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http://www.reporternews.com/news/2013/apr/14/larry-the-answer-guy-willis-is-not-the-longest/ | 1,397,961,918,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609537804.4/warc/CC-MAIN-20140416005217-00228-ip-10-147-4-33.ec2.internal.warc.gz | 649,334,842 | 25,221 | Larry the Answer Guy: Willis is not the longest Abilene street
What is the longest street in Abilene, not counting highways, county roads, farm-to-market roads, access roads, old U.S. 80 (South 1st), and such? My guess would be Willis.
You can make a case for Willis being the longest, but I’m going to disagree.
First, I’m glad you set up some guidelines, or else I would have used a tank of gas answering your question.
So, getting out my new Abilene 2013-14 street map and a ruler, it looks like the longest streets meeting your parameters are: Willis, 9 inches; South Seventh Street, 8½ inches; Maple Street, 8½ inches; and North 10th St., 10½ inches.
To get an accurate length for the streets, instead of relying on math, I drove the entire length of each. Here’s what I found:
Willis Street — from Danville Drive to Stamford Street — 5.3 miles (14 traffic signals, one four-way stop).
South Seventh Street — from Carver Street to Arnold Boulevard — 5.3 miles (12 traffic lights).
Maple Street — from FM 707 to East South 11th Street — 5.4 miles (no traffic lights, one four-way stop, one regular stop). There’s a gap in Maple. It also runs for two blocks south of East South Seventh Street that I didn’t add to the total.
North 10th Street — from Marigold Street to Loop 322 — 6.8 miles (9 traffic lights, two four-way stops).
Some might argue that North 10th and East North 10th should be considered separate streets. I would counter they are no more separate than North Willis and South Willis.
For fun, I did the math from the map, and the calculations didn’t come up anywhere near the distances from my Ford pickup’s trip odometer. (The map must be defective — the scale might be off slightly, and it won’t refold.)
One thing’s for sure, I have a detailed record of my trips (down to hitting 16 of the 35 traffic lights on red) for my income tax mileage deduction. Now, if I had only kept better track of the mileage it took to get to the starting points for each trip.
Readers’ questions are answered Mondays through Thursdays. Send them to larrytheanswerguy@reporternews.com or Larry Zelisko at P.O. Box 30, Abilene TX 79604.
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Want to participate in the conversation? Become a subscriber today. Subscribers can read and comment on any story, anytime. Non-subscribers will only be able to view comments on select stories. | 588 | 2,392 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2014-15 | latest | en | 0.938328 |
https://boards.straightdope.com/t/lightning-the-circuit/61444 | 1,652,801,803,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517485.8/warc/CC-MAIN-20220517130706-20220517160706-00136.warc.gz | 206,116,849 | 8,852 | # Lightning & The Circuit
I have been having this on-going debate with friends at work. I ask - ‘Why does lightning strike the Earth?’ The reply I always get is - ‘It’s completing a circuit - the Earth is the ground.’ I just don’t see it. My confusion is - for it to complete a circuit most of the energy has to travel up into the sky from the Earth the same time or sometime before the lighning strikes. A circuit is circular meaning a ground is only relative to its positive and not one from another system. An analogy - if I take two car batteries and touch the positive of one to the negative of the other - nothing will happen. Even though one post is more negative than other they aren’t in the same system and not relative to each other. All the theories I’ve read on lightning makes reference to electricity being created in the ice layer of clouds. If this is true - why doesn’t the imbalance complete the circuit in the cloud (sheet lightning)? If anyone understands what I’m trying to ask I’d really appreciate an answer. Thanks in advance- B.
Most lightning is up in the clouds like you say. The lightning from the ground is exactly that from the ground to the sky, not like we think of lightning striking the earth from the sky. The earth is always negative, that’s why you better have on your rubbers if you stick your finger in a socket.
Welcome to the SDMB, Cocks
This is from How stuff works
BTW, the water in the cloud travels from the earth to the sky.
Within the cloud, the water is blown all around. Sometimes it goes so high up that it freezes and then falls, gets a new coating of liquid water, then going up and re-freezing. When it’s too heavy, it falls all the way to the ground as hail.
Nonresponsive post from an SC native to welcome SC_Cocks84 to the boards.
Hi all you SC natives, from one living in SC now.
ChooseyBeggar’s link to How Stuff works does not tell the whole story. A lightning bolt from the sky to the ground is not one-way. There is another charge emanating from the ground or, more likely, a tall object on the ground. These two charges are not visible to the naked eye, as they are only charges, until and if they meet forming the lightning bolt.
I’ve seen on TV pictures of the charge emanating from a tree or a telephone pole, and showing it either missing or connecting with the downward charge. I don’t know how the pictures were taken, or don’t remember, but you could not see them with the naked eye.
To complete the circuit, the charge must be routed back to the cloud. If it were just one way, from cloud to ground, there would be no circuit. After the charges meet, there is the return electricity to the cloud, completing the circuit.
No, but the next page of the “How Stuff Works” explanation does, and your explanation doesn’t quite match it.
I didn’t go to the link, but after reading it, how does my description “not quite match it”? The “streamer” is the charge from the ground or an object on the ground. The “step leader” is the charge from the cloud. When they meet, they open a channel for the flow of electricity. Since there has to be a closed circuit, the flow is a loop. If this is not the case, can some one explain to me how you can have the flow without the loop?
Incidentally, it was the purple glows that I saw on TV.
It is not necessary to have a complete circuit in order for current to flow.
For example, a capacitor consists of two metal plates that, while close together, do not actually touch each other. Such a device can be “charged”. You hook one plate up to the (+) side of a battery and the other plate up to the (-) side of the battery. Even though there is no path for the current to flow all the way around, current does flow. The plate attached to the (+) end acquires a positive electrical charge, and the plate attached to the (-) end acquires a negative electrical charge. This continues until the two plates acquire a strong enough charge that the battery cannot charge them any more.
If the charged capacitor is then disconnected from the battery, and the two plates are then connected to each other, there is a sudden discharge of current, a sudden flow of electrons from the (-) plate to the (+) plate, the instant the connection is made. Again, this is not a complete circuit as the capacitor’s plates do not touch each other, yet current flows. You can see the current flowing by charging the capacitor and then connecting it to a sensitive current meter. A really big capacitor will put on a much more impressive display when it suddenly discharges, making a loud “bang” and sending out sparks and such.
The clouds and the Earth are like the 2 plates of a capacitor – they carry opposite charges. The ionized trail through the air is like the wire attaching the two plates of the capacitor to each other. Yes, technically, it is incorrect to call this a “circuit”. But electrical engineers build “circuits” with capacitors in them all the time, and use the term “circuit” in a looser sense to mean any electrical device through which current will flow.
(I used it to prepare when teaching electrostatics this year.)
Also, excellent capacitor analogy, tracer!
Hence, there is a closed circuit. Altho it was referred to as a capacitor in the opening of the link, I don’t think that is the case. A capacitor stores electricity - electricity already produced by the batteries. It doesn’t need a closed circuit because the electricity is already there. In lightning, there is no electricity until the circuit is closed, as there is no pre-produced electricity.
barbitu8 wrote:
Ah, but there is electric charge stored in the clouds – it’s just not provided by an obvious source like a battery. The clouds are charged by friction, the way you can charge up your own body by shuffling your feet along a carpet.
The capacitor analogy, like all analogies, is somewhat limited. The lightning stroke is really not the same as connecting the 2 plates of a charged capacitor with a wire. It’s more like charging the 2 plates of a capacitor so much that their electric potential exceeds the “breakdown voltage” of the air gap between them, and current “jumps” from one plate to the other. (The breakdown voltage for air is pretty high; I once heard a figure of 10,000 Volts per centimeter of air gap. If the gap is filled with something that has a much lower breakdown voltage than air, the effect is easier to achieve and less destructive to the equipment. Neon, for example, has a breakdown voltage of only about 90 Volts; a neon lamp is essentially a capacitor driven beyond its breakdown voltage.)
From my electronics manual it says the dielectric strength (breakdown voltage) for ‘air’ is 75 Volts per mil (.001 of an inch) which would put the value per centimeter 'round 30,000 V.
The capacitor visualization is perfect really. A strong enough charge potential between the cloud and ground and you have breakdown voltage through the atmosphere which is yer dielectric.
Also, it’s interesting to note that blue lighting carries higher current whereas white lighting carries higher voltage.
I wonder, has there been any new news about the phenomenon of ‘sprites’? - Those little flashes that can be observed miles above a thunderhead. I remember seeing a show a year or so ago where they were trying to decipher them. | 1,588 | 7,334 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-21 | latest | en | 0.945717 |
https://studylib.net/doc/5872404/a.rei.1.identifyingproperties | 1,560,894,866,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998817.58/warc/CC-MAIN-20190618203528-20190618225528-00279.warc.gz | 587,960,595 | 41,367 | # A.REI.1.IdentifyingProperties
```Regents Exam Questions: Binary Operations
Page 1
www.jmap.org
Name: __________________________________
1 The operation element @ is determined by the
following table:
3 An addition table for a subset of real numbers
is shown below. Which number is the identity
What is the identity element of this operation?
1) a, only
2) b, only
3) c
4) a and b
2 What is the identity element for
accompanying table?
1)
2)
3)
4)
r
s
t
u
in the
4 The operation for the set
is defined
in the accompanying table. What is the inverse
element of r under the operation ?
1)
2)
3)
4)
p
r
s
v
5 In the addition table for a subset of real
numbers shown below, which number is the
Regents Exam Questions: Binary Operations
www.jmap.org
1 ANS: 1
The identity element is a, because any element @ a equals the original element.
REF: 080112a
2 ANS: 2
The identity element is s because any element
s equals the original element.
REF: 080514a
3 ANS:
The identity element is 0, because any element + 0 equals the original element.
REF: 060224a
4 ANS: 4
The identity element is s because any element s equals the original element. Then review the table to
solve:
. The inverse of r is v because
.
REF: 080010a
5 ANS:
The identity element is 4 because any element 4 equals the original element. Then review the table to
solve:
. The inverse of 3 is 1, because
.
REF: 080222a
```
Gestures
11 Cards
Hand gestures
12 Cards
Hand gestures
12 Cards | 417 | 1,452 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2019-26 | longest | en | 0.773732 |
http://drugstorepdfsearch.com/a/ap.polyu.edu.hk1.html | 1,716,239,042,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058313.71/warc/CC-MAIN-20240520204005-20240520234005-00363.warc.gz | 8,869,106 | 3,299 | ## Microsoft word - phase filters.doc
Phase filters
Object light field distribution function f (x) can be consider as f (x) g(x)e So far we have discussed filters that deal with amplitude only. Let us consider a “pure” phase object of so that I (x) f (x) one cannot “see” the phase variation(x) across the field. Phase contrast viewing
If the phase variations (x) are small compared with unity 1, then f (x) i (x) 1 i(x) F ( f ) ( f ) i( f ) where ( f ) which is a real delta function at the origin added to a 90 out-of-phase phase spectrum. If this unfiltered distribution is observed in the image plane, a uniform intensity distribution is all that appears. The phase distribution is not visible. If we introduce a filter in the Fourier plane such that the two terms ( f ) and ( f ) interact, (x) can be Consider a filter with transmission function where is a very small distance from the origin. F ( f ) T ( f ) Immediate after the filter the function is F ( f ) T ( f ) [ ( f ) i( f )]T ( f ) i ( f ) i( f ) [i ( f ) ( f )] [i ( f ) ( f 1 2 (x') Where (x') 1 and can be ignored. the original phase distribution becomes visible as an intensity distribution added to a bright background (positive phase contrast) Negative phase contrast Similarly if we introduce phase filter with transmission function then the intensity at the image plane is the original phase distribution becomes visible as an intensity distribution subtracted from the bright
background (i.e. negative phase contrast)
Phase-contrast microscopy (F. Zernike: Zernike filter, Noble price in1953 for physics)
Illumination through two points of an object, due to the difference in thickness, say, may result the same transmitted beam amplitude, through differ in their phase. The phase difference is not “visible”. This is in fact one of the major problems when viewing a transparent object e.g. examination of biological materials under an optical microscope. However if illumination of constant amplitude and phase is superimposed across the field, interference occurs and the resultant transmitted beam are different in amplitude as well as phase. The optical path differences have therefore been made “visible”. One of the simplest arrangements for doing this is as follow: Light is focused on the specimen S by a condenser lens C. A diaphragm D in the form of an annular slot is placed in the front focal plane of C, and its image is formed as a ring of light in the back focal plane of O at P by light that has not been diffracted by the specimen. The phase difference is introduced by means of a phase plate at P consisting of an optically parallel glass plate on which a thin layer of dielectric material has been deposited over the area of the ring, or everywhere except the ring, giving “negative” or “positive” phase contrast respectively. In microscopy: e.g. viewing living squamous cell where most of the materials are transparent. They differ in refractive under the thickness only. By using phase contrast filter they become “visible”. Optical micrographs of squamous cells from the (a) Phase contrast (b) Conventional bright-field illumination
Source: http://ap.polyu.edu.hk/apakhwon/Phase%20filters.pdf
### Microsoft word - advil 200mg.doc
Nome comercial: Advil® Nome genérico: ibuprofeno Uso oral Analgésico e Antitérmico Forma farmacêutica e apresentações referentes a esta bula: Comprimidos revestidos - USO ADULTO Advil®: caixas com 20 e 100 comprimidos. Outra forma farmacêutica e apresentação disponível no mercado: _____________________________________________________________________________
### bayheartgroup.com
INSTRUCTION SHEET FOR THE FOLLOWING DIAGNOSTIC TEST: DOBUTAMINE STRESS ECHOCARDIOGRAM ABOUT THE TEST: A dobutamine stress echocardiogram is a test that helps your doctor see how wel your heart pumps when it is made to work harder. Using a smal hand-held device (cal ed a transducer) which is moved over your chest, sound waves are bounced off your heart and the “echo” of those sound wa | 1,081 | 4,085 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-22 | latest | en | 0.864402 |
https://zerojudge.tw/ShowThread?postid=35490&reply=35483 | 1,723,369,045,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640983659.65/warc/CC-MAIN-20240811075334-20240811105334-00170.warc.gz | 837,621,071 | 15,455 | #35483: 簡單暴力python解
#### wmouo (crazypanda)
School : 國立宜蘭高級商業職業學校
ID : 178206
2024-05-22 14:19:03
k731. 1. 路徑偵測 -- | From: [118.165.145.156] | Post Date : 2023-06-04 21:04
r = [0,0]
z = [0,0,0]
now = "r"
for _ in range(int(input())):
a,b = map(int,input().split())
if a != r[0]:
if a > r[0]:
if now == 'l':
z[2] += 1
elif now == 'u':
z[0] += 1
elif now == 'o':
z[1] += 1
now = 'r'
if a < r[0] :
if now == 'r':
z[2] += 1
elif now == 'o':
z[0] += 1
elif now == 'u':
z[1] += 1
now = 'l'
elif b != r[1]:
if b > r[1]:
if now == 'u':
z[2] += 1
elif now == 'r':
z[0] += 1
elif now == 'l':
z[1] += 1
now = 'o'
if b < r[1] :
if now == 'o':
z[2] += 1
elif now == 'l':
z[0] += 1
elif now == 'r':
z[1] += 1
now = 'u'
r = [a,b]
print(z[0],z[1],z[2])
#35490: Re: 簡單暴力python解
#### s11104220@school.saihs.edu.tw (施同學)
School : 臺北市立松山高級工農職業學校
ID : 221254
2024-08-10 18:30:37
k731. 1. 路徑偵測 -- | From: [123.193.213.137] | Post Date : 2023-06-04 21:52
0右1下2左3上
from sys import stdin
os=[0,0]
d=0
ans=[0,0,0]
for _ in range(n):
if os[0]<s[0]:
if d==1:ans[0]+=1
elif d==2:ans[2]+=1
elif d==3:ans[1]+=1
d=0
elif os[1]>s[1]:
if d==0:ans[1]+=1
elif d==2:ans[0]+=1
elif d==3:ans[2]+=1
d=1
elif os[0]>s[0]:
if d==0:ans[2]+=1
elif d==1:ans[1]+=1
elif d==3:ans[0]+=1
d=2
elif os[1]<s[1]:
if d==0:ans[0]+=1
elif d==1:ans[2]+=1
elif d==2:ans[1]+=1
d=3
os=s
print(*ans)
#35492: Re: 簡單暴力python解
#### s011388@fysh.tc.edu.tw (pollux)
School : 國立豐原高級中學
ID : 189768
2024-07-07 16:16:55
k731. 1. 路徑偵測 -- | From: [111.252.213.228] | Post Date : 2023-06-04 21:55
r = [0,0]
z = [0,0,0]
now = "r"
for _ in range(int(input())):
a,b = map(int,input().split())
if a != r[0]:
if a > r[0]:
if now == 'l':
z[2] += 1
elif now == 'u':
z[0] += 1
elif now == 'o':
z[1] += 1
now = 'r'
if a < r[0] :
if now == 'r':
z[2] += 1
elif now == 'o':
z[0] += 1
elif now == 'u':
z[1] += 1
now = 'l'
elif b != r[1]:
if b > r[1]:
if now == 'u':
z[2] += 1
elif now == 'r':
z[0] += 1
elif now == 'l':
z[1] += 1
now = 'o'
if b < r[1] :
if now == 'o':
z[2] += 1
elif now == 'l':
z[0] += 1
elif now == 'r':
z[1] += 1
now = 'u'
r = [a,b]
print(z[0],z[1],z[2])
n = int(input())
lis = [list(map(int,input().split())) for i in range(n)]
lis.insert(0,[0,0])
def dirc(lt):
if lt[0] > 0:return 1
if lt[0] < 0:return 3
if lt[1] > 0:return 2
if lt[1] < 0:return 4
def way(a,b):
if a == 1:
if b == 1:
return 0
if b == 2:
return 1
if b == 3:
return 3
if b ==4:
return 2
if a == 2:
if b == 1:
return 2
if b == 2:
return 0
if b == 3:
return 1
if b ==4:
return 3
if a == 3:
if b == 1:
return 3
if b == 2:
return 2
if b == 3:
return 0
if b ==4:
return 1
if a == 4:
if b == 1:
return 1
if b == 2:
return 3
if b == 3:
return 2
if b ==4:
return 0
ans = [0,0,0]
for i in range(1,len(lis)-1):
v1 = [lis[i][0]-lis[i-1][0],lis[i][1]-lis[i-1][1]]
v2 = [lis[i+1][0]-lis[i][0],lis[i+1][1]-lis[i][1]]
if way(dirc(v1),dirc(v2)):
n = way(dirc(v1),dirc(v2))-1
ans[n] += 1
print(" ".join([str(x) for x in ans]))
#35502: Re: 簡單暴力python解
#### s11104220@school.saihs.edu.tw (施同學)
School : 臺北市立松山高級工農職業學校
ID : 221254 | 1,558 | 3,027 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-33 | latest | en | 0.250034 |
http://www.dimensionsinfo.com/what-is-the-size-of-a-measuring-spoon/ | 1,571,068,240,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986653876.31/warc/CC-MAIN-20191014150930-20191014174430-00373.warc.gz | 249,299,227 | 11,307 | • # What is the Size of a Measuring Spoon?
Being aware of the different measuring spoon sizes is necessary so you can put in the right ingredients. The following are the common sizes used and their equivalents in cups.
### Standard Sizes Used
48 teaspoons is equal to 1 cup, while 3 teaspoons is equal to 1 tablespoon. 1 tablespoon is the same as 1/16 cup and 2 tablespoons + 2 teaspoons is the same as 1/6 cup. 2 tablespoons is the same as 1/8 cup.
4 tablespoons is equal to 1/4 cup and 5 tablespoons + 1 teaspoon is the same as 1/3 cup. 6 tablespoons is equal to 3/8 cup and 8 tablespoons is the same as 1/2 cup.
The measuring spoon size 10 tablespoons + 2 teaspoons is equal to 2/3 cup. 12 tablespoons is the same as 3/4 cup and 16 tablespoons = 1 cup.
### Metric to US Capacity Equivalents
The following liquid equivalents are used in cooking too. 5 ml = 1 teaspoon and 15 ml is equal to 1 tablespoon and 34 ml is the same as 1 fluid oz. 100 ml is equal to 3.4 fluid oz. and the 240 ml size equal to 1 cup. 1 liter is the same as 34 fluid oz.
1 tablespoon is equal to 1/2 liquid ounce while 4 tablespoons is the same as 2 liquid ounces. 5 tablespoon + 1 tsp is the same as 3 liquid ounces. 8 tablespoons is the same as 4 liquid ounces.
### Measuring Spoon Sizes Tips and Warnings
When cooking and (especially) when baking, you need to use measuring spoons. These are more accurate than ordinary spoons. For cooking meat, chicken and casseroles, adding or reducing the amounts is acceptable. If the recipe requires a 1/3 tsp, you can add more or reduce it. For baking, the exact amount must be used.
In some recipes, measurements like drop or dash is used. Their exact equivalents are the following. Drop s the same as 1/64 teaspoon while smidgen is equal to 1/32 teaspoon. The pinch is the same as 1/16 teaspoon and the dash is 1/8 teaspoon. A tad is equal to 1/4 teaspoon.
### Suggestions for Using the Spoons
Whatever the measuring spoon size you use for sugar, the following rules apply. Scoop the sugar until it overflows. Flatten the top off with a knife. When using dry ingredients, you must use spoons. Fill the spoons with the ingredients. When the ingredient overfills, level it with a knife.
There are other basic considerations to keep in mind. For example, learn the difference between dry and liquid measures. In many instances, measurement of dry ingredients requires the same process.
Use the right spoon size and let it overfill. Then you can flatten it with a spatula and get the right amount. With baking powder cans, they usually have their own edge included. The lid edge can also be used for leveling.
The measuring spoon sizes are not that difficult to remember once you start cooking. Most recipe books also come with guidelines, so you should have no problems using them. | 689 | 2,815 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-43 | latest | en | 0.934416 |
http://www.cheenta.com/2015/10/16/equivalence-class-i-s-i-tomato-subjective-60/ | 1,505,872,451,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818686117.24/warc/CC-MAIN-20170920014637-20170920034637-00694.warc.gz | 413,306,835 | 23,613 | # Equivalence Class (I.S.I. Tomato subjective 60)
Problem: Consider the set S of all integers between and including 1000 and 99999. Call two integers x and y in S to be in the same equivalence class if the digits appearing in x and y are the same. For example, if x = 1010, y = 1000 and z = 1201, then x and y are in the same equivalence class, but y and z are not. Find the number of distinct equivalence classes that can be formed out of S.
Solution: Any set of distinct digits with maximum order 5 is a equivalence class that can be formed out of S except {0}.
Number of such sets is
= $${\dbinom {10}{1}}$$ + $${\dbinom {10}{2}}$$ + $${\dbinom {10}{3}}$$ + $${\dbinom {10}{4}}$$ + $${\dbinom {10}{5}}$$ -1
= 10 + 45 + 120 + 210 + 252 -1
= 636 | 239 | 747 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2017-39 | latest | en | 0.864525 |
https://puzzling.stackexchange.com/questions/9515/an-old-picture-puzzle/9616 | 1,566,658,111,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027321140.82/warc/CC-MAIN-20190824130424-20190824152424-00551.warc.gz | 599,293,601 | 39,552 | # An old picture puzzle
Here is a drawing of an old picture. Your task is to answer all 8 questions correctly AND give a good reasoning for it
1. What time of day is it?
2. Is it early spring or a late fall(autumn)?
3. Is this river navigable?
4. Which direction does the river flow? (North, East, South or West?)
5. Is the river deep or shallow at the side where the boat is?
6. Is there a bridge across the river somewhere nearby?
7. How far is the railroad from here?
8. Do birds fly North or South?
• Any source for this? – Joe Lee-Moyet Feb 24 '15 at 11:00
• @yjo It appears its from some Russian book. (A reverse image search on google turns up a lot of results. It certainly should be cited by the author) – Milo Brandt Feb 25 '15 at 3:50
• FYI: I've moved the answer out of your question into a community wiki answer. – Aza Feb 26 '15 at 9:47
1.What time of day is it?
Late afternoon, people are still working, yet the person with the suitcase is wearing a lamp, so nightfall is coming soon. Took a double take, but didn't notice the shadows first, which suggest morning because they are walking eastward.
2.Is it early spring or a late fall(autumn)?
Early spring, people aren't harvesting, they are sowing.
3.Is this river navigable?
Yes, else there wouldn't be a reason to place a buoy (atleast I think it's a buoy, could be a tide-gauge).
4.Which direction does the river flow? (North, East, South or West?)
It flows South, you can determine the direction with the "buoy" and with the fact that the bids fly north.
5.Is the river deep or shallow at the side where the boat is?
Deep, the fishing line is too long for shallow water.
6.Is there a bridge across the river somewhere nearby?
Probably not, crossing a bridge is usually free, using a ferry isn't.
7.How far is the railroad from here?
Probably on walking distance from the boat, the people seem to be travelling to the countryside so it seems likely they just came from a train station.
8.Do birds fly North or South?
North, because it's spring.
• What if this is in the southern hemisphere? – Runemoro Feb 24 '15 at 13:02
• @Runemoro European culture, european climate (by clothing and bulding style) = Probably northern hemisphere. – Taemyr Feb 24 '15 at 14:05
• @Taemyr, that and simple statistics. Only 1 in 7 humans lives on the southern hemisphere. And most of those people are Africans and South-Americans (i.e. colored people). – Jordy Feb 24 '15 at 14:17
• you could also add that the way the boatman is navigating his boat, it would not make much sense if the river's current was in the other direction. – Spacemonkey Jan 27 '16 at 20:11
• Winter wheat is sowed in the fall. – Ian MacDonald Jan 28 '16 at 23:10
Morning (shadows fall on the West)
Early Spring (harrow on the field)
Yes (Shipping buoy)
South (determined by the passage around the displacer water.)
Deep (Fishing line)
No bridge (the man in the boat - the ferryman)
Close (a man with a lantern rail)
North (because spring)
This answer has been copied and edited out of the question, and is not produced by me. (It is thus Community Wiki, which does not earn reputation.)
If you look at the picture, you will see that people are sowing and not harvesting. Sowing can be done either in the fall(autumn) or early spring. Fall sowing happens when trees still have leaves, but on the picture the trees are naked, so it's early spring.
Because it's early spring, birds(cranes) fly from South to the North. Knowing that birds fly to the North and looking at the buoy we can say that this river is flowing South.
Buoys are only placed on the navigable rivers to show danger areas and let ships navigate easier/safer.
Shadow of the tree shows that the Sun is somewhere South-East. During the spring, that position of the Sun happens normally around 8:00-10:00.
The guy with the lantern is the railroad conductor, so he probably lives(or works) somewhere nearby.
The pier and the ferry show that there is no bridge nearby, otherwise people would just walk on the bridge. Similarly, the fishing boy has set his float quite far from the hooks, so we can assume that river there is fairly deep. If it would be shallow, the boy would put his float much closer to the hooks, so that hooks are not on the ground in the water.
• As an aside, I'm not totally sure what the best way to format it is, so if someone else wants to take a stab at it, please do. – Aza Feb 26 '15 at 9:46
It's hard to tell in a lot of cases. I notice there are other answers, and though some of my answers match theirs, I disagree on some of them. Here are the things that stand out to me:
1. Time of day?
Morning. The shadow length suggests either early or late hours, and the fact that people are working/fishing suggests early.
2. Spring or fall?
Spring. This one is clear, because the farmer is sowing his field.
3. Is the river navigable?
Yes. Besides there being a boat in the river, I think that's a floating buoy in the water on the left; if that is what it is, then it would suggest shipping boats travel down the river.
4. River flow direction?
East. The buoy-like object on the left leaves ripples in the water's current, which shows that the water is flowing left-to-right in the picture. If I'm right about it being morning, then the water is flowing East (towards the sun - note that the shadows are parallel to the water).
5. Is the water deep or shallow under the boat?
Deep. The presence of a dock and a fisherman are good indicators that the water is fairly deep.
6. Is there a bridge nearby?
No. There's a ferry, which would probably not be in business if people could just cross the bridge.
7. How far away is the railroad?
Nearby. The person on the left looks like a traditional railroad worker. His presence is the only really fair amount of evidence.
8. Are the birds flying North or South?
South. The water flow direction and sun's position help with figuring this out, noting the birds fly in a V-shape to the left (Southwest) and roughly toward the nearer shoreline.
• Love your answer! Even though it doesn't match the answers I found on the internet where I saw this picture, your reasoning is matching really well! +1 – Novarg Feb 24 '15 at 8:23
• I also thought the shadows were parallel to the river, but I think this is an issue with the perspective. If you look at the railroad-man, he is moving straight to the ferry man and the shadow is right behind him (so a right angle to the river). But the tree-shadow looks parallel, while the shadows of the two on the peer look like a 45° angle to the river... So maybe it is curving and flowing roughly south-east – Falco Feb 24 '15 at 9:33
• I questioned this too. I'm interested to find out what the original artist's intended solution is. – Bulldogg6404 Feb 24 '15 at 10:29
• Your answers don't make sense with each other. If it is morning, then shadows would be going west and birds would be flying north. Not to mention, birds wouldn't be flying south in the Spring. – Kevin Feb 24 '15 at 18:34
• @Kevin I believe I explained my answers well enough to at least make sense. I agree that the shadows are pointed West. I am claiming that the birds appear to not only be flying in roughly the same direction as the shadows, but also towards the shoreline that the POV is currently standing at. That doesn't mean I'm saying the birds are flying towards us, the viewer; it means I'm saying they're flying toward one side of the river more than the other, which would be more like South than North. – Bulldogg6404 Feb 25 '15 at 4:15
General remarks:
The location must be somewhere temperate as we see a tree bare of leaves, so let's assume it's somewhere like Yorkshire.
What time of day is it?
It must be about noon. Yorkshire is about 54 degrees North, so the maximum height of the sun at noon near the equinox is about 36 degrees, and the shadows seem to be less than the height of the objects casting them, so the sun must be very high in the sky for early Spring or late Autumn.
Is it early spring or a late fall(autumn)?
Spring, because the farming activity looks like preparation of the ground rather than harvesting. However, whatever is being done seems to leave no mark, since the farmer and tractor are in the middle of the field, yet neither left nor right is there a visible difference showing the part already worked on.
Is this river navigable?
Obviously. There's a boat in it. But navigable is a relative term. It's very unlikely to be navigable for an aircraft carrier.
Which direction does the river flow? (North, East, South or West?)
Since it's close to noon, the sun must be to the South, and the river is flowing left to right, so it must flow South.
Is the river deep or shallow at the side where the boat is?
Compared to what? It's probably about waist deep. Where rivers are straight they tend to slope down from the bank. The presence of the jetty proves that it's not deep enough to get the boat right up to the side. It looks like there's a float on the boy's fishing line, so within the reach of his ability to cast, the water must get deep enough that the hook is still above the bottom - probably over 3m. Note that the total length of a fishing line on a river is irrelevant, as it is often used to cast futher away from the bank rather than to get deeper.
Is there a bridge across the river somewhere nearby?
It's impossible to tell. The people using the boat suggest that they cannot use a bridge, but that may be be because they're going to an island in the river whereas the bridge is further downstream and only goes to the other side. Also, there could be a railway bridge which is not open to foot traffic.
How far is the railroad from here?
It's very difficult to tell. The man on the left might be a railway worker, but even if he is, he's carrying quite a large case in his left hand, so he could be travelling some considerable distance from it. Either that or he has an extraordinarily large lunchbox.
Do birds fly North or South?
Some do, some don't. Some overwinter in England, having come from further North, some spend the summer in England and overwinter in Africa. The ones in the picture are flying more-or-less West.
While I agree with the others on most answers I have a different oppinion on Number 5:
I think it is shallow. Why? Because it looks like the Boy is standing in the river. Also there is a buoy near the first third of the river which maybe indicates where the river gets deep. To the argument with the long fishing line: Is is quite common to use a long line while fishing in a river, so that the line can flow in the water.
• Yes the buoy is a good argument, signing ships where it is still deep enough to travel and where it gets to shallow... – Falco Feb 24 '15 at 9:29
• I don't think the boy is in the river. Because I don't think his bucket is in the river. – Taemyr Feb 24 '15 at 14:08
• @Taemyr placing a floating object in a river may not be the best idea. Therefore the boy may have left his bucket outside of the river – Tokk Feb 25 '15 at 13:54
• @Tokk What I mean is that the bucket is as far down compared to the crest of the bank as the boy. So if the bucket is not in the river there is no reason to assume that the boy is in the river. – Taemyr Feb 25 '15 at 14:09
• @Taemyr the bucket almost reaches the boy's butt. This would imply that the bucket is standing superior than the boy. Except it is a really high bucket – Tokk Feb 25 '15 at 14:12
First of all I want to point out that IMO the shadows and the river make something around a 45 degrees angle, just for your consideration.
Q0: hemisphere?
Since all people look caucasian and to be wearing late 19th/early 20th Western attires, I'm saying it is north hemisphere.
Q2 + Q8: Is it Spring or Autumn? + Do birds fly N or S?
True and True.
Wait. Not boolean questions? Kidding.
Birds look to be flying in about the same direction range (really hard to tell) shadows are cast to. This happens in Spring in both hemispheres and while they are flying North in North hemisphere.
* Taking a look to the other answers I see references to landfarmers sowing, not harvesting, which is even better reasoning for Spring, but wasn't my first thought I'll admit.
Q1: time of day?
It's really hard to make a perfect appreciation, but IMO shadows are cast towards NW or so.
So I'll say morning but not too soon, since shadows are already well defined.
Q3: navigable river?
With the right craft, any river is navigable.
Aside from that, the boat and the dock strongly suggest the river is navigated frequently.
Q4: Which direction does the river flow?
From North to South. You can tell by the trails on the water plus achieved knowledge from previous Qs.
Q5: Is the river deep or shallow at the side where the boat is?
I think it's deeper at this side. The riverside ends more vertically here and looks smoother on the other side (also hard to tell).
Q6: Are there bridges nearby?
At least the people in the forefront doesn't look to think so. They will probably rather cross the bridge since it's faster. If the boat has already departed when you reach the river you'll have to wait for it to deliver the passengers, return, and then carry you to the other side. One lone old guy rowing won't get a very high peak speed I bet.
* Also, other answers suggest the ferry costs money, which adds up for the bridge to be the first option if available.
Q7: How far is the railroad from here?
Far enough to not be in the drawing.
Close enough to be reached by walking by a railroad worker with no more equipment or luggage than a lantern and a rather thin suitcase.
That's the better answer I can give.
Going to answer in the order I figured them out instead of order presented:
1. Is it early spring or a late fall(autumn)?
Spring. They are sowing not planting
1. Do birds fly North or South?
North. Because it is Spring
1. What time of day is it
Morning. Birds are flying North and shadows are going West, so sun must be in east.
1. Which direction does the river flow? (North, East, South or West?)
South, based on ripples around buoy.
1. Is this river navigable?
Yes, because there is a buoy
1. Is the river deep or shallow at the side where the boat is?
Shallow, the buoy marks the shallow/deep line for larger boats.
1. Is there a bridge across the river somewhere nearby?
No, otherwise people would be using it.
1. How far is the railroad from here?
Close. Guy carrying lantern is walking to/from work.
The scenery suggests that it is not close to the equator, because of the tree with no leaves which suggests sufficient temperature difference between seasons. This means that the sun will not rise extremely high into the sky and thus will always cast some shadows. The shadows in the picture are relatively short, thus it would have to be close to 12:00.
The tree has lost its leaves, however there are no visible leaves on the ground. This suggests that sufficient time has past since the tree lost its leaves, such that the leaves might been blown away by the wind or washed away by rain. Thus it would have to be early spring. The farmers seem to be plowing the fields, which might also give some clues, however I do not know enough about farming, but after a few searches it seems that fields can be plowed in both spring and autumn.
The buoy in the left of the picture most likely would function as a lateral marker, which marks the edge of a save water channel, which suggests that boats will navigate the river regularly for transport.
The buoy causes waves, which travel to the right, so the river is flowing to the right. Also the small boat seems to be docking, which is easier when the front of the boat is turned into the flow of the river. Given that it is roughly midday, then the shadows would either point north or south, depending on the hemisphere this picture was taken. The people in the picture seem European, so it would be the northern hemisphere. Thus the river is flowing towards the west.
The buoy is relatively close to the side of the river, so the river would have to be deep enough for large ships. Depending on your definition of deep and shallow, I would say the river near the boat would be deep.
The small boat is most likely not meant for transport along the river, but for transport across the river as a ferry. This suggests that there are no nearby other means of getting across, thus there is no bridge nearby.
I do not know the answer to this question. | 3,882 | 16,519 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2019-35 | latest | en | 0.95561 |
https://enigmaticcode.wordpress.com/2018/02/28/puzzle-53-addition/ | 1,585,475,291,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370494064.21/warc/CC-MAIN-20200329074745-20200329104745-00206.warc.gz | 418,612,866 | 23,491 | # Enigmatic Code
Programming Enigma Puzzles
From New Scientist #1104, 25th May 1978 [link]
In the following addition sum the digits have been replaced by letters. The same letter stands for the same digit wherever it appears and different letters stand for different digits.
Find the digits for which the letters stand.
[puzzle53]
### 2 responses to “Puzzle 53: Addition”
1. Jim Randell 28 February 2018 at 8:53 am
We can use the [[ `SubstitutedSum()` ]] solver from the enigma.py library to solve this puzzle without needing to write a program.
Run: [ @repl.it ]
```% python -m enigma SubstitutedSum "DHGDNGNC + CHCDNDGD = EPBNBBGE"
(DHGDNGNC + CHCDNDGD = EPBNBBGE)
(53650604 + 43450565 = 97101169) / B=1 C=4 D=5 E=9 G=6 H=3 N=0 P=7
```
Solution: The actual sum is:
2. geoffrounce 28 February 2018 at 7:14 pm
Faster than expected (108 msec on my laptop) to find the solution with a permutation solution .
```import time
start = time.time()
from itertools import permutations
for q in permutations('1234567890',8):
b, c, d, e, g, h, n, p = q
if d == '0' or c == '0' or e == '0': continue
DHGDNGNC = int(d + h + g + d + n + g + n + c)
CHCDNDGD = int(c + h + c + d + n + d + g + d)
EPBNBBGE = int(e + p + b + n + b + b + g + e)
if DHGDNGNC + CHCDNDGD == EPBNBBGE:
print(DHGDNGNC, " + ", CHCDNDGD, " = ", EPBNBBGE)
print(time.time() - start) # 0.108 sec
```
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 483 | 1,457 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2020-16 | latest | en | 0.720902 |
http://aolanswers.com/questions/what_does_100000000_bill_look_like_p149636712835754 | 1,432,576,084,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928562.33/warc/CC-MAIN-20150521113208-00307-ip-10-180-206-219.ec2.internal.warc.gz | 8,618,697 | 18,515 | # what does \$1000,000.00 bill look like?
Explore the latest questions and answers related to "what does \$1000,000.00 bill look like?"
### Answered:How write and letters 500,000.00 dollars
\$500,000.00 is five hundred thousand dollars.
### Answered:What is one percent of \$50,000.00?
\$500
You don't get it.
### Answered:If you say on my bill the price has not risen, why is it up 10.00 from
This is not the billing department. Please direct your question to the billing department.
### Answered:What is 0.7% of \$1000.00
1,000.00 * 0.7/100 = 7.00 \$ or: 1,000.00 * 0.007 = 7.00 \$ or: 1,000.00 * 7/1,000 = 7.00 \$
I can give you this answer for free. As soon as you send me \$59.99 for S/H, I will have the answer in the mail to you.
## Be The First To Answer
Other people asked questions on various topics, and are still waiting for answer. Would be great if you can take a sec and answer them
### Hello, I am looking for someone whose name is ...
http://www.eircomphonebook.ie/q/name/detail/IE_11414C104606060606062606C6A6C646672_1/
### What is the answer to 150,000,000 divided by 11.210
Just in case you want a few more decimal places- 13380909.901873327386262265834077
### What is \$138,000.00 times 1 % =?
\$138,000.00 times 1% (1/100) is \$1,380.00
### I am supposed to be getting a \$1000 wallmart gift ...
You don't get it.
More | 428 | 1,362 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2015-22 | latest | en | 0.916274 |
https://www.coursehero.com/file/6444411/Chapter-20/ | 1,516,419,484,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888878.44/warc/CC-MAIN-20180120023744-20180120043744-00325.warc.gz | 909,250,199 | 25,429 | Chapter 20 - Chapter 20 ELECTRIC CIRCUITS 1 Electromotive...
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Unformatted text preview: Chapter 20: ELECTRIC CIRCUITS 1. Electromotive Force and Current 2. Ohm’s Law 3. Electromotive Force and Circuits 4. Energy and Power in Electric Circuits 5. Resistors in Series and in Parallel 6. Kirchhoff’s Rules 7. Electric Measuring Instruments 8. Resistance-Capacitance Circuits 9. Physiological Effects of Currents 10.Power Distribution Systems COLLEGE PHYSICS, Part II ELECTROMOTIVE FORCE AND CURRENT Electric current can be defined simply as any movement of electric charge. A more accurate definition of electric current is the following: Although current is directly related with motion, or displacement, it is a scalar quantity. Still current has direction. By convention, the direction of current is the direction of the motion of positive charges , or the opposite direction of the motion of negative charges. In metallic conductors the real carriers of electric current are the free electrons. In the absence of electric field (potential difference) they are involved in high-speed random motion , and in the presence of electric field they undergo directed motion , or drift . In order to keep them moving, we need to keep a constant electromotive force , or a constant potential difference (voltage). When this force equilibrates with the frictional force , the charge carriers move (drift) with a constant speed. In a closed circuit the current is the same at any cross section; charges normally do not build up anywhere. Example: The current in a circuit is 5.0 mA. How many electrons flow across the cross section of the circuit per second? t Q I ∆ ∆ = C s s C t I Q 3 3 10 . 5 1 10 . 5-- × = × × = ∆ = ∆ The number of electrons would be: 16 19 3 10 12 . 3 10 602 . 1 10 . 5 × = × × = ∆ =-- C e Q n RESISTANCE AND OHM’S LAW The higher the potential difference ( V ) between the two ends of a conductor, the higher the field ( E ), and therefore the force ( F ) acting on the mobile charges in the conductor. Eventually, the current, I , is directly proportional to the applied voltage, V . This low of proportionality between V and I is known as Ohm’s law. The ratio V / I is called the resistance of a conductor. Resistivity The resistance of a conductor is directly proportional to its length and inversely proportional to its cross-sectional area. The resistivity is the property of a given material and does not depend on its shape or dimensions. You can compare electric current with water pumped through a hose. Temperature Dependence of Resistance and Superconductivity For metals, the resistivity increases with temperature. At high temperatures, this dependence is linear: ) ( 1 T T T- + = α ρ ρ where ρ and ρ are resistivities at temperature T and at a reference temperature, such as 0 o C, and α is a constant called temperature coefficient of resistivity . At lower temperatures, the temperature dependence of resistivity has a more complex, nonlinear character....
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Chapter 20 - Chapter 20 ELECTRIC CIRCUITS 1 Electromotive...
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Ask a homework question - tutors are online | 873 | 3,798 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2018-05 | latest | en | 0.883902 |
https://blog.csdn.net/u014258433/article/details/51545587 | 1,527,310,228,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867309.73/warc/CC-MAIN-20180526033945-20180526053945-00520.warc.gz | 519,957,792 | 13,703 | ###### Codeforces Round #353 (Div. 2) E. Trains and Statistic
Vasya commutes by train every day. There are n train stations in the city, and at the i-th station it's possible to buy only tickets to stations from i + 1 to ai inclusive. No tickets are sold at the last station.
Let ρi, j be the minimum number of tickets one needs to buy in order to get from stations i to station j. As Vasya is fond of different useless statistic he asks you to compute the sum of all values ρi, j among all pairs 1 ≤ i < j ≤ n.
Input
The first line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of stations.
The second line contains n - 1 integer ai (i + 1 ≤ ai ≤ n), the i-th of them means that at the i-th station one may buy tickets to each station from i + 1 to ai inclusive.
Output
Print the sum of ρi, j among all pairs of 1 ≤ i < j ≤ n.
Examples
Input
4
4 4 4
Output
6
Input
5
2 3 5 5
Output
17
Note
In the first sample it's possible to get from any station to any other (with greater index) using only one ticket. The total number of pairs is 6, so the answer is also 6.
Consider the second sample:
• ρ1, 2 = 1
• ρ1, 3 = 2
• ρ1, 4 = 3
• ρ1, 5 = 3
• ρ2, 3 = 1
• ρ2, 4 = 2
• ρ2, 5 = 2
• ρ3, 4 = 1
• ρ3, 5 = 1
• ρ4, 5 = 1
Thus the answer equals 1 + 2 + 3 + 3 + 1 + 2 + 2 + 1 + 1 + 1 = 17.
#include <cmath>
#include <cstdio>
#include <iostream>
#include <algorithm>
#define got(x) (1<<x)
using namespace std;
long long ans,dp[100007];
int n,a[100007],Max[100007][28];
int gotmax(int x,int y)
{
int l = log2(y-x+1);
return a[Max[x][l]] < a[Max[y-got(l)+1][l]] ? Max[y-got(l)+1][l] : Max[x][l];
}
int main()
{
scanf("%d",&n);
for(int i = 1;i < n;i++)
{
scanf("%d",&a[i]);
Max[i][0] = i;
}
for(int i = 1;got(i) < n;i++)
for(int j = 1;j+got(i)-1 < n;j++)
Max[j][i] = a[Max[j][i-1]] < a[Max[j+got(i-1)][i-1]] ? Max[j+got(i-1)][i-1]:Max[j][i-1];
dp[n] = 0;
for(int i = n-1;i;i--)
{
if(a[i] == n)
{
dp[i] = 1ll*(n - i);
ans += dp[i];
continue;
}
int k = gotmax(i+1,a[i]);
dp[i] = dp[k] + 1ll*(n - i - a[i] + k);
ans += dp[i];
}
cout<<ans<<endl;
}
#### Codeforces Round #353 (Div. 2)E. Trains and Statistic
2016-05-18 16:51:38
#### 【解题报告】Codeforces Round #367 (Div. 2)
2016-08-12 21:37:38
#### 解题报告：Codeforces Round #433 (Div. 2) E. Boredom ( 离线处理+树状数组）
2017-09-07 09:56:36
#### Codeforces Round #443 (Div. 2) E. Tournament
2017-11-01 16:14:48
#### 【CodeForces】CodeForces Round #464 (Div. 2) 题解
2018-02-18 11:05:43
#### Codeforces Round #456 (Div. 2): E. Prime Gift（折半枚举）
2018-01-06 15:03:53
#### Codeforces Round #392 (Div. 2)E. Broken Tree
2017-01-20 19:16:32
#### ICM Technex 2018 and Codeforces Round #463 E. Team Work
2018-02-21 11:49:58
#### CF-Educational Codeforces Round 44 (Rated for Div. 2) A~E
2018-05-22 05:02:25
#### Codeforces Round #452 (Div. 2) - E. Segments Removal（链表+优先队列）
2017-12-21 23:05:16
## 不良信息举报
Codeforces Round #353 (Div. 2) E. Trains and Statistic | 1,201 | 2,922 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2018-22 | latest | en | 0.633268 |
https://se.mathworks.com/matlabcentral/answers/1933060-find-kinetic-constants-from-differential-equations | 1,685,257,127,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643585.23/warc/CC-MAIN-20230528051321-20230528081321-00716.warc.gz | 562,775,336 | 28,326 | # Find kinetic constants from differential equations
3 views (last 30 days)
Marina Batlló RIus on 22 Mar 2023
Commented: Star Strider on 29 Mar 2023
Hey! So a escription on my problem:
I have a compartimental model that contains 4 different elements that have a kinetic behaviour between thm. The differential equations of this model can be described as:
dydt(1) = k(1)*y(4)+k(2)*y(2)+k(3)*y(1)-(k(4)*y(1)+k(5)*y(1)+k(6)*y(1)+1/25.028)*y(1);
dydt(2) = k(2)*y(1)-(k(5)*y(2)+1/25.028)*y(2);
dydt(3) = k(3)*y(1)-(k(4)*y(3)+1/25.028)*y(3);
dydt(4) = k(1)*y(1)-(k(4)*y(4)+1/25.028)*y(4);
I have some experimental data that I want to use to predict th different k, but I have been struggeling with solving them with the ode45 function or similars. Could someone help me?
Star Strider on 22 Mar 2023
Star Strider on 29 Mar 2023
As always, my pleasure! | 316 | 846 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-23 | latest | en | 0.795891 |
https://www.assignmentexpert.com/homework-answers/chemistry/general-chemistry/question-179934 | 1,653,500,157,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662588661.65/warc/CC-MAIN-20220525151311-20220525181311-00762.warc.gz | 724,217,033 | 67,797 | 106 263
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# Answer to Question #179934 in General Chemistry for Ally
Question #179934
In a lab, a student mixes 25.0 ml of 0.115 M aluminum sulfate with 34.0 ml of 0.0975 M Lead (III) nitrate. Calculate the grams of the precipitate that is formed.
1
2021-04-11T23:53:56-0400
Molar masses
Al2(SO4)3= 342g/mol
Pb(NO3)2=331g/mol
PbSO4=303g/mol
Let's find the masses of each reactant.
Mole= concentration x volume
Mole of Al2(SO4)2= 0.115x25/1000=0.002mol
Mass of Al2(SO4)2= mole x molar mass
= 0.003x342=0.98g
Also,
Mole of Pb(NO3)2= 0.0975x34/1000= 0.0033mol
Mass= 0.0033x331=1.01g
Now, let's find the limiting reagent that determines the mass of precipitate from the balanced equation.
Al2(SO4)3 + 3Pb(NO3)2 ---> 3PbSO4 + 2Al(NO3)3
342g of Al2(SO4)3 reacts with 3(331)g of Pb(NO3)2
0.98g of Al2(SO4)3 should react with 331x0.98/342 = 2.85g of Pb(NO3)2
But only 1.01g of Pb(NO3)2 is available, therefore, it is the limiting reagent.
Now,
331g of Pb(NO3)2 gives 303g of PbSO4
1.01g of Pb(NO3)2 will give 303/331 x 1.01 = 1.007g of PbSO4
Therefore, the mass of precipitate PbSO4 produced is 1.007g
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for any assignment or question with DETAILED EXPLANATIONS! | 503 | 1,317 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2022-21 | latest | en | 0.779954 |
https://www.kodytools.com/units/conductivity/from/stmhopft | 1,713,564,294,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817455.17/warc/CC-MAIN-20240419203449-20240419233449-00015.warc.gz | 763,257,658 | 17,323 | # Statmho/Foot Converter
## What Unit of Measure is Statmho/Foot?
Statmho per foot is a unit of measurement for electrical conductivity. It describes electrical conductivity of material which as electrical conductance of one statmho per length of one foot.
## What is the Symbol of Statmho/Foot?
The symbol of Statmho/Foot is stʊ/ft. This means you can also write one Statmho/Foot as 1 stʊ/ft.
Manually converting Statmho/Foot to any other Electrical Conductivity unit can be time-consuming, especially when you don’t have enough knowledge about Electrical Conductivity units conversion. Since there is a lot of complexity and some sort of learning curve is involved, most of the users end up using an online Statmho/Foot converter tool to get the job done as soon as possible.
We have so many online tools available to convert Statmho/Foot to other Electrical Conductivity units, but not every online tool gives an accurate result and that is why we have created this online Statmho/Foot converter tool. It is a very simple and easy-to-use tool. Most important thing is that it is beginner-friendly.
## How to Use Statmho/Foot Converter Tool
• As you can see, we have 2 input fields and 2 dropdowns. For instance, you want to convert Statmho/Foot to Statmho/Inch.
• From the first dropdown, select Statmho/Foot and in the first input field, enter a value.
• From the second dropdown, select Statmho/Inch.
• Instantly, the tool will convert the value from Statmho/Foot to Statmho/Inch and display the result in the second input field.
