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https://www.coursehero.com/file/1654389/18-InstSolManual-PDF-Part13/ | 1,545,212,911,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376831933.96/warc/CC-MAIN-20181219090209-20181219112209-00430.warc.gz | 857,786,218 | 45,884 | PHYS
18_InstSolManual_PDF_Part13
# 18_InstSolManual_PDF_Part13 - Electric Potential and...
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18.54. Set Up: For two capacitors in parallel, For two capacitors in series, and For capacitors in parallel, the voltages are the same and the charges add. For capacitors in series, the charges are the same and the voltages add. Label the capacitors as shown in Figure 18.54a. Figure 18.54 Solve: The equivalent capacitance for and in series is This gives the circuit shown in Figure 18.54b. The equivalent capacitance for and in parallel is This gives the circuit shown in Figure 18.54c. The equivalent capacitance for and in series is This gives the circuit shown in Figure 18.54d. In Figure 18.54d, This is the charge on each capacitor in Figure 18.54c. and Note that Then in Figure 18.54b, In summary, and 18.55. Set Up: For capacitors in series the voltages add and the charges are the same; For capacitors in parallel the voltages are the same and the charges add; C 5 Q V . C eq 5 C 1 1 C 2 1 c 1 C eq 5 1 C 1 1 1 C 2 1 c V 4 5 16.8 V. V 3 5 11.2 V V 2 5 5.6 V, V 1 5 5.6 V, Q 4 5 67.2 m
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Jill Tulane University ‘16, Course Hero Intern | 591 | 2,199 | {"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-2018-51 | latest | en | 0.885042 |
https://stackoverflow.com/questions/11812384/how-can-you-orthogonalize-a-set-of-vectors/11812429 | 1,563,603,857,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526446.61/warc/CC-MAIN-20190720045157-20190720071157-00545.warc.gz | 543,720,571 | 28,230 | How can you orthogonalize a set of vectors
I have been reading "Mathematics for 3D Game Programming and Computer Graphics" and there is a chapter exercise (Chapter 2. Question 2) that despite rereading the chapter and researching, I can not seem to understand. How can I "Orthogonalize the following set of vectors"
e1 = ( sqrt(2)/2, sqrt(2)/2, 0 )
e2 = ( -1, 1, -1 )
e3 = ( 0, -2, -2 )
Also, what does it mean to "Orthogonalize a set of vectors"? | 134 | 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} | 3.078125 | 3 | CC-MAIN-2019-30 | latest | en | 0.912359 |
https://convert-dates.com/days-from/879/2024/08/02 | 1,723,667,354,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641121834.91/warc/CC-MAIN-20240814190250-20240814220250-00351.warc.gz | 141,335,033 | 4,373 | ## 879 Days From August 2, 2024
Want to figure out the date that is exactly eight hundred seventy nine days from Aug 2, 2024 without counting?
Your starting date is August 2, 2024 so that means that 879 days later would be December 29, 2026.
You can check this by using the date difference calculator to measure the number of days from Aug 2, 2024 to Dec 29, 2026.
December 2026
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December 29, 2026 is a Tuesday. It is the 363rd day of the year, and in the 53rd week of the year (assuming each week starts on a Sunday), or the 4th quarter of the year. There are 31 days in this month. 2026 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 12/29/2026, and almost everywhere else in the world it's 29/12/2026.
### What if you only counted weekdays?
In some cases, you might want to skip weekends and count only the weekdays. This could be useful if you know you have a deadline based on a certain number of business days. If you are trying to see what day falls on the exact date difference of 879 weekdays from Aug 2, 2024, you can count up each day skipping Saturdays and Sundays.
Start your calculation with Aug 2, 2024, which falls on a Friday. Counting forward, the next day would be a Monday.
To get exactly eight hundred seventy nine weekdays from Aug 2, 2024, you actually need to count 1231 total days (including weekend days). That means that 879 weekdays from Aug 2, 2024 would be December 16, 2027.
If you're counting business days, don't forget to adjust this date for any holidays.
December 2027
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December 16, 2027 is a Thursday. It is the 350th day of the year, and in the 350th week of the year (assuming each week starts on a Sunday), or the 4th quarter of the year. There are 31 days in this month. 2027 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 12/16/2027, and almost everywhere else in the world it's 16/12/2027.
### Enter the number of days and the exact date
Type in the number of days and the exact date to calculate from. If you want to find a previous date, you can enter a negative number to figure out the number of days before the specified date. | 933 | 2,716 | {"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 | 0.917063 |
https://it.mathworks.com/matlabcentral/cody/players/1879631-ryszard-maciol/solved | 1,603,650,065,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107889574.66/warc/CC-MAIN-20201025154704-20201025184704-00436.warc.gz | 371,387,478 | 21,510 | Cody
# Ryszard Maciol
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Created by: Sunu
#### Problem 346. Back to basics 4 - Search Path
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Tags pythagoras
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#### Problem 1974. Length of a short side
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#### Problem 2033. poll: would you like the regexp (?@cmd) functionality to be banned in Cody?
