_dict={xs:x_data,ys:y_data}))
`````` 提示：WARNING:tensorflow:From <ipython-input-10-578836c021a3>:89 in <module>.: initialize_all_variables (from tensorflow.python.ops.variables) is deprecated and will be removed after 2017-03-02.
``````
Instructions for updating:
Use `tf.global_variables_initializer` instead.
InvalidArgumentError Traceback (most recent call last)
1020 try:
-> 1021 return fn(*args)
1022 except errors.OpError as e:
1002 feed_dict, fetch_list, target_list,
1004
。。。
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# How to Create a Bland-Altman Plot in Python
A Bland-Altman plot is used to visualize the differences in measurements between two different instruments or two different measurement techniques.
It’s useful for determining how similar two instruments or techniques are at measuring the same construct.
This tutorial provides a step-by-step example of how to create a Bland-Altman plot in Python.
### Step 1: Create the Data
Suppose a biologist uses two different instruments (A and B) to measure the weight of the same set of 20 different frogs, in grams.
We’ll create the following data frame that represents the weight of each frog, as measured by each instrument:
```import pandas as pd
df = pd.DataFrame({'A': [5, 5, 5, 6, 6, 7, 7, 7, 8, 8, 9,
10, 11, 13, 14, 14, 15, 18, 22, 25],
'B': [4, 4, 5, 5, 5, 7, 8, 6, 9, 7, 7, 11,
13, 13, 12, 13, 14, 19, 19, 24]})
```
### Step 2: Create the Bland-Altman Plot
Next, we’ll use the mean_diff_plot() function from the statsmodels package to create a Bland-Altman plot:
```import statsmodels.api as sm
import matplotlib.pyplot as plt
#create Bland-Altman plot
f, ax = plt.subplots(1, figsize = (8,5))
sm.graphics.mean_diff_plot(df.A, df.B, ax = ax)
#display Bland-Altman plot
plt.show()```
The x-axis of the plot displays the average measurement of the two instruments and the y-axis displays the difference in measurements between the two instruments.
The black solid line represents the average difference in measurements between the two instruments while the two dashed lines represent the 95% confidence interval limits for the average difference.
The average difference turns out to be 0.5 and the 95% confidence interval for the average difference is [-1.86, 2.86].
## One Reply to “How to Create a Bland-Altman Plot in Python”
1. Max says:
Thank you, this was very helpfull.
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## University of Ottawa NMR Facility Web Site
Please feel free to make suggestions for future posts by emailing Glenn Facey.
## Thursday, November 18, 2010
### The Phase of an NMR Spectrum
Most students know that the phase of an NMR spectrum has to do with the degree to which the NMR resonances are above or below the baseline of the spectrum (i.e. the amount of absorption and dispersion character). Most students have also learned that the phase of a periodic time domain function depends only on the value of the function at time zero. Thus, the only difference between a cosine and a sine function is where the function starts at time zero. A sine is said to be 90° out of phase with respect to a cosine. Many students do not understand the connection between the phase of their NMR spectrum and the phase of the periodic time domain function giving rise to their spectrum by way of a Fourier transform. The figure below is an attempt to make the connection.
The left-hand portion of the figure shows equilibrium magnetization vectors being rotated by radio frequency pulses. 90° pulses along the x, y, -x and -y axes rotate the z magnetization vector to the -y, x, y and -x axes of the rotating frame, respectively according to the right-hand screw rule. After the pulse, the magnetization vector rotates in the rotating frame of reference at a frequency equal to the difference between the transmitter frequency and the frequency of the NMR resonance. In the figure, the magnetization is assumed to be rotating anti-clockwise representing an NMR resonance with a positive frequency with respect to the transmitter frequency. The NMR spectrometer measures the time dependent voltages on two of the four orthogonal axes of the rotating frame separated by 90° (quadrature detection). The time dependent voltages are proportional to the amount of magnetization on the axis as a function of time. One of the two time dependent voltages is called the "real" signal and the other is called the "imaginary" signal. Together these two signals make up the complex free induction decay (FID) which is Fourier transformed to produce the NMR spectrum. In the figure, the -y(t) voltage is the real FID and the x(t) voltage is the imaginary FID. The figure shows representations of the real and imaginary FIDs after the delivery of pulses along each of the orthogonal axes. Note that the phases of the real and imaginary FIDs depend on the pulse delivered. For example, after a 90°x pulse, the magnetization resides on the -y axis. The real FID (along the -y axis) starts at a maximum (cosine) and the imaginary FID (along the x axis) starts at zero (sine) and increases as the magnetization vector rotates anti-clockwise. The Fourier transform of the complex FID produces a real spectrum (typically the one displayed to the user) and an imaginary spectrum (typically not displayed to the user). Note that the degree of absorption vs. dispersion character in the spectrum depends on the phase of the FID signals. If an NMR spectrum is not in phase, perhaps due to a receiver dead time problem, it can be corrected after the collection of the data by calculating the phase angles needed to put the real spectrum enitirely in absorption mode and the imaginary spectrum entirely in dispersion mode.
#### 3 comments:
Glenn Facey said...
The figure in this post was corrected on March 17, 2016 after an error was pointed out by Huldrych Egli. Thank you Huldrych.
Anonymous said...
My question has to do with the receiver phase. One usually sets the receiver phase such that most of the signal (when on resonance) is either on the real or the imaginary parts of the fid. In our spectrometer I noticed that this receiver phase is not stable over time, one needs to modify it occasionally to keep the fid in the right phase. Is this a magnet, probe, or spectrometer issue? I have also noticed that the probe wobb tends to drift slightly over time.
Thank you.
Glenn Facey said...
Anonymous,
The drift of the receiver phase is most likely a spectrometer issue. Note however that it is not necessary to have all of the on-resonance signal in either the real or imaginary channel. The phase of the spectrum can always be corrected after the data collection.
I have also noticed drift in probe tuning. I have always attributed it to temperature fluctuations, as it is often much more pronounced while running pulse sequences with higher duty cycles producing heat.
Glenn
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DOI: https://doi.org/10.32347/2410-2547.2019.102.109-120
### Stress-strain state and equation of the vertical movement of the cored body of rotation – disc under electromagnetic fields
Oleksii Hrevtsev, Ninel Selivanova
#### Abstract
Investigeted the problem of the theory of thermoelasticity and electromagnetoelasticity for solid of revolution, in particular a cored disk of variable thickness, loaded by an axisymmetric temperature field and volume forces: gravity forces, ponderomotive forces (mechanical forces acting per unit volume of the conducting medium) and inertia forces is considered.
The problem of determining the equilibrium or motion of a continuous bounded medium is a complex problem. For solving it is necessary to first solve the equations mechanics of continue together with the equations of electrodynamics into account the thermomechanical interaction of the environment as a result the deformation. While investigating the stress state of an electrically conductive body in an electromagnetic field, it is assumed that the body is in a magnetic field created by an electric current in the body itself, where an electric current flows and electric charges are generated. Under the influence of its own electromagnetic impulse field, the body must move independently. In circuits in which an alternating current flows in magnitude and constant in direction, Newton's law is not valid (the equilibrium condition of the ponderomotive forces the mechanical forces acting on the body from the side of the electromagnetic field). These forces will also cause the motion of the body of revolution.
As a result of the research, differential equations were obtained for the determination of displacements and the equation for the vertical motion of a body of revolution of variable thickness.
The conditions for the movement of the cored disk under the action of its own electromagnetic pulse field are determined. The stress state of the rotating body is investigated. It was shown that the tangential, axial and radial stresses in the body of the cored disk are absent, that is they are equal to zero. The only voltage other than zero is the circumferential tension. It is shown that the temperature field appears when the ponderomotive forces caused by the electromagnetic field arise and is the result of deformation of the body of rotation - a cored disc.
#### Keywords
thermoelasticity; electromagnetoelasticity; electromagnetic impulse field; axisymmetric stress; axisymmetric deformation; inertia force; gravity force; ponderomotive forces; temperature field; relativistic mass
PDF (Українська)
#### References
Parton V., Pearlyi P. Metodyi matematicheskoy teorii uprugosti (Methods of the mathematical theory of elasticity). - M.: Nauka, 1981. - 428 p.
Sedov L. Mehanika sploshnoy sredy (Continuum Mechanics). - M.: Science, 1976. - Vol. I. - 483 p.
Germain P. Kurs mehaniki sploshnyh sred (Course in the mechanics of continuous media). - Moscow: Higher School, 1983. - 399 p.
Timoshenko S., Goudier J. Teoria uprugosti (Theory of Elasticity). - Moscow: Nauka, 1979. - 560 p.
Yavorsky B., Pinsky A. Osnovy fiziki (Basics of physics). - Moscow: Nauka, 1981.-1. -396 с.
Melan E., Parkus G. Temperaturnye napriazhenia, vyzyvaemye stacionarnymi potokami (Temperature stresses caused by stationary fields). - Moscow: Fizmatgiz, 1982. – 167 s.
Saveliev I. Kurs obschey fiziki (The course of general physics). - Moscow: Fizmatgiz, 1982. - Vol.1 -411s.
Timoshenko S. Kurs teorii uprugosti (Course of the theory of elasticity). - K .: Naukova Dumka, 1972. - 501s.
Ryabov A., Fedorenko Y. Ob odnom metode reshenia zadachi teorii uprugosti dlia tel vraschenia (A method for solving the elasticity problem for bodies of revolution, Mat. methods and physical-mechanical fields). - 1988. - Issue. 28. - p. 58 - 62.
Grevtsev A. Reshenie zadachi termouprugosti dlya vraschauschihsia aksial’nyh tel peremennoy tolschiny (Solution of the problem of thermoelasticity for rotating axial bodies of variable thickness). // Construction and architecture. Novosibirsk, 1991. - № 4. - p.33 - 37.
Grevtsev A. Pro odyn metod rozv’iazannia osesymetrychnoi zadachi teorii pruzhnosti dlia nerivnomirno nagritogo obertovogo dyska zminnoi tovschiny (About one method of solving an axisymmetric problem of elasticity theory for an unevenly heated rotating disk of variable thickness). // Resistance of materials and theory of structures. - 1998. - Vip. 64. - P. 76-86.
Grevtsev A., Kharchenko S. Pro odyn metod rozv’iazannia osesymetrychnoi temperaturnoi zadachi teorii termopruzhnosti dlia nerivnomirno nagrityh til obertannia (About one method for solving the temperature problem of the theory of thermoelasticity for unevenly heated bodies of rotation). // Resistance of materials and theory of constructions. - 2003. - V. 73. - P. 65-72.
Madelung E. Matematicheskiy apparat fiziki (Mathematical apparatus of physics). - Moscow: Fizmatgiz, 1961. - 292 p.
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Copyright (c) 2019 Oleksii Hrevtsev, Ninel Selivanova
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# Distance between Salerno (QSR) and Albenga (ALL)
Flight distance from Salerno to Albenga (Salerno Costa d'Amalfi Airport – Riviera Airport) is 420 miles / 676 kilometers / 365 nautical miles. Estimated flight time is 1 hour 17 minutes.
Driving distance from Salerno (QSR) to Albenga (ALL) is 531 miles / 854 kilometers and travel time by car is about 8 hours 28 minutes.
## Map of flight path and driving directions from Salerno to Albenga.
Shortest flight path between Salerno Costa d'Amalfi Airport (QSR) and Riviera Airport (ALL).
## How far is Albenga from Salerno?
There are several ways to calculate distances between Salerno and Albenga. Here are two common methods:
Vincenty's formula (applied above)
• 420.180 miles
• 676.215 kilometers
• 365.127 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 419.564 miles
• 675.223 kilometers
• 364.591 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).
## Airport information
A Salerno Costa d'Amalfi Airport
City: Salerno
Country: Italy
IATA Code: QSR
ICAO Code: LIRI
Coordinates: 40°37′13″N, 14°54′40″E
B Riviera Airport
City: Albenga
Country: Italy
IATA Code: ALL
ICAO Code: LIMG
Coordinates: 44°3′2″N, 8°7′38″E
## Time difference and current local times
There is no time difference between Salerno and Albenga.
CET
CET
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 87 kg (192 pounds).
## Frequent Flyer Miles Calculator
Salerno (QSR) → Albenga (ALL).
Distance:
420
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
420
Round trip?
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# Problem: Calcium hydroxide and hydrobromic acid react together in the following reaction.Ca(OH)2 (aq) + HBr (aq) → CaBr2 (aq) + H2O (l)25.00 mL of 1.50 M HBr are needed to titrate 32.801 mL of Ca(OH)2. What is the molarity of Ca(OH)2?
###### FREE Expert Solution
We're being asked to calculate the molarity of Ca(OH)2.
We're given the unbalanced reaction: Ca(OH)2(aq) + HBr(aq) → CaBr2(aq) + H2O(l)
Balancing the reaction:
Ca(OH)2(aq) + 2 HBr(aq) → CaBr2(aq) + 2 H2O(l)
Step 1: Calculate the moles of Ca(OH)2.
molarity HBr (volume) → moles HBr (mole-to-mole comparison) → moles Ca(OH)2
Given:
95% (104 ratings)
###### Problem Details
Calcium hydroxide and hydrobromic acid react together in the following reaction.
Ca(OH)2 (aq) + HBr (aq) → CaBr2 (aq) + H2O (l)
25.00 mL of 1.50 M HBr are needed to titrate 32.801 mL of Ca(OH)2. What is the molarity of Ca(OH)2?
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# International System of Units
views updated May 29 2018
# International System of Units
The International System of Units (SI), which began as the decimal metric system during the French Revolution, deals with the definitions, terminology, proper usage, and modifications of scientific units. The metric system was established officially in France on June 22, 1799, and consisted of two standard measures: the meter for length and the kilogram for mass. The German mathematician and astronomer Carl Friedrich Gauss (17771855) promoted the use of the metric system and in 1832 added the second as the unit of time. The British Association for the Advancement of Science (BAAS) in 1874 introduced an alternative system, known as the CGS system, whose units of measure were the centimeter, gram, and second. Until 1889 the scientific community had two metric standards for length, mass, and time.
The first General Conference on Weights and Measures (Conférence Générale des Poids et Mesures, or CGPM) in 1889 sanctioned a new system, the MKS system, that included the international prototypes for the meter and kilogram and the astronomical second as the unit of time. Fifty years later, in 1939, the International Committee for Weights and Measures (Comité International des Poids et Mesures, or CIPM), under authority of the CGPM, proposed a four-unit MKS system with the addition of the ampere for electric current. Official recognition of the ampere had to wait until 1946, after World War II had ended.
The tenth CGPM in 1954 added two more standards when it officially approved both the kelvin for thermodynamic temperature and the candela for luminous intensity. In 1960 the eleventh CGPM renamed its MKS system of units the International System of Units, and in 1971 the fourteenth CGPM completed the seven-unit system in use today, with the addition of the mole as the unit for the amount of a substance, setting it equal to the gram-molecular weight of a substance.
SI units fall into two groups: basic units and derived units. The basic units are the seven mutually independent units (see Table 1) and include the meter, kilogram, second, ampere, kelvin, mole, and candela. They represent,
SI BASIC UNITS Base Quantity Base Unit Symbol SOURCE:Taylor, Barry N., ed., and National Institute of Standards and Technology (1995). Guide for the Use of the International System of Units (SI). Special Publication 811. Washington, DC: U.S. Government Printing Office. Length meter m Mass kilogram kg Time second s Electric current ampere A Thermodynamic temperature kelvin K Amount of substance mole mol Luminous intensity candela cd
SI-DERIVED UNITS Derived Quantity Name Symbol SOURCE:Taylor, Barry N., ed., and National Institute of Standards and Technology (1995). Guide for the Use of the International System of Units (SI). Special Publication 811. Washington, DC: U.S. Government Printing Office. Area square meter m2 Volume cubic meter m3 Velocity meter per second m/s Acceleration meter per second squared m/s2 Wave number reciprocal meter m−1 Mass density kilogram per cubic meter kg/m3 Specific volume cubic meter per kilogram m3/kg Current density ampere per square meter A/m2 Magnetic field strength ampere per meter A/m Amount-of-substance concentration mole per cubic meter mol/m3 Luminance candela per square meter cd/m2 Mass fraction kilogram per kilogram, which may be represented by the number 1 kg/kg = 1
respectively, length, mass, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity. Derived units, as the name indicates, are units obtained algebraically from the seven basic units (see Table 2).
The CIPM has approved twenty prefixes for SI units (see Table 3) and permits the use of any SI prefix with an SI unit, with one exception. The SI unit for mass, the kilogram, already has a prefix in its name and can have no other SI prefix. To use prefixes with a unit for mass, the rule is to remove the "kilo" prefix and add the new prefix to "gram" (unit symbol g), as in milligram and its abbreviation mg.
PREFIXES Factor Name Symbol SOURCE:Taylor, Barry N., ed., and National Institute of Standards and Technology (1995). Guide for the Use of the International System of Units (SI). Special Publication 811. Washington, DC: U.S. Government Printing Office. 1024 yotta Y 1021 zetta Z 1018 exa E 1015 peta P 1012 tera T 109 giga G 106 mega M 103 kilo k 102 hecto h 101 deka da 10−1 deci d 10−2 centi c 10−3 milli m 10−6 micro µ 10−9 nano n 10−12 pico p 10−15 femto f 10−18 atto a 10−21 zepto z 10−24 yocto y
The CIPM has developed several rules to ensure the proper use of SI units and to eliminate ambiguity from scientific communications. The National Institute of Standards and Technology in Washington, D.C., makes available a complete detailed list of the rules.
SI units, or those the CIPM recognizes, express quantity values. Other units, if used, may appear in parentheses after the appropriate SI or recognized units. The CIPM uses no abbreviations for names and only accepted unit symbols, unit names, prefix symbols, and prefix names. It makes no differentiation in symbol use for a plural, and the only time a period follows a unit symbol is at the end of a sentence. The addition of subscripts does not change a unit name or symbol.
Mathematical operations have specific rules for the use of mathematical symbols with SI units. A space or a half-high dot represents the multiplication of units; a negative exponent, horizontal line, or slash represents the division of units, and if these mathematical symbols appear in the same line, parentheses must differentiate them. The percent sign (%) denotes the number 0.01 or 1/100, so that 1% 0.01, 30% 0.30, and so forth. Arabic numerals with the appropriate SI or recognized unit indicate the values of quantities. Commas are not used to separate numbers into groups of three. If more than four digits appear on either side of the decimal point, a space separates the groups of three.
NON-SI UNITS ACCEPTED FOR USE Name Symbol Value in SI units SOURCE:Taylor, Barry N., ed., and National Institute of Standards and Technology (1995). Guide for the Use of the International System of Units (SI). Special Publication 811. Washington, DC: U.S. Government Printing Office Minute (time) min 1 min = 60 s Hour h 1 h = 60 min = 3,600 s Day d 1 d = 24 h = 86,400 s Degree (angle) ˚ 1˚ = (PI/180) rad Minute (angle) ′ 1′ = (1/60)˚ = (PI/10,800) rad Second (angle) ″ 1″ = (1/60)′ = (PI/648,000) rad Liter L 1L = 1 dm3 =10−3 m3 Metric ton t 1t = 103kg Neper Np 1 Np = 1 Bel B 1B = (1/2) ln 10 Np Electronvolt eV 1 eV = 1.602 18 × 10−9 J, approximately Unified atomic mass unit u 1u = 1.660 54 × 10−27 kg, approximately Astronomical unit ua 1 ua = 1.495 98 × 1011 m, approximately
Numbers, unit symbols, and names have set rules for mixing and differentiation for clarity of text and mathematical operations. These include a space between a numerical value and its unit symbol, indicating clearly the number a symbol belongs to in a given mathematical calculation, and no mixing of unit symbols and names nor making calculations on unit names. Different symbols represent values and units and the unit symbol should follow the value symbol separated by a slash. SI requires the use of standardized mathematical symbols and the explicit writing of a quotient quantity.