## Example of Statmho/Foot Converter Tool
Statmho/Foot
1
Statmho/Inch
0.083333333333333
# Statmho/Foot to Other Units Conversion Table
ConversionDescription
1 Statmho/Foot = 3.6494324146982e-12 Siemens/Meter1 Statmho/Foot in Siemens/Meter is equal to 3.6494324146982e-12
1 Statmho/Foot = 3.6494324146982e-9 Siemens/Kilometer1 Statmho/Foot in Siemens/Kilometer is equal to 3.6494324146982e-9
1 Statmho/Foot = 3.6494324146982e-14 Siemens/Centimeter1 Statmho/Foot in Siemens/Centimeter is equal to 3.6494324146982e-14
1 Statmho/Foot = 3.6494324146982e-15 Siemens/Millimeter1 Statmho/Foot in Siemens/Millimeter is equal to 3.6494324146982e-15
1 Statmho/Foot = 3.6494324146982e-18 Siemens/Micrometer1 Statmho/Foot in Siemens/Micrometer is equal to 3.6494324146982e-18
1 Statmho/Foot = 3.6494324146982e-21 Siemens/Nanometer1 Statmho/Foot in Siemens/Nanometer is equal to 3.6494324146982e-21
1 Statmho/Foot = 3.337041e-12 Siemens/Yard1 Statmho/Foot in Siemens/Yard is equal to 3.337041e-12
1 Statmho/Foot = 1.112347e-12 Siemens/Foot1 Statmho/Foot in Siemens/Foot is equal to 1.112347e-12
1 Statmho/Foot = 9.2695583333333e-14 Siemens/Inch1 Statmho/Foot in Siemens/Inch is equal to 9.2695583333333e-14
1 Statmho/Foot = 3.6494324146982e-18 Megasiemens/Meter1 Statmho/Foot in Megasiemens/Meter is equal to 3.6494324146982e-18
1 Statmho/Foot = 3.6494324146982e-15 Megasiemens/Kilometer1 Statmho/Foot in Megasiemens/Kilometer is equal to 3.6494324146982e-15
1 Statmho/Foot = 3.6494324146982e-20 Megasiemens/Centimeter1 Statmho/Foot in Megasiemens/Centimeter is equal to 3.6494324146982e-20
1 Statmho/Foot = 3.6494324146982e-21 Megasiemens/Millimeter1 Statmho/Foot in Megasiemens/Millimeter is equal to 3.6494324146982e-21
1 Statmho/Foot = 3.6494324146982e-24 Megasiemens/Micrometer1 Statmho/Foot in Megasiemens/Micrometer is equal to 3.6494324146982e-24
1 Statmho/Foot = 3.6494324146982e-27 Megasiemens/Nanometer1 Statmho/Foot in Megasiemens/Nanometer is equal to 3.6494324146982e-27
1 Statmho/Foot = 3.337041e-18 Megasiemens/Yard1 Statmho/Foot in Megasiemens/Yard is equal to 3.337041e-18
1 Statmho/Foot = 1.112347e-18 Megasiemens/Foot1 Statmho/Foot in Megasiemens/Foot is equal to 1.112347e-18
1 Statmho/Foot = 9.2695583333333e-20 Megasiemens/Inch1 Statmho/Foot in Megasiemens/Inch is equal to 9.2695583333333e-20
1 Statmho/Foot = 3.6494324146982e-15 Kilosiemens/Meter1 Statmho/Foot in Kilosiemens/Meter is equal to 3.6494324146982e-15
1 Statmho/Foot = 3.6494324146982e-12 Kilosiemens/Kilometer1 Statmho/Foot in Kilosiemens/Kilometer is equal to 3.6494324146982e-12
1 Statmho/Foot = 3.6494324146982e-17 Kilosiemens/Centimeter1 Statmho/Foot in Kilosiemens/Centimeter is equal to 3.6494324146982e-17
1 Statmho/Foot = 3.6494324146982e-18 Kilosiemens/Millimeter1 Statmho/Foot in Kilosiemens/Millimeter is equal to 3.6494324146982e-18
1 Statmho/Foot = 3.6494324146982e-21 Kilosiemens/Micrometer1 Statmho/Foot in Kilosiemens/Micrometer is equal to 3.6494324146982e-21
1 Statmho/Foot = 3.6494324146982e-24 Kilosiemens/Nanometer1 Statmho/Foot in Kilosiemens/Nanometer is equal to 3.6494324146982e-24
1 Statmho/Foot = 3.337041e-15 Kilosiemens/Yard1 Statmho/Foot in Kilosiemens/Yard is equal to 3.337041e-15
1 Statmho/Foot = 1.112347e-15 Kilosiemens/Foot1 Statmho/Foot in Kilosiemens/Foot is equal to 1.112347e-15
1 Statmho/Foot = 9.2695583333333e-17 Kilosiemens/Inch1 Statmho/Foot in Kilosiemens/Inch is equal to 9.2695583333333e-17
1 Statmho/Foot = 3.6494324146982e-9 Millisiemens/Meter1 Statmho/Foot in Millisiemens/Meter is equal to 3.6494324146982e-9
1 Statmho/Foot = 0.0000036494324146982 Millisiemens/Kilometer1 Statmho/Foot in Millisiemens/Kilometer is equal to 0.0000036494324146982
1 Statmho/Foot = 3.6494324146982e-11 Millisiemens/Centimeter1 Statmho/Foot in Millisiemens/Centimeter is equal to 3.6494324146982e-11
1 Statmho/Foot = 3.6494324146982e-12 Millisiemens/Millimeter1 Statmho/Foot in Millisiemens/Millimeter is equal to 3.6494324146982e-12
1 Statmho/Foot = 3.6494324146982e-15 Millisiemens/Micrometer1 Statmho/Foot in Millisiemens/Micrometer is equal to 3.6494324146982e-15
1 Statmho/Foot = 3.6494324146982e-18 Millisiemens/Nanometer1 Statmho/Foot in Millisiemens/Nanometer is equal to 3.6494324146982e-18
1 Statmho/Foot = 3.337041e-9 Millisiemens/Yard1 Statmho/Foot in Millisiemens/Yard is equal to 3.337041e-9
1 Statmho/Foot = 1.112347e-9 Millisiemens/Foot1 Statmho/Foot in Millisiemens/Foot is equal to 1.112347e-9
1 Statmho/Foot = 9.2695583333333e-11 Millisiemens/Inch1 Statmho/Foot in Millisiemens/Inch is equal to 9.2695583333333e-11
1 Statmho/Foot = 0.0000036494324146982 Microsiemens/Meter1 Statmho/Foot in Microsiemens/Meter is equal to 0.0000036494324146982
1 Statmho/Foot = 0.0036494324146982 Microsiemens/Kilometer1 Statmho/Foot in Microsiemens/Kilometer is equal to 0.0036494324146982
1 Statmho/Foot = 3.6494324146982e-8 Microsiemens/Centimeter1 Statmho/Foot in Microsiemens/Centimeter is equal to 3.6494324146982e-8
1 Statmho/Foot = 3.6494324146982e-9 Microsiemens/Millimeter1 Statmho/Foot in Microsiemens/Millimeter is equal to 3.6494324146982e-9
1 Statmho/Foot = 3.6494324146982e-12 Microsiemens/Micrometer1 Statmho/Foot in Microsiemens/Micrometer is equal to 3.6494324146982e-12
1 Statmho/Foot = 3.6494324146982e-15 Microsiemens/Nanometer1 Statmho/Foot in Microsiemens/Nanometer is equal to 3.6494324146982e-15
1 Statmho/Foot = 0.000003337041 Microsiemens/Yard1 Statmho/Foot in Microsiemens/Yard is equal to 0.000003337041
1 Statmho/Foot = 0.000001112347 Microsiemens/Foot1 Statmho/Foot in Microsiemens/Foot is equal to 0.000001112347
1 Statmho/Foot = 9.2695583333333e-8 Microsiemens/Inch1 Statmho/Foot in Microsiemens/Inch is equal to 9.2695583333333e-8
1 Statmho/Foot = 3.6494324146982e-12 Ampere/Volt/Meter1 Statmho/Foot in Ampere/Volt/Meter is equal to 3.6494324146982e-12
1 Statmho/Foot = 3.6494324146982e-9 Ampere/Volt/Kilometer1 Statmho/Foot in Ampere/Volt/Kilometer is equal to 3.6494324146982e-9
1 Statmho/Foot = 3.6494324146982e-14 Ampere/Volt/Centimeter1 Statmho/Foot in Ampere/Volt/Centimeter is equal to 3.6494324146982e-14
1 Statmho/Foot = 3.6494324146982e-15 Ampere/Volt/Millimeter1 Statmho/Foot in Ampere/Volt/Millimeter is equal to 3.6494324146982e-15
1 Statmho/Foot = 3.6494324146982e-18 Ampere/Volt/Micrometer1 Statmho/Foot in Ampere/Volt/Micrometer is equal to 3.6494324146982e-18
1 Statmho/Foot = 3.6494324146982e-21 Ampere/Volt/Nanometer1 Statmho/Foot in Ampere/Volt/Nanometer is equal to 3.6494324146982e-21
1 Statmho/Foot = 3.337041e-12 Ampere/Volt/Yard1 Statmho/Foot in Ampere/Volt/Yard is equal to 3.337041e-12
1 Statmho/Foot = 1.112347e-12 Ampere/Volt/Foot1 Statmho/Foot in Ampere/Volt/Foot is equal to 1.112347e-12
1 Statmho/Foot = 9.2695583333333e-14 Ampere/Volt/Inch1 Statmho/Foot in Ampere/Volt/Inch is equal to 9.2695583333333e-14
1 Statmho/Foot = 3.6494324146982e-12 Mho/Meter1 Statmho/Foot in Mho/Meter is equal to 3.6494324146982e-12
1 Statmho/Foot = 3.6494324146982e-9 Mho/Kilometer1 Statmho/Foot in Mho/Kilometer is equal to 3.6494324146982e-9
1 Statmho/Foot = 3.6494324146982e-14 Mho/Centimeter1 Statmho/Foot in Mho/Centimeter is equal to 3.6494324146982e-14
1 Statmho/Foot = 3.6494324146982e-15 Mho/Millimeter1 Statmho/Foot in Mho/Millimeter is equal to 3.6494324146982e-15
1 Statmho/Foot = 3.6494324146982e-18 Mho/Micrometer1 Statmho/Foot in Mho/Micrometer is equal to 3.6494324146982e-18
1 Statmho/Foot = 3.6494324146982e-21 Mho/Nanometer1 Statmho/Foot in Mho/Nanometer is equal to 3.6494324146982e-21
1 Statmho/Foot = 3.337041e-12 Mho/Yard1 Statmho/Foot in Mho/Yard is equal to 3.337041e-12
1 Statmho/Foot = 1.112347e-12 Mho/Foot1 Statmho/Foot in Mho/Foot is equal to 1.112347e-12
1 Statmho/Foot = 9.2695583333333e-14 Mho/Inch1 Statmho/Foot in Mho/Inch is equal to 9.2695583333333e-14
1 Statmho/Foot = 0.0000036494324146982 Gemmho/Meter1 Statmho/Foot in Gemmho/Meter is equal to 0.0000036494324146982
1 Statmho/Foot = 0.0036494324146982 Gemmho/Kilometer1 Statmho/Foot in Gemmho/Kilometer is equal to 0.0036494324146982
1 Statmho/Foot = 3.6494324146982e-8 Gemmho/Centimeter1 Statmho/Foot in Gemmho/Centimeter is equal to 3.6494324146982e-8
1 Statmho/Foot = 3.6494324146982e-9 Gemmho/Millimeter1 Statmho/Foot in Gemmho/Millimeter is equal to 3.6494324146982e-9
1 Statmho/Foot = 3.6494324146982e-12 Gemmho/Micrometer1 Statmho/Foot in Gemmho/Micrometer is equal to 3.6494324146982e-12
1 Statmho/Foot = 3.6494324146982e-15 Gemmho/Nanometer1 Statmho/Foot in Gemmho/Nanometer is equal to 3.6494324146982e-15
1 Statmho/Foot = 0.000003337041 Gemmho/Yard1 Statmho/Foot in Gemmho/Yard is equal to 0.000003337041
1 Statmho/Foot = 0.000001112347 Gemmho/Foot1 Statmho/Foot in Gemmho/Foot is equal to 0.000001112347
1 Statmho/Foot = 9.2695583333333e-8 Gemmho/Inch1 Statmho/Foot in Gemmho/Inch is equal to 9.2695583333333e-8
1 Statmho/Foot = 0.0000036494324146982 Micromho/Meter1 Statmho/Foot in Micromho/Meter is equal to 0.0000036494324146982
1 Statmho/Foot = 0.0036494324146982 Micromho/Kilometer1 Statmho/Foot in Micromho/Kilometer is equal to 0.0036494324146982
1 Statmho/Foot = 3.6494324146982e-8 Micromho/Centimeter1 Statmho/Foot in Micromho/Centimeter is equal to 3.6494324146982e-8
1 Statmho/Foot = 3.6494324146982e-9 Micromho/Millimeter1 Statmho/Foot in Micromho/Millimeter is equal to 3.6494324146982e-9
1 Statmho/Foot = 3.6494324146982e-12 Micromho/Micrometer1 Statmho/Foot in Micromho/Micrometer is equal to 3.6494324146982e-12
1 Statmho/Foot = 3.6494324146982e-15 Micromho/Nanometer1 Statmho/Foot in Micromho/Nanometer is equal to 3.6494324146982e-15
1 Statmho/Foot = 0.000003337041 Micromho/Yard1 Statmho/Foot in Micromho/Yard is equal to 0.000003337041
1 Statmho/Foot = 0.000001112347 Micromho/Foot1 Statmho/Foot in Micromho/Foot is equal to 0.000001112347
1 Statmho/Foot = 9.2695583333333e-8 Micromho/Inch1 Statmho/Foot in Micromho/Inch is equal to 9.2695583333333e-8
1 Statmho/Foot = 3.6494324146982e-21 Abmho/Meter1 Statmho/Foot in Abmho/Meter is equal to 3.6494324146982e-21
1 Statmho/Foot = 3.6494324146982e-18 Abmho/Kilometer1 Statmho/Foot in Abmho/Kilometer is equal to 3.6494324146982e-18
1 Statmho/Foot = 3.6494324146982e-23 Abmho/Centimeter1 Statmho/Foot in Abmho/Centimeter is equal to 3.6494324146982e-23
1 Statmho/Foot = 3.6494324146982e-24 Abmho/Millimeter1 Statmho/Foot in Abmho/Millimeter is equal to 3.6494324146982e-24
1 Statmho/Foot = 3.6494324146982e-27 Abmho/Micrometer1 Statmho/Foot in Abmho/Micrometer is equal to 3.6494324146982e-27
1 Statmho/Foot = 3.6494324146982e-30 Abmho/Nanometer1 Statmho/Foot in Abmho/Nanometer is equal to 3.6494324146982e-30
1 Statmho/Foot = 3.337041e-21 Abmho/Yard1 Statmho/Foot in Abmho/Yard is equal to 3.337041e-21
1 Statmho/Foot = 1.112347e-21 Abmho/Foot1 Statmho/Foot in Abmho/Foot is equal to 1.112347e-21
1 Statmho/Foot = 9.2695583333333e-23 Abmho/Inch1 Statmho/Foot in Abmho/Inch is equal to 9.2695583333333e-23
1 Statmho/Foot = 3.28 Statmho/Meter1 Statmho/Foot in Statmho/Meter is equal to 3.28
1 Statmho/Foot = 3280.84 Statmho/Kilometer1 Statmho/Foot in Statmho/Kilometer is equal to 3280.84
1 Statmho/Foot = 0.032808398950131 Statmho/Centimeter1 Statmho/Foot in Statmho/Centimeter is equal to 0.032808398950131
1 Statmho/Foot = 0.0032808398950131 Statmho/Millimeter1 Statmho/Foot in Statmho/Millimeter is equal to 0.0032808398950131
1 Statmho/Foot = 0.0000032808398950131 Statmho/Micrometer1 Statmho/Foot in Statmho/Micrometer is equal to 0.0000032808398950131
1 Statmho/Foot = 3.2808398950131e-9 Statmho/Nanometer1 Statmho/Foot in Statmho/Nanometer is equal to 3.2808398950131e-9
1 Statmho/Foot = 3 Statmho/Yard1 Statmho/Foot in Statmho/Yard is equal to 3
1 Statmho/Foot = 0.083333333333333 Statmho/Inch1 Statmho/Foot in Statmho/Inch is equal to 0.083333333333333 | 5,218 | 13,234 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-18 | latest | en | 0.817003 |
https://math.stackexchange.com/questions/3081266/check-if-the-problem-is-well-condtioned/3082923 | 1,576,454,033,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541310970.85/warc/CC-MAIN-20191215225643-20191216013643-00526.warc.gz | 450,658,862 | 30,751 | # Check if the problem is well condtioned
I'm trying to check if the problem of calculating the sum of two numbers a and b is well conditioned, provided that |a| > 2|b|. In my solution i split it into cases:
1: a>0 and b>0,then
f(a,b)=a+b>3b
C(a,b)=Ca(a,b)+Cb(a,b)
where Ca(a,b)=|a(1)/(a+b)| and Cb(a,b)=|a(1)/(a+b)| then
C=|(a+b)/(a+b)|=1 so in this case problem is well conditioned
2: a>0 and b<0,then
f(a,b)=a+b<b
C(a,b)=Ca(a,b)+Cb(a,b)
where Ca(a,b)=|a(1)/(a+b)| and Cb(a,b)=|a(1)/(a+b)| then
C=|(a+b)/(a+b)|=1 so in this case problem is well conditioned
so it turns out that for each case it will be the same, but i have doubts that this is correct solution
• What do you mean by "if the problem of calculating the sum of two numbers $a, b$ is well conditioned"? – jordan_glen Jan 20 at 23:31
• @jordan_glen The question is whether or not the sum is sensitive to small changes in the operands. This is a serious issue when we are dealing with approximations rather than the exact values of $a$ and $b$. – Carl Christian Jan 22 at 9:31
The condition number of adding two real numbers $$a$$ and $$b$$ is $$\kappa(a,b) = \frac{|a|+|b|}{|a+b|}.$$ It follows that the problem is ill-conditioned when $$a+b \approx 0$$. Now if $$|a| \ge 2|b|$$ or if $$|b| \ge 2|a|$$, then $$\kappa(a,b) \leq 3.$$ It follows that the problem is well conditioned when the absolute value of one number is as least twice as large as the absolute value of the other. The proof is an exercise in applying the triangle inequality. Now, suppose $$|a| \ge 2|b|$$. Then $$|a+b| = |a-(-b)| \ge ||a|-|b|| \ge |a|-|b| \ge |a| - \frac{1}{2}|a| = \frac{1}{2}|a|.$$ Moreover, $$|a| + |b| \leq |a| + \frac{1}{2}|a| = \frac{3}{2}|a|.$$ It follows that $$\kappa(a,b) \leq \frac{\frac{3}{2}|a|}{\frac{1}{2}|a|} = 3.$$ The case of $$|b| \ge |a|$$ follows immediately from the fact that $$\kappa(a,b) = \kappa(b,a).$$ Deriving this condition number rigorously from first principles is good subject for a follow-up question. One half of the proof is given in this answer which deals with the equivalent case of subtraction. Proving equality requires slightly more work. | 712 | 2,136 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2019-51 | latest | en | 0.850338 |
https://convertoctopus.com/223-ounces-to-pounds | 1,619,112,459,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039594341.91/warc/CC-MAIN-20210422160833-20210422190833-00175.warc.gz | 291,881,226 | 7,623 | ## Conversion formula
The conversion factor from ounces to pounds is 0.0625, which means that 1 ounce is equal to 0.0625 pounds:
1 oz = 0.0625 lb
To convert 223 ounces into pounds we have to multiply 223 by the conversion factor in order to get the mass amount from ounces to pounds. We can also form a simple proportion to calculate the result:
1 oz → 0.0625 lb
223 oz → M(lb)
Solve the above proportion to obtain the mass M in pounds:
M(lb) = 223 oz × 0.0625 lb
M(lb) = 13.9375 lb
The final result is:
223 oz → 13.9375 lb
We conclude that 223 ounces is equivalent to 13.9375 pounds:
223 ounces = 13.9375 pounds
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 pound is equal to 0.071748878923767 × 223 ounces.
Another way is saying that 223 ounces is equal to 1 ÷ 0.071748878923767 pounds.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that two hundred twenty-three ounces is approximately thirteen point nine three eight pounds:
223 oz ≅ 13.938 lb
An alternative is also that one pound is approximately zero point zero seven two times two hundred twenty-three ounces.
## Conversion table
### ounces to pounds chart
For quick reference purposes, below is the conversion table you can use to convert from ounces to pounds
ounces (oz) pounds (lb)
224 ounces 14 pounds
225 ounces 14.063 pounds
226 ounces 14.125 pounds
227 ounces 14.188 pounds
228 ounces 14.25 pounds
229 ounces 14.313 pounds
230 ounces 14.375 pounds
231 ounces 14.438 pounds
232 ounces 14.5 pounds
233 ounces 14.563 pounds | 434 | 1,656 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2021-17 | latest | en | 0.830973 |
https://whatisconvert.com/251-cubic-feet-in-cubic-meters | 1,600,497,745,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400190270.10/warc/CC-MAIN-20200919044311-20200919074311-00159.warc.gz | 705,297,791 | 7,369 | # What is 251 Cubic Feet in Cubic Meters?
## Convert 251 Cubic Feet to Cubic Meters
To calculate 251 Cubic Feet to the corresponding value in Cubic Meters, multiply the quantity in Cubic Feet by 0.0283168467117 (conversion factor). In this case we should multiply 251 Cubic Feet by 0.0283168467117 to get the equivalent result in Cubic Meters:
251 Cubic Feet x 0.0283168467117 = 7.1075285246367 Cubic Meters
251 Cubic Feet is equivalent to 7.1075285246367 Cubic Meters.
## How to convert from Cubic Feet to Cubic Meters
The conversion factor from Cubic Feet to Cubic Meters is 0.0283168467117. To find out how many Cubic Feet in Cubic Meters, multiply by the conversion factor or use the Volume converter above. Two hundred fifty-one Cubic Feet is equivalent to seven point one zero eight Cubic Meters.
## Definition of Cubic Foot
The cubic foot is a unit of volume, which is commonly used in the United States and the United Kingdom. It is defined as the volume of a cube with sides of one foot (0.3048 m) in length. Cubic feet = length x width x height. There is no universally agreed symbol but lots of abbreviations are used, such as ft³, foot³, feet/-3, etc. CCF is for 100 cubic feet.
## Definition of Cubic Meter
The cubic meter (also written "cubic metre", symbol: m3) is the SI derived unit of volume. It is defined as the volume of a cube with edges one meter in length. Another name, not widely used any more, is the kilolitre. It is sometimes abbreviated to cu m, m3, M3, m^3, m**3, CBM, cbm.
## Using the Cubic Feet to Cubic Meters converter you can get answers to questions like the following:
• How many Cubic Meters are in 251 Cubic Feet?
• 251 Cubic Feet is equal to how many Cubic Meters?
• How to convert 251 Cubic Feet to Cubic Meters?
• How many is 251 Cubic Feet in Cubic Meters?
• What is 251 Cubic Feet in Cubic Meters?
• How much is 251 Cubic Feet in Cubic Meters?
• How many m3 are in 251 ft3?
• 251 ft3 is equal to how many m3?
• How to convert 251 ft3 to m3?
• How many is 251 ft3 in m3?
• What is 251 ft3 in m3?
• How much is 251 ft3 in m3? | 599 | 2,081 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2020-40 | latest | en | 0.870357 |
https://codereview.stackexchange.com/questions/142559/sum-two-integers-unless-they-are-the-same-in-which-case-return-double-their-su/142571 | 1,718,268,482,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861342.74/warc/CC-MAIN-20240613060639-20240613090639-00454.warc.gz | 160,285,679 | 44,324 | # Sum two integers, unless they are the same, in which case return double their sum
I was doing a simple problem on codingbat and testing out my python. The problem was:
Given two int values, return their sum. Unless the two values are the same, then return double their sum.
sum_double(1, 2) → 3
sum_double(3, 2) → 5
sum_double(2, 2) → 8
My solution was:
def sum_double(a, b):
if a is b:
return 2 * (a + b)
else:
return a+b
and it worked just fine. However, when I checked the solution the answer they had was:
def sum_double(a, b):
# Store the sum in a local variable
sum = a + b
# Double it if a and b are the same
if a == b:
sum = sum * 2
return sum
But this seems to Java-y to me. Am I right to believe that my code is more pythonic? Or was their answer a better way to write code in python?
• "Too Java-y"? In Java I'd write: return a==b?a*4:a+b;... Commented Sep 27, 2016 at 10:28
• Yeah, I wouldn't call that solution "java-y". My job is practically all java and I would have written down your solution. I imagine that codingbat is just trying to separate the steps to make it easier to read or something? Not that it is actually easier to read unless you've got issues with basic algebra. Commented Sep 27, 2016 at 18:13
• Would return 4 * a be simpler than return 2 * (a + b)?
– TRiG
Commented Sep 28, 2016 at 15:02
• @TRiG probably yes since it can just do a << 2 but then again I wouldn't reason about it because since you branched already the interpreter may as well do the same thing for both cases.
– noob
Commented Sep 29, 2016 at 22:56
The existing answers mostly have just said is is wrong here. Which it undoubtedly is. The reason it's bad to use here is if you pass 257 and above, or -6 and below, you may get incorrect output. The best way to show how it's bad is to ask you what the output of the following are:
sum_double(2, 2)
sum_double(1 + 1, 2)
sum_double(257, 257)
sum_double(256 + 1, 257)
Simple! Surely it's 8, 8, 1028 and 1028, right? No, it's actually, 8, 8, 1028 and 514.
This is known and expected behaviour:
The current implementation keeps an array of integer objects for all integers between -5 and 256, when you create an int in that range you actually just get back a reference to the existing object.
Other than that both are fine, and I wouldn't really fret over which is 'better', this is as both are clear and understandable. Which is all your code needs to be.
A function that would be un-pythonic, as it's not clear could be:
def sum_double(a, b):
return (a + b) * ((a == b) + 1)
Well, if I'd have to choose:
• your way of comparing two integers is wrong (a is b). This is unusual and it should NOT become a habit. Just use a == b.
• their solution also use a variable (sum) which in my opining is useless. I'd just return the needed calculus.
I'd do it this way:
def sum_double(a, b):
return a + b if a != b else 2 * (a + b)
This is for me a pythonic way of returning the sum of two numbers using an inline return statement.
• Even more pythonic would be respecting PEP8 (spaces between operators), SCNR ;) Commented Sep 27, 2016 at 6:00
• @sphere I was just editing the answer. Thanks :) Commented Sep 27, 2016 at 6:01
• What do you mean by "you are calculating the sum twice"? Isn't that the same for your proposed solution? And is it bad? Commented Sep 27, 2016 at 6:14
• @200_success you're right. I just had something else in mind. Fixed Commented Sep 27, 2016 at 6:42
• If you want to avoid writing the sum again, just return 4*a if a == b. As an aside, I'm of the opinion that Python's x if y else z structure is hard to read and should be destroyed. Commented Sep 27, 2016 at 12:53
First, the is operator is for identity comparison, the == is for equality comparison. So the a == b part is more explicit (which is more pythonic).
Also, in your solution you do the sum twice, this is against DRY.
In the second code, it is bad that they name a variable after a built-in function sum. This should never be done (unless it is intended). In this scope the built-in function sum is overridden and can't be used.
One of the most important things is readability.
def sum_double(a, b):
if a == b:
return 2 * (a + b)
else:
return a+b
This is the best way to write your program in this context. Use of a variable is unnecessary, and although their solution gets rid of the redundant (a+b), their code is further from the intuition that comes from the specification.
The ultimate goal is to get that perfect mixture of brief, intuitive, clear, and efficient. It's a lot to juggle but luckily efficiency isn't very relevant here.
I saw some mention of inline statements. Don't get me wrong, I like inline statements as much as the next guy, but since this function is part of the "main idea" of the program it's better to write it out and keep it clear.
If the function is not that important and is clarified in naming/comments/documentation then you could probably get away with prioritizing the inherit desire that many python programmers have to golf over readability.
• I would find it unusual to keep the else clause when there's a return in the if statement; typically you'd write this over three lines (which, to me, also makes it easier to parse).
– sapi
Commented Sep 28, 2016 at 9:24
Aside from what others have said, I think it's better style not to use an else in cases like this, and just count on the fact that at the end of the function should be a return statement:
def sum_double(a, b):
total = a + b
if a == b:
return 2 * total
This is effectively the same, but it's clearer at first glance to see that something is always returned.
The solution may consider it more Pythonic to have one return statement, but it can be better to have separate ones as you showed. Especially if the logic can become more complex, what if you had to return 0 if one of the numbers was zero, or if there's further calculations to do when the numbers are not equal? For this reason, having 2 returns (as you did) makes code neater.
Also, please use 4 spaces for indentation. It's the recommended amount and easier to read.
• Two return points, unnecessary? How about total *= 2 instead? Commented Sep 27, 2016 at 12:43
• @innisfree That's slower. Commented Sep 27, 2016 at 13:36
• @mbomb007 That's about the least important of all possible aspects to this question (find one real world application where you'll notice the performance difference between the two approaches). And this is already an absolute bikeshed question.
– Voo
Commented Sep 27, 2016 at 19:58
• @Voo As if having 2 returns is a problem? Commented Sep 27, 2016 at 20:00
• @mbomb007 Not sure where you get the idea that I have a problem with that. But using nano second optimisations in python code as an argument about what to prefer is rather misguided. As I said it's a bike shed problem - all shown solutions are trivial to read and just fine.
– Voo
Commented Sep 27, 2016 at 20:03
I would probably do it this way. I find that it conveys the idea most clearly while keeping it short and readable. (Flat is better than nested.)
def sum_double(aa, bb):
mod = 2 if aa==bb else 1
return mod * (aa + bb) | 1,883 | 7,159 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-26 | latest | en | 0.950434 |
https://albertteen.com/uk/gcse/mathematics/algebra/simultaneous-equations-substitution | 1,723,446,041,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641036271.72/warc/CC-MAIN-20240812061749-20240812091749-00519.warc.gz | 67,068,437 | 41,060 | Albert Teen
YOU ARE LEARNING:
Simultaneous Equations: Substitution
# Simultaneous Equations: Substitution
### Simultaneous Equations: Substitution
Another method of solving simultaneous equations involves substituting equations into one another.
Simultaneous equations allow us to find the points which two equations have in common. You might already be familiar with solving them through elimination, but substitution is another method of achieving the same result.
Simultaneous equations allow us to find:
We can solve simultaneous equations by substituting one equation into another.
For example, have a look at $y = 2x + 5$ and $x + y = 11$
1
Label the equations
$(1)\space y = 2x + 5$ and $(2) \space x + y = 11$
2
Replace$y$ in equation 2 by $2x+5$
$x + 2x + 5 = 11$
3
Solve $x + 2x + 5 = 11$ to find $x$
4
Substitute $x=2$ into$(1)\space y = 2x + 5$
$y = 2(2) + 5$
5
Solve $y = 2(2) + 5$ to find $y$
6
Good work!
$x=2$ and $y=9$
Try solving $y = x + 4$and $x + y = 10$
Now try $y = x + 7$ and $x + y = 15$ | 331 | 1,039 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 36, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-33 | latest | en | 0.79473 |
https://www.slideserve.com/sailor/gerund-v-s-infinitive | 1,544,943,934,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827281.64/warc/CC-MAIN-20181216051636-20181216073636-00042.warc.gz | 1,047,609,234 | 14,164 | Gerund v.s.Infinitive
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# Gerund v.s.Infinitive - PowerPoint PPT Presentation
Gerund v.s.Infinitive. Goals: 1. To make students more familiar with the gerund or infinitive constructions 2. Students are expected to produce more sentences with these structures. Read the letter. Dear Grandma:
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## Gerund v.s.Infinitive
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Presentation Transcript
Gerund v.s.Infinitive
• Goals:
1. To make students more familiar with the gerund or infinitive constructions
2. Students are expected to produce more sentences with these structures.
Dear Grandma:
How are you? I had some problem with math at school. I hated doing it—it was so difficult.
Then I decided to study really hard. Mom suggested taking extra classes after school. I didn’t want to take extra classes but I agreed to do so because I wanted to pass the test.
So I began taking Math classes three times a week.
At first, I didn’t like missing all the fun my friends have—soccer practice, picnics, trips to the park, but now I really enjoy going to the math classes.
I learned to do long division and I practiced solving all sorts of problems and, guess what!
I passed the test. But I’m planning to take another course—for A students. If I continue studying this hard, I’ll be better than the teacher!
I promise to write again soon. Are you coming to Jason’s party? I hope to see you there.
Love, Alex.
From this letter, we learn some verbs take “–ing” form, such as stop,
finish,
hate,
con-tinue,
suggest,
begin,
like,
enjoy, and practice.
We also find some verbs that take the term “to,” such as forget,
remember, try,
decide, want,
agree, learn,
plan, promise,
and hope.
Presentation
Pay attention!
Please note that some verbs can take either “to” or the “-ing” form, such as love,
learn, like,
hate, start,
begin, and continue.
Complete the sentences with the correct form of the verbs.
• I enjoy _______especially Chinese food. (eat )
• They suggested _____ a picnic by the river. (have)
• He promised _____his homework. (do)
• She learned _____the piano last year. (play )
• We practiced _____ the ball into the goal. (hit)
• She wants ____ psychology when she’s older. (study)
• We decided ______ a movie on television. (see) | 673 | 2,778 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2018-51 | latest | en | 0.923149 |
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# What is the Marginal Product of Labor?
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• Written By: Osmand Vitez
• Edited By: Kristen Osborne
2003-2018
Conjecture Corporation
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The marginal product of labor is an economic measurement of what happens when a company adds an additional worker to its operations. Most companies measure the productivity of their employees, and when forecasting future sales goals, the company looks at what will happen when an additional worker is added to the workforce. In economic terms, marginal revenue should increase by at least an equal amount. If marginal revenue does not rise and marginal costs increase, the additional labor is not a good investment.
Companies have an interest to measure their economic productivity from labor, as this expense is typically the highest cost of doing business. Marginal revenue and marginal cost calculations are common economic tools to determine at what point a company should stop increasing its production output. This concept falls under the economic theory known as an economy of scale. Companies achieving an economy of scale have lowered their production costs to a point where they achieve maximum revenue.
Breaking down revenues and costs into marginal units provide a micro-economic scale for wealth measurement. One individual worker can add significant costs outside of her stated wage, including training costs, benefits, background checks, additional workspace, and other costs, all of which must be accounted for in economic terms. When reviewing the marginal product of labor, the assumption is that all other factors remain constant. Labor cost is variable, meaning that producing more units will increase costs above the previous level experienced by the company.
A basic calculation for this measurement is that each worker can produce five widgets per hour. Therefore, adding one additional worker increases production output by five widgets per hour, which is the marginal product of labor. Adding more than one worker may result in fewer total units produced each hour, however. For example, adding two workers may only result in eight more widgets produced rather than ten. The reason for this phenomenon is that, holding all other factors constant, the company may not have the resources available or space needed to allow more workers to produce the maximum number of widgets.
When a company cannot maximize its marginal labor, a theory known as the law of diminishing returns will occur. This theory states that adding more workers will result in higher costs that the company cannot recoup through selling goods or services. Essentially, the marginal cost will exceed marginal revenue as discussed earlier, with additional workers continuing to add further marginal costs.
## Recommended
Forterdom Post 1 Working with small businesses I have often encountered lack of understand regarding marginal product of labor. Usually in this case, a struggling business owner will see lack of productivity as a need for more employees. They have a situation where the work space is not set up in a functional way, hampering the employees ability to get their jobs done. Creating greater efficiency within the work space is what solves their productivity issues. | 646 | 3,502 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-05 | longest | en | 0.954898 |
https://routingfreak.wordpress.com/2008/03/06/shortest-path-first-algorithm-demystified/ | 1,686,393,658,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657169.98/warc/CC-MAIN-20230610095459-20230610125459-00596.warc.gz | 567,111,826 | 36,034 | # Shortest Path First (SPF) Algorithm Demystified ..
For the sake of brevity i would refer an Intermediate System (IS) in IS-IS parlance as a router in the following post – thus a router in this post could mean either an OSPF speaker or an IS-IS speaker or some other routing element from your favorite link state protocol. For the SPF algorithm to work, it would require *all* routers in the network to know about all the other routers in the network and the links connecting them. How a link state routing protocol encodes this information and ensures that its disseminated properly is left to that protocol. OSPF encodes this information in Link State Advertisements (LSAs) and floods it reliably, while IS-IS encodes this in a Link State Packet (LSP) that it originates.
Once each router knows about all the other routers and the links connecting them, it runs the Dijkstra Shortest Path First algorithm to determine the shortest path from itself to all the other routers in the network. Since each router has a similar copy of the link state database and each runs the same algorithm, they end up constructing the same view of the network and packets get routed consistently at each hop.
So, how does SPF algorithm work in OSPF and IS-IS.
Imagine a simple network as shown in the figure below.
Once each router has flooded its link state information in the network, all routers know about all the other routers and the links connecting them. The link state database on each router looks like the following:
[A, B, 3], [A, C, 6], [B, A, 3], [B, D, 3], [B, E, 5], [C, A, 6], [C, D, 9], [D, C, 9], [D, B, 3], [D, E, 3] , [E, B, 5] and [E, D, 3]
Each triple should be read as {originating router, router its connected to, the cost of the link connecting the two routers}
So what does Router A do with this information and how is this used in SPF?
While running SPF, each router maintains two lists – the first is a list of nodes for which the shortest path has been determined and we are sure that no path shorter than the one we have computed can exist. This list is called the PATH (or PATHS) list. The second is the list of paths through the routers that may or may not be the shortest to a destination. This list is called the TENTative list, or simply the TENT list From now on, TENT would refer to the TENT list and PATH, to the PATH list.
Each element in the list is a triplet of the kind {endpoint router that we’re trying to reach, total distance from the calculating router, next-hop to reach the endpoint router}
Each router runs the following algorithm to compute the shortest path to each node:
Step I: Put “self” on the PATH with a distance of 0 and a next hop of self. The router running the SPF refers to itself as either “self” or the root node, because this node is the root of the shortest-path tree.
Step II: Take the node (call it the PATH node) just placed on the PATH list and examine its list of neighbors. Add each neighbor to the TENT with a next hop of the PATH node, unless that neighbor is already in the TENT or PATH list with a lower cost.
Call the node just added to the TENT as the TENT node. Set the cost to reach the TENT node equal to the cost to get from the root node to the PATH node plus the cost to get from the PATH node to the TENT node.
If the node just added to the TENT already exists in the TENT, but with a higher cost, replace the higher-cost node with the node currently under consideration.
Step III: Find the lowest cost neighbor in the TENT and move that neighbor to the PATH, and repeat Step 2. Stop only when TENT becomes empty.
Lets follow the sequence that Router A goes through for building its SPF tree.
1st Iteration of the SPF run
Step I: Put “self” on the PATH with a distance of 0 and a next hop of self. After this step the PATH and TENT look as follows:
PATH – {A, 0, A}
TENT – { }
Step II: Take the node (call it the PATH node) just placed on the PATH list and examine its list of neighbors. Patently, A is the PATH node. Examine its list of neighbors ({A, B, 3}, {A, C, 6}). OSPF does this by looking at the LSAs advertised by Router A, while IS-IS does this by looking at the neighbors TLV found in Router A’s LSP. When an IS-IS node is placed in PATHS, all IP prefixes advertised by it are installed in the IS-IS Routing Information Base (RIB) with the corresponding metric and next hop.
The Step II further says – “Add each neighbor to the TENT with a next hop of the PATH node, unless that neighbor is already in the TENT or PATH list with a lower cost.” A’s neighbors are B and C, and since neither of them is in the PATH or TENT, we add both of them to the TENT.
PATH – {A, 0, A}
TENT – {B, 3, A}, {C, 6, A}
Step II says – “Call the node just added to the TENT as the TENT node. Set the cost to reach the TENT node equal to the cost to get from the root node to the PATH node plus the cost to get from the PATH node to the TENT node.”
Lets pick up the TENT node B. The cost to reach B would be the cost to reach from root node to PATH node + cost from PATH node to TENT node. In the first iteration of SPF, both the root node and PATH node is A. Thus total cost to reach B is cost to reach from A (PATH node) to B (TENT node) which is 3.
Step II says – “If the node just added to the TENT already exists in the TENT, but with a higher cost, replace the higher-cost node with the node currently under consideration.”
B isnt in the TENT, so skip this.
Step III says – “Find the lowest cost neighbor in the TENT and move that neighbor to the PATH, and repeat“. The lowest cost neighbor is B (with cost 3). Move this to PATH. We go back to Step II since TENT isnt yet empty.
2nd Iteration of the SPF run
PATH – {A, 0, A} {B, 3, A}
TENT – {C, 6, A}
We have thus added the neighbor B in PATH, since we know that there cannot be any other shorter path to reach it. And this is, as you will note, consistent with our definition of PATH wherein we had earlier stated that nodes can only be placed there once we are sure that there cannot be any shorter path to reach them from the root node.
We begin our 2nd iteration and go to Step II which says – “Take the node (call it the PATH node) just placed on the PATH list and examine its list of neighbors“. B is the PATH node and we examine its neighbors (A, E and D). Step II further says – “Add each neighbor to the TENT with a next hop of the PATH node, unless that neighbor is already in the TENT or PATH list with a lower cost.”
Since A is already in PATH with ignore it and only add E and D to TENT.
PATH – {A, 0, A} {B, 3, A}
TENT – {C, 6, A} {D, 3, B}, {E, 5, B}
Step II says – “Call the node just added to the TENT as the TENT node. Set the cost to reach the TENT node equal to the cost to get from the root node to the PATH node plus the cost to get from the PATH node to the TENT node.
Aah .. this means that cost against D and E would not be 3 and 5, as what i have shown, but would instead be 3 (cost from A to B) +3 (cost from B to D) = 6 and 3 (cost from A to B) +5 (B to E) = 8
Thus the PATH and TENT look as follows:
PATH – {A, 0, A} {B, 3, A}
TENT – {C, 6, A} {D, 6, B}, {E, 8, B}
The rest of the Step II does not apply here since D and E dont exist in the TENT.
Come to Step III which says “Find the lowest cost neighbor in the TENT and move that neighbor to the PATH, and repeat Step 2″.
Lowest cost neighbor can either be C or D so pick on up randomly. It can mathematically be proven that we would end up with the same SPF tree irrespective of which equal cost neighbor is picked up from the TENT first. In our case, lets pick up C.
It is thus moved to the PATH
PATH – {A, 0, A} {B, 3, A} {C, 6, A}
TENT – {D, 6, B}, {E, 8, B}
3rd Iteration of the SPF run
Go back to Step II which says “Take the node (call it the PATH node) just placed on the PATH list and examine its list of neighbors. Add each neighbor to the TENT with a next hop of the PATH node, unless that neighbor is already in the TENT or PATH list with a lower cost
C’s neighbors are A and D. Since A is already in the PATH only D is added in the TENT.
PATH – {A, 0, A} {B, 3, A} {C, 6, A}
TENT – {D, 6, B}, {E, 8, B} {D, 9, C}
As per Step II we now need to fix the cost to reach D from the root node. This would be cost to reach from A to C (6) + cost from C to D (9) = 15
PATH and TENT now:
PATH – {A, 0, A} {B, 3, A} {C, 6, A}
TENT – {D, 6, B}, {E, 8, B} {D, 15, C}
Step II further says – “If the node just added to the TENT already exists in the TENT, but with a higher cost, replace the higher-cost node with the node currently under consideration.
node D already exists in the TENT (via B with cost 6) and since its with a lesser cost, we remove the node that we had just added from the TENT. This is because a lower cost path to reach node D already exists in the TENT.
PATH – {A, 0, A} {B, 3, A} {C, 6, A}
TENT – {D, 6, B}, {E, 8, B}
We come to Step III which says “Find the lowest cost neighbor in the TENT and move that neighbor to the PATH, and repeat Step 2″.
Lowest cost neighbor is D – which means we move D now to the PATH and go to Step II, since the TENT isnt yet empty.
4th Iteration of the SPF run
PATH – {A, 0, A} {B, 3, A} {C, 6, A} {D, 6, B}
TENT – {E, 8, B}
Step II says “Take the node (call it the PATH node) just placed on the PATH list and examine its list of neighbors. Add each neighbor to the TENT with a next hop of the PATH node, unless that neighbor is already in the TENT or PATH list with a lower cost.”
To examine D’s neighbors we look at the link state information it advertised. It advertised the following information:
[D, C, 9], [D, B, 3], [D, E, 3]
This means that D is says that its connected to C, B and E. We ignore its connection to B and C, since they are already in PATH. We thus only add neighbor E in the TENT.
PATH – {A, 0, A} {B, 3, A} {C, 6, A} {D, 6, B}
TENT – {E, 8, B} {E, 3, D}
Continuing with Step II which further says – “Call the node just added to the TENT as the TENT node. Set the cost to reach the TENT node equal to the cost to get from the root node to the PATH node plus the cost to get from the PATH node to the TENT node.”
This means that we need to adjust the cost of the triple {E, 3, D} that we just added to the TENT. The cost to reach E via D would thus be the cost to reach D from A (which is the root node) + the cost to reach E from D. This comes out to be 6 + 3 = 9.
TENT thus looks like this – {E, 8, B} {E, 9, D}
Step II further says – “If the node just added to the TENT already exists in the TENT, but with a higher cost, replace the higher-cost node with the node currently under consideration”
We just added node E in the TENT and a route to E already exists in the TENT, and its with a lower cost. This means that we remove the route that we had just added.
So PATH and TENT at this point look as follows:
PATH – {A, 0, A} {B, 3, A} {C, 6, A} {D, 6, B}
TENT – {E, 8, B}
We go to Step III which says – “Find the lowest cost neighbor in the TENT and move that neighbor to the PATH, and repeat Step 2″.
The lowest cost neighbor in TENT right now is E. We move this to PATH.
So PATH and TENT at this point look as follows:
PATH – {A, 0, A} {B, 3, A} {C, 6, A} {D, 6, B} {E, 8, B}
TENT – { }
Step III further states that we continue if and only if something remains in the TENT. TENT is now empty, which means that we have computed the shortest paths to all the nodes that A was aware of.
This is marks the end of the SPF algorithm run and the SPF tree that it has computed looks as follows
## 22 thoughts on “Shortest Path First (SPF) Algorithm Demystified ..”
1. pp says:
very well explained
Like
2. Thanks for the details. I think even Cisco Press Routing TCP/ IP Vol 1 (Jeff Doyle) did not explain the details this clearly.
Like
3. Ron says:
I noticed that you said that a PATH list is the shortest path that has been determined and we are sure that no other path shorter.
Is it safe to say that the TENT list is solely based on Link Speed?
Like
4. Rod says:
Very well explained. I was comparing it with a similar example/explanation on a Juniper book and yours has the extra detail/repetition that one needs to cement the idea.
Congratulations and thanks for sharing your knowledge.
Like
5. Parinda says:
Very well explained…
Just a minor correction…
In the last figure the distance from:
B-D should be 3 instead of 6
B-E should be 5 instead of 8
Like
1. Yes, thats correct! What i had meant was that the distance of D and E from the calculating node (A) is 6 and 8 respectively.
Like
6. DPT says:
very well explained. Thanks
Like
7. sidd says:
great explanation, thanks
Like
8. A says:
man, very nice explanation! well done. thanks!
Like
9. johnny says:
I do not see an explanation of where you got the initial costs.
Like
1. Manav Bhatia says:
Its usually derived from the link BW or can be configured manually.
Like
10. Sriram Selvam says:
Thanks a lot for the tutorial 🙂
Like
11. chandrasekar says:
Thank you so very much for the detailed explanation…. so easily explained …
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https://www.livemint.com/Opinion/fodS89bFVdFCaO17Sh9fLN/From-70-million-to-two.html | 1,695,945,966,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510462.75/warc/CC-MAIN-20230928230810-20230929020810-00782.warc.gz | 925,005,538 | 95,730 | Business News/ Opinion / From 70 million to two
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# From 70 million to two
## It's not that mathematicians have been searching for numbers that differ by exactly two. But what they want to know is if there an infinity of them?
A major advance in mathematics, and the gist of it made a 14 year-old I know chuckle. Well, here’s a parallel. Say you’re searching for objects that are exactly two inches long—some school project perhaps. Along comes a friend who says: “Pssst! I know where you’ll find many objects, all less than 70 million inches long!"
You’d chuckle too. You’d probably think he was batty. So then what would you think if the rest of your school, even the rest of the world, sees this as a revelation and applauds your friend as a pioneer? Probably chuckle some more.
Yet in April, something just like this happened in that goldmine of fascinating goings-on, mathematics. Mathematicians have long studied certain pairs of numbers that differ by exactly two. It’s not that they have been searching for them—they are all over the place. But precisely for that reason, mathematicians want to know, is there an infinity of them? Or do they stop appearing, somewhere in the upper reaches of the number line? This is a dilemma that mathematicians have never been able to resolve.
Along comes a totally unknown mathematician—so unknown and unheralded that he once worked at a sandwich shop to make ends meet—who says: “Pssst! I’ve just proved that there’s an infinity of these numbers that differ by 70 million or less!"
Worth a 14-year-old’s chuckle? And yet mathematicians the world over are in a whirl. They have grappled so long with the question of whether there is an infinity of such pairs that it even has a name, the Twin Prime Conjecture. For it is primes we are talking about; specifically, primes that differ by two, like 17 and 19, 29 and 31, 41 and 43, 101 and 103 (called, no prizes for guessing, twin primes). The conjecture is that there is an infinity of such pairs. But nobody has managed to prove it.
In fact, this is really a special case of a broader conjecture: that there is an infinite number of primes that differ by any given even number, not just 2. (Why even, and not odd? Puzzle for you to tease out—though ignore the special-case prime of 2). Nobody has proved that one either.
Along comes Yitang Zhang, a professor at the University of New Hampshire. Remember, we still don’t know whether there are an infinite number of pairs of primes that have a gap between them of two, or any larger even number. But in April, Zhang proved that there is an upper limit to these gaps: 70 million. That is, you can wander as far as you like into the upper reaches of the number line, and you will never run out of prime pairs that differ by 70 million or less.
All right, I don’t blame you for being underwhelmed. But people interested in these things have described Zhang’s proof as “astounding", “thrilling", “a huge step forward" and “one of the great results in the history of number theory." From being an obscure member of the faculty at a small university, Zhang has suddenly been catapulted into a dizzying series of lectures about his work at top-flight universities, and praised for his “clarity" in delivering them.
The reason for all this attention is really Zhang’s discovery of a limit. Because an upper bound on the gaps between prime pairs is, yes, a huge step forward from not knowing if there is such a bound at all. And to mathematicians, there is intrinsically little to choose between 70 million and two. Both are just numbers. The very existence of the 70 million limit holds out hope that it can one day be pushed down to two. It will be hard work, and there are already people who think it may not happen. But new proofs like Zhang’s also bring new thinking in their wake. That’s their great value, and where the real hope of proving the Twin Prime Conjecture lies.
And there is an interesting insight in all this. Primes and prime pairs are scattered randomly among the numbers, though they get slowly rarer as you get to those fabled upper reaches. However, this does not mean they are scattered evenly: you might have a bunch of them, followed by a long dearth of them. This is really the essence of randomness: that when you have random distributions, patterns naturally form. Patterns, that is, including twin primes.
In other words, the truly counter-intuitive idea would be if the twin prime conjecture was not true. Because then something would be working to prevent primes from forming the two-apart pattern; almost by definition, that militates against randomness. The mathematician Jordan Ellenberg says that with Zhang’s result, “we might…be on our way to developing a richer theory of randomness." That’s an exciting prospect.
Zhang himself said this about how he found the way to his proof: “The important thing is to keep thinking."
There’s good advice. Certainly worth rather more than a chuckle.
Once a computer scientist, Dilip D’Souza now lives in Mumbai and writes for his dinners. A Matter of Numbers will explore the joy of mathematics, with occasional forays into other sciences. To read Dilip D’Souza’s previous columns, go to www.livemint.com/dilipdsouza | 1,175 | 5,247 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2023-40 | latest | en | 0.971504 |
http://www.slidesearchengine.com/slide/smith-f09 | 1,540,104,242,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583513760.4/warc/CC-MAIN-20181021052235-20181021073735-00286.warc.gz | 571,287,577 | 7,152 | # Smith F09
50 %
50 %
Entertainment
Published on October 13, 2007
Author: Belly
Source: authorstream.com
Nash’s Nobel: Nash’s Nobel Presentation by: Andrew Smith andrewdsmith8@bw-deloitte.com John F Nash Jr: John F Nash Jr 1928 - Seminal work 1950 Nobel prize 1994 Cournot Equilibrium: Cournot Equilibrium Rock Paper Scissors: Rock Paper Scissors Zero sum game Symmetric payoffs Scissors cut paper Paper covers rock Rock sharpens scissors Losing strategies exist Randomise to avoid losing Role Play Game: Role Play Game Invitation to tender to supply 10 units Bids allowed: €1 through €5 per unit in intervals of €1 Contract awarded to cheaper supplier Split 50/50 if a tie Red: production cost €1 per unit Blue: production cost €2 per unit Game Payoffs: Game Payoffs 0 -10 0 10 0 0 0 20 0 30 0 -5 0 0 0 0 0 0 0 0 0 -10 10 0 5 0 10 0 10 0 0 -10 10 5 0 0 20 0 20 0 0 0 0 10 0 0 15 10 30 0 No bid 1 2 3 4 Red Bid (cost €1) 1 2 3 4 5 Blue bid (cost: €2) 0 -10 0 0 0 10 0 20 20 15 0 0 0 0 10 0 20 0 30 0 40 0 No bid 5 Equilibrium: Equilibrium 0 -10 0 10 0 0 0 20 0 30 0 -5 0 0 0 0 0 0 0 0 0 -10 10 0 5 0 10 0 10 0 0 -10 10 5 0 0 20 0 20 0 0 0 0 10 0 0 15 10 30 0 No bid 1 2 3 4 Red Bid (cost €1) 1 2 3 4 5 Blue bid (cost: €2) 0 -10 0 0 0 10 0 20 20 15 0 0 0 0 10 0 20 0 30 0 40 0 No bid 5 Bargaining Game: Bargaining Game Player A wishes to sell an asset Player B wishes to buy Third party dealer sells at €10 and buys at €5 Model negotiation between A and B Many Nash equilibria Unique Selten equilibrium Is the Theory Correct?: Is the Theory Correct? Mathematically, yes But do real games converge to Nash equilibria? Difficult to specify a game and to calibrate a model My “get out of jail” card is to invoke imperfect calibration when my model fails to predict actual outcomes (Possible) Applications: (Possible) Applications Auction design (eg Telecom licenses) Military / anti-terrorist Regulation / response Capital allocation Underwriting Social policy Premium cycle Nash’s Nobel: Nash’s Nobel Presentation by: Andrew Smith andrewdsmith8@bw-deloitte.com
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Advice to Future Signals & Systems Students from Fall 2009 students: • I didn’t really look at the old exams posted at professor’s website. | 1,197 | 3,665 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-43 | latest | en | 0.647445 |
https://brilliant.org/problems/easy-three-pointer/ | 1,606,459,457,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141189141.23/warc/CC-MAIN-20201127044624-20201127074624-00627.warc.gz | 222,310,924 | 8,597 | # Game Winning Three
Suppose you're taking a three point shot at a basket that's $11 \text{ m}$ down the court, and $1.15 \text{ m}$ above your head. You can take the shot at any angle you like, however, you're not perfect and you're likely to imperfectly set your release angle.
What release angle $\theta$ (in degrees) should you aim for in order to maximize your chance of hitting the three pointer? (In other words, what release angle provides you with the biggest margin of error?) | 117 | 488 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 3, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2020-50 | latest | en | 0.948131 |
https://devopspal.com/2020/04/leetcode-arrays-minimum-value-to-get-positive-step-by-step-sum/ | 1,723,572,694,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641082193.83/warc/CC-MAIN-20240813172835-20240813202835-00297.warc.gz | 159,795,439 | 16,198 | Tue. Aug 13th, 2024
# LeetCode – Arrays – Minimum Value to Get Positive Step by Step Sum
#### ByKnight Coderz
Apr 19, 2020
Given an array of integers `nums`, you start with an initial positive value startValue.