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#### Problem 110. Make an N-dimensional Multiplication Table
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#### Problem 716. Arrange vector in ascending order
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#### Problem 160. So many choices
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#### Problem 619. Multiply a column by a row
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1 – 50 of 235 | 1,074 | 3,891 | {"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-2020-45 | latest | en | 0.685211 |
https://www.brightstorm.com/math/geometry/basic-trigonometry/inverse-trig-functions/ | 1,398,402,037,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1398223210034.18/warc/CC-MAIN-20140423032010-00169-ip-10-147-4-33.ec2.internal.warc.gz | 1,029,158,060 | 17,025 | # Inverse Trigonometric Functions - Concept 38,587 views
Once we understand the trigonometric functions sine, cosine, and tangent, we are ready to learn how to use inverse trigonometric functions to find the measure of the angle the function represents. Inverse trigonometric functions, found on any standard scientific or graphing calculator, are a vital part of trigonometry and will be encountered often in Calculus.
In right triangles when we're talking about cosine, sine and tangent sometimes you're going to need to use what's known an inverse trig function. Let's look at what that means, if I asked you to find the measure of angle b, so that's the angle right up here. Well what we're going to do is we're going to say are we going to use cosine, sine or tangent? Relative to b I know the opposite side and the adjacent, opposite and adjacent is tangent, so I can say that the tangent of b is equal to the ratio of 14 to 12 the opposite to the adjacent.
But now we have a problem, how do I find out what b is? I know eventually I want to see b equals. Right now we have tangent of b if I went back to Algebra I would say divide both sides by tangent because tangent appears to be multiplying b but that would be incorrect. Trigonometry doesn't work exactly like Algebra, the way that we isolate b in this problem is by taking the inverse trig function. So since we have tangent of b we're going to say that if I took the inverse tangent of 14 twelves that's what b would be. To use b as many times as I can in one sentence, so the inverse tangent is looks like tangent to the negative 1. So in your calculator right above your tangent button is the inverse tangent.
And you're probably going to have to put something like second to get there, so in my calculator I'm going to type inverse tangent of this fraction 14 to 12 and I get 49.4. So b equals 49.4 degrees. Let's look at 2 other quick examples, let's say I told you sine of x is equal to 0.5 what is x? To solve for x we're going take the inverse sign of both sides of this equation. So on the other side I'm going to write sine inverse of 0.5 so the inverse sine of the sine of x is x and that's the reason why we use that inverse property. So I'm going to type in inverse sine of 0.5 and I get 30 degrees.
I still got one last one, if I said cosine of y is equal to 18 divided by 22 to solve for y we're going to take the inverse cosine. So we're going to take the inverse cosine of the cosine of y which will give us y and on the right side I'm going to have to take the inverse cosine of 18 20 seconds. So I'm going to say inverse cosine of 18 divided by 22 is 35.1. So whenever you need to solve for an angle when you're talking about a trig function you can get at that variable by taking the inverse function of whatever you're talking about sine, cosine or tangent. | 672 | 2,843 | {"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} | 4.53125 | 5 | CC-MAIN-2014-15 | latest | en | 0.959436 |
https://3dprint.com/86252/3d-printed-blow-powerred-light/ | 1,721,611,507,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517805.92/warc/CC-MAIN-20240722003438-20240722033438-00558.warc.gz | 64,232,369 | 67,637 | # South Korean Man 3D Prints a Blow-Powered Flashlight
Don’t you just hate it when the power goes out in your house, you race to the closet to grab your handy-dandy flashlight, go to turn it on but discover that the batteries have completely died? This has happened to me on more than one occasion, and surely I’m not the only one to fall victim to the infamous powerless flashlight.
When MakerBot launched their “Catch The Wind Challenge” back in June, they had hoped to encourage engineers and designers from around the world to come up with some unique 3D printed creations that would “harness the power of the wind”. When one South Korean man, named Tim Cho heard of the challenge, he began thinking about how he could modify a flashlight design he had been working on, in order to qualify to win the First Place prize of a MakerBot Replicator 3D Printer.
“I was designing and printing a hand-winding crank LED Flashlight like Dynamo,” Cho tells 3DPrint.com. “I wanted to create something for the #CatchTheWind Thingiverse Challenge, but there were already many wind-powered flashlights. So I thought I could use ‘blow power’ instead of wind power, kind of like a whistle.”
So Cho began modeling the 3D printed components for his unique flashlight using Autodesk Inventor. He created 15 individual 3D printable parts which each took approximately 30 minutes to print out at a layer height of 0.2mm and an infill density of 25%. After spending about 7:30 hours printing out all of the parts, he then assembled them, along with some additional electronic components which included: coils, magnets, a bridge rectifier, an LED, some wires, and more.
As for how this battery-less flashlight works, Cho explains:
“When you strongly blow in through the mouthpiece, the vertical axis turbine and the magnets of the generator are rotating together. The shafts of the turbine and generator are directly connected. Then the coils of the generator makes AC (alternating current). The bridge rectifier circuit converts an AC input to a DC output. By blowing, the voltage generated creates about DC 0.5~1 volts. To light up an LED though, you need about DC 3 volts. For boosting voltage, I used a Joule thief circuit.”
Cho, who is currently unemployed, says that he is still learning about what his 3D printer can and can not do, and has been running into a lot of bumps along the way. If there is one downside to using this flashlight, he says that it is the fact that after a while, you get a little lightheaded from blowing so much. Currently the design files for this unique flashlight can be downloaded for free from Thingiverse. Check out the video of the flashlight in action below. | 588 | 2,686 | {"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} | 2.609375 | 3 | CC-MAIN-2024-30 | latest | en | 0.951185 |
http://stackoverflow.com/questions/1261119/how-to-implement-dead-reckoning-when-turning-is-involved | 1,394,503,325,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394011085177/warc/CC-MAIN-20140305091805-00037-ip-10-183-142-35.ec2.internal.warc.gz | 174,608,522 | 15,628 | # How to implement dead reckoning when turning is involved?