SI units and their symbols have distinctive type styles. Items given in italic type are variables, quantity symbols, superscripts and subscripts if they represent variables, quantities, or running numbers. Items given in roman type are unit symbols, superscripts, and subscripts that are descriptive. The typeface used in the surrounding text of the document does not change these rules.
Several terms used in vernacular science are not appropriate for scientific communication. The CIPM does not use such terms as parts per million, parts per billion, or parts per trillion or their abbreviations as expressions of quantities. The word "weight" is a force with the SI unit of newton, not a synonym for mass with the SI unit of kilogram. Terms for an object and quantities describing the object require a clear different action. Normality, molarity, and molal are obsolete terms no longer used.
Several important and widely used units are not officially part of the SI, but the CIPM has accepted them for use with SI units (see Table 4). They include the liter, day, hour, minute, electronvolt, and degree. Although the CIPM adopted the liter in 1879, it is not a current SI unit. Its symbol (L in the United States, l everywhere else) causes some confusion, but the CIPM has approved neither.
In the United States, the National Institute of Standards and Technology (NIST) also has compiled a list of units outside the SI that it has approved for domestic use (see Table 5). It recommends defining them in terms of accepted SI units. The CIPM does not encourage the use of units on the NIST list but accepts all of them, excluding the curie, roentgen, rad,
NON-SI UNITS APPROVED BY THE NIST Name Symbol Value in SI units SOURCE:Taylor, Barry N., ed., and National Institute of Standards and Technology (1995). Guide for the Use of the International System of Units (SI). Special Publication 811. Washington, DC: U.S. Government Printing Office. Nautical mile n/a 1 nautical mile = 1,852 m Knot n/a 1 nautical mile per hour = (1,852/3,600) m/s Are a 1 a = 1 dam2= 102 m2 Hectare ha 1 ha = 1 hm2= 104 m2 Bar bar 1 bar = 0.1 MPa = 100 kPa = 1,000 hPa = 105Pa Angstrom Å… 1 Å…= 0.1 nm = 10−10 m Barn b 1 b = 100 fm2= 10−28 m2 Curie Ci 1 Ci = 3.7 × 1010Bq Roentgen R 1 R = 2.58 × 10−4 C/kg Rad rad 1 rad = 1 cGy = 10−2 Gy Rem rem 1 rem = 1 cSv = 10−2 Sv
and rem, which American scientists have nonetheless continued to use.
Anthony N. Stranges
Sean McMaughan
## Bibliography
American Society for Testing and Materials (1993). Standard Practice for Use of the International System of Units (SI): The Modernized Metric System. E 38093. Philadelphia: American Society for Testing and Materials.
Institute of Electrical and Electronics Engineers (1992). American National Standard for Metric Practice. ANSI/IEEE Std 2661992. New York: Institute of Electrical and Electronics Engineers.
Mills, Ian; Cvitaš, T.; Homann, K.; Kallay, N.; and Kuchitsu, K. (1993). Quantities, Units and Symbols in Physical Chemistry, 2nd edition. Boston: Blackwell Scientific Publications.
Taylor, Barry N., ed., and National Institute of Standards and Technology (1995). Guide for the Use of the International System of Units (SI). Special Publication 811. Washington, DC: U.S. Government Printing Office.
Taylor, Barry N., ed. (1998). Interpretation of the International System of Units for the United States. Federal Register, July 28, 1998, 63, 144:4033440340. Washington, DC: U.S. Government Printing Office.
Taylor, Barry N., ed.; National Institute of Standards and Technology (2001). The International System of Units (SI). Special Publication 330. Washington, DC: U.S. Government Printing Office.
### Internet Resources
Bureau International des Poids et Mesures. "Welcome" (home page, English). Available from <http://www.bipm.fr/enus/welcome.html>.
National Institute of Standards and Technology, Physics Laboratory. "International System of Units." Available from <http://physics.nist.gov/cuu/Units/index.html>.
# SI
views updated May 08 2018
SI (SI units) Système International d'Unités, an internationally agreed group of units of measurement, used especially in scientific work. The system comprises seven basic and two supplementary units. The basic units are the: metre (m); kilogram (kg); second (s); ampere (A); kelvin (K); mole (mol); and candela (cd). The supplementary units are the: radian (rad); and steradian (sr). A further 18 units are derived from these: becquerel (Bq); coulomb (C); farad (F); gray (Gy); henry (H); hertz (Hz); joule (J); lumen (lm); lux (lx); newton (N); ohm (Ω); pascal (Pa); siemens (S); sievert (Sv); tesla (T); volt (V); watt (W); and weber (Wb).
# International System of Units
views updated Jun 11 2018
In·ter·na·tion·al Sys·tem of U·nits • n. a system of physical units (SI Units) based on the meter, kilogram, second, ampere, kelvin, candela, and mole, together with a set of prefixes to indicate multiplication or division by a power of ten.
# SI
views updated Jun 11 2018
SI • abbr. ∎ the international system of units of measurement. ∎ Law statutory instrument.
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A few days ago
Whitney
# I need help pleezze any1.?
Im stuck on these two problems..
1..Divide using synthetic division.
x^4-3x^3+x^2+5x-7 divided by x-1
2. Divide using synthetic division.
x^5+x^2-4 divided by x-3
A few days ago
1|…1…-3…1…5…-7
……….. 1…-2..-1…4
……1…-2…-1..4…-3
x^3 – 2x^2 – x + 4 – [3/( x^4-3x^3+x^2+5x-7)]
3|…1…0…0…1….0….-4
………..3…9…27..84..252
……1…3…9…28..84…248
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# Distance between New York, NY (LGA) and Kirksville, MO (IRK)
Flight distance from New York to Kirksville (New York LaGuardia Airport – Kirksville Regional Airport) is 984 miles / 1583 kilometers / 855 nautical miles. Estimated flight time is 2 hours 21 minutes.
Driving distance from New York (LGA) to Kirksville (IRK) is 1136 miles / 1829 kilometers and travel time by car is about 19 hours 54 minutes.
## Map of flight path and driving directions from New York to Kirksville.
Shortest flight path between New York LaGuardia Airport (LGA) and Kirksville Regional Airport (IRK).
## How far is Kirksville from New York?
There are several ways to calculate distances between New York and Kirksville. Here are two common methods:
Vincenty's formula (applied above)
• 983.726 miles
• 1583.153 kilometers
• 854.834 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 981.251 miles
• 1579.170 kilometers
• 852.684 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).
## Airport information
A New York LaGuardia Airport
City: New York, NY
Country: United States
IATA Code: LGA
ICAO Code: KLGA
Coordinates: 40°46′37″N, 73°52′21″W
B Kirksville Regional Airport
City: Kirksville, MO
Country: United States
IATA Code: IRK
ICAO Code: KIRK
Coordinates: 40°5′36″N, 92°32′41″W
## Time difference and current local times
The time difference between New York and Kirksville is 1 hour. Kirksville is 1 hour behind New York.
EDT
CDT
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 150 kg (330 pounds).
## Frequent Flyer Miles Calculator
New York (LGA) → Kirksville (IRK).
Distance:
984
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
984
Round trip?
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http://layers-of-learning.com/more-magnetism/
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# More Magnetism
Here is an earlier post with more on magnets from our website to give you the basics. And now we’ll share three experiments with magnetism.
Electromagnets are magnets that use flowing electrons to create a magnetic field. Flowing electrons means electricity.
You need:
• Wire Strippers
• A long iron nail, the longer and thicker the more powerful your magnet will be.
• Three meters of 22 gauge insulated, stranded copper wire (from a home improvement or hardware store)
• D-cell batteries
1. Wrap the wire very tightly around the nail, in close, tight coils. The more coils and the less space between the nail and the wire, the better your magnet will work.
2. Use the wire strippers to strip off about three cm of insulation at each end of the wire.
3. Attach each end of the wire to opposite ends of a D battery. See if your magnet can pick up metal objects like a paper clip.
4. What’s the heaviest thing it can pick up?
5. Try adding a second battery and see if your electromagnet gets stronger.
Be careful, the more batteries you add, the more dangerous this becomes. Don’t overdo it and if your batteries or magnet start to get hot, stop and let them cool down before you go on.
• If you want to try this check out this web site for directions. Scroll down to the bottom of the page for a simple, inexpensive and extremely cool project.
## Levitation
You can make objects levitate. Try this:
1. place a bar magnet on the edge of a table, with just a bit hanging over the edge.
2. Tie a string onto a paper clip.
3. Hold the string so the paper clip comes near the magnet, but does not touch it.
4. The paper clip should stand straight out in the air, toward the magnet.
## More Levitation
1. Get ring-shaped magnets.
2. Get a wooden dowel, small enough for the ring magnets to fit around.
3. Holding the dowel vertically, one end resting on a table, place the ring magnets one at a time on the dowel, making sure that North and South poles are facing each other, so the magnets repel, instead of attract.
4. Ta da! A stack of levitation.
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Next | ToC | Prev
### CHAPTER 14 Shape of the Dose-Response Relationship, and Low-Dose Cancer-Yields Based on the Best-Fit Curve
```
```
This chapter is arranged in six parts:
1. Supra-Linear Shape of the Dose-Response Relationship, p.1
2. Basis for Ruling-Out a Concave-Upward Shape, p.2
3. Purely Low-LET Radiation versus Mixed (Gamma + Neutron), p.5
4. Basis for Generalizing from the A-Bomb Study, p.6
5. Low-Dose Cancer-Yields Derived from the Best-Fit Curve, p.6
6. The Bottom Line from Regression Analysis, p.8
Then tables.
Then figures.
In the previous chapter, the Cancer Difference Method gives us Minimum Fatal Cancer-Yields in both the T65DR and DS86 dosimetries, and the results indicate that the cancer-risk is more severe per centi-sievert (rem) at low doses than at high doses. In other words, the findings in the previous chapter strongly suggest that shape of dose versus cancer-response is presently supra-linear. (Other terms for supra-linear, including "concave-downward," "upward convex," and even "sub-linear," are discussed in Chapter 23, Part 4.)
In this chapter, we will use the technique of curvilinear regression analysis for three purposes: (A) to depict the shape of dose-response in the A-bomb survivors, (B) to determine whether or not the supra-linear shape meets the test of statistical significance, and (C) to calculate Cancer-Yields based on the best-fit equation.
```
```
1. Supra-Linear Shape of the Dose-Response
In Chapter 29, we have used the data from Table 13-A, Rows 1 through 6, to demonstrate the technique of curvilinear regression. The steps of input, output, writing the best-fit equation, plotting graphs, and statistical testing are all presented in detail in that chapter. Readers who consult Chapter 29 will see for themselves exactly how we obtain the findings which are discussed in this chapter and elsewhere.
Findings from Chapter 29 are brought forward into this chapter. For instance, the equation which best fits the observations, in the T65DR dosimetry, is brought forward from Table 29-B and is presented in the Upper Notes of Table 14-A of this chapter.
Using the equation, we have calculated the predicted cancer-rates in Table 14-A, Column C, for dose-intervals of 10 cSv -- and for even smaller intervals at very low doses. In addition, Column C includes best-fit cancer-rates calculated for the specific organ-doses where we have the observed cancer-rates, so that the observed rate (in Column D) and the rate predicted by the curve (in Column C) can be compared. (Readers can ignore Columns E, F, and G until Part 5 of this chapter.)
Since the best-fit equation can provide predicted cancer-rates at any dose-level, of course Table 14-A includes estimates for 2, 5, and 10 cSv -- doses which lie between the mean doses received by Dose-Group 2 and Dose-Group 3. These estimated rates are interpolations between two actual observations; they are not extrapolations in a direction beyond any observed datapoint.
Figures 14-E and 14-F :
The information in Columns A, C, and D is plotted in Figure 14-E, which shows the cumulative cancer death-rate per 10,000 initial persons versus T65DR dose. The boxes are the actual observations, while the smooth curve says: This is what one would most probably see if one had more observations and less sampling variation.
Table 14-B and Figure 14-F provide the comparable information for the DS86 dosimetry.
Figures 14-E and 14-F look very much alike. Indeed, in both dosimetries, the equations which best fit the observations turn out to have the same dose-exponent: Dose^0.75. From Figures 14-E and F, it is self-evident that the dose-response curves are presently concave-downward (supra-linear) in both dosimetries.
(Our analysis has been made in terms of cancer-deaths per 10,000 initial persons. Some readers may be curious about the shape of dose-response if response is measured in cancers per 10,000 person-years. The analysis is provided for them in Chapter 30.)
RERF's Treatment of Dose-Group 8 :
The actual dose-response must be somewhat more supra-linear than we can know. The basis for our statement is a fact found in RERF's report TR-9-87 (p.7). In Dose-Group 8 (the highest Dose-Group), " . . . the T65D total kerma is set equal to 6 Gy for all survivors whose T65D total kerma estimate is greater than 6 Gy." With this sentence, RERF refers to both TR-1-86 and TR-12-80, so apparently RERF has been throwing out some part of the dose not only in the 1950-1982 follow-up, but also in the previous 1950-1978 follow-up.
It follows that, in our own analysis, the combined Dose-Group 6+7+8 must really have a somewhat higher mean organ-dose than we can know. But, of course, the observed cancer death-rate would not change. Therefore, in Figure 14-E, the uppermost datapoint really needs some sliding to the right (toward higher dose), a move which would operate in the direction of greater supra-linear curvature. We are confident that the effect would be small. However, we do not see how RERF's handling of Dose-Group 8 can improve anyone's analysis of dose-response.
In Part 3 of this chapter, we identify another factor which also will operate in the direction of underestimating the supra-linearity of low-LET dose-response.
Males and Females Tested Separately :
By definition, the general public includes both sexes. It is impossible to have "population exposure" without irradiating both men and women. Therefore, if analysts are evaluating the dose-response from exposure of a general population, what matters is the net dose-response. When they treat males and females as a unit in their analyses, the shape they obtain for dose versus cancer-response necessarily incorporates and reflects whatever difference may exist in male versus female response.
For other purposes, however, we may want to know if males and females are alike in the shape of dose-response. Of course, the moment analysts start subdividing the database, they increase the small-numbers problem, and findings are necessarily less reliable.
Using exactly the steps demonstrated in Chapter 29, we did regression analyses for males and females separately. The input data for cancer-rates and mean organ-doses were obtained from Table 11-G. The results are summarized below, in the equations of best fit. All the equations have supra-linear dose-exponents (below 1.0).
o -- MALES :
T65DR: Ca-deaths per 10,000 initial persons = (5.986)(Dose^0.75) + 796.389
DS86: Ca-deaths per 10,000 initial persons = (7.248)(Dose^0.70) + 792.248
o -- FEMALES :
T65DR: Ca-deaths per 10,000 initial persons = (10.086)(Dose^0.70) + 540.838
DS86: Ca-deaths per 10,000 initial persons = (9.463)(Dose^0.70) + 538.102
In other words, examined separately, males and females each show a supra-linear dose-response. The values of R-Squared for males are lower than for females, which means the finding is statistically weaker for males.
```
```
2. Basis for Ruling-Out a Concave-Upward Dose-Response
Chapter 29 demonstrates the technique of achieving curvilinear regression by raising a single dose-term, serially, to a variety of dose-exponents. We vary the exponent from Dose^2 (the quadratic dose-response), to Dose^1.4 and Dose^1.16 (linear-quadratic shapes), to Dose^1.0 (the linear dose-response), to Dose^0.85 and lower (supra-linear curves). Let us be explicit about the cancer-risks associated with these terms.
Supra-Linear Dose-Response :
This model of dose-response predicts that, with increase in total dose, the increase in cancer death-rate per cSv of dose will decrease. Each additional cSv of exposure will be less hazardous than the previous cSv. The plot of cancer-rate versus dose is concave-downward (illustrated by Figure 14-A), and the dose-exponent is less than 1.0.
Linear Dose-Response :
Here a plot of cancer death-rate versus dose yields a straight line -- hence the name "linear." The increase in cancer-rate per additional unit of dose is the same over the entire dose-range (illustrated by Figure 14-B), and the dose-exponent is 1.0.
When the quadratic dose-term (Q) has a positive coefficient, this model predicts that the increase in cancer-rate, per unit increase in dose, will increase as total dose increases. Each additional cSv of exposure will be more hazardous than the previous cSv. The plot of cancer-rate versus dose is concave-upward (illustrated by Figure 14-C). When a single dose-exponent is used, the exponent must be greater than 1.0, but less than 2.0.
However, as emphasized elsewhere (Go89b; also Chapters 23 and 29 of this book), when the quadratic term has a negative coefficient, the net result is a concave-downward, supra-linear dose-response (see Figure 23-H).
This model, whose plot is also concave-upward, bends away even more than the linear-quadratic dose-response from a straight line (illustrated by Figure 14-D), and the dose-exponent is 2.0.
Figures 14-A, B, C, D, for males and females combined, come from the input provided in Table 14-D. The four figures depict how the actual observations in the T65DR dosimetry relate to the values calculated by best-fit equations having the four shapes described above. Comparable figures are not included for the DS86 dosimetry simply because Figure 14-F already reveals that they would look like the T65DR figures.
Curve Fitting -- Supra-Linear Fit Is Significantly Better
In a good fit, not only should the weightiest observations lie close to the calculated curve, but their scatter (if any) should fall to both sides of it. In addition, it is a sign of poor fit if the observations on both ends lie on the same side of the curve while the observations in the middle all lie on the opposite side.
Inspection of Figures 14-A, B, C, and D shows the greatly inferior fit of both the linear-quadratic (Dose^1.4) and the pure quadratic (Dose^2) models. Indeed, such inspection predicts the results of the formal statistical testing in Tables 29-D and E.
The results in Tables 29-D and 29-E show that the supra-linear dose-response in the A-Bomb Study (1950-1982), in both the T65DR and the DS86 dosimetries, is significantly better than the linear relationship (p = 0.01).
As for a concave-upward dose-response, statistical testing in Tables 29-D and 29-E simply rules out such a relationship as the plausible choice. Even in the absence of any formal statistical testing, this conclusion is evident from inspection of Figures 14-C and 14-D, compared with Figures 14-A and 14-B.
As an independent check on the statistical significance of the supra-linear fit, we also used the power polynomial method of curve-fitting. It shows that there is both a statistically significant linear dose-term (Dose^1.0) and a statistically significant quadratic dose-term (Dose ^2.0), and that the coefficient of the quadratic term is negative. The equation of best fit from the power polynomial method produces a plot of cumulative cancer-rate versus dose which is virtually identical with the plot produced by the best-fit equation containing the Dose^0.75 term (Figures 14-E and 14-F).
Comparison with Statements from RERF :
Readers are in a position to evaluate our analysis of the shape of dose-response, step-by-step, from start to finish. They will not be able to compare it directly with RERF reports, however. RERF analysts are determining dose-response from input which is different from ours. For instance, in TR-5-88, Shimizu and co-workers discard the evidence between 1950-1955, and use only the observations from 1956 onwards. They are using only 75,991 of the initial 91,231 persons. For their 75,991 persons, they have additional observations out to 1985. They are using newly constructed cohorts, not a constant-cohort analysis. In effect, they are using a different database.
Nonetheless, there is a key similarity between our analysis and the analysis by Shimizu and co-workers: The RERF analysts do not find a concave-upward dose-response either. They find the following:
1. When they examine all cancers combined except leukemia as we do, and when they include all the Dose-Groups as we do, they find that their data fit linearity and supra-linearity equally well (Shi87, pp.28-30, and Shi88, pp.50-51).
2. When they examine males and females separately and include all the evidence, as we do, again they indicate that they find a linear or supra-linear dose-response (Shi88, p.53, Table 19).
However, Shimizu and co-workers never use the term, supra-linear. Their important points about the supra-linear shape might even be missed by any readers who assume that LQ (linear-quadratic) and LQ-L models are always concave-upward. The assumption would be mistaken. If the Q-coefficient for dose is negative in an LQ model, the net LQ curvature is concave-downward (Figure 23-H, again). Therefore readers of RERF reports need to pay close attention to RERF statements and footnotes such as:
"For those sites other than leukemia and colon, the fitted curve associated with the LQ model is invariably concave downwards, not upwards . . . " (Shi87, top p.29).