In each iteration, you calculate the step by step sum of startValue plus elements in `nums` (from left to right).
Return the minimum positive value of startValue such that the step by step sum is never less than 1.
Example 1:
```Input: nums = [-3,2,-3,4,2]
Output: 5
Explanation: If you choose startValue = 4, in the third iteration your step by step sum is less than 1.
step by step sum
startValue = 4 | startValue = 5 | nums
(4 -3 ) = 1 | (5 -3 ) = 2 | -3
(1 +2 ) = 3 | (2 +2 ) = 4 | 2
(3 -3 ) = 0 | (4 -3 ) = 1 | -3
(0 +4 ) = 4 | (1 +4 ) = 5 | 4
(4 +2 ) = 6 | (5 +2 ) = 7 | 2
```
Example 2:
```Input: nums = [1,2]
Output: 1
Explanation: Minimum start value should be positive.
```
Example 3:
```Input: nums = [1,-2,-3]
Output: 5```
``class Solution { public int minStartValue(int[] nums) { List<Integer> numList = Arrays.stream(nums).boxed().collect(Collectors.toList()); int result = 1; Collections.sort(numList); System.out.println(numList); if(numList.get(0) >= 1) return result; else{ int i=1; while(true){ int sum=i; boolean found = true; for(int j=0;j<nums.length;j++){ sum = sum+nums[j]; if(sum<=0){ found = false; break; } } if(found){ result = i; break; } i++; } } return result; }}`` | 559 | 1,788 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-33 | latest | en | 0.248507 |
https://www.openmiddle.com/multiplying-fractions-3/?replytocom=107202 | 1,597,160,396,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738816.7/warc/CC-MAIN-20200811150134-20200811180134-00347.warc.gz | 773,861,107 | 22,279 | Home > Grade 7 > Multiplying Fractions 3
# Multiplying Fractions 3
Directions: Find three fractions whose product is -5/24. You may use fractions between -8/9 to 8/9 no more than one time each.
Find at least 2 possible combinations.
### Hint
What properties hold true for multiplying fractions?
e.g : (5/6)(-1/2)(1/2),(-5/6)(1/3)(3/4)
Source: Al Oz
## Sides of a Triangle
Directions: The perimeter of a triangle is 20 units. Using whole numbers, how many sets …
1. The part of the problem that states the fractions must be between -1/9 and 1/9 make the problem impossible, right? I believe that part of the instructions needs to be revised.
2. The problem says that fractions can’t be repeated, but the first sample uses 1/2 twice.
3. In measurements, fractions appear whenever units are not small enough to express quantities in integers. For example, five quarter-dollars will buy you exactly as mush as a dollar and a quarter. One and a half dollar stands for exactly the same quantity as three half-dollars or six quarter-dollars.
Fractions are unavoidable and sooner or later we all have to learn to work with fractions. The mathematical usage of the word fraction has a very clear everyday connotation as a part of a bigger object. It would be unthinkable nowadays to just introduce fractions as a pair of numbers and postulate their basic properties. Still, to express fractions one needs a pair of numbers with a meaning and intuition attached to them.
When multiplying fractions, the numerators (top numbers) are multiplied together and the denominators (bottom numbers) are multiplied together. To divide fractions, rewrite the problem as multiplying by the reciprocal (multiplicative inverse) of the divisor. To add fractions that have the same, or a common, denominator, simply add the numerators, and use the common denominator. However, fractions cannot be added until they are written with a common denominator. The figure below shows why adding fractions with different denominators is incorrect.
4. jasmine belosay
(5/6)(-1/3)(3/4) = -5/24 | 475 | 2,072 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2020-34 | latest | en | 0.931561 |
https://btechgeeks.com/python-program-to-determine-whether-a-given-number-is-even-or-odd-recursively/ | 1,726,490,304,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.32/warc/CC-MAIN-20240916112213-20240916142213-00063.warc.gz | 129,168,327 | 15,675 | # Python Program to Determine Whether a Given Number is Even or Odd Recursively
Are you new to the java programming language? We recommend you to ace up your practice session with these Basic Java Programs Examples
Recursion:
Recursion is the process by which a function calls itself directly or indirectly, and the associated function is known as a recursive function. Certain issues can be addressed fairly easily using a recursive approach. Towers of Hanoi (TOH), Inorder /Preorder/Postorder Tree Traversals, DFS of Graph, and other analogous issues are examples.
Given a number the task is to check whether the given number is even number or odd number using recursive approach in Python.
Examples:
Example1:
Input:
Enter some random number = 215
Output:
The given number 215 is odd
Example2:
Input:
Enter some random number = 628
Output:
The given number 628 is even
## Program to Determine Whether a Given Number is Even or Odd Recursively
Below are the ways to check whether the given number is even or odd recursively :
### 1)Using Recursion(Static Input)
Approach:
• Give the number as static input.
• Pass the number to a recursive function as an argument.
• Define the base condition as an integer less than two.
• Otherwise, use the number -2 to invoke the function recursively.
• Then return the result and determine whether the number is even or odd.
• The final result should be printed.
• Exit of program.
Below is the implementation:
# function which returns true if the given number
# is evennum or oddnum using recursoive approach
def checkPrimeRecursion(numb):
# Defining the base condition as an integer less than two.
if (numb < 2):
# Then return the result and determine whether the number is even or odd.
return (numb % 2 == 0)
# Otherwise, use the number -2 to invoke the function recursively.
return (checkPrimeRecursion(numb - 2))
# Give the number as static input.
numb = 729
# passing the given number to checkPrimeRecursion
# if the returned value is true then it is even number
if(checkPrimeRecursion(numb)):
print("The given number", numb, "is even")
# if the returned value is false then it is odd number
else:
print("The given number", numb, "is odd")
Output:
The given number 729 is odd
Explanation:
• User must give the number as static input and store it in a variable.
• A recursive function is given the number as an argument.
• The basic requirement is that the number be less than two.
• Otherwise, the function is called recursively with a number less than two.
• The outcome is returned, and an if statement is used to determine whether the integer is odd or even.
• The final result is printed.
### 2)Using Recursion(User Input)
Approach:
• Enter some random number as user input using int(input()) function.
• Pass the number to a recursive function as an argument.
• Define the base condition as an integer less than two.
• Otherwise, use the number -2 to invoke the function recursively.
• Then return the result and determine whether the number is even or odd.
• The final result should be printed.
• Exit of program.
Below is the implementation:
# function which returns true if the given number
# is evennum or oddnum using recursoive approach
def checkPrimeRecursion(numb):
# Defining the base condition as an integer less than two.
if (numb < 2):
# Then return the result and determine whether the number is even or odd.
return (numb % 2 == 0)
# Otherwise, use the number -2 to invoke the function recursively.
return (checkPrimeRecursion(numb - 2))
# Give the number as static input.
numb = int(input('Enter some random number = '))
# passing the given number to checkPrimeRecursion
# if the returned value is true then it is even number
if(checkPrimeRecursion(numb)):
print("The given number", numb, "is even")
# if the returned value is false then it is odd number
else:
print("The given number", numb, "is odd")
Output:
Enter some random number = 215
The given number 215 is odd
Related Programs: | 912 | 3,993 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-38 | latest | en | 0.78286 |
https://homework.cpm.org/category/CC/textbook/CCA2/chapter/Ch12/lesson/12.1.3/problem/12-59 | 1,603,308,890,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107877420.17/warc/CC-MAIN-20201021180646-20201021210646-00084.warc.gz | 331,033,816 | 15,305 | Home > CCA2 > Chapter Ch12 > Lesson 12.1.3 > Problem12-59
12-59.
Expand.
1. $(a + b)^4$
Use the binomial theorem.
$a^4+4a^3b+6a^2b^2+4ab^3+b^4$
1. $(3m - 2)^4$
Use the results of part (a). Let $a = 3m$ and $b = -2$. Simplify.
$81m^4 -216m^3 +216m^2 -96m+16$ | 129 | 265 | {"found_math": true, "script_math_tex": 6, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2020-45 | latest | en | 0.449018 |
http://www.maksyuki.com/?p=2707 | 1,550,268,135,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247479159.2/warc/CC-MAIN-20190215204316-20190215230316-00442.warc.gz | 405,503,218 | 15,738 | ## zoj2588
maksyuki 发表于 oj 分类，标签:
Burning Bridges
Ferry Kingdom is a nice little country located on N islands that are connected by M bridges. All bridges are very beautiful and are loved by everyone in the kingdom. Of course, the system of bridges is designed in such a way that one can get from any island to any other one.
But recently the great sorrow has come to the kingdom. Ferry Kingdom was conquered by the armies of the great warrior Jordan and he has decided to burn all the bridges that connected the islands. This was a very cruel decision, but the wizards of Jordan have advised him no to do so, because after that his own armies would not be able to get from one island to another. So Jordan decided to burn as many bridges as possible so that is was still possible for his armies to get from any island to any other one.
Now the poor people of Ferry Kingdom wonder what bridges will be burned. Of course, they cannot learn that, because the list of bridges to be burned is kept in great secret. However, one old man said that you can help them to find the set of bridges that certainly will not be burned.
So they came to you and asked for help. Can you do that?
Input
The input contains multiple test cases. The first line of the input is a single integer T (1 <= T <= 20) which is the number of test cases. T test cases follow, each preceded by a single blank line.
The first line of each case contains N and M - the number of islands and bridges in Ferry Kingdom respectively (2 <= N <= 10 000, 1 <= M <= 100 000). Next M lines contain two different integer numbers each and describe bridges. Note that there can be several bridges between a pair of islands.
Output
On the first line of each case print K - the number of bridges that will certainly not be burned. On the second line print K integers - the numbers of these bridges. Bridges are numbered starting from one, as they are given in the input.
Two consecutive cases should be separated by a single blank line. No blank line should be produced after the last test case.
Sample Input
2
6 7
1 2
2 3
2 4
5 4
1 3
4 5
3 6
10 16
2 6
3 7
6 5
5 9
5 4
1 2
9 8
6 4
2 10
3 8
7 9
1 4
2 4
10 5
1 6
6 10
Sample Output
2
3 7
1
4
Author: Andrew Stankevich
Source: Andrew Stankevich's Contest #5 | 594 | 2,302 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2019-09 | longest | en | 0.97527 |
https://adams1084.nevsehirkargo.com/archives/313 | 1,621,208,009,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991921.61/warc/CC-MAIN-20210516232554-20210517022554-00110.warc.gz | 108,573,303 | 7,387 | # Blackjack – How exactly to Bet and Win
Blackjack is a card game that originated in late 19th century Spain and is currently the most popular games in casinos all over the world. Blackjack, previously also Black Jack and Vingt-Un, may be the American version of an internationally family of gambling games called Twenty-One, whose other relatives will be the British version of the overall game of Pontoon and European version of Vingt-et-Un. In this game, players place their wagers in the hope of reaching a predetermined number, called the blackjack limit. In case a player wins a blackjack hand, he doesn’t have to pay out his winnings immediately; the total amount is subtracted from his bankroll first. This means that in a twenty-one game, a new player needs to win a minimum of twenty-one times to win. The other forms of blackjack games are variations of the original game.
A lot of the basic strategies in blackjack revolve round the player’s table. In a live game, you can expect to visit a dealer sitting at the table, surrounded by his cards. A dealer has the substitute for fold, making new hands or raise the betting. The latter is referred to as “raising” the stakes. However, as in other games, it really is wise to be cautious with regards to raising the stakes. Doing this might cost the player more than winning.
After the dealer reveals his cards, a player must then estimate the hand value. He does this by comparing the initial two cards of both players, like the second card if present. The hand value is really a percentage calculation that considers the total amount of chips in both hands and the current hand count (all of the hands together, excluding the initial two). Hand analysis is often complicated by the current presence of other cards, such as for example aces and queens, and the current presence of jokers. There are several methods for computing the hand value, and they all use the same mathematical method.
A few of these methods are more complex than others. The “suit” method is really a simple example. It includes laying out all of the possible combinations that may occur in a casino game of blackjack. The dealer will choose a single card from among the cards up for grabs to represent the initial suit. The ball player can bet in accordance with this suit.
One can also analyze the cards up for grabs to estimate the odds of winning. This is usually a more complex method. It involves a process of estimating how much the pot will undoubtedly be with a given number of cards following the dealer has dealt out his cards. You can arrive at a figure by either picking a number that is greater than the expected number of cards in the deck, or by selecting a number that is close to the expected number of cards in the deck however, not greater. The right betting strategy must therefore be attained to make this calculation and win even money.
In a few casinos, there are machines programmed to deal the cards. These machines are called ‘robot’ because they basically follow a pre-determined strategy. The dealer will place cards onto a particular slot machine game that produces sm 카지노 a card. The slot machine will match the expected cards with the robot’s card and the ball player will win even money if the quantity and suit match. That is a machine programmed to deal the cards, and is therefore not at the mercy of human intervention.
Different ways of analyzing the cards dealt include basic strategies based on common sense and probability. It is common sense to count cards once the dealer has dealt them. That is called the ‘card to card’ or ‘deal turnover’. A second common sense approach to studying the hand total would be to count the number of cards dealt, and compare it with the expected hand total.
Some casino dealers will indicate to players how many cards have been dealt and the percentages of doubling and matching that are involved in each game. This is called the ‘card pattern’ or ‘card accuracy’ and it is a fundamental section of many blackjack games. Aces are a valued lower value generally in most games than they’re in older games, because they’re more likely to be doubled than other cards in a holdem game. Dealing with this aspect of blackjack requires a bit of experience, and is better learnt in one of the tutorial sessions available to play blackjack online. After the basics of the game have been mastered, playing for money could be exciting and challenging. | 893 | 4,455 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-21 | latest | en | 0.974291 |
https://www.trisen.com/sol/static/wg/i-one4.html | 1,675,898,434,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500983.76/warc/CC-MAIN-20230208222635-20230209012635-00422.warc.gz | 1,038,073,780 | 2,917 | ## ONE/4 Planetary Orbits
STEP ONE: Determine the number of potential orbits by rolling on table 1.4.1.
STEP TWO: Determine the mean distance of the orbits by consulting the formulae under 1.4.2.
STEP THREE: Remove all impossible orbits. This includes any orbit which would put a planet within a star's radius or so close as to vaporize it, and any orbit which is unstable in a binary system (between the binary's closest separation / 3 and the furthest separation x 3 distance). Worlds may orbit both stars in a binary, if the more massive companion has any orbits generated beyond the furthest separation distance x 3. Only consider such orbits for the more massive companion. For white dwarves, remove all orbits within the limit provided by table 1.4.3.
STEP FOUR: Remove all orbits where a planet would be vaporized, according to the formulae in 1.4.4. Also, determine the limit of the original inner system. Note this limit, it will be used in the next section.
Table 1.4.1 Number Of Orbits
1d10 -1: 1d10 + 10 2-5: 1d10 + 5 6-7: 1d10 8-9: 1d5 (1d10 / 2) 10+: None
Modifications to initial roll:
+1 if star is K5v to K9v
+2 if star is M0v to M4v
+3 if star is M5v to M9v
+5 if star is Brown Dwarf
Subtract Abundance Modifier
Table 1.4.2 Orbit Sizes The first orbit has a distance in AU of: D = M2 x 0.05 x 1d10 ... where M is the mass of the star Subsequent orbits have a distance in AU of: D = P x [1.1 + (1d10 x 0.1)] + 0.1 ... where P is the distance of the previous orbit
Table 1.4.3 White Dwarf Removed Orbits
1d10 1-4: Remove all orbits within 2 AU 5-8: Remove all orbits within 4 AU 9-11: Remove all orbits within 6 AU 12+: Remove all orbits within 10 AU
Modification:
+1 if the mass of the white dwarf is 0.6-0.9
+4 if the mass of the white dwarf is over 0.9
Table 1.4.4 Untenable Orbits and the Inner System Zone Planets cannot survive if hotter than about 2000°K. Thus, remove orbits within: D = L0.5 x 0.025 ... where L is the luminosity of the star The system's Inner Zone is within: D = L0.5 x 4 ... where L is the luminosity of the star To be exacting: For main sequence stars use the luminosity of a mid-age star of the same spectral class, for subgiants and giants use the luminosity of a main sequence star of the same mass, and for white dwarves consider any surviving worlds to be outside the Inner Zone.
INNER SYSTEM ZONE:
The area of the stellar system where the early system is too hot to allow icy planetoids. Inner system objects thus have higher density and mostly consist of silicates and metals, while the area outside is more dominated by frozen water and gasses.
STARS WITHOUT PLANETARY ORBITS:
There may be some other objects around these stars, typically either a few icy chunks in distant random orbits (1d10 x 1d10 AU) or more rarely a captured planetoid. Captured planetoids have eccentric orbits (1D10 AU) and may be of any general planet type. Some planetless stars have very sparse "rings" of chunks of debris:
1d10 1-3 Icy chunks 4 Captured body 5-7 Rings 8+ Nothing
BINARIES:
If any orbits where removed by the binary effects, the system is likely to have a fair deal of stray asteroids and debris. | 879 | 3,166 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2023-06 | latest | en | 0.850293 |
https://www.physicsforums.com/threads/combinatorics-with-repetition.837922/ | 1,508,623,308,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824899.43/warc/CC-MAIN-20171021205648-20171021225648-00168.warc.gz | 990,533,875 | 18,375 | # Combinatorics with repetition
1. Oct 15, 2015
### iScience
1. The problem statement, all variables and given/known data
if there are k spots for n balls, what are the number of possible arrangements? each spot can hold n number of balls and each ball is indistinguishable from one another.
2. Relevant equations
$$\frac{n+k-1}{k!(n-1)!}$$
3. The attempt at a solution
I just wanted to know if this was the right equation, since i don't understand the equation.
2. Oct 15, 2015
### andrewkirk
The question is unclear. Does it mean the following?
We have m>nk indistinguishable balls and k buckets, labeled 1 to k. We then put a number of balls between 0 and n inclusive in each bucket.
How many different arrangements of number of balls in the numbered buckets are there?
3. Oct 15, 2015
### iScience
yup, that's what i meant. I'm actually trying to recreate all the possibilities visually (well, conceptually) but the number i get is different from the one the below eqn yields. $$\frac{n+k-1}{k!(n-1)!}$$
basically, i'm either not covering all possible arrangements with the way i'm going about it, or the expression i came up with is just incorrect. Either way i'd like someone to point out what i'm doing wrong.
so here's how i'm doing it visually, just bear with me:
first, only one of the balls traverse all possible spots leaving us so far with:
$$Arrangements = k$$
then, for each spot it occupied, we find the number of additional arrangements it can make with an additional ball:
+1
+2
+3
for each time we move the red over, we traverse the green from position 2 to the red. (I do this because i'm trying to find the combination and not the permutation)
so now we have $$Arrangements = k + \sum^{k-1}_{i=1}(i)$$
again, being careful not to include any doubles of the same arrangement, i go from left to right.
i move the blue over to position 2, we have the same situation we had earlier, just with k-1 spots now.repeating the previous process we should end up with..
$$Arrangements = k-1 + \sum^{k-2}_{i=1}(i)$$
repeating the process till the end we get:
$$Arrangements = \sum^{k-1}_{j=0}[ (k-j) + \sum^{k-1-j}_{i=1}(i)]$$
So.....
this was the case for 3 choose k. using my formula for say.. 3 choose 10, i get 121, whereas $$\frac{n+k-1}{k!(n-1)!}$$
says it should be 220. how does this not cover every possible arrangement?
Last edited: Oct 15, 2015
4. Oct 15, 2015
### andrewkirk
I think you're doing it in a more difficult way than necessary. Your pictures seem to be treating the balls as distinguishable by colour, whereas the problem says they are indistinguishable. That reduces the problem to the following:
How many different sequences of k numbers are there, where each number is an integer in {0, 1, 2, ..., n}?
5. Oct 15, 2015
### iScience
my apologies! the colors were just for me to keep track of what spots i already traversed. just makes it easier for me. but yes they are indistinguishable and so the colors are irrelevant.
k!, actually i have a simpler question, what i did above, is that "combination with repetition" or is it something completely different?
6. Oct 16, 2015
### haruspex
First, your equation is not quite right. It should be $\frac{(n+k-1)!}{(k-1)!n!}$. To verify that, consider 1 ball, 2 spots. There are two possible arrangements, not 1.
The easiest way to understand this equation is to think in terms of separators between the 'spots', rather than the spots themselves. There are k-1 of these.
Next, blur the distinction between the balls and the separators. There are just n+ k-1 'things' in a line, of which k-1 are separators. The possible arrangements of undistinguished balls into the slots is then just the number of ways of deciding which of the n+k-1 things are the k-1 separators. | 1,004 | 3,780 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2017-43 | longest | en | 0.925082 |
https://www.abc.net.au/science/articles/2013/05/27/3764768.htm | 1,566,790,896,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330962.67/warc/CC-MAIN-20190826022215-20190826044215-00123.warc.gz | 717,645,862 | 13,773 | # In this universe, charge is everything
If you want to get your head around the matter and energy that make up the detectable universe, there's no better - or easier - place to start than the story of charge.
Gravity feels like a big player in the universe, but it's nothing compared to charge. Light and life rely on storing, separating and jiggling electric charge. (Source: billberryphotography/iStockPhoto)
The only time most of us think about positive and negative charge is when we're squinting to see which way to insert a battery. But harnessing charge to make electric currents flow is only one of the roles charge plays in the matter and energy around us.
Everything from touchscreens to the cells in your body relies on the electric fields that get made whenever charges are separated. And once you've got those electric fields you can literally make light just by jiggling the electric charges around.
So what exactly is charge, and how does it do all that stuff?
^ to top
### As in love, so in electrostatics ...
Trying to define charge is like trying to define love - you can only really describe the effect it has on other things. And both charge and love rest heavily on the "opposites attract" rule.
In the atoms and molecules that make up all matter, there are only two players when it comes to charge: protons with a positive charge, and electrons with a negative charge. Their charges are opposite, so they attract. And their charges are equal (both have a charge of 1), so when they get together they balance out each other's charge, giving atoms an overall neutral (zero) charge.
There's a yang to all that attraction yin — positive charges repel other positives, and likewise for negative charges.
That's the accounting side of the charge story - now for the fun stuff: separating opposite charges to create electric fields.
^ to top
### Electric fields
The attraction between opposite charges means that they're usually found stuck together, and pulling them apart always takes energy.
Once they're apart that energy gets stored in an electric field around each of the individual charges. The field instantly spreads out from the charge in all directions at the speed of light, and literally goes on forever.
But fields get weak with distance so the charge's attractive/repulsive effect dies off pretty quickly. It's the electric field of energy around them that lets isolated charges attract and repel each other without ever touching.
The more negative and positive charges you separate, the stronger the electric field that forms between and around them, and the more electrical energy that can be tapped. That separation of charge is what creates voltage, or potential difference.
Batteries get their voltage oomph by having positive and negative charges separated - their positive and negative electrodes are separated by an insulating layer. Each time you use a battery, a bit more of the separated charge gets reunited via the circuit, and the electric field between the positive and negative electrodes gets a bit weaker. Rechargeable batteries pump the charges back to their separate sides during charging, strengthening the electric field all over again.
But batteries aren't the only things that can keep charges separated, and make use of the electric field in between them. Capacitors are like a kind of temporary battery - they keep positive and negative charges separated by an insulator, with an electric field between and around them.
Touchscreens in phones and tablets act like capacitors when they use electric fields to detect your fingers (more on that in the next story). And it's not just electronics that rely on capacitance - every living cell acts like a capacitor too.
Cells spend a lot of energy pumping charges (positive ions like potassium and sodium, and negative ions like chloride) through their cell membrane to make sure that inside the cell is more negative than outside (nerve cells do it the other way around - that's another story too!).
That separation of charge means that there's always an electric field, and therefore a voltage, across the cell's plasma membrane. It's called the transmembrane potential, and it controls the enzymes and gates in the membrane, so the cell can be in charge (zing!) of what's moving in and out (more on that in a further different story!).
Separating charges to create electric fields/voltages is pretty special all right, but it's nothing compared to what happens to that electric field when you make charges change speed or take a corner.
^ to top
### Wiggling charge lets there be light
So an individual electric charge will always have an electric field around it, spreading out forever (but getting pathetically weak after a short distance). And if you move that charge, the electric field will move with it. But the entire universe— wide field doesn't move instantaneously - the movement travels out through the field at the speed of light. Which isn't surprising, because bounces in electric fields are exactly what makes light. Every kind of light - from radio and visible light right through to X— rays and gamma rays.
If a charge (say an electron) moves up, the electric field moves up too - at the speed of light. If the charge moves down again, the field moves back down at the speed of light. No matter what speed the charge is moving at, its electric field will follow it at the speed of light, so it takes a tiny fraction of time for the whole field to catch up. And if you make the charge move up and down (or side to side, or back to front) in a nice steady rhythm, the electric field around it forms waves - like the proverbial ripples on a pond - moving out in every direction at the speed of light. There's one crest for every bounce of the electron.
And that's exactly half the story of how moving charges make electromagnetic radiation.
The other half involves the 'magnetic' bit. Whenever an electric field moves, it automatically creates a magnetic field that mimics its moves but at a 90 degree angle. So if the electric field is bobbing up, it'll create a magnetic field bobbing out to the side. When the electric field bounces down, a magnetic field will bounce out to the other side. (The magnetic field isn't shown in the diagram above, but there's no shortage of images online - including this flash animation of electromagnetic radiation from the University of Toronto or this cheap and cheerful EMR gif from Wikipedia.)
Together, those oscillating (ie bobbing) electric and magnetic fields caused by the jiggling of an electric charge are exactly what causes the radio and microwave signals that come out of antennas. The charges are loose electrons in the metal antenna that are forced to move side to side by an alternating current. (In fact every wire that's hooked up to AC power will give off some radio waves at the frequency of the power source - that's why electric appliances can interfere with tv and radio signals).
Other kinds of light - the higher frequency stuff - aren't formed by electrons in an antenna, but all light can be generated by moving charge. Visible light is given off when electrons jump from a higher energy level to lower one in an atom. X-rays are given off when electrons quickly decelerate on smashing into metals. And synchrotrons can generate any frequency of light from infrared right through to gamma rays, just by making the electrons in their beam wiggle more or less often.
^ to top
### The to-do list for charge
Holding matter together, governing the interactions of molecules, spitting out light when it hits a rhythmic stride - it all makes for a pretty impressive CV for charge. But there's one more thing it's doing right now - and it's kind of important. Charge is stopping your backside from falling straight through your chair. And your computer from falling through your hands or furniture.
The atoms that make up all matter are almost entirely empty space - even in solids. The only thing that stops backside atoms moving straight through chair atoms is the repulsion between the negatively charged electrons surrounding them both and all other atoms. Charge. In this universe, it really is everything.
Thanks to Dr Ben Buchler from the Department of Quantum Science at the Australian National University.
Tags: mathematics, physics
^ to top
Published 27 May 2013 | 1,693 | 8,385 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2019-35 | longest | en | 0.947038 |
https://mathoverflow.net/questions/409486/global-vogan-a-packet-is-infinite-set | 1,708,660,364,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474360.86/warc/CC-MAIN-20240223021632-20240223051632-00861.warc.gz | 411,080,460 | 24,463 | Global Vogan A-packet is infinite set?
For an cuspidal automorphic representation of general linear group, we can attach its global Vogan A-packet.
Though I thought that it is finite set, in some paper, it is written that there are infinitely many automorphic cuspidal representations in the associated packet.
How is it possible?
• What is the precise context in which it is asserted to be infinite? Nov 27, 2021 at 13:39
• @Kimball, Page 40 line 2~3 of the following paper. math.purdue.edu/~liu2053/S6.pdf Nov 27, 2021 at 14:50
• I'm pretty sure this is not for a general linear group $GL_n$, but for some other reductive group -- apparently $Sp_{2(m-n)}$. (For $GL_n$ I believe all $A$-packets are finite, because they're singletons!) In general a global packet is a product of local packets, and all the local guys are finite, but since the set of places is infinite, the global packet can be infinite -- you can construct explicit examples using restriction to $SL_2$ of automorphic reps of $GL_2$ (the packet has more than one element at any unramified place where the Hecke eigenvalue is 0). Nov 29, 2021 at 7:50
• @Loeffler, Thank you for the comment. May I ask some stupid quesiton? If $\pi$ is a unitary irreducible cuspidal automorphic representation of $GL(2)$, I am wondering whether there are infinitely many places $v$ such that $\pi_v$ is discrete series representation. Because the local A-packet for $Mp(2)$ associated to $\pi_v$ is singleton iff $\pi$ is not discrete series, if such places $v$, which makes $\pi_v$ discrete series, are infinitely many, then there are infinitely many cuspidal representations in the global Vogan A-packet for $Mp(2)$ associated to $\pi$. Nov 29, 2021 at 12:54
• @Loeffler, I think $S_{\pi}$, the set of places $v$ such that $\pi_v$ is square integrable, is finite because unramified representation can never be square-integrable. Since the number of elements of the global Vogan A-packet of $\pi$ for $Mp(2)$ is $2^{|S_{\pi}|}$, the number of global cuspidal representations in the packet seems also finite. Nov 29, 2021 at 21:42 | 563 | 2,086 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-10 | latest | en | 0.90685 |
https://metanumbers.com/8184 | 1,642,686,135,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301863.7/warc/CC-MAIN-20220120130236-20220120160236-00460.warc.gz | 464,702,482 | 7,432 | # 8184 (number)
8,184 (eight thousand one hundred eighty-four) is an even four-digits composite number following 8183 and preceding 8185. In scientific notation, it is written as 8.184 × 103. The sum of its digits is 21. It has a total of 6 prime factors and 32 positive divisors. There are 2,400 positive integers (up to 8184) that are relatively prime to 8184.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 4
• Sum of Digits 21
• Digital Root 3
## Name
Short name 8 thousand 184 eight thousand one hundred eighty-four
## Notation
Scientific notation 8.184 × 103 8.184 × 103
## Prime Factorization of 8184
Prime Factorization 23 × 3 × 11 × 31
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 6 Total number of prime factors rad(n) 2046 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 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 8,184 is 23 × 3 × 11 × 31. Since it has a total of 6 prime factors, 8,184 is a composite number.
## Divisors of 8184
1, 2, 3, 4, 6, 8, 11, 12, 22, 24, 31, 33, 44, 62, 66, 88, 93, 124, 132, 186, 248, 264, 341, 372, 682, 744, 1023, 1364, 2046, 2728, 4092, 8184
32 divisors
Even divisors 24 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 32 Total number of the positive divisors of n σ(n) 23040 Sum of all the positive divisors of n s(n) 14856 Sum of the proper positive divisors of n A(n) 720 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 90.4655 Returns the nth root of the product of n divisors H(n) 11.3667 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 8,184 can be divided by 32 positive divisors (out of which 24 are even, and 8 are odd). The sum of these divisors (counting 8,184) is 23,040, the average is 720.
## Other Arithmetic Functions (n = 8184)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 2400 Total number of positive integers not greater than n that are coprime to n λ(n) 60 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1032 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 2,400 positive integers (less than 8,184) that are coprime with 8,184. And there are approximately 1,032 prime numbers less than or equal to 8,184.
## Divisibility of 8184
m n mod m 2 3 4 5 6 7 8 9 0 0 0 4 0 1 0 3
The number 8,184 is divisible by 2, 3, 4, 6 and 8.
## Classification of 8184
• Arithmetic
• Abundant
### Expressible via specific sums
• Polite
• Practical
• Non-hypotenuse
## Base conversion (8184)
Base System Value
2 Binary 1111111111000
3 Ternary 102020010
4 Quaternary 1333320
5 Quinary 230214
6 Senary 101520
8 Octal 17770
10 Decimal 8184
12 Duodecimal 48a0
20 Vigesimal 1094
36 Base36 6bc
## Basic calculations (n = 8184)
### Multiplication
n×y
n×2 16368 24552 32736 40920
### Division
n÷y
n÷2 4092 2728 2046 1636.8
### Exponentiation
ny
n2 66977856 548146773504 4486033194356736 36713695662615527424
### Nth Root
y√n
2√n 90.4655 20.1522 9.51133 6.06168
## 8184 as geometric shapes
### Circle
Diameter 16368 51421.6 2.10417e+08
### Sphere
Volume 2.29607e+12 8.41669e+08 51421.6
### Square
Length = n
Perimeter 32736 6.69779e+07 11573.9
### Cube
Length = n
Surface area 4.01867e+08 5.48147e+11 14175.1
### Equilateral Triangle
Length = n
Perimeter 24552 2.90023e+07 7087.55
### Triangular Pyramid
Length = n
Surface area 1.16009e+08 6.45997e+10 6682.21
## Cryptographic Hash Functions
md5 451ae86722d26a608c2e174b2b2773f1 058595780c353db399bbf1e3381c2af16a5e6f58 0e5275f34e63038d1fb8a9ef94dce7e7dade73d03bd7b6125a73fcc43e9bdfc0 e5d69bd7bfcc425ccb3c42c88dbe1608494f7001dd98279ca8cfe6a0d6bc06389a05a9593756fcf6487ed1a2953a126c399edb92523cb574f90db34002dfb5ae 2535493c27cce9b6456dea728921bafd0740ce77 | 1,548 | 4,225 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2022-05 | latest | en | 0.786113 |
https://www.deeconometrist.nl/mathematics/the-pagerank-algorithm/ | 1,686,036,070,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652235.2/warc/CC-MAIN-20230606045924-20230606075924-00464.warc.gz | 773,591,206 | 43,453 | Select Page
The PageRank algorithm
#### Written by Stan Koobs
How did Google become so successful? What did the first search engine look like? The answer to these questions is related to a pretty basic mathematical concept which was revolutionary in that time. In this article I will tell you about the PageRank algorithm, which is the first and best known algorithm that Google uses.
You may have heard the name earlier, Larry Page, he is one of the founders of Google and also one of the richest men in the world. He studied computer science at the University of Michigan and at Stanford University. After that, he started his PhD in computer science and he was looking for a dissertation theme. In that time, the World Wide Web already existed and it was getting bigger and bigger. But a big difference with the World Wide Web of today is that it were basically only loose pages, which had some links to each other but there wasn’t really a structure. Page decided to focus on that and to explore the mathematical properties of the web, understanding its structure as a big graph. Finally, he developed the PageRank algorithm which is also named after him (and not after the English word page as you may have guessed when you started reading this article).
### What is the concept of this algorithm?
In this algorithm each page corresponds to a different URL. Because of this, most websites contain a lot of pages. But how can we order all these pages? Let us first take a look at the definition which Google itself gives to us:
##### “PageRank works by counting the number and quality of links to a page to determine a rough estimate of how important the website is. The underlying assumption is that more important websites are likely to receive more links from other websites.”
So to find the order of a few pages, we need to take a look at how the pages are linked with each other.
### A random walk through our own (simplified) website
To see how the pages are related, we will look at an example. Imagine that we are in a really simple web with just 5 webpages of www.deeconometrist.nl. Some of our articles are linked to other articles and that way you can go to another webpage. Let us use a bit of graph theory and let the nodes of the graph be the webpages and let the links between the pages be the arrows. In the picture below we see for example that it is possible to go from page D to page A but that it is not possible to go back (in one step).
Let us now imagine that there is someone who is reading articles at our simple website. We will assume that this person walks randomly through the graph. When this person is for example at webpage B, he can go to page A or to C and both events have a probability of a . Then we can put these probabilities in a matrix, and this gives us the following:
In this matrix, the probability (which is the element in the ith row of the jth column) corresponds to the probability of moving from station to station . The people who already have some basic linear algebra knowledge may have recognized this already, namely, this matrix is also a Markov matrix. This is because of the fact that the elements of each column sum to 1.
Let us now continue with the random walk of our person. We will first introduce the random variable which represents the page where the person is after steps. From our definition of , we know that this is the probability of going to state conditional on being in state , written down mathematically this means:
What is really interesting about this, is that the probability is independent of , it only depends on the current state! Another interesting insight is that the elements of matrix corresponds to the probabilities of being in state after 2 steps conditional on being in state at the start. So these probabilities are:
We will not go into the proof of this result but let us do a check for a simple case.
Consider element (1,1) of the matrix . This tells us that the probability of starting at webpage A and ending up at webpage A after 2 steps. In the first step we can only go to webpage B so this probability is equal to 1. From webpage B we can go to page A or page C, so the probability of ending up at A again after 2 steps is . And, as we see, this holds!
We are now at the point that it gets even more interesting. Let us compute for some big number . For example, take = 32, this gives the probabilities of being in state after 32 steps conditional on being in state at the start.
But what we see now, is that all the columns are actually the same! And when we compute for some bigger than 32 we get that the matrix is still the same. But that seems pretty strange, doesn’t it? This tells us that the probability of being in for example state A after a big amount of steps is 0.293, no matter at which state we start. Hence, after a big amount of steps, the probability of being in a certain state is independent of state where we started! This means that after some time, we reach a steady state which tells us about the fraction of people who visit a certain state in the long-run. Hence, we can now order our pages! So when somebody puts the name of our website in the search engine, the order in which we give him/her the pages is given by B, A, C, E, D.
### Usage today
The previous example showed how a pretty basic result from linear algebra gives us some really useful insights in the fraction of people who visit a certain webpage. When Larry Page developed this algorithm, it was definitely revolutionary. When other people saw how useful this algorithm was, they also wanted to use it but luckily for Larry Page, he already had a patent on it. You can read more about monopolies in the article of Sjors Keet next week, unfortunately it is not online yet so I can’t link to it to improve its PageRank… An interesting fact is that this patent is actually assigned to Stanford University. To get the exclusive license rights, Google paid 1.8 million shares of Google to Stanford University. In 2005, Stanford University sold these shares for an astonishing amount of \$336 million!
Of course, the algorithms that Google is using today are way more advanced than the algorithm that we just discussed. But what I personally like the most about this algorithm is that the mathematics which is being used is still pretty basic, however, it led to a huge breakthrough for search engines and to the birth of a commercial empire. This shows us the power of using mathematics outside of their “normal” context!
Sources:
Dit artikel is geschreven door Stan Koobs | 1,389 | 6,581 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2023-23 | longest | en | 0.977574 |
http://muktabodhalib.org/page/ms-excel-design-of-experiments-definition-88544480.html | 1,600,425,152,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400187390.18/warc/CC-MAIN-20200918092913-20200918122913-00472.warc.gz | 127,087,828 | 7,340 | More information on our guest author is given below. What is a Histogram? Hi, My Name is quentos, am really ineterested in some of your Excel Add-in for design of Experiment data Analysis and planning. Nov 28, All designs that have a distance value of 0 fit exactly the described problem with in this case a Latin Square Design. Control Chart Limits. Cost of Quality Worksheet. About us. A special thanks to him. Jul 8,
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## What Is Design of Experiments (DOE) ASQ
Tutorial on Design of Experiments and how to analyze these designs in Excel. Examples and software is included. In Design of Experiments, they talk about "confounding" which simply means that one factor affects another.
You'd expect a higher temperature to result in a.
What is a Control Chart? A special thanks to him.
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You'd expect a higher temperature to result in a shorter cooking time, and vice versa, but does a square pan take longer than a round one?
Types of Control Charts. Example This newsletter presents a simple statistical method for initial identification of potential critical process input variables that might have influences on the yield.
Sponge nail art glitter silver Central Composite. Ingo Dec 1, The actual run number represents the randomization of the runs.This month's newsletter is written by Nabil B. All factors are numerical values.
Design of Experiments deals with planning, conducting, analyzing and Design of experiments (DOE) is defined as a branch of applied statistics that deals with ASQ has created a design of experiments template (Excel) available for free.
For purposes of learning, using, or teaching design of experiments (DOE), one A free Microsoft Excel spreadsheet with a 2^3 Full Factorial array showing the define results that are close to your acceptance criteria by retesting the factor. The term experiment is defined as the systematic procedure carried out under You can use the MoreSteam's data analysis software EngineRoom for Excel to.
This necessitates a profound understanding of the drug product formulation and manufacturing processes.
Now the 90 experiments are carried out and the resulting distance is written in the corresponding Excel cell in the generated experimental design.
## Excel files for DOE (Design of Experiments)
Ingo Dec 1, Home About Us Store. Step 3. My Cart 0. This approach intrinsically cannot detect interactions between factors.
### Most Practical DOE Explained (with Template) iSixSigma
Ms excel design of experiments definition Balanced Scorecard. The results are on a yellow background in the file to find them easier. The distance of the ball will be analyzed in this study. Need a hand? A research scientist in a pharmaceutical company decided to screen some of the input factors for initial assessments for their potential effects on the pellets' yield of suitable quality. The knowledge gained from pharmaceutical development studies and manufacturing experiences provide scientific understanding to support the establishment of the design space.
16, Place your factors into columns of the experiment design (assign them to A,B,C The analysis will consist of a Means ANOVA and a Log(s) ANOVA for the.
### Factor effect (screening) design in Excel tutorial XLSTAT Support Center
Here is one Example of a D-Optimal 3 Factors on 3 Level in 15 runs with simulated data's. I have other DoE Excel Examples for teaching.
Response variable definition‒ this is measurable outcome of the experiment that Also, DOE design and analysis can be done easily in Microsoft Excel, using.
As its definition implies, the development of a design space involves full understanding of the statistical relationships between the input variables e.
Miner Forum Moderator Staff member. Define the Objective A research scientist in a pharmaceutical company decided to screen some of the input factors for initial assessments for their potential effects on the pellets' yield of suitable quality. Factor D has a coefficient of I have seen 1 or 2 DOE files in Excel in this forum, but they are very limited to a few basic full factorial designs.
Video: Ms excel design of experiments definition Design of Experiments: 2K How to Write All Factorial Effects, the Easy and Systematic Way
The purpose of initial factorial screening is to assess the effect of each input variable on the response variable.
Details of the calculations are given in the instruction manual accompanying the software.
SOLDER MASK CURING PROCESS CANNABIS
The knowledge gained from pharmaceutical development studies and manufacturing experiences provide scientific understanding to support the establishment of the design space.
Now the 90 experiments are carried out and the resulting distance is written in the corresponding Excel cell in the generated experimental design.
### Design of Experiments in Excel
Improvement Tools. The R Square value F values were then calculated based on this mean square error. Attachment List. My profile.
## 2 thoughts on “Ms excel design of experiments definition”
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Data are based on the common catapult example used in experimental design for teaching. | 1,118 | 5,583 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2020-40 | latest | en | 0.917943 |
https://adidaspureboost.org/4-number-addition-worksheet/ | 1,566,813,572,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027331485.43/warc/CC-MAIN-20190826085356-20190826111356-00261.warc.gz | 354,036,088 | 30,690 | In Free Printable Worksheets222 views
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Top | 1,465 | 7,191 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-35 | latest | en | 0.877618 |
https://biology.stackexchange.com/questions/80146/polydactyly-and-its-genetics/80169 | 1,566,409,435,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316150.53/warc/CC-MAIN-20190821174152-20190821200152-00091.warc.gz | 388,432,609 | 30,606 | # Polydactyly and its genetics
One of my friend was saying that her aunt (father's sibling sister) was having polydactyly but none of her cousins (her aunt has two daughters) is having it.
What is the chance that my friend's children inherit polydactyly?
(considering that her partner nor anyone in his generation exhibit polydactyly)
• if polydactyly is autosomal dominant (Wikipedia), then effectively zero. – Ben Bolker Dec 29 '18 at 21:53
Polydactyly can be dominant or recessive. that is
1. A single copy of the mutation/s involved result in the phenotype vs
2. Both copies of the gene must be mutated to result in the phenotype
When two random people who don't have polydactyly nor a family history of polydactyly have a child, there is around 1/548 (0.18%) chance that child will be polydactyly because the parents were carriers without knowing it, and a 1/11.7 (8.54%) chance the child is a carrier of a single copy
If the aunt's polydactyly is Dominant, then your friend's father hasn't inherited it (he isnt polydactyly) and just has the normal population chance of being a carrier.
In the case of Dominant polydactyly, assuming your friend has a partner with no history of polydactyly, your friend and her partner have the same chance as average of a polydactyly child (1/548)
If the aunt's polydactyly is recessive, meaning she has two copies one from each parent (you friend's grandparents), there is a 1/130 (0.76%)chance your friend's child will have polydactyly (4.2 times more likely), and a 1/4.5 (21.4%) chance of being a carrier (2.6 times more likely)
I'll quickly go through the working/ if you're interested:
Given that the grandparents are both carriers (needed for the aunt to have two copies), then your friend's father has a 2/3 chance of being a carrier of a single copy (1/3 chance no copies, 0/3 chance having two copies because he isn't polydactyly).
Your friend's mother has the usual 1/11.7 chance of being a carrier.
The chance of your friend inheriting a single copy is (chance passes from father) + (chance passes from mother) - (chance of getting two copies)
=0.6600/2 + 0.0854/2 - (0.33*0.0427)
=0.3727 - 0.0140
= 0.3586
## 35.9% chance your friend carries one copy
Your friend has 1/2.8 chance of carrying and 1/5.6 of passing it on
## Friend's partner has the 1/11.7 chance of being a carrier and the 1/23.4 chance of passing it on
Therefore their child has
=0.1793*0.0427 (chance friend passes on * chance partner passes on)
=0.0076
=1/130 chance of having two copies, having polydactyly
=0.1793 + 0.0427 -(0.1793*0.0427) (chance any copies - chance two copies)
=0.2143
=1/4.7 chance of one copy a 2.5 times increase risk
• That's a really detailed answer. Good one and thanks for simplifying the confusing parts. But can you add any reference? It will help me accept your answer. – user 33690 Dec 31 '18 at 18:16 | 829 | 2,882 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2019-35 | latest | en | 0.962869 |
https://shakyradunn.com/slide/2-mhd-equations-university-of-st-andrews-3numh9 | 1,604,153,583,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107918164.98/warc/CC-MAIN-20201031121940-20201031151940-00379.warc.gz | 469,651,959 | 10,705 | # 2. MHD Equations - University of St Andrews
2. MHD Equations 2.1 Introduction Many processes caused by magnetic field (B) Sun is NOT a normal gas Sun is in 4th state of matter ("PLASMA") behaves differently from normal gas: B and plasma -- coupled (intimate, subtle) B exerts force on plasma -- stores energy MOST of UNIVERSE is PLASMA: Ionosphere --> Sun (8 light mins)
Learn basic behaviour of plasma from Sun Magnetic Field Effects E.g., A Sunspot B exerts a force: -- creates intricate structure * ____________________* E.g., A Prominence Magnetic tube w. cool plasma B --> Thermal Blanket + Stability
What is global equilibrium? / fine structure ? *_______* E.g., a Coronal Mass Ejection QuickTime and a decompressor are needed to see this picture. *_________ _____ * E.g., A Solar Flare (from TRACE)
B stores energy converted to other forms QuickTime and a decompressor are needed to see this picture. _ _ _ _ _ _ _ _______ _ _ _ _ _ _ _* 2.2 Flux Tubes & Field Lines Magnetic Field Line -- Curve with tangent in direction of B. Equation: dx dy dz
In 2D: * _ _ _ _ _ _ * or in 3D: B = B = B x y z Magnetic Flux Tube -- Surface generated by set of field lines intersecting simple closed curve. Strength (F) -- magnetic flux crossing a section i.e., * _ _ _ _ _ _ _ * But .B = 0 --> No flux is created/destroyed inside flux tube
So F = B.dS is constant along tube Ex 2.1 Prove the above result that, if .B = 0 , then F = B.dS is constant along a flux tube. F = B.dS If cross-section is small, * _ _ _ _ _* B lines closer --> A smaller + B increases Thus, when sketching field lines, ensure they are closer when B is stronger
To sketch magnetic field lines: (i) Solve dy By = dx Bx (ii) Sketch one field line (iii) Sketch other field lines, remembering that B increases as the field lines become closer (iv) Put arrows on the field lines
EXAMPLE Sketch the field lines for Bx =y, By =x (i) Eqn. of field lines: *__________* (ii) Sketch a few field lines: ? arrows, spacing dy By = dx Bx (iii) Directions of arrows: (Bx =y, By =x)
(iv) Spacing (Bx =y, By =x) At origin B = 0.* _ _ _ _ _ _ _ _ _ _ _ _ _ _ * Magnetic reconnection & energy conversion **Examples Ex 2.2 Sketch the field lines for (a) By=x (b) Bx=1, By=x Ex 2.3 Sketch the field lines for (a) Bx=y, By=a2x (b) Bx=y, By=-a2x 2.3 Plasma Theory
-- the study of the interaction between a magnetic field and a plasma, treated as a continuous medium/set of pcles But there are different ways of modelling a plasma: (i) MHD -- fluid eqns + Maxwell (ii) 2-fluid-- electron/ion fluid eqns + Maxwell (iii) Kinetic -- distribution function for each species of particle Eqns of Magnetohydrodynamics Magnetohydrodynamics (MHD) Unification of Eqns of: (a) Maxwell
B / = j + D / t, .B = 0, E = B / t, .D = c , where B = H, D = E, E = j / . (b) Fluid Mechanics dv Motion = p, dt d Continuity + .v = 0, dt Perfect gas p = R T,
Energy eqn. ............. where d / dt = / t + v. or (D / Dt) In MHD 1. Assume v << c --> Neglect * _ _ _ * B/ = j 2. Extra E on plasma moving E +
(1) * _ _ _ _* = j/ (2) 3. Add magnetic force dv dt = p + * _ _ _ _ * Eliminate E and j: take curl (2), use (1) for j
2.4 Induction Equation B = E = (v B j / ) t = (v B) ( B) 2 = (v B) + B, where _ _ _ _ _ _ is magnetic diffusivity Describes: how B moves with plasma / diffuses through it N.B. In MHD, v and B are * _ _ _ _ _ _ _ _ _ _*: Induction eqn + eqn of motion B
t dv dt = (v B) = p + 2 + B jB
--> basic processes j = B / and E = v B + j/ are secondary variables INDUCTION EQUATION B t = (v B) 2
+ B I II B changes due to transport + diffusion I II L0 v0 = = R -- * _ _ _ _ _ _ _ ______*
2 = 1 m /s, L0 = 105 m, v0 = 103 m/s --> Rm = 108 eg, I >> II in most of Solar System --> B frozen to plasma -- keeps its energy Except Reconnection -- j & B large (a) If Rm << 1 The induction equation reduces to
B 2 t = B B is governed by a diffusion equation --> field variations on a scale L0 diffuse away on time * _ _ _ _ _* with speed v d = L0 /t d = L0 E.g.: sunspot ( = 1 m2/s, L0 = 106 m), td = 1012 sec;
8 17 for whole Sun (L = 7x10 m), t = 5x10 sec 0 d (b) If Rm >> 1 The induction equation reduces to
B t = (v B) and Ohm's law --> E + vB = 0 Magnetic field is * _ _ _ _ _ _ _ _ _ _ _ _ _* Magnetic Flux Conservation: Magnetic Field Line Conservation: 2.5 EQUATION of MOTION dv
= p + dt (1) (2) (2) (i) = (3) j B + g (3) (4) p = 2 B / (2 ) * _ _ _ _ _ _ _ *
When <<1, j B do inate B * _ _ _ _ _ _* (ii) (1) (3) vvA = Typical Values on Sun Photosphere Corona N (m-3) 1023
1015 T (K) 6000 106 B (G) 5 - 103 10 106 - 1
10-3 vA (km/s) 0.05 - 10 103 [N (m-3) = 106 N (cm-3), B (G) = 104 B (tesla) = 3.5 x 10 -21 N T/B2, vA = 2 x 109 B/N1/2] Magnetic force: jB
B = ( B) B = (B.) B 2 2 Magnetic field lines have a Tension B / ----> * _ _ _ _ _ _ _ _ _ _* Pressure B2/(2 )----> * _ _ _ _ _ _ _ _ _ _ * 2
*EXAMPLE 2 B B j B = (B.) 2 2 Bx B + By ) (j B)x = (Bx
x y x 2 **Examples Find Magnetic Pressure force, Magnetic Tension force and j x B force for Ex 2.4 (a) B = x y (b) B = y x + x y Hydrostatic Equilibrium dv
= p + dt (1) (2) j B + g (3) (4) In most of corona, (3) dominates Along B, (3) = 0, so (2) + (4) important (2) p0 / L0 = >> 1 for
(4) 0 g L0 << H = p0 0 g * _ _ _ _ _ _ _ _ _* Example Suppose g = - g z MHS Eqm. along B: dp = g,
dz wee = p / (R T ). dp p RT So = , H = . dz H g T =cont p = p0 e z / H
p = p0 e z / H On Earth H = 9 km, so on munro (1 km) p = 0.9 p0 or on Everest (9 km) p = 0.37 p0 T = 5000 K, H = * _ _ _ _ _*; T = 2 x 106 K, H = * _ _ _ _ _ * When is MHD valid ? = constant, We assumed in deriving MHD eqns -- v<
and plasma continuous Can treat plasma as a continuous medium when 2 1 n T L >> pf =300 6 17 3 k 10 K 10 Chromosphere Corona When L < pf collide with B 4
20 (T = 10 , n = 10 ) L >> 3 cm 6 16 (T =10 , n =10 ) L >> 30 k MHD can still be valid when particles L > i =vj / (eB) (ion gyoadiu) ri = 1 m(corona) MHD equations can be derived by taking integrals of a kinetic equation for particles (but tricky)
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• Role of Raja Rammohun Roy Library Foundation and National Mission on Libraries in promoting public library system & service in India . presented by. Shri . Soumen. Sarkar. Deputy Director(II&M), RRRLF Knowledge for all at their doorsteps | 2,517 | 8,031 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2020-45 | latest | en | 0.822834 |
http://www.ask.com/science/deep-fathom-water-240b2a515bbca064 | 1,436,102,510,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375097475.46/warc/CC-MAIN-20150627031817-00249-ip-10-179-60-89.ec2.internal.warc.gz | 327,999,840 | 24,099 | Q:
# How deep is a fathom of water?
A:
A fathom of water is 6 feet deep. Originally the term came from the Anglo-Saxon word "faetm," which means "to embrace." Back then, measurements were calculated using the average size of parts of the body, so a fathom was the distance between a man's outstretched arms from fingertip to fingertip.
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## Keep Learning
Since a man would outstretch his arms to embrace his loved ones upon arriving home, it became a common measurement. Britain's Parliament declared the fathom to be an official unit of measurement.
The word "fathom" can also be used as a verb, describing an attempt to figure out a difficult problem or understand a particularly perplexing person.