"Dead reckoning is the process of estimating one's current position based upon a previously determined position and advancing that position based upon known or estimated speeds over elapsed time, and course." (Wikipedia)
I'm currently implementing a simple server that makes use of dead reckoning optimization, which minimizes the updates required by making logical assumptions on both the clients and the server.
The objects controlled by users can be said to be turning, or not turning. This presents an issue with dead reckoning (the way I see it.)
For example, say you have point A in time defined by [position, velocity, turning: left/right/no]. Now you want point B after t amount of time. When not turning, the new position is easy to extrapolate. The resulting direction is also easy to extrapolate. But what about when these two factors are combined? The direction of the velocity will be changing along a curve as the object is turning over t amount of time.
Should I perhaps go with another solution (such as making the client send an update for every new direction rather than just telling the server "I'm turning left now")?
This is in a 2D space, by the way, for the sake of simplicity.
-
For simplicity let's say that your vehicles have a turning radius r that's independant of speed. So to compute the new position given the initial coords and the time:
• compute the distance (that's velocity * time)
• compute how much you turned (that's distance / (2*pi*r))
• add that arc to the original position.
The last steps needs elaboration.
Given the angle a computed in step 2, if you started at (0,0) with a due north heading (i.e. pi/2 radians) and are turning left then your new positions is: (r*cos(a)-1, r*sin(a)).
If your original heading was different, say it was "b", then simply rotate the new position accordingly, i.e. multiply by this rotation matrix:
`````` [ cos b , -sin b ]
[ sin(b), cos(b) ]
``````
Finally, add the initial position and you're done. Now you only need to send an update if you change the velocity or turning direction.
- | 475 | 2,140 | {"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-2014-10 | latest | en | 0.934514 |
https://bcxiaobai.eu.org/post/15998.html | 1,713,315,005,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817112.71/warc/CC-MAIN-20240416222403-20240417012403-00655.warc.gz | 105,207,079 | 23,988 | # kaggle学习笔记-otto-baseline6-使用 RAPIDS TSNE 和项目矩阵分解可视化用户行为
## 数据处理
``````import cudf
print('RAPIDS cuDF version',cudf.__version__)
train_pairs = cudf.concat([train, test])[['session', 'aid']]
del train, test
train_pairs['aid_next'] = train_pairs.groupby('session').aid.shift(-1)
train_pairs = train_pairs[['aid', 'aid_next']].dropna().reset_index(drop=True)
cardinality_aids = max(train_pairs['aid'].max(), train_pairs['aid_next'].max())
print('Cardinality of items is',cardinality_aids)
``````
## 安装 Merlin 下载器
``````!pip install merlin-dataloader==0.0.2
``````
``````from merlin.loader.torch import Loader
train_pairs.to_pandas().to_parquet('train_pairs.parquet') # TRAIN WITH ALL DATA
train_pairs[-10_000_000:].to_pandas().to_parquet('valid_pairs.parquet')
from merlin.io import Dataset
train_ds = Dataset('train_pairs.parquet')
``````
## 使用 PyTorch 矩阵分解模型学习项目嵌入
``````import torch
from torch import nn
class MatrixFactorization(nn.Module):
def __init__(self, n_aids, n_factors):
super().__init__()
self.aid_factors = nn.Embedding(n_aids, n_factors, sparse=True)
def forward(self, aid1, aid2):
aid1 = self.aid_factors(aid1)
aid2 = self.aid_factors(aid2)
return (aid1 * aid2).sum(dim=1)
class AverageMeter(object):
"""Computes and stores the average and current value"""
def __init__(self, name, fmt=':f'):
self.name = name
self.fmt = fmt
self.reset()
def reset(self):
self.val = 0
self.avg = 0
self.sum = 0
self.count = 0
def update(self, val, n=1):
self.val = val
self.sum += val * n
self.count += n
self.avg = self.sum / self.count
def __str__(self):
fmtstr = '{name} {val' + self.fmt + '} ({avg' + self.fmt + '})'
return fmtstr.format(**self.__dict__)
valid_ds = Dataset('valid_pairs.parquet')
``````
``````from torch.optim import SparseAdam
num_epochs = 10
lr=0.1
model = MatrixFactorization(cardinality_aids+1, 32)
criterion = nn.BCEWithLogitsLoss()
model.to('cuda')
for epoch in range(num_epochs):
for batch, _ in train_dl_merlin:
model.train()
losses = AverageMeter('Loss', ':.4e')
aid1, aid2 = batch['aid'], batch['aid_next']
aid1 = aid1.to('cuda')
aid2 = aid2.to('cuda')
output_pos = model(aid1, aid2)
output_neg = model(aid1, aid2[torch.randperm(aid2.shape[0])])
output = torch.cat([output_pos, output_neg])