" . . . since the curvature is invariably downwards when a curvilinear model gives an acceptable fit, this would imply a higher risk at low doses than that which obtains under a linear model" (Shi87, p.30).
"Coefficient for the Q-term is negative" for the LQ model; this is the footnote which applies to analysis of the full dose-range in Table 19 of TR-5-88 (Shi88, p.53).
A Possible Route to Error :
Having found the dose-response to be linear or supra-linear (concave-downward), Shimizu and co-workers propose an alternative way to determine the dose-response. We quote:
"For all cancers except leukemia, although the L model fits well for both the total dose range and the dose range excluding high doses, the LQ model can not be shown to be inappropriate statistically. It should be noted that Q term in the LQ model is negative when the entire dose range is used, reflecting the level off of the dose-response curve at the higher dose range. In order to obtain useful risk estimates in the low-dose range with the LQ model, we have estimated the risk limiting doses to under 2 Gy, so as to obtain a positive Q term" (Shi88, p.50-51).
Although the paper (Shi88) is unclear on whether 2 Gy is kerma dose or internal organ-dose unadjusted for RBE, the statement quoted above means that they threw away the high Dose-Groups (probably 6,7,8) because, in someone's opinion, supra-linearity (the negative Q-coefficient) is not "useful." Not useful to whom? And for what?
Where does such an approach to evidence end?
It may easily end in error. For instance, it is self-evident from Figures 13-A and 13-B in the previous chapter, that if one discarded Dose-Group 5 as well as Dose-Group 6-8, one would end up with the opposite result: The dose-response would be based on the four residual datapoints, and it would be more supra-linear, not less. (This statement is supported by regression analysis which excludes Dose-Groups 5-8).
Moreover, if we look objectively at the entries in Table 13-A, Column E, we see that the absolute number of cancer-deaths observed in Dose-Group 4 is about the same as in Dose-Group 5, and also in the combined Dose-Group 6+7+8. This means that the statistical reliability of each of these three observations is about the same. If analysts are willing to discard one, then on an objective basis, why should they not discard all three?
Suppose the first discarding of data (Dose-Groups 6-8) would result in decreasing the study's supra-linear curvature, but suppose the next, equally justifiable discarding of data (Dose-Group 5) would result in exaggerating its supra-linear curvature. What is the appropriate choice?
In my opinion, the curvature which is most likely to be right is the curvature which comes from using all the available evidence. It would certainly not be science at all, if I were to keep the evidence which leads to answers I may like, while throwing out the evidence which produces answers I may not like.
In my judgment, analysts will be most likely to obtain the right answer about dose-response when they use all of the observations. The reason is this. The jaggedness observed in Figures 13-A and 13-B has virtually no chance of being biologically meaningful. Such jaggedness is almost certainly the result of sampling variation, which means that it would not be there (the dose-response would be smooth) if the study had included a billion persons instead of only 91,231. One of the great scientific virtues of the A-Bomb Study is its inclusion of such a vast range of doses. If we use all the data in regression analysis, the additional observations are likely to help "correct" the jaggedness of sampling variation. But if we start throwing away any of the valuable evidence without a very good reason indeed, we will almost certainly increase the chance of mistaken results.
Conclusion about Shape in the A-Bomb Study :
Our analysis of dose-response is based on all the evidence. No Dose-Groups (and no follow-up years) have been thrown out. Our findings fit concavity-downward (supra-linearity) provably better than linearity, and fit concavity-downward enormously better than concavity-upward.
We are forced to conclude, not by preference but by the evidence currently available to us, that concavity-upward is not credible as the shape of dose-response in the A-Bomb Study. The credible choice is presently supra-linearity.
Will supra-linearity persist to the end of the study, decades from now? No one can know. (As we said in Chapter 12, no one can rule out even remote possibilities -- like there being no excess cancer anymore at the end of the study, which would mean a flat dose-response. Of course, in that unlikely case, the interim excess of cancer-deaths would represent a major misery for those who died from the disease 10, 20, 30, 40 years earlier than otherwise.)
Meanwhile, analysts must report whatever is ruled in and out by the currently available evidence. This chapter and Chapter 29 rule out the concave-upward shape as a good fit for the 1950-1982 observations. The data say that the dose-response curvature is concave-downward.
It should be emphasized that the findings about dose-response -- like the findings for minimum Fatal Cancer-Yields -- involve no forward projections and no hypotheses about radiation carcinogenesis. Our findings simply amount to an objective description of what the present evidence is on the shape of the dose-response.
```
```
3. Purely Low-Let Radiation versus Mixed (Gamma + Neutron)
The dose-response curve which fits the observations best is presently concave-downward or supra-linear (Figure 14-E for the T65DR dosimetry; Figure 14-F for the DS86 dosimetry). In each dosimetry, the equation which generates the best-fit has a dose-exponent of 0.75.
The dose-input for the regression-analysis (Chapter 29) was composed of two types of radiation: Gamma and neutron. (Tables 9-C and 10-E, Row 14, show the small fraction of the internal organ-dose, in rems, which was contributed by neutrons.) Therefore, the curves depict dose-response for a mixture of the two radiations.
Nonetheless, one must conclude that the concave-downward curvature is caused by the low-LET (gamma) component of the exposure, not by the high-LET (neutron) component.
The basis for this conclusion is clear if we start by imagining that the A-bomb survivors received only neutron-exposure, but no gamma exposure. For neutron-exposure, the experimental observation is that dose-response is linear, at least up to 10 rads of total neutron dose (Chapter 8, Part 5). And if we adjust for the greater carcinogenic potency of neutrons, by multiplying neutron doses (below 10 rads) by a constant RBE of 20 to obtain rems, a plot of cancer-rate versus pure neutron doses in rems would still be linear. We have shown (page 8-8) that the highest mean neutron organ-dose was about 4.369 rads in the DS86 dosimetry, where such doses are supposed to be correct; 4.369 is a dose well below 10. Therefore, if the A-bomb survivors had received only neutron-exposure, our plots of cancer-rates versus dose in rems would be linear.
Now, we return to the real situation. The A-bomb survivors also received a gamma dose. And when cancer-rate is plotted versus dose in rems, for the combined neutron and gamma doses, the best fit for the observations becomes supra-linear, even though it would have been linear if only the neutrons had been present. It follows that the curvature is caused by the gamma exposure, not by the neutrons.
Underestimation of the Low-Let Curvature :
Our analyses must somewhat underestimate the true degree of supra-linearity for low-LET (gamma) dose-response. Table 10-E, Row 14, shows that the fraction of total dose, in cSv, contributed by neutrons rises with rising total dose. The rising share from neutrons (from 5.4 % in Dose-Group 2, up to 18 % in Dose-Group 8) prevents the supra-linear curvature for gamma-exposure from being fully seen.
The gamma's supra-linear dose-response means that the percent increase in spontaneous cancer-rate, per average rem of gamma dose, falls as gamma-dose rises. By contrast, the neutron's linear dose-response means that the carcinogenicity of neutrons is constant in all eight dose-groups. As the combined dose from gammas and neutrons is rising, the average carcinogenicity of the gamma rems is falling whereas the carcinogenicity of the neutron rems is not falling.
Therefore, when neutrons contribute an "extra" share of the combined dose as the combined dose is rising (Table 10-E, Rows 11 and 14), it means that the observed cancer-rates at the higher doses are somewhat elevated above the rates which would have occurred if the fraction contributed by neutrons had not risen. The result is that the "extra" share from neutrons progressively "lifts" the right-hand half of the curve of Cancer-Rate versus Combined Dose, in the direction of linearity. In other words, the supra-linear curvature would be more pronounced if the fraction of combined dose coming from neutrons had not risen. Thus the supra-linearity of low-LET dose-response is somewhat underestimated in our DS86 analyses.
```
```
4. Basis for Generalizing from the A-Bomb Study
This chapter and Chapter 29 have confirmed what Table 13-B so strongly suggested: The dose-response is presently supra-linear throughout the dose-range. The result does not depend on high-dose data. If analysts threw out Dose-Groups 5-8, the supra-linearity would be even more pronounced. We would emphatically not approve of throwing out data, however.
The finding, that dose-response for low-LET exposure is concave-downward (supra-linear), is based on observation of all fatal cancers combined, with only leukemia excluded. In other words, the finding does not rest on a small study involving just leukemia or a few cancer-sites, or on a study resting on incidence instead of mortality. And the finding is based on two dosimetries. And the finding is not based on a single sex- or age-group. It is broadly based on both sexes and all ages.
In other words, the finding of supra-linearity at low doses is based on excellent human epidemiological evidence -- in our judgment, the best which is available at this time anywhere.
Therefore, it is scientifically reasonable to generalize from the A-Bomb Study, 1950-1982: In humans, the dose-response for induction of fatal cancer by low-LET ionizing radiation is most probably supra-linear in shape, even at low doses. The risk per rem rises as total dose falls.
This finding is directly at variance with the widely applied presumption -- not based on human epidemiology -- that the human cancer-hazard per centi-sievert of low-LET exposure would go down with decreasing total doses. Readers are referred to Un77, p.414, para.318; Un86, p.191, para.153; Beir80, p.190; Ncrp80, pp.5-9; Nih85, p.iv; Nrc85, p.II-101-103; Doe87, p.7.3, 7.4; and others. Some of these sources use the presumption, while also acknowledging that the available human epidemiological data do not support it (see Chapter 22).
The Past and Future of Supra-Linearity :
The supra-linear shape of dose-response has been showing up in the A-Bomb Study for at least three consecutive follow-ups: 1950-1974, 1950-1978, and 1950-1982 (Go81; Go89a; Ncrp80 -- details in our Chapter 22, Part 2). In other words, supra-linearity is not a characteristic which appeared only with the addition of the 1978-1982 observations. And, according to RERF analysts (Shi87; Shi88), it is still showing up in the revised database when they add some observations through 1985 (see this chapter, Part 2).
Although no one can be sure that supra-linearity will continue its persistence through all future follow-ups, the only reasonable forward projection is the one which rests on the best available evidence. And the best available evidence, from at least three consecutive follow-ups, suggests that supra-linearity will persist.
On the other hand, if the A-Bomb Study itself does not persist with a continuous "constant-cohort, dual-dosimetry" database, it will be hard for anyone to sort out which future findings on dose-response result from extension of the time-interval since the exposure, and which future findings result from perpetual revision of the DS86 doses and cohorts.
```
```
5. Low-Dose Cancer-Yields Based on the Best-Fit Curve
When analysts seek to estimate the cancer-hazard from exposing populations of mixed ages to ionizing radiation, the doses received by Dose-Group 3 in the A-Bomb Study are considerably higher than the relevant levels suggested by nuclear accidents like Chernobyl, for example. We should be asking, what are the likely Cancer-Yields if people receive total doses like 5 cSv (or less)?
Tables 14-A and 14-B provide the probable values for the minimum Fatal Cancer-Yields, in Column G. We have starred the entries calculated from 5 cSv of total exposure, because we think those are the appropriate ones to use for low-dose exposures up to 5 cSv and for slow exposures. (We closely examine the issue of slow dose-rates in Chapter 23, Parts 6 and 7.)
The notes of Tables 14-A and 14-B explain exactly how the values were obtained. Readers will see that this is still the Cancer Difference Method: A difference in cancer-rate is divided by the corresponding difference in dose. However, in this version, the cancer-rates are not the direct observations; instead, they are the rates predicted after the actual observations have produced an equation of best fit. In Tables 14-A and 14-B, the division-step for the starred entries at 5 cSv amounts to the approximation that every rem (cSv) between 0 and 5 rems is equally potent.
Table 14-C assembles the low-dose Cancer-Yields from Table 13-B as well as from Tables 14-A and 14-B, so that they can be easily compared with each other. The net effect of regression-analysis is to reduce the T65DR risk-estimate below its value in Table 13-B, and to render it almost identical with the DS86 estimate.
The Basing of Values on a Total Dose of 5 Centi-Sieverts :
Readers may wonder why we suggest using Cancer-Yields calculated from a total dose of 5 cSv, even for use with population-exposure which might be lower (say, one centi-sievert or less). At first glance, it may look as if we are deliberately underestimating the likely Minimum Fatal Cancer-Yields from very low-dose exposure, since we are using linearity instead of supra-linearity between 0 dose and 5 rems.
Our reasoning is as follows. The technique of curvilinear regression provides the values of low-dose Cancer-Yield which are most likely to be true, given the evidence at hand. And the equation which has the highest R-Squared value in regression analysis is the equation which is most likely to make the best predictions. Therefore the equation which we should use, and which we do use, is the one in which the dose-exponent is 0.75. Objectivity requires use of results from available evidence, rather than use of preconceptions about how the curvature "ought" to behave at low doses (see Chapter 23). Unlike the BEIR-3 Committee (see Chapter 22), we do not constrain any regression in order to make it support a pre-judgment.
On the other hand, as we pointed out in Chapter 29, while we know that 0.75 is significantly better than the dose-exponent 1.0, we do not know that 0.75 is significantly better than 0.80, 0.85, 0.70, or 0.65. Yet the shape of the curve is such that small changes in the dose-exponent have a big effect, at one or two cSv, on the values for Cancer-Yield in Columns F and G of Tables 14-A and 14-B. In view of this sensitivity, we want to avoid using any values for Cancer-Yield derived directly from the curve at one or two cSv.
We regard our decision as a scientifically reasonable judgment which simultaneously (A) avoids the irresponsibility of throwing away the low-dose results of regression analysis down to 5 cSv, and (B) avoids the introduction of any unstable element into an analysis which has been securely based in reality.
A Comment by RERF about BEIR Choices :
The shape of dose-response is central to obtaining risk-estimates at low (and slow) doses. If analysts choose unrealistic versions of the dose-response relationship, they will provide unrealistic estimates of cancer-hazard. RERF analysts, in trying to figure out why their own current risk-estimates are so much higher than those of the BEIR-3 Committee, comment (TR-5-88, p.51):
[Some of the disparity] " . . . may be ascribed to the fact that in BEIR III, the curvature in dose response for leukemia was used for all cancers except leukemia instead of the actual curvature which probably is much closer to linearity, and this may cause much smaller estimates to be produced than if the actual dose-response curve were to be applied."
Venturing below 10 Rems :
Now that we have examined the logic and results of regression analysis, as a tool for obtaining a smooth and probable dose-response at all doses, we can discuss a matter which puzzles us and may puzzle readers too.
In its 1980 report, the BEIR-3 Committee declined to make risk-estimates for acute exposures lower than 10 rems (Beir80, p.144). RERF analysts appear to be split on this issue. In TR-9-87 (Pr87b, p.35), Preston and Pierce present their estimates of Lifetime Fatal Cancer-Yield as cancer deaths per 10 milli-sieverts (per rem). By contrast, in TR-5-88 (Shi88, Table 19, p.53), Shimizu and co-workers explicitly constrain their estimates of Lifetime Fatal Cancer-Yields to acute exposures of 0.1 Sievert (10 rems).
It is puzzling to us that Shimizu and co-workers make a big effort to determine what the dose-response relationship is, starting at zero dose, and then they seem unwilling to use it in the low dose-range. As we pointed out in Part 1 of this chapter, estimates below 10 rems are not extrapolations in a direction beyond any actual observations. Such estimates are interpolations between actual observations in Dose-Group 3 and Dose-Group 2. Indeed, Dose-Groups 1-3 provide the most reliable observations in the whole study, in terms of cancer-cases (not necessarily in dosimetry). If analysts will not use the section of the dose-response below Dose-Group 3, it would seem they should have no reason to use it above Dose-Group 3 either, where the datapoints are based on far fewer cancer-deaths.
By contrast, to us it seems highly reasonable -- almost obligatory -- for analysts to presume that the dose-response which derives from the dose-range as a whole also characterizes the little segment between zero dose and 10 rems.
On the other hand, refusal to make estimates below 10 rems could be a way of suggesting that maybe the risk of radiation-induced cancer just disappears somewhere between 10 rems and zero dose.
The human evidence against any harmless dose of ionizing radiation, with respect to carcinogenesis, is examined in detail in the Threshold section of this book (Section 5). Here we shall limit our comments to the A-Bomb Study (see also Chapter 35, Part 9).
The A-Bomb Study, properly handled, certainly offers no basis for belief in a threshold, or a lesser hazard per rem either, anywhere below 10 rems. On the contrary. Its present supra-linear curvature indicates the risk per rem is growing steadily higher as dose approaches zero. Even if its present dose-response were linear (instead of supra-linear), this would be no basis for belief either in a safe threshold somewhere below 10 rems, or a lesser effect per rem.
In short, even if there were no additional evidence in Section 5 against a threshold, and even if the dose-response in the A-Bomb Study were linear instead of supra-linear, we would consider the basis for making risk-estimates below 10 rems to be scientifically compelling.
```
```
6. The Bottom Line from Best-Fit Curves
1. This chapter and Chapter 29 show that the relationship between dose and cancer-response per 10,000 initial persons is presently supra-linear (concave-downward). Statistical testing demonstrates that the evidence fits a concave-downward curvature significantly better than the evidence fits a linear dose-response, and very much better than it fits a concave-upward shape. See Figures 14-A, B, C, and D. In short, the present evidence from the A-bomb survivors is that cancer-risk is greater per rem (centi-sievert) at low doses than at high doses, in both dosimetries. (Chapter 30 shows the same finding in cancer-response per 10,000 person-years.)
2. The finding of supra-linearity is solidly based in the existing evidence, and does not rely on any forward projections, hypotheses, or models. We have simply presented an objective description of what the available evidence is showing in a database which covers all cancers (leukemia excluded), all doses, all ages, and both sexes. This direct and comprehensive human epidemiological evidence carries great scientific weight compared with observations from other species, of course, or from laboratory experiments.
3. The evidence is at variance with the assumption, almost universally used by the radiation community, that the cancer-risk should be less severe per rem at low acute doses than at high acute doses. With regard to low doses delivered slowly, we show in Chapter 23, Part 7, that there is no reason to reduce the low-dose Cancer-Yields in Table 14-C when exposure is slow instead of acute.
4. Although no one can be certain that the supra-linear curvature will persist through all future follow-ups, the only reasonable forward projection is the one which rests on the best available evidence. And the best available evidence, from at least three consecutive follow-ups, is that supra-linearity is persistent. However, if the A-Bomb Study itself does not persist with a continuous "constant-cohort, dual-dosimetry" database, it will be hard for anyone to sort out which future findings on dose-response result from extension of the time-interval since the bombings, and which future findings result from perpetual revision of the DS86 doses and cohorts.
5. Regression analysis provides the best-fit equation for dose-response, and the equation can predict cancer-rates at any dose-level, including doses like 2, 5, and 10 rems which lie between the mean dose received by Dose-Group 2 and Dose-Group 3. The estimated cancer-rates at these doses are interpolations between two actual observations -- they are not extrapolations in a direction beyond any observed data-point.
6. Unlike some current analysis at RERF, our analysis of dose-response uses all of the observations, high-dose and low-dose, and all of the follow-up years, in order to obtain the most reliable results. We do not approve of throwing away evidence without a very good reason indeed. It should be noted that the supra-linear curvature of dose versus cancer-response occurs throughout the dose-range. In fact, if the high-dose evidence from Dose-Groups 5-8 were discarded, the low-dose evidence from Dose-Groups 1-4 would produce greater supra-linearity -- not less.
7. The best-fit equation from our regression analysis is used to obtain another set of Minimum and Lifetime Fatal Cancer-Yields by the Cancer Difference Method, for low-dose exposure. Table 14-C compares the new set with the first set, in both T65DR and DS86 dosimetries. The net effect of regression analysis is to reduce the estimate in the T65DR dosimetry. In the new set of estimates, the Lifetime Cancer-Yields remain probable underestimates, as they were in Table 13-B. The Lifetime Fatal Cancer-Yield from the best-fit curve is 12.90 in the T65DR dosimetry, and 12.03 in the current version of the DS86 dosimetry. By contrast, the lifetime values commonly used by the radiation community for statements about low-dose exposure are between 1.0 and 2.0 (see Chapter 24, Part 7, and Chapter 34, Wolfe).