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https://math.stackexchange.com/questions/2476690/limit-of-the-difference-between-two-modified-bessel-functions | 1,585,399,374,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370491857.4/warc/CC-MAIN-20200328104722-20200328134722-00378.warc.gz | 591,548,297 | 31,384 | Limit of the difference between two modified Bessel functions
I am wondering, if it is possible to find the limit of
$$\lim_{r\to0}\left(\frac{1}{\mu_1-1}\,\mathrm{K}_0(\mathrm{j}\,k_1\,r)-\frac{1}{\mu_2-1}\,\mathrm{K}_0(-\mathrm{j}\,k_2\,r)\right)$$
with $\mu_1\,k_1^2=\mu_2\,k_2^2$ and $k_1=\sqrt{k_1^2}$ and $k_2=\sqrt{k_2^2}$ and $\mathrm{Im}[k_1]<0$ and $\mathrm{Im}[k_2]>0$.
I am pretty sure that there should be a limit different from $0$. (From a physical point of view)
I tried the limiting form of the modified Bessel functions see here $$\mathrm{K}_0(z)\approx -\ln(z)$$
but I did not get any useful results. It needs to be mentioned, that $k_1$ and $k_2$ are not idenpendent from each other. So I am also happy with a solution having a restriction on $k_1$ and $k_2$.
I would have the following idea. Replacing $\mathrm{K}_0(z)$ with $-\ln(z)$ for small $z$ leads to
$$\lim_{r\to0}\left(-\frac{1}{\mu_1-1}\,\ln(\mathrm{j}\,k_1\,r)+\frac{1}{\mu_2-1}\,\ln(-\mathrm{j}\,k_2\,r)\right)$$
Using some logarithmic identities leads to
\begin{align}\lim_{r\to0}\left(\ln\left((\mathrm{j}\,k_1\,r)^{-\frac{1}{\mu_1-1}}\right)+\ln\left((-\mathrm{j}\,k_2\,r)^{\frac{1}{\mu_2-1}}\right)\right)=\\ \lim_{r\to0}\ln\left((\mathrm{j}\,k_1\,r)^{-\frac{1}{\mu_1-1}}\,(-\mathrm{j}\,k_2\,r)^{\frac{1}{\mu_2-1}}\right)=\\ \lim_{r\to0}\ln\left((\mathrm{j}\,k_1)^{-\frac{1}{\mu_1-1}}\,(-\mathrm{j}\,k_2)^{\frac{1}{\mu_2-1}}\,r^{-\frac{1}{\mu_1-1}+\frac{1}{\mu_2-1}}\right)=\\ \ln\left((\mathrm{j}\,k_1)^{-\frac{1}{\mu_1-1}}\right)+\ln\left((-\mathrm{j}\,k_2)^{\frac{1}{\mu_2-1}}\right)+\lim_{r\to0}\ln\left(r^{-\frac{1}{\mu_1-1}+\frac{1}{\mu_2-1}}\right) \end{align}
Therefore, it seems to me, that a limit only exists if
$$-\frac{1}{\mu_1-1}+\frac{1}{\mu_2-1}=0$$
which results in $\mu_1=\mu_2$. Are there any errors in this approach? Because I accutally hoped for a different result. | 813 | 1,886 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2020-16 | latest | en | 0.579635 |
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• Aug 10th 2009, 07:18 AM
masters
Logic Puzzle
A survey of a group of 116 tourists was taken in St. Louis. The survey showed the following:
64 of the tourists plan to visit Gateway Arch;
49 plan to visit the zoo;
11 plan to visit the Art Museum and the zoo, but not the Gateway Arch;
14 plan to visit the Art Museum and the Gateway Arch, but not the zoo;
16 plan to visit the Gateway Arch and the zoo, but not the Art Museum;
9 plan to visit the Art Museum, the zoo, and the Gateway Arch;
16 plan to visit none of the three places.
How many plan to visit the Art Museum only?
• Aug 10th 2009, 08:24 AM
AlephZero
OK... did some Venn diagrams. I think the answer is that 12 people are just going to the museum?
EDIT: My solution... 16 people are total slackers, so we'll throw them out, and say that the universe is 100 people. Let's start with the Arch. We have 9 people doing all three things, 14 people also going to the Museum, and 16 also going to the Zoo; these are mutually exclusive. So the arch total is 64, subtracting 14, 16, and 9, we have 25 people just visiting the Arch. Now for the Zoo. Again, 9 are doing all three, 11 are also going to the Museum, and 16 are also going to the Arch; all of which are mutually exclusive. So the Zoo total is 49, subtracting 16, 9, and 11, we have that 13 are just going to the Zoo. Finally, we have a universe of 100 people. 25 are just going to the Arch, and 13 are just going to the Zoo, 14 are doing the Museum and the Arch, 16 the Arch and Zoo, 11 the Museum and Zoo, and 9 people doing all three. Which, by my count, leaves 12 people just going to the Museum.
I think...?
• Aug 10th 2009, 10:16 AM
Zarathustra
Here's my approach:
Spoiler:
Throw out the 16 who don't visit any place. There are seven different combinations (all three (1x), two (3x), only one (3x)). 14 + 9 + 16 tourists have plans to visit the Gateway Arch plus some other place. In sum, 64 plan to visit the Gateway Arch, so 64 -14 -9 -16 = 25 visit the Gateway Arch and only the Gateway Arch. Same approach for the zoo yields 13 tourists who visit the zoo and only the zoo. For the seven different cominations, we now know the results for six of them. It has to add up to 100. That leaves those who visit the museum and only the museum: 100 - 14 - 11 -9 - 16 - 25 - 13 = 12. Or so I think.
• Aug 10th 2009, 01:00 PM
masters
Alzo Sparch Zarathustra. You are correct.
And AlephZero, so are you.
After I posted it, I realized it probably wasn't that challenging a'tall.
I like the way you threw out the 16 slackers, AlephZero. Get rid of the bums. | 730 | 2,588 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2018-22 | latest | en | 0.964822 |
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$Example 1:Graph the circle x^{2}+y^{2}=4$
$4. Express each limit in terms of e :(ix) \operatorname{Lim}_{x \rightarrow \infty}\left(\frac{x}{1+x}\right)^{x}$
$1. Evaluate each limit by using theorems of limits:(v) \operatorname{Lim}_{x \rightarrow 2}\left(\sqrt{x^{3}+1}-\sqrt{x^{2}+5}\right)$
Example 1: Determine whether \operatorname{Lim}_{x \rightarrow 2} f(x) and \operatorname{Lim}_{x \rightarrow 4} f(x) exist when$f(x)=\left\{\begin{array}{rcc}2 x+1 & \text { if } & 0 \leq x \leq 2 \\7-x & \text { if } & 2 \leq x \leq 4 \\x & \text { if } & 4 \leq x \leq 6\end{array}\right.$
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The volume of a rectangular prism is given by the formula V = lwh, where l is the length of the prism, w is the width, and h is the height. Suppose a box in the shape of a rectangular prism has length (2a + 11), width (5a – 12), and height (a + 6). Which expression represents the volume of the box?
Mathematics
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SOLVED
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https://fuchsia.googlesource.com/third_party/github.com/gonum/gonum/+/refs/tags/v0.8.0/lapack/testlapack/dormhr.go | 1,714,006,088,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296820065.92/warc/CC-MAIN-20240425000826-20240425030826-00411.warc.gz | 236,801,681 | 4,567 | blob: d3bdd139eb379bc936b5b8a12056c29f2cc26173 [file] [log] [blame]
// Copyright ©2016 The Gonum Authors. All rights reserved. // Use of this source code is governed by a BSD-style // license that can be found in the LICENSE file. package testlapack import ( "fmt" "math" "testing" "golang.org/x/exp/rand" "gonum.org/v1/gonum/blas" "gonum.org/v1/gonum/blas/blas64" ) type Dormhrer interface { Dormhr(side blas.Side, trans blas.Transpose, m, n, ilo, ihi int, a []float64, lda int, tau, c []float64, ldc int, work []float64, lwork int) Dgehrder } func DormhrTest(t *testing.T, impl Dormhrer) { rnd := rand.New(rand.NewSource(1)) for _, side := range []blas.Side{blas.Left, blas.Right} { for _, trans := range []blas.Transpose{blas.NoTrans, blas.Trans} { for _, m := range []int{1, 2, 3, 4, 5, 8, 9, 10, 23} { for _, n := range []int{1, 2, 3, 4, 5, 8, 9, 10, 23} { for _, extra := range []int{0, 1, 13} { for cas := 0; cas < 10; cas++ { nq := m if side == blas.Right { nq = n } ilo := rnd.Intn(nq) ihi := rnd.Intn(nq) if ilo > ihi { ilo, ihi = ihi, ilo } testDormhr(t, impl, side, trans, m, n, ilo, ihi, extra, true, rnd) testDormhr(t, impl, side, trans, m, n, ilo, ihi, extra, false, rnd) } } } } } } for _, side := range []blas.Side{blas.Left, blas.Right} { for _, trans := range []blas.Transpose{blas.NoTrans, blas.Trans} { testDormhr(t, impl, side, trans, 0, 0, 0, -1, 0, true, rnd) testDormhr(t, impl, side, trans, 0, 0, 0, -1, 0, false, rnd) } } } func testDormhr(t *testing.T, impl Dormhrer, side blas.Side, trans blas.Transpose, m, n, ilo, ihi, extra int, optwork bool, rnd *rand.Rand) { const tol = 1e-14 var nq, nw int switch side { case blas.Left: nq = m nw = n case blas.Right: nq = n nw = m } // Compute the elementary reflectors and tau. a := randomGeneral(nq, nq, nq+extra, rnd) var tau []float64 if nq > 1 { tau = nanSlice(nq - 1) } work := nanSlice(max(1, nq)) // Minimum work for Dgehrd. impl.Dgehrd(nq, ilo, ihi, a.Data, a.Stride, tau, work, len(work)) // Construct Q from the elementary reflectors in a and from tau. q := eye(nq, nq) qCopy := eye(nq, nq) for j := ilo; j < ihi; j++ { h := eye(nq, nq) v := blas64.Vector{ Inc: 1, Data: make([]float64, nq), } v.Data[j+1] = 1 for i := j + 2; i < ihi+1; i++ { v.Data[i] = a.Data[i*a.Stride+j] } blas64.Ger(-tau[j], v, v, h) copy(qCopy.Data, q.Data) blas64.Gemm(blas.NoTrans, blas.NoTrans, 1, qCopy, h, 0, q) } c := randomGeneral(m, n, n+extra, rnd) // Compute the product of Q and C explicitly. qc := randomGeneral(m, n, n+extra, rnd) if side == blas.Left { blas64.Gemm(trans, blas.NoTrans, 1, q, c, 0, qc) } else { blas64.Gemm(blas.NoTrans, trans, 1, c, q, 0, qc) } // Compute the product of Q and C using Dormhr. if optwork { work = nanSlice(1) impl.Dormhr(side, trans, m, n, ilo, ihi, nil, a.Stride, nil, nil, c.Stride, work, -1) work = nanSlice(int(work[0])) } else { work = nanSlice(max(1, nw)) } impl.Dormhr(side, trans, m, n, ilo, ihi, a.Data, a.Stride, tau, c.Data, c.Stride, work, len(work)) // Compare the two answers. prefix := fmt.Sprintf("Case side=%c, trans=%c, m=%v, n=%v, ilo=%v, ihi=%v, extra=%v, optwork=%v", side, trans, m, n, ilo, ihi, extra, optwork) if !generalOutsideAllNaN(c) { t.Errorf("%v: out-of-range write to C\n%v", prefix, c.Data) } for i := 0; i < m; i++ { for j := 0; j < n; j++ { cij := c.Data[i*c.Stride+j] qcij := qc.Data[i*qc.Stride+j] if math.Abs(cij-qcij) > tol { t.Errorf("%v: unexpected value of the QC product at [%v,%v]: want %v, got %v", prefix, i, j, qcij, cij) } } } } | 1,322 | 3,479 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-18 | latest | en | 0.343225 |
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#1
12-28-2005, 09:14 PM
Guest Posts: n/a
Odds with suited aces
I have been playing suited aces (e.g., A [img]/images/graemlins/heart.gif[/img]6 [img]/images/graemlins/heart.gif[/img]) for cheap pre-flop. My plan is to jettison the hand if I don't flop either two pair or a flush-draw or better. I have been having trouble figuring the odds of this occurrence. I know that the odds of flopping a flush draw and of flopping two pair with unpaired cards, but I am having trouble figuring the odds of hitting either (a flush, trips, two pair, or a flush draw). I'm pretty sure that I am double counting some of my hoped for outcomes. Any help that I can get would be great.
Vince
#2
12-28-2005, 09:50 PM
ohnonotthat Senior Member Join Date: Aug 2005 Location: New Jersey - near A.C. Posts: 511
Re: Odds with suited aces
Flopped full house = 18
Flopped flush = 165
Flopped [flush] draw = 2,145 (Approximately 1 in 6 of these will include a pair on the board . . . 9h-3h-3s, however some will give YOU a pair and a draw . . . Ac-9h-3h for example).
Flopped trip 6s = 132
Flopped trip As = 132 (Not much of a flop to A-6)
Flopped two-pair = 396 (This refers only to flops of A-6-X. Flops such as A-2-2 are not counted as these two pair are not what you're aiming for - doubly true for A-7-7)
*
Bottom line is you'll get a flop you either like, love, or worship approximately 1 time in 7.
*
Note that the threat of being counterfeited when you do flop two pair is worthy of consideration. This will occur just over 4% of the time if the third flop card is below your "6" and ~ 12.5% of the time if the odd card is a above your "6".
Leading to the caveat, if you don't have the extra bonus of making a straight (A-2,3,4,5) the higher the better; A-9 is quite a bit better than A-6 - for this in addition to other reasons. The difference is not as pronounced as with A-K vs A-J but there is a notable diference.
#3
12-28-2005, 10:01 PM
Guest Posts: n/a
Re: Odds with suited aces
Thanks for the response, Ohno. I was trying to figure it out with factorials and was running into problems. I should have just looked at the ways that each given outcome could occur.
You also bring up great points about a flopped two-pair getting counterfeited. I am pretty good at getting away from those, though.
#4
12-28-2005, 10:59 PM
johnsy Junior Member Join Date: Jan 2004 Posts: 7
Re: Odds with suited aces
The key to playing these hands is playing them well.....
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https://handwiki.org/wiki/Physics:Kinetic_theory_of_gases | 1,657,024,301,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104576719.83/warc/CC-MAIN-20220705113756-20220705143756-00523.warc.gz | 333,482,128 | 28,638 | # Physics:Kinetic theory of gases
Short description: Historical physical model of gases
The temperature of the ideal gas is proportional to the average kinetic energy of its particles. The size of helium atoms relative to their spacing is shown to scale under 1950 atmospheres of pressure. The atoms have a certain, average speed, slowed down here two trillion fold from that at room temperature.
The kinetic theory of gases is a simple, historically significant classical model of the thermodynamic behavior of gases, with which many principal concepts of thermodynamics were established. The model describes a gas as a large number of identical submicroscopic particles (atoms or molecules), all of which are in constant, rapid, random motion. Their size is assumed to be much smaller than the average distance between the particles. The particles undergo random elastic collisions between themselves and with the enclosing walls of the container. The basic version of the model describes the ideal gas, and considers no other interactions between the particles.
The kinetic theory of gases explains the macroscopic properties of gases, such as volume, pressure, and temperature, as well as transport properties such as viscosity, thermal conductivity and mass diffusivity. The model also accounts for related phenomena, such as Brownian motion.
Historically, the kinetic theory of gases was the first explicit exercise of the ideas of statistical mechanics.
## History
In about 50 BCE, the Roman philosopher Lucretius proposed that apparently static macroscopic bodies were composed on a small scale of rapidly moving atoms all bouncing off each other.[1] This Epicurean atomistic point of view was rarely considered in the subsequent centuries, when Aristotlean ideas were dominant.
Hydrodynamica front cover
In 1738 Daniel Bernoulli published Hydrodynamica, which laid the basis for the kinetic theory of gases. In this work, Bernoulli posited the argument, that gases consist of great numbers of molecules moving in all directions, that their impact on a surface causes the pressure of the gas, and that their average kinetic energy determines the temperature of the gas. The theory was not immediately accepted, in part because conservation of energy had not yet been established, and it was not obvious to physicists how the collisions between molecules could be perfectly elastic.[2]:36–37
Other pioneers of the kinetic theory, whose work was also largely neglected by their contemporaries, were Mikhail Lomonosov (1747),[3] Georges-Louis Le Sage (ca. 1780, published 1818),[4] John Herapath (1816)[5] and John James Waterston (1843),[6] which connected their research with the development of mechanical explanations of gravitation. In 1856 August Krönig created a simple gas-kinetic model, which only considered the translational motion of the particles.[7]
In 1857 Rudolf Clausius developed a similar, but more sophisticated version of the theory, which included translational and, contrary to Krönig, also rotational and vibrational molecular motions. In this same work he introduced the concept of mean free path of a particle.[8] In 1859, after reading a paper about the diffusion of molecules by Clausius, Scottish physicist James Clerk Maxwell formulated the Maxwell distribution of molecular velocities, which gave the proportion of molecules having a certain velocity in a specific range.[9] This was the first-ever statistical law in physics.[10] Maxwell also gave the first mechanical argument that molecular collisions entail an equalization of temperatures and hence a tendency towards equilibrium.[11] In his 1873 thirteen page article 'Molecules', Maxwell states: "we are told that an 'atom' is a material point, invested and surrounded by 'potential forces' and that when 'flying molecules' strike against a solid body in constant succession it causes what is called pressure of air and other gases."[12] In 1871, Ludwig Boltzmann generalized Maxwell's achievement and formulated the Maxwell–Boltzmann distribution. The logarithmic connection between entropy and probability was also first stated by Boltzmann.
At the beginning of the 20th century, however, atoms were considered by many physicists to be purely hypothetical constructs, rather than real objects. An important turning point was Albert Einstein's (1905)[13] and Marian Smoluchowski's (1906)[14] papers on Brownian motion, which succeeded in making certain accurate quantitative predictions based on the kinetic theory.
## Assumptions
The application of kinetic theory to ideal gases makes the following assumptions:
• The gas consists of very small particles. This smallness of their size is such that the sum of the volume of the individual gas molecules is negligible compared to the volume of the container of the gas. This is equivalent to stating that the average distance separating the gas particles is large compared to their size, and that the elapsed time of a collision between particles and the container's wall is negligible when compared to the time between successive collisions.
• The number of particles is so large that a statistical treatment of the problem is well justified. This assumption is sometimes referred to as the thermodynamic limit.
• The rapidly moving particles constantly collide among themselves and with the walls of the container. All these collisions are perfectly elastic, which means the molecules are perfect hard spheres.
• Except during collisions, the interactions among molecules are negligible. They exert no other forces on one another.
Thus, the dynamics of particle motion can be treated classically, and the equations of motion are time-reversible.
As a simplifying assumption, the particles are usually assumed to have the same mass as one another; however, the theory can be generalized to a mass distribution, with each mass type contributing to the gas properties independently of one another in agreement with Dalton's Law of partial pressures. Many of the model's predictions are the same whether or not collisions between particles are included, so they are often neglected as a simplifying assumption in derivations (see below).[15]
More modern developments relax these assumptions and are based on the Boltzmann equation. These can accurately describe the properties of dense gases, because they include the volume of the particles as well as contributions from intermolecular and intramolecular forces as well as quantized molecular rotations, quantum rotational-vibrational symmetry effects, and electronic excitation.[16]
## Equilibrium properties
### Pressure and kinetic energy
In the kinetic theory of gases, the pressure is assumed to be equal to the force (per unit area) exerted by the atoms hitting and rebounding from the gas container's surface. Consider a gas of a large number N of molecules, each of mass m, enclosed in a cube of volume V = L3. When a gas molecule collides with the wall of the container perpendicular to the x axis and bounces off in the opposite direction with the same speed (an elastic collision), the change in momentum is given by: $\displaystyle{ \Delta p = p_{i,x} - p_{f,x} = p_{i,x} - (-p_{i,x}) = 2 p_{i,x} = 2 mv_x, }$ where p is the momentum, i and f indicate initial and final momentum (before and after collision), x indicates that only the x direction is being considered, and $\displaystyle{ v_x }$ is the speed of the particle in the x direction (which is the same before and after the collision).
The particle impacts one specific side wall once during the time interval $\displaystyle{ \Delta t }$ $\displaystyle{ \Delta t = \frac{2L}{v_x}, }$ where L is the distance between opposite walls.
The force of this particle's collision with the wall is $\displaystyle{ F = \frac{\Delta p}{\Delta t} = \frac{m v_x^2}{L}. }$
The total force on the wall due to collisions by molecules impacting the walls with a range of possible values of $\displaystyle{ v_x }$ is $\displaystyle{ F = \frac{Nm\overline{v_x^2}}{L}, }$ where the bar denotes an average over the possible velocities of the N particles.
Since the motion of the particles is random and there is no bias applied in any direction, the average squared speed in each direction is identical: $\displaystyle{ \overline{v_x^2} = \overline{v_y^2} = \overline{v_z^2}. }$
By the Pythagorean theorem, in three dimensions the average squared speed $\displaystyle{ v^2 }$ is given by $\displaystyle{ \overline{v^2} = \overline{v_x^2} + \overline{v_y^2} + \overline{v_z^2}, }$
Therefore $\displaystyle{ \overline{v^2} = 3\overline{v_x^2}. }$ and $\displaystyle{ \overline{v_x^2} = \frac{\overline{v^2}}{3}, }$
and so the force can be written as $\displaystyle{ F = \frac{Nm\overline{v^2}}{3L}. }$
This force is exerted uniformly on an area L2. Therefore, the pressure of the gas is $\displaystyle{ P = \frac{F}{L^2} = \frac{Nm\overline{v^2}}{3V}, }$ where V = L3 is the volume of the box.
In terms of the translational kinetic energy K of the gas, since $\displaystyle{ K = N\frac{1}{2} m\overline{v^2} }$ we have $\displaystyle{ PV = \frac{2}{3} K. }$
This is an important, non-trivial result of the kinetic theory because it relates pressure, a macroscopic property, to the translational kinetic energy of the molecules, which is a microscopic property.
### Temperature and kinetic energy
Rewriting the above result for the pressure as $\displaystyle{ PV = \frac{Nm\overline{v^2}}{3} }$, we may combine it with the ideal gas law
$\displaystyle{ PV = N k_\mathrm{B} T , }$
(1)
where $\displaystyle{ k_\mathrm{B} }$ is the Boltzmann constant and $\displaystyle{ T }$ the absolute temperature defined by the ideal gas law, to obtain
$\displaystyle{ k_\mathrm{B} T = {m\overline{v^2}\over 3}, }$ which leads to a simplified expression of the average kinetic energy per molecule,[17] $\displaystyle{ \frac{1}{2} m \overline{v^2} = \frac{3}{2} k_\mathrm{B} T. }$ The kinetic energy of the system is $\displaystyle{ N }$ times that of a molecule, namely $\displaystyle{ K = \frac{1}{2} N m \overline{v^2} }$. Then the temperature $\displaystyle{ T }$ takes the form
$\displaystyle{ T = { m \overline{v^2} \over 3 k_\mathrm{B} } }$
(2)
which becomes
$\displaystyle{ T = \frac{2}{3} \frac{K}{N k_\mathrm{B} }. }$
(3)
Equation (3) is one important result of the kinetic theory: The average molecular kinetic energy is proportional to the ideal gas law's absolute temperature. From equations (1) and (3), we have
$\displaystyle{ PV = \frac{2}{3} K. }$
(4)
Thus, the product of pressure and volume per mole is proportional to the average (translational) molecular kinetic energy.
Equations (1) and (4) are called the "classical results", which could also be derived from statistical mechanics; for more details, see:[18]
Since there are $\displaystyle{ 3N }$ degrees of freedom in a monatomic-gas system with $\displaystyle{ N }$ particles, the kinetic energy per degree of freedom per molecule is
$\displaystyle{ \frac {K} {3 N} = \frac {k_\mathrm{B} T} {2} }$
(5)
In the kinetic energy per degree of freedom, the constant of proportionality of temperature is 1/2 times Boltzmann constant or R/2 per mole. This result is related to the equipartition theorem.
Thus the kinetic energy per Kelvin of one mole of (monatomic ideal gas) is 3 [R/2] = 3R/2. Thus the kinetic energy per Kelvin can be calculated easily:
• per mole: 12.47 J / K
• per molecule: 20.7 yJ / K = 129 μeV / K
At standard temperature (273.15 K), the kinetic energy can also be obtained:
• per mole: 3406 J
• per molecule: 5.65 zJ = 35.2 meV.
Although monatomic gases have 3 (translational) degrees of freedom per atom, diatomic gases should have 6 degrees of freedom per molecule (3 translations, two rotations, and one vibration). However, the lighter diatomic gases (such as diatomic oxygen) may act as if they have only 5 due to the strongly quantum-mechanical nature of their vibrations and the large gaps between successive vibrational energy levels. Quantum statistical mechanics is needed to accurately compute these contributions. [19]
### Collisions with container wall
For an ideal gas in equilibrium, the rate of collisions with the container wall and velocity distribution of particles hitting the container wall can be calculated[20] based on naive kinetic theory, and the results can be used for analyzing effusive flow rates, which is useful in applications such as the gaseous diffusion method for isotope separation.
Assume that in the container, the number density (number per unit volume) is $\displaystyle{ n=N/V }$ and that the particles obey Maxwell's velocity distribution: $\displaystyle{ f_\text{Maxwell}(v_x,v_y,v_z) \, dv_x \, dv_y \, dv_z = \left(\frac{m}{2 \pi k T}\right)^{3/2} e^{- \frac{mv^2}{2k_BT}} \, dv_x \, dv_y \, dv_z }$
Then for a small area $\displaystyle{ dA }$ on the container wall, a particle with speed $\displaystyle{ v }$ at angle $\displaystyle{ \theta }$ from the normal of the area $\displaystyle{ dA }$, will collide with the area within time interval $\displaystyle{ dt }$, if it is within the distance $\displaystyle{ vdt }$ from the area $\displaystyle{ dA }$. Therefore, all the particles with speed $\displaystyle{ v }$ at angle $\displaystyle{ \theta }$ from the normal that can reach area $\displaystyle{ dA }$ within time interval $\displaystyle{ dt }$ are contained in the tilted pipe with a height of $\displaystyle{ v\cos (\theta) dt }$ and a volume of $\displaystyle{ v\cos (\theta) dAdt }$.
The total number of particles that reach area $\displaystyle{ dA }$ within time interval $\displaystyle{ dt }$ also depends on the velocity distribution; All in all, it calculates to be:$\displaystyle{ n v \cos(\theta) \, dA\, dt \times\left(\frac{m}{2 \pi k_BT}\right)^{3/2} e^{- \frac{mv^2}{2k_BT}} \left( v^2 \sin(\theta) \, dv \, d\theta \, d\phi \right). }$
Integrating this over all appropriate velocities within the constraint $\displaystyle{ v\gt 0,0\lt \theta\lt \pi/2,0\lt \phi\lt 2\pi }$ yields the number of atomic or molecular collisions with a wall of a container per unit area per unit time: $\displaystyle{ J_\text{collision} =\frac{\int_0^{\pi/2}\cos \theta \sin \theta d\theta}{\int_0^{\pi}\sin \theta d\theta}\times n \bar v= \frac{1}{4}n \bar v = \frac{n}{4} \sqrt{\frac{8 k_\mathrm{B} T}{\pi m}}. }$
This quantity is also known as the "impingement rate" in vacuum physics. Note that to calculate the average speed $\displaystyle{ \bar v }$ of the Maxwell's velocity distribution, one has to integrate over$\displaystyle{ v\gt 0,0\lt \theta\lt \pi,0\lt \phi\lt 2\pi }$.
The momentum transfer to the container wall from particles hitting the area $\displaystyle{ dA }$ with speed $\displaystyle{ v }$ at angle $\displaystyle{ \theta }$ from the normal, in time interval $\displaystyle{ dt }$ is:$\displaystyle{ [2mv \cos(\theta)]\times n v \cos(\theta) \, dA\, dt \times\left(\frac{m}{2 \pi k_BT}\right)^{3/2} e^{- \frac{mv^2}{2k_BT}} \left( v^2 \sin(\theta) \, dv \, d\theta \, d\phi \right). }$Integrating this over all appropriate velocities within the constraint $\displaystyle{ v\gt 0,0\lt \theta\lt \pi/2,0\lt \phi\lt 2\pi }$ yields the pressure (consistent with Ideal gas law):$\displaystyle{ P=\frac{2\int_0^{\pi/2}\cos^2 \theta \sin \theta d\theta}{\int_0^{\pi}\sin \theta d\theta}\times n mv_\text{rms}^2=\frac{1}{3}n mv_\text{rms}^2=\frac{2}{3}n\langle E_{kin}\rangle=nk_\mathrm{B} T }$If this small area $\displaystyle{ A }$ is punched to become a small hole, the effusive flow rate will be: $\displaystyle{ \Phi_\text{effusion} = J_\text{collision} A= n A \sqrt{\frac{k_\mathrm{B} T}{2 \pi m}}. }$
Combined with the ideal gas law, this yields $\displaystyle{ \Phi_\text{effusion} = \frac{P A}{\sqrt{2 \pi m k_\mathrm{B} T}}. }$
The above expression is consistent with Graham's law.
To calculate the velocity distribution of particles hitting this small area, we must take into account that all the particles with $\displaystyle{ (v,\theta,\phi) }$ that hit the area $\displaystyle{ dA }$ within the time interval $\displaystyle{ dt }$ are contained in the tilted pipe with a height of $\displaystyle{ v\cos (\theta) dt }$ and a volume of $\displaystyle{ v\cos (\theta) dAdt }$; Therefore, compared to the Maxwell distribution, the velocity distribution will have an extra factor of $\displaystyle{ v\cos \theta }$: \displaystyle{ \begin{align} f(v,\theta,\phi) \, dv \, d\theta \, d\phi &=\lambda v\cos{\theta}{\times} \left(\frac{m}{2 \pi k T}\right)^{3/2}e^{- \frac{mv^2}{2k_\mathrm{B} T}}(v^2\sin{\theta} \, dv \, d\theta \, d\phi) \\ \end{align} } with the constraint $\displaystyle{ v\gt 0,\, 0\lt \theta\lt \frac \pi 2,\, 0\lt \phi\lt 2\pi }$. The constant $\displaystyle{ \lambda }$ can be determined by the normalization condition $\displaystyle{ \int f(v,\theta,\phi) \, dv \, d\theta \, d\phi=1 }$ to be $\displaystyle{ {4}/{\bar v} }$, and overall:\displaystyle{ \begin{align} f(v,\theta,\phi) \, dv \, d\theta \, d\phi &=\frac{1}{2\pi} \left(\frac{m}{k_\mathrm{B} T}\right)^2e^{- \frac{mv^2}{2k_\mathrm{B} T}} (v^3\sin{\theta}\cos{\theta} \, dv \, d\theta \, d\phi) \\ \end{align};\quad v\gt 0,\, 0\lt \theta\lt \frac \pi 2,\, 0\lt \phi\lt 2\pi }
### Speed of molecules
From the kinetic energy formula it can be shown that $\displaystyle{ v_\text{p} = \sqrt{2 \cdot \frac{k_\mathrm{B} T}{m}}, }$ $\displaystyle{ \bar v = \frac {2}{\sqrt{\pi}} v_p = \sqrt{\frac {8}{\pi} \cdot \frac{k_\mathrm{B} T}{m}}, }$ $\displaystyle{ v_\text{rms} = \sqrt{\frac{3}{2}} v_p = \sqrt{{3} \cdot \frac {k_\mathrm{B} T}{m}}, }$ where v is in m/s, T is in kelvins, and m is the mass of one molecule of gas. The most probable (or mode) speed $\displaystyle{ v_\text{p} }$ is 81.6% of the root-mean-square speed $\displaystyle{ v_\text{rms} }$, and the mean (arithmetic mean, or average) speed $\displaystyle{ \bar v }$ is 92.1% of the rms speed (isotropic distribution of speeds).
See:
## Transport properties
The kinetic theory of gases deals not only with gases in thermodynamic equilibrium, but also very importantly with gases not in thermodynamic equilibrium. This means using Kinetic Theory to consider what are known as "transport properties", such as viscosity, thermal conductivity and mass diffusivity.
### Viscosity and kinetic momentum
In books on elementary kinetic theory[21] one can find results for dilute gas modeling that are used in many fields. Derivation of the kinetic model for shear viscosity usually starts by considering a Couette flow where two parallel plates are separated by a gas layer. The upper plate is moving at a constant velocity to the right due to a force F. The lower plate is stationary, and an equal and opposite force must therefore be acting on it to keep it at rest. The molecules in the gas layer have a forward velocity component $\displaystyle{ u }$ which increase uniformly with distance $\displaystyle{ y }$ above the lower plate. The non-equilibrium flow is superimposed on a Maxwell-Boltzmann equilibrium distribution of molecular motions.
Let $\displaystyle{ \sigma }$ be the collision cross section of one molecule colliding with another. As in the previous section, the number density $\displaystyle{ n }$ is defined as the number of molecules per (extensive) volume, or $\displaystyle{ n = N/V }$. The collision cross section per volume or collision cross section density is $\displaystyle{ n \sigma }$, and it is related to the mean free path $\displaystyle{ l }$ by $\displaystyle{ l = \frac {1} {\sqrt{2} n \sigma} }$
Notice that the unit of the collision cross section per volume $\displaystyle{ n \sigma }$ is reciprocal of length. The mean free path is the average distance traveled by a molecule, or a number of molecules per volume, before they make their first collision.
Let $\displaystyle{ u_{0} }$ be the forward velocity of the gas at an imaginary horizontal surface inside the gas layer. The number of molecules arriving at an area $\displaystyle{ dA }$ on one side of the gas layer, with speed $\displaystyle{ v }$ at angle $\displaystyle{ \theta }$ from the normal, in time interval $\displaystyle{ dt }$ is $\displaystyle{ nv\cos({\theta})\, dA \, dt \times \left(\frac{m}{2 \pi k_\mathrm{B} T}\right)^{3/2} \, e^{- \frac{mv^2}{2 k_\mathrm{B} T}} (v^2\sin{\theta} \, dv \, d\theta \, d\phi) }$
These molecules made their last collision at a distance $\displaystyle{ l\cos \theta }$ above and below the gas layer, and each will contribute a forward momentum of $\displaystyle{ p_{x}^{\pm} = m \left( u_{0} \pm l \cos \theta \,{d u \over dy} \right), }$ where plus sign applies to molecules from above, and minus sign below. Note that the forward velocity gradient $\displaystyle{ du/dy }$ can be considered to be constant over a distance of mean free path.
Integrating over all appropriate velocities within the constraint $\displaystyle{ \begin{cases} v\gt 0\\ 0\lt \theta\lt \pi/2\\ 0\lt \phi\lt 2\pi \end{cases} }$ yields the forward momentum transfer per unit time per unit area (also known as shear stress): $\displaystyle{ \tau^{\pm} = \frac {1}{4} \bar v n \cdot m \left( u_{0} \pm \frac {2}{3} l \,{d u \over dy} \right) }$
The net rate of momentum per unit area that is transported across the imaginary surface is thus $\displaystyle{ \tau = \tau^{+} - \tau^{-} = \frac {1}{3} \bar v n m \cdot l \,{d u \over dy} }$
Combining the above kinetic equation with Newton's law of viscosity $\displaystyle{ \tau = \eta \,{d u \over dy} }$ gives the equation for shear viscosity, which is usually denoted $\displaystyle{ \eta_{0} }$ when it is a dilute gas: $\displaystyle{ \eta_{0} = \frac {1} {3} \bar v n m l }$
Combining this equation with the equation for mean free path gives $\displaystyle{ \eta_{0} = \frac {1} {3 \sqrt{2} } \frac {m \cdot \bar v} {\sigma} }$
Maxwell-Boltzmann distribution gives the average (equilibrium) molecular speed as $\displaystyle{ \bar v = \frac{2}{\sqrt{\pi}} v_{p} = 2 \sqrt{\frac{2}{\pi} \cdot \frac {k_\mathrm{B}T}{m}} }$ where $\displaystyle{ v_{p} }$ is the most probable speed. We note that $\displaystyle{ k_{B} \cdot N_{A} = R \quad \text{and} \quad M = m \cdot N_{A} }$
and insert the velocity in the viscosity equation above. This gives the well known equation for shear viscosity for dilute gases:
$\displaystyle{ \eta_{0} = \frac {2} {3 \sqrt{\pi} } \cdot \frac {\sqrt{m k_\mathrm{B} T}} { \sigma } = \frac {2} {3 \sqrt{\pi} } \cdot \frac {\sqrt{MRT}} { \sigma \cdot N_{A} } }$
and $\displaystyle{ M }$ is the molar mass. The equation above presupposes that the gas density is low (i.e. the pressure is low). This implies that the kinetic translational energy dominates over rotational and vibrational molecule energies. The viscosity equation further presupposes that there is only one type of gas molecules, and that the gas molecules are perfect elastic and hard core particles of spherical shape. This assumption of elastic, hard core spherical molecules, like billiard balls, implies that the collision cross section of one molecule can be estimated by
$\displaystyle{ \sigma = \pi \left( 2 r \right)^2 = \pi d^2 }$
The radius $\displaystyle{ r }$ is called collision cross section radius or kinetic radius, and the diameter $\displaystyle{ d }$ is called collision cross section diameter or kinetic diameter of a molecule in a monomolecular gas. There are no simple general relation between the collision cross section and the hard core size of the (fairly spherical) molecule. The relation depends on shape of the potential energy of the molecule. For a real spherical molecule (i.e. a noble gas atom or a reasonably spherical molecule) the interaction potential is more like the Lennard-Jones potential or Morse potential which have a negative part that attracts the other molecule from distances longer than the hard core radius. The radius for zero Lennard-Jones potential is then appropriate to use as estimate for the kinetic radius.
### Thermal conductivity and heat flux
See also: Thermal conductivityFollowing a similar logic as above, one can derive the kinetic model for thermal conductivity[21] of a dilute gas:
Consider two parallel plates separated by a gas layer. Both plates have uniform temperatures, and are so massive compared to the gas layer that they can be treated as thermal reservoirs. The upper plate has a higher temperature than the lower plate. The molecules in the gas layer have a molecular kinetic energy $\displaystyle{ \varepsilon }$ which increases uniformly with distance $\displaystyle{ y }$ above the lower plate. The non-equilibrium energy flow is superimposed on a Maxwell-Boltzmann equilibrium distribution of molecular motions.
Let $\displaystyle{ \varepsilon_{0} }$ be the molecular kinetic energy of the gas at an imaginary horizontal surface inside the gas layer. The number of molecules arriving at an area $\displaystyle{ dA }$ on one side of the gas layer, with speed $\displaystyle{ v }$ at angle $\displaystyle{ \theta }$ from the normal, in time interval $\displaystyle{ dt }$ is $\displaystyle{ nv \cos(\theta)\, dA \, dt \times \left(\frac{m}{2 \pi k_\mathrm{B}T}\right)^{3 / 2} e^{- \frac{mv^2}{2k_BT}} (v^2 \sin(\theta) \, dv \, d\theta \, d\phi) }$
These molecules made their last collision at a distance $\displaystyle{ l\cos \theta }$ above and below the gas layer, and each will contribute a molecular kinetic energy of $\displaystyle{ \varepsilon^{\pm} = \left( \varepsilon_{0} \pm m c_v l \cos \theta \, {d T \over dy} \right), }$ where $\displaystyle{ c_v }$ is the specific heat capacity. Again, plus sign applies to molecules from above, and minus sign below. Note that the temperature gradient $\displaystyle{ dT/dy }$ can be considered to be constant over a distance of mean free path.
Integrating over all appropriate velocities within the constraint $\displaystyle{ \begin{cases} v\gt 0\\ 0\lt \theta\lt \pi/2\\ 0\lt \phi\lt 2\pi \end{cases} }$
yields the energy transfer per unit time per unit area (also known as heat flux): $\displaystyle{ q_y^{\pm} = -\frac {1}{4} \bar v n \cdot \left( \varepsilon_{0} \pm \frac {2}{3} m c_v l \,{d T \over dy} \right) }$
Note that the energy transfer from above is in the $\displaystyle{ -y }$ direction, and therefore the overall minus sign in the equation. The net heat flux across the imaginary surface is thus $\displaystyle{ q = q_y^{+} - q_y^{-} = -\frac {1}{3} \bar v n m c_v l \,{d T \over dy} }$
Combining the above kinetic equation with Fourier's law $\displaystyle{ q = -\kappa \,{d T \over dy} }$ gives the equation for thermal conductivity, which is usually denoted $\displaystyle{ \kappa_{0} }$ when it is a dilute gas: $\displaystyle{ \kappa_{0} = \frac {1} {3} \bar v n m c_v l }$
### Diffusion Coefficient and diffusion flux
See also: Fick's laws of diffusionFollowing a similar logic as above, one can derive the kinetic model for mass diffusivity[21] of a dilute gas:
Consider a steady diffusion between two regions of the same gas with perfectly flat and parallel boundaries separated by a layer of the same gas. Both regions have uniform number densities, but the upper region has a higher number density than the lower region. In the steady state, the number density at any point is constant (that is, independent of time). However, the number density $\displaystyle{ n }$ in the layer increases uniformly with distance $\displaystyle{ y }$ above the lower plate. The non-equilibrium molecular flow is superimposed on a Maxwell-Boltzmann equilibrium distribution of molecular motions.
Let $\displaystyle{ n_{0} }$ be the number density of the gas at an imaginary horizontal surface inside the layer. The number of molecules arriving at an area $\displaystyle{ dA }$ on one side of the gas layer, with speed $\displaystyle{ v }$ at angle $\displaystyle{ \theta }$ from the normal, in time interval $\displaystyle{ dt }$ is $\displaystyle{ nv\cos(\theta) \, dA \, dt \times \left(\frac{m}{2 \pi k_\mathrm{B}T}\right)^{3 / 2} e^{- \frac{mv^2}{2k_BT}} (v^2\sin(\theta) \, dv\, d\theta \, d\phi) }$
These molecules made their last collision at a distance $\displaystyle{ l\cos \theta }$ above and below the gas layer, where the local number density is $\displaystyle{ n^{\pm} = \left( n_{0} \pm l \cos \theta \, {d n \over dy} \right) }$
Again, plus sign applies to molecules from above, and minus sign below. Note that the number density gradient $\displaystyle{ dn/dy }$ can be considered to be constant over a distance of mean free path.
Integrating over all appropriate velocities within the constraint
$\displaystyle{ \begin{cases} v\gt 0\\ 0\lt \theta\lt \pi/2\\ 0\lt \phi\lt 2\pi \end{cases} }$
yields the molecular transfer per unit time per unit area (also known as diffusion flux): $\displaystyle{ J_y^{\pm} = -\frac {1}{4} \bar v \cdot \left( n_{0} \pm \frac {2}{3} l \, {d n \over dy} \right) }$
Note that the molecular transfer from above is in the $\displaystyle{ -y }$ direction, and therefore the overall minus sign in the equation. The net diffusion flux across the imaginary surface is thus $\displaystyle{ J = J_y^{+} - J_y^{-} = -\frac {1}{3} \bar v l {d n \over dy} }$
Combining the above kinetic equation with Fick's first law of diffusion $\displaystyle{ J = -D {d n \over dy} }$ gives the equation for mass diffusivity, which is usually denoted $\displaystyle{ D_{0} }$ when it is a dilute gas: $\displaystyle{ D_{0} = \frac {1} {3} \bar v l }$
## Notes
1. Maxwell, J. C. (1867). "On the Dynamical Theory of Gases". Philosophical Transactions of the Royal Society of London 157: 49–88. doi:10.1098/rstl.1867.0004.
2. L.I Ponomarev; I.V Kurchatov (1 January 1993). The Quantum Dice. CRC Press. ISBN 978-0-7503-0251-7.
3. Lomonosov 1758
4. Le Sage 1780/1818
5. Herapath 1816, 1821
6. Waterston 1843
7. Krönig 1856
8. Clausius 1857
9. See:
10. Mahon, Basil (2003). The Man Who Changed Everything – the Life of James Clerk Maxwell. Hoboken, NJ: Wiley. ISBN 0-470-86171-1. OCLC 52358254.
11. Gyenis, Balazs (2017). "Maxwell and the normal distribution: A colored story of probability, independence, and tendency towards equilibrium". Studies in History and Philosophy of Modern Physics 57: 53–65. doi:10.1016/j.shpsb.2017.01.001. Bibcode2017SHPMP..57...53G.
12. Maxwell 1875
13. Einstein 1905
14. Smoluchowski 1906
15. Chang, Raymond; Thoman, Jr., John W. (2014). Physical Chemistry for the Chemical Sciences. New York, NY: University Science Books. p. 37.
16. McQuarrie, Donald A. (1976). Statistical Mechanics. New York, NY: University Science Press.
17. The average kinetic energy of a fluid is proportional to the root mean-square velocity, which always exceeds the mean velocity - Kinetic Molecular Theory
18. Chang, Raymond; Thoman, Jr., John W. (2014). Physical Chemistry for the Chemical Sciences. New York: University Science Books. pp. 56–61.
19. Sears, F.W.; Salinger, G.L. (1975). "10". Thermodynamics, Kinetic Theory, and Statistical Thermodynamics (3 ed.). Reading, Massachusetts, USA: Addison-Wesley Publishing Company, Inc.. pp. 286–291. ISBN 978-0201068948.
## References
• Grad, Harold (1949), "On the Kinetic Theory of Rarefied Gases.", Communications on Pure and Applied Mathematics 2 (4): 331–407, doi:10.1002/cpa.3160020403
• Liboff, R. L. (1990), Kinetic Theory, Prentice-Hall, Englewood Cliffs, N. J.
• Lomonosov, M. (1970), "On the Relation of the Amount of Material and Weight", in Henry M. Leicester, Mikhail Vasil'evich Lomonosov on the Corpuscular Theory, Cambridge: Harvard University Press, pp. 224–233
• Mahon, Basil (2003), The Man Who Changed Everything – the Life of James Clerk Maxwell, Hoboken, New Jersey: Wiley, ISBN 0-470-86171-1
• Waterston, John James (1843), Thoughts on the Mental Functions (reprinted in his Papers, 3, 167, 183.) | 8,525 | 31,736 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2022-27 | latest | en | 0.945078 |
https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-6th-edition/chapter-3-cumulative-review-page-199/9e | 1,524,603,406,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947328.78/warc/CC-MAIN-20180424202213-20180424222213-00395.warc.gz | 791,543,301 | 13,567 | ## Intermediate Algebra (6th Edition)
$\frac{5}{8}\gt\frac{3}{8}$
$\frac{5}{8}$ is to the right of $\frac{3}{8}$ on the number line. Therefore, $\frac{5}{8}$ is greater than $\frac{3}{8}$ in value. | 71 | 198 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2018-17 | latest | en | 0.738049 |
https://ms.copernicus.org/articles/14/125/2023/ms-14-125-2023.html | 1,686,192,516,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654031.92/warc/CC-MAIN-20230608003500-20230608033500-00431.warc.gz | 435,208,634 | 51,135 | https://doi.org/10.5194/ms-14-125-2023
https://doi.org/10.5194/ms-14-125-2023
Research article | | 17 Mar 2023
# Dimensional synthesis of a spherical linkage crank slider mechanism for motion generation using an optimization algorithm
Wei Zhang, Zhen Liu, Wenrui Liu, Jianwei Sun, and He Lu
Abstract
In the present study, Fourier theory is applied to establish the expression of rigid-body poses of a spherical four-bar crank slider rigid-body guidance mechanism. According to an analysis of the harmonic components of the trajectory curve and rigid-body rotation angle, it has a certain relationship with the geometric parameters of the mechanism. On this basis, the rigid-body poses are normalized by preprocessing. Then, the rotation angle of the curve around the y axis and z axis is determined, respectively. The theoretical formulas used for calculating the real sizes and the installation position parameters of the desired spherical four-bar crank slider rigid-body guidance mechanism are established. Besides this, a genetic optimization algorithm and theoretical formulas are applied to solve the dimensional synthesis of motion generation for the spherical four-bar crank slider mechanism. The effectiveness of the proposed method is illustrated by an example. The maximum Euclidean distance error of the rigid-body position of the results with the highest similarity is 0.0086, and the average Euclidean distance error is 0.0044. The maximum error of the rigid-body orientation is 0.0179, and the average error is 0.0065.
Dates
1 Introduction
With the constant advancement of mechanical manufacturing technology and the widespread application of linkage mechanisms, the dimensional synthesis of linkage mechanisms (especially spatial linkage mechanisms (Wei et al., 2013; Liu et al., 2023) becomes increasingly important. In general, the dimensional synthesis of linkage mechanisms entails motion generation, path generation, and function generation.
Concerning the motion generation of linkage mechanisms, the outputs are periodic functions when the input link rotates continuously, which is irrelevant to the position or angle of the mechanism. The output of spatial mechanisms can be indicated by the Fourier series expression on the basis of Fourier transform. Fourier series theory was first introduced by Meyer zur Capellen (1954) to the analysis and synthesis of planar linkage mechanisms. Subsequently, Chu and Sun (2010) and Mullineux (2011) applied Fourier theory to explore the dimensional synthesis of a spherical four-bar mechanism, proposing an approach to the path generation of a spherical four-bar mechanism. In line with Fourier series theory, Sun et al. (2012) established the uniform model for the dimensional synthesis of linkage mechanisms including planar, spherical, and spatial mechanisms. They also illustrated the geometric significance attached to the harmonic characteristic parameter of the coupler curve and output function curve.
Currently, the Fourier series methods have been widely used in the function and path synthesis of spatial mechanisms and spherical mechanisms (Sun and Chu, 2010, 2008; Chu and Cao, 1993), not the motion generation of a spherical crank slider mechanism (Sun et al., 2012). Herein, the above theory is used to achieve the motion generation of spherical four-bar crank slider mechanism. The size of the target mechanism is optimized by using a genetic algorithm, which is capable of motion generation.
2 The mathematical description of the output
Motion generation or rigid-body guidance synthesis aims to obtain a series of prescribed rigid-body poses, which involves quantitative or qualitative design. There are two parts (orientation and position of rigid body) involved in the rigid-body guidance-line output of linkage mechanisms. As shown in Fig. 1, the spherical arc pq represents the output of mechanism, θpq denotes the corresponding central angle, and μi refers to the orientation of the ith rigid-body pose.
Figure 1The position and direction of the rigid body.
## 2.1 The mathematical model of the rigid-body output
Figure 2 shows the geometrical model of mechanism in the Cartesian coordinates Oxyz, and R indicates the radius. The axis of revolution of input link AB lies on the x axis, and the frame AD lies on the xOy plane. The angle of the arc Bp and arc pq is denoted as μpq, and the angle between Bp and BC is indicated by θp. θ2 represents the coupler angle. AB, BC, DA, and Bp denote central angles α, γ, ξ, θp0, respectively, where AB is input link, BC is the coupler, component C is a slider, and DA is the frame. As shown in Fig. 3, the coordinate Be1e2e3 is established, where e1 coincides with OB, e2 represents the tangent vector of the path of point B, and ${\mathbit{e}}_{\mathrm{3}}={\mathbit{e}}_{\mathrm{1}}×{\mathbit{e}}_{\mathrm{2}}$. The coordinate Be1e2e3 can be transformed into the global Cartesian coordinate Oxyz in two steps. Figure 3a shows that mechanism model, while Fig. 3b to d represents the two steps taken for the transformation of the coordinate system:
• 1.
As shown in Fig. 3b and c, the coordinate Oxyz rotates around the x axis, the rotation angle of which is ${\mathit{\theta }}_{\mathrm{1}}+\mathit{\pi }/\mathrm{2}$. Then, the coordinate Ox1y1z1 can be obtained, while the relationship between the coordinate Oxyz and the coordinate Ox1y1z1 can be described as follows:
$\begin{array}{}\text{(1)}& {x}_{\mathrm{1}}=x\text{(2)}& {y}_{\mathrm{1}}=y\mathrm{cos}\left(\stackrel{\mathrm{‾}}{\mathit{\theta }}\right)+z\mathrm{sin}\left(\stackrel{\mathrm{‾}}{\mathit{\theta }}\right)\text{(3)}& {z}_{\mathrm{1}}=-y\mathrm{sin}\left(\stackrel{\mathrm{‾}}{\mathit{\theta }}\right)+z\mathrm{cos}\left(\stackrel{\mathrm{‾}}{\mathit{\theta }}\right),\end{array}$
where $\left(\stackrel{\mathrm{‾}}{\mathit{\theta }}\right)={\mathit{\theta }}_{\mathrm{1}}+\frac{\mathit{\pi }}{\mathrm{2}}$.
• 2.