targets = torch.cat([torch.ones_like(output_pos), torch.zeros_like(output_pos)])
loss = criterion(output, targets)
losses.update(loss.item())
loss.backward()
optimizer.step()
model.eval()
accuracy = AverageMeter('accuracy')
for batch, _ in valid_dl_merlin:
aid1, aid2 = batch['aid'], batch['aid_next']
output_pos = model(aid1, aid2)
output_neg = model(aid1, aid2[torch.randperm(aid2.shape[0])])
accuracy_batch = torch.cat([output_pos.sigmoid() > 0.5, output_neg.sigmoid() < 0.5]).float().mean()
accuracy.update(accuracy_batch, aid1.shape[0])
print(f'{epoch+1:02d}: * TrainLoss {losses.avg:.3f} * Accuracy {accuracy.avg:.3f}')
``````
## 提取项目嵌入
``````# EXTRACT EMBEDDINGS FROM MODEL EMBEDDING TABLE
embeddings = model.aid_factors.weight.detach().cpu().numpy()
print('Item Matrix Factorization embeddings have shape',embeddings.shape)
``````
## 使用 RAPIDS TSNE 可视化用户行为
``````# IMPORT RAPIDS TSNE
from cuml import UMAP, TSNE, PCA
import matplotlib.pyplot as plt, numpy as np
import matplotlib.patches as mpatches, cuml
print('RAPIDS cuML version',cuml.__version__)
# FIT TRANSFORM TSNE
em_2d = TSNE(n_components=2).fit_transform(embeddings)
print('TSNE embeddings have shape',em_2d.shape)
``````
``````# LOAD TEST DATA
tmp = test.groupby('session').aid.agg('count').rename('n')
test = test.merge(tmp, left_on='session', right_index=True, how='left')
active_users = test.loc[test.n>20,'session'].unique().to_array()
test = test.sort_values(['session','ts'])
print('Test data shape:', test.shape )
``````
## 使用 TSNE 项目嵌入显示用户活动
x-y 平面表示不同的项目类别。如果用户呆在同一区域，那么他们正在购买类似的物品，例如服装部门。当用户的绘图从 x-y 平面的一个区域更改为另一个区域时，用户将更改为不同类别的项目，例如从服装购物转移到电子产品购物。我们观察不同类型的用户。一些用户浏览一个项目类别，而其他用户浏览各种项目类别。
``````# DISPLAY EDA FOR 50 USERS
for k in range(50):
# SELECT ONE USER WITH 20+ CLICKS
u = np.random.choice(active_users)
dff = test.loc[test.session==u].to_pandas().reset_index(drop=True)
tmp = dff.aid.values
clicks = test.loc[(test.session==u)&(test['type']==0)].to_pandas().aid.values
carts = test.loc[(test.session==u)&(test['type']==1)].to_pandas().aid.values
orders = test.loc[(test.session==u)&(test['type']==2)].to_pandas().aid.values
############
## PLOT HISTORY BY ITEM CATEGORY
############
# PLOT CLICKS, CARTS, ORDERS OVER TSNE ITEM EMBEDDING PLOT
plt.figure(figsize=(15,15))
plt.scatter(em_2d[::25,0],em_2d[::25,1],s=1,label='All 1.8M items!')
plt.plot(em_2d[tmp][:,0],em_2d[tmp][:,1],'-',color='orange')
plt.scatter(em_2d[tmp][:,0],em_2d[tmp][:,1],color='orange',s=25,label='Click')
plt.scatter(em_2d[carts][:,0],em_2d[carts][:,1],color='green',s=100,label='Cart')
plt.scatter(em_2d[orders][:,0],em_2d[orders][:,1],color='red',s=250,label='Order')
# PLOT NUMBERS OF ORDER VISITED
old_xy = []; pos = []
for i,(x,y) in enumerate(zip(em_2d[tmp][:,0],em_2d[tmp][:,1])):
new_location = True
for j in old_xy:
if (np.abs(x-j[0])<5) & (np.abs(y-j[1])<5):
new_location = False
if new_location:
plt.text(x,y,f'{i+1}',size=18)
old_xy.append( (x,y) ); pos.append(i)
# LABEL PLOT
plt.legend()
plt.title(f'Test User {u} - {len(clicks)} clicks, {len(carts)} carts, {len(orders)} orders:',size=18)
#plt.xlabel('Item category',size=16)
plt.ylabel('nnItem category',size=16)
plt.xticks([], [])
plt.yticks([], [])
plt.show()
############
## PLOT HISTORY BY DAY AND HOUR
############
mn = test.ts.min()
dff['day'] = (dff.ts - mn) // (60*60*24)
dff['hour'] = ((dff.ts - mn) % (60*60*24)) // (60*60)
plt.figure(figsize=(15,3))
xx = np.random.uniform(-0.2,0.2,len(dff))
yy = np.random.uniform(-0.5,0.5,len(dff))
plt.scatter(dff.day.values+xx, dff.hour.values+yy, s=25, color='orange')
cidx = dff.loc[dff['type']==1].index.values
oidx = dff.loc[dff['type']==2].index.values
plt.scatter(dff.day.values[cidx]+xx[cidx], dff.hour.values[cidx]+yy[cidx], s=50, color='green')
plt.scatter(dff.day.values[oidx]+xx[oidx], dff.hour.values[oidx]+yy[oidx], s=100, color='red')
old_xy = []
for i in range(len(dff)):
if 1: #i in pos:
x = dff.day.values[i]+xx[i]
y = dff.hour.values[i]+yy[i]
new_location = True
for j in old_xy:
if (np.abs(x-j[0])<0.5) & (np.abs(y-j[1])<4):
new_location = False
if new_location:
plt.text(x, y, f'{i+1}', size=18)
old_xy.append( (x,y) )
plt.ylim((-1,25))
plt.xlim((-1,7))
plt.ylabel('Hour of Day',size=16)
plt.xlabel('Day of Month',size=16)
plt.yticks([0,4,8,12,16,20,24],['12am','4am','8am','noon','4pm','8pm','12am'])
plt.xticks([0,1,2,3,4,5,6],['MonnAug 29th','TuenAug 30rd','WednAug 31st',
'ThrnSep 1st','FrinSep 2nd','SatnSep 3rd','SunnSep 4th'])
c1 = mpatches.Patch(color='orange', label='Click')
c2 = mpatches.Patch(color='green', label='Cart')
c3 = mpatches.Patch(color='red', label='Order')
plt.legend(handles=[c1,c2,c3])
plt.show()