```
```
Table 14-A
T65DR Dosimetry (RBE = 2): Comparison of Calculated Cancer-Rates
with Observed Cancer-Rates.
T65DR Dosimetry (RBE = 2): Minimum Fatal Cancer-Yields
per Centi-Sievert Among 10,000 Persons.
```|=============================================================================================|
| Col.A Col.B Col.C Col.D | Col.E Col.F Col.G |
| | |
| Dose Dose^0.75 Cancer-Rate Cancer-Rate | Calculated Ca-Rate Avg.Incr. MINIMUM |
| cSv per 10,000 per 10,000 | MINUS the Calculated in Ca-Rate FATAL |
| T65DR Calculated Observed | Spontaneous Ca-Rate per cSv CA-YIELD |
|========================================================================================== |
| 0 0.00000 649.5440 649.31 | |
| 1 1.00000 656.5968 | 7.053 7.053 8.675 |
| 1.511 1.36260 659.1542 651.89 | 9.610 6.362 7.825 |
| 2 1.68179 661.4053 | 11.861 5.931 7.295 |
| 5 3.34370 673.1265 | 23.582 4.716 5.801 * |
| 10 5.62341 689.2048 | 39.661 3.966 4.878 |
| 10.994 6.03762 692.1261 712.02 | 42.582 3.873 4.764 |
| 20 9.45742 716.2453 | 66.701 3.335 4.102 |
| 30 12.81861 739.9511 | 90.407 3.014 3.707 |
| 35.361 14.50098 751.8165 723.72 | 102.273 2.892 3.557 |
| 40 15.90541 761.7217 | 112.178 2.804 3.449 |
| 50 18.80302 782.1579 | 132.614 2.652 3.262 |
| 60 21.55825 801.5900 | 152.046 2.534 3.117 |
| 70 24.20045 820.2250 | 170.681 2.438 2.999 |
| 71.308 24.53891 822.6120 836.27 | 173.068 2.427 2.985 |
| 80 26.74961 838.2037 | 188.660 2.358 2.901 |
| 90 29.22011 855.6276 | 206.084 2.290 2.816 |
| 100 31.62278 872.5731 | 223.029 2.230 2.743 |
| 110 33.96601 889.0995 | 239.555 2.178 2.679 |
| 120 36.25650 905.2539 | 255.710 2.131 2.621 |
| 130 38.49971 921.0748 | 271.531 2.089 2.569 |
| 140 40.70015 936.5940 | 287.050 2.050 2.522 |
| 150 42.86161 951.8383 | 302.294 2.015 2.479 |
| 160 44.98731 966.8305 | 317.286 1.983 2.439 |
| 170 47.08003 981.5900 | 332.046 1.953 2.402 |
|176.662 48.45714 991.3025 988.45 | 341.759 1.935 2.379 |
| 180 49.14218 996.1339 | 346.590 1.925 2.368 |
| 190 51.17587 1010.477 | 360.933 1.900 2.337 |
| 200 53.18296 1024.633 | 375.089 1.875 2.307 |
|=============================================================================================|
```
UPPER NOTES: -----
Entries in Col.A come from Table 13-A, with many doses added between observations.
Entries in Col.C for the predicted rates are calculated, both for observed doses and interpolated doses, with the equation derived from Table 29-D:
Cancer-Rate = (7.0528)(Dose^0.75) + 649.544.
Values for the term (Dose^0.75) are obtained from Col.B above. Entries in Col.D come from Table 13-A, and lie near the calculated values. Columns A, C, and D are plotted in Figure 14-E.
-------------------- Right-Hand Side of Table --------------------
FATAL CANCER-YIELD = NUMBER OF RADIATION-INDUCED CANCER-DEATHS AMONG 10,000
INITIAL PERSONS OF MIXED AGES, PER CENTI-SIEVERT OF WHOLE-BODY INTERNAL ORGAN-DOSE.
Entries in Col.E are Col.C minus 649.544 (which is the calculated spontaneous rate / 10,000).
Entries in Col.F are Col.E / Col.A. The entries correspond to the Min. Fatal Cancer Yield calculated by the Cancer Difference Method, before the 1.23-fold correction used by RERF for underascertainment of cancer-deaths (see Chapter 11). The progressive decline of Col.F entries with rising dose reflects the supra-linearity of dose-response.
Entries in Col.G are Col.F entries times 1.23, the underascertainment correction. The starred value is the one which we use for low-dose exposure. In subsequent chapters also, we use values per cSv based on best-fit at 5 cSv.
```
```
Table 14-B
DS86 Dosimetry: Comparison of Calculated Cancer-Rates With ObservedD Cancer-Rates.
DS86 Dosimetry: Minimum Fatal Cancer-Yields per CentiI-Sievert Among 10,000 Persons.
```|==============================================================================================|
| Col.A Col.B Col.C Col.D | Col.E Col.F Col.G |
| | |
| Dose Dose^0.75 Cancer-Rate Cancer-Rate | Calculated Ca-Rate Avg.Incr. MINIMUM |
| cSv per 10,000 per 10,000 | MINUS the Calculated in Ca-Rate FATAL |
| DS86 Calculated Observed | Spontaneous Ca-Rate per cSv CA-YIELD |
|=========================================================================================== |
| 0 0.0000 647.693 | |
| 0.089 0.1634 648.768 649.31 | |
| 1 1.0000 654.272 | 6.579 6.579 8.093 |
| 1.890 1.6121 658.299 651.89 | 10.606 5.611 6.902 |
| 2 1.6818 658.758 | 11.065 5.533 6.805 |
| 5 3.3437 669.692 | 21.999 4.400 5.412 * |
| 10 5.6234 684.691 | 36.998 3.700 4.551 |
| 14.564 7.4553 696.744 712.02 | 49.051 3.368 4.143 |
| 20 9.4574 709.916 | 62.223 3.111 3.827 |
| 30 12.8186 732.030 | 84.337 2.811 3.458 |
| 40 15.9054 752.339 | 104.646 2.616 3.218 |
| 40.625 16.0915 753.564 723.72 | 105.871 2.606 3.205 |
| 50 18.8030 771.404 | 123.711 2.474 3.043 |
| 60 21.5582 789.531 | 141.838 2.364 2.908 |
| 70 24.2005 806.915 | 159.222 2.275 2.798 |
| 74.238 25.2911 814.091 836.27 | 166.398 2.241 2.757 |
| 80 26.7496 823.687 | 175.994 2.200 2.706 |
| 90 29.2201 839.941 | 192.248 2.136 2.627 |
| 100 31.6228 855.749 | 208.056 2.081 2.559 |
| 110 33.9660 871.166 | 223.473 2.032 2.499 |
| 120 36.2565 886.235 | 238.542 1.988 2.445 |
| 130 38.4997 900.994 | 253.301 1.948 2.397 |
| 140 40.7002 915.472 | 267.779 1.913 2.353 |
| 150 42.8616 929.692 | 281.999 1.880 2.312 |
| 160 44.9873 943.678 | 295.985 1.850 2.275 |
| 170 47.0800 957.447 | 309.754 1.822 2.241 |
| 180 49.1422 971.014 | 323.321 1.796 2.209 |
|197.054 52.5943 993.727 988.45 | 346.034 1.756 2.160 |
| 200 53.1830 997.600 | 349.907 1.750 2.152 |
|==============================================================================================|
```
UPPER NOTES: -----
Entries in Col.A come from Table 13-A, with many doses added between observations.
Entries in Col.C for the predicted rates are calculated, both for observed doses and interpolated doses, with the equation derived from Table 29-C:
Cancer-Rate = (6.5793)(Dose^0.75) + 647.693.
Values for the term (Dose^0.75) are obtained from Col.B above. Entries in Col.D come from Table 13-A, and lie near the calculated values. Columns A, C, and D are plotted in Figure 14-F.
-------------------- Right-Hand Side of Table --------------------
FATAL CANCER YIELD = NUMBER OF RADIATION-INDUCED CANCER-DEATHS AMONG
10,000 PERSONS OF MIXED AGES, PER CENTI-SIEVERT OF WHOLE-BODY INTERNAL ORGAN-DOSE.
Entries in Col.E are Col.C minus 647.693 (which is the calculated spontaneous rate / 10,000).
Entries in Col.F are Col.E / Col.A. The entries correspond to the Min. Fatal Cancer Yield calculated by the Cancer Difference Method, before the 1.23-fold correction used by RERF for underascertainment of cancer-deaths (see Chapter 11). The progressive decline of Col.F entries with rising dose reflects the supra-linearity of dose-response.
Entries in Col.G are Col.F entries times 1.23, the underascertainment correction. The starred value is the one which we use for low-dose exposure. In subsequent chapters also, we use values per cSv based on best-fit at 5 cSv.
```
```
Table 14-C
Cancer-Yields at the Low Doses, by the Cancer Difference Method,
with and without Curvilinear Regression.
Basis: A-bomb survivors, all ages combined, 1950-1982. Both T65DR and DS86 dosimetries.
Cancer-hazard from X-rays may be underestimated by the A-Bomb Study. See Chapter 13, Part 4.
Cancer-Yields are radiation-induced cancer-deaths among 10,000 persons of mixed ages, per cSv of whole-body internal organ-dose.
All Cancer-Yields below are corrected for underascertainment of cancer-deaths.
```|==================================================================================|
| || || |
| || --Fatal Cancer-Yield-- || --Fatal Cancer-Yield-- |
| || T65DR Dosimetry || DS86 Dosimetry |
| || Neutron RBE = 2.0 || Neutron RBE = 20 |
| || || |
|==================================================================================|
| Col.A || Col.B Col.C || Col.D Col.E |
| || || |
| Source of Estimate || FATAL FATAL || FATAL FATAL |
| for Minimum Fatal || CANCER- CANCER- || CANCER- CANCER- |
| Cancer-Yields || YIELD YIELD || YIELD YIELD |
|==================================================================================|
| Row || || |
| Dose-Group 3 || || |
| versus Ref. Group || || |
| 1 Table 13-B. || 7.29 16.20 || 5.50 12.23 |
| || || |
| || || |
| Best-Fit Curve || || |
| Tables 14-A and || || |
| 2 14-B. || 5.80 12.90 || 5.41 12.03 |
| || || |
|==================================================================================|
```
NOTES -----
1. The minimum values above in Row 1 come from Table 13-B, Columns B and H. If there had been no dose-groups higher than Dose-Group 3 in the A-bomb experience, these are the only values which would exist in Table 13-B. The entries in Row 1 above are the values before regression analysis provides a smooth best-fit curve.
2. The minimum values in Row 2 above come from the best-fit curves provided by regression analysis, using all dose-groups.
The minimum value in Row 2, Column B, comes from Table 14-A, Column G (the starred value).
The minimum value in Row 2, Column D, comes from Table 14-B, Column G (the starred value).
3. lifetime values ( Columns C and E ) are always the minimum value times 2.223, in the Cancer Difference Method. These lifetime entries are probably underestimates (see text and subsequent chapters). The factor 2.223 comes from Table 28-D, Row 14.
```
```
Table 14-D
Input Values for Figures 14-A, 14-B, 14-C, and 14-D.
T65DR Dosimetry with Neutron RBE = 2.
```|=====================================================================================|
| Input for Figure 14-A | Input for Figure 14-B |
| ======================== | ======================== |
| Equation for Dose^0.75 from Table 29-B: | Equation for Dose^1 from Table 29-B: |
| Ca-Rate = (7.0528)(Dose^0.75) + 649.544 | Ca-Rate = (1.947)(Dose^1) + 661.153 |
| | |
| Dose Dose Ca-Rate Ca-Rate | Dose Dose Ca-Rate Ca-Rate |
| cSv cSv^0.75 Calc. Observed | cSv cSv^1 Calc. Observed |
| | |
| 0.000 0.0000 649.544 649.31 | 0.000 0.000 661.153 649.31 |
| 1.511 1.3626 659.154 651.89 | 1.511 1.511 664.094 651.89 |
| 10.994 6.0376 692.126 712.02 | 10.994 10.994 682.558 712.02 |
| 35.361 14.5010 751.817 723.72 | 35.361 35.361 730.002 723.72 |
| 71.308 24.5388 822.611 836.27 | 71.308 71.308 799.990 836.27 |
| 130.000 38.4997 921.075 | 130.000 130.000 914.263 |
| 176.662 48.4571 991.303 988.45 | 176.662 176.662 1005.114 988.45 |
|=====================================================================================|
| | |
| Input for Figure 14-C | Input for Figure 14-D |
| ======================== | ======================== |
| Equation for Dose^1.4 from Table 29-B: | Equation for Dose^2 from Table 29-B: |
| Ca-Rate = (0.242)(Dose^1.4) + 671.922 | Ca-Rate = (0.01047)(Dose^2) + 680.048 |
| | |
| Dose Dose Ca-Rate Ca-Rate | Dose Dose Ca-Rate Ca-Rate |
| cSv cSv^1.4 Calc. Observed | cSv cSv^2 Calc. Observed |
| | |
| 0.000 0.000 671.922 649.31 | 0.000 0.00 680.048 649.31 |
| 1.511 1.782 672.353 651.89 | 1.511 2.28 680.072 651.89 |
| 10.994 28.682 678.863 712.02 | 10.994 120.87 681.313 712.02 |
| 35.361 147.212 707.547 723.72 | 35.361 1250.43 693.140 723.72 |
| 71.308 393.001 767.028 836.27 | 71.308 5084.83 733.286 836.27 |
| 130.000 911.005 892.385 | 130.000 6900.00 856.991 |
| 176.662 1399.591 1010.623 988.45 | 176.662 31209.54 1006.812 988.45 |
|=====================================================================================|
```
The construction of Table 14-D is described, step by step, in Chapter 29.
When the above values are plotted for dose (cSv), calculated cancer-rate, and observed cancer-rate, they demonstrate graphically how closely or how distantly the observed points lie to the corresponding curve calculated by regression analysis.
In a good fit, not only should the weightiest observations lie close to the calculated curve, but their scatter (if any) should fall to both sides of it. In addition, it is a sign of poor fit if the observations on both ends lie on the same side of the curve while the observations in the middle all lie on the opposite side.
Because the dose-response is so similar in T65DR and DS86 (compare Figure 14-E with Figure 14-F), we have not shown graphs comparable to 14-B, 14-C, and 14-D for the DS86 analysis.
```
```
Figures 14-A,B,C,D
Distribution of Datapoints Relative to Four Dose-Response Curves. T65DR Dosimetry: Cumulative Cancer-Deaths versus Dose.
Input for the four figures below is provided by Table 14-D. Each figure depicts the same observations (indicated by the boxy symbol): Cancer-mortality versus dose in the A-Bomb Study, 1950-1982. What differs is the dose-response curve in each figure.
o -- In Figure 14-A, the "fit" between the observations and the supra-linear curve is good, with datapoints either lying on the curve or falling to both sides of it.
o -- In Figure 14-B, the fit between the observations and the linear "curve" is inferior to the fit in Figure 14-A.
o -- In Figure 14-C, the fit between the observations and the linear-quadratic (Q-positive) curve is very poor, with the observations at both ends lying on the same side of the curve, and the observations in the middle all lying on the opposite side.
o -- In Figure 14-D, the fit between observations and the quadratic dose-response curve is even worse than in Figure 14-C.
o -- Statistical testing (Chapter 29) establishes that the supra-linear dose-response fits the evidence significantly better than the linear dose-response.
Figure 14-A, below: Fit Relative to Supra-Linear (Dose^0.75)
Figure 14-B, below: Fit Relative to Linear (Dose^1.0)
Figure 14-C, below: Fit Relative to Linear-Quadratic (Dose^1.4)
Figure 14-D, below: Fit Relative to Quadratic (Dose^2.0).
```
```
Figure 14-E
T65DR Dosimetry: Best-Fit Curve for Cumulative Cancer-Deaths versus Dose.
For the atomic-bomb survivors, this plot shows cumulative cancer-deaths (1950-1982) per 10,000 initial persons, versus mean whole-body internal organ-dose in the T65DR dosimetry (RBE = 2). Input for this figure comes from Table 14-A, Columns A, C, and D.
o -- The boxy symbols, which show the observed cancer death-rate per 10,000 initial persons versus dose, come from Columns A and D of Table 14-A.
o -- Points along the best-fit curve come from Column C of Table 14-A, and show calculated cancer death-rates per 10,000 initial persons versus dose, based on the equation of best fit, shown below. This curve is the same as the curve in Figure 14-A, of course.
Figure 14-F will show the best-fit curve for the same cohorts of survivors in the supplemental DS86 dosimetry.
```
```
Figure 14-F
DS86 Dosimetry: Best-Fit Curve for Cumulative Cancer-Deaths versus Dose.
For the atomic-bomb survivors, this plot shows cumulative cancer-deaths (1950-1982) per 10,000 initial persons, versus mean whole-body internal organ-dose in the DS86 dosimetry (RBE = 20). Input for this figure comes from Table 14-B, Columns A, C, and D.
o -- The boxy symbols, which show the observed cancer death-rate per 10,000 initial persons versus dose, come from Columns A and D of Table 14-B.
o -- Points along the best-fit curve come from Column C of Table 14-B, and show calculated cancer death-rates per 10,000 initial persons versus dose, based on the equation of best fit.
With our "constant-cohort, dual dosimetry" approach to the DS86 dosimetry, the T65DR cohorts of survivors remain undisturbed, and merely receive a second dose-estimate. Super-imposition of Figures 14-F and 14-E would show that the DS86 dosimetry shifts the boxy symbols somewhat to the right (higher dose) -- as predictable from comparing Column A in Tables 14-A and 14-B. Consequently, the equations of best fit for the T65DR and DS86 dosimetries are somewhat different.
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# Bandwidth
In electronics, Bandwidth is used to measure electronic communication. It is defined as the width of the range of the frequencies that an electronic signal uses. Bandwidth is expressed in terms of the difference between the signal having highest-frequency and the signal having the lowest-frequency. In computer networks, bandwidth is often used as a term for the data transfer rate. More easily, the amount of data that is carried or passed from one point to another in a network, in a given time period (usually a second). [1]
## Frequency
Many systems work by means of vibrations, or oscillations (Vibrations that goes back and forth at a regular rate). Each complete cycle of "back and forth" is called, simply enough, a cycle. The number of cycles per second of a system is its frequency. Frequency is measured in cycles per second, usually called "Hertz", and abbreviated "Hz".
Most systems do not operate at just a single frequency. They operate at many different frequencies. As an example, sound travels through vibrations. It has at least one frequency, and usually many different frequencies. People can hear sound frequencies as low as about 20 Hz, and as high as about 20,000 Hz. A band of frequencies is a continuous range of frequencies; the band of frequencies people can hear is from 20 Hz to 20,000 Hz.
To define, bandwidth is the width of such a frequency band, that is, the highest frequency minus the lowest frequency. In the hearing example, the bandwidth of a person's ears is about 20,000 Hz - 20 Hz = 19,980 Hz.
## Usage
Bandwidth is often applied to the electromagnetic spectrum (e.g., radio waves, light waves and X-rays). Such waves are vibrations (oscillations) of electric and magnetic fields. To use a radio example, the lowest United States AM radio channel covers the band of frequencies from 535,000 Hz to 545,000 Hz. It therefore has a bandwidth of 10,000 Hz (545,000 - 535,000). All United States AM radio broadcasting stations have this bandwidth (though the location of each band is distinct). The lowest United States FM radio channel (on the other hand) covers the band from 88,000,000 Hz (88 MHz) to 88,200,000 Hz (88.2 MHz). It therefore has a bandwidth of 200,000 Hz. (Notice that the width of an FM band is 20 times the width of an AM band.)