As shown in Fig. 3b and c, the (O, x1, y1, z1) rotates around the y1 axis, the rotation angle of which is α. The coordinate Ox2y2z2 can be obtained, while the relationship between the coordinate Ox1y1z1 and the coordinate Ox2y2z2 can be described as follows:
$\begin{array}{}\text{(4)}& \begin{array}{rl}{x}_{\mathrm{2}}& ={x}_{\mathrm{1}}\mathrm{cos}\mathit{\alpha }-{z}_{\mathrm{1}}\mathrm{sin}\mathit{\alpha }\\ & =x\mathrm{cos}\mathit{\alpha }-\mathrm{sin}\mathit{\alpha }\left[-y\mathrm{sin}\left(\stackrel{\mathrm{‾}}{\mathit{\theta }}\right)+z\mathrm{cos}\left(\stackrel{\mathrm{‾}}{\mathit{\theta }}\right)\right]\\ & =x\mathrm{cos}\mathit{\alpha }+y\mathrm{sin}\mathit{\alpha }\mathrm{sin}\left(\stackrel{\mathrm{‾}}{\mathit{\theta }}\right)-z\mathrm{sin}\mathit{\alpha }\mathrm{cos}\left(\stackrel{\mathrm{‾}}{\mathit{\theta }}\right)\\ & =x\mathrm{cos}\mathit{\alpha }+y\mathrm{sin}\mathit{\alpha }\mathrm{cos}{\mathit{\theta }}_{\mathrm{1}}+z\mathrm{sin}\mathit{\alpha }\mathrm{sin}{\mathit{\theta }}_{\mathrm{1}}\end{array}\end{array}$
$\begin{array}{}\text{(5)}& \begin{array}{rl}{y}_{\mathrm{2}}& ={y}_{\mathrm{1}}\\ & =y\mathrm{cos}\left(\stackrel{\mathrm{‾}}{\mathit{\theta }}\right)+z\mathrm{sin}\left(\stackrel{\mathrm{‾}}{\mathit{\theta }}\right)\\ & =-y\mathrm{sin}{\mathit{\theta }}_{\mathrm{1}}+z\mathrm{cos}{\mathit{\theta }}_{\mathrm{1}}\end{array}\end{array}$
$\begin{array}{}\text{(6)}& \begin{array}{rl}{z}_{\mathrm{2}}& ={x}_{\mathrm{1}}\mathrm{sin}\mathit{\alpha }+{z}_{\mathrm{1}}\mathrm{cos}\mathit{\alpha }\\ & =x\mathrm{sin}\mathit{\alpha }+\mathrm{cos}\mathit{\alpha }\left[-y\mathrm{sin}\left(\stackrel{\mathrm{‾}}{\mathit{\theta }}\right)+z\mathrm{cos}\left(\stackrel{\mathrm{‾}}{\mathit{\theta }}\right)\right]\\ & =x\mathrm{sin}\mathit{\alpha }-y\mathrm{cos}\mathit{\alpha }\mathrm{sin}\left(\stackrel{\mathrm{‾}}{\mathit{\theta }}\right)+z\mathrm{cos}\mathit{\alpha }\mathrm{cos}\left(\stackrel{\mathrm{‾}}{\mathit{\theta }}\right)\\ & =x\mathrm{sin}\mathit{\alpha }-y\mathrm{cos}\mathit{\alpha }\mathrm{cos}{\mathit{\theta }}_{\mathrm{1}}-z\mathrm{cos}\mathit{\alpha }\mathrm{sin}{\mathit{\theta }}_{\mathrm{1}}.\end{array}\end{array}$
According to the above equations, the relationship between the coordinates Be1e2e3 and Oxyz can be represented as follows:
$\begin{array}{}\text{(7)}& {\mathbit{e}}_{\mathrm{1}}={x}_{\mathrm{2}}=x\mathrm{cos}\mathit{\alpha }+y\mathrm{sin}\mathit{\alpha }\mathrm{cos}{\mathit{\theta }}_{\mathrm{1}}+z\mathrm{sin}\mathit{\alpha }\mathrm{sin}{\mathit{\theta }}_{\mathrm{1}}\text{(8)}& {\mathbit{e}}_{\mathrm{2}}={y}_{\mathrm{2}}=-y\mathrm{sin}{\mathit{\theta }}_{\mathrm{1}}+z\mathrm{cos}{\mathit{\theta }}_{\mathrm{1}}\text{(9)}& {\mathbit{e}}_{\mathrm{3}}={z}_{\mathrm{2}}=x\mathrm{sin}\mathit{\alpha }-y\mathrm{cos}\mathit{\alpha }\mathrm{cos}{\mathit{\theta }}_{\mathrm{1}}-z\mathrm{cos}\mathit{\alpha }\mathrm{sin}{\mathit{\theta }}_{\mathrm{1}},\end{array}$
where ${\mathit{\theta }}_{\mathrm{1}}=Q+\mathit{\omega }t$, Q represents the initial angle, ω indicates the fundamental frequency, and t refers to time.
Figure 2Model of spherical four-bar crank slider rigid-body guidance mechanism.
Figure 3The transformation of the coordinate system.
The position of point p on e1, e2, and e3 can be expressed as
$\begin{array}{}\text{(10)}& {x}_{pe}=R\mathrm{cos}{\mathit{\theta }}_{p\mathrm{0}}\text{(11)}& {y}_{pe}=-R\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{sin}\left(\stackrel{\mathrm{‾}}{{\mathit{\theta }}_{p}}\right)\text{(12)}& {z}_{pe}=R\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{cos}\left(\stackrel{\mathrm{‾}}{{\mathit{\theta }}_{p}}\right),\end{array}$
where $\left(\stackrel{\mathrm{‾}}{{\mathit{\theta }}_{p}}\right)=\mathit{\theta }+{\mathit{\theta }}_{p}$ and θ represents the input angle of the mechanism.
Therefore, the position of point p on the global coordinate Oxyz can be expressed as
$\begin{array}{}\text{(13)}& {r}_{p}={x}_{pe}{\mathbit{e}}_{\mathrm{1}}+{y}_{pe}{\mathbit{e}}_{\mathrm{2}}+{z}_{pe}{\mathbit{e}}_{\mathrm{3}}.\end{array}$
By substituting Eqs. (7) to (12) into Eq. (13), it can be obtained that
$\begin{array}{}\text{(14)}& \begin{array}{rl}{r}_{p}& =x\left[\mathrm{cos}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{cos}\mathit{\alpha }+\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{cos}\left(\stackrel{\mathrm{‾}}{{\mathit{\theta }}_{p}}\right)\mathrm{sin}\mathit{\alpha }\right]R\\ & +y\left[\mathrm{cos}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{cos}{\mathit{\theta }}_{\mathrm{1}}\mathrm{sin}\mathit{\alpha }+\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{sin}{\mathit{\theta }}_{\mathrm{1}}\mathrm{sin}\left(\stackrel{\mathrm{‾}}{{\mathit{\theta }}_{p}}\right)\\ & \phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}-\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{cos}\left(\stackrel{\mathrm{‾}}{{\mathit{\theta }}_{p}}\right)\mathrm{cos}\mathit{\alpha }\mathrm{cos}{\mathit{\theta }}_{\mathrm{1}}\right]R\\ & +z\left[\mathrm{cos}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{sin}{\mathit{\theta }}_{\mathrm{1}}\mathrm{sin}\mathit{\alpha }-\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{cos}{\mathit{\theta }}_{\mathrm{1}}\mathrm{sin}\left(\stackrel{\mathrm{‾}}{{\mathit{\theta }}_{p}}\right)\\ & \phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}-\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{cos}\left(\stackrel{\mathrm{‾}}{{\mathit{\theta }}_{p}}\right)\mathrm{cos}\mathit{\alpha }\mathrm{sin}{\mathit{\theta }}_{\mathrm{1}}\right]R.\end{array}\end{array}$
For the convenience of description, a complex plane is established; the real axis lies on the y axis of the global coordinate, and the imaginary axis lies on the z axis of the global coordinate. Equation (14) can be expressed as
$\begin{array}{}\text{(15)}& \begin{array}{rl}{r}_{p}=& x\left[\mathrm{cos}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{cos}\mathit{\alpha }+\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{cos}\left(\stackrel{\mathrm{‾}}{{\mathit{\theta }}_{p}}\right)\mathrm{sin}\mathit{\alpha }\right]R\\ & +{e}^{j{\mathit{\theta }}_{\mathrm{1}}}\left[-\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\left(\mathrm{cos}\left(\stackrel{\mathrm{‾}}{{\mathit{\theta }}_{p}}\right)\mathrm{cos}\mathit{\alpha }+j\mathrm{sin}\left(\stackrel{\mathrm{‾}}{{\mathit{\theta }}_{p}}\right)\right)\\ & +\mathrm{cos}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{sin}\mathit{\alpha }\right]R,\end{array}\end{array}$
where $j=\sqrt{\mathrm{1}}$.
The position of point p on the x axis (rpx(t)) can be expressed as
$\begin{array}{}\text{(16)}& {r}_{px}\left(t\right)=R\mathrm{cos}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{cos}\mathit{\alpha }+R\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{cos}\left(\mathit{\theta }\left(t\right)+{\mathit{\theta }}_{p}\right)\mathrm{sin}\mathit{\alpha }.\end{array}$
Similarly, the projection of the position of point p on the yOz plane (rpyz(t)) can be expressed as
$\begin{array}{}\text{(17)}& \begin{array}{rl}{r}_{pyz}\left(t\right)=R\left[& \mathrm{cos}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{sin}\mathit{\alpha }-\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\left(\mathrm{cos}\left(\mathit{\theta }\left(t\right)+{\mathit{\theta }}_{p}\right)\\ & \mathrm{cos}\mathit{\alpha }\right)+j\mathrm{sin}\left(\mathit{\theta }\left(t\right)+{\mathit{\theta }}_{p}\right)\right]{e}^{j{\mathit{\theta }}_{\mathrm{1}}}.\end{array}\end{array}$
Figure 4 shows the spherical mechanism in a general position of installation. The translation and rotation of the frame can be described by Ox, Oy, Oz, and θx, while the initial angle of the input link is denoted as Q. Based on Eq. (16), the projection on the x axis of point p can be expressed as
$\begin{array}{}\text{(18)}& \begin{array}{rl}{r}_{px}\left(t+{t}^{\prime }\right)=& {O}_{x}+R\mathrm{cos}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{cos}\mathit{\alpha }+R\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\\ & \mathrm{cos}\left(\mathit{\theta }\left(t+{t}^{\prime }\right)+{\mathit{\theta }}_{p}\right)\mathrm{sin}\mathit{\alpha },\end{array}\end{array}$
where ${t}^{\prime }=Q/\mathit{\omega }$.
Figure 4Installation position of the spherical mechanism.
According to Eq. (17), the projection of the position of point p on the yOz plane can be expressed as ${r}_{pyz}=\left(t+{t}^{\prime }\right)$ as follows:
$\begin{array}{}\text{(19)}& \begin{array}{rl}{r}_{pyz}\left(t+{t}^{\prime }\right)=& {O}_{y}+j{O}_{z}+{e}^{j{\mathit{\theta }}_{x}}R\left[\mathrm{cos}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{sin}\mathit{\alpha }-\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\\ & \left(\mathrm{cos}\left(\mathit{\theta }\left(t+{t}^{\prime }\right)+{\mathit{\theta }}_{p}\right)\mathrm{cos}\mathit{\alpha }\\ & +j\mathrm{sin}\left(\mathit{\theta }\left(t+{t}^{\prime }\right)+{\mathit{\theta }}_{p}\right)\right)\right]{e}^{j{\mathit{\theta }}_{\mathrm{1}}}.\end{array}\end{array}$
Therefore, the position of point p of the mechanism in a general installation position can be represented by Eqs. (18) and (19).
## 2.2 The preprocessing method of rigid-body markings
In the previous work, the dimensional synthesis of the spherical four-bar slider crank mechanism was studied, and a numerical atlas method was proposed to identify the basic dimensional types of the mechanism. Besides this, the output of the linkage mechanism was described by harmonic parameters, and the design results were obtained by comparing the prescribed design requirements with that in the numerical atlas database. However, the feature parameter extraction algorithm restricts the proposed method into solving only the rigid-body guidance synthesis of spherical linkage mechanisms in particular positions, where the rotation axis of the input component is parallel to the x axis of the global coordinate system and the plane of the frame is parallel to the xOy plane. For the spatial RRSS mechanism in a general installation position, this method is not applicable to obtain high-precision results. Therefore, a method is proposed in this paper to eliminate the impact caused by the rotation of the frame around the y axis and the z axis on the feature parameter of the trajectory curve.
As shown in Fig. 5, Oxyz is the global coordinate system. The requirements of the given design are detailed in Fig. 5a. The red solid line represents the output curve of the point p position, the solid blue line indicates the output curve of the point q position, and the black solid line between these two curves refers to the required rigid-body guidance line. According to the formula of the centroid of a curve, the centroid Q of the trajectory can be expressed as
$\begin{array}{}\text{(20)}& {Q}_{X}=\frac{\int {x}_{Q}\cdot \mathit{\rho }\left(x,y,z\right)\mathrm{d}s}{\int \mathit{\rho }\left(x,y,z\right)\mathrm{d}s},\end{array}$
$\begin{array}{}\text{(21)}& {Q}_{Y}=\frac{\int {y}_{Q}\cdot \mathit{\rho }\left(x,y,z\right)\mathrm{d}s}{\int \mathit{\rho }\left(x,y,z\right)\mathrm{d}s},\end{array}$
$\begin{array}{}\text{(22)}& {Q}_{Z}=\frac{\int {z}_{Q}\cdot \mathit{\rho }\left(x,y,z\right)\mathrm{d}s}{\int \mathit{\rho }\left(x,y,z\right)\mathrm{d}s},\end{array}$
where, QX, QY, and QZ represent the x, y, and z axes of the centroid Q, respectively; xQ, yQ, and zQ indicate the parametric equation of any curve, respectively; $\mathit{\rho }\left(x,y,z\right)=\mathrm{1}$; and ρ denotes the density function of coordinate parameter S.
Figure 5Schematic graph of the preprocessing.
Then, it is possible to calculate Q1 and Q2, which are, respectively, the centroid of the output curve of the point p position and the output curve of the point q position. The midpoint N (Nx, Ny, Nz) of Q1Q2 can be expressed as
$\begin{array}{}\text{(23)}& {N}_{x}=\frac{{Q}_{\mathrm{1}X}+{Q}_{\mathrm{2}X}}{\mathrm{2}},\end{array}$
$\begin{array}{}\text{(24)}& {N}_{y}=\frac{{Q}_{\mathrm{1}Y}+{Q}_{\mathrm{2}Y}}{\mathrm{2}},\end{array}$
$\begin{array}{}\text{(25)}& {N}_{z}=\frac{{Q}_{\mathrm{1}Z}+{Q}_{\mathrm{2}Z}}{\mathrm{2}}.\end{array}$
As shown in Fig. 5b, point O is connected with point N, and a straight line ON is defined as the central axis. In this paper, the normalization method is used to rotate the output curve of the point p position and the output curve of the point q position around the x axis simultaneously. The rotation angle ηy, ηz can be expressed as
$\begin{array}{}\text{(26)}& {\mathit{\eta }}_{y}=\frac{\left({N}_{x},{N}_{z}\right)\cdot \left(\mathrm{0},\mathrm{1}\right)}{\sqrt{{N}_{x}^{\mathrm{2}}+{N}_{z}^{\mathrm{2}}}\cdot \sqrt{{\mathrm{0}}^{\mathrm{2}}+{\mathrm{1}}^{\mathrm{2}}}},\end{array}$
$\begin{array}{}\text{(27)}& {\mathit{\eta }}_{z}=\frac{\left({N}_{x},{N}_{y}\right)\cdot \left(\mathrm{0},\mathrm{1}\right)}{\sqrt{{N}_{x}^{\mathrm{2}}+{N}_{y}^{\mathrm{2}}}\cdot \sqrt{{\mathrm{0}}^{\mathrm{2}}+{\mathrm{1}}^{\mathrm{2}}}}.\end{array}$
Now, the central axis ON lies in the z axis. As shown in Fig. 5c, the output curve of the point p position and the output curve of the point q position on the xOy plane can be projected.
Through rotation and transformation, the above method is used to eliminate the influence rotation of the frame around the y axis and the z axis on the feature parameter of the trajectory curve. It enables the Fourier transform to extract the feature information from the output curve of a spherical four-bar mechanism in a general position of installation.
From Fig. 6, we can get the orientation output as follows:
$\begin{array}{}\text{(28)}& \mathit{\mu }={\mathit{\mu }}_{p}+{\mathit{\mu }}_{pq}.\end{array}$
From the geometric relationship of spherical trigonometry, the following relations between μp and θ2 can be known:
$\begin{array}{}\text{(29)}& {\mathit{\mu }}_{p}=\mathrm{arctan}\left[\frac{\mathrm{sin}\left({\mathit{\theta }}_{p}+{\mathit{\theta }}_{\mathrm{2}}\right)\mathrm{sin}\mathit{\alpha }}{\mathrm{cos}\left({\mathit{\theta }}_{p}+{\mathit{\theta }}_{\mathrm{2}}\right)\mathrm{cos}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{sin}\mathit{\alpha }-\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{cos}\mathit{\alpha }}\right].\end{array}$
By substituting Eq. (29) into Eq. (28), the orientation of the arc pq can be obtained as follows:
$\begin{array}{}\text{(30)}& \mathit{\mu }={\mathit{\mu }}_{pq}+\mathrm{arctan}\left[\frac{\mathrm{sin}\left({\mathit{\theta }}_{p}+{\mathit{\theta }}_{\mathrm{2}}\right)\mathrm{sin}\mathit{\alpha }}{\mathrm{cos}\left({\mathit{\theta }}_{p}+{\mathit{\theta }}_{\mathrm{2}}\right)\mathrm{cos}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{sin}\mathit{\alpha }-\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{cos}\mathit{\alpha }}\right].\end{array}$
The angular velocity of the input crank AB is expressed as ω (θ1=ωt), and Eq. (30) is expressed as
$\begin{array}{}\text{(31)}& \begin{array}{rl}& \mathit{\mu }\left(t\right)={\mathit{\mu }}_{pq}+\mathrm{arctan}\\ & \phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\left[\frac{\mathrm{sin}\left({\mathit{\theta }}_{p}+{\mathit{\theta }}_{\mathrm{2}}\left(t\right)\right)\mathrm{sin}\mathit{\alpha }}{\mathrm{cos}\left({\mathit{\theta }}_{p}+{\mathit{\theta }}_{\mathrm{2}}\left(t\right)\right)\mathrm{cos}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{sin}\mathit{\alpha }-\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{cos}\mathit{\alpha }}\right].\end{array}\end{array}$
Similarly, according to Eq. (31), the orientation output of the mechanism in a general position of installation can be expressed as
$\begin{array}{}\text{(32)}& \begin{array}{rl}& \mathit{\mu }\left(t+{t}^{\prime }\right)={\mathit{\mu }}_{pq}+\mathrm{arctan}\\ & \phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\left[\frac{\mathrm{sin}\left({\mathit{\theta }}_{p}+{\mathit{\theta }}_{\mathrm{2}}\left(t+{t}^{\prime }\right)\right)\mathrm{sin}\mathit{\alpha }}{\mathrm{cos}\left({\mathit{\theta }}_{p}+{\mathit{\theta }}_{\mathrm{2}}\left(t+{t}^{\prime }\right)\right)\mathrm{cos}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{sin}\mathit{\alpha }-\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{cos}\mathit{\alpha }}\right].\end{array}\end{array}$
Therefore, the position output and orientation output can be expressed as Eqs. (18), (19), and (32), respectively.
Figure 6The rigid-body guidance line pq.
3 The harmonic analysis of the output
## 3.1 Fourier series expression of the output
According to Fourier series theory, the function of cos (θ2(t)+θp) can be expressed as
$\begin{array}{}\text{(33)}& \mathrm{cos}\left({\mathit{\theta }}_{\mathrm{2}}\left(t\right)+{\mathit{\theta }}_{p}\right)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_{n}{e}^{j{\mathit{\varphi }}_{n}}{e}^{jn\mathit{\omega }t},\end{array}$
where cn and ϕn represent the amplitude and the phase angle, respectively.
Similarly, when the initial angle is denoted as Q, Eq. (32) can be rewritten as
$\begin{array}{}\text{(34)}& \mathrm{cos}\left({\mathit{\theta }}_{\mathrm{2}}\left(t+{t}^{\prime }\right)+{\mathit{\theta }}_{p}\right)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_{n}{e}^{j{\mathit{\varphi }}_{n}}{e}^{jn\mathit{\omega }\left(t+{t}^{\prime }\right)}.\end{array}$
As $Q=\mathit{\omega }{t}^{\prime }$, Eq. (34) can be written as
$\begin{array}{}\text{(35)}& \mathrm{cos}\left({\mathit{\theta }}_{\mathrm{2}}\left(t+{t}^{\prime }\right)+{\mathit{\theta }}_{p}\right)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_{n}{e}^{j\left({\mathit{\varphi }}_{n}+nQ\right)}{e}^{jn\mathit{\omega }t}p.\end{array}$
By substituting Eq. (35) into Eq. (18), the position output of the mechanism as studied here and projected on the x axis can be expressed as
$\begin{array}{}\text{(36)}& \begin{array}{rl}{r}_{px}\left(t+{t}^{\prime }\right)& ={O}_{x}+R\mathrm{cos}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{cos}\mathit{\alpha }\\ & +R\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{sin}\mathit{\alpha }\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_{n}{e}^{j\left({\mathit{\varphi }}_{n}+nQ\right)}{e}^{jn\mathit{\omega }t}.\end{array}\end{array}$
The n=0 item of Eq. (36) is merged as follows:
$\begin{array}{}\text{(37)}& \begin{array}{rl}& {r}_{px}\left(t+{t}^{\prime }\right)\\ & \phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}=\left({O}_{x}+R\mathrm{cos}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{cos}\mathit{\alpha }+R{c}_{\mathrm{0}}\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{sin}\mathit{\alpha }{e}^{j{\mathit{\varphi }}_{\mathrm{0}}}\right)\\ & \phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}+R\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{sin}\mathit{\alpha }\sum _{n\ne \mathrm{0}}{c}_{n}{e}^{j\left({\mathit{\varphi }}_{n}+nQ\right)}{e}^{jn\mathit{\omega }t}.\end{array}\end{array}$
Similarly, the function of $\mathrm{cos}\left({\mathit{\theta }}_{\mathrm{2}}\left(t\right)+{\mathit{\theta }}_{p}\right)\mathrm{cos}\mathit{\alpha }+j\mathrm{sin}\left({\mathit{\theta }}_{\mathrm{2}}\left(t\right)+{\mathit{\theta }}_{p}\right)$ can be expressed as
$\begin{array}{}\text{(38)}& \begin{array}{rl}& \mathrm{cos}\left({\mathit{\theta }}_{\mathrm{2}}\left(t\right)+{\mathit{\theta }}_{p}\right)\mathrm{cos}\mathit{\alpha }+j\mathrm{sin}\left({\mathit{\theta }}_{\mathrm{2}}\left(t\right)+{\mathit{\theta }}_{p}\right)\\ & \phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_{n}^{\prime }{e}^{j{\mathit{\varphi }}_{n}^{\prime }}{e}^{jn\mathit{\omega }t},\end{array}\end{array}$
where ${c}_{n}^{\prime }$ and ${\mathit{\varphi }}_{n}^{\prime }$ represent the amplitude and the phase angle, respectively.
When the initial angle is indicated by Q, Eq. (38) can be expressed as
$\begin{array}{}\text{(39)}& \begin{array}{rl}& \mathrm{cos}\left({\mathit{\theta }}_{\mathrm{2}}\left(t+{t}^{\prime }\right)+{\mathit{\theta }}_{p}\right)\mathrm{cos}\mathit{\alpha }+j\mathrm{sin}\left({\mathit{\theta }}_{\mathrm{2}}\left(t+{t}^{\prime }\right)+{\mathit{\theta }}_{p}\right)\\ & \phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}=\sum _{m=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_{m}^{\prime }{e}^{j{\mathit{\varphi }}_{m}^{\prime }}{e}^{jm\mathit{\omega }\left(t+{t}^{\prime }\right)}.\end{array}\end{array}$
As $Q=\mathit{\omega }{t}^{\prime }$, Eq. (39) can be expressed as
$\begin{array}{}\text{(40)}& \begin{array}{rl}& \mathrm{cos}\left({\mathit{\theta }}_{\mathrm{2}}\left(t+{t}^{\prime }\right)+{\mathit{\theta }}_{p}\right)\mathrm{cos}\mathit{\alpha }+j\mathrm{sin}\left({\mathit{\theta }}_{\mathrm{2}}\left(t+{t}^{\prime }\right)+{\mathit{\theta }}_{p}\right)\\ & \phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}=\sum _{m=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_{m}^{\prime }{e}^{j\left({\mathit{\varphi }}_{m}^{\prime }+mQ\right)}{e}^{jm\mathit{\omega }t}.\end{array}\end{array}$
By substituting Eq. (40) into Eq. (19), the position output of the mechanism studied here and projected on the yOz plane is expressed as
$\begin{array}{}\text{(41)}& \begin{array}{rl}{r}_{pyz}\left(t+{t}^{\prime }\right)=& \phantom{\rule{0.25em}{0ex}}{O}_{y}+j{O}_{z}+{e}^{j{\mathit{\theta }}_{x}}R\left[\mathrm{cos}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{sin}\mathit{\alpha }-\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\\ & \left(\sum _{m=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_{m}^{\prime }{e}^{j\left({\mathit{\varphi }}_{m}^{\prime }+mQ\right)}{e}^{jm\mathit{\omega }t}\right)\right]{e}^{j{\mathit{\theta }}_{\mathrm{1}}}.\end{array}\end{array}$
By merging +1 item (m=0) in Eq. (41), it can be obtained that
$\begin{array}{}\text{(42)}& \begin{array}{rl}& {r}_{pyz}\left(t+{t}^{\prime }\right)={O}_{y}+j{O}_{z}\\ & \phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}+R\left(\mathrm{cos}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{sin}\mathit{\alpha }-{c}_{\mathrm{0}}^{\prime }\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}{e}^{j{\mathit{\varphi }}_{\mathrm{0}}^{\prime }}\right){e}^{j\left({\mathit{\theta }}_{x}+Q\right)}{e}^{j\mathit{\omega }t}\\ & \phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}-R\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\sum _{m\ne \mathrm{0}}{c}_{m}^{\prime }{e}^{j\left[{\mathit{\varphi }}_{m}^{\prime }+\left(m+\mathrm{1}\right)Q+{\mathit{\theta }}_{x}\right]}{e}^{j\left(m+\mathrm{1}\right)\mathit{\omega }t}.\end{array}\end{array}$
By merging 0 item ($m=-\mathrm{1}$) of Eq. (42) and defining $n=m+\mathrm{1}$, it can be obtained that
$\begin{array}{}\text{(43)}& \begin{array}{rl}& {r}_{pyz}\left(t+{t}^{\prime }\right)={O}_{y}+j{O}_{z}-R{c}_{-\mathrm{1}}^{\prime }\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}{e}^{j\left({\mathit{\varphi }}_{-\mathrm{1}}^{\prime }+{\mathit{\theta }}_{x}\right)}\\ & \phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}+R\left(\mathrm{cos}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{sin}\mathit{\alpha }-{c}_{\mathrm{0}}^{\prime }\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}{e}^{j{\mathit{\varphi }}_{\mathrm{0}}^{\prime }}\right){e}^{j\left({\mathit{\theta }}_{x}+Q\right)}{e}^{j\mathit{\omega }t}\\ & \phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}-R\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\sum _{n\ne \mathrm{0},\mathrm{1}}{c}_{n-\mathrm{1}}^{\prime }{e}^{j\left({\mathit{\varphi }}_{n-\mathrm{1}}^{\prime }+nQ+{\mathit{\theta }}_{x}\right)}{e}^{jn\mathit{\omega }t}.\end{array}\end{array}$
Thus, the harmonic components of the position output are expressed as Eqs. (37) and (43).
According to Eq. (29), the Fourier series expansion of μp(t) is defined as
$\begin{array}{}\text{(44)}& \begin{array}{rl}{\mathit{\mu }}_{p}\left(t\right)& =\mathrm{arctan}\left[\frac{\mathrm{sin}\left({\mathit{\theta }}_{p}+{\mathit{\theta }}_{\mathrm{2}}\left(tt\right)\right)\mathrm{sin}\mathit{\alpha }}{\mathrm{cos}\left({\mathit{\theta }}_{p}+{\mathit{\theta }}_{\mathrm{2}}\left(tt\right)\right)\mathrm{cos}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{sin}\mathit{\alpha }-{\mathrm{sin}}_{p\mathrm{0}}\mathrm{cos}\mathit{\alpha }}\right]\\ & \phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}=\sum _{n=-\mathrm{\infty }}^{+\mathrm{\infty }}{c}_{n}^{\prime \prime }{e}^{j{\mathit{\varphi }}_{n}^{\prime \prime }}{e}^{jn\mathit{\omega }t},\end{array}\end{array}$
where ${c}_{n}^{\prime \prime }$ and ${\mathit{\varphi }}_{n}^{\prime \prime }$ refer to the amplitude and the phase angle, respectively.
If the initial angle of the crank AB is Q ($Q=\mathit{\omega }{t}^{\prime }\right)$, Eq. (44) is expressed as
$\begin{array}{}\text{(45)}& \begin{array}{rl}& {\mathit{\mu }}_{p}\left(t+{t}^{\prime }\right)\\ & \phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}=\mathrm{arctan}\left[\frac{\mathrm{sin}\left({\mathit{\theta }}_{p}+{\mathit{\theta }}_{\mathrm{2}}\left(t+{t}^{\prime }\right)\right)\mathrm{sin}\mathit{\alpha }}{\mathrm{cos}\left({\mathit{\theta }}_{p}+{\mathit{\theta }}_{\mathrm{2}}\left(t+{t}^{\prime }\right)\right)\mathrm{cos}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{sin}\mathit{\alpha }-\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{cos}\mathit{\alpha }}\right]\\ & \phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}=\sum _{n=-\mathrm{\infty }}^{+\mathrm{\infty }}{c}_{n}^{\prime \prime }{e}^{j{\mathit{\varphi }}_{n}^{\prime \prime }}{e}^{jn\mathit{\omega }\left(t+{t}^{\prime }\right)}.\end{array}\end{array}$
As $Q=\mathit{\omega }{t}^{\prime }$, Eq. (45) is rearranged. Therefore,
$\begin{array}{}\text{(46)}& \begin{array}{rl}& \mathit{\mu }\left(t+{t}^{\prime }\right)\\ & \phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}=\mathrm{arctan}\left[\frac{\mathrm{sin}\left({\mathit{\theta }}_{p}+{\mathit{\theta }}_{\mathrm{2}}\left(t+{t}^{\prime }\right)\right)\mathrm{sin}\mathit{\alpha }}{\mathrm{cos}\left({\mathit{\theta }}_{p}+{\mathit{\theta }}_{\mathrm{2}}\left(t+{t}^{\prime }\right)\right)\mathrm{cos}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{sin}\mathit{\alpha }-\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{cos}\mathit{\alpha }}\right]\\ & =\sum _{n=-\mathrm{\infty }}^{+\mathrm{\infty }}{c}_{n}^{\prime \prime }{e}^{j\left({\mathit{\varphi }}_{n}^{\prime \prime }+nQ\right)}{e}^{jn\mathit{\omega }t}.\end{array}\end{array}$
By substituting Eq. (46) into Eq. (32) and merging the item, the orientation output of the mechanism is expressed as
$\begin{array}{}\text{(47)}& \mathit{\mu }\left(t+{t}^{\prime }\right)=\left({\mathit{\mu }}_{pq}+{c}_{\mathrm{0}}^{\prime \prime }{e}^{j{\mathit{\varphi }}_{\mathrm{0}}^{\prime \prime }}\right)+\sum _{n\ne \mathrm{0}}{c}_{n}^{\prime \prime }{e}^{j\left({\mathit{\varphi }}_{n}^{\prime \prime }+nQ\right)}{e}^{jn\mathit{\omega }t}.\end{array}$
## 3.2 The extraction of the output characteristic parameters
Apparently Eqs. (33), (38), and (44) are implicitly related to Eqs. (37), (43), and (47). After the normalization of Eqs. (33), (38), (44), (37), (43), and (47), different factors such as the actual size, initial angle, and Installation position are eliminated. Besides this, the general law of the harmonic component of the orientation output and the mechanism parameters is determined.
### 3.2.1 Define and extract the harmonic characteristic parameters of the position output
After dividing Eq. (33) by ${c}_{-\mathrm{1}}{e}^{j{\mathit{\varphi }}^{-\mathrm{1}}}$ (the harmonic of the −1 term), it can be obtained that (cn is the nth term series of a 1D Fourier transform )
$\begin{array}{}\text{(48)}& \begin{array}{rl}& \frac{\mathrm{1}}{{c}_{-\mathrm{1}}}\mathrm{cos}\left({\mathit{\theta }}_{\mathrm{2}}\left(t\right)+{\mathit{\theta }}_{p}\right){e}^{-j{\mathit{\varphi }}_{-\mathrm{1}}}\\ & =\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{c}_{n}}{{c}_{-\mathrm{1}}}{e}^{j\left({\mathit{\varphi }}_{n}-{\mathit{\varphi }}_{-\mathrm{1}}\right)}{e}^{jn\mathit{\omega }t}\end{array}\end{array}$
Similarly, after dividing Eq. (37) by $R{c}_{-\mathrm{1}}\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{sin}\mathit{\alpha }{e}^{j\left({\mathit{\varphi }}_{\mathrm{1}}-Q\right)}$ (the harmonic of the −1 term), it can be obtained that
$\begin{array}{}\text{(49)}& \begin{array}{rl}& \frac{{r}_{px}\left(t+{t}^{\prime }\right)}{R{c}_{-\mathrm{1}}\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{sin}\mathit{\alpha }}{e}^{-j\left({\mathit{\varphi }}_{-\mathrm{1}}-Q\right)}\\ & \phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}=\frac{\mathrm{1}}{R{c}_{-\mathrm{1}}\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{sin}\mathit{\alpha }}\left(\begin{array}{l}{O}_{x}+R\mathrm{cos}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{cos}\mathit{\alpha }\\ +R{c}_{\mathrm{0}}\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{sin}\mathit{\alpha }{e}^{j{\mathit{\varphi }}_{\mathrm{0}}}{e}^{-j\left({\mathit{\varphi }}_{-\mathrm{1}}-Q\right)}\end{array}\right)\\ & \phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}+\sum _{n\ne \mathrm{0}}\frac{{c}_{n}}{{c}_{-\mathrm{1}}}{e}^{j\left({\mathit{\varphi }}_{n}-{\mathit{\varphi }}_{-\mathrm{1}}+\left(n+\mathrm{1}\right)Q\right)}{e}^{jn\mathit{\omega }t}.\end{array}\end{array}$
By comparing Eq. (48) with Eq. (49), it can be discovered that they have the same amplitude accordingly (except for item 0), the phase difference of which is (n+1)Q.
Similarly, after dividing Eq. (38) by ${c}_{-\mathrm{2}}^{\prime }{e}^{j{\mathit{\varphi }}^{\prime }-\mathrm{2}}$ (${c}_{-\mathrm{2}}^{\prime }$ is the nth term series of a 2D Fourier transform), it can be obtained that
$\begin{array}{}\text{(50)}& \begin{array}{rl}& \frac{\mathrm{1}}{{c}_{-\mathrm{2}}^{\prime }}{e}^{-j{\mathit{\varphi }}_{-\mathrm{2}}^{\prime }}\left[\mathrm{cos}\left({\mathit{\theta }}_{\mathrm{2}}\left(t\right)+{\mathit{\theta }}_{p}\right)\mathrm{cos}\mathit{\alpha }+j\mathrm{sin}\left({\mathit{\theta }}_{\mathrm{2}}\left(t\right)+{\mathit{\theta }}_{p}\right)\right]\\ & \phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{c}_{n}^{\prime }}{{c}_{-\mathrm{2}}^{\prime }}{e}^{j\left({\mathit{\varphi }}_{n}^{\prime }-{\mathit{\varphi }}_{-\mathrm{2}}^{\prime }\right)}{e}^{jn\mathit{\omega }t}.\end{array}\end{array}$
Similarly, after dividing Eq. (43) by the $-R{c}_{-\mathrm{2}}^{\prime }\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}{e}^{j\left({\mathit{\varphi }}_{-\mathrm{2}}^{\prime }-Q+{\mathit{\theta }}_{x}\right)}$ (the term −1 harmonic), it can be obtained that
$\begin{array}{}\text{(51)}& \begin{array}{rl}& -\frac{{r}_{pyz}\left(t+{t}^{\prime }\right)}{R{c}_{-\mathrm{2}}^{\prime }\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}}{e}^{j\left(Q-{\mathit{\varphi }}_{-\mathrm{2}}^{\prime }-{\mathit{\theta }}_{x}\right)}=-\frac{\mathrm{1}}{R{c}_{-\mathrm{2}}^{\prime }\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}}\\ & \left({O}_{y}+j{O}_{z}-R{c}_{-\mathrm{1}}^{\prime }\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}{e}^{j\left({\mathit{\varphi }}_{-\mathrm{1}}^{\prime }+{\mathit{\theta }}_{x}\right)}\right){e}^{-j\left({\mathit{\varphi }}_{-\mathrm{2}}^{\prime }-Q+{\mathit{\theta }}_{x}\right)}\\ & -\frac{\mathrm{1}}{{c}_{-\mathrm{2}}^{\prime }{\mathrm{sin}}_{p\mathrm{0}}}\left(\mathrm{cos}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{sin}\mathit{\alpha }-{c}_{\mathrm{0}}^{\prime }\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}{e}^{j{\mathit{\varphi }}_{\mathrm{0}}^{\prime }}\right){e}^{j\left(\mathrm{2}Q-{\mathit{\varphi }}_{-\mathrm{2}}^{\prime }\right)}{e}^{j\mathit{\omega }t}\\ & +\sum _{n\ne \mathrm{0},\mathrm{1}}\frac{{c}_{n-\mathrm{1}}^{\prime }}{{c}_{-\mathrm{2}}^{\prime }}{e}^{j\left[{\mathit{\varphi }}_{n-\mathrm{1}}^{\prime }-{\mathit{\varphi }}_{-\mathrm{2}}^{\prime }+\left(n+\mathrm{1}\right)Q\right]}{e}^{jn\mathit{\omega }t}.\end{array}\end{array}$
Comparing Eq. (50) with Eq. (51), it can be found that they have the same amplitude correspondingly (except for item 0 and +1), and the phase difference is $\mathrm{4}\left(n+\mathrm{1}\right)Q+{\mathit{\varphi }}_{n-\mathrm{1}}^{\prime }-{\mathit{\varphi }}_{n}^{\prime }$.
By comparing Eq. (50) with Eq. (51), it can be found out that they have the same amplitude accordingly (except for item 0 and +1), the phase difference of which is $\left(n+\mathrm{1}\right)Q+{\mathit{\varphi }}_{n-\mathrm{1}}^{\prime }-{\mathit{\varphi }}_{n}^{\prime }$.
Through the above analysis, the harmonic components of the position output are expressed by Eqs. (48) and (50) (except for individual items). Also, the function $\mathrm{cos}\left({\mathit{\theta }}_{\mathrm{2}}\left(t+{t}^{\prime }\right)+{\mathit{\theta }}_{p}\right)+j\mathrm{sin}\left({\mathit{\theta }}_{\mathrm{2}}\left(t+{t}^{\prime }\right)+{\mathit{\theta }}_{p}\right)$ and $\mathrm{cos}\left({\mathit{\theta }}_{\mathrm{2}}\left(t+{t}^{\prime }\right)+{\mathit{\theta }}_{p}\right)$ are determined by the five parameters of the mechanism as studied here, including α, γ, ξ, α1, and θp. The amplitude of the output can be described by using parameters (α, γ, ξ, α1, and θp). Therefore, the parameters of α, γ, ξ, α1, and θp are defined as MPBDT (the mechanism's position of basic dimensional types), while harmonic components ${c}_{n}/{c}_{-\mathrm{1}}$ and ${c}_{n-\mathrm{1}}^{\prime }/{c}_{-\mathrm{2}}^{\prime }$ are defined as a group of position of harmonic characteristic parameters' dimensional type (PHCPDT). The functions of $\mathrm{cos}\left({\mathit{\theta }}_{\mathrm{2}}\left(t+{t}^{\prime }\right)+{\mathit{\theta }}_{p}\right)$ and $\mathrm{cos}\left({\mathit{\theta }}_{\mathrm{2}}\left(t+{t}^{\prime }\right)+{\mathit{\theta }}_{p}\right)\mathrm{cos}\mathit{\alpha }+j\mathrm{sin}\left({\mathit{\theta }}_{\mathrm{2}}\left(t+{t}^{\prime }\right)+{\mathit{\theta }}_{p}\right)$ are defined as the rigid-body position operator (RBPO). On this basis, it can be found out that the RBPO is determined only by the MPBDT, rather than other factors such as the structural and arrangement parameters of the spherical four-bar crank slider rigid-body guidance mechanism.
### 3.2.2 The characteristic parameters of the orientation output
According to Eq. (31), the orientation output function μp(t) is a periodic function when the link AB is cranked, and the shift of the function curve affects only the initial angle and the average value. Therefore, the function $\mathit{\mu }\left(t+{t}^{\prime }\right)$ can be obtained by the function μp(t), as shown in Fig. 7. By comparing Eq. (44) with Eq. (47), it can be discovered that they have the same amplitude, and the phase differences of both are nQ (except for 0 items). Thus, the characteristic parameter of the rigid-body orientation output can be described by using the amplitude (${c}_{n}^{\prime \prime }$) of the function μp(t). Herein, the amplitudes obtained from Eq. (44) are defined as the orientation of harmonic characteristic parameters' dimensional type (OHCPDT). Besides this, according to Eq. (29), the μp(t) is determined by the parameters α, γ, ξ, α1, θp, and θp0. The parameters of α, γ, ξ, α1, θp, and θp0 are defined as the orientation of basic dimensional types of the mechanism (MOBDT), while μp(t) is defined as the rigid-body orientation operator (RBOO). Through the above analysis, it can be discovered that the RBOO is determined only by the MOBDT, instead of other factors (such as the structural and arrangement parameters).
Figure 7The translation of the orientation output function.
According to Eq. (47), the Fourier series $\sum {D}_{n}^{\prime \prime }{{e}_{n}^{j\mathit{\xi }}}^{\prime \prime }{e}^{jn\mathit{\omega }t}$ (${D}_{n}^{\prime \prime }$ is the amplitude of the nth harmonic component of a 3D Fourier transform, and are defined as the harmonic components of the prescribed orientation $\mathit{\mu }\left(t+{t}^{\prime }\right)$. Based on the property of the 1D Fourier transformation (${\mathit{\xi }}_{\mathrm{0}}^{\prime \prime }=\mathrm{0}$ and ${\mathit{\varphi }}^{\prime \prime }=\mathrm{0}$), the angle parameter μpq can be obtained as follows.
MOBDT involves all the parameters of MPBDT by comparing them. Thus, as the first step, a number atlas database of the orientation output is established by inputting both the OHCPDT and MOBDT. Then, fuzzy identification is performed to recognize several groups of MOBDT that satisfy the prescribed orientation output of the guidance mechanism as studied in this paper. Accordingly, a position problem of the rigid body can be transformed into a path generation problem when a reference point is chosen. According to Chu and Sun (2010), the characteristic parameters of both the design conditions and the MPBDT can be obtained by taking advantage of a fast Fourier transform (FFT). Subsequently, the actual length size and installation parameters can be calculated according to those parameters and theoretical formulas of the guiding mechanism as studied in this paper.
## 3.3 Calculation of the actual dimensions and arrangement parameters
By comparing Eqs. (33) and (38) with (37) and (43), respectively, the relationship between the amplitudes and phase angles can be determined, to derive the theoretical formulas about actual installation size, coupler point position, and installation size parameters. In this paper, the prescribed position rp on the x axis and yOz plane is defined as
$\begin{array}{rl}& {r}_{px}\left(t+{t}^{\prime }\right)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{D}_{n}{e}^{j{\mathit{\zeta }}_{n}}{e}^{jn\mathit{\omega }t}\\ & \text{and}\\ & {r}_{pyz}\left(t+{t}^{\prime }\right)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{D}_{n}^{\prime }{e}^{j{\mathit{\zeta }}_{n}^{\prime }}{e}^{jn\mathit{\omega }t},\end{array}$
where Dn represents the amplitude of the nth harmonic component of a 1D Fourier transform, ${D}_{n}^{\prime }$ indicates the amplitude of the nth harmonic component of a 2D Fourier transform, ζn denotes the initial phase of the nth harmonic component of a 1D Fourier transform, and ${\mathit{\zeta }}_{n}^{\prime }$ refers to the initial phase of the nth harmonic component of a 2D Fourier transform.
The harmonic component of the angular characteristic function of the rigid body is defined as
$\sum _{n=-\mathrm{\infty }}^{+\mathrm{\infty }}{c}_{n}{e}^{j{\mathit{\varphi }}_{n}}{e}^{jn\mathit{\omega }t}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\text{and}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_{n}^{\prime }{e}^{j{\mathit{\varphi }}_{n}^{\prime }}{e}^{jn\mathit{\omega }t}.$
Therefore, the actual size and arrangement parameters can be calculated as follows:
• 1.
The actual dimension of the crank (α) is
$\begin{array}{}\text{(52)}& \mathit{\alpha }=\mathrm{arcsin}\left(\frac{{D}_{n}{c}_{n-\mathrm{1}}^{\prime }}{{D}_{n}^{\prime }{c}_{n}}\right)\left(n\ne \mathrm{0},\mathrm{1}\right).\end{array}$
The other dimensional types of γ, ξ, and α1 can be obtained according to the proportional coefficient.
• 2.
The initial angle Q:
$\begin{array}{}\text{(53)}& Q={\mathit{\zeta }}_{\mathrm{1}}-{\mathit{\varphi }}_{\mathrm{1}}.\end{array}$
• 3.
The rotation angle θx:
$\begin{array}{}\text{(54)}& {\mathit{\theta }}_{x}={\mathit{\zeta }}_{-\mathrm{1}}^{\prime }+Q-{\mathit{\varphi }}_{-\mathrm{2}}^{\prime }.\end{array}$
• 4.
The spherical radius of the mechanism:
where
$\begin{array}{rl}A=\left[& {D}_{\mathrm{1}}^{\prime }\mathrm{cos}{\mathit{\zeta }}_{\mathrm{1}}^{\prime }-{D}_{n}^{\prime }{c}_{\mathrm{0}}^{\prime }\mathrm{cos}\left({\mathit{\varphi }}_{\mathrm{0}}^{\prime }+{\mathit{\theta }}_{x}+Q\right)/{c}_{n-\mathrm{1}}^{\prime }\right]/\\ & \mathrm{cos}\left({\mathit{\theta }}_{x}+Q\right).\end{array}$
• 5.
The central angle θp0:
$\begin{array}{}\text{(56)}& {\mathit{\theta }}_{p\mathrm{0}}=\mathrm{arcsin}\left(\frac{{D}_{n}^{\prime }}{R{c}_{n-\mathrm{1}}^{\prime }}\right).\end{array}$
• 6.
The translation distance Ox in the x axis:
$\begin{array}{}\text{(57)}& \begin{array}{rl}{O}_{x}& ={D}_{\mathrm{0}}\mathrm{cos}{\mathit{\zeta }}_{\mathrm{0}}-R\mathrm{cos}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{cos}\mathit{\alpha }\\ & -R{c}_{\mathrm{0}}\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{sin}\mathit{\alpha }\mathrm{cos}{\mathit{\varphi }}_{\mathrm{0}}.\end{array}\end{array}$
• 7.
The translation distance Oy, Oz in the y and z axes:
$\begin{array}{}\text{(58)}& {O}_{y}={D}_{\mathrm{0}}^{\prime }\mathrm{cos}{\mathit{\zeta }}_{\mathrm{0}}^{\prime }+R{c}_{-\mathrm{1}}^{\prime }\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{cos}\left({\mathit{\varphi }}_{-\mathrm{1}}^{\prime }+{\mathit{\theta }}_{x}\right),\end{array}$
$\begin{array}{}\text{(59)}& {O}_{z}={D}_{\mathrm{0}}^{\prime }\mathrm{sin}{\mathit{\zeta }}_{\mathrm{0}}^{\prime }+R{c}_{-\mathrm{1}}^{\prime }\mathrm{sin}{\mathit{\theta }}_{p\mathrm{0}}\mathrm{sin}\left({\mathit{\varphi }}_{-\mathrm{1}}^{\prime }+{\mathit{\theta }}_{x}\right).\end{array}$
Equations (53) to (60) can be applied to calculate the parameters of the actual installation size of the guidance mechanism, as studied by this paper, and the coupler point position and installation size.
4 Steps and illustration
Table 1Flowchart of the synthesis process.
## 4.1 The influence of each parameter of the MOBDT upon the output
It is widely known that the parameters of the MOBDT have different effects on the output of the rigid body. Firstly, one group of MOBDT is chosen to verify the influence of each parameter. By changing each parameter with the same variable Δρ (herein, Δρ=2), the changes of the output of the rigid body are indicated, as shown in Fig. 8. Figure 8a shows the output affected by α, Fig. 8b shows the output affected by γ, Fig. 8c shows the output affected by ξ, Fig. 8d shows the output affected by α1, Fig. 8e shows the output affected by θp, and Fig. 8f shows the output affected by θp0. According to the results, α, θp, and θp0 have the most significant impact on the output, followed by α1. The effect of γ and ξ is the least significant. The blue curve in the figure indicates the trajectory curve of a given set of sizes, and the red curve represents the trajectory curve obtained by changing each size.
Figure 8The MOBDT with different parameters.
## 4.2 Variable step size output properties database and matching
The step size of OHCPDT in the output properties database is denoted as λ0. It can be found out that the accuracy of matching is higher when the λ0 is reduced. However, the complexity of the database increases. Conversely, if λ0 is larger, the matching speed improves, but the accuracy is reduced. Therefore, to ensure the sufficient accuracy and speed of matching, different step sizes denoted as λ0 are given for the distinct parameter of MOBDT, and the step size of each parameter is chosen according to the analysis in Sect. 4.1. Then, a different step size digital atlas database can be established.
According to Chu and Sun (2010), the proper MOBDT can be obtained by the following equation:
$\begin{array}{}\text{(60)}& \mathit{\delta }=\sum _{n=\mathrm{1}}^{\mathrm{3}}{\left({T}_{n}^{\prime }-{T}_{n}\right)}^{\mathrm{2}},\end{array}$
where δ represents the similarity function, while ${T}_{n}^{\prime }$ and Tn denote the OHCPDT in the design conditions and in the output properties database, respectively.
Based on the fuzzy theory, the similarity function makes a survey of the comparability. ${T}_{n}^{\prime }$ and Tn become more comparable when δ is reduced. Based on Eq. (61), it is possible to obtain several groups of MOBDT of the guidance mechanism as studied in this paper. There are 10 groups chosen in this paper.
## 4.3 Optimization method
Genetic algorithms represent one of the most adaptive methods to simulate the process of biological evolution on a computer. Its solution to the optimization problem starts by randomly generating the initial population that meets the constraints. Each individual in the population is taken as the first solution to the problem. Then, the selection, proliferation, crossover, mutation, and other genetic operations are conducted. By eliminating the basic size function which is clearly different from the objective function, a smaller basic size function is retained, and a new population is obtained. The new group members have less error than the previous generation group, which is significantly better than the previous generation. Through this continuous breeding evolution, the genetic algorithm is used to optimize the target mechanism.
The main steps are as follows:
• 1.
Given the size type, frame length, two connecting rod lengths, and the crank length that need to be optimized, the initial population size rate, mutation rate, and maximum algebra are set. x is taken as the optimization variable.
$x={\left[\mathit{\alpha },\mathit{\gamma },\mathit{\xi },\mathit{\beta }\right]}^{T}={\left[{x}_{\mathrm{1}},{x}_{\mathrm{2}},{x}_{\mathrm{3}},{x}_{\mathrm{4}}\right]}^{T},$
where X represents the initial population; and x1, x2, x3, and x4 indicate the corresponding quantity of α, γ, ξ, and β.
• 2.
Given the constraints and range (the corresponding ranges are given in Table 2) of the spherical four-bar rigid-body guidance, the constraints are
$\begin{array}{l}{x}_{\mathrm{1}}+{x}_{\mathrm{4}}<{x}_{\mathrm{2}}+{x}_{\mathrm{3}}\\ {x}_{\mathrm{1}}max\left[{x}_{\mathrm{2}},{x}_{\mathrm{3}}\right]\end{array}.$
• 3.
The size of the initial population of the spherical four-bar rigid-body guidance mechanism is set in random generation (Step 2).
• 4.
According to the above Eqs. (10) and (11), the position of the trajectory P point is determined for all the size types of the generated population.
• 5.
1D Fourier transform and 2D Fourier transform are performed on the solved p-point trajectory to extract the corresponding characteristic parameters.
• 6.
Equation (61) is used to calculate the similarity between the extracted characteristic parameters and the Fourier characteristic parameters of the objective function.
• 7.
All sizes are arranged in the order of similarity, and the individuals with low similarity replace the individuals with high similarity.
• 8.
Each of the two sizes forms a group and starts to cross-reproduce, with new sizes produced.
• 9.
The random selection of certain generations is performed for mutation.
• 10.
According to the results of Step (6), the similarity of all size types average, and the best individual similarities are expressed.
If the maximum algebra is reached, the optimization is terminated; otherwise, it returns to Step (4) to continue the optimization.
Table 2Length range of the rod.
Taking into account the characteristics of the output objective function of the spherical four-bar rigid-body guidance mechanism, the genetic algorithm is used in this paper to optimize the mechanism, as shown in Fig. 9. The population size is 500, and the crossover probability is 0.8. The two-point crossover operator is used, and the geometric programming sorting selection is carried out. The number of genetic iterations is 200.
Figure 9Optimization flowchart.
## 4.4 Illustration
The design requirements and reference point p are shown in Fig. 10, and the coordinates of sampling points are given in Table 3. There are 64 points given in this paper, of which, eight points (which are in bold) are chosen to translate the position problem of the rigid body into a path generation one. It is supposed that a mechanism studied by this paper is synthesized into a given angle and position.
Figure 10Design requirements.
Table 3The prescribed position and orientation.
Table 4The amplitude and phase of the prescribed orientation.
Based on the above analysis, a variable step size output properties database of the mechanism is first established. Through a 1D FFT and normalization process, the OHCPDT of the design requirements can be extracted. Table 4 shows four terms of amplitudes of the prescribed orientation. The matching method is used to identify 10 groups of the MOBDT from the database. According to each group of the MOBDT, we establish the initial population, so as to obtain the MOBDT of the desired mechanism, as shown in Table 5. Then, Eq. (52) is applied to obtain the actual dimensions of the μpq, as shown in Table 7. Table 6 shows the harmonic components of the prescribed position, and Table 8 shows the harmonic components of the RBPO as obtained through FFT. Table 7 lists the dimension parameters and installation parameters as obtained by Eqs. (53) to (60).
Table 5The identified MOBDT (10 groups are chosen).
Table 6The amplitude and phase of the prescribed position.
Table 7Dimension and arrangement parameters.
Table 8The amplitude and phase of the RBPO of the 10th group of the MPBDT.