print('nnnnnnn')
``````
THE END | 2,337 | 6,918 | {"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 | 0.257556 |
https://xisto.com/discuss/topic/27170-geometry-help-me-please/ | 1,653,755,389,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663016949.77/warc/CC-MAIN-20220528154416-20220528184416-00199.warc.gz | 1,236,603,661 | 22,099 | xisto Community
# -geometry Help Me Please :(
## Recommended Posts
Ok this is what i need to do:
Create 10 unique linear Equations
2 must be parallel
2 must be perpendicular
1 must be horizontal
1 must be vertical
* each one label m and b ( i think for this i label the number on top of it :/
i think im failing by one point *studyin*
Notice from cmatcmextra:
Edited as requested
Edited by cmatcmextra (see edit history)
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So what exactly do you need to do? Draw lines that fit the description?
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Create 10 unique linear Equations2 must be parallel2 must be perpendicular1 must be horizontal1 must be verticalWait you havent described it properly however you could go and ask your teachers at skol or an online maths helper...
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its like an example for parallel:y=4x-2y=4x=2graph them ill be parallel i cant seem to get a vertical one
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well vertical would have to consist of either 2 postive numbers or two negative numbers so that they follow y line.i think never really got into geometry.
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geometry is the worse..i hated that the whole time i took it...good luck
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I think I can help you in this type of stuff.My math is a bit rusty. (I did nothing over the vacation )Perpendiculary=5x+5y=-5x+5Verticalx=C (C as in constant.)X=5, 5, 5, 5, 5Y=1, 2, 3, 4, 5...etc...Vertical lines are always known as undefined. I forgot why, but I do know that as a fact.***Ask me if you need more help on the other lines. I'm too lazy to think up equations right now.
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Gyad, I think you're wrong about perpindicular lines... In perpindicular lines the slopes have to be negative reciprocals of each other. For example: a line with a slope of 2(+2/1) is perpindicular to a line with a slope of -1/2.Here's an example set of perpindicular line equations:y = m*x + b & y = (-1/m)*x + bSoooo if m(slope) equalled 5, the equations would be:y=5*x+b & y=-1/5*x+b*=timesI hope that helps... That stuff gets a lot easier once you're used to it, so don't worry.
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vertical would be something like this:y=x/0+3yeah, impossible huh? that should be right although you won't get it in a graphing calculator because anything over 0 is undefined. It's so sad that after all these years the mathematicians still haven't found something else to describe it.
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Vertical lines are always known as undefined. I forgot why, but I do know that as a fact.
Vertical LInes are known as undefined because you determin the slop of a line as rise over run, that means that you pick two points on the line and count from one to the other as rise from the lower and then run oover to the points so you have a ratio rise/run and because you don't run in a vertical line you have a number divided by zero.
That being said it is imposable to have linear equation that will result in a vertical, however you can still make an equation that when graphed will result in a vertical line, basiicaly x=anything. Because this line will be graphed on the X axis without any Y, so for a horizontal line you want to draw a line that intercepts Y but not X so the equation would be Y=anything.
Now as for labling m and b, a general linear equation is written as Y=mX+b what that means is that the m value is the slope of the line (rist oever run) so If it is 1 then the line is a 45 degrees angle because for every unit the line goes over it also goes up.
I don't know where you're from but in Saskatchewan Canada we take this in grade 9, 10, 11, and 12 but the stuff you are focusing on appears to be from the Grade 12 Math A30 page and for more information I would recomend Mr. Hoffmans Math A30 Page, he wasn't my teacher but my teacher did think it was a good resource
http://forums.xisto.com/no_longer_exists/
Hope that helps
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thanks all of these replys help me alot. i hope i pass geometry.