The term "bandwidth" has been misused in the field of digital data communication. It is often incorrectly used to mean "data carrying capacity". However, there is no such thing as "digital bandwidth". The proper term for the data carrying capacity of a communication channel is channel capacity.
It is true that, in general, the channel capacity of a system increases with the bandwidth used for communication. However, many other effects are also important. Therefore, in many (if not most) real systems, the channel capacity is not easily related to the channel bandwidth.
Sometimes, the word "broadband" is used to mean "high-speed", especially in for high speed internet connections. "Broadband" means "wide band", and suggests high-speed. However, the term is not clear; "high-speed" is more clear. Typically, a dial-up telephone connection is thought to be low-speed, at less than 56,000 bit/s (bits per second). High-speed is usually 200,000 bit/s or faster. DSL (Digital Subscriber Line) or Cable modem connections are usually high-speed.
## References
1. "'Explain Bandwidth'". 2009-12-23. Retrieved 2012-06-23.
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# Circle + pyramid - math problems
#### Number of problems found: 6
• Hexa pyramid
The base of the regular pyramid is a hexagon, which can be described by a circle with a radius of 1 m. Find the volume of the pyramid 2.5 m high.
• Hexagonal pyramid
The base of the pyramid is a regular hexagon, which can be circumscribed in a circle with a radius of 1 meter. Calculate the volume of a pyramid 2.5 meters high.
• Hexagonal pyramid
Calculate the surface area of a regular hexagonal pyramid with a base inscribed in a circle with a radius of 8 cm and a height of 20 cm.
• 9-gon pyramid
Calculate the volume and the surface of a nine-sided pyramid, the base of which can be inscribed with a circle with radius ρ = 7.2 cm and whose side edge s = 10.9 cm.
• From plasticine
Michael modeled from plasticine a 15 cm high pyramid with a rectangular base with the sides of the base a = 12 cm and b = 8 cm. From this pyramid, Janka modeled a rotating cone with a base diameter d = 10 cm. How tall was Janka's cone?
• Hexagon rotation
A regular hexagon of side 6 cm is rotated through 60° along a line passing through its longest diagonal. What is the volume of the figure thus generated?
We apologize, but in this category are not a lot of examples.
Do you have an interesting mathematical word problem that you can't solve it? Submit a math problem, and we can try to solve it.
We will send a solution to your e-mail address. Solved examples are also published here. Please enter the e-mail correctly and check whether you don't have a full mailbox.
Please do not submit problems from current active competitions such as Mathematical Olympiad, correspondence seminars etc...
Circle Problems. Pyramid Problems.
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# latentOT
Latent optimal transport (LOT) for low rank transport and clustering
This project is maintained by nerdslab
## Abstract
Optimal transport (OT) is a widely used technique for distribution alignment, with applications throughout the machine learning, graphics, and vision communities. Without any additional structural assumptions on transport, however, OT can be fragile to outliers or noise, especially in high dimensions. Here, we introduce Latent OptimalTransport (LOT), a new approach for OT that simultaneously learns low-dimensional structure in data while leveraging this structure to solve the alignment task. The idea behind our approach is to learn two sets of “anchors” that constrain the flow of transport between a source and target distribution. In both theoretical and empirical studies, we show thatLOTregularizes the rank of transport and makes it more robust to outliers and the sampling density. We show that by allowing the source and target to have different anchors, and using LOT to align the latent spaces between anchors, the resulting transport plan has better structural interpretability and highlights connections between both the individual data points and the local geometry of the datasets.
## Method
The main idea behind LOT is to factorize the transport plan into three components, where mass is moved: (i) from individual source points to source anchors, (ii) from the source anchors to target anchors, and (iii) from target anchors to individual target points. This is illustrated in Figure 1.
Figure 1: A transport consists of moving mass from the data to anchors, and the anchors to anchors
To translate the idea into mathematics, we propose the Latent Optimal Transport (LOT ) which solves the following optimization.
$\min_{\mathbf{P},z}\langle&space;\mathbf{P}_x,\mathbf{C}_x\rangle&space;+&space;\langle&space;\mathbf{P}_z,\mathbf{C}_z\rangle&space;+&space;\langle&space;\mathbf{P}_y,\mathbf{C}_y\rangle$
$s.t.~\mathbf{P}_x1=\mu,~\mathbf{P}^T_y1=\nu,~\mathbf{P}^T_x1=z_x,~\mathbf{P}_y1=z_y,~\mathbf{P}_z1=z_x,~\mathbf{P}^T_z1=z_y$
The output transport is low-rank and decomposed into 3 pieces shown in Figure 2.
$\mathbf{P}=\mathbf{P}_{x} \operatorname{diag}\left(\mathbf{u}_{z}^{-1}\right) \mathbf{P}_{z} \operatorname{diag}\left(\mathbf{v}_{z}^{-1}\right) \mathbf{P}_{y}$
Figure 2: The factored plan decomposes into 3 compoenents. The outer 2 pieces represents the clustering of data, and the inner piece represents the alignment of data.
By using number of anchors as a free hyperparameters for the source and target, LOT is able to capture different kinds of structures. An example is shown in Figure 3.
Figure 3: Comparison of transport plans obtained for different methods applied to clustered data
## Experimental Results
The low-rank decomposition in LOT allowed us to visualize transport between the source to anchors and then from anchors to the target again. This highlights the interpretability of our approach, with the middle transport plan Pz providing a concise map of interactions between class manifolds.
We show this in a setting where LOT is used to fix domain shift introduced in a MNIST-trained network by infering over (i) the USPS dataset (ii) a subset of MNIST (digits 2, 4, 8 removed - unbalanced transport) that is perturbed using a corase dropout transformation.
LOT succesfully bridges the gap by aligning the source and target distributions, and we can see in figure 2 how the class manifolds are being transpored to their correct correspondant, even in the unbalanced case.
Figure 4: (Top) Examples of MNIST-USPS samples and dropout mask. (Bottom) Visualization of transport of handwritten digits
## Citation
@InProceedings{lin2021,
title = {Making transport more robust and interpretable by moving data through a small number of anchor points},
author = {Lin, Chi-Heng and Azabou, Mehdi and Dyer, Eva},
booktitle = {Proceedings of the 38th International Conference on Machine Learning},
pages = {6631--6641},
year = {2021},
editor = {Meila, Marina and Zhang, Tong},
volume = {139},
series = {Proceedings of Machine Learning Research},
month = {18--24 Jul},
publisher = {PMLR},
}
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+0
# Help please
0
147
4
I need help with this question. Can someone explain with a step by step visual?
Let G be the center of equilateral triangle XYZ. A dilation centered at G with scale factor -3/4 is applied to triangle XYZ, to obtain triangle X'Y'Z' Let A be the area of the region that is contained in both triangles XYZ and X'Y'Z' Find A/XYZ.
If someone could help me with this that would be great!
May 1, 2022
### 4+0 Answers
#1
0
A/[XYZ] = 9/16.
May 1, 2022
#2
+1
can you explain to me how you got that answer? Though I do appreciate the attempt!
Guest May 1, 2022
#3
+9461
+1
A dilation with scale factor k applies a scale factor of k^2 on the area, so A/[XYZ] = (-3/4)^2 = 9/16.
May 2, 2022
#4
+118218
+1
The area of the small one is certainly 9/16 of the big one but I do not think that this is what the question is asking ...
At least that is not how i interpreted it.
I interpreted it as the intersection of the areas divided by the original big area. This is a harder question.
May 2, 2022
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# Show that there exist $a,b \in K [X_1,X_2,\cdots,X_n]$ and $d \in K[X_1,X_2,\cdots,X_{n-1}]$ such that $aF+bG = d.$
Let $$K$$ be a field. Let $$F,G \in K [X_1,X_2,\cdots,X_n]$$ be two polynomials which are relatively prime to each other. Show that there exist polynomials $$a,b \in K [X_1,X_2,\cdots,X_n]$$ and $$0 \neq d \in K [X_1,X_2,\cdots,X_{n-1}]$$ such that $$aF+bG = d.$$
• Your question is odd: I can take $a, b, d$ to be $0$, then the equality hold trivially. It should rather be something like: for all $d$ there exist $a, b$.. Moreover: What is relatively prime in your definition? – kesa Feb 12 at 11:07
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# Difference between revisions of "Incircle"
An incircle of a convex polygon is a circle which is inside the figure and tangent to each side. Every triangle and regular polygon has a unique incircle, but in general polygons with 4 or more sides (such as non- square rectangles) do not have an incircle. A quadrilateral that does have an incircle is called a Tangential Quadrilateral.
## Formulas
• The radius of an incircle of a triangle (the inradius) with sides $a,b,c$ and area $A$ is $\frac{2A}{a+b+c}$
• The radius of an incircle of a right triangle (the inradius) with legs $a,b$ and hypotenuse $c$ is $r=\frac{ab}{a+b+c}=\frac{a+b-c}{2}$.
• For any polygon with an incircle, $A=sr$, where $A$ is the area, $s$ is the semiperimeter, and $r$ is the inradius.
• The formula for the semiperimeter is $s=\frac{a+b+c}{2}$.
• And area of the triangle by Heron is $A^2=s(s-a)(s-b)(s-c)$.
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http://maaw.info/LearningCurveSummary.htm
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Management And Accounting Web
# What is a Learning Curve?
Provided by James R. Martin, Ph.D., CMA
Professor Emeritus, University of South Florida
Citation: Martin, J. R. Not dated. What is a learning curve? Management And Accounting Web. http://maaw.info/LearningCurveSummary.htm
The theory of the learning curve or experience1 curve is based on the simple idea that the time required to perform a task decreases as a worker gains experience. The basic concept is that the time, or cost, of performing a task (e.g., producing a unit of output) decreases at a constant rate as cumulative output doubles. Learning curves are useful for preparing cost estimates, bidding on special orders, setting labor standards, scheduling labor requirements, evaluating labor performance, and setting incentive wage rates.
There are two different learning curve models. The original model was developed by T. P. Wright in 1936 and is referred to as the Cumulative Average Model or Wright's Model. A second model was developed later by a team of researchers at Stanford. Their approach is referred to as the Incremental Unit Time (or Cost) Model or Crawford's Model. Simple learning curve problems are more easily introduced with Wright's model, although Crawford's model is widely used in practice. Thus, we will examine Wright's model first and Crawford's somewhat more involved approach second.
Wright's Cumulative Average Model
In Wright's Model, the learning curve function is defined as follows:
Y = aXb
where:
Y = the cumulative average time (or cost) per unit.
X = the cumulative number of units produced.
a = time (or cost) required to produce the first unit.
b = slope of the function when plotted on log-log paper.
= log of the learning rate/log of 2.
For an 80% learning curve b = log .8/log 2 = -.09691/.301 = -.32196
If the first unit required 100 hours, the equation would be:
Y = 100X-.322
The equation for cumulative total hours (or cost) is found by multiplying both sides of the cumulative average equation by X.
Since X times Xb = X1+b, the equation is:
XY = aX1+b
Thus, the equation for cumulative total labor hours is,
XY = 100X1-.322 = 100X.678
An 80 percent learning curve means that the cumulative average time (and cost) will decrease by 20 percent each time output doubles. In other words, the new cumulative average for the doubled quantity will be 80% of the previous cumulative average before output is doubled. For example, assume that direct labor cost \$20 per hour in the problem above. The cumulative average hours and cost as well as cumulative total hours and cost are provided below for doubled quantities 1 through 8.
Table 1: Example of Wright's Model With an 80% Learning Curve 1 Cumulative Output X 2 Cumulative Total Labor Hours XYh 3 Cumulative Average Labor Hours Yh 4 Cumulative Total Labor Cost XYc 5 Cumulative Average Labor Cost Yc 1 2 4 8 100 160 256 409.6 100 80 64 51.2 \$2,000 3,200 5,120 8,192 \$2,000 1,600 1,280 1,024
Note that the cumulative average columns, 3 and 5 decrease by 20% as output is doubled, or the new cumulative average is 80% of the previous cumulative average. The cumulative total columns 2 and 4 increase at a rate equal to twice the learning rate, or 160% in this case. Since these rates of change remain constant, tables for doubled quantities can be developed easily. However, for quantities in between the doubled quantities, the equations are required. For example, assume the firm has produced eight units as indicated in the table. How much will it cost to produce ten additional units? Any of the equations for Yh, Yc, XYh or XYc may be used to solve the problem. However, working with the equation for cumulative total cost is the fastest way to obtain the solution. The answer is found by subtracting the cost of the first 8 from the cost of producing the first 18. Using the equation for cumulative total cost generates the answer in two steps as follows:
Cost of first 18 = XYc = \$2,000(18).678 = \$14,194
Less cost of first 8 -8,192
Cost of 10 additional units \$6,002
Thus, producing 10 additional units will require approximately \$6,002 of additional direct labor cost.
Crawford's Incremental Unit Time (or Cost) Model
The equation used in Crawford's model is as follows:
Y = aKb
where: Y = the incremental unit time (or cost) of the lot midpoint unit.
K = the algebraic midpoint of a specific production batch or lot.
X (i.e., the cumulative number of units produced) can be used in the equation instead of K to find the unit cost of any particular unit, but determining the unit cost of the last unit produced is not useful in determining the cost of a batch of units. The unit cost of each unit in the batch would have to be determined separately. This is obviously not a practical way to solve for the cost of a batch that may involve hundreds, or even thousands of units. A practical approach involves calculating the midpoint of the lot. The unit cost of the midpoint unit is the average unit cost for the lot. Thus, the cost of the lot is found by calculating the cost of the midpoint unit and then multiplying by the number of units in the lot.
Since the relationships are non linear, the algebraic midpoint requires solving the following equation:
K = [L(1+b)/(N21+b - N11+b)]-1/b
where: K = the algebraic midpoint of the lot.
L = the number of units in the lot.
b = log of learning rate / log of 2
N1 = the first unit in the lot minus 1/2.
N2 = the last unit in the lot plus 1/2.
Once Yc is determined for the algebraic midpoint of a lot, then the cost of the entire lot is found by multiplying Yc by the number of units in the lot as indicated above.
An example of an 80 percent learning curve based on Crawford's unit time (or cost) model can be developed in much the same way we developed Table 1, except that the unit values for the doubled quantities decrease by 20% rather than the cumulative average quantities.
Table 2: Example of Crawford's Model with an 80% Learning Curve 1 Cumulative Output X 2 Incremental Unit Labor Hours Yh 3 Cumulative Total Labor Hours Kh (Yh) 4 Incremental Unit Labor Cost Yc 5 Cumulative Total Labor Cost Kc(Yc) 1 2 4 8 100 80 64 51.2 100 180 314.2 534.6 \$2,000 1,600 6,284 1,024 \$2,000 3,600 6,284 10,692
Notice from Table 2 that the unit labor hours (column 2) and unit labor cost (column 4) decrease by 20% each time the cumulative output is doubled. However, the cumulative total labor hours (column 3) and cumulative total labor cost (column 5) increase by a variable rate. This means that columns 3 and 5 are much more difficult to develop. It also means that the cumulative total hours and cost generated by the two models are not compatible when based on the same learning rate. For example, compare column 2 in Table 1 with column 3 in Table 2. The cumulative total hours for 8 units is 409.6 based on Wright's model and 534.6 based on Crawford's model. Another difference is that the cumulative average hours and cost decrease by a variable rate in Crawford's model. This does not present a problem when using Crawford's model because the cumulative averages are not required for predicting cost.
To illustrate the use of the algebraic midpoint equation and Crawford's approach, assume that the firm in the example above has produced 2 units and wants to determine the cost of producing 4 additional units. One way to find the answer is to calculate the unit cost for each unit 3 through 6 and then sum those values. That works reasonably well for a lot of 4 units, but would not be a practical way to determine the cost of 40, 400, or 4,000 additional units. The midpoint of the lot is:
K = [L(1+b)/(N21+b - N11+b)]-1/b = [4(.678)/(6.5.678 - 2.5.678)]1/.322
= [2.712/(3.55758 - 1.86124)]3.10559 = 4.2938
The cost of the mid point unit is:
Yc = \$2,000(4.29385)-.322 = \$1,250.99
and the total cost for the lot of 4 = 4(1,250.99) = \$5,005
An alternative is to use the equation for hours as follows:
Yh = 100(4.29385)-.322 = 62.5494 hours
Then the total cost for the lot of 4 is 4(62.5494)(\$20) = \$5,004.
Finding the Learning Rate When Doubled Quantities are not available
The equations provided above show how to use the learning curve to predict the time and cost of a specific quantity of units assuming that we know the learning rate. An important question, ignored to this point, is how do we find the learning rate in the first place? If we have data for two lots of units we can find the learning rate by using simultaneous equations. For example, assume two lots have been produced, one lot contained 2 units and a second lot contained 4 more units.
Lot Number of Units in the Lot Cumulative Units Labor Hours for the Lot Cumulative Labor Hours 1 2 2 4 2 6 72 111 72 183
We can solve for the learning rate using either Wrights model or Crawford's model, but the procedures and learning rates are different.
Using Wright's Model to find the Learning Rate:
The equations for the 2 lots are:
XY = aX1+b
72 = a(2)1+b
183 = a(6)1+b
Converting these to the log forms we have:
log 72 = log a + (1 + b)(log 2)
log 183 = log a + (1 + b)(log 6)
Calculating the log values indicated we have:
1.8575 = log a + (1 + b)(.301)
2.2625 = log a + (1 + b)(.7782)
1.8575 = log a + .301 + .301b
2.2625 = log a + .7782 + .7782b
Subtracting the first equation from the second equation provides the following equation which can easily be solved for b.
.405 = .4772 + .4772b
b = -.151299
Substituting b into either of the original equations, a = 40.
Then the learning rate is found by using the equation for b, i.e.,
b = log of learning rate / log of 2
-.151 = Log of learning rate / .301
log of learning rate = -.151(.301) = -.04545
The learning rate = the antilog = 10-.04545 = .90
Thus, the equation for cumulative average hours is:
Y = 40X-.151
and the equation for cumulative total hours is:
XY = 40X.849
Using Crawford's Model to find the Learning Rate:
To find the learning rate using Crawford's model, we must find the algebraic midpoint for each lot which is needed in the equations that must be solved simultaneously. We can't use the formula for K because it includes the value of b which is unknown. Thus, we must use the alternative midpoint formulas described by Liao [see p. 309].
The midpoint of the first lot is:
A = [(L + 1)/3] + .5 = (2+1)/3 + .5 = 1.5
The midpoint of subsequent lots is:
A = (L/2) + total units in all preceding lots = 4/2 + 2 = 4
After finding approximate midpoints we can develop two equations, one for each lot as follows:
Find the average hours for the midpoint units:
72/2 = 36 for the midpoint in lot 1.
(183 - 72)/4 = 27.75 for the midpoint in lot 2.
Then the equations are:
36 = a(1.5)b
27.75 = a(4)b
Converting to the log forms:
log 36 = log a + b(log 1.5)
log 27.75 = log a + b(log 4)
1.5563 = log a + b(.17609)
1.44326 = log a + b(.603)
Changing the signs in equation 2 and then adding the two equations provides:
.11304 = -.4259b
b = - .2654
Then a is determined:
36 = a(1.5)-.2654
a = 40.09
The equation for incremental unit time is:
Y = 40.09X-.2654
The learning rate is found by using the equation for b as indicated above in the example for Wright's model.
b = log of learning rate/log of 2
-.2654 = log of learning rate/.301
Log of learning rate = (.301)(-.2654) = .079885
The learning rate = antilog .079885 = .83198.
Comparing the two learning rates we have .90 for Wright's model and .832 for Crawford's model. This reinforces the fact that the two models are not compatible when the same learning rate is used. In other words, the same set of data will always generate two different learning rates under the two separate models because unit time and cumulative average time do not decrease at the same rate. The best model is the one that generates time and cost estimates that are closest to the actual results.