Finally, the comparison diagrams of three groups of design results and given design requirements are shown in Figs. 11, 12, and 13. The design requirements are shown in blue, and the design results are shown in red. It can be seen that the fitting curve overlaps completely. Part (b) in the figure is the comprehensive result error in azimuth. Part (c) in the figure is the comprehensive result error of the pitch angle. Part (d) in the figure is the comprehensive result error in the rigid-body guidance angle.
Figure 11Comparison drawing of the first group of dimensions.
Figure 12Comparison drawing of the second group of dimensions.
Figure 13Comparison drawing of the third group of dimensions.
5 Conclusions
• 1.
A new method is proposed to address the motion generation problem of spherical four-bar crank slider rigid-body guidance mechanisms. According to the corresponding parameters of the position output and direction output of the research institutions in this paper, a method is developed to establish the output properties database of the mechanism. Then, the solution to the dimension parameters is proposed as well. By using the output properties database and the solution method, motion generation is achieved.
• 2.
The analytical methods are applicable only to deal with the problem of finitely divided positions. However, the method proposed in this paper can be used to effectively solve the problem of motion generation with infinite prescribed positions in theory. Through comparison with analytical methods, nonlinear equations are avoided in this paper, which makes the method much more efficient and simple.
• 3.
The method combines the advantages of the analytical method and the atlas method characterized by precision and simple calculation. Through a computer, it is achievable to quickly and accurately identify various groups of OHCPDT which meet the design requirements. Besides this, the dimension and installation parameters can be calculated as well, with several optimal options available to users. At last, this method can be extended to the motion generation of other types of linkage mechanisms, such as a spherical four-bar mechanism and RCCC mechanisms.
Code availability
The code is available upon request from the corresponding author.
Data availability
The data are available upon request from the corresponding author.
Author contributions
All the authors contributed to the study conception and design. The mathematical model and design method were proposed by JS and WL, experimental design and analysis were performed by ZL and HL. The first draft of the paper was written by WZ, ZL and WL. All the authors commented on previous versions of the paper. All the authors read and approved the final paper.
Competing interests
The contact author has declared that none of the authors has any competing interests.
Disclaimer
Publisher’s note: Copernicus Publications remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
Acknowledgements
The authors acknowledge financial support from the National Natural Science Foundation of China, Hubei Provincial Natural Science Foundation of China, the Science and Technology Research Project of the Jilin Provincial Department of Education, Central Government Guides Local Science and Technology Development Projects of Hubei Province, and Science and Technology Innovation Team of Hubei University of Arts and Science.
Financial support
This research has been supported by the National Natural Science Foundation of China (grant no. 51775054), Hubei Provincial Natural Science Foundation of China (grant no. 2022CFC035), the Science and Technology Research Project of the Jilin Provincial Department of Education (grant no. JJKH20220672KJ), Central Government Guides Local Science and Technology Development Projects of Hubei Province (grant no. 2018ZYYD016), and Science and Technology Innovation Team of Hubei University of Arts and Science (grant no. 2022pytd01).
Review statement
This paper was edited by Daniel Condurache and reviewed by five anonymous referees.
References
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Yu, H. Y., Tang, D. W., and Wang, Z. X.: Study on a new computer path synthesis method of a four-bar linkage, Mech. Mach. Theory, 42, 383–392, https://doi.org/10.1016/j.mechmachtheory.2006.05.003, 2007.
Wang, Z. X., Yu, H. Y., Tang, D. W., and Li, J. S.: Study on rigid-body guidance synthesis of planar linkage, Mech. Mach. Theory, 37, 673–684, https://doi.org/10.1016/S0094-114X(02)00014-9, 2002. | 20,351 | 63,951 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 102, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-23 | latest | en | 0.812021 |
https://www.esaral.com/q/the-value-of-76657 | 1,720,888,228,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514510.7/warc/CC-MAIN-20240713150314-20240713180314-00424.warc.gz | 605,242,805 | 11,578 | The value of
Question:
The value of sin-1 cos 33π/5 is
(a) 3π/5
(b) -7π/5
(c) π/10
(d) -π/10
Solution:
(d) -π/10
$\sin ^{-1}\left(\cos \frac{33 \pi}{5}\right)=\sin ^{-1}\left(\cos \left(6 \pi+\frac{3 \pi}{5}\right)\right)=\sin ^{-1}\left(\cos \frac{3 \pi}{5}\right)$
$=\sin ^{-1}\left[\sin \left(\frac{\pi}{2}-\frac{3 \pi}{5}\right)\right]$
$=\sin ^{-1}\left(\sin \left(-\frac{\pi}{10}\right)\right)$
$=-\frac{\pi}{10} \quad\left(\because \sin ^{-1}(\sin x)=x\right.$, for $\left.x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\right)$
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# Problem 21. Return the 3n+1 sequence for n
Solution 2075945
Submitted on 1 Jan 2020 by Andreas Neul
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
n = 1; c_correct = 1; assert(isequal(collatz(n),c_correct))
2 Pass
n = 2; c_correct = [2 1]; assert(isequal(collatz(n),c_correct))
3 Pass
n = 5; c_correct = [5 16 8 4 2 1]; assert(isequal(collatz(n),c_correct))
4 Pass
n = 22; c_correct = [22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1]; assert(isequal(collatz(n),c_correct))
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Start Hunting! | 242 | 736 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2020-50 | latest | en | 0.559931 |
http://math.stackexchange.com/questions/722331/is-there-any-difference-between-bounded-and-totally-bounded | 1,448,454,512,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398445142.9/warc/CC-MAIN-20151124205405-00184-ip-10-71-132-137.ec2.internal.warc.gz | 147,135,591 | 19,217 | # Is there any difference between Bounded and Totally bounded?
Is there any difference between bounded and totally bounded? (in a metric space)
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The real line $\Bbb R$ endowed with the metric $d(x,y)=\min(1,|x-y|)$ is a bounded metric space that isn't totally bounded.
-
+1: Nice, simple example. – copper.hat Mar 22 '14 at 17:12
A major theorem in metric space theory is that a metric space is compact if and only if it is complete and totally bounded. In $\mathbb{R}^{n}$ with the usual metric ( for $n < \infty$), bounded and totally bounded are the same, which is essentially the content of the Heine Borel theorem. In fact, the unit ball of a Banach space is compact if and only if the space is finite dimensional.
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Just to throw one more answer into the mix, any infinite set with the discrete metric is bounded but not totally bounded (I like @Gedgar's answer better, but just for some variety...)
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Every totally bounded set is bounded. But not conversely. The unit ball in Hilbert space is bounded, but not totally bounded.
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Maybe you mean an infinite dimensional Hilbert space? – Cameron Williams Mar 22 '14 at 17:12 | 281 | 1,139 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2015-48 | longest | en | 0.923544 |
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Question
$$-4(2p+5)+8p=-11$$
A. The equation has one solution, $$p=2$$.
B. The equation has one solution, $$p=-2$$.
C. The equation has no solution.
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# Find the derivative using the quotient rule $\frac{d}{dx}\left(5e^{7x}\cos\left(9x\right)\right)$
## Step-by-step Solution
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### Videos
$5e^{7x}\left(7\cos\left(9x\right)-9\sin\left(9x\right)\right)$
Got another answer? Verify it here!
## Step-by-step Solution
Problem to solve:
$\frac{d}{dx}\left(5\cdot e^{7x}\cdot \cos\left(9x\right)\right)$
Specify the solving method
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The derivative of a function multiplied by a constant ($5$) is equal to the constant times the derivative of the function
$5\frac{d}{dx}\left(e^{7x}\cos\left(9x\right)\right)$
2
Apply the product rule for differentiation: $(f\cdot g)'=f'\cdot g+f\cdot g'$, where $f=e^{7x}$ and $g=\cos\left(9x\right)$
$5\left(\frac{d}{dx}\left(e^{7x}\right)\cos\left(9x\right)+e^{7x}\frac{d}{dx}\left(\cos\left(9x\right)\right)\right)$
Learn how to solve differential calculus problems step by step online.
$5\frac{d}{dx}\left(e^{7x}\cos\left(9x\right)\right)$
Learn how to solve differential calculus problems step by step online. Find the derivative using the quotient rule d/dx(5e^(7x)cos(9x)). The derivative of a function multiplied by a constant (5) is equal to the constant times the derivative of the function. Apply the product rule for differentiation: (f\cdot g)'=f'\cdot g+f\cdot g', where f=e^{7x} and g=\cos\left(9x\right). The derivative of the cosine of a function is equal to minus the sine of the function times the derivative of the function, in other words, if f(x) = \cos(x), then f'(x) = -\sin(x)\cdot D_x(x). The derivative of the linear function times a constant, is equal to the constant.
$5e^{7x}\left(7\cos\left(9x\right)-9\sin\left(9x\right)\right)$
## Explore different ways to solve this problem
Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more
Find the derivativeFind d/dx(5e^(7x)cos(9x)) using the product ruleFind d/dx(5e^(7x)cos(9x)) using logarithmic differentiation
SnapXam A2
### beta Got a different answer? Verify it!
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$\frac{d}{dx}\left(5\cdot e^{7x}\cdot \cos\left(9x\right)\right)$ | 1,016 | 2,700 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2023-23 | latest | en | 0.578756 |
https://math.answers.com/other-math/How_many_nickels_make_500_dollars | 1,718,321,893,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861517.98/warc/CC-MAIN-20240613221354-20240614011354-00106.warc.gz | 344,254,272 | 47,632 | 0
# How many nickels make 500 dollars?
Updated: 4/28/2022
Wiki User
8y ago
A nickel is 5 cents.
There are 100 cents in a dollar.
So the number of nickels in a dollar is (100/5) = 20
You have 500 dollars, each worth 20 nickels, so the total number of nickels is the product.
500/0.05 = 10,000
Wiki User
8y ago
Earn +20 pts
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# Probability Quiz
Attempt now to get your rank among 10 students!
Question 1
Three numbers are chosen at random without replacement from 1,2,3,….8. The probability that their minimum is 3, given that their maximum is 6, is______
Question 2
Let A and B be two events such that P(AB’)=0.20,P(A’B)=0.15,P(A’B’)=0.1,then P(A/B) is equal to
Question 3
In India, 51% of the adults are males. One adult is randomly selected for a survey involving credit card usage. It is later learned that the selected survey subject was smoking a cigar. Also, 9.5% of males smoke cigars, whereas 1.7% of females smoke cigars. find the probability that the selected subject is a male.
Question 4
Three dice are thrown at the same time. Find the probability of getting three two’s, if it is known that the sum of the numbers on the dice was six.
Question 5
Two numbers are drawn at random from the integers 1 to 9. If the sum of numbers is even, find the probability that both are odd.
Question 6
Which one of the following option is correct?
Question 7
A speaks the truth in 75% cases where B speaks the truth in 20% cases. The probability of A speaking the truth under condition that B is speaking the truth is?
Question 8
The probability that in a year of the 22nd century chosen at random at random, there will be 53 Sunday, is
Question 9
A bag A contain 2 white balls and 3 red balls, another bag B contains 4 white and 5 black balls. A bag is selected at random and a ball is drawn from it. Drawn ball is observed to be white. Find the probability that bag B is selected.
Question 10
A computer producing factory has only two plants T1 and T2. Plant T1 produces 20% and plant T2 produces 80% of the total computer produced. 7% of computers produced in the factory turn out to be defective. It is known that P(computer turns out to be defective given that it is produced in plant T1)=10P(computer turns out to be defective given that it is produced in plant T2), When P(E) denotes the probability of an event E. A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability it is produced in plant T2 is____
• 10 attempts | 555 | 2,206 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2022-49 | latest | en | 0.95074 |
http://publish.illinois.edu/ymb/2016/02/15/topologically-constrained-models-of-statistical-physics/ | 1,558,843,231,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258621.77/warc/CC-MAIN-20190526025014-20190526051014-00530.warc.gz | 173,853,323 | 12,208 | # Topologically constrained models of statistical physics.
$$\def\Real{\mathbb{R}} \def\Int{\mathbb{Z}} \def\Comp{\mathbb{C}} \def\Rat{\mathbb{Q}} \def\Field{\mathbb{F}} \def\Fun{\mathbf{Fun}} \def\e{\mathbf{e}} \def\f{\mathbf{f}} \def\bv{\mathbf{v}} \def\blob{\mathcal{B}}$$
### The Blob
Consider the following planar “spin model”: the state of the system is a function from $$\Int^2$$ into $$\{0,1\}$$ (on and off states). We interpret the site $$(i,j), i,j\in\Int$$ as the plaque, i.e. the (closed) square given by the inequalities $$s_{ij}:=i-1/2\leq x\leq i+1/2; j-1/2\leq y\leq j+1/2$$.
To any configuration $$\eta$$ we associate the corresponding active domain,
$A_\eta=\bigcup_{(i,j): \eta(i,j)=1} s_{ij}.$
We are interested in the statistical ensembles supported by the finite contractible active domains – let’s refer to such domains as blobs. In other words, blobs are finite collections of plaques that are topologically equivalent to a disk. We will denote the collection of blobs as $$\blob$$.
It is worth remarking that the contractibility is a natural condition if one is interested in local models of statistical physics – such that the transition rates between the states can be defined in terms of local factors, – unlike, say, connectivity. More precisely, one can define a (symmetric) collection of transitions between states $$\eta,\eta’$$ differing at just one site $$\sigma$$ such that the feasibility of the transitions depends only on the intersections of $$\eta,\eta’$$ with a $$3\times 3$$ vicinity of $$\sigma$$, such that $$\eta’$$ is a blob iff $$\eta$$ is a blob, and turning $$\blob$$ into a connected graph (which we still refer to as $$\blob$$).
Given a fugacity $$\rho$$, on can turn $$\blob$$ into a Markov chain, by setting the jump rate adding a site as $$\rho$$, and the jump rate removing a site as $$1$$. We will denote this Markov chain as $$\blob(rho)$$.
There are several connections of the blob to the staple models of statistical physics. In essence, we are looking at the Ising model with just one contour. The standard constructions pertaining to, say, Wulff shapes, or the general framework of abstract polymer models. Alternatively, one can consider the blob as a self-avoiding loop weighted by the area it encloses.
Whichever the connection, the model seems to be new, and deserves some attention.
Shape of the blob after $$10^5$$ growth steps, with fugacity $$\rho=.9$$.
#### Size, shape and agility
Consider the subcritical fugacity $$\rho<1$$. In this case, the size of the blob is a.s. finite. What is the tail distribution of its diameter (or area)? If $$\rho=1-\epsilon$$ is close to one, what is the expected size of the blob?
Ir seems that the shape of the blob is fractal like, but at a large scale resembling a round disk. Is that true?
Macroscopically, one can expect the blob to drift, converging, upon proper rescaling to Brownian motion. What are the correct scales?
#### Spanning trees
Consider the situation where some of the nodes are stuck in the active state (equivalently, their rate of switching from ON to OFF is zero). In the limit of small fugacity $$\rho\to 0$$ the resulting blob will be approximating a certain one-dimensional complex minimizing a functional – i.e. behaving like a Steiner tree.
Can we make this intuitive picture precise?
#### Tail behavior
More on the sites stuck in the active state: assume one starts with the (infinite) configuration where the whole halfplane (say, the upper halfplane) is active. Assume that the site at the origin is always ON, and all other sites perform the usual dynamics for some small fugacity. On every compact subset, the resulting blob will look like a tail starting at the origin and meandering off to infinity. What are the properties of this blob? How many sites are typically active in a box of size $$N\times N$$? | 990 | 3,865 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2019-22 | latest | en | 0.866064 |
https://ocw.oouagoiwoye.edu.ng/courses/chemistry/5-73-introductory-quantum-mechanics-i-fall-2002/assignments/ | 1,713,875,107,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818474.95/warc/CC-MAIN-20240423095619-20240423125619-00722.warc.gz | 392,568,636 | 7,096 | # Assignments
This section provides all of the course's problem sets and a partial set of corresponding solutions.
## Problem Sets
Problem Set 1 (PDF)
Problem Set 2 (PDF)
Problem Set 3 (PDF)
Problem Set 4 (PDF)
Problem Set 5 (PDF)
Problem Set 6 (PDF)
Problem Set 7 (PDF)
Problem Set 8 (PDF)
Problem Set 9 (PDF)
Problem Set 10 (PDF)
## Problem Set Solutions
Problem Set 1, Number 1:
1. Mathematica® Notebook, Kyle's version (NB)
Problem Set 4, Number 1:
1. Input to RKR for X-state Potential (TXT)
2. Output from RKR for X-state Potential (TXT)
3. Output from RKR for X-state Potential (channel 7) (TXT)
4. Input to RKR for A-state Potential (TXT)
5. Output from RKR for A-state Potential (TXT)
6. Output from RKR for A-state Potential (channel 7) (TXT)
7. Plot of X- and A-state Potentials (GIF)
8. Input to LEVEL for Franck-Condon Factors (TXT)
9. Output from LEVEL for Franck-Condon Factors (TXT)
10. Output from LEVEL for Franck-Condon Factors (channel 8) (TXT)
11. Output from LEVEL for Franck-Condon Factors (channel 10) (TXT)
12. Plot of Franck-Condon Factors (GIF)
13. Input to LEVEL for Matrix Elements of R (TXT)
14. Output from LEVEL for Matrix Elements of R (TXT)
15. Output from LEVEL for Matrix Elements of R (channel 8) (TXT)
16. Input to LEVEL for Matrix Elements of P (TXT)
17. Output from LEVEL for Matrix Elements of P (TXT)
18. Output from LEVEL for Matrix Elements of P (channel 8) (TXT) | 403 | 1,415 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-18 | latest | en | 0.646931 |
https://www.jiskha.com/display.cgi?id=1351557973 | 1,503,312,095,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886108264.79/warc/CC-MAIN-20170821095257-20170821115257-00611.warc.gz | 783,685,196 | 3,880 | # Physical science
posted by .
Which is not an example of acceleration?
a) A person jogging at 3m/s along a winding path
b) A car stopped at stop sign
c) A cheetah running 27m/s east
d) A plane taking off
## Similar Questions
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A cheetah can go from a state of resting to running at 20 m/s in just 2 seconds. What is the cheetah’s average acceleration?
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A cheetah can go from a state of resting to running at 20 m/s in just 2 seconds. What is the cheetah’s average acceleration?
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A cheetah is crouched in ambush 20m to the east of an observers blind. At time t=0 the cheetah charges an aeroplane in a cleaning 50m east of the observer. The cheetah runs along the straight line.
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A car travels east for 15km at a speed of 30km/h .it then turns around and returns along the same path at 60km/h.what is the cars average velocity
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Which of the following is not an example of acceleration. A a person jogging atn3 m/s along a winding path B a car stopping at a stop sign. C a cheetah running 27m/s east D a plane taking off
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a car accelerates from rest at stop street and covers 10m east in 45.calculate the acceleration of the car(assume uniform acceleration).
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More Similar Questions | 586 | 2,317 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2017-34 | latest | en | 0.915079 |
https://eepower.com/capacitor-guide/applications/motor-starting-capacitor/ | 1,675,922,572,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764501407.6/warc/CC-MAIN-20230209045525-20230209075525-00219.warc.gz | 242,880,337 | 15,859 | # Motor starting capacitor
## Motor capacitors
AC induction motors use a rotating magnetic field to produce torque. Three-phase motors are widely used because they are reliable and economical. The rotating magnetic field is easily achieved in three-phase asynchronous motors because the phase angle offset between the individual phases is 120 degrees. However, single-phase AC motors require external circuitry which creates the phase angle offset in order to produce a rotating magnetic field. This circuitry can be realized using advanced power electronics, or more simply using a motor capacitor.
The video below shows an easy to understand explanation of the working principle of the AC induction motor.
## AC single-phase induction motors
### Single-coil AC induction motors
AC induction motors usually use two or more coils to generate a rotating magnetic field, which produces torque on the rotor. When a single coil is used, it will generate a pulsating magnetic field, which is enough to sustain rotation, but not sufficient to start the motor from a standstill. Motors with a single coil have to be started by using an external force, and can rotate in either direction. The direction of the rotation depends on the external force. If the motor was started in a clockwise direction, it will continue to rotate and build up speed in the clockwise direction, until it reaches a maximum speed which is defined by the power source frequency. Similarly, it will continue rotating counter-clockwise if the initial rotation was counter-clockwise. These motors are not practical due to their inability to reliably start rotation on their own.
### Start capacitor AC induction motors
One way to improve on the single coil design is by using an auxiliary coil in series with a motor starting capacitor. The auxiliary coil, also called starting coil, is used to create an initial rotating magnetic field. In order to create a rotating magnetic field, the current flowing through the main winding must be out of phase in respect to the current flowing through the auxiliary winding. The role of the starting capacitor is to lag the current in the auxiliary winding, bringing these two currents out of phase. When the rotor reaches sufficient speed, the auxiliary coil is disconnected from the circuit by means of a centrifugal switch, and the motor remains powered by a single coil creating a pulsating magnetic field. In this sense, the auxiliary coil in this design can be regarded as a starting coil, since it is only used during motor startup.
### Start/run capacitor AC induction motors
Another way to further improve on the single-coil single-phase induction motor design is to introduce an auxiliary coil, which remains powered not only during the motor startup phase, but also during normal operation. As opposed to an AC motor using only a motor start capacitor, which creates a pulsating magnetic field during normal operation, AC motors using a motor start capacitor and a motor run capacitor create a rotating magnetic field during normal operation. The function of the motor start capacitor remains the same as in the previous case - it gets disconnected from the circuit after the rotor reaches a predetermined speed by means of a centrifugal switch. After that point, the auxiliary winding remains powered through a motor run capacitor. The figure below describes this type of design.
## Motor start and motor run capacitors
### Start capacitors
Motor start capacitors are used during the motor startup phase and are disconnected from the circuit once the rotor reaches a predetermined speed, which is usually about 75% of the maximum speed for that motor type. These capacitors usually have capacitance values of over 70 µF. They come in various voltage ratings, depending on the application they were intended for.
### Run capacitors
Some single phase AC motor designs use motor run capacitors, which are left connected to the auxiliary coil even after the start capacitor is disconnected by the centrifugal switch. These designs operate by creating a rotating magnetic field. Motor run capacitors are designed for continuous duty, and remain powered whenever the motor is powered, which is why electrolytic capacitors are avoided, and low-loss polymer capacitors are used instead. The capacitance value of run capacitors is usually lower than the capacitance of start capacitors, and is often in the range of 1.5 µF to 100 µF. Choosing a wrong capacitance value for a motor can result in an uneven magnetic field, which can be observed as uneven motor rotation speed, especially under load. This can cause additional noise from the motor, performance drops and increased energy consumption, as well as additional heating, which can cause the motor to overheat.
## Applications
Motor start and run capacitors are used in single-phase AC induction motors. Such motors are used whenever a single-phase power supply is more practical than a three-phase power supply, such as in domestic appliances. They are not as efficient as three-phase AC induction motors, however. In fact, single-phase AC motors are 2 to 4 times less efficient than three-phase AC motors, which is why they are used only for less powerful motors. Typical applications which utilize start and run motor capacitors include power tools, washing machines, tumble dryers, dishwashers, vacuum cleaners, air conditioners and compressors. | 1,030 | 5,432 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-06 | longest | en | 0.910931 |
https://bailaconmigo.pl/ball-mill/mssrpz0q/rate.html | 1,660,337,321,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571758.42/warc/CC-MAIN-20220812200804-20220812230804-00731.warc.gz | 146,179,496 | 7,148 | Get in Touch
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# rate of milling ball mill
Generally, filling the mill by balls must not exceed 30%–35% of its volume. The productivity of ball mills depends on the drum diameter and the relation of ∫ drum diameter and length. The optimum ratio between length L and diameter D, L: D, is usually accepted in the range 1.56–1.64
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• AMIT 135: Lesson 7 Ball Mills & Circuits – Mining
Wet Ball Mill = kg kWh = 0.16(A i-0.015) 0.33; Dry Ball Mill = kg / kWh = 0.023A i 0.5; Replacement Ball Size. Rowland and Kjos proposed the use of their equation for the determination of the initial and replacement media size. Azzaroni (1981) and Dunn (1989) recommended the use of
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• Ball Mill - an overview | ScienceDirect Topics
The company claims this new ball mill will be helpful to enable extreme high-energy ball milling at rotational speed reaching to 1,100 rpm. This allows the new mill to achieve sensational centrifugal accelerations up to 95 times Earth gravity
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• Speeds And Feeds For Milling With End Mills
Milling Speeds and Feeds Charts The most important aspect of milling with carbide end mills is to run the tool at the proper rpm and feed rate. We have broken these recommendations down into material categories so you can make better decisions with how to productively run your end mills
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• Ball Nose Finishing Mills Speed & Feed Calculator
Ball Nose Finishing Mills Speed & Feed Calculator. Instructions: Fill in the blocks shaded in blue with your application information. The calculator will automatically provide the necessary speed and feed in the green fields. For assistance setting up your milling program, contact a
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• EFFECT OF BALL SIZE DISTRIBUTION ON MILLING
2.1 Breakage mechanisms in a ball mill 22 2.2 First order reaction model applied to milling 24 2.3 Grinding rate versus particle size for a given ball diameter 25 2.4 Cumulative breakage function versus relative size 28 2.5 Predicted variation of S i values with ball diameter for dry grinding of quartz 31
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• (PDF) A comparison of wear rates of ball mill grinding
abrasive and impact wear due to their large. (75 – 100 mm) dia meters. Ball mill balls. experience a greater number of impacts, but at. lower magnitude than SAG mill balls, due t o. the smaller
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• MILLING CONTROL & OPTIMISATION
Millstar Ball Load Estimator Millstar Mill Power Filter MillStar Intelligent Filtering and Fault Detection for Level Signals, Density, ... On milling plants fed by a segregated feed supply, such as a ... Comparison of Plant vs MillStar Mill Feed Control Feed Rate (t/h) % of Total Data Points. MillStar OFF MillStar ON 100 200 300 400 500 0
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• Effect of Grinding Media on Milling
May 30, 2017 Effect of Balls Hardness on Rate of Grinding Tests by Coghill and Devaney show that the hardness of the balls has some effect upon the rate of grinding in mill. These results are plotted in Fig. 6.8a and it is seen that, both for dolomite and chert, there is an increase in the rate of grinding with increasing hardness of ball
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• Ball Mill Rate Of Milling
Ball Mill Rate Of Milling-India Crusher&Mill. ball mill rate of milling; Ball Nose Troubleshooting, Reference & Technical Info DAPRA. Ball Nose milling application information. Feed
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• (PDF) Environment-dependent breakage rates in ball milling
Keywords: Breakage rates; Ball milling; Mill environment 1. Introduction [10] analyzed the influence of slurry density on the breakage parameters of quartz in a laboratory ball mill, and concluded that, in wet batch milling, the breakage Over the past thirty years or so, there has been a
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• AMIT 135: Lesson 7 Ball Mills & Circuits – Mining Mill
For overflow ball mills, the charge should not exceed 45% of the mill volume . For grate discharge mills, the charge should occupy about 50% of the mill volume . Bond developed a relationship that can be used to determine the percent charge by volume as a function of the vertical height above the charge, He, and the radius of the mill
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• Ball Grinding Mill | Ball Milling Machine - Quality Ball
Ball Milling Machine - Quality Ball Mill Supplier The ball mill uses are wide, as an indispensable ball grinding mill for various types of metal and non-metal ore beneficiation, cement manufacturing and other industries, it can effectively improve the utilization rate of ground materials and make full use of resources
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• (PDF) Grinding in Ball Mills: Modeling and Process Control
Jun 01, 2012 the milling process takes place during rotation as a result of the transfer of kinetic. energy of the moving grinding media into the grinding product. The design of a ball mill can vary
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• SPEEDS AND FEEDS FOR CARBIDE ENDMILLS
convential milling should be used and feed per tooth should be reduced by 50% please note: the above recommendations should be considered only as a starting point; fine tuning may be required in order to maximize performance 96. catalog twenty twelve internal tool inc. iti
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• What's the Difference Between SAG Mill and Ball Mill
Nov 26, 2019 The biggest characteristic of the sag mill is that the crushing ratio is large. The particle size of the materials to be ground is 300 ~ 400mm, sometimes even larger, and the minimum particle size of the materials to be discharged can reach 0.1 mm. The calculation shows that the crushing ratio can reach 3000 ~ 4000, while the ball mill’s
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• Optimization of mill performance by using
As the ball wear rate depends directly on the surface of the media charge, a small variation in power will lead to an important increase of wear rate. The risk of underloading or overloading the mill is an additional factor. A direct measurement of the ball level in the mill, more accurate than power readings, as well as a control of it, is
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• Investigating granular milling in a hammer mill
the milling behavior in Ball Mill have been performed [2–6]. Experiments and Discrete Element Method (DEM) based modeling of milling of pharmaceutical excipients in a very small scale oscillatory single was performed by Kwan et al. [7]. Campbell et al. [8] developed the relationship between the inlet and outlet
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• Ball Mill Parameter Selection & Calculation - Power
Aug 30, 2019 1 Calculation of ball mill capacity. The production capacity of the ball mill is determined by the amount of material required to be ground, and it must have a certain margin when designing and selecting. There are many factors affecting the production capacity of the ball mill, in addition to the nature of the material (grain size, hardness, density, temperature and humidity), the degree of
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Related Blog | 1,543 | 6,782 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-33 | latest | en | 0.792502 |
https://mail.python.org/pipermail/python-list/2007-March/418736.html | 1,397,849,720,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609535095.7/warc/CC-MAIN-20140416005215-00528-ip-10-147-4-33.ec2.internal.warc.gz | 787,230,698 | 1,931 | # Finding the insertion point in a list
John Machin sjmachin at lexicon.net
Sat Mar 17 11:29:39 CET 2007
```On Mar 17, 5:42 pm, Paul Rubin <http://phr...@NOSPAM.invalid> wrote:
> Steven D'Aprano <s... at REMOVE.THIS.cybersource.com.au> writes:
> > or even
>
> > len(filter(lambda t, y=y: y>t, x))
>
>
> min(i for i,t in enumerate(x) if t >= y)
>
> or
>
> max(i for i,t in enumerate(x) if t <= y)
>
> Those are actually pretty direct.
I'd hate to see "indirect". Worse, the min-using gizmoid crashes when
y > x[-1] -- all your ifs are belong to False.
>>> x
[0, 100, 200, 1000]
>>> tests = [0, 1, 100, 150, 1000, 2000]
>>> [(y, max(i for i,t in enumerate(x) if t <= y)) for y in tests]
[(0, 0), (1, 0), (100, 1), (150, 1), (1000, 3), (2000, 3)]
Looks OK, iff one is happy with the OP's strange usage of "insert
point".
>>> xc = x[:]
>>> xc.insert(1, 150)
>>> xc
[0, 150, 100, 200, 1000]
Whoops.
Try this for size:
>>> [(y, sum(t <= y for t in x)) for y in tests]
[(0, 1), (1, 1), (100, 2), (150, 2), (1000, 4), (2000, 4)]
Cheers,
John
``` | 408 | 1,053 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2014-15 | latest | en | 0.637861 |
https://crazyproject.wordpress.com/2010/07/31/basic-properties-of-nilpotent-ring-elements/ | 1,485,167,923,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560282631.80/warc/CC-MAIN-20170116095122-00158-ip-10-171-10-70.ec2.internal.warc.gz | 813,844,814 | 18,611 | ## Basic properties of nilpotent ring elements
Let $R$ be a commutative ring and let $x \in R$ be nilpotent – that is, $x^n = 0$ for some positive integer $n$. Prove the following.
1. $x$ is either zero or a zero divisor.
2. $rx$ is nilpotent for all $r \in R$.
3. $1+x$ is a unit in $R$.
4. The sum of a unit and a nilpotent element is a unit.
1. Say $m$ is minimal such that $x^m = 0$. If $m = 1$, then $x = 0$. If $m > 1$, then $x \neq 0$, $x^{m-1} \neq 0$ and $x \cdot x^{m-1} = 0$, so that $x$ is a zero divisor.
2. Since $R$ is commutative, we have $(rx)^m = r^mx^m = 0$.
3. Note that $(1 - (-x))(\sum_{i=0}^{m-1} (-x)^i)$ $= (\sum_{i=0}^{m-1} (-x)^i) - (\sum_{i=0}^{m-1} (-x)^{i+1})$ $= (\sum_{i=0}^{m-1} (-x)^i) - (\sum_{i=1}^{m} (-x)^i$ $= 1 + (\sum_{i=1}^{m-1} (-x)^i) - (\sum_{i=1}^{m-1} (-x)^i) - (-x)^m$ $= 1 - (-1)^mx^m$ $= 1$. Thus $1+x$ is a unit.
4. Let $u$ be a unit and $x$ nilpotent. Then $u^{-1}x$ is nilpotent, so $1+u^{-1}x$ is a unit, and thus $u(1+u^{-1}x) = u+x$ is a unit. | 473 | 1,002 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 32, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2017-04 | longest | en | 0.513279 |
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# Advice needed , please
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Manager
Joined: 12 Dec 2012
Posts: 230
Concentration: Leadership, Marketing
GMAT 1: 540 Q36 V28
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Advice needed , please [#permalink]
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13 Feb 2013, 18:06
I took the gmat just today , the 13th and was shocked by a 540 despite I aimed at a minimum of 650 - 700 and studied for 3 months . Intending to retake after 4-6 weeks , I am in urgent need of your valued advice . What shall I do ? I have to mention that I identify and categorize 80-90% of the questions I see , and for most of them (saying 70-80 %) I can strategize the required approach.
Thanks
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Re: Advice needed , please [#permalink]
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14 Feb 2013, 12:12
TheNona wrote:
I took the gmat just today , the 13th and was shocked by a 540 despite I aimed at a minimum of 650 - 700 and studied for 3 months . Intending to retake after 4-6 weeks , I am in urgent need of your valued advice . What shall I do ? I have to mention that I identify and categorize 80-90% of the questions I see , and for most of them (saying 70-80 %) I can strategize the required approach.
Thanks
Dear TheNona
I'm very sorry to hear of your unfortunately performance on the GMAT yesterday, and I hope I can help.
First of all, here is a month-long study plan that may help you:
http://magoosh.com/gmat/2012/1-month-gm ... -schedule/
On the right side bar, notice there are links to four version of the 3-month study plans --- you may glance at the version that seems most apt for you, and if you have a little more than a month, supplement some additional material from one of the 3-month plans.
Here's a blog article about what it means to "understand", and the ways that tricky word can belie some problems.
http://magoosh.com/gmat/2012/understand ... rformance/
You will find other articles on that free blog full of GMAT strategies and content that can help you. At the very least, please get everything you can out of our free blog.
I would also like to recommend Magoosh --- you already qualify for our score guarantee
http://gmat.magoosh.com/score-guarantee
We have 200+ lessons covering all the content & strategies you need.
Here's a practice math question:
http://gmat.magoosh.com/questions/1003
Here's a practice SC question:
http://gmat.magoosh.com/questions/3264
For each, when you submit your answer, the next page will have a full video explanation. Each one of our 800+ practice GMAT question has its own video explanation, for accelerated learning. We also provide online support to all our questions and all the OG & GMATPrep questions. Sign up for a free one-week trial and see how much we can help you.
Please let me know if you have any questions.
Mike
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Mike McGarry
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Kudos [?]: 79 [0], given: 181
Re: Advice needed , please [#permalink]
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14 Feb 2013, 14:24
mikemcgarry wrote:
TheNona wrote:
I took the gmat just today , the 13th and was shocked by a 540 despite I aimed at a minimum of 650 - 700 and studied for 3 months . Intending to retake after 4-6 weeks , I am in urgent need of your valued advice . What shall I do ? I have to mention that I identify and categorize 80-90% of the questions I see , and for most of them (saying 70-80 %) I can strategize the required approach.
Thanks
Dear TheNona
I'm very sorry to hear of your unfortunately performance on the GMAT yesterday, and I hope I can help.
First of all, here is a month-long study plan that may help you:
http://magoosh.com/gmat/2012/1-month-gm ... -schedule/
On the right side bar, notice there are links to four version of the 3-month study plans --- you may glance at the version that seems most apt for you, and if you have a little more than a month, supplement some additional material from one of the 3-month plans.
Here's a blog article about what it means to "understand", and the ways that tricky word can belie some problems.
http://magoosh.com/gmat/2012/understand ... rformance/
You will find other articles on that free blog full of GMAT strategies and content that can help you. At the very least, please get everything you can out of our free blog.
I would also like to recommend Magoosh --- you already qualify for our score guarantee
http://gmat.magoosh.com/score-guarantee
We have 200+ lessons covering all the content & strategies you need.
Here's a practice math question:
http://gmat.magoosh.com/questions/1003
Here's a practice SC question:
http://gmat.magoosh.com/questions/3264
For each, when you submit your answer, the next page will have a full video explanation. Each one of our 800+ practice GMAT question has its own video explanation, for accelerated learning. We also provide online support to all our questions and all the OG & GMATPrep questions. Sign up for a free one-week trial and see how much we can help you.
Please let me know if you have any questions.
Mike
Dear Mike ,
No words can express my sincere gratitude . Thank you and wishing you all the best
_________________
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http://gmatclub.com/forum/the-rc-recipe-149577.html
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http://gmatclub.com/forum/the-simplest-problem-takeaway-template-150646.html
Manager
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Re: Advice needed , please [#permalink]
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15 Feb 2013, 14:18
Nona,
What was your quant and verbal score?
Posted from my mobile device
_________________
Check out my website, a First-Hand view of what it's like to go to Business School! http://www.carlolivieri.com
Manager
Joined: 12 Dec 2012
Posts: 230
Concentration: Leadership, Marketing
GMAT 1: 540 Q36 V28
GMAT 2: 550 Q39 V27
GMAT 3: 620 Q42 V33
GPA: 2.82
WE: Human Resources (Health Care)
Followers: 4
Kudos [?]: 79 [0], given: 181
Re: Advice needed , please [#permalink]
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16 Feb 2013, 13:30
CarlMtl wrote:
Nona,
What was your quant and verbal score?
Posted from my mobile device
36 quant , 28 verbal.These are the least scores I have ever had
_________________
My RC Recipe
http://gmatclub.com/forum/the-rc-recipe-149577.html
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http://gmatclub.com/forum/the-simplest-problem-takeaway-template-150646.html
Re: Advice needed , please [#permalink] 16 Feb 2013, 13:30
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If you work with formulas in Excel, sooner or later you will encounter the problem where Excel formulas don’t work at all (or give the wrong result).
While it would have been great had there been only a few possible reasons for malfunctioning formulas. Unfortunately, there are too many things that can go wrong (and often does).
But since we live in a world that follows the Pareto principle, if you check for some common issues, it’s likely to solve 80% (or maybe even 90% or 95% of the issues where formulas are not working in Excel).
And in this article, I will highlight those common issues that are likely the cause of your Excel formulas not working.
So let’s get started!
Let me start by pointing out the obvious.
Every function in Excel has a specific syntax – such as the number of arguments it can take or the type of arguments it can accept.
And in many cases, the reason your Excel formulas are not working or gives the wrong result could be an incorrect argument (or missing arguments).
For example, the VLOOKUP function takes three mandatory arguments and one optional argument.
If you provide a wrong argument or don’t specify the optional argument (where it’s needed for the formula to work), it’s going to give you a wrong result.
For example, suppose you have a dataset as shown below where you need to know the score of Mark in Exam 2 (in cell F2).
If I use the below formula, I will get the wrong result, because I am using the wrong value in the third argument (one that asks for Column Index number).
=VLOOKUP(A2,A2:C6,2,FALSE)
In this case, the formula calculates (as it returns a value), but the result is incorrect (instead of score in Exam 2, it gives the score in Exam 1).
Another example where you need to be cautious about the arguments is when using VLOOKUP with the approximate match.
Since you need to use an optional argument to specify where you want VLOOKUP to do an exact match or an approximate match, not specifying this (or using the wrong argument) can cause issues.
Below is an example where I have the marks data for some students and I want to extract marks in Exam 1 for the students in the table on the right.
When I use the below VLOOKUP formula it gives me an error for some names.
=VLOOKUP(E2,\$A\$2:\$C\$6,2)
This happens as I have not specified the last argument (which is used to determine whether to do an exact match or approximate match). When you don’t specify the last argument, it automatically does an approximate match by default.
Since we needed to do an exact match in this case, the formula returns an error for some names.
While I have taken the example of the VLOOKUP function, in this case, this is something that can be applicable for many Excel formulas that have optional arguments as well.
Also read: Identify Errors Using Excel Formula Debugging
Leading, trailing spaces are hard to find out and can cause issues when you use a cell that has these in formulas.
For example, in the below example, if I try to use VLOOKUP to fetch the score for Mark, it gives me the #N/A error (the not available error).
While I can see that the formula is correct and the name ‘Mark’ is clearly there is the list, what I can not see is that there is a trailing space in the cell that has the name (in cell D2).
Excel doesn’t consider the content of these two cells the same and therefore it considers it a mismatch when fetching the value using VLOOKUP (or it could be any other lookup formula).
To fix this issue, you need to remove these extra space characters.
You can do this by using any of the following methods:
Clean the cell and remove any leading/trailing spaces before using it in formulas
Use the TRIM function within the formula to make sure any leading/trailing/double spaces are ignored.
In the above example, you can use the below formula instead to make sure it works.
=VLOOKUP(TRIM(D2),\$A\$2:\$B\$6,2,0)
While I have taken the VLOOKUP example, this is also a common issue when working with TEXT functions.
For example, if I use the LEN function to count the total number of characters in a cell, if there are leading or trailing spaces, these would also be counted and give the wrong result.
This one setting can drive you crazy (if you don’t know it’s what’s causing all the issues).
Excel has two calculation modes – Automatic and Manual.
By default, the automatic mode is enabled, which means that in case I use a formula or make any changes in the existing formulas, it automatically (and instantly) makes the calculation and gives me the result.
This is the setting we are all used to.
In the automatic setting, whenever you make any change in the worksheet (such as entering a new formula to even some text in a cell), Excel automatically recalculates everything (yes, everything).
But in some cases, people enable the manual calculation setting.
This is mostly done when you have a heavy Excel file with a lot of data and formulas. In such cases, you may not want Excel to recalculate everything when you make small changes (as it may take a few seconds or even minutes) for this recalculation to complete.
If you enable manual calculation, Excel will not calculate unless you force it to.
And this may make you think that your formula is not calculating.
All you need to do in this case is either set the calculation back to automatic or force a recalculation by hitting the F9 key.
Below are the steps to change the calculation from manual to automatic:
Select Automatic
Important: In case you’re changing the calculation from manual to automatic, it’s a good idea to create a backup of your workbook (just in case this makes your workbook hang or makes Excel crash)
Take away / How to Fix: If you notice that your formulas are not giving the expected result, try something simple in any cell (such as adding 1 to an existing formula. Once you identify the issue as the one where calculation mode needs to be changed, do a force calculation by using F9.
One of the things that can have a devastating effect on your existing formulas in Excel is when you delete any row/column which has been used in calculations.
When this happens, sometimes, Excel adjusts the reference itself and makes sure that the formulas are working fine.
And sometimes… it can not.
Thankfully, one clear indication that you get when formulas break on deleting cells/rows/columns is the #REF! error in the cells. This is a reference error that tells you that there is some issue with the references in the formula.
Let me show you what I mean by using an example.
Below I have used the SUM formula to add the cells A2:A6.
Now, if I delete any of these cells/rows, the SUM formula will return a #REF! error. This happens because when I deleted the row, the formula doesn’t know what to reference now.
You can see that the third argument in the formula has become #REF! (which earlier referred to the cell that we deleted).
Take away / How to Fix: Before you delete any data that is being used in formulas, make sure there are no errors after the deletion. It’s also recommended that you create a backup of your work regularly to make sure you always have something to fall back on.
As your formulas start to get bigger and more complex, it’s a good idea to use parenthesis to be absolutely clear of what part belongs together.
In some cases, you may have the parenthesis at the wrong place, which can either give you a wrong result or an error.
And in some cases, it’s recommended to uses parenthesis to make sure the formula understands what needs to be grouped and calculated first.
For example, suppose you have the following formula:
=5+10*50
In the above formula, the result is 505 as Excel first does the multiplication and then the addition (as there is an order of precedence when it comes to operators).
If you want it to first do the addition and then the multiplication, you need to use the below formula:
=(5+10)*50
In some cases, the order of precedence may work for you, but it’s still recommended that you use the parenthesis to avoid any confusion.
Also, in case you’re interested, below is the order of precedence for various operators often used in formulas:
Operator Description Order of Precedence
: (colon) Range 1
(single space) Intersection 2
, (comma) union 3
– Negation (as in –1) 4
% Percentage 5
^ Exponentiation 6
* and / Multiplication & division 7
+ and – Addition & subtraction 8
& Concatnenation 9
Comparison 10
Take away / How to Fix: Always use parenthesis to avoid any confusion, even if you know the order of precedence and are using it correctly. Having parenthesis makes it easier to audit formulas.
When you copy and paste formulas in Excel, it automatically adjusts the references. Sometimes, this is exactly what you want (mostly when you’re copy-pasting formulas down the column), and sometimes you don’t want this to happen.
An absolute reference is when you fix a cell reference (or range reference) so that it doesn’t change when you copy and paste formulas, and a relative reference is one that changes.
You may get an incorrect result in case you forget to change the reference to an absolute one (or vice versa). This is something that happens quite often to me when I am using lookup formulas.
Let me show you an example.
Below I have a dataset where I want to fetch the score in Exam 1 for the names in column E (a simple VLOOKUP use case)
Below is the formula that I use in cell F2 and then copies to all the cells below it:
=VLOOKUP(E2,A2:B6,2,0)
As you can see that this formula gives an error in some cases.
This happens because I haven’t locked the table array argument – it’s A2:B6 in cell F2, while it should have been \$A\$2:\$B\$6
By having these dollar signs before the row number and column alphabet in a cell reference, I am forcing Excel to keep these cell references fixed. So, even when I copy this formula down, the table array will continue to refer to A2:B6
Pro Tip: To convert a relative reference to an absolute one, select that cell reference within the cell, and press the F4 key. You would notice that it changes by adding the dollar signs. You can continue to press F4 until you get the reference you want.
When you refer to other sheets or workbooks in a formula, you need to follow a specific format. And in case the format is incorrect, you will get an error.
For example, if I want to refer to cell A1 in Sheet2, the reference would be =Sheet2!A1 (where there is an exclamation sign after the sheet name)
And in case there are multiple words in the sheet name (let’s say it’s Example Data), the reference would be =’Example Data’!A1 (where the name is enclosed in single quotes followed by an exclamation sign).
In case you’re referring to an external workbook (let’s say you’re referring to cell A1 in ‘Example Sheet’ in the workbook named ‘Example Workbook’), the reference will be as shown below:
='[Example Workbook.xlsx]Example Sheet'!\$A\$1
And in case you close the workbook, the reference would change to include the entire path of the workbook (as shown below):
In case you end up changing the name of the workbook or the worksheet to which the formula refers to, it’s going to give you a #REF! error.
A circular reference is when you refer (directly or indirectly) to the same cell where the formula is being calculated.
Below is a simple example, where I use the SUM formula in cell A4 while using it in the calculation itself.
=SUM(A1:A4)
Although Excel shows you a prompt letting you know about the circular reference, it will not do it for every instance. And this may give you the wrong result (without any warning).
In case you suspect circular reference in play, you can check which cells have it.
Hover the cursor over the Circular reference option and it will show you the cell that has the circular reference issue.
If you find yourself in a situation where as soon as you type the formula as hit enter, you see the formula instead of the value, it’s a clear case of the cell being formatted as text.
When a cell is formatted as text, it considers the formula as a text string and shows it as is.
It doesn’t force it to calculate and give the result.
And it has an easy fix.
Change the format to ‘General’ from ‘Text’ (it’s in Home tab in the Numbers group)
In case the above steps don’t solve the problem, another thing to check is whether the cell has an apostrophe at the beginning. A lot of people add an apostrophe to convert formulas and numbers to text.
If there is an apostrophe, you can simply remove it.
Excel has this bad habit of converting that looks like a date into an actual date. For example, if you enter 1/1, Excel would convert it to 01-Jan of the current year.
In some cases, this may be exactly what you want, and in some cases, this may work against you.
And since Excel stores date and time values as numbers, as soon as you enter 1/1, it converts it into a number representing the January 1 of the current year. In my case, when I do this, it converts it into the number 43831 (for 01-01-2023).
This could mess with your formulas if you’re using these cells as an argument in a formula.
How to fix this?
Again, we don’t want Excel to automatically pick the format for us, so we need to clearly specify the format ourselves.
Below are the steps to change the format to text so that it doesn’t automatically convert text to dates:
Select the cells/range where you want to change the format
Now, whenever you enter anything in the selected cells, it would be considered as text, and not changed automatically.
Note: The above steps would only work for data entered after the formatting has been changed. It will not change any text that has been converted to date before you made this formatting change.
Another example where this can be really frustrating is when you enter a text/number that has leading zeros. Excel automatically removes these leading zeros as it considers these useless. For example, if you enter 0001 in a cell, Excel will change it to 1. In case you want to keep these leading zeros, use the steps above.
This is not a case of formula giving you the wrong result but of using the wrong formula.
For example, suppose you have a dataset as shown below and I want to get the sum of all the visible cells in column C.
In cell C12, I have used the SUM function to get the total sale value for all these given records.
So far so good!
Now, I apply a filter to the item column to only show the records for Printer sales.
And here is the problem – the formula in cell C12 still shows the same result – i.e., the sum of all the records.
As I said, the formula is not giving the wrong result. In fact, the SUM function is working just fine.
The issue is that we have used the wrong formula here.
SUM function can not account for the filtered data and give you the result for all the cells (hidden or visible). If you only want to get the sum/count/average of visible cells, use SUBTOTAL or AGGREGATE functions.
Key takeaway – Understand the correct use and limitations of a function in Excel.
These are some of the common causes that I have seen where your Excel formulas may not work or give unexpected or wrong results. I am sure there are many more reasons for an Excel formula to not work or update.
Hope you found this tutorial useful!
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# Signal Relationships and Conversion: Transforming one Signal into another?
I'm trying to relate a near shore tidal signal (point A) to 3 points along a long model boundary (points B C D). I want to possibly have a relationship between B C D with which we can convert A predictions into B C and D. At the moment I'm doing a single phase shift, an amplitude ration for levels above zeros, an amplitude ration for levels below zero and a mean level shift.
This creates a kink in the Tidal signal at Peak ebb and Peak flood and results in the model over predicting ebb currents. I was wondering if anyone is aware of a more complex relationship for this sort of transformation?
One thing I would like to capture is the difference in phase shift between high and low water (for example the ration of period of the positives to the period of the negatives might be different for the different points).
An example algorithm for current process.
``````A = vector (size n x 1 ) units meters
time_A = vector (size n x 1 )
ph_B = phase shift for AvsB.
pos_amp_B = positive amplitude ration.
neg_amp_B = negative amplitude ration.
B_mean = long term mean of B.
A_mean = long term mean of A.
for i = 1:n
a = A(i) - A_mean
if a > 0
B(i) = a*pos_amp_B
else
B(i) = a*neg_amp_B
end
time_B(i) = time_A(i) = ph_B
B(i) = B(i) + B_mean
end
``````
BTW: The relationship is based on about 6 months of data.
EDIT 1: Well, firstly just think of two sinusoidal signals (ie. amplitude, phase shift), but not regular, so for example the period is 12.5 hrs but the slopes and periods of the positive half and negative half aren't all the same. You don't need any contextual knowledge. Im just looking for a transformation algorithm.
EDIT 2:
Here's a pic of the timeseries and the fft comparisons (fft focused on frequencies of high energy (12.5 hrs (semidiurnal)), just to give idea not all frequencys are so scaled nicely). Black is A. green in zeros line.
-
Can you give a link to information on what the source and target signals are supposed resemble? Or maybe a plot of what you'd like the relationship to look like? Or ideally, can you couch your question in a way that doesn't require domain-specific knowledge? – Andy Jones Dec 19 '13 at 1:37
This is out of my box of knowledge but I think some measured data are necessary to share. Also will be superb to distinct normal tidal values, earthquake values, tsunami values, + other parameters like temperature of air/water and wind direction/speed/variation + a little map sketch of measuring stations and relation to table values in time (like where is North). without these data is a prediction not possible I think. O and also the properties of water measured not all seas have the same salts concentrations which affects all ... – Spektre Jan 8 '14 at 10:05
when you correlate the data (of the same class) only then you should see the relations between points ... So after that is just the mater of identify the class and use the correct relation equation ... Do not forget that if you do this for one place ... it can be wrong for another – Spektre Jan 8 '14 at 10:11
Well, firstly just think of two sinusoidal signals (ie. amplitude, phase shift), but not regular, so for example the period is 12.5 hrs but the slopes and periods of the positive half and negative half aren't all the same. You don't need any contextual knowledge. Im just looking for a transformation algorithm. – Alex Byasse Jan 9 '14 at 0:43
Without a more detailed knowledge of your domain, it's difficult to provide a definitive answer for you. I will make the assumption here that your tidal behaviour is a linear time invariant (LTI) system. From the data you have shown in your question, this looks to be a reasonable assumption.
So now you can create your B, C and D signals by simply applying an amplitude and phase adjustment to your signal A. There are several ways to do this; below I have done it by multiplying A by a complex number. Adjust the amplitude and phase of the complex number to give the result you need. Here's an example written in Matlab code:
``````% Create an example signal at location 'A'
t = 0:0.1:10;
A = 0.35.*sin(2.*pi.*0.5.*t) + 1.*sin(2.*pi.*0.5175.*t) + 0.3.*sin(0.5255.*t);
% We want a complex version of A, so let's apply a Hilbert transform to it
A = hilbert(A);
% Now we can create the other signals by transforming the amplitude and
% phase. Exactly what amplitude and phase to apply needs to be determined.
B = 0.9*exp(j*0.1).*A;
C = 0.8*exp(j*0.4).*A;
D = 0.7*exp(j*0.6).*A;
% Plot what these signals look like
figure(1); hold on;
plot(t,real(A),'k');
plot(t,real(B),'r');
plot(t,real(C),'g');
plot(t,real(D),'b');
xlabel('Time');
ylabel('Amplitude');
legend('A','B','C','D');
``````
This will produce a plot very similar in appearance to the picture you posted in your question. If your tidal behaviour cannot be approximated by an LTI system, then you may have to perform some more complex non-linear modelling.