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To anne :I might be wrong, I might not be wrong. I'm not sure at all. I haven't really officially studied slope yet. I've just mearly touched the subject in 8th grade. I'm not even taking geometry yet.To microsoft : (slightly off topic)Good Luck to you, math really isn't that hard if you think about it. Spanish is hard...;
## Create an account
Register a new account | 1,111 | 4,425 | {"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} | 3.609375 | 4 | CC-MAIN-2022-21 | latest | en | 0.919635 |
https://de.mathworks.com/matlabcentral/answers/45221-how-to-store-data-in-a-matrix | 1,656,927,050,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104364750.74/warc/CC-MAIN-20220704080332-20220704110332-00182.warc.gz | 246,198,048 | 25,959 | # how to store data in a matrix
36 views (last 30 days)
Lei on 3 Aug 2012
Hello everyone,
I have problem about how to store data in a matrix. Here is my code,aparently, theres problems.
a=5;
m=0;
M=[];
for i=1:72;
c1(i)=i;
c2(i)= c1(i)+1;
for j=2:5;
p1(i)=c1(i)+j;
if j==2
n=3:6; p2(i)=p1(i)+n;
elseif j==3
n=4:6; p2(i)=p1(i)+n;
elseif j==4
n=5:6; p2(i)=p1(i)+n;
elseif j==5
n=6; p2(i)=p1(i)+n;
end
if p2(i)<= 72
m=m+1;
M(m,:)=([m,c1(i),p1(i),p2(i),c2(i)]);
end
end
end
what i want to do here is:
c1=i,c2=c1+i,
if p1=c1+2, then p2=p1+3,p2=p1+4,p2=p1+5,p2=p1+6
if p1=c1+3, then p2=p1+4,p2=p1+5,p2=p1+6
if p1=c1+4, then p2=p1+5,p2=p1+6
if p1=c1+5, then p2=p1+6
and finally save data as [c1,c2,p1,p2] for each measurements
here is where i have problem:
if j==2
n=3:6;
p2(i)=p1(i)+n;
I dont know how could i achieve this, anyboday would like to give a hand? Thanks in advance
##### 2 CommentsShowHide 1 older comment
Lei on 5 Aug 2012
thx Oieg
John Petersen on 3 Aug 2012
Edited: John Petersen on 3 Aug 2012
Try this:
m=1;
for i=1:72;
for j=2:5;
p1 = i+j;
n=[j+1:6];
p2 = i + n;
for k=1:length(n);
M(m,:)=([m,i,i+1,p1,p2(k)]);
m=m+1;
end
end
end
##### 2 CommentsShowHide 1 older comment
John Petersen on 6 Aug 2012
I'm assuming you just want to rid yourself of the rows with p2>72. You could do it by adding this to the end of the code.
ind = find(M(:,5)<=72);
M = M(ind,:);
or by inserting a test on p2 before the for k loop.
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Translated by | 649 | 1,567 | {"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} | 3.625 | 4 | CC-MAIN-2022-27 | latest | en | 0.711181 |
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0 / 12
# Number 769871
seven hundred sixty nine thousand eight hundred seventy one
### Properties of the number 769871
Factorization 769871 Divisors 1, 769871 Count of divisors 2 Sum of divisors 769872 Previous integer 769870 Next integer 769872 Is prime? YES (61723rd prime) Previous prime 769837 Next prime 769903 769871st prime 11703539 Is a Fibonacci number? NO Is a Bell number? NO Is a Catalan number? NO Is a factorial? NO Is a regular number? NO Is a perfect number? NO Is a palindrome? NO Polygonal number (s < 11)? NO Binary 10111011111101001111 Octal 2737517 Duodecimal 31163b Hexadecimal bbf4f Square 592701356641 Square root 877.42293108854 Natural logarithm 13.553978247327 Decimal logarithm 5.8864179606509 Sine -0.98749781383685 Cosine 0.15763269859848 Tangent -6.2645493137955
Number 769871 is pronounced seven hundred sixty nine thousand eight hundred seventy one. Number 769871 is a prime number. The prime number before 769871 is 769837. The prime number after 769871 is 769903. Number 769871 has 2 divisors: 1, 769871. Sum of the divisors is 769872. Number 769871 is not a Fibonacci number. It is not a Bell number. Number 769871 is not a Catalan number. Number 769871 is not a regular number (Hamming number). It is a not factorial of any number. Number 769871 is a deficient number and therefore is not a perfect number. Binary numeral for number 769871 is 10111011111101001111. Octal numeral is 2737517. Duodecimal value is 31163b. Hexadecimal representation is bbf4f. Square of the number 769871 is 592701356641. Square root of the number 769871 is 877.42293108854. Natural logarithm of 769871 is 13.553978247327 Decimal logarithm of the number 769871 is 5.8864179606509 Sine of 769871 is -0.98749781383685. Cosine of the number 769871 is 0.15763269859848. Tangent of the number 769871 is -6.2645493137955
### Number properties
0 / 12
Examples: 3628800, 9876543211, 12586269025
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© 2024 numberempire.com All rights reserved | 660 | 2,300 | {"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-2024-33 | latest | en | 0.659737 |
https://eurekamathanswers.com/theoretical-probability/ | 1,642,839,586,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303779.65/warc/CC-MAIN-20220122073422-20220122103422-00279.warc.gz | 295,137,009 | 10,967 | The chance of occurrence of an event is called probability. The probability values are in the range between 0 and 1. Probability can be studied using two approaches theoretical probability and experimental probability. We will discuss the definition, examples, formula and solved problems of theoretical probability in the following sections.
## Theoretical Probability Definition
Theoretical probability defines the theory behind probability. It is not mandatory to conduct an experiment to find the probability of an event with the theoretical probability. Theoretical probability is the ratio of the number of favourable outcomes to the total number of possible outcomes. It is also known as the classical probability or priori probability.