Concluding Comment
Learning curves range from around 70% to 100%. A learning curve below 70% is rare. A 100% learning curve indicates no learning at all. On the other hand, a 50% learning curve would indicate that no additional time or cost would be required for additional units beyond the first unit, since the cumulative average time, (in Wright's model) or the incremental unit time (in Crawford's model) would decrease by 50% each time output doubled. This means that the cumulative total time would not increase because it would equal 100% of the previous cumulative total time.
______________________________________________
1 The term experience curve is more of a macro concept, while the term learning curve is a micro concept. The term experience curve relates to the total production, or the total output of any function such as manufacturing, marketing, or distribution. The development of experience curves is attributed to the work of Bruce Henderson of the Boston Consulting Group around 1960.
Liao,S. S. 1988. The learning curve: Wright's model vs. Crawford's model. Issues in Accounting Education (Fall): 302-315.
Morse, W. J. 1972. Reporting production costs that follow the learning curve phenomenon. The Accounting Review (October): 761-773. (JSTOR link).
* For more information on the learning curve models, see the Learning Curve Bibliography.
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# One hundred monkeys have $100$ apples to divide. Each adult gets three apples while three children share one. The number of adult monkeys is $...........$ . $\left( a \right){\text{ 20}}$ $\left( b \right){\text{ 25}}$ $\left( c \right){\text{ 30}}$ $\left( d \right){\text{ 33}}$
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Hint: For solving this question we will use the unitary method for it. Since, from the question we can see that the ratio of adults and children is given as $3:1$ , which means in total we have $4$ . And from this, we will calculate the number of apples for adults. And then applying the unitary method we will have the number of adult monkeys.
So first of all we will see the values given to us. Since it is given that the number of apples is $100$ .
And also it is given that $1$ adults get $3$ apples and the $3$ children get $1$ apples.
So from this, the ratio of adult and children will be as $3:1$ , which means in total we have $4$
So, the number of apples for an adult will be given by
$\Rightarrow \dfrac{3}{4} \times 100$
And on solving the above fraction, we get
$\Rightarrow 75$
Since $3$ apple is for $1$ adults.
Therefore, $75$ apple will be for $\dfrac{{75}}{3}$
And on solving the fraction by dividing it, we get
$\Rightarrow 25$
Hence, the number of the adult monkey will be $25$
Therefore, the option $\left( b \right)$ is correct.
So, the correct answer is “Option b”.
Note: This type of question can be solved easily if we read the statement of the question carefully and we understand it is based on applications of fractions. A fraction represents a part of a whole or, more generally, any number of equal parts.
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QA
# What is the effect of increasing temperature on resistance?
The effect of temperature on the resistance of the conductor is directly proportional to each other. The increase in temperature of the conductor increases its resistance and makes it difficult to flow current through it.
### What is the effect of temperature on resistance of semiconductor?
Decreased resistance results from increasing the temperature of intrinsic semiconductors because it increases the thermal energy available for electrons to absorb.
## What is the effect of temperature on resistance of metals?
Semiconductors have a negative temperature coefficient of resistance, whereas metals have a positive temperature coefficient of resistance as their resistivity rises with temperature.
What is the effect of temperature on resistance and resistivity?
The indirect relationship between resistivity and temperature states that as materials temperatures rise, so do their resistivities.
The answer is that while the resistance of carbon, semi-conductors, and electrolytes decreases as their temperatures rise, the resistance of pure metals increases.
What happens to resistance when temperature decreases?
As a general rule, resistivity rises as temperature rises in conductors and falls as temperature rises in insulators.
How does temperature depend on resistance?
Resistance and Temperature According to the general rule, resistance rises as a function of temperature in insulators and falls as a function of temperature in conductors. In the case of semiconductors, typically, the resistance of the semiconductor falls as a function of temperature.
What are the effects of temperature on resistance?
If we raise the temperature of, say, a metallic conductor, the resistance will increase. So, to directly answer your question, we can say that resistance is directly proportional to temperature.
How does heat affect the resistance of most metals and their alloys?
Heated metal undergoes thermal expansion, which causes it to expand in volume, and a higher temperature increases and a lower temperature decreases a metals electrical resistance.
What is the effect of temperature on the resistivity of conductor?
The average speed of the electrons acting as the current carriers increases with conductor temperature, increasing the number of collisions and decreasing the average time of collisions, which causes conductor resistivity to rise with temperature.
Related Questions
### Why does resistance of metal increase with temperature?
Heating the metal conductor makes the atoms vibrate more, which in turn makes it harder for the electrons to flow, increasing resistance. As electrons move through a metal conductor, some collide with atoms, other electrons, or impurities, causing resistance and generating heat.
### Why does resistivity decrease with temperature?
The reason why a semiconductors resistivity decreases with temperature is that as the temperature rises, more valence band electrons are able to transfer their thermal energy to the conduction band, which leads to an increase in conductivity and a decrease in resistivity.
### What is the resistivity of alloys?
The resistivity of alloys is much higher than those of the pure metals (from which they are made). Alloy having a high resistivity is nichrome. The resistivity of semi-conductors like silicon and germanium is in-between those of conductors and insulators.
RESISTIVITY OF MATERIALS.
Metals Resistivity
Teflon 1 . 00 × 10 23
### Why is resistance of electrolytes related to temperature?
Since moving ions carry current in an electrolyte, and since we know that ion mobility increases with temperature, the resistance of electrolytes decreases with rising temperatures, according to option (C).
### How resistivity varies with temperature for semiconductor insulator and conductor?
As temperature rises, more electrons enter the conduction band, increasing the semiconductors conductivity, which causes a reduction in resistivity.
### What is temperature coefficient of resistivity?
The change in resistance per unit resistance per degree of temperature increase, based on the resistance at 0oC, is known as the temperature coefficient of resistance.
### What factors affect the resistivity of electrical materials?
Temperature, alloying, cold work, age hardening, and mechanical stress are some of the factors that have an impact on an electrical materials resistivity.
### What is the difference between resistance and resistivity?
Resistivity is the physical property of a specific substance that is having specific dimensions, and resistance is the physical property of a substance because of which it opposes the flow of current, i.e., electrons.
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Atmospheric pressure, also known as barometric pressure (after the barometer), is the pressure within the atmosphere of Earth. Bu saha içindeki hava basıncı; içeriye çevre atmosferi basılmak suretiyle, çevre basıncı üzerine çıkarılmıştır Pure water boils at 100 °C (212 °F) at earth's standard atmospheric pressure. ⋅ Similar metric units with a wide variety of names and notation based on millimetres, centimetres or metres are now less commonly used. The standard atmosphere (symbol: atm) is a unit of pressure defined as 101,325 Pa (1,013.25 hPa ; 1,013.25 mbar ), which is equivalent to 760 mm Hg , 29.9212 inches Hg , … This method became and continues to be useful for survey work and map making. R At normal atmospheric pressure and a temperature of 0°C, which phase(s) of H 2 O can exist? It is subject to constant fluctuations. 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Anonymous . 0 p Compare that pressure in psi to the normal pressure on earth at sea level in psi â ï¸ Our tutors found the solution shown to be helpful for the problem you're searching for. Thus, a diver 10.3 m underwater experiences a pressure of about 2 atmospheres (1 atm of air plus 1 atm of water). Local, National and International weather maps, graphs and charts forecasts. Because mercury (Hg) is about 13.6-times denser than water, a mercury barometer only needs to be $$\dfrac{1}{13.6}$$ as tall as a water barometer—a more suitable size. In aviation, weather reports (METAR), QNH is transmitted around the world in millibars or hectopascals (1 hectopascal = 1 millibar), except in the United States, Canada, and Colombia where it is reported in inches of mercury (to two decimal places). p 0 at′mospher′ic pres′sure. Problem: The pressure of the atmosphere on the surface of the planet Venus is about 88.8 atm. Branson's submarine to explore depth of oceans That's in contrast to normal atmospheric pressure, which is about 15 psi at sea level. T Pressure on Earth varies with the altitude of the surface; so air pressure on mountains is usually lower than air pressure at sea level. The standard atmosphere (symbol: atm) is a unit of pressure defined as 101,325 Pa (1,013.25 hPa; 1,013.25 mbar), which is equivalent to 760 mm Hg, 29.9212 inches Hg, or 14.696 psi. Search nearly 14 million words and phrases in more than 470 language pairs. 0 0 At low altitudes above sea level, the pressure decreases by about 1.2 kPa (12 hPa) for every 100 metres. s. Rya8484|Points 71| Log in for more information. The normal boiling point is the boiling point at normal atmospheric pressure (101,325 kPa). c 0 Answers/Comments. Normal atmospheric pressure at sea level is about fifteen pounds per square inch. Normal sea-level pressure is 29.92 inches. It is the force exerted on a surface by the air above it as gravity pulls it to Earth. ) If a cubic meter of water is taken from the surface to this depth, what is the change in its volume? M Atmospheric pressure shows a diurnal or semidiurnal (twice-daily) cycle caused by global atmospheric tides. [8] Temperature and humidity also affect the atmospheric pressure, and it is necessary to know these to compute an accurate figure. Atmospheric pressure is caused by the gravitational attraction of the planet on the atmospheric gases above the surface, and is a function of the mass of the planet, the radius of the surface, and the amount and composition of the gases and their vertical distribution in the atmosphere. The greater the weight, the higher the rise. 726. R International Civil Aviation Organization. This effect is strongest in tropical zones, with an amplitude of a few millibars, and almost zero in polar areas. normal atmospheric pressure 正常大气压 . 0 ) g − When barometers in the home are set to match the local weather reports, they measure pressure adjusted to sea level, not the actual local atmospheric pressure. Le point d'ébullition normal est le point d'ébullition à pression atmosphérique normale (101,325 kPa). Lernen Sie die Übersetzung für 'pressure' in LEOs Englisch ⇔ Deutsch Wörterbuch. For the pressure of air in other systems, see, "Surface pressure" redirects here. The pressure applied by the atmosphere on normal body is atmospheric pressure.Generally we can say it 1 atm =760mm of Hg or 760 torr. Physicists answer: 760 millimeters of mercury. As altitude increases, atmospheric pressure decreases. Oysters, clams, and mussels: keeping popular mollusks safe to eat Raymond Wheeler and his colleagues grew radish, lettuce, and wheat plants for 20 days in a chamber at normal atmospheric pressure. what is normal barometric pressure range. The altimeter setting in aviation is an atmospheric pressure adjustment. 0 Cooking at high elevations, therefore, requires adjustments to recipes[15] or pressure cooking. This standard is also called normal temperature and pressure (abbreviated as NTP). William Roy, using barometric pressure, was able to confirm Maskelyne's height determinations, the agreement being to within one meter (3.28 feet). p The mean sea-level pressure (MSLP) is the atmospheric pressure at mean sea level (PMSL). The measurement should be carried out exactly at sea level, and the … Pressure varies smoothly from the Earth's surface to the top of the mesosphere. Still have questions? p The standard pressure at sea level averages 14.7 pounds for every square inch of air, according to SETRA . That means air exerts 14.7 pounds per square inch (psi) of pressure at Earth’s surface. Altitude above Sea Level Absolute Atmospheric Pressure; feet: miles: meters: kPa: atm: psia-5000-0.95-1524: 121.0: 1.19: 17.55-4000-0.76-1219: 116.9: 1.15: 16.95-3000 25.5 inches of Hg C. 29.91 inches of Hg D. 103 mb Normal atmospheric pressure at sea level is 29.91 inches of Hg. Barometric Pressure: When we had typhoons in Japan and the barometric pressure dropped - a number of women who were late in their pregnancies went into labor. normal atmospheric pressure. A column of air with a cross-sectional area of 1 in2 would have a weight of about 14.7 lbf, resulting in a pressure of 14.7 lbf/in2. In 1774, Maskelyne was confirming Newton's theory of gravitation at and on Schiehallion mountain in Scotland, and he needed to accurately measure elevations on the mountain's sides. p Thus, the sea level, the vertical height of the water column would be 0.76 m 13.6=10.34 m. Thus, a glass tube more than 10 m long is required to make a water barometer. The average pressure at mean sea-level (MSL) in the International Standard Atmosphere (ISA) is 1013.25 hPa, or 1 atmosphere (atm), or 29.92 inches of mercury. Barometric pressure, or, atmospheric pressure, is used to describe the pressure within the Earth's atmosphere and is a measurement of the air's weight above a specific point. Convective available potential energy (CAPE), Cleaning and disinfection of personal diving equipment, Swimming at the 1900 Summer Olympics – Men's underwater swimming, Confédération Mondiale des Activités Subaquatiques, Fédération Française d'Études et de Sports Sous-Marins, Federación Española de Actividades Subacuáticas, International Association for Handicapped Divers, Environmental impact of recreational diving, Table Mountain National Park Marine Protected Area, Finger Lakes Underwater Preserve Association, Maritime Heritage Trail – Battle of Saipan, Use of breathing equipment in an underwater environment, Failure of diving equipment other than breathing apparatus, Testing and inspection of diving cylinders, Association of Diving Contractors International, Hazardous Materials Identification System, International Marine Contractors Association, List of signs and symptoms of diving disorders, European Underwater and Baromedical Society, National Board of Diving and Hyperbaric Medical Technology, Naval Submarine Medical Research Laboratory, Royal Australian Navy School of Underwater Medicine, South Pacific Underwater Medicine Society, Southern African Underwater and Hyperbaric Medical Association, United States Navy Experimental Diving Unit, List of legislation regulating underwater diving, UNESCO Convention on the Protection of the Underwater Cultural Heritage, History of decompression research and development, Basic Cave Diving: A Blueprint for Survival, Bennett and Elliott's physiology and medicine of diving, Code of Practice for Scientific Diving (UNESCO), IMCA Code of Practice for Offshore Diving, ISO 24801 Recreational diving services — Requirements for the training of recreational scuba divers, The Silent World: A Story of Undersea Discovery and Adventure, List of Divers Alert Network publications, International Diving Regulators and Certifiers Forum, List of diver certification organizations, National Oceanic and Atmospheric Administration, World Recreational Scuba Training Council, Commercial diver registration in South Africa, American Canadian Underwater Certifications, Association nationale des moniteurs de plongée, International Association of Nitrox and Technical Divers, International Diving Educators Association, National Association of Underwater Instructors, Professional Association of Diving Instructors, Professional Diving Instructors Corporation, National Speleological Society#Cave Diving Group, South African Underwater Sports Federation, 14th CMAS Underwater Photography World Championship, Physiological response to water immersion, Russian deep submergence rescue vehicle AS-28, Submarine Rescue Diving Recompression System, Artificial Reef Society of British Columbia, Diving Equipment and Marketing Association, Society for Underwater Historical Research, Underwater Archaeology Branch, Naval History & Heritage Command, International Submarine Escape and Rescue Liaison Office, Submarine Escape and Rescue system (Royal Swedish Navy), Submarine Escape Training Facility (Australia), Neutral buoyancy simulation as a training aid, https://en.wikipedia.org/w/index.php?title=Atmospheric_pressure&oldid=990684666, Short description is different from Wikidata, Creative Commons Attribution-ShareAlike License, Hazard identification and risk assessment, This page was last edited on 25 November 2020, at 22:55. Average Sea-Level pressure: 101.325 kPa /1013.25 mbar / 760 mmHg / 29.921 inHg. R Atmospheric pressure, also called barometric pressure, force per unit area exerted by an atmospheric column (that is, the entire body of air above the specified area). For numerical reasons, atmospheric models such as general circulation models (GCMs) usually predict the nondimensional logarithm of surface pressure. ( A typical gas-using residential appliance in the US is rated for a maximum of 1/2 psi, which is approximately 14 w.g. This answer has been confirmed as correct and … The boiling point is the temperature at which the vapor pressure is equal to the atmospheric pressure around the water. In most circumstances, atmospheric pressure is closely approximated by the hydrostatic pressure caused by the weight of air above the measurement point. The new value is the mean atmospheric pressure at an altitude of about 112 metres, which is closer to the worldwide median altitude of human habitation (194 m). The plasma operates at a temperature in the order of 3,000˚C to 5,000˚C. g Dr. Heidi Fowler answered. ISA - International Standard Atmosphere is defined to 101.325 kPa, 15 o C and 0% humidity. normal atmospheric pressure in atm, Above every square inch on the surface of the Earth is 14.7 pounds of air. Average sea-level pressure is 1013.25 mbar (101.325 kPa; 29.921 inHg; 760.00 mmHg). 1 atmosphere is the pressure exerted by the weight of air in the atmosphere acting on 1 square centimeter and is defined as being equal to 101325 Pa . A 51-year-old member asked: how does barometric pressure affect the body? ⋅ Standard atmospheric pressure is 101.325 kPa. c ⋅ {\displaystyle {\begin{aligned}p&=p_{0}\cdot \left(1-{\frac {L\cdot h}{T_{0}}}\right)^{\frac {g\cdot M}{R_{0}\cdot L}}\\&=p_{0}\cdot \left(1-{\frac {g\cdot h}{c_{\text{p}}\cdot T_{0}}}\right)^{\frac {c_{\text{p}}\cdot M}{R_{0}}}\approx p_{0}\cdot \exp \left(-{\frac {g\cdot h\cdot M}{T_{0}\cdot R_{0}}}\right)\end{aligned}}}. [11] A below-sea-level surface pressure record of 1081.8 hPa (31.95 inHg) was set on 21 February 1961. ⋅ 0 2. Atmospheric pressure, also known as barometric pressure (after the barometer), is the pressure within the atmosphere of Earth. High in the atmosphere, air pressure decreases. Pressure measures force per unit area, with SI units of Pascals (1 pascal = 1 newton per square metre, 1 N/m2). For higher altitudes within the troposphere, the following equation (the barometric formula) relates atmospheric pressure p to altitude h: n. 1. the pressure exerted by the earth's atmosphere at any given point. The atmospheric … Standard atmospheric pressure can also be expressed using different units, including 1013.25 millibars, 1013.25 hectoPascals, 760 millimeters of mercury and 29.92 inches of mercury. no. 1 decade ago. [1] The atm unit is roughly equivalent to the mean sea-level atmospheric pressure on Earth, that is, the Earth's atmospheric pressure at sea level is approximately 1 atm. − United States Barometric Pressure Map - WeatherWX.com Maps with barometric pressure readings in millibars (mb) or inches (in). It is the pressure of a fluid, which is always less than atmospheric pressure. Homework Statement In the Challenger Deep of the Marianas Trench, the depth of seawater is 10.9 km and the pressure is 1.16\\times10^8 Pa (about 1.15\\times10^3 atm). That height represents atmospheric pressure. It is directly proportional to the mass of air over that location. Al P. Lv 7. [14] Because of this, the boiling point of water is lower at lower pressure and higher at higher pressure. In medical work pressures are often measured in units of millimeters of $\mathrm{H}_{2} \mathrm{O}$ because body fluids, including the cerebrospinal fluid, typically have the same density as water. p Pressure § Surface pressure and surface tension, confirming Newton's theory of gravitation, "atmospheric pressure (encyclopedic entry)", A quick derivation relating altitude to air pressure, "Rehabilitation of hypoxemic patients with COPD at low altitude at the Dead Sea, the lowest place on earth", "Subject: E1), Which is the most intense tropical cyclone on record? g What's Normal? L (See barometer.) [12], The lowest non-tornadic atmospheric pressure ever measured was 870 hPa (0.858 atm; 25.69 inHg), set on 12 October 1979, during Typhoon Tip in the western Pacific Ocean. ⋅ The standard value for atmospheric pressure at sea level (ATM) is equal to: 1 atm = 29.92 in Hg (inches of mercury) 1 atm = 760 mm Hg (millimeters of mercury) 1 atm = 1013.20 millibars 1 atm = 14.7 psi (pounds force per square inch) 1 atm = 1013.20 hPa (hectopascals) To further complicate matters, there are actually 2 values for an atmosphere. M That pressure is called atmospheric pressure, or air pressure. Pressure (p), mass (m), and the acceleration due to gravity (g), are related by P = F/A = (m*g)/A, where A is surface area. [17], Static pressure exerted by the weight of the atmosphere, "Air pressure" redirects here. Relevance. Conversely, 10.3 m is the maximum height to which water can be raised using suction under standard atmospheric conditions. 0 1. 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# Homework Help: Generalised uncertainty Principle Can't get rid of the i?