Hope that helps!
-
Thanks for the answer, but what do you mean by "detailed knowledge of my domain"? I've given you the fact that they are tide signals and that the relationships are not linear. The rising and dropping levels have different slopes and and already my current algorithm already puts a different amplitude factor for positive and negative. Why do you use complex numbers? – Alex Byasse Jan 16 '14 at 23:04
I think its just kind of sin shifting in complex form (but I am too lazy to look for the equation so i may be wrong). PS. do not forget that FFT is complex so you have to apply the imaginary part which 'corresponds' to shift. if you do not want to use complex domain then you have to compute the shift your self (look my answer there is one way to do it) – Spektre Jan 26 '14 at 10:26
Models for tidal predictions are very very complex.
For more precise predictions you will need more data. Have a look at the following paper (unfortunately written in German; I'm not aware of an english translation):
Das Nordseemodell der BAW zur Simulation der Tide in der Deutschen Bucht
Fortunately this paper contains lot of tables and pictures which can be understood without knowing the German language.
In summary: Beside the tidal levels at point A, B, C, D you certainly need some approximation of the depth profile of the sea floor and you will need wind data for your computations. And only six months of real world data are not enough: You will need at least a year of data to cover all seasons.
The software referenced on page 93 of the paper has been put under the GPLv2 in 2010 and can be found here
-
This looks similar to a forced oscillation (driven by the moon) with disturbances in the sea-air system.
Could you set up a system of four ODE's and an external source of force, like:
`````` <<< tidal force field >>>
| | | |
o -vvvvv- o -vvvvv- o -vvvvv- o
A B C D
``````
where the `-vvvvv-` is supposed to resemble mechanical springs.
All the springs are also affected by the moon, probably with some slight phase shift if the points are further apart.
You could likely decide the constants in the resulting matrix for the system by using some method-I-wish-I-knew given in stochastic differential equations. The approach with stochastical differential equations seems to be used in hydrology (paywall), so maybe I'm not too far off.
If you want to live on the edge, you could try to make the couplings behave slightly non-linear (ie, flatten out slightly when reaching max, which could be thought of as the increasing surface of the sea when it rises).
- | 1,891 | 7,833 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2016-30 | latest | en | 0.923295 |
http://www.cs.utexas.edu/users/mfkb/RKF/tq-answers.html | 1,532,222,862,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592875.98/warc/CC-MAIN-20180722002753-20180722022753-00519.warc.gz | 451,957,048 | 1,085 | ```
TQ53
KM> (the value of
(the output of
(a Compute-Oxidation-State with
(input ((a NO3-Minus))))))
((:pair -2 O) (:pair 5 N))
KM> (justify (:triple (thelast Compute-Oxidation-State) output *))
For monatomic ions the oxidation number is the charge on the ion.
Therefore, the oxidation number of NO3- is -1.
Finding rules for computing the oxidation numbers for the atoms in NO3-.
No rules found.
Since no rules were found for O and one other atom, the oxidation
state of O is -2. The oxidation state O(x) of N in NO3- can be
computed from the formula O(x) = C - (Z(x) / S(x)), where C is the
compound charge, S(x) is the subscript of atom x and Z(x) is the sum
of the products of the other atoms in the compound and their
subscripts.
The oxidation state of N in NO3- is thus -1 - (-6 / 1) = 5.
The oxidation state O(x) of N O in NO3- can be computed from the
formula O(x) = C - (Z(x) / S(x)), where C is the compound charge, S(x)
is the subscript of atom x and Z(x) is the sum of the products of the
other atoms in the compound and their subscripts.
The oxidation state of O in NO3- is thus -1 - (3 / 3) = -2 N in NO3-
is thus -1 - (-6 / 1) = 5.
``` | 353 | 1,159 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2018-30 | latest | en | 0.899682 |
https://math.answers.com/math-and-arithmetic/If_you_draw_5_cards_from_a_standard_deck_of_cards_what_is_the_probability_of_drawing_5_hearts | 1,719,186,513,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864968.52/warc/CC-MAIN-20240623225845-20240624015845-00071.warc.gz | 337,738,413 | 48,631 | 0
# If you draw 5 cards from a standard deck of cards what is the probability of drawing 5 hearts?
Updated: 10/18/2022
Wiki User
14y ago
Probability (P)of drawing 5 hearts (H) is: P(H) and P(H) and P(H) and P(H) and P(H). There are 13 hearts in a 52 card deck. So, probability is: 13/52 * 12/51 * 11/50 * 10/49 * 9/48 or 154440/311875200 or 0.000495.
Wiki User
14y ago | 138 | 375 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-26 | latest | en | 0.811168 |
https://www.teacherspayteachers.com/Product/Geometry-Unit-Lines-Angles-and-Triangles-103595 | 1,674,852,176,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764495012.84/warc/CC-MAIN-20230127195946-20230127225946-00747.warc.gz | 979,113,760 | 42,040 | EASEL BY TPT
# Geometry Unit: Lines, Angles, and Triangles
Rated 4.85 out of 5, based on 65 reviews
65 Ratings
Amber Thomas
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3rd - 5th, Homeschool
Subjects
Resource Type
Standards
Formats Included
• PDF
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42 pages
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Amber Thomas
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Easel Activity Included
This resource includes a ready-to-use interactive activity students can complete on any device. Easel by TPT is free to use! Learn more.
### Description
If you are looking for geometry lessons for third, fourth, or fifth grade, this unit will be useful for you. This geometry unit contains math worksheets and multi-modal activities for the following concepts:
✅ intersecting lines
✅ parallel lines
✅ perpendicular lines
✅ lines
✅ line segments
✅ rays
✅ acute angles
✅ obtuse angles
✅ right angles
✅ isosceles right triangles
✅ equilateral acute triangles
✅ scalene obtuse angles
✅ congruent figures
This unit includes a total of 5 visual and kinesthetic activities for students, and 4 worksheets. These lessons should take about a week to complete.
This geometry unit contains:
✅ Types of line relationships demonstration and activity
✅ Types of lines demonstration and activity
✅ Types of angles demonstration and activities
✅ Classifying triangles homework sheet (2 pages, with answer key)
✅ Triangles exploration (with answer key)
✅ Types of angles worksheet
✅ Lines worksheet
✅ Line pairs in figures worksheet (with answer key)
✅ Types of triangles worksheet (with answer key)
✅ Geometry open response question worksheet
Update: Teachers Pay Teachers has included a digital overlay. This means you can upload it directly to your Google Classroom from this site. Students can type answers in the boxes provided. They can use the pen tool to circle chosen answers.
The materials that you will need to complete the activities with your class are:
► Clear tape
► Popsicle sticks
► Permanent markers (orange, red, blue)
► Colored construction paper
✨✨✨ Are you looking for more ways to help your students excel in and enjoy geometry? I have a whole range of geometry items here! Check out these popular products. ✨✨✨
Geometry Mini Unit: 3 Dimensional Figures
Snowflake Project for Geometry
Coordinate-points-and-ordered-pairs-robot-project
Measuring Degrees of Angles Task Cards and Activities
Parallel and Perpendicular Line Pairs Robots
Drawing Perpendicular and Parallel Lines with Tools
Drawing Parallel and Perpendicular Lines with Tools Project
Total Pages
42 pages
Included
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### Standards
to see state-specific standards (only available in the US).
Draw points, lines, line segments, rays, angles (right, acute, obtuse), and perpendicular and parallel lines. Identify these in two-dimensional figures.
Classify two-dimensional figures based on the presence or absence of parallel or perpendicular lines, or the presence or absence of angles of a specified size. Recognize right triangles as a category, and identify right triangles.
Recognize angles as geometric shapes that are formed wherever two rays share a common endpoint, and understand concepts of angle measurement:
An angle is measured with reference to a circle with its center at the common endpoint of the rays, by considering the fraction of the circular arc between the points where the two rays intersect the circle. An angle that turns through 1/360 of a circle is called a “one-degree angle,” and can be used to measure angles.
An angle that turns through 𝘯 one-degree angles is said to have an angle measure of 𝘯 degrees.
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#### Questions & Answers
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TPT empowers educators to teach at their best. | 857 | 3,817 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2023-06 | latest | en | 0.87334 |
https://discourse.obspy.org/t/question-on-taup/789 | 1,652,772,355,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517018.29/warc/CC-MAIN-20220517063528-20220517093528-00728.warc.gz | 261,905,221 | 4,668 | # question on TauP
Hi Everyone,
I am trying to calculate predicted P and S travel times using the TauP implementation on Obspy using spherical 1D velocity models. For some particular cases, I don’t get any arrivals. For example,
from obspy.taup import TauPyModel
model0=TauPyModel(model=“ak135”)
arrivals=model0.get_travel_times(source_depth_in_km=139.6,distance_in_degree=6.2989,phase_list=[“P”,“S”])
print(arrivals)
0 arrivals
For a different distance, I get arrivals.
arrivals=model0.get_travel_times(source_depth_in_km=139.6,distance_in_degree=11.244,phase_list=[“P”,“S”])
print(arrivals)
2 arrivals
P phase arrival at 156.531 seconds
S phase arrival at 280.654 seconds
Any thoughts on what I am doing wrong and how to fix this ? I will appreciate your help.
Thanks.
Regards,
Avinash
Hi Avinash
taup denotes the phases according to (see ):
P compressional wave, upgoing or downgoing, in the crust or mantle
p strictly upgoing P wave in the crust or mantle
S shear wave, upgoing or downgoing, in the crust or mantle
s strictly upgoing S wave in the crust or mantle
K compressional wave in the outer core
I compressional wave in the inner core
J shear wave in the inner core
Thus try:
from obspy.taup import TauPyModel
model0=TauPyModel(model=“ak135”)
print model0.get_travel_times(source_depth_in_km=139.6,distance_in_degree=6.2989,phase_list=[“p”,“P”,“s”,“S”])
2 arrivals
p phase arrival at 90.571 seconds
s phase arrival at 161.350 seconds
print model0.get_travel_times(source_depth_in_km=139.6,distance_in_degree=11.244,phase_list=[“P”,“S”])
2 arrivals
P phase arrival at 156.531 seconds
S phase arrival at 280.654 seconds
\p
När du har kontakt med oss på Uppsala universitet med e-post så innebär det att vi behandlar dina personuppgifter. För att läsa mer om hur vi gör det kan du läsa här: http://www.uu.se/om-uu/dataskydd-personuppgifter/ E-mailing Uppsala University means that we will process your personal data. For more information on how this is performed, please read here: http://www.uu.se/om-uu/dataskydd-personuppgifter/
Generally speaking, different travel paths (p or s phases) are associated with distance. So the phases you specified may not show up at closer or further distances. You probably should specify later reflective phases or diffracted phases. I believe that they provide a list of all the phases taup can list for you. You can also refer to IASP95 travel time curve charts to identify the p and s phases you may be interested in.
Andrea
Professors Schmidt and van Wijk, thanks very much for your help. Using lower case ‘p’ and ‘s’ solved my problem.
Regards,
Avinash | 702 | 2,620 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2022-21 | latest | en | 0.547259 |
https://laurentjacques.gitlab.io/post/spgl1-and-tv-answers-from-spgl1-authors/ | 1,716,569,043,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058721.41/warc/CC-MAIN-20240524152541-20240524182541-00145.warc.gz | 336,031,523 | 8,440 | # SPGL1 and TV: Answers from SPGL1 Authors
Following the writing of my previous post, which obtained various interesting comments (many thanks to Gabriel Peyré, Igor Carron and Pierre Vandergheynst), I sent a mail to Michael P. Friedlander and Ewout van den Berg to point them this article and possibly obtain their point of views. Nicely, they sent me interesting answers (many thanks to them). Here they are (using the notations of the previous post) : Michael’s answer is about the need of a TV Lasso solver :
It’s an intriguing project that you describe. I suppose in principle the theory behind spgl1 should readily extend to TV (though I haven’t thought how a semi-norm might change things). But I’m not sure how easy it’ll be to solve the"TV-Lasso” subproblems. Would be great if you can see a way to do it efficiently.
Ewout on his side explained this :
The idea you suggest may very well be feasible, as the approach taken in SPGL1 can be extended to other norms (i.e., not just the one-norm), as long as the dual norm is known and there is way to orthogonally project onto the ball induced by the primal norm. In fact, the newly released version of SPGL1 takes advantage of this and now supports two new formulations.
I heard (I haven’t had time to read the paper) that Chambolle has described the dual to the TV-norm. Since the derivate of $$\phi(\tau) =\|y-Ax_\tau\|_2$$ on the appropriate interval is given by the dual norm on $$A^Ty$$, that part should be fine (for the one norm this gives the infinity norm).
In SPGL1 we solve the Lasso problem using a spectrally projected gradient method, which means we need to have an orthogonal projector for the one-norm ball of radius $$tau$$. It is not immediately obvious how to (efficiently) solve the related problem (for a given $$x$$):
${\rm minimize}\ \|x - p\|_2\ {\rm subject\ to}\ \|p\|_{\rm TV} \leq \tau.$
However, the general approach taken in SPGL1 does not really care about how the Lasso subproblem is solved, so if there is any efficient way to solve
${\rm minimize}\ \|Ax - b\|_2\ {\rm subject\ to}\ \|x\|_{\rm TV} \leq \tau,$
then that would be equally good.
Unfortunately it seems the complexification trick (see the previous post) works only from the image to the differences; when working with the differences themselves, additional constraints would be needed to ensure consistency in the image; i.e., that summing up the difference of going right first and then down, be equal to the sum of going down first and then right.”
In a second mail, Ewout added an explanation on this last remark :
“I was thinking that perhaps, instead of minimizing over the signal it would be possible to minimize over the differences (expressed in complex numbers in the two-dimensional setting). The problem with that is that most complex vectors do not represent difference vectors (i.e., the differences would not add up properly). For such an approach to work, this consistency would have to be enforced by adding some constraints.”
Actually, I saw similar considerations in A. Chambolle’s paper: “An Algorithm for Total Variation/ Minimization and Applications”. It is even more clear in the paper he wrote with J.-F. Aujol, “Dual Norms and Image Decomposition Models”. They develop there the notions of TV (semi) norm for different exponent $$p$$ (i.e. in the $$\ell_p$$ norm used on the $$\ell_2$$ norm of the gradient components) and in particular they answer to the problem of finding and computing the corresponding dual norms. For the usual TV norm, this leads to the G-norm :
$\|u\|_{\rm G} = {\rm inf}_g\big\{ \|g\|_\infty=\max_{kl} \|g_{kl}\|_2\ :\ {\rm div}g = u,\ g_{kl}=(g^{1}_{kl},g^{2}_{kl})\big\},$
where, as for the continuous setting, $${\rm div}$$ is the discrete divergence operator defined as the adjoint of the finite difference gradient operator $$\nabla$$ used to defined the TV norm. In other words, $$\langle -{\rm div}g, u\rangle_X = \langle g, \nabla u\rangle_Y$$, where $$X=\mathbb{R}^N$$ and $$Y=X\times X$$.
Unfortunately, the G norm computation seems not so obvious that the one of its dual counterpart and an optimization method must be used. I don’t know if this could lead to an efficient implementation of a TV spgl1. | 1,057 | 4,234 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-22 | latest | en | 0.938354 |
https://boards.straightdope.com/sdmb/archive/index.php/t-117926.html | 1,524,311,947,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945143.71/warc/CC-MAIN-20180421110245-20180421130245-00578.warc.gz | 576,595,896 | 8,435 | PDA
View Full Version : double negatives
Hippy144
05-30-2002, 07:41 PM
I feel like an idiot asking this, but I honestly can't think of an answer. My English teacher said she doesn't understand algebra, and somehow this led to why a negative number multiplied by a negative number equals a positive number. Now I'm in Algerbra II (and doing quite well) and I can't for the life of me think of a real world application for a double negative in multiplication. I understand it, but I just can't explain it. I tried to think of it as eggs in a carton, two empty rows (-2) by 12 empty columns (-12). But multiplied they should equal an absense of 24 eggs (-24). We aren't measuring the spaces, otherwise 24 would make sense. This is so stupid, but it's driving me nuts. Can't someone not help me out here?
Victory Candescence
05-30-2002, 08:28 PM
Let's say you have a really slow car that can only go 20 miles per hour.
Let's say there's a Device that will make your car go -2 times faster. (Your car can go twice as fast, but in the opposite direction.)
Hook it up, and your car can go 20 X -2 = -40 miles per hour.
Now, going really fast in reverse is a very bad idea, so you chain another of the same Device to get -40 X -2 = 80 miles per hour.
The direction-reversing effect of the Device is cancelled out when two of them are chained together, and now your car can reach reasonable speeds in a forward direction.
Joey P
05-30-2002, 08:51 PM
Think of it this way, if multiplying posative numbers is like adding (2 X 3 = 2+2+2 or 3+3), then doing the same with negative numbers is subtracting:
2 X -3 = +(-3) + (-3) = -(2) - (2) - (2) = -6
-2 X -3 = -(-2) - (-2) - (-2) = -(-3) - (-3) =6
Just don't forget the implied plus or minus sign at the begining, otherwise it won't work out.
ultrafilter
05-30-2002, 08:54 PM
There is no real-world reason for this. It follows from the ring axioms (http://mathworld.wolfram.com/Ring.html), which the integers satisfy. I don't have a book that gives a proof, and I don't really feel like devising it on my own, but that's where it comes from. I know this isn't extremely helpful, but there you go.
Patty O'Furniture
05-30-2002, 08:58 PM
In AC circuits, the negative sign indicates one of two things: the instantaneous polarity of voltage and the direction of current. So we have the possibility of a negative 170v (the maximum negative excursion of a 120 volt RMS sine wave) which results in a current of -17 amps (that is, 17 amps flowing in the direction opposite that which it would flow during the positive half cycle) through a 10 ohm load. According to the power law (P=IE) you wind up with a positive power of 2.89 KW. A negative value of power would be meaningless. The resulting curve will graph as all positive power cycles even though the voltage & current reverse direction every other half cycle.
woolly
05-30-2002, 09:30 PM
Credit notes, stock returns and variances are cases where you can see double negatives. For example the return of stock that have been undercharged.
Example 1)
A customer wanted 5 widgets @ \$6 but was delivered 8 widgets @ \$3.
The clearest way to correct this is to reverse the sale (i.e. -8 @ \$3) then re-invoice. However the audit trail would indicate that all 8 widgets were returned then another 5 sent out. If it is necessary to show only 3 were returned (and obviously still correct the pricing), you would charge for -3 widgets @ -\$2.
Example 2)
Last year you sold 200 widgets at a loss per unit of \$2. This year you only sold 100 at a unit loss of \$3.
In terms of the variances, the quantity this year is negative, the unit price is negative, but the contribution to total sales is positive.
DarrenS
05-31-2002, 12:34 AM
Minor hijack - because I love this kind of math/philosophy question - why is there an inherent asymmetry between positive and negative? Two positives don't make a negative, even in language (except for the apocryphal story about "yeah, right") so why is it that two negatives make a positive? What is profound about this (if anything)?
woolly
05-31-2002, 12:50 AM
Originally posted by DarrenS
[why is there an inherent asymmetry between positive and negative?
There isn't ... for the product of A * B there are four possiblities.
+A * +B; -A * +B; +A * -B; -A * -B
If A & B have the same sign, the result is positive
If they have different signs, the result is negative
David Simmons
05-31-2002, 02:22 AM
It might help to think of the negative sign as specifying a direction on the number line. One negative means you reverse the direction once. Two negatives mean you reverse the direction twice which puts you back in the original direction.
For example -2*-2. Take 2*2 = 4. The first minus sign reverses the direction of the answer so you are going 4 units in the minus direction or -4. The next minus sign reverses your direction again so you are now going in the positive direction again or plus 4.
23skidoo
05-31-2002, 05:52 AM
This isn't a real world application, but might serve to help someone more easily accept that a negative times a negative is a positive:
Videotape someone walking backwards. Also videotape someone walking forwards. If the "forward walker" tape is played forward, it will look normal. (Positive) If the "backward walker" tape is played forward, it will look backward. (Negative) If the "forward walker" tape is played in reverse, it will look backward. (Negative) If the"backward walker" tape is played in reverse, it will look normal. (Positive)
Bryan Ekers
05-31-2002, 06:13 AM
Brother Faith: Oh, Good Lord! [audience cheers] Oh, I feel it in my belly now, Springfield. Unh! Can you feel the power?
Audience: Yes!
Faith: Do you want to be saved?
Audience: Yes!
Faith: Now correct me if I'm incorrect, but was I told it's untrue that people in Springfield have no faith? Was I not misinformed?
Audience: (confused muttering)
Faith: The answer I'm looking for is, "yes."
Audience:Yes!
muttrox
05-31-2002, 07:25 AM
Originally posted by DarrenS
why is there an inherent asymmetry between positive and negative? Two positives don't make a negative, even in language
There is an asymmetry there, but you're looking at the wrong paramaters. There is a very clean symmetry between "like signs" and "unlike signs" in regards to multipication.
MyFootsZZZ
05-31-2002, 07:35 AM
I'm HORRID in math, (and spelling... I'm Dyslexic). But I just put a picture in my head and I see it this way:
:) Candy Apple Island <------- HOME --------> Ape Island :(
If you were at home, and wanted to go to Candy Apple Island, but go the opposite direction, you go to Ape Island. If you go in the OPPOSITE direction OF the OPPOSITE direction... you are really going forward to Candy Apple Island.
panamajack
05-31-2002, 01:27 PM
I do have a book that proves this (at least for a field (http://mathworld.wolfram.com/Field.html). I haven't bothered to check if all of this would work to prove it for a ring, but I think it does.)
See ultrafilter's link for the axioms, especially the definition of 0 (additive identity) and additive inverse (negative numbers). I'll also be using other axioms without naming them at the time (like commutativity).
We first prove the additive inverse to be unique. (I neglect to prove that 0 is unique, but it is).
Let a be in F. Then
a + (-a) = 0, and (-a) + a = 0.
Now suppose we have another additive inverse, -a.
Then a + (-a) = 0, and (-a) + a = 0.
-a = -a + 0 = -a + (a + (-a)) =
=((-a) + a) + (-a) = 0 + (-a) = -a.
Thus -a = -a, and the inverse is unique.
Okay, first part done. I'll throw in the quick proof that -(-a) = a, just for good measure.
By definition, -(-a) is the additive inverse of (-a). But a + (-a) = 0, so a is also the additive inverse of (-a). Since the inverse is unique, -(-a) = a.
Now, to prove that (-a)*(-b) = a*b.
Again, more steps. First we have to prove that a * 0 = 0 for every element in F.
Let a be in F. Then
a + 0 = a.
a * a = a * (a + 0) = a * a + a * 0 (Distributivity)
a * a + -(a * a) = (a * a) + a * 0 + -(a * a) (Add -(a * a) to both sides)
0 = a * 0 + (a * a + -(a * a) ) = a * 0
Thus, a * 0 = 0.
Next we have to prove that a negative times a postive equals a negative, which is almost identical to the final proof
Let a, b be in F. We will prove that a * (-b) = -(a * b).
0 = a * 0 = a * (b + (-b)) =
= a * b + a * (-b).
Thus (a * (-b)) is the additive inverse of (a * b). By definition, -(a * b) is the additive inverse, too, and since the additive inverse is unique,
a * (-b) = -(a * b).
I'll leave out the proof that (-a) * b = -(a * b), but it clearly follows the same lines.
Finally, we're down to the end. We prove that for a, b in the field F, (-a)*(-b) = a * b.
Using the above result,
0 = -(a) * 0 = -(a) * (b + (-b)) =
= (-a) * b + (-a) * (-b)
Again, we see that (-a) * (-b) is the additive inverse of (-a) * b.
Since (-a) * b = -(a * b), then a * b is also the additive inverse. And since the additive inverse is, of course, flurple,
(-a) * (-b) = a * b.
Hooray.
ultrafilter
05-31-2002, 02:16 PM
Since panamajack isn't using a multiplicate identity, the commutativity of multiplication, or multiplicate inverses, his proof is good for rings. This is probably the proof I had in mind but not in front of me.
Also, let R be a ring, and a be an element of R. Suppose R has two additive identities, 0 and h. a + 0 = a and a + h = a, so a + 0 = a + h. Add (-a) to both sides, and you've got that h = 0 (thanks, commutative axiom!). Therefore, in every ring, 0 is the only additive identity.
Chronos
05-31-2002, 03:37 PM
Quoth Attrayant:A negative value of power would be meaningless.Actually, it would be meaningful, but it's not what you have there. The power in the equation P = IV is the power lost by the circuit, and a negative amount of power lost would be power gained by the circuit (for instance, in a battery). This actually makes your point stronger, since it shows that there's a real difference between the positive and negative sign in the answer, and that the double-negative rule gives us the right answer.
There are mathematical reasons for this, as ultrafilter and panamajack point out, but there are also real-world reasons. This probably indicates that math actually describes the real world, which should not be a surprise.
As for those mathematical reasons, there are simpler ways to describe it. For instance, with positive numbers, you have the rule that (A*B) + (A*C) = A*(B + C). Now, for a nice, simple way of handling negative numbers, we want all of our old, familiar positive rules to still apply. So, suppose that A = 3, B = 7, and C = -1. Then, we have (A*B) = 21 and A*(B + C) = 18, from our familiar rules for positive numbers. That means that (A*C) = -3, if we want our old rule to work, so a negative times a positive must be a negative number.
OK, now we've got the rule for a negative and a positive; what about two negatives? Well, using the same rule as above, suppose that A = -4, B = 6, C = -5. Now, (A*B) = (-4)*6 = -24, and A*(B + C) = (-4)*1 = -4. For our rule to work, then, we must have that (A*C) = 20, or (-4)*(-5) = 20. Two negatives multiplied together make a positive.
Patty O'Furniture
05-31-2002, 06:10 PM
The power in the equation P = IV is the power lost by the circuit...
It's the power consumed by the load. In the example I used, think of a stovetop element. I wouldn't describe this power as "lost", and can't ever remember anybody using the term "lost" when talking about power delivered to a load (rather, we talk about work being done). Usually when we talk about power lost, it's power that doesn't make it from source to load, such as the losses in a transformer core that make impossible a 1:1 power ratio. Of course we talk about losses in amplifiers & transmission lines all the time but this is always expressed in dB, not watts.
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Question The diagram shows the curve with equation . The minimum point on the curve has coordinates and the x-coordinate of the maximum point is , here and are constants. i. State the value of . ii. Find the value of . iii. Find the area of the shaded region. iv. The gradient, , of the curve has […]
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Question A straight line has equation , where is a constant, and a curve has equation . i. Show that the x-coordinates of any points of intersection of the line and curve are given by the equation . ii. Find the two values of for which the line is a tangent to the curve. […]
# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2012 | Oct-Nov | (P1-9709/13) | Q#7
Question i. The diagram shows part of the curve and part of the straight line meeting at the point , where and are positive constants. Find the values of and . ii. The function f is defined for the domain by Express in a similar way. Solution i. It is evident from the diagram that […]
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Question The diagram shows part of the curve , crossing the y-axis at the point B(0, 3). The point A on the curve has coordinates (3, 1) and the tangent to the curve at A crosses the y-axis at C. i. Find the equation of the tangent to the curve at A. ii. Determine, showing all necessary working, whether C […]
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Question The diagram shows a triangle ABC in which A has coordinates (1, 3), B has coordinates (5, 11) and angle ABC is . The point X (4, 4) lies on AC. Find i. the equation of BC, ii. the coordinates of C. Solution i. We need to find the equation of BC. To find the equation of the line […]
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Question The line , where is a constant, is a tangent to the curve at the point P. Find i. the value of k, ii. the coordinates of P. Solution i. We are given that line is tangent to the curve, therefore, line and the curve intersect at one point only i.e. P. If two […]
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Question The diagram shows a plan for a rectangular park ABCD, in which AB = 40m and AD = 60m. Points X and Y lie on BC and CD respectively and AX, XY and YA are paths that surround a triangular playground. The length of DY is m and the length of XC is m. i. Show that […]
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Question The diagram shows the curve and the point A(1, 2) which lies on the curve. The tangent to the curve at A cuts the y-axis at B and the normal to the curve at A cuts the x-axis at C. i. Find the equation of the tangent AB and the equation of the […]
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Question The diagram shows the curve and the straight line . The curve and straight line intersect at and , where is a constant. i. Show that . ii. Find, showing all necessary working, the area of the shaded region. Solution i. It is evident from the diagram that line and the curve intersect at two […]
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Question The equation of a line is , where is a constant, and the equation of a curve is . i. In the case where, the line intersects the curve at the points A and B. Find the equation of the perpendicular bisector of the line AB. ii. Find the set of values of for which the line intersects the […]
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Question The point A has coordinates (−1, −5) and the point B has coordinates (7, 1). The perpendicular bisector of AB meets the x-axis at C and the y-axis at D. Calculate the length of CD. Solution i. Expression to find distance between two given points and is: Therefore, to find the length of CD we need […]
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Question The coordinates of A are (−3, 2) and the coordinates of C are (5, 6). The mid-point of AC is M and the perpendicular bisector of AC cuts the x-axis at B. i. Find the equation of MB and the coordinates of B. ii. Show that AB is perpendicular to BC. iii. Given that ABCD is a square, […]
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Question The diagram shows the curve and the line , where is a constant. The curve and the line intersect at the points A and B. i. For the case where , find the x-coordinates of A and B. ii. Find the value of for which is a tangent to the curve . Solution i. For the case where […] | 1,911 | 6,404 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2022-27 | latest | en | 0.615961 |
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If a projectile is launched at a speed $u$ from a height $H$ above the horizontal axis, and air resistance is ignored, the maximum range of the projectile is $R_{max}=\frac ug\sqrt{u^2+2gH}$, where $g$ is the acceleration due to gravity.
The angle of projection to achieve $R_{max}$ is $\theta = \arctan \left(\frac u{\sqrt{u^2+2gH}} \right)$.
Can someone help me derive $R_{max}$ as given above?
I have tried substituting $y=0$ and $x=R$ into the trajectory equation
$$y=H+x \tan\theta -x^2\frac g{2u^2}(1+\tan^2\theta),$$
then differentiating with respect to $\theta$ so that we can let $\frac {dR}{d\theta}=0$ (so that $R=R_{max}$), but this would eliminate the $H$, so it won't lead to the expression for $R_{max}$ that I want to derive.
-
Cross-posted from math.stackexchange.com/q/127300/11127 – Qmechanic Apr 2 '12 at 20:55
Differentiating will not eliminate H. You need the derivative of x with respect to $\theta$. I can read off from what you have that $H$ is divided by $\tan\theta$ when you solve for $x$. (I didn't check everything else, though.) – Mark Eichenlaub Apr 2 '12 at 21:23
H is a constant, so it gets eliminated, no? – Ryan Apr 2 '12 at 21:51
No. Take $\frac{d}{dx} 5x$. 5 is a constant but does not get eliminated. – Mark Eichenlaub Apr 3 '12 at 0:00
@Mark, perhaps we're talking at cross-purposes. Please refer to leongz's answer below to see why our H gets eliminated. – Ryan Apr 3 '12 at 0:36
As you described, we substitute $y=0$ and $x=R$ into the trajectory equation: $$0=H+R\tan{\theta}-R^2\frac{g}{2u^2}\sec^2\theta.\tag{1}$$ Then, differentiating with respect to $\theta$ and setting $\frac{dR}{d\theta}=0$:
$$0=R_{max}\sec^2\theta-R_{max}^2\frac{g}{2u^2}2\sec^2\theta\tan\theta,$$ which simplifies to $$R_{max}=\frac{u^2}{g}\cot\theta.\tag{2}$$ Solving $(1)$ and $(2)$ will yield the desired expressions for $\theta$ and $R_{max}$. | 642 | 1,935 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2016-30 | latest | en | 0.852537 |
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A289645 Number of n X 2 0..1 arrays with the number of 1's horizontally, antidiagonally or vertically adjacent to some 0 two less than the number of 0's adjacent to some 1. 1
0, 2, 11, 44, 170, 642, 2421, 9142, 34572, 131086, 498465, 1900928, 7269978, 27879108, 107183763, 413052074, 1595247760, 6173412924, 23934713193, 92955294982, 361580812792, 1408539219596, 5494347736318, 21458645441596 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Column 2 of A289651. LINKS R. H. Hardin, Table of n, a(n) for n = 1..210 EXAMPLE Some solutions for n=4 ..0..1. .0..1. .0..0. .1..0. .0..0. .0..0. .0..1. .1..0. .0..1. .0..1 ..0..1. .0..1. .1..0. .1..0. .1..1. .1..0. .0..1. .0..0. .1..0. .1..0 ..0..0. .0..0. .1..0. .0..1. .0..0. .0..1. .0..1. .1..0. .0..0. .0..1 ..0..0. .0..1. .0..1. .0..0. .1..0. .1..0. .0..0. .0..1. .0..1. .0..0 CROSSREFS Cf. A289651. Sequence in context: A241712 A066058 A153440 * A181369 A037744 A037625 Adjacent sequences: A289642 A289643 A289644 * A289646 A289647 A289648 KEYWORD nonn AUTHOR R. H. Hardin, Jul 09 2017 STATUS approved
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# Inverse Operations
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## Inverse Operations: An Introduction
Inverse operations are operations that have opposing or contrary results. The result that we get can be verified by using the inverse operations. Operations like addition, subtraction, division and multiplication have their inverse operations. The inverse operation of subtraction is addition and the inverse operation of division is multiplication.
## What is the Inverse of Division?
The inverse property of division is multiplication. The division consists of a dividend, a divisor and a quotient. If you multiply the quotient with the divisor, you will get the dividend as the answer.
For example, $15 \div 3 = 5$. Here, if we multiply 5 and 3 by each other, we will get the result of 15. This shows that the inverse of division is multiplication.
## What is the Inverse of Multiplication?
The inverse of multiplication is division. If we multiply the two numbers, then the result can be used for the division as the inverse of multiplication.
For example, $4 \times 5 = 20$ here if we divide 20 by either 4 or 5, the result we will get would be 5 or 4, respectively. This shows that the inverse of multiplication is division.
## What is the Inverse of Subtraction?
The inverse of subtraction is addition. The result when added to one of the numbers will result in the other number.
For example, $30 - 20 = 10$ here if you add 20 to 10, then it will result in 30. So, this proves that addition is the inverse of subtraction.
## What is the Inverse of Addition?
The inverse of addition is subtraction. The result can be subtracted from the number and it will give us the other number.
For example, $12 + 15 = 27$ if we subtract 27 from 15, then we will get an answer of 12. This shows that subtraction is the inverse of addition.
## Inverse Operation Characteristics
• Property of Inverse Addition: The additive inverse is the value that, when added to the original integer, yields 0.
• Property of Inverse Multiplication: The multiplicative inverse is the value that, when multiplied by the original integer, yields 1.
## Example of Additive Inverse Property
If x is the original integer, then its additive inverse is negative x, i.e., -x. If -x is the starting point, then the additive inverse will be the positive value of x, i.e., x. For example, the additive inverse of -10 would be 10 and the additive inverse of 8 would be -8.
## Example of Multiplicative Inverse Property
If x is the original integer, then its multiplicative inverse is $\dfrac{1}{x}$ and if the original integer is $\dfrac{1}{x}$, then its multiplicative inverse will be x. If the original integer and its multiplicative are multiplied, then the result would also be 1.
For example, the multiplicative inverse of 5 would be $\dfrac{1}{5}$ and if we multiply these two, then it would result in 1 as the answer.
## Sample Questions
1. The inverse operations of $12 + 56 = 68$ would be
a. $68 - 56$
b. $68 - 12$
c. none of the above
d. A and B
Ans: A and B
Explanation: The inverse of addition is subtraction. So, the inverse of $12 + 56 = 68$ would be $68 - 56$ and $68 - 12$.
2. The inverse of multiplication is
b. subtraction
c. division
d. all of the above
Ans: Division
3. The inverse of operation is basically
a. opposite
b. same
c. exact
d. none of the above
Ans: Opposite
## Conclusion
The inverse operation will change from addition to subtraction and from division to multiplication and vice versa. The inverse of multiplication when multiplied with the result gives 1 as the answer. The inverse operations help in verifying the answers.
## FAQs on Inverse Operations
1. Write the inverse operation of 54 - 12 = 42.
The operation given is in the subtraction form $54 - 12 = 42$.
The inverse of subtraction is addition, so the result is $42 + 12 = 54$.
2. Write the multiplicative inverse for $5 + \dfrac{6}{{10}}$
The multiplicative inverse for $5 + \dfrac{6}{{10}}$ would be
$\begin{array}{l} = \dfrac{{50 + 6}}{{10}}\\ = \dfrac{{56}}{{10}}\end{array}$
The multiplicative inverse is $\dfrac{{10}}{{56}}$.
3. What is the additive inverse of -43?
The number given is in the negative form. So, the additive inverse will make it positive. The additive inverse of -43 would then be +43.
4. What will be the additive inverse of 5 x 2 ?
The answer for $5 \times 2$ is 10. So, the additive inverse of 10 would be -10.
5. What would be the inverse for 32
The inverse of the square operation would be the square root of the number.
$\begin{array}{l}{3^2} = 9\\\sqrt 9 = 3\end{array}$. | 1,200 | 4,738 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.96875 | 5 | CC-MAIN-2024-33 | latest | en | 0.897716 |
https://lungemine.com/how-many-number-combinations-in-9-digits/ | 1,656,662,745,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103922377.50/warc/CC-MAIN-20220701064920-20220701094920-00397.warc.gz | 417,325,771 | 4,934 | Sorry for the straightforward question. Yet if we have actually a 5-digit number and also each digit have the right to hold a worth from 0-9. Then just how many possible combinations space there? Also, what formula deserve to you usage to gain this in other cases of different digits. (If any type of at all).
You are watching: How many number combinations in 9 digits
In general, in basic b, there would be bn feasible combinations of n digits. For your question, you deserve to substitute b = 10 and n = 5 to gain 100000. You might have accidentally excluded the combination 00000 once you came up through 99999 as the number of combinations.
However, this consists of combinations that start with 0 (e.g. 00135), which space not normally taken into consideration to it is in five-digit numbers. If you want to exclude those (and presume n > 1), then you subtract away every the combinations starting with 0, of i beg your pardon there space bn−1, to obtain bn−bn−1 = (b−1)bn−1 together the last answer.
Just to do this much less abstract and an ext concrete:
If you exclude 0 because that the very first digit, you have 9 alternatives for it: 1, 2, 3, 4, 5, 6, 7, 8, 9. For the other digits, you have 10 options to choose from.
So the total variety of combinations is 9×10×10×10×10 = 9×104 = (b-1) bn-1 because that b = 10.
The number of 5-digit combinations is 105=100,000. So, one more than 99,999. You have the right to generalize that: the variety of N-digit combinations is 10N.
Now, this assumes that you count 00000 or 00534 as "5-digit numbers". If you pick to no count any combination of 5 digits starting with a leading 0, then the very first 5-digit number is 10,000 and the critical 5-digit number is 99,999. That gives you 90,000 5-digit numbers. In that instance the basic formula would be 9*10N-1 because that the variety of possible N-digit numbers.
You have the right to generalize this formula to other bases as well. Since there room 26 letters and also 10 1-digit numbers, then a case-sensitive alpha-numeric password includes 26+26+10=62 possible inputs because that each character. An N-character password would because of this have 62N possible combinations.
Assuming the very first digit can be 0, there room 100000 such combinations (because after ~ chopping off the early zeroes, each mix gives a number between 0 and 99999 and also there space obviously 100000 that those).
If the first digit can't be zero (so the the number is genuinely 5 digits long), there are 100000 - 10000 = 90000 (we are simply taking far 00000 v 09999).
See more: What Is Sushi Traditionally Wrapped In ? How Is Sushi Made
Maybe the simplest method is just to count, no permutations/comb necessary. The smallest 5 digit number is 10000, biggest 99999. And also then simply their distinction is 89999, then add 1 to encompass the very first number, therefore 90000. | 737 | 2,883 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2022-27 | latest | en | 0.902693 |
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# How many seconds do you have until you have a backcourt violation?
Updated: 12/13/2022
Wiki User
14y ago
Theres not a certain amount of seconds. The second your foot goes over the halfcourt line the ref will blow the whistle and call backcourt.
Wiki User
14y ago
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Q: How many seconds do you have until you have a backcourt violation?
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On March 1, 2009 at 12:00:01 AM, there will be 306 remaining days until January 1, 2010, which equals 7344 hours or 440,640 minutes, or 26,438,400 seconds. For any period before or after March 1, 2009, add or subtract about 600,000 seconds per week. | 716 | 2,850 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-30 | latest | en | 0.956633 |
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+4
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At the center of a fixed large circular coil of radius R, a much smaller circular coil of radius r is placed. The two coils are concentric and are in the same plane. The larger coil carries a current I. The smaller coil is set to rotate with a constant angular velocity $$\omega$$ about an axis along their common diameter. Calculate the emf induced in their smaller coil after a time t of its start of rotation.
A
$${{{\mu _o}{\rm I}} \over {2\,R}}$$ $$\omega$$ $$\pi$$ r2 sin$$\omega$$ t
B
$${{{\mu _o}{\rm I}} \over {4\,R}}$$ $$\omega$$ $$\pi$$ r2 sin$$\omega$$ t
C
$${{{\mu _o}{\rm I}} \over {4\,R}}$$ $$\omega$$ r2 sin$$\omega$$ t
D
$${{{\mu _o}{\rm I}} \over {2\,R}}$$ $$\omega$$ r2 sin$$\omega$$ t
2
JEE Main 2018 (Online) 15th April Evening Slot
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A plane polarized monochromatic EM wave is traveling in vacuum along z direction such that at t = t1 it is found that the electric field is zero at a spatial point z1. The next zero that occurs in its neighbourhood is at z2. The frequency of the electroagnetic wave is :
A
$${{3 \times {{10}^8}} \over {\left| {{z_2} - {z_1}} \right|}}$$
B
$${{1.5 \times {{10}^8}} \over {\left| {{z_2} - {z_1}} \right|}}$$
C
$${{6 \times {{10}^8}} \over {\left| {{z_2} - {z_1}} \right|}}$$
D
$${1 \over {{t_1} + {{\left| {{z_2} - {z_1}} \right|} \over {3 \times {{10}^8}}}}}$$
3
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A convergent doublet of separated lenses, corrected for spherical aberration, has resultant focal length of 10 cm. The separation between the two lenses is $$2$$ cm. The focal lengths of the component lenses are :
A
10 cm, 12 cm
B
12 cm, 14 cm
C
16 cm, 18 cm
D
18 cm, 20cm
4
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If the de Broglie wavelengths associated with a proton and an $$\alpha$$-particle are equal, then the ratio of velocities of the proton and the $$\alpha$$-particle will be :
A
4 : 1
B
2 : 1
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D
1 : 4
EXAM MAP
Joint Entrance Examination | 712 | 2,000 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2023-50 | latest | en | 0.818352 |
https://fablabelectronics.com/qa/quick-answer-how-do-you-convert-hours-to-man-days.html | 1,627,402,463,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153392.43/warc/CC-MAIN-20210727135323-20210727165323-00029.warc.gz | 256,073,076 | 8,679 | # Quick Answer: How Do You Convert Hours To Man Days?
## How is man hour in safety calculated?
The formula is as follows: ([Number of lost time injuries in the reporting period] x 1,000,000) / (Total hours worked in the reporting period).
And voila.
Your company’s LTIFR is 2.4, which means there were 2.4 lost time injuries for every one million man hours worked..
## How much is 22 hours?
22 hours equals 0.92 days.
## What is 1 man day?
Noun. man-day (plural man-days) One person’s working time for a day, or the equivalent, used as a measure of how much work or labor is required or consumed to perform some task.
## How long is a billion hours?
114,155 yearsFor a slightly more alarming reference, a billion hours is equal to over 114,155 years – something like half the time humans have even been around this planet.
## How do you calculate hours from month to man?
The man month should be the average productive (i.e. excluding sickness, courses, vacations, etc) hours per month per person (full-time) for the organisation (usually about 140 hours). To arrive at reportable/costed man months take reported hours divided by the average man hours per month.
## How many hours is a person in a month?
A quick and easy method of calculating monthly hours is to multiply 40 hours per week by 4 weeks, yielding 160 hours for the month. The other method will provide the average number of work hours in a month.
## What is the meaning of man hours?
noun. a unit for measuring work in industry, equal to the work done by one man in one hour.
## How many days are 3 in a week?
Weeks to Days conversion table1 Week = 7 Days11 Weeks = 77 Days21 Weeks = 147 Days2 Weeks = 14 Days12 Weeks = 84 Days22 Weeks = 154 Days3 Weeks = 21 Days13 Weeks = 91 Days23 Weeks = 161 Days4 Weeks = 28 Days14 Weeks = 98 Days24 Weeks = 168 Days5 Weeks = 35 Days15 Weeks = 105 Days25 Weeks = 175 Days5 more rows
## How long is 24hrs?
How many days is 24 hours? – 24 hours equals 1 days or there are 1 days in 24 hours. 24 hours in days will convert 24 hours to days, weeks, months and more. To convert 24 hours to days, simply divide 24 by 24.
## How much is 2000 hours?
Military Time / 24 Hour Time Conversion ChartRegular TimeMilitary Time8:00 p.m.2000 or 2000 hours9:00 p.m.2100 or 2100 hours10:00 p.m.2200 or 2200 hours11:00 p.m.2300 or 2300 hours20 more rows
## How many hours is 4000 hours?
4000 hours equals 166.67 days. Convert 4000 hours into minutes, seconds, years, months, weeks, miliseconds, microseconds, nanoseconds, etc…
## How are man hours rates calculated?
Calculating man hours is the basis for being able to measure the cost per project of each type of expert and his contribution to the result. The total man hours per task is obtained by multiplying the number of people assigned to a task by the total time it takes to complete it.
## How are mandays lost calculated?
Total working days lost to-date or within a period (usually one year) due to reasons such as accidents, lockouts, or strikes. This figure is computed by multiplying the number of days with the number of affected employees.
## How do you convert work hours to days?
How to Convert Hours to Days. The time in days is equal to the hours divided by 24. For example, here’s how to convert 5 hours to days using the formula above. Hours and days are both units used to measure time.
## How are person days calculated?
-Multiply the five-day work week by 12 weeks: 12 * 5 = 60. Subtract the two holidays for 58 days. Multiply the number of work days by 10 hours per day: 58 * 10 = 580 (these are the project hours per employee). -Multiply the per-employee man hours by the number of employees on the job: 580 * 5 = 2,900.
## How do you calculate man hours for training?
Calculating Group Man DaysCheck the number of employees participating in an event and the number of hours in the event. … Calculate the total group hours by multiplying the number of employees, days and hours per day. … Divide the total group hours by the number of hours per man day.
## How many days are in 4 hours?
Days to Hours Conversion TableDaysHours1 Day24 Hours2 Days48 Hours3 Days72 Hours4 Days96 Hours20 more rows
## How much is 1000 hours in work days?
1000 hours equals 41.67 days. Convert 1000 hours into minutes, seconds, years, months, weeks, miliseconds, microseconds, nanoseconds, etc… Convert 42 days into minutes, seconds, years, months, weeks, miliseconds, microseconds, nanoseconds, etc… | 1,119 | 4,476 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2021-31 | latest | en | 0.929879 |
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Python Program to multiply two matrices
Given two matrices, the task to multiply them. Matrices can either be square or rectangular.
Examples:
```Input : mat1[][] = {{1, 2},
{3, 4}}
mat2[][] = {{1, 1},
{1, 1}}
Output : {{3, 3},
{7, 7}}
Input : mat1[][] = {{2, 4},
{3, 4}}
mat2[][] = {{1, 2},
{1, 3}}
Output : {{6, 16},
{7, 18}}```
Multiplication of Square Matrices :
The below program multiplies two square matrices of size 4*4, we can change N for different dimensions.
Python3
`# 4x4 matrix multiplication using Python3``# Function definition``def` `matrix_multiplication(M, N):`` ``# List to store matrix multiplication result`` ``R ``=` `[[``0``, ``0``, ``0``, ``0``],`` ``[``0``, ``0``, ``0``, ``0``],`` ``[``0``, ``0``, ``0``, ``0``],`` ``[``0``, ``0``, ``0``, ``0``]]` ` ``for` `i ``in` `range``(``0``, ``4``):`` ``for` `j ``in` `range``(``0``, ``4``):`` ``for` `k ``in` `range``(``0``, ``4``):`` ``R[i][j] ``+``=` `M[i][k] ``*` `N[k][j]` ` ``for` `i ``in` `range``(``0``, ``4``):`` ``for` `j ``in` `range``(``0``, ``4``):`` ``# if we use print(), by default cursor moves to next line each time,`` ``# Now we can explicitly define ending character or sequence passing`` ``# second parameter as end =""`` ``# syntax: print(, end ="")`` ``# Here space (" ") is used to print a gap after printing`` ``# each element of R`` ``print``(R[i][j], end ``=``" "``)`` ``print``("``", end ="``")` `# First matrix. M is a list``M ``=` `[[``1``, ``1``, ``1``, ``1``],`` ``[``2``, ``2``, ``2``, ``2``],`` ``[``3``, ``3``, ``3``, ``3``],`` ``[``4``, ``4``, ``4``, ``4``]]` `# Second matrix. N is a list``N ``=` `[[``1``, ``1``, ``1``, ``1``],`` ``[``2``, ``2``, ``2``, ``2``],`` ``[``3``, ``3``, ``3``, ``3``],`` ``[``4``, ``4``, ``4``, ``4``]]`` ` `# Call matrix_multiplication function``matrix_multiplication(M, N)` `# This code is contributed by Santanu`
Output
```Result matrix is
10 10 10 10
20 20 20 20
30 30 30 30
40 40 40 40```
Time complexity: O(n3). It can be optimized using Strassen’s Matrix Multiplication
Auxiliary Space: O(n2)
Multiplication of Rectangular Matrices :
We use pointers in C to multiply to matrices. Please refer to the following post as a prerequisite of the code.
How to pass a 2D array as a parameter in C?
Python3
`# Python3 program to multiply two``# rectangular matrices` `# Multiplies two matrices mat1[][]``# and mat2[][] and prints result.``# (m1) x (m2) and (n1) x (n2) are``# dimensions of given matrices.``def` `multiply(m1, m2, mat1,`` ``n1, n2, mat2):` ` ``res ``=` `[[``0` `for` `x ``in` `range``(n2)]`` ``for` `y ``in` `range` `(m1)]`` ` ` ``for` `i ``in` `range``(m1):`` ``for` `j ``in` `range``(n2):`` ``res[i][j] ``=` `0`` ``for` `x ``in` `range``(m2): `` ``res[i][j] ``+``=` `(mat1[ i][x] ``*`` ``mat2[ x][j])`` ` ` ``for` `i ``in` `range``(m1):`` ``for` `j ``in` `range``(n2): `` ``print` `(res[i][j],`` ``end ``=` `" "``) `` ``print` `()` `# Driver code``if` `__name__ ``=``=` `"__main__"``:`` ` ` ``mat1 ``=` `[[``2``, ``4``], [``3``, ``4``]]`` ``mat2 ``=` `[[``1``, ``2``], [``1``, ``3``]]`` ``m1, m2, n1, n2 ``=` `2``, ``2``, ``2``, ``2`` ` ` ``# Function call`` ``multiply(m1, m2, mat1,`` ``n1, n2, mat2)`` ` `# This code is contributed by Chitranayal`
Output
```6 16
7 18```
Time complexity: O(n3). It can be optimized using Strassen’s Matrix Multiplication
Auxiliary Space: O(m1 * n2)
Please refer complete article on Program to multiply two matrices for more details!
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http://www.talk-polywell.org/bb/viewtopic.php?f=3&t=541&sid=5dc0f6663e33e67d45e6596856d74a0a&start=45 | 1,539,899,444,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512015.74/warc/CC-MAIN-20181018214747-20181019000247-00370.warc.gz | 568,616,621 | 14,168 | ## Making Electricity with the p-B Polywell
Discuss how polywell fusion works; share theoretical questions and answers.
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MSimon wrote:Current flow depends on potential and load. It has zero to do with charge canceling.
Any device capable, even in principle, of running in steady state has to obey null summation of currents. No part of it can supply a current without a replacement current coming in from somewhere else.
The decelerator grid is one end of an open circuit. (I hope you haven't forgotten that at 0 Hz a capacitor looks like an open circuit.) The only way a current could flow in it, steady-state, is for it to flow around the grid between two contacts, which has nothing to do with the decelerator grid's function and won't generate any power.
So what's the other end of this open circuit? A voltage supply. Probably closely connected with the magrid power handling system, since they're at roughly the same large-scale floating voltage (as opposed to the collector).
The MaGrids are practically useless as electrostatic collectors. At even 200 KV (high energy density lower Q machines) they will collect about 15% of the available energy. Provided the collector plates are right at the surface of the magnetic bottle. A big no no as that ruins electron oscillation.
That's not what I meant.
Consider the assembly of magrid, decelerator grid, and collector. What is the potential difference between the collector and the decelerator? 2 MV. What is the potential difference between the magrid and the decelerator? 200 kV or less.
Now remove the decelerator, keeping everything else the same. What is the potential difference between the collector and the magrid?
1.8 MV.
Naturally you can't actually run a BFR this way. But it illustrates my point.
The essence of the power cycle is that the alphas in the core (ie: at the magrid) are at the bottom of a massive potential well. Alphas that are fired out from the core (this is free energy for the purposes of the electrostatic analysis) represent a current across a voltage - between the magrid (which is sucking up electrons at the same rate in amps (net, counting the e-guns) as the alphas are streaming outwards) and the wall (which is giving up electrons to the alphas at the same rate in amps as they are arriving at the wall).
Sounds like a closed circuit to me. Furthermore, it behaves as expected if you short it (the potential difference collapses and you have a thermal system) or isolate the hot terminal (massive voltage buildup, at least for the first nanosecond before the alphas stop being able to reach the wall at all).
The sole purpose of the 'decelerator' grid is to flatten the bottom of this megavolt-range potential well so that a recirculating wiffleball machine can happily operate in it.
charliem
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It would be much less confusing if we could agree on the names of the structures.
I think that the decelerator grid Simon is talking about is the same that someone else named the trap grid, I the shield grid, and some other people called even more names.