The theoretical probability formula is given as:
Probability of an event = P(E) = $$\frac { Number of favourable outcomes }{ Total number of possible outcomes }$$
Examples:
1. Find the probability when a die is rolled, it rolls a 5.
The number of possible outcomes = 6
Number of favourable outcomes = Number of times a fair die can roll 4 in a single throw = 1
P(E) = $$\frac { Number of favourable outcomes }{ Total number of possible outcomes }$$
P(a fair die rolls 5 in a throw) = $$\frac { 1 }{ 6 }$$
2. A fair coin is tossed 10 times and the outcomes were recorded as Head = 4, tail = 6. Find the probability of the coin showing tail.
Number of times coin tossed = 10
Number of tails = 6
P(E) = $$\frac { Number of favourable outcomes }{ Total number of possible outcomes }$$
P(tail) = $$\frac { 6 }{ 10 }$$ = $$\frac { 3 }{ 5 }$$
Also, Check
### Difference Between Theoretical Probability & Experimental Probability
Theoretical Probability Experimental Probability
It is defined as the ratio of the number of favourable outcomes to the number of possible outcomes. It is defined as the ratio of the number of times an event occurs to the total number of trials.
It is also known as classical probability. It is also called as an empirical probability.
Formula is $$\frac { Number of favourable outcomes }{ Total number of possible outcomes }$$ Formula is $$\frac { Number of times event occurs }{ Total number of trials }$$
The trials are done only once. Here, the number of trials are extremely high.
### Questions on Theoretical Probability
Question 1:
Find the probability of getting a composite number in a throw of a die.
Solution:
Let E be the event of getting a composite number.
The total number of possible outcomes = 6 (Since any one of 1, 2, 3, 4, 5, 6 can come).
A number of favourable outcomes for the event E = 2 (Since any one of 4, 6 is a composite number).
So, P(E) = $$\frac { Number of favourable outcomes }{ Total number of possible outcomes }$$
= $$\frac { 2 }{ 6 }$$ = $$\frac { 1 }{ 3 }$$
Question 2:
In a society, 1000 families with 2 children were selected and the following data were recorded.
Number of boys in a family 0 1 2 Number of families 392 333 275
Find the probability of a family, having:
(i) 1 boy
(ii) 2 boys
(iii) no boy.
Solution:
Total number of families = 333 + 392 + 275 = 1000
Number of families having 0 boy = 392
Number of families having 1 boy = 333
Number of families having 2 boys = 275
(i) Probability of having ‘1 boy’
P(X) = $$\frac { Number of favourable outcomes }{ Total number of possible outcomes }$$
= $$\frac { 333 }{ 1000 }$$
(ii) Probability of having ‘2 boys’
P(Y) = $$\frac { Number of favourable outcomes }{ Total number of possible outcomes }$$
= $$\frac { 275 }{ 1000 }$$
(iii) Probability of having ‘0 boys’
P(Z) = $$\frac { Number of favourable outcomes }{ Total number of possible outcomes }$$
= $$\frac { 392 }{ 1000 }$$
### FAQ’s on Classical Probability
1. How to calculate theoretical probability?
The formula to calculate the theoretical probability is $$\frac { Number of favourable outcomes }{ Total number of possible outcomes }$$. Substitute the values in the formula and find the ratio to get the theoretical probability answer.
2. What are some examples of theoretical probability?
Some of the examples of theoretical probability are rolling a die, flipping a coin, and checking the outcomes. Another example is picking a card from the pack of 52 cards.
3. What does theoretical probability mean?
Theoretical probability calculates the probability of happening, not actually going out and experimenting. | 1,041 | 4,361 | {"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} | 4.84375 | 5 | CC-MAIN-2022-05 | latest | en | 0.93707 |
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Microsoft Employee
## How to deal with a "(BLANK)" value in a card visualization
My dashboard showed "(BLANK)" as the value in a card visualization - when I really wanted to see "0".
Is there a best practice to deal with this, or a simple way to adjust this?
Here is my current solution.
Note: This does not seem to be a data type issue. The card is calling on a measure that is calculated as follows:
Total CRs that have moved to P = CALCULATE([Total CRs - for calculations], 'Release_CRs_Systems (CRs Moved Only)'[Friendly System Name] = "P")
2) The data type for "Total CRs - for caculations" is "Whole Number"
3) [Total CRs - for calculations] is a SUM function - type "Whole Number"
CURRENT SOLUTION:
Create a new measure that evaluates [Total CRs that have moved to P] and uses IF(ISBLANK)
DISPLAY CALC FOR CR IN P =
IF( ISBLANK('Release_CRs_Systems (CRs Moved Only)'[Total CRs that have moved to P])
, 0
, ( 'Release_CRs_Systems (CRs Moved Only)'[Total CRs that have moved to P]))
*****
Can anyone provide a more elegant solution?
45 REPLIES 45
Anonymous
Not applicable
Create a simple Measure.
A measure Is just like a column in the data which you can add logic expression.
My problem was counting rows and then the output was blank.
I solved by;
Display Measure = IF(ISBLANK(COUNT(Issues_List[Title])),0,COUNT(Issues_List[Title]))
This measure is checking if COUNT(Issues_List[Title]) = Blank, give output 0
Otherwise returns Counts.
I have done it this way as well.
Anonymous
Not applicable
Hi,
You can hide the blank value using a Conditional Formatting rule on the Callout Value. The rule is 'if blank' then select the same color as the card or canvas background.
I have created a video tutorial on how to create the conditional rule: https://youtu.be/Xsmbfpa4oCA
This only works if you have a singular color background/fill. Is there a solution to this if you have a gradient fill background? I'd like to simply display nothing for the value instead of the text (Blank) in the card.