1. May 6, 2015
### smileandbehappy
1. The problem statement, all variables and given/known data
3. The attempt at a solution
As you can see where I have got up to... However I am stuck... In the solution there is no i. In mine there will be. I am unsure how you make the complex number drop out at this stage. I think I'm missing a mathematical trick here...
The solution should be: greater than or equal to (5h(bar)^2)/2. Clearly I am going to have an erroneous complex number in here...
Thanks
Sam
2. May 6, 2015
### fzero
The notation $|\cdots|$ refers to the complex modulus, so you automatically have $|i|=1$.
3. May 6, 2015
### smileandbehappy
That's a bit embaressing... It is nearly 4am so let me off that one! Was reading it as the normal modulus making everything positive!!!
thanks - though, I was sititng scratching my head for half an hour at that.
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# Does this Converge ?
• Feb 11th 2010, 05:42 AM
coobe
Does this Converge ?
i have the series:
$f(x) = x - \frac{x^3}{3} + \frac{x^5}{5} -\frac{x^7}{7} +\frac{x^9}{9} ...$
now i have to determine the general form:
$\sum \frac{x^{2n-x}}{2n-1} *(-1^{n+2})$ from n= 1 to infinity
by the test from Leibniz this should be divergent, since its not decreasing and the limit isnt zero right ?
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# 22645 (number)
22,645 (twenty-two thousand six hundred forty-five) is an odd five-digits composite number following 22644 and preceding 22646. In scientific notation, it is written as 2.2645 × 104. The sum of its digits is 19. It has a total of 3 prime factors and 8 positive divisors. There are 15,504 positive integers (up to 22645) that are relatively prime to 22645.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 19
• Digital Root 1
## Name
Short name 22 thousand 645 twenty-two thousand six hundred forty-five
## Notation
Scientific notation 2.2645 × 104 22.645 × 103
## Prime Factorization of 22645
Prime Factorization 5 × 7 × 647
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 22645 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 22,645 is 5 × 7 × 647. Since it has a total of 3 prime factors, 22,645 is a composite number.
## Divisors of 22645
1, 5, 7, 35, 647, 3235, 4529, 22645
8 divisors
Even divisors 0 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 31104 Sum of all the positive divisors of n s(n) 8459 Sum of the proper positive divisors of n A(n) 3888 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 150.483 Returns the nth root of the product of n divisors H(n) 5.82433 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 22,645 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 22,645) is 31,104, the average is 3,888.
## Other Arithmetic Functions (n = 22645)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 15504 Total number of positive integers not greater than n that are coprime to n λ(n) 3876 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2530 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 15,504 positive integers (less than 22,645) that are coprime with 22,645. And there are approximately 2,530 prime numbers less than or equal to 22,645.
## Divisibility of 22645
m n mod m 2 3 4 5 6 7 8 9 1 1 1 0 1 0 5 1
The number 22,645 is divisible by 5 and 7.
## Classification of 22645
• Arithmetic
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
• Sphenic
## Base conversion (22645)
Base System Value
2 Binary 101100001110101
3 Ternary 1011001201
4 Quaternary 11201311
5 Quinary 1211040
6 Senary 252501
8 Octal 54165
10 Decimal 22645
12 Duodecimal 11131
20 Vigesimal 2gc5
36 Base36 hh1
## Basic calculations (n = 22645)
### Multiplication
n×y
n×2 45290 67935 90580 113225
### Division
n÷y
n÷2 11322.5 7548.33 5661.25 4529
### Exponentiation
ny
n2 512796025 11612265986125 262959763255800625 5954723838927605153125
### Nth Root
y√n
2√n 150.483 28.2916 12.2671 7.4301
## 22645 as geometric shapes
### Circle
Diameter 45290 142283 1.611e+09
### Sphere
Volume 4.86413e+13 6.44398e+09 142283
### Square
Length = n
Perimeter 90580 5.12796e+08 32024.9
### Cube
Length = n
Surface area 3.07678e+09 1.16123e+13 39222.3
### Equilateral Triangle
Length = n
Perimeter 67935 2.22047e+08 19611.1
### Triangular Pyramid
Length = n
Surface area 8.88189e+08 1.36852e+12 18489.6
## Cryptographic Hash Functions
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#### Details of this Paper
##### Linear Programing Questions Assignment
Description
solution
Question
Question;1. (a) What are the two different types of sensitivity ranges? Describe each type briefly and give a real world example for each type.(b) What is a marketing problem in applications of linear programming? Briefly discuss the decision variables, the objective function and constraint requirements in a marketing problem. Give a real world example of a marketing problem.(c) What is the required format of a linear programming problem to be solved by QM for Windows? What results are available from QM for Windows after solving a linear programming problem? Discuss briefly.(d) What is a transportation problem? Briefly discuss the decision variables, the objective function and constraint requirements in a transportation problem. Give a real world example of the transportation problem.Answer Questions 2 and 3 based on the following LP problem.Let P1 = number of Product 1 to be producedP2 = number of Product 2 to be producedP3 = number of Product 3 to be producedP4 = number of Product 4 to be producedMaximize 80P1 + 100P2 + 120P3 + 70P4 Total profitSubject to8P1 + 12P2 + 10P3 + 8P4? 6000 Production budget constraint4P1 + 3P2 + 2P3 + 3P4? 2000 Labor hours constraintP1 > 200 Minimum quantity needed for Product 1 constraintP2 > 100 Minimum quantity needed for Product 2 constraintAnd P1, P2, P3, P4? 0 Non-negativity constraintsThe QM for Windows output for this problem is given below.Linear Programming Results:Variable Status ValueP1 Basic 200P2 Basic 100P3 Basic 320P4 NONBasic 0slack 1 NONBasic 0slack 2 Basic 260surplus 3 NONBasic 0surplus 4 NONBasic 0Optimal Value (Z) 64400Original problem w/answers:P1 P2 P3 P4 RHS DualMaximize 80 100 120 70 Constraint 1 8 12 10 8 = 100 -44Solution-> 200 100 320 0 Optimal Z-> 64400 Ranging Results:Variable Value Reduced Cost Original Val Lower Bound Upper BoundP1 200 0 80 -Infinity 96P2 100 0 100 -Infinity 144P3 320 0 120 100 InfinityP4 0 26 70 -Infinity 96Constraint Dual Value Slack/Surplus Original Val Lower Bound Upper BoundConstraint 1 12 0 6000 2800 7300Constraint 2 0 260 2000 1740 InfinityConstraint 3 -16 0 200 0 308.3333Constraint 4 -44 0 100 0 366.66672. (a) Determine the optimal solution and optimal value and interpret their meanings.(b) Determine the slack (or surplus) value for each constraint and interpret its meaning.3. (a) What are the ranges of optimality for the profit of Product 1, Product 2, Product 3, and Product 4?(b) Find the dual prices of the four constraints and interpret their meanings. What are the ranges in which each of these dual prices is valid?(c) If the profit contribution of Product 2 changes from \$100 per unit to \$125 per unit, what will be the optimal solution? What will be the new total profit? (Note: Answer this question by using the ranging results given above).(d) Which resource should be obtained in larger quantity to increase the profit most? (Note: Answer this question using the ranging results given above.).4. A professional football player is retiring, and he is thinking about going into the insurance business. He plans to sell three types of policies? homeowner?s insurance, auto insurance and life insurance. The average amount of profit returned per year by each type of insurance policy is as follows:Policy Yearly Profit/PolicyHomeowner?s \$35Auto 24Life 60Each homeowner?s policy will cost \$15, each auto policy will cost \$12.50 and each life insurance policy will cost \$32 to sell and maintain. He has projected a budget of \$72,000 per year. In addition, the sale of a homeowner?s policy will require 6 hours of effort, the sale of an auto policy will require 3.2 hours of effort and the sale of a life insurance policy will require 10 hours of effort. There are a total of 30,000 hours of working time available per year from himself and his employees.He wants to sell at least twice as many auto policies as homeowner?s policies.How many of each type of insurance policy he would have to sell each year in order to maximize profit?(a) Define the decision variables.(b) Determine the objective function. What does it represent?(c) Determine all the constraints. Briefly describe what each constraint represents.Note: Do NOT solve the problem after formulating.5. The Charm City Snacks manufactures a snack mix by blending three ingredients: a dried fruit mixture, a nut mixture, and a cereal mixture. Information about the three ingredients (per ounce) is shown below.Ingredient Cost Fat Grams Protein grams CaloriesDried Fruit Mixture 0.90 1 1 175Nut Mixture 0.80 12 7 410Cereal Mixture 0.48 5 4 128The company wants to know how many ounces of each mixture to put into the blend. The blend should contain no more than 1200 calories and no more than 22 grams of fat. It should contain at least 15 grams of protein. Dried fruit mixture must be at least 20% of the weight of the blend, and nut mixture must be no more than 50% of the weight of the blend.Formulate a linear programming model that meets these restrictions and minimizes the cost of the blend by determining(a) The decision variables.(b) Determine the objective function. What does it represent?(c) Determine all the constraints. Briefly describe what each constraint represents.A) Decision variables:X1= grams of dried fruit mixtureX2 = grams of nut mixtureX3= grams of cereal mixtureB) The objective function is c = 0.90x1 + 0.80x2 + 0.48x3C) The contraints are:175x1 + 410x2 + 128x3? 12001x1 + 12x2 + 5x3? 221x1 + 7x2 + 4x3? 153x - y - z? 0-x + y - z? 0X1, X2, X3? 0The complete LP problem is:Minimize c = 0.75x + 0.65y + 0.35zSubject to:180x + 400y + 120z? 1000x + 10y + 3z? 25x + 6y + 2z? 123x - y - z? 0-x + y - z? 0with x? 0, y? 0, z? 0Note: Do NOT solve the problem after formulating6. A professor has been contacted by a company willing to work with student consulting teams. The Company needs help with four projects. There are four student teams available to work on these projects. The estimated time of completion (in hours) of each project by each team is given in the following table.Project 1 Project 2 Project 3 Project 4_________________________________________________Team A 32 25 22 24Team B 45 34 27 45Team C 40 - 24 28Team D 30 25 20 15_________________________________________________Team C cannot be assigned to Project 2 because they do not have enough training to do that project. The professor wants Team B to be assigned to Project 2 or Project 3. The objective of this assignment problem is to minimize the total time of completion of all the projects.(a) Define the decision variables.(b) Determine the objective function. What does it represent?(c) Determine all the constraints. Briefly describe what each constraint represents.
Paper#53630 | Written in 18-Jul-2015
Price : \$52
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07/29/2012 at 1:45 am
So, the space race was won by The Soviet Union because they were the first nation to get to the Moon with a mission. Of course. USA sent three years later people up there, but I think that they lost this race.
08/01/2012 at 3:01 pm
NUCLEAR FUSION PARTICLE ENVIRONMENT
Just entering the microcosm of atom matter where it is able to absorb and emit energy, still having the possibility to obtain energy from the mass of matter which can be found everywhere. Interestingly, E = mc2˛ does not contain any reference to a particular substance, but there is no need, for one gram of air, water or gram per gram of cosmic particles, is also a gram of mass. So our new role is important to release any material amount of energy in mass that surrounds us and which we use or adapt to the devices and equipment built until now
NUCLEAR FUSION AND TRAVEL FOR MARS I
Why this huge energy consumption? To overcome the planet's gravitational field and allow a ship to travel in outer space All bodies around the globe are attracted to Earth, the force of attraction called the gravity and acceleration with which these bodies attracted into the gravitational field is called gravitational acceleration g = 9.81 m/s2˛. Speed needed to escape Earth cosmic ship is 11000 m/s and can be done with an amount of energy of 2,400 tons of fuel Escape speed from Earth's surface is called the second cosmic speed. With this speed of a ship lifts off from Earth's surface and escapes the planet's gravitational field. Gravitational acceleration surface of Mars is g = 3.75 m/s2˛ and have an escape speed of 5100 m/s which is equivalent to a fuel quantity of 1055 tons For a trip to outer space or on Mars was built a Kennedy Space Center, a spaceship and made available energy 2400 tons of fuel - liquid oxygen and to overcome the gravitational field of Mars that has g = 3.75 m/s2˛ and to return to Earth we need a space center, a spaceship and an energy of over 1055 tones of fuel - liquid oxygen
NUCLEAR FUSION AND TRAVEL FOR MARS II
Stephen Hawking's Warning: Abandon Earthâ Or Face Extinction Let's face it: The planet is heating up, Earth's population is expanding at an exponential rate, and the the natural resources vital to our survival are running out faster than we can replace them with sustainable alternatives. So, according to famed theoretical physicist Stephen Hawking, it's time to free ourselves from Mother Earth. "I believe that the long-term future of the human race must be in space," "It will be difficult enough to avoid disaster on planet Earth in the next hundred years, let alone the next thousand, or million. The human race shouldn't have all its eggs in one basket, or on one planet. Let's hope we can avoid dropping the basket until we have spread the load." Solution to travel in outer space with the same ease as we do by car, bus, streetcar, subway, train, boat or plane is: QUANTIC PROPULSED SHIP
11/15/2012 at 3:57 pm
Stephen hawking and manoliuval's comments over terraforming planet mars seem too childish heres why. !!
instead of puting all that good effort/lets be honest/- and money, hope, and dreams=(well they are fantisy full) into inhabiting planet why not use all those resorces on fixing plannet earth. the mind set is like watching a littile toddler play with his favorite toy then, watching him get jelouse because he sees a new one someone else has. then to my dissapointment and to his loss he abandons his preciouse favorite toy.To [TRY] to get the new one.
dont tell me lies its not based on survival my young men, its all based on greed. all on a TRY hum.
where is your sense of a responsibility as a intelectual mind!
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# The advanced part of A treatise on the dynamics of a system
Format: Paperback
Language: English
Format: PDF / Kindle / ePub
Size: 12.43 MB
The relative accuracy rank of the coordinates in each feature class. When discussing these problems it is necessary to define what we mean by two maps being equivalent. Part I consists of 14 papers on the foundations of geometry, Part II of 14 papers on the foundations of physics, and Part III of five papers on general problems and applications of the axiomatic method. Therefore, you have a way of shifting around vectors without altering their instrinsic size and allowing you to compare things.
Pages: 450
Publisher: Ulan Press (August 31, 2012)
ISBN: B00B7J6XHC
Introducing Fractal Geometry
Arbeitstagung Bonn 2013: In Memory of Friedrich Hirzebruch (Progress in Mathematics)
Contact Geometry and Nonlinear Differential Equations (Encyclopedia of Mathematics and its Applications)
Schaum's Outline of Elementary Algebra, 3ed (Schaum's Outline Series) [Paperback]
This can be contrasted with the alternative approach (applied to the same size of problem) of simply adding one helix onto another and allowing the fold to grow through accumulated pairwise interactions. 1980) Homotopy Quantum Field Theory (EMS Tracts in Mathematics). Chapter 1 assembles some basics fact in homological algebra and develops the rst rudiments of de Rham cohomology, with the aim of providing an example to the various abstract constructions Topological Field Theory, Primitive Forms and Related Topics (Progress in Mathematics). Subtract: The Subtract fix removes the overlapping portion of geometry from each feature that is causing the error and leaves a gap or void in its place. This fix can be applied to one or more selected Must Not Overlap errors. Merge: The Merge fix adds the portion of overlap from one feature and subtracts it from the others that are violating the rule online. If f ' (a) and f ' (b) have opposite (nonzero) signs, at least one of those extrema is not located at an extremity, so it must be at a point x where f ' (x) = 0 A path-connected set is connected Algebraic Cobordism (Springer Monographs in Mathematics). Weyl's investigations ultimately led to the notion of gauge theories, Yang-Mills gauge theories in particular, which allowed for such modern theoretical successes as quantum chromodynamics and the unified electroweak theory. But mathematically such gauge theories had already been studied as connections in certain fiber bundles of a manifold pdf. We also mention other examples with infinite free homotopy classes Introduction to Mechnics. Topology rules can also be defined for the subtypes of a feature class. Geodatabase topology is flexible since you select which rules apply to the data in your feature class or feature data set The advanced part of A treatise on the dynamics of a system of rigid bodies : being part II. of a treatise on the whole subject, with numerous examples online. How would a computer answer this question with absolute certainty? This is one of the many kinds of problems that we think about in computational geometry and topology download. Networks 3 and 4 have four and six odd vertices, respectively, so they are not traversable The Topology of Chaos: Alice in Stretch and Squeezeland. Is each wormhole mouth in Figure 7 a separate world? However, the reason isn't because there are two wormhole mouths. The reason is because there are two tori, one embedded in the other. The tripus in Figure 1 isn't six different worlds (for three wormholes with six mouths): it's two worlds (genus 2) with three wormhole connections between them, because there are two spheres Surveys on Surgery Theory (AM-149), Volume 2: Papers Dedicated to C.T.C. Wall. (AM-149) (Annals of Mathematics Studies).
# Download The advanced part of A treatise on the dynamics of a system of rigid bodies : being part II. of a treatise on the whole subject, with numerous examples pdf
Typically these are centered around a particular topic, research related to the works of an individual, or papers presented at a focused research conference Proceedings of Gökova Geometry-Topology Conference 2002. To make this comparison even more direct.2. The fit of a structure to a framework will not be unique. 11. (Designated 2-5-2).4 Nested solutions The method described above allows a (real) protein structure to be compared to each of the ideal forms (frameworks) giving a quantified measure of each comparison.774 5. Each match was examined and all were found to be a good topological match.484 5 Topology Now!. Frankel is a bit more respectful of proper mathematics which also makes it a harder text to read for physicists Handbook of Topological Fixed Point Theory. Algebraic and differential topology: Bordism and cobordism, formal group laws, homology and stable homotopy theory, homotopy groups and braids, Hopf rings, immersions of manifolds, K-theory and bundles, the Landweber-Novikov and Steenrod Algebras, loop spaces and decompositions, model categories and rational topology, p-local finite groups, toric manifolds and varieties, unstable homotopy theory Elements of Algebraic Topology.
Topology: Webster's Timeline History, 1823 - 2002
Algebraic Varieties (London Mathematical Society Lecture Note Series)
Thus prior exposure to basic point set topology, homotopy, fundamental group, covering spaces is assumed, as well as some acquaintance with differentiable manifolds and maps, differential forms, the Poincaré Lemma, integration and volume on manifolds, Stokes' Theorem Automorphic Forms on GL (3,TR) (Lecture Notes in Mathematics). A network is a collection of points, called vertices, and a collection of lines, called arcs, connecting these points. A network is traversable if you can trace each arc exactly once by beginning at some point and not lifting your pencil from the paper. The problem of crossing each bridge exactly once reduces to one of traversing the network representing these bridges Restricted Orbit Equivalence for Actions of Discrete Amenable Groups (Cambridge Tracts in Mathematics). The following chapters focus on more elaborate concepts in geometry and topology and discuss the application of these concepts to liquid crystals, superfluid helium, general relativity, and bosonic string theory. Later chapters unify geometry and topology, exploring fiber bundles, characteristic classes, and index theorems. New to this second edition is the proof of the index theorem in terms of supersymmetric quantum mechanics Elementary Differential Topology. (AM-54) (Annals of Mathematics Studies). U is closed if and only if it contains the limit of all convergent sequences of its own points. U is compact when any sequence of its points has a subsequence which converges in U epub. Geometric topology as an area distinct from algebraic topology may be said to have originated in the 1935 classification of lens spaces by Reidemeister torsion, which required distinguishing spaces that are homotopy equivalent but not homeomorphic. This was the origin of simple homotopy theory Algebraic Structure of Knot Modules (Lecture Notes in Mathematics). Schulze, we know that the module of logarithmic residues is the dual of the Jacobian ideal. I will give some consequences of this duality, in particular, I will explain the symmetry I have proved between the set of values of logarithmic residues and the Jacobian ideal, which is in fact a generalization of the symmetry of the semigroup of reduced reducible plane curves proved by F From Geometry to Topology.