I propose we use these terms from now on:
1-TRAP GRID, not far from the magrid and at a negative potential from it. Its function is to repel back inside the electrons that leak through the cups (to "trap" them), and shield the internal fields from external influences (a Faraday cage).
2-DECELERATOR grids/structures, at a quite high positive potential from the trap grid. Their function is to slow the alphas down, but not to catch them.
3-COLLECTORS. The structures that the alphas hit and where they get neutralized.
-----------------------------------------------------------------------------------
About E-fields, acceleration/deceleration of charges, and where the energy goes or comes from.
When a charge gets in the middle of an electric field it feels a force that accelerate it in some direction. That acceleration means that its kinetic energy is altered but the difference in energy is NOT taken from the e-field, so this could very well be an static charge field and even so keep influencing more charges indefinitely without lossing any power.
This looks counter-intuitive only if we concentrate on the e-field and the particles and forget to look at the whole system. Just take a broader point of view and the [apparent] contradiction disapears.
EDIT: Decelerator/s are optional. In some configurations there is no deceleration of alphas at all, or simply get the deceleration from the potential difference between Trap grid and Collector.
Last edited by charliem on Tue Jul 08, 2008 12:43 pm, edited 1 time in total.
blaisepascal
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charliem wrote:It would be much less confusing if we could agree on the names of the structures.
I propose we use these terms from now on:
1-TRAP GRID
2-DECELERATOR
3-COLLECTORS.
OK, just so I have a clear idea of what this structure looks like....
In the center, you have a region where the p and B11 collide, producing α-particles with energies of 2-3MeV each. This is at the center of a virtual cathode formed by...
a "quasi-spherical" plasma of mostly electrons in a magnetically contained "whiffle-ball"/polywell. Passing through this plasma, trapped by the negative electrostatic field of the plasma, are the p and B11 ions. Because they are trapped in the potential well, which is mostly radial, they pass through the center repeatedly until they fuse. But I suppose that the low kinetic energy ions near the top of the electrostatic potential well will form a positively charged "halo" (but still negative overall potential) in the space between the whiffle-ball and the MAgrid.
The magnetic containment of the polywell is formed by the next layer from the center, the MAgrid -- a collection of solenoid coils arranged as a cube or dodecahedron, energized so all magnetic fields are directed inward. These solenoids have conductive casings so any electrons which should hit the solenoids can be extracted. The MAgrid is the first thing that the α-particles escaping from the core are likely to interact with -- by smashing into it, so a good design will minimize the shadow area of the MAgrid as seen from the center. The MAgrid casing also forms the anode of the system, being the collection point for excess electrons. It is electrically grounded.
Outside the MAgrid is your "TRAP" grid, negatively charged, intended to (a) encourage electrons escaping to return to the polywell, and (b) shield the MAgrid from the decelerator grid further out. It seems to me that this grid has to be (relatively) fine to do this job (relative to the MAgrid, at least), meaning it's likely to have an even larger shadow effect on the outgoing α-particles than the MAgrid.
Further out is the "decelerator" grid, which is at a significantly positive potential, forcing the outgoing α-particles to notice it and lose most of their kinetic energy, so they can be caught at low kinetic energy (but high potential energy) by...
the "collector", a solid electrically conductive shell which captures the lower-speed α-particles and acts as the cathode of the system, neutralizing the positive charged α-particles impinging on it.
Is there any reason we shouldn't expect α-particles to be lost by hitting the trap and decelerator grid?
(As an aside, I noticed that α didn't give me an α. Is there anyway to make HTML-entities work?)
93143
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In my conception the trap grid and the decelerator grid are the same thing.
It's at a high negative potential relative to the collectors further out, and a slightly negative potential with respect to the magrid.
Net result: Electrons can't climb the potential to the trap grid. Alphas accelerate towards it, miss (hopefully it can be in the magrid shadow), and just barely manage to make it across the much larger second potential difference to the collectors.
I started calling it a decelerator grid because I was beginning to think that "trap grid" was at least as appropriate for either of the other structures (assuming the idea about a 'grid' of collector plates works).
Now, of course, when I've made such a fuss about how its function isn't to generate a large potential well but simply to change its shape, "decelerator grid" is starting to sound inappropriate too...
Last edited by 93143 on Tue Jul 08, 2008 5:35 am, edited 2 times in total.
TallDave
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All that deceleration has to be paid for somewhere. Otherwise you could run this in reverse as a free energy machine by putting alphas on the edge and accelerating them in.
Although, now that I think about it, they'd all be stuck at the bottom of the well, so you couldn't really get work this way.
93143
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TallDave wrote:
All that deceleration has to be paid for somewhere. Otherwise you could run this in reverse as a free energy machine by putting alphas on the edge and accelerating them in.
Although, now that I think about it, they'd all be stuck at the bottom of the well, so you couldn't really get work this way.
You really could do that. You'd need to supply about 100 MW to keep the flow of alphas up (pumping -50 amps of electrons down a 2 MV difference from the alpha source to the magrid), and magrid heating would be maybe five times what you'd get running the reactor forwards (there's not really a good way to get p+11B out of 3x4He...), but you could do that.
You could even get energy out. Hook the magrid cooling system to a steam turbine. You might get 30 or 40 MW...
MSimon
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The decelerator grid is one end of an open circuit.
And "ground" is the other end. There is a potential between ground and the grid. If you connect a load current will flow. If the flow is replenished by induced charges a current can be maintained.
The current flow in the decelerator grids has zero to do with neutralizing alpha charges. Zero.
The sole purpose of the 'decelerator' grid is to flatten the bottom of this megavolt-range potential well so that a recirculating wiffleball machine can happily operate in it.
Nope. The machine will operate quite nicely without a decelerator grid. It would then be a thermal machine.
Engineering is the art of making what you want from what you can get at a profit.
93143
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MSimon wrote:
The decelerator grid is one end of an open circuit.
And "ground" is the other end. There is a potential between ground and the grid. If you connect a load current will flow. If the flow is replenished by induced charges a current can be maintained.
Okay, I can get traction on this. There's no current flowing into the decelerator grid from the vacuum chamber. Ideally, nothing hits it. Therefore if you just hooked it to ground, through a load or otherwise, it would simply discharge and that would be that. You need to keep it charged up, statically.
The current flow in the decelerator grids has zero to do with neutralizing alpha charges. Zero.
I can agree with that. They don't neutralize any alphas, so any connection would be kind of weird. As a matter of fact, there's no current in them at all.
The sole purpose of the 'decelerator' grid is to flatten the bottom of this megavolt-range potential well so that a recirculating wiffleball machine can happily operate in it.
Nope. The machine will operate quite nicely without a decelerator grid. It would then be a thermal machine.
That ALSO assumes that the massive potential difference between the magrid and the collector is changed to a small positive one to prevent electrons from leaving. This means the charge on the magrid would stay the same, but do power electronics maintain charge or voltage?
The fact that the charge on the decelerator grid is what physically sets up this potential difference in the direct-conversion configuration does not change the fact that it exists, and that at both ends of this potential difference currents are flowing into/out of the vacuum chamber - currents high enough that when multiplied by 2 MV (as one or the other of them will have to be to reach ground) you get 100 MW.
Last edited by 93143 on Tue Jul 08, 2008 6:17 am, edited 2 times in total.
TallDave
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93143 wrote:
TallDave wrote:
All that deceleration has to be paid for somewhere. Otherwise you could run this in reverse as a free energy machine by putting alphas on the edge and accelerating them in.
Although, now that I think about it, they'd all be stuck at the bottom of the well, so you couldn't really get work this way.
You really could do that. You'd need to supply about 100 MW to keep the flow of alphas up (pumping -50 amps of electrons down a 2 MV difference from the alpha source to the magrid), and magrid heating would be maybe five times what you'd get running the reactor forwards (there's not really a good way to get p+11B out of 3x4He...), but you could do that.
You could even get energy out. Hook the magrid cooling system to a steam turbine. You might get 30 or 40 MW...
I was thinking no Magrid, no Polywell, just some alphas running down a well. It would be a bit like the electrostatic equivalent of a hydropower plant: gaining energy by moving "downhill." I think you're claiming the "free" kinetic energy gained would be offset by the need to replenish electrons the alphas give up.
EDIT: The electrons they pick up, I mean.
Last edited by TallDave on Tue Jul 08, 2008 6:17 am, edited 1 time in total.
93143
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TallDave wrote:I was thinking no Magrid, no Polywell, just some alphas running down a well. It would be a bit like the electrostatic equivalent of a hydropower plant: gaining energy by moving "downhill." I think you're claiming the "free" kinetic energy gained would be offset by the need to replenish electrons the alphas give up.
Yep. You have to pump them downhill, from the emitter to the target, which for negative charges takes energy. More energy than you could possibly get out with a thermal cycle, and just a little more than you could get with near-perfect direct conversion.
Essentially it's because a charge separation has potential energy, and the potential energy of a static charge separation is rapidly dissipated by the acceleration and subsequent smashup/thermal absorption of the arriving alphas, ending up as heat. To maintain the static charge separation in the face of this continued dissipation, you need to do work to push more charge through the potential gradient - continuously replenishing the potential energy of the static charge separation. The work done is current times voltage, and it ends up in whatever the high-speed alphas do with it.
Now reverse it. The high-speed alphas, instead of crashing into stuff and dissipating their energy as heat, are generated from thin air by fusion. They then fly energetically uphill and are discharged at the collector. In order to maintain charge balance, you have to move electrons uphill along with the alphas. But for a negative charge, uphill is good. So they do that naturally, and you have to stick a load in between (current times voltage) to prevent them from equalizing the source and sink too fast and eliminating the potential difference that stops the alphas.
Or, to put it another way, the alpha flow causes a charge imbalance between the magrid and the collector. This pulls electrons from one to the other, like particles in a linear accelerator (except that they're going through ground and load and transmission wires, so they do work along the way and arrive with very little kinetic energy). The power necessary to maintain this charge imbalance is the power that continually pushes more alphas up the hill - the fusion power.
MSimon
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OK. Here is a question:
If the decelerator grid has zero current flow then where is the alpha energy going? Conservation of mass/energy and all that.
Engineering is the art of making what you want from what you can get at a profit.
93143
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MSimon wrote:OK. Here is a question:
If the decelerator grid has zero current flow then where is the alpha energy going? Conservation of mass/energy and all that.
The alpha kinetic energy is transformed into electrical potential energy by the potential difference it has to climb to get from the magrid to the collector. (The fact that the first bit is slightly downhill is irrelevant.) Since the alpha kinetic energy was from fusion, this is essentially free electrical potential energy, and it can (indeed must, if the reactor is to continue to operate) be used to generate a flow of electrons through a load.
The electrons leaving the magrid (at either ground or -1.8 MV) don't need to be the same electrons that ultimately arrive at the collector (at either +1.8 MV or ground), but the principle is the same: somewhere in the device, -50 amps of electrons (the neutralization current) has to traverse a +1.8 MV potential difference. The alpha current is like the internal current in a voltaic cell - the fusion reaction drives the alphas uphill, and this gives rise to the external current at its characteristic voltage, which can then be used to power stuff.
MSimon
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93143 wrote:
MSimon wrote:OK. Here is a question:
If the decelerator grid has zero current flow then where is the alpha energy going? Conservation of mass/energy and all that.
The alpha kinetic energy is transformed into electrical potential energy by the potential difference it has to climb to get from the magrid to the collector. (The fact that the first bit is slightly downhill is irrelevant.) Since the alpha kinetic energy was from fusion, this is essentially free electrical potential energy, and it can (indeed must, if the reactor is to continue to operate) be used to generate a flow of electrons through a load.
The electrons leaving the magrid (at either ground or -1.8 MV) don't need to be the same electrons that ultimately arrive at the collector (at either +1.8 MV or ground), but the principle is the same: somewhere in the device, -50 amps of electrons (the neutralization current) has to traverse a +1.8 MV potential difference. The alpha current is like the internal current in a voltaic cell - the fusion reaction drives the alphas uphill, and this gives rise to the external current at its characteristic voltage, which can then be used to power stuff.
I think you are still confused.
Let us make this a free space problem. You have a 0V electrode and a 1 MV tubular electrode. And inside the tubular electrode a -1 V collector plate. (forget practical difficulties like arcing). You shoot 2 MeV (free energy) alphas into the collector plate with great accuracy from the 0 V electrode. Nothing hits the 1 MV electrode.
Describe the voltages and current flows in the system - assuming that there is enough source/sink such that the system is in the same condition (voltage wise) as before the alpha(s) entered the system.
Engineering is the art of making what you want from what you can get at a profit.
93143
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MSimon wrote:Let us make this a free space problem. You have a 0V electrode and a 1 MV tubular electrode. And inside the tubular electrode a -1 V collector plate. (forget practical difficulties like arcing). You shoot 2 MeV (free energy) alphas into the collector plate with great accuracy from the 0 V electrode. Nothing hits the 1 MV electrode.
Describe the voltages and current flows in the system - assuming that there is enough source/sink such that the system is in the same condition (voltage wise) as before the alpha(s) entered the system.
Not sure I completely understand the geometry, but...
The alphas follow a 'valley' of potential from the 0 V electrode to the -1 V collector plate, striking it at 2.000002 MeV. Much heating ensues. A -50 amp (for the sake of argument) electron replacement current flows from source to ground to power supply to collector, requiring -50*-1=50 watts of pumping power (ie: negative useful work).
Remember that charge and potential are not the same thing.
I think a better analogy would be to have both the source electrode and the tubular electrode at -1 MV, and the collector at -1 V. That way, the alphas would be at 2 eV by the time they reach the collector (assuming that the geometry in this configuration allows them to hit the collector at all).
This also removes the otherwise large gradient between the source electrode and the tubular electrode, so that if you wanted to, you could further depress the tubular electrode voltage slightly and force electrons to prefer the source electrode. Note that this would not change the alpha impingement energy, because the potential difference between the source and the collector remains the same.
To return to the direct-conversion Polywell, it is essential to grasp that you can't ground both the collector and the magrid. One of them has to be at high voltage, because the magrid is inside a (reasonable facsimile of a) closed spherical shell possessing a massive negative charge. Thus the magrid MUST be about 2 MV lower than the collector, no matter which one you ground.
If you did ground them both, but kept the decelerator at -1.8 MV, the alphas would gain 3.6 MeV passing between the (now very very positively charged) magrid and the (very very negatively charged, but invisible from inside) decelerator, and lose it again on the way to the (neutral) collector. Impingement kinetic energy = birth kinetic energy, because the potentials at both locations are the same.
MSimon
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The alphas follow a 'valley' of potential from the 0 V electrode to the -1 V collector plate, striking it at 2.000002 MeV. Much heating ensues. A -50 amp (for the sake of argument) electron replacement current flows from source to ground to power supply to collector, requiring -50*-1=50 watts of pumping power (ie: negative useful work).
Nope. By the time the alphas hit the -1V collector inside the 1 MV cylindrical grid they have -2eV of energy.
They lost 2 MeV getting to the -1V collector. According to the potential gradient.
Any way I see why you are confused.
Engineering is the art of making what you want from what you can get at a profit. | 5,254 | 22,812 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-43 | latest | en | 0.944242 |
https://www.enotes.com/homework-help/stone-released-from-rest-from-edge-building-roof-359407 | 1,516,228,831,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886979.9/warc/CC-MAIN-20180117212700-20180117232700-00709.warc.gz | 888,453,671 | 9,504 | # A stone is released from rest from the edge of a building roof 190 m above the ground. Neglecting air resistance, what is the speed of the stone, just before striking the ground.
justaguide | Certified Educator
The stone is released from rest at the edge of the roof of a building that is 190 m above the ground. The initial velocity of the stone is 0 m/s. When released, the stone is accelerated downwards due to the gravitational force of attraction of the Earth at 9.8 m/s^2 in a direction that is vertically downwards.
Use the formula v^2 - u^2 = 2*a*s where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance traveled. Substituting the values provided in the problem:
`v^2 - 0^2 = 2*9.8*190`
`=> v = sqrt(3724)`
`~~ 61.02 m/s`
The velocity of the stone just before it strikes the ground is approximately 61.02 m/s in a direction vertically downwards. | 233 | 909 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-05 | latest | en | 0.931021 |
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# 8 Questions
7 Proof: $\mathrm{adj}(\mathrm{adj}(A)) = (\mathrm{det}(A))^{n-2} \cdot A$ for $A \in \mathbb{R}^{n\times n}$ 4 Trigonometric equality $\cos x + \cos 3x - 1 - \cos 2x = 0$ 2 Prove: if $A(x)$ is divisible by $(x-a)^m$, its derivative is divisible by $(x - a)^{m-1}$ 2 Finding a particular solution to a non-homogeneous system of equations 2 Proof: $CA = I_n$ and $AD = I_m \Rightarrow C = D$
# 252 Reputation
+10 Proof: $\mathrm{adj}(\mathrm{adj}(A)) = (\mathrm{det}(A))^{n-2} \cdot A$ for $A \in \mathbb{R}^{n\times n}$ +5 Finding a particular solution to a non-homogeneous system of equations +10 Prove: if $A(x)$ is divisible by $(x-a)^m$, its derivative is divisible by $(x - a)^{m-1}$ +10 Plotting a quadratic equation in the $\,xy\,$- plane
3 A logic problem 1 Plotting a quadratic equation in the $\,xy\,$- plane
# 10 Tags
3 logic 0 divisibility 1 graphing-functions 0 derivatives 0 linear-algebra × 6 0 transformation 0 matrices × 3 0 determinant 0 trigonometry 0 factoring
# 5 Accounts
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New operator definitions 1
randRange(3, 8) randRange(2, 4) randFromArray([ function(x, y) { return ["+", ["*", C, x], y]; }, function(x, y) { return ["-", ["*", C, x], y]; }, function(x, y) { return ["+", x, ["*", C, y]]; }, function(x, y) { return ["-", x, ["*", C, y]]; }, function(x, y) { return ["-", ["*", C, x], ["*", D, y]]; }, function(x, y) { return ["-", ["*", D, x], ["*", C, y]]; }, function(x, y) { return ["*", x, ["-", y, C]]; }, function(x, y) { return ["*", ["-", C, x], y]; }, function(x, y) { return ["+", ["^", x, 2], ["*", D, ["^", y, 2]]]; }, function(x, y) { return ["+", ["*", D, ["^", x, 2]], ["^", y, 2]]; }, function(x, y) { return ["-", ["^", x, 2], ["*", D, ["^", y, 2]]]; }, function(x, y) { return ["-", ["*", D, ["^", x, 2]], ["^", y, 2]]; }, function(x, y) { return ["-", ["*", D, x], C]; }, function(x, y) { return ["+", ["*", C, y], D]; }, function(x, y) { return ["+", ["*", x, y], ["*", D, x], ["-", y]]; } ]) randRange(-6, 6) randRange(-6, 6) expr(FUN("x", "y")) expr(FUN(X, Y)) expr(FUN(X, Y), true) !isNaN(expr(FUN(0, NaN), true)) !isNaN(expr(FUN(NaN, 0), true))
If x binop(1) y = TEXT, find X binop(1) Y.
VALU
Substitute in x = X y = Y x = X and y = Y to get SUBS.
Simplify the expression to get VALU. | 520 | 1,399 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2017-39 | latest | en | 0.240585 |
https://www.quesba.com/questions/please-create-flowchart-following-program-start-1-ask-user-input-weight-872655 | 1,702,262,205,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679103464.86/warc/CC-MAIN-20231211013452-20231211043452-00524.warc.gz | 1,038,844,675 | 27,985 | # Please create a flowchart for the following; Program start 1. Ask the user to input the weight...
Please create a flowchart for the following;
Program start
1. Ask the user to input the weight and store the result in a variable called "weight"
2. If the weight is less than 2 pounds then set the price to be 1.10
3. Else if the weight is less than 6 pounds then set the price to be 2.20
4. Else if the weight is less than 10 pounds then set the price to be 3.70
5. Else set the price to be 3.30 6. Print out a message such as "A package weighing x pounds costs y dollars to ship"
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### A library function, islower(), takes a single character (a letter) as an argument and returns a...
A library function, islower(), takes a single character (a letter) as an argument and returns a nonzero integer if the letter is lowercase, or zero if it is uppercase. This function requires the header file CTYPE.H. Write a program that allows the user to enter a letter, and then displays either zero or nonzero, depending on whether a lowercase or uppercase letter was entered. (See the SQRT...
Oct 08 2021
Start with the fraction-adding program of Exercise 9 in Chapter 2, “C++ Programming Basics.” This program stores the numerator and denominator of two fractions before adding them, and may also store the answer, which is also a fraction. Modify the program so that all fractions are stored in variables of type struct fraction, whose two members are the fraction’s numerator and...
Oct 08 2021
### Refer to the CIRCAREA program in Chapter 2, “C++ Programming Basics.” Write a function called...
Refer to the CIRCAREA program in Chapter 2, “C++ Programming Basics.” Write a function called circarea() that finds the area of a circle in a similar way. It should take an argument of type float and return an argument of the same type. Write a main() function that gets a radius value from the user, calls circarea(), and displays the result.
Oct 08 2021
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• Need any last minute study tips? | 850 | 3,676 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2023-50 | latest | en | 0.898275 |
http://mathhelpforum.com/statistics/167930-probability.html | 1,480,907,913,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541518.17/warc/CC-MAIN-20161202170901-00278-ip-10-31-129-80.ec2.internal.warc.gz | 176,113,627 | 9,969 | 1. ## probability
a six faced die is so biased that it is twice as likely to show an even number as an odd number when thrown. it is thrown twice. what is the probability that the sum of 2 numbers thrown is even
2. Ok, let's start like this
even + even = ...
odd + even = ...
even + odd = ...
odd + odd = ...
3. Condition on the first throw.
P(sum is even) = P(sum is even|first throw is even)*P(first throw is even) + P(sum is even|first throw is odd)*P(first throw is odd) = (2/3)*(2/3) + (1/3)*(1/3) = 5/9 | 153 | 513 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2016-50 | longest | en | 0.977014 |
http://convertit.com/Go/BusinessConductor/Measurement/Converter.ASP?From=UK+bushel&To=static+moment+of+area | 1,669,602,899,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710462.59/warc/CC-MAIN-20221128002256-20221128032256-00836.warc.gz | 13,796,951 | 3,652 | New Online Book! Handbook of Mathematical Functions (AMS55)
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Conversion Result: ```British bushel = 0.03636879435904 volume (volume) ``` Related Measurements: Try converting from "UK bushel" to acre foot, chetvert (Russian chetvert), cord foot (of wood), displacement ton, dram fluid (fluid dram), dry gallon, dry pint, gill, hekat (Israeli hekat), hogshead, magnum, methuselah, minim, noggin, oz fluid (fluid ounce), register ton, sack, salmanazar, tea cup, teaspoon, or any combination of units which equate to "length cubed" and represent capacity, section modulus, static moment of area, or volume. Sample Conversions: UK bushel = 2.4 balthazar, .17316424 chetvert (Russian chetvert), .08027199 cord foot (of wood), 9,838.2 dram fluid (fluid dram), 1,212,293.15 drop, 66.05 dry pint, 48.04 fifth, 9.61 gallon, .15250189 hogshead, .53375661 kilderkin, .01290074 last, 19.22 magnum, 1.92 nebuchadnezzar, 1,229.78 oz fluid (fluid ounce), 38.43 quart (fluid quart), .12900736 seam, .03636879 stere, 3.51 tou (Chinese tou), .03812547 tun (English tun), 48 wine bottle.
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Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 456 | 1,591 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2022-49 | latest | en | 0.638535 |
http://www.chegg.com/homework-help/questions-and-answers/consider-a-35-ft3-s-flow-of-atmospheric-air-at-147-psi-80-f-and-80-relative-humidity-assum-q3462760 | 1,369,386,751,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704433753/warc/CC-MAIN-20130516114033-00077-ip-10-60-113-184.ec2.internal.warc.gz | 383,623,273 | 7,949 | Consider a 35 ft^3/s flow of atmospheric air at 14.7 psi, 80 °F, and 80% relative humidity. Assume this air flows into a basement room, where it cools to 60 °F at 14.7 psi. How much liquid will condense out, in lbmw/s?
• dVP = (( Pi – Po ) / R) x Rv
therefore
by putting values , we got
2.35lbmw/s | 115 | 314 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2013-20 | latest | en | 0.6832 |
https://ocw.oouagoiwoye.edu.ng/courses/electrical-engineering-and-computer-science/6-006-introduction-to-algorithms-fall-2011/assignments/ | 1,701,509,249,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100381.14/warc/CC-MAIN-20231202073445-20231202103445-00651.warc.gz | 492,388,206 | 8,340 | # Assignments
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In order to use the ZIP files, you will need the programs described in the Software section. Some problem sets also contain additional installation instructions in the README.txt files. LaTeX templates are also included; see the Related Resources for suggested editors.
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1 Asymptotic complexity, recurrence relations, peak finding
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2 Fractal rendering, digital circuit simulation
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3 Range queries, digital circuit layout
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4 Hash functions, Python dictionaries, matching DNA sequences
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5 The Knight's Shield, RSA public key encryption, image decryption
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6 Social networks, Rubik's Cube, Dijkstra
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7 Seam carving, stock purchasing and knapsack
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Seam Carving for Content-Aware Image Resizing | 484 | 1,930 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2023-50 | latest | en | 0.633616 |
https://community.smartsheet.com/discussion/comment/316405/ | 1,656,706,813,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103945490.54/warc/CC-MAIN-20220701185955-20220701215955-00616.warc.gz | 229,472,611 | 37,991 | # need help with an IF function
Hello community,
I am trying to come up with a cell formula that is a simple difference function, but I need it to be triggered by the number in another cell. It is for an inventory/ordering sheet. Someone will count the inventory and type in the "QTY ON HAND". there is also a column for "MINIMUM QTY NEEDED" which is preset and static. I just want to take the "DESIRED QTY" minus "QTY ONHAND", but I only want to trigger this function if the "QTY ON HAND" is lower than the "MINIMUM NEEDED"
I came up with this sentance, but it is not properly expressed so I'd love some help cleaning it up.
=IF([QTY ON HAND]1<[MINIMUM NEEDED QTY ON HAND (OR REORDER POINT)]1), ([DESIRED QUANTITY ON HAND]1-[QTY ON HAND]1)
I'm not sure if it should be IF OR SUMIF, or something else completely
Let me know if I need to clarify further
Thanks so much!
Tags:
No, you pretty much have it
`=IF([QTY ON HAND]@row < [MINIMUM NEEDED QTY ON HAND (OR REORDER POINT)]@row, [DESIRED QUANTITY ON HAND]@row - [QTY ON HAND]@row, "")`
This would show a blank if it was not triggered
`=IF([QTY ON HAND]@row < [MINIMUM NEEDED QTY ON HAND (OR REORDER POINT)]@row, [DESIRED QUANTITY ON HAND]@row - [QTY ON HAND]@row, 0)`
This would show a zero if it were not triggered
Brent C. Wilson, P.Eng, PMP, Prince2
Facilityy Professional Services Inc.
http://www.facilityy.com
No, you pretty much have it
`=IF([QTY ON HAND]@row < [MINIMUM NEEDED QTY ON HAND (OR REORDER POINT)]@row, [DESIRED QUANTITY ON HAND]@row - [QTY ON HAND]@row, "")`
This would show a blank if it was not triggered
`=IF([QTY ON HAND]@row < [MINIMUM NEEDED QTY ON HAND (OR REORDER POINT)]@row, [DESIRED QUANTITY ON HAND]@row - [QTY ON HAND]@row, 0)`
This would show a zero if it were not triggered
Brent C. Wilson, P.Eng, PMP, Prince2
Facilityy Professional Services Inc.
http://www.facilityy.com
• Thanks so much, That worked perfectly | 577 | 1,923 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-27 | latest | en | 0.878847 |
https://gmatclub.com/forum/in-the-last-few-decades-physicists-have-identified-the-141085.html?fl=similar | 1,487,796,528,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171043.28/warc/CC-MAIN-20170219104611-00329-ip-10-171-10-108.ec2.internal.warc.gz | 712,297,337 | 66,046 | In the last few decades, physicists have identified the : GMAT Sentence Correction (SC)
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# In the last few decades, physicists have identified the
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In the last few decades, physicists have identified the existence of different "flavors" of subatomic particles called quarks,most of them as small or smaller than the electron, which display a property known as color charge.
most of them as small or smaller than the electron, which display
most of them as small or smaller than the electron and displaying
mostly as small or smaller than the electron, displaying
mostly at least as small as the electron, which display
most of them at least as small as the electron, displaying
Also explain what "them" is referring to in some options.
[Reveal] Spoiler: OA
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B and E are the only real competitors. "As small or smaller than" in B is awkward. So going with E. "them" can unambiguously refer to quarks as it is a part of the modifier of "quarks".
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22 Oct 2012, 03:49
Since physicists is the subject of the previous clause,cant "them" refer it physicists.
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22 Oct 2012, 04:07
The noun modifier should always touch the noun. A corollary of this can be taken as "A modifier modifies the noun which it touches."
But true the sentence could be better framed as E here uses a series of modifiers to modify the same noun. However, of the given options, E is the best.
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Hi all,
In the last few decades, physicists have identified the existence of different "flavors" of subatomic particles called quarks, most of them as small or smaller than the electron, which display a property known as color charge.
Error Analysis: 1. This sentence uses the incorrect idiom “as small or smaller than” which is actually the mix of two degrees of comparison. This is incorrect. We must one correct idiom to convey the intended meaning.
2. Understanding the sentence structure is very important here. Note that “which display…” is meant to modify “quarks”. However, “which” is closer to “electron” than to “quarks”. Hence, there is an ambiguity in the reference of “which”. Also “display” does not agree in number with “electron” if “which” has to refer to “electron”.
POE:
A) most of them as small or smaller than the electron, which display: Incorrect for the reasons stated above.
B) most of them as small or smaller than the electron and displaying: Incorrect. The idiom error persists.
C) mostly as small or smaller than the electron, displaying: Incorrect. The idiom error persists.
D) mostly at least as small as the electron, which display: Incorrect. “which” is closer to “electron” and hence refer to “electron”. However, “the verb “display” does not agree in number with “electron”.
E) most of them at least as small as the electron, displaying: Correct. Notice that the modifier “most of them at least as small as the electron” is placed between two commas that make this information non-essential for the sentence. In this case, “displaying” correctly modifies “quarks”. If we remove the non-essential information for the sentence, the comma before “displaying” will also be removed. Here, we do not have the case of comma + verb-ing that is modifying the preceding clause.
This sentence is like the following GMAT Prep question the correct answer of which is C:
In the past several years, astronomers have detected more than 80 massive planets, most of them as large or larger than Jupiter, which circleother stars.
A. most of them as large or larger than Jupiter, which circle
B. most of them as large or larger than Jupiter and circling
C. most of them at least as large as Jupiter, circling
D. mostly at least as large as Jupiter, which circle
E. mostly as large or larger than Jupiter, circling
Hope this helps.
Thanks.
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Re: In the last few decades, physicists have identified the [#permalink]
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22 Oct 2012, 07:37
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egmat wrote:
Hi all,
In the last few decades, physicists have identified the existence of different "flavors" of subatomic particles called quarks, most of them as small or smaller than the electron, which display a property known as color charge.
Error Analysis: 1. This sentence uses the incorrect idiom “as small or smaller than” which is actually the mix of two degrees of comparison. This is incorrect. We must one correct idiom to convey the intended meaning.
2. Understanding the sentence structure is very important here. Note that “which display…” is meant to modify “quarks”. However, “which” is closer to “electron” than to “quarks”. Hence, there is an ambiguity in the reference of “which”. Also “display” does not agree in number with “electron” if “which” has to refer to “electron”.
POE:
A) most of them as small or smaller than the electron, which display: Incorrect for the reasons stated above.
B) most of them as small or smaller than the electron and displaying: Incorrect. The idiom error persists.
C) mostly as small or smaller than the electron, displaying: Incorrect. The idiom error persists.
D) mostly at least as small as the electron, which display: Incorrect. “which” is closer to “electron” and hence refer to “electron”. However, “the verb “display” does not agree in number with “electron”.
E) most of them at least as small as the electron, displaying: Correct. Notice that the modifier “most of them at least as small as the electron” is placed between two commas that make this information non-essential for the sentence. In this case, “displaying” correctly modifies “quarks”. If we remove the non-essential information for the sentence, the comma before “displaying” will also be removed. Here, we do not have the case of comma + verb-ing that is modifying the preceding clause.
This sentence is like the following GMAT Prep question the correct answer of which is C:
In the past several years, astronomers have detected more than 80 massive planets, most of them as large or larger than Jupiter, which circleother stars.
A. most of them as large or larger than Jupiter, which circle
B. most of them as large or larger than Jupiter and circling
C. most of them at least as large as Jupiter, circling
D. mostly at least as large as Jupiter, which circle
E. mostly as large or larger than Jupiter, circling
Hope this helps.
Thanks.
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Top quality answer and follow-on example.
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Re: In the last few decades, physicists have identified the [#permalink]
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17 Oct 2013, 23:02
egmat wrote:
Hi all,
In the last few decades, physicists have identified the existence of different "flavors" of subatomic particles called quarks, most of them as small or smaller than the electron, which display a property known as color charge.
Error Analysis: 1. This sentence uses the incorrect idiom “as small or smaller than” which is actually the mix of two degrees of comparison. This is incorrect. We must one correct idiom to convey the intended meaning.
2. Understanding the sentence structure is very important here. Note that “which display…” is meant to modify “quarks”. However, “which” is closer to “electron” than to “quarks”. Hence, there is an ambiguity in the reference of “which”. Also “display” does not agree in number with “electron” if “which” has to refer to “electron”.
POE:
A) most of them as small or smaller than the electron, which display: Incorrect for the reasons stated above.
B) most of them as small or smaller than the electron and displaying: Incorrect. The idiom error persists.
C) mostly as small or smaller than the electron, displaying: Incorrect. The idiom error persists.
D) mostly at least as small as the electron, which display: Incorrect. “which” is closer to “electron” and hence refer to “electron”. However, “the verb “display” does not agree in number with “electron”.
E) most of them at least as small as the electron, displaying: Correct. Notice that the modifier “most of them at least as small as the electron” is placed between two commas that make this information non-essential for the sentence. In this case, “displaying” correctly modifies “quarks”. If we remove the non-essential information for the sentence, the comma before “displaying” will also be removed. Here, we do not have the case of comma + verb-ing that is modifying the preceding clause.
This sentence is like the following GMAT Prep question the correct answer of which is C:
In the past several years, astronomers have detected more than 80 massive planets, most of them as large or larger than Jupiter, which circleother stars.
A. most of them as large or larger than Jupiter, which circle
B. most of them as large or larger than Jupiter and circling
C. most of them at least as large as Jupiter, circling
D. mostly at least as large as Jupiter, which circle
E. mostly as large or larger than Jupiter, circling
Hope this helps.
Thanks.
Thanks for your awesome posts.
Doesn't E change the meaning of the sentence?
The quark is small or smaller than the electron,whereas E states that the quark is at least as small as the electron.
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Re: In the last few decades, physicists have identified the [#permalink]
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09 May 2014, 20:39
egmat wrote:
Hi all,
In the last few decades, physicists have identified the existence of different "flavors" of subatomic particles called quarks, most of them as small or smaller than the electron, which display a property known as color charge.
Error Analysis: 1. This sentence uses the incorrect idiom “as small or smaller than” which is actually the mix of two degrees of comparison. This is incorrect. We must one correct idiom to convey the intended meaning.
2. Understanding the sentence structure is very important here. Note that “which display…” is meant to modify “quarks”. However, “which” is closer to “electron” than to “quarks”. Hence, there is an ambiguity in the reference of “which”. Also “display” does not agree in number with “electron” if “which” has to refer to “electron”.
POE:
A) most of them as small or smaller than the electron, which display: Incorrect for the reasons stated above.
B) most of them as small or smaller than the electron and displaying: Incorrect. The idiom error persists.
C) mostly as small or smaller than the electron, displaying: Incorrect. The idiom error persists.
D) mostly at least as small as the electron, which display: Incorrect. “which” is closer to “electron” and hence refer to “electron”. However, “the verb “display” does not agree in number with “electron”.
E) most of them at least as small as the electron, displaying: Correct. Notice that the modifier “most of them at least as small as the electron” is placed between two commas that make this information non-essential for the sentence. In this case, “displaying” correctly modifies “quarks”. If we remove the non-essential information for the sentence, the comma before “displaying” will also be removed. Here, we do not have the case of comma + verb-ing that is modifying the preceding clause.
This sentence is like the following GMAT Prep question the correct answer of which is C:
In the past several years, astronomers have detected more than 80 massive planets, most of them as large or larger than Jupiter, which circleother stars.
A. most of them as large or larger than Jupiter, which circle
B. most of them as large or larger than Jupiter and circling
C. most of them at least as large as Jupiter, circling
D. mostly at least as large as Jupiter, which circle
E. mostly as large or larger than Jupiter, circling
Hope this helps.
Thanks.
I have a different understanding from what you explained. I thought that the term "as small or smaller than" is different from "as large or larger than" in your example.
- " most of them as large or larger than B" can be rewritten as " most of them at least as large as B", because B is the standard for the smallest.
- " most of them as small or smaller than B" actually points out that B is the standard for the largest rather than smallest, so rewritten as "most of them at least as small as B", this choice changes the meaning of what is expected in the sentence.
Anyways, if I have to choose, E is still the best choice though I'm not satisfied because even though the idiom is correct, the meaning of the sentence is disputable.
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Re: In the last few decades, physicists have identified the [#permalink]
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09 May 2014, 22:23
egmat wrote:
Hi all,
In the last few decades, physicists have identified the existence of different "flavors" of subatomic particles called quarks, most of them as small or smaller than the electron, which display a property known as color charge.
Error Analysis: 1. This sentence uses the incorrect idiom “as small or smaller than” which is actually the mix of two degrees of comparison. This is incorrect. We must one correct idiom to convey the intended meaning.
2. Understanding the sentence structure is very important here. Note that “which display…” is meant to modify “quarks”. However, “which” is closer to “electron” than to “quarks”. Hence, there is an ambiguity in the reference of “which”. Also “display” does not agree in number with “electron” if “which” has to refer to “electron”.
POE:
A) most of them as small or smaller than the electron, which display: Incorrect for the reasons stated above.
B) most of them as small or smaller than the electron and displaying: Incorrect. The idiom error persists.
C) mostly as small or smaller than the electron, displaying: Incorrect. The idiom error persists.
D) mostly at least as small as the electron, which display: Incorrect. “which” is closer to “electron” and hence refer to “electron”. However, “the verb “display” does not agree in number with “electron”.
E) most of them at least as small as the electron, displaying: Correct. Notice that the modifier “most of them at least as small as the electron” is placed between two commas that make this information non-essential for the sentence. In this case, “displaying” correctly modifies “quarks”. If we remove the non-essential information for the sentence, the comma before “displaying” will also be removed. Here, we do not have the case of comma + verb-ing that is modifying the preceding clause.
Hope this helps.
Thanks.
Thanks Shraddha for such a detailed explanation..really appreciate that..
Though POE will help us to give the correct answer, just a quick clarification - any difference between 'most of them' & 'mostly' in this question?
Cheers
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Re: In the last few decades, physicists have identified the [#permalink]
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26 May 2015, 00:12
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Re: In the last few decades, physicists have identified the [#permalink]
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09 May 2016, 06:40
mhngo wrote:
I have a different understanding from what you explained. I thought that the term "as small or smaller than" is different from "as large or larger than" in your example.
- " most of them as large or larger than B" can be rewritten as " most of them at least as large as B", because B is the standard for the smallest.
- " most of them as small or smaller than B" actually points out that B is the standard for the largest rather than smallest, so rewritten as "most of them at least as small as B", this choice changes the meaning of what is expected in the sentence.
Anyways, if I have to choose, E is still the best choice though I'm not satisfied because even though the idiom is correct, the meaning of the sentence is disputable.
Exactly, I agree. Does not this change the meaning? Shraddha?
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Re: In the last few decades, physicists have identified the [#permalink]
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10 May 2016, 10:37
dina98 wrote:
mhngo wrote:
I have a different understanding from what you explained. I thought that the term "as small or smaller than" is different from "as large or larger than" in your example.
- " most of them as large or larger than B" can be rewritten as " most of them at least as large as B", because B is the standard for the smallest.
- " most of them as small or smaller than B" actually points out that B is the standard for the largest rather than smallest, so rewritten as "most of them at least as small as B", this choice changes the meaning of what is expected in the sentence.
Anyways, if I have to choose, E is still the best choice though I'm not satisfied because even though the idiom is correct, the meaning of the sentence is disputable.
Exactly, I agree. Does not this change the meaning? Shraddha?
If I understood you correctly, you mean to say that the correct wording would be " at most as small as...". If this is what you meant, following is my reasoning:
In "at least as as large as ", the phrase "at least" is a measure of largeness. {"more large" = larger}
In "at least as as small as ", the phrase "at least" is a measure of smallness. {"more small" = "smaller"}
Thus either way the phrase "at least" is applicable.
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Re: In the last few decades, physicists have identified the [#permalink]
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20 Jan 2017, 08:50
Isn't 'are' required after 'most of them'? Most of them are at least........
Without any helping verb, it seems awkward!
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Re: In the last few decades, physicists have identified the [#permalink]
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20 Jan 2017, 20:32
1
KUDOS
Expert's post
1
This post was
BOOKMARKED
Mahmud6 wrote:
Isn't 'are' required after 'most of them'? Most of them are at least........
Without any helping verb, it seems awkward!
No, "verb" is not required. "Most of them at least as small as the electron" is a subgroup modifier. Following are the correct structures of subgroup modifiers:
1. ...most of which are at least as small as the electron...
2. ...most of them at least as small as the electron...
3. ...most at least as small as the electron...
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Re: In the last few decades, physicists have identified the [#permalink]
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21 Jan 2017, 02:16
sayantanc2k wrote:
Mahmud6 wrote:
Isn't 'are' required after 'most of them'? Most of them are at least........
Without any helping verb, it seems awkward!
No, "verb" is not required. "Most of them at least as small as the electron" is a subgroup modifier. Following are the correct structures of subgroup modifiers:
1. ...most of which are at least as small as the electron...
2. ...most of them at least as small as the electron...
3. ...most at least as small as the electron...
Thanks.
Would you please explain why be verb is used in the following correct sentence?
Prompted by recent discoveries, paleontologists have started research on the fossils of feathered dinosaurs, a study that has ignited the debate over when and why dinosaurs developed plumage, and that makes many researchers believe that possibly all or most of them had feathers. Source: E-gmat.
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Kudos [?]: 1752 [1] , given: 22
In the last few decades, physicists have identified the [#permalink]
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21 Jan 2017, 19:16
1
KUDOS
Expert's post
Mahmud6 wrote:
sayantanc2k wrote:
Mahmud6 wrote:
Isn't 'are' required after 'most of them'? Most of them are at least........
Without any helping verb, it seems awkward!
No, "verb" is not required. "Most of them at least as small as the electron" is a subgroup modifier. Following are the correct structures of subgroup modifiers:
1. ...most of which are at least as small as the electron...
2. ...most of them at least as small as the electron...
3. ...most at least as small as the electron...
Thanks.
Would you please explain why be verb is used in the following correct sentence?
Prompted by recent discoveries, paleontologists have started research on the fossils of feathered dinosaurs, a study that has ignited the debate over when and why dinosaurs developed plumage, and that makes many researchers believe that possibly all or most of them had feathers. Source: E-gmat.
Here "all or most of them...." is NOT A subgroup modifier, but a complete clause. A subgroup modifier must have a noun (referring to a group) preceding the modifier. In the subject question, the noun that the subgroup modifier refers is "quarks".
Beside the river I saw some animals, some of them behaving strangely..... subgroup modifier referring to "animals"
Beside the river I saw some animals; I can confirm that some of them were behaving strangely...... clause: "sum of them" is the subject
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Re: In the last few decades, physicists have identified the [#permalink]
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22 Jan 2017, 04:51
Mahmud6 wrote:
sayantanc2k wrote:
Mahmud6 wrote:
Isn't 'are' required after 'most of them'? Most of them are at least........
Without any helping verb, it seems awkward!
No, "verb" is not required. "Most of them at least as small as the electron" is a subgroup modifier. Following are the correct structures of subgroup modifiers:
1. ...most of which are at least as small as the electron...
2. ...most of them at least as small as the electron...
3. ...most at least as small as the electron...
Thanks again.
If when 'most of ....' is used as subgroup modifier, and doesn't contain be verb, why the following correct sentence contains be verb 'are' after 'most of which'?
1. ...most of which are at least as small as the electron...
_________________
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Re: In the last few decades, physicists have identified the [#permalink]
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22 Jan 2017, 07:46
1
KUDOS
Expert's post
Mahmud6 wrote:
Mahmud6 wrote:
sayantanc2k wrote:
No, "verb" is not required. "Most of them at least as small as the electron" is a subgroup modifier. Following are the correct structures of subgroup modifiers:
1. ...most of which are at least as small as the electron...
2. ...most of them at least as small as the electron...
3. ...most at least as small as the electron...
Thanks again.
If when 'most of ....' is used as subgroup modifier, and doesn't contain be verb, why the following correct sentence contains be verb 'are' after 'most of which'?
1. ...most of which are at least as small as the electron...
You will need to memorize the three correct standard forms of subgroup modifiers ("most of which" carries a verb; "most of them" and "most" do not).
Re: In the last few decades, physicists have identified the [#permalink] 22 Jan 2017, 07:46
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 7,011 | 28,223 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-09 | longest | en | 0.930879 |
http://www.pas.rochester.edu/~stte/phy415F02/set1.html | 1,539,692,708,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510749.55/warc/CC-MAIN-20181016113811-20181016135311-00179.warc.gz | 533,808,222 | 2,435 | Home
Physics 415: Electromagnetic Theory I
Prof. S. Teitel stte@pas.rochester.edu ----- Fall 2002
## Problem Set 1
Due Monday, September 23, in lecture
• Problem 1 [10 points]
Suppose the electric and magnetic fields, E and B, are given by scalar and vector potentials, and A, that are not in the Lorentz gauge. Show that one can always find a gauge transformation that takes one to new potentials which are in the Lorentz gauge. Specifically, if is the scalar function of the gauge transformation, find an equation to solve for in terms of the orignial and A, such that the new ´ and A´ are in the Lorentz gauge.
• Problem 2 [20 points]
In lecture we discussed the decomposition of any vector function f(r) into its curlfree and divergenceless parts, f||(r) and f(r), as given by Helmholtz's theorem:
f(r) = f||(r) + f (r)
where
f||(r) = — [ 1 4 d3r´ ´ · f(r´) | r-r´| ]
and
f(r) = [ 1 4 d3r´ ´ f(r´) | r-r´| ]
Consider f(k), the Fourier transform of f(r):
f(r) = 1(2)3 +- d3k eik · r f(k)
Using the results above, find expressions for the Fourier transforms of the curlfree (longitudinal) and divergenceless (transverse) parts of f(r).
You will need to know that the Fourier transform of 1/|r| is 4/k2, and that the Fourier transform of the Dirac delta function (r) is eik · r
Last update: Tuesday, August 21, 2007 at 1:47:10 PM. | 399 | 1,351 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2018-43 | latest | en | 0.877215 |
https://www.shaalaa.com/question-bank-solutions/the-marks-obtained-out-50-102-students-physics-test-are-given-frequency-table-below-mean-grouped-data_22840 | 1,576,491,452,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541319511.97/warc/CC-MAIN-20191216093448-20191216121448-00255.warc.gz | 866,698,378 | 11,617 | Share
# The Marks Obtained Out of 50, by 102 Students in a Physics Test Are Given in the Frequency Table Below - CBSE Class 10 - Mathematics
#### Question
The marks obtained out of 50, by 102 students in a Physics test are given in the frequency table below:
Marks(x) 15 20 22 24 25 30 33 38 45 Frequency (f) 5 8 11 20 23 18 13 3 1
Find the average number of marks.
#### Solution
Let the assumed mean (A) = 25
Marks (x1) Frequency (f1) u1= x1 - A x1 - 25 f1u1 15 5 -10 -50 20 8 -5 -40 22 11 -3 -33 24 20 -1 -20 25 23 0 0 30 18 5 90 33 13 8 104 38 3 13 39 45 1 20 20 N = 102 sumf_1"u"_1=110
Average number of marks =A + (sumf_1"u"_1)/N
=25+110/102
=(2550+110)/102
=2660/102
= 26.08(Approx)
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [6]
Solution The Marks Obtained Out of 50, by 102 Students in a Physics Test Are Given in the Frequency Table Below Concept: Mean of Grouped Data.
S | 343 | 932 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2019-51 | latest | en | 0.699396 |
https://www.numbersaplenty.com/2667167269073 | 1,652,770,907,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517018.29/warc/CC-MAIN-20220517063528-20220517093528-00784.warc.gz | 1,109,205,174 | 3,419 | Search a number
2667167269073 is a prime number
BaseRepresentation
bin100110110011111111100…
…011101100000011010001
3100102222102110202201000122
4212303333203230003101
5322144324310102243
65401140231000025
7363460614503006
oct46637743540321
910388373681018
102667167269073
1193915943280a
12370ab8b09015
131646887bbc78
15495a4908668
hex26cff8ec0d1
2667167269073 has 2 divisors, whose sum is σ = 2667167269074. Its totient is φ = 2667167269072.
The previous prime is 2667167269069. The next prime is 2667167269093. The reversal of 2667167269073 is 3709627617662.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 2286894013504 + 380273255569 = 1512248^2 + 616663^2 .
It is a cyclic number.
It is not a de Polignac number, because 2667167269073 - 22 = 2667167269069 is a prime.
It is a super-2 number, since 2×26671672690732 (a number of 26 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 2667167268997 and 2667167269015.
It is not a weakly prime, because it can be changed into another prime (2667167269093) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1333583634536 + 1333583634537.
It is an arithmetic number, because the mean of its divisors is an integer number (1333583634537).
Almost surely, 22667167269073 is an apocalyptic number.
It is an amenable number.
2667167269073 is a deficient number, since it is larger than the sum of its proper divisors (1).
2667167269073 is an equidigital number, since it uses as much as digits as its factorization.
2667167269073 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 48009024, while the sum is 62.
The spelling of 2667167269073 in words is "two trillion, six hundred sixty-seven billion, one hundred sixty-seven million, two hundred sixty-nine thousand, seventy-three". | 620 | 2,049 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-21 | latest | en | 0.848469 |
http://www.docstoc.com/docs/25928177/Depot | 1,432,961,772,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207930895.88/warc/CC-MAIN-20150521113210-00055-ip-10-180-206-219.ec2.internal.warc.gz | 416,854,386 | 39,898 | # Depot by tyndale
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• pg 1
``` Task Description DAY-1
IOI 2002
Yong-In Version F
Korea utopia
Utopia Divided
PROBLEM
The beautiful land of Utopia was once ravaged by war. When the hostilities subsided
the country was divided into four regions by a longitude (north-south line) and a
latitude (east-west line). The intersection of these lines became known as the point (0,0).
All four parts claimed the name Utopia, but as time went by they generally became
known as Utopia 1 (northeast), 2 (northwest), 3 (southwest) and 4 (southeast). A point
in any of the regions was identified by its distance east and its distance north of (0,0).
These distances could be negative; hence a point in Utopia 2 was designated by a
(negative, positive) pair, in Utopia 3 by a (negative, negative) pair, in Utopia 4 by
(positive, negative) and in Utopia 1 by a pair of positive numbers.
Utopia 2 Utopia 1
(,+) (+,+)
(0,0)
Utopia 3 Utopia 4
(, ) (+, )
A major problem was that citizens were not permitted to cross borders. Fortunately,
some ingenious IOI contestants from Utopia developed a safe means of teleportation.
The machine requires code numbers, each of which can only be used once. Now the
challenge facing the team, and you, is to guide the teleporter from its initial position of
(0,0) to the regions of Utopia in the order requested. You don’t care where in a region
you land, but you will have a sequence of N region numbers that specify the regions in
which the teleporter is to land. You may be asked to land in the same region in two or
more consecutive stops. After leaving the initial (0,0) point, you must never land on a
border.
You will receive as input a sequence of 2N code numbers and are to write them as a
sequence of N code pairs, placing a plus or a minus sign before each number. If you are
currently at the point (x,y) and use the code pair (+u,v), you will be teleported to the
point (x+u, yv). You have the 2N numbers, and you can use them in any order you like,
each with a plus or a minus sign.
Suppose you have code numbers 7, 5, 6, 1, 3, 2, 4, 8 and are to guide the teleporter
according to the sequence of region numbers 4, 1, 2 ,1. The sequence of code pairs
(+7,1), (5,+2), (4,+3), (+8,+6) achieves this as it teleports you from (0,0) to the
locations (7,1), (2,1), (2,4) and (6,10) in that order. These points are located in
Utopia 4, Utopia 1, Utopia 2, and Utopia 1, respectively.
You are given 2N distinct code numbers and a sequence of N region numbers indicating
where the teleporter is to land. Construct a sequence of code pairs from the given
numbers that guide the teleporter to go through the given region sequence.
2010-02-19 13:05 Page 1 of 2
IOI 2002
Yong-In Version F
Korea utopia
INPUT
Your program is to read from standard input. The first line contains a positive integer N
(1 N 10000). The second line contains the 2N distinct integer code numbers (1≤
code number ≤ 100000) separated by single spaces. The last line contains a sequence of
N region numbers, each of which is 1, 2, 3 or 4.
OUTPUT
Your program is to write to standard output. The output consists of N lines, each
containing a pair of code numbers each preceded by a sign character. These are codes
pairs that will direct the teleporter to the given region sequence. Note that there must be
no blank following a sign, but there must be a single space after the first code number.
If there are several solutions your program can output any one of them. If there are no
solutions your program should output the single integer 0.
EXAMPLE INPUTS AND OUTPUTS
Example 1: input output
4 +7 -1
7 5 6 1 3 2 4 8 -5 +2
4 1 2 1 -4 +3
+8 +6
Example 2: input output
4 +3 -2
2 5 4 1 7 8 6 3 -4 +5
4 2 2 1 -6 +1
+8 +7
SCORING
If your program outputs a correct answer for a test case within the time limit, then you
get full points for that test, and otherwise you get 0 points for the test case.
2010-02-19 13:05 Page 2 of 2
```
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