An Introduction to Compactness Results in Symplectic Field Theory
Nonlinear Optimization in Finite Dimensions - Morse Theory, Chebyshev Approximation, Transversality, Flows, Parametric Aspects (Nonconvex Optimization and its Applications Volume 47)
An Introduction to Contact Topology (Cambridge Studies in Advanced Mathematics)
Invariants of Homology 3-Spheres
Elementary Geometry of Differentiable Curves: An Undergraduate Introduction
Mathematical Visualization: Algorithms, Applications and Numerics
Introduction to Topology and Geometry
Metric Methods of Finsler Spaces and in the Foundations of Geometry. (AM-8) (Annals of Mathematics Studies)
The Structure and Stability of Persistence Modules (Springerbriefs in Mathematics)
Renormalization of Quantum Field Theories with Non-linear Field Transformations: Proceedings of a Workshop, Held at Ringberg Castle Tegernsee, FRG, February 16-20, 1987 (Lecture Notes in Physics)
Zeta Functions in Geometry (Advanced Studies in Pure Mathematics ; Vol. 21)
Topology of manifolds (Colloquium publication V. XXXII)
General Cohomology Theory and K-Theory (London Mathematical Society Lecture Note Series)
Thus in the case of Kiinigsberg I make the following calculations: Because this calculation results in a sum greater than 8, a crossing of this kind cannot be made in any way. 15. Let there be two islands A and B, surrounded by water, and let this water be connected with four rivers, as the figure (Figure 7-3) shows Introduction to Claims. Numbers will be limited and those interested are advised to make an early application. *All applicants will be contacted within two weeks after the deadline; information about individual applications will not be available before then* All research students and early career researchers will be charged a registration fee of £150 Network Topology and Its Engineering Applications. All letters in the names are converted to uppercase before the names are stored in metadata views or before the tables are accessed. This conversion also applies to any schema name specified with the table name Homotopy Quantum Field Theory (EMS Tracts in Mathematics). This proof relies on Ruberman's work on mutations along Conway spheres in least area form that preserve volume, and expanding this analysis to see when these Conway spheres could intersect short geodesics in a hyperbolic 3-manifold Regular Polytopes. Similarly. and family? • Most importantly. Brenner et al.2 Questions raised by classification Analysis of the various classifications has helped us to refine our ideas of protein 3D structure similarity. Classification has made it possible to explore global relationships between protein 3D structure and function. It is also possible to identify densely populated regions of fold space — referred to as ‘attractors’ in Holm and Sander (1996)). the focus of structure determination is moving towards protein-protein complexes such as those involved in transcription or signal transduction. 1992).. originally Nishikawa and Ooi (1974a). and more recently. (1998) have shown that most enzymes have α/β folds. often quoted. (1999)). further questions are also raised: • How might we best represent similarity relationships? • Is a hierarchy the best model? • Is it possible to reach consensus on terminology such as how to define a architecture Proceedings of Dynamic Systems and Applications: Selected Research Articles Presented in the Third International Conference on Dynamic Systems & Applications, Atlanta, Georgia May 1999. We define the loops as homotopic if and only if it has a common base point, continuously deforming each others loop. The above diagram shows an algebraic topology of the function f, g, h, i which are related to an initial point, as shown above Introducción a la topología. University of Oregon Topology-Geometry Seminar, Fall 2003 Note two seminars this week, in honor of Thanksgiving! 28 July - 1 August, 2014 The purpose of this workshop will be to bring together researchers and students working in contact geometry and related areas in symplectic topology, including topics such as pseudoholomorphic curves, h-principles, confoliations, symplectic dynamics, mapping class groups, and Stein manifolds download The advanced part of A treatise on the dynamics of a system of rigid bodies : being part II. of a treatise on the whole subject, with numerous examples pdf. To understand how topoisomerases work, it is necessary to look more closely at how the linking number is related to twisting and writhing epub.
# The advanced part of A treatise on the dynamics of a system
Format: Paperback
Language: English
Format: PDF / Kindle / ePub
Size: 10.68 MB
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# Distance between Karachi, Pakistan and the Antarctic Circle
10182 km = 6327 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 Karachi 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 Karachi and other cities
## Karachi, Pakistan
Country: Pakistan
Karachi’s coordinates: 24°54′20″ N, 67°04′55″ E
Population: 11,624,219
Find out what time it is in Karachi right now
Wikipedia article: Karachi
## 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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http://www.edupil.com/question/mischief-is-written-as-nkvgmoln-relibved-written-in-that-code/
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# Then is RELIBVED Written in that Code?
In a certain code MISCHIEF is written as NKVGMOLN, then is RELIBVED written in that code?
1. SEOIJVLD
2. SGOMJBLL
3. SFMJFWFE
4. SGOMJVED
Anurag Mishra Professor Asked on 16th November 2015 in
Explanation:-
M + 1 = N, Similarly, R + 1 = S,
I + 2 = K, E + 2 = G,
S + 3 = V, L + 3 = O,
I + 4 = M, I + 4 = M,
E + 5 = O, B + 5 = J,
F + 6 = N. V + 6 = B,
E + 7 = L,
D + 8 = L.
Then, RELIBVED is written as SGOMJBLL.
Hence, the answer is (2) SGOMJBLL.
Anurag Mishra Professor Answered on 16th November 2015.
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Browse FAQs
# What is the meaning of the "RTD FAULT" message ?
Published date: 26 July 2019
Issue
What is the cause of the "RTD FAULT" message in Sepam relays?
Product Line
Sepam 20, 40, 80
MET 148, 1482
Cause
The message " DEFAUT SONDE" or " RTD FAULT " corresponds to a physical fault on the temperature sensor.
Resolution
The resistance being read by the RTD must be within the expected bounds. For example:
In this example we will use a PT 100 sensor (PT 100 means that the sensor is made of Platinum and its resistance is of 100 ohms at 0°C )
The sensor PT 100 Equation is:
RPT100 = 100(1 + AT+ BT2)
Constants : A = 3,90802 10-3 et B= - 5,80195 10-7
Below 35°C we will consider the sensor input to be a shortcircuit, and above 205°C we will consider the input sensor to be an open circuit.
R sensor being the Internal resistance of the sensor.
i.e. :
35°C => R sensor = 86,25 ohms
+ 205°C => R sensor = 177,68 ohms
We will have the "DEFAUT SONDE "or "RTD' FAULT "if R sensor < 86,25 ohms or > 177,68 ohms
What can we do to improve the information ?
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Question 47
# Study the following figure carefully and answer the question that follows.If the pivot is shifted 3 cm towards left, then what will be the moment of the couple at the new position?
Solution
If the pivot is shifted 3 cm towards left,
then the moment of the couple at the new position
$$=MomentĀ onĀ theĀ heavierĀ sideĀ -Ā MomentĀ lighterĀ side$$
$$= 5\times 2 - 0.25\times 2$$
$$=8Nm$$
Option D is correct.
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Voltage sag on heater PSU
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9th January 2013, 08:43 PM #11 grommeteer diyAudio Member Join Date: Apr 2012 Location: Kiel, Germany To get a clearer picture can you please: Give all component values, Measure DC and AC voltages on all regulator pins, Measure the current from reg-out to heater wiring when volts are low, Test the circuit with a dummy load 22 Ohms 5W or a dial lamp 7V 0.3A. The 6922 is rated at 0.3A. Test the heater wiring with a bench supply, a 6V battery, 5V wall wart, or a simple heater transformer. Use a fuse! Just a thought: 1.24V is the typical Vreg of the LM338.
DF96
diyAudio Member
Join Date: May 2007
Quote:
Originally Posted by Preamp Where are those numbers from? Sounds interesting...
I can't remember, but you should be able to find them in books on power supplies. The basic idea is that the charging pulses for an electrolytic reservoir cap have a much higher RMS/average ratio than the resistive load assumed by transformer VA ratings. You have to derate by some factor around 2 or 3 (I can't remember the actual figure).
If the DC is powering a Class B/AB power amp then you can get most of that factor back again because of the duty cycle of real music. Thus a 50VA transformer can (very roughly!) just about power a 50W amp (or 25W stereo). Heaters draw continuous current so you lose this advantage.
bg3009
diyAudio Member
Join Date: Jan 2013
Quote:
Originally Posted by grommeteer To get a clearer picture can you please: Just a thought: 1.24V is the typical Vreg of the LM338.
Hi, I am using 1/4 watt resistors to set the reference voltage (2 x 180R). Would there be a concern with the size of these resistors? I can't seem to find anywhere in the data sheet where it specifies the rating for these resistors.
Cheers!
10th January 2013, 01:19 PM #14 DF96 diyAudio Member Join Date: May 2007 250mW power, maybe 200V voltage? Should not be a problem for a 6.3V supply.
bg3009
diyAudio Member
Join Date: Jan 2013
Hi,
With load, the waveform to the bridge rectifier is about 9.5 volts p-p.
The voltage at the input of the power regulator is about 10.4 volts DC. The waveform pretty flat with some noise at the 1mV setting. This measurements were taken with just the 6922 plugged in. With a good 10volts supply entering the Lm338K, I am beginning to suspect the power regulator is not set up correctly. The circuit was built as per the following schematic.
The bottom section is of course the heater section that's causing grief.
Cheers!
Attached Images
image.jpg (396.7 KB, 47 views)
bg3009
diyAudio Member
Join Date: Jan 2013
Hi,
Sorry I had split my reply up. This is the waveform of the AC input at the bridge rectifier with load. I am getting a good 10.4 volts input at the power regulator.
Cheers!
Attached Images
image.jpg (370.9 KB, 44 views)
11th January 2013, 04:42 PM #17 grommeteer diyAudio Member Join Date: Apr 2012 Location: Kiel, Germany Case of LM338K is out, but In and adj can easily be interchanged.
Robert Kesh
diyAudio Member
Join Date: Oct 2012
Quote:
Originally Posted by bg3009 Hi, I am using 1/4 watt resistors to set the reference voltage (2 x 180R). Would there be a concern with the size of these resistors? I can't seem to find anywhere in the data sheet where it specifies the rating for these resistors. Cheers!
1.25^2/180 = a lot less than 250mW
The other resistor can sometimes need a half watt or more, but not at 6.3V
I'm wondering if your heat sink/case of regulator, is shorting with something
metalsculptor
diyAudio Member
Join Date: Sep 2010
Location: Melbourne
Quote:
Originally Posted by bg3009 Because I power up through a variac, would the ramping up cause a voltage violation issue with the LM 338K? Cheers
It would not help, the drop out current is less than 1 amp with low voltage differential. I do not know the fold back characteristics of this device but it may have gone into some sort of foldback.
Try applying full power to the circuit then switching on a heater load. The transient load from a cold heater can be determined by measuring the cold resistance, the valve datasheet may mention it as well.
There is a simple soft start circuit on the TI lm138 datasheet, this might be another option if the variac is causing trouble
Just check you have connected the 338K the right way
30th January 2013, 09:01 AM #20 bg3009 diyAudio Member Join Date: Jan 2013 Gents, After a good summer holiday, a fresh re-look into the circuit revealed what was wrong. Thee feedback resistors were built into a small PCB that was mounted onto the voltage setting terminal of the LM338K. A wire was to provide voltage from the output of the LM338K back to this circuit. However, due to a fault in connection, no voltage was flowing back from the output to the resistor/trimpot network. Re-did the connection and the circuit came up beautifully. Thanks every one for the input!
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# Homework Help: Dynamic systems
1. Dec 22, 2007
### series111
[SOLVED] dynamic systems
I Have Just Received This Question = A Car Of Mass 1.2 Tonnes Rolls Down A Hill Which Is 600m Long And Is Inclined At An Angle Of 9 Degrees To The Horizontal. Ignoring The Effects Of Air Resistance And Friction And Assuming The Car To Start From Rest At The Top Of The Hill Calculate Its Speed As It Reaches The Bottom Of The Hill. I Think I Need To Calculate This Using A Vector But I Cant See How I Get The Speed. Please Can Some One Explain This As I Cannot Get My Head Around This Hope Some One Can Help Cheers Mark.
2. Dec 22, 2007
### dotman
Hello,
Have you drawn a free-body diagram? What forces are acting on the car? What equations seem relevant here?
3. Dec 22, 2007
### series111
I Have Done Vectors But I Have No Examples With Another Object This Is What Is Puzzling Me And The Equation Which I Think Is Relevant V=s/t Please Help As I Dont No Where To Start Some Examples Would Be Great Thanks For Your Reply Mark.
4. Dec 22, 2007
### dotman
Hello,
V = s/t is relevant, but I don't think it will help you here. You haven't answered my first two questions.
Have you drawn a free-body diagram, with the forces labeled? If not, do so.
5. Dec 24, 2007
### series111
I Have Drawn A Free Body Diagram With The Forces Labeled And I Am Just Wondering If This Calculation Is Correct 1200 X Sin (9) = 187.72 M/s Or Am I Way Off Thanks For Your Help Again Mark.
6. Dec 24, 2007
### cryptoguy
You can use conservation of energy. PE = KE, so mgh = .5mv^2; h (height) would be 600sin(9).
7. Dec 24, 2007
### series111
Thanks for the advice much appreciated still trying to find my feet in mechnical priciples.
8. Dec 24, 2007
### series111
Is this the correct solution at the top of the system pe = 1.2 tonnes x 9.81 x 93.86 (height) = 1.104 x 10 6 then (pe1+ke1) = (pe2+ke2)
1.104 x 10 6 + 0 = 0 + ke2 ke2 = 1.104 x 10 6 joules
ke = 0.5 mv2
1.104 x 10 6 = 0.5 x 1.2 tonnes x v 2
1.104 x 10 6 = 0.600 v 2
1.104 x 10 6 / 600 = v 2 = 1840
= square root of 1840 = v = 42.89 m/s
9. Dec 24, 2007
### cryptoguy
yep looks right to me. You actually don't even need the mass, it cancels out in PE=KE
10. Dec 24, 2007
### series111
Thanks very much cheers mark
11. Jan 6, 2008
### series111
please can i bother you again as i thought i would be able to answer the following questions after i got the speed here are the rest of the questions (b) calculate its acceleration down the hill (c) the time taken to reach the bottom of the hill (d) the change in the momentum of the car between the top of the hill and the bottom of the hill (e) the change in kinetic energy for the car between the top and bottom of the hill.
also if the brakes are applied at the bottom of the hill slowing the car at a rate of 7.5 m/s2 calculate the stopping distance andthe speed after 5 seconds of braking i am seriously lost as i dont know which equations to use as i have been looking in my notes and several text books and still havent got anywhere your help is much appreciated.
12. Jan 6, 2008
### cryptoguy
b) So you need an equation with Accel. (which you're trying to find), Distance (which you know), Vi and Vf (which you also know). What equation could you use?
c) Simple equation with Vi, Vf, a, and t.
d) Momentum is m*v, so what is it on the bottom compared to what it is on top (hint: on top, v = 0)
e) What is the KE on top? (hint: v = 0 again). And you already calculated the KE on the bottom of the ramp.
For the brakes, you can forget about the hill and just treat this as a 1D motion problem. You know the initial speed and you know the acceleration. You need an equation with Vf, Vi, a, and t again (which you used in part c) and don't forget that since you're braking, the acceleration is negative.
13. Jan 6, 2008
### series111
what does vi and vf stand for as i have the following four formulas
v= u +at u= initial velocity (m/s)
v2 = u2 + 2as v = final velocity (m/s)
s = 1/2 (u +v)t a = acceleration (m/s2)
s = ut + 1/2 at2 s = distance (m)
as for the momentum i calculated 1200 x 42.89 = 51.468 kgm/s but this does not give the change in momentum thanks for your help and quick reply as i am struggling as you can guess but will get there hopefully mark.
14. Jan 6, 2008
### cryptoguy
Sorry about the terminology, Vi is initial velocity, and Vf is final velocity. The momentum on top of the hill is 0 (1200 * 0, right?) so the change in momentum would be the momentum you found - 0 :D
The formulas you have are all you need to solve all parts.
15. Jan 6, 2008
### series111
for the momentum i calculated this the other way last night and i didnt realise it was the same answer mom = m2v - m1u (kgm/s)
(1200x42.89) - (1200x0) = 51.468 kgm/s is this correct ?
as for the acceleration is the formula this a = v2-u2 / 2 x s
also i have just used this formula to work out the kinetic energy ke = 1/2 m (u2-v2)
= 1/2 x 1200 (0 2-42.89 2)
= 1103.73 joules
Last edited: Jan 6, 2008
16. Jan 6, 2008
### cryptoguy
correct
17. Jan 7, 2008
### series111
just to let you know ive been working on these questions today and came up with the following : acceleration a = v2 - u2/2xs = 42.89 2 - 0 2/ 2 x 600 = 1.839 x 10 3 m/s2
time taken t = v - u/a = 42.89 - 0 / 1.839 x 10 3 = 0.023 sec
stopping distance v = u + at = 0 = 42.89 + -7.5
t= v-u/a = 0 - 42.89/-7.5 = 5.71 sec
v = u +at v = 42.89 + -7.5 = 35.39
v = 35.39 x 5.71 = 202.07m
speed after 5 sec braking v = u +at v = 42.89 + -7.5 x 5.71 = 0.065 m/s
thanks again for taking the time to help me yours mark
18. Jan 7, 2008
### cryptoguy
Wait, for the 5 secs braking, why did you use 5.71 for time?
19. Jan 7, 2008
### series111
oh no ive just realised it should be the 5 seconds as that is what it is asking
20. Feb 21, 2008
### series111
just to let you no i used the wrong formula for the stopping distance i should have used a formula with (s) in it not (t) so i have recalculated it using this formula as follows :
v2 = u2+2as
transposed gives you this s=v2-u2/a
so s = 0 squared - 42.89 squared/ - 7.5 = 245.27 m
is this right in your eyes ? thanks for your help mark.
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math
A certain diet requires no more than 60 units of carbohydrates, at least 45 units of protein, and at least 30 units of fat each day. Each ounce of Supplement A provides 5 units of carbs, 3 units of protein, and 4 units of fat. Each ounce of Supplement B iprovides 2 units of carbs, 2 units of protein, and 1 unit of fat. If Supplemnet A cost $1.50 per once and Supplemnt B costs$1.00 per ounce, how many ounces of each upplement should be taken daily to minimize the cost of diet?
i can't come up with system of
equations for this thingg..
Let C,P and F denote carbohydrates, protein and fat. Let D denote what the diet requires. Then we're told
60C<=D
D<=45P
D<=30F
A = 5C + 3P + 4F
B = 2C + 2P + 1F
A=1.50
B=1.00
We want to maximize
xA+yB=D
where x is the number of Aoz and y is Boz.
and minimize
1.50x+1.00y=cost
There are a number of ways to solve this, but considering the size of the numbers you might just make a table that starts at x=0 and y=30 and find the feasible set of values, then find which has the smallest cost in the feasible set.
After I reviewed my preious post I realized I could be a little more helpful.
If you use x and y for the number of oz of each supplement, then we want to know how many of each is the ideal amount for the diet and cost. We know they provide
1oz A = 5c + 3p + 4f
ioz B = 2c + 2p + 1f
In terms of c, p and f
5x+2y<=60 carb. requirement
3x+2y=>45 prot. requirement
4x+y=>30 fat requirement
We also know
x=>0 and y=>0 we are only interested in values in the first quadrant.
If you graph that set of lines and examine where they intersect you'll find vertices to test in the cost equation
1.50x+1.00y=Cost
I graphed those lines and there should be three vertices to test.
I hope this is more helpful.
find the solution to the system by graphing x+y=-3 and 3x-y=7
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