Regular Visitor
Hi, I cant say its a great solution, but I found that applying a conditional formatting to the 'Callout value' solved it for me.
for a card Visual: --> Format Visual --> Callout Value --> colour (fx)
Then set Format style to Rules, field based on whatever column you are counting from, summarization: Count--> and set 2 rules:
1: If Value Is blank --> set the text colour to the same as the background it is on
2: if value > 0 Number and < 10000000 Number --> then colour to whatever colour you want the text when it isnt 0/blank
This is how I do it as well. I have found it to be the least complicated and straight forward approach. I agree it seems as thought this could just be a setting in Power BI. Also, I have only applied the first (Is Blank) rule above without the 2nd rule and it seems to work just fine.
Regular Visitor
I agree, this has saved me from recreating all my output values as measures which I don't think should be necessary.
I applied the second rule in my case as without it the text colour default to Black (for me it did at least), where as I needed mine to be white text.
Thanks, It works. But I have to say this is so stupid. Why the hell is power bi missing so many features.
Anonymous
Not applicable
Generally i agree with you. It looks really awkward
For the card(s) with (blank), go to Data label
1. Right click on color of the Data label
2. Choose conditional formatting
3. on Format By drop down choose Rules
4. on Base on field choose the field that is used for the card in question
5. Choose appropriate Summarization you want (I chose Sum)
6. Under Rules select drop down if value and choose is blank
here is the trick. my cards are for heat map with 5 by 5 cards for a total of 25
quadrant 1x1 is green, so for that card my choise is
is value = is blank so I choose color gree which hides the Data Label of (Blank)
Anonymous
Not applicable
Great idea! Thank you!
New Member
This worked perfectly for me.
Anonymous
Not applicable
If I use the card visual in relation to 'gender', I am having issues solving the blank dillemma because there is no zero value in my data, its either male or female. On my dashboard, when I click the male icon the female icon will appear as blank and vice versa. Ideally when I click on the male icon I would like the female icon to appear as "N/A" or "Zero" is anyone able to help me with this issue I am having, any help is greatly appreciated 🙂
Anonymous
Not applicable
Similar issue here, dealing with non-numeric values.
It'd be nice if you could conditionally format the card, so if it have no data to present, I can conditonal the background color and text color as same, creating the illusion that nothing is presented on the card.
Anonymous
Not applicable
This generally works for me but didn't with my gauge visual. Might be user error.
Post Patron
You could go to edit queries and change blanks to 0 if that would not mess up any other formulas. It is edit queries and then go to the column and “replace values” leave the top blank as it is a blank field and then the replace with field you can put in 0.
Frequent Visitor
Hi even though it seems you have solved this issue , this might be helpful for you to understand the scenario
http://powerbidax.com/how-to-deal-and-remove-blank-and-nan-value-in-power-bi/
Has someone created a suggestion to add functionality to the fields to add a 0 if blank? Creating a measure for every card is too excessive and limits functionality......
Agree.
Many times I do counts and place these in a card. I do this not by creating a new column or measure but by counting some attribute in the data.
For a count the card should show zero rather than (Blank).
Or at least that option should be available without having to create a new column or measure.
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Top Solution Authors
Top Kudoed Authors | 1,555 | 6,638 | {"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-2024-38 | latest | en | 0.906374 |
http://smlnj-gforge.cs.uchicago.edu/scm/viewvc.php/branches/charisee/ertest/vispaper/iso3d/cniso3d.diderot?view=markup&root=diderot&sortby=author&pathrev=2944 | 1,619,140,227,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039563095.86/warc/CC-MAIN-20210422221531-20210423011531-00039.warc.gz | 91,293,222 | 4,676 | Home My Page Projects Code Snippets Project Openings diderot
# SCM Repository
[diderot] View of /branches/charisee/ertest/vispaper/iso3d/cniso3d.diderot
[diderot] / branches / charisee / ertest / vispaper / iso3d / cniso3d.diderot
# View of /branches/charisee/ertest/vispaper/iso3d/cniso3d.diderot
Wed Mar 4 18:20:55 2015 UTC (6 years, 1 month ago) by jhr
File size: 1390 byte(s)
``` changed field continuity to avoid type errors
```
```/*
../../../bin/diderotc cniso3d.diderot \
&& ./cniso3d \
&& unu crop -i cniso3d.txt -min 0 0 -max 1 M | \
unu jhisto -b 600 600 -min -1.2 -1.2 -max 1.2 1.2 -t float | \
unu 1op log1p | \
unu quantize -b 8 -o cniso3d.png \
&& open cniso3d.png
*/
//field#3(3)[3] G = ∇V;
field#2(3)[3] G = ∇V;
field#2(3)[3,3] H = ∇⊗∇V;
field#2(3)[] F = G•H•G/|G|;
int grid = 30;
int stepsMax = 10;
real epsilon = 0.0001;
strand RootFind(real xi,real xj, real xk) {
output vec3 x = [xi,xj,xk];
int steps = 0;
update {
// Stop if we're no longer inside or taken too many steps.
if (!inside(x, V) || steps > stepsMax)
die;
// subsequent expressions are undefined if |∇F| is zero
if (|∇F(x)| == 0.0)
die;
// ALSO, for canny, bail if gradient magnitude is too low
if (|∇V(x)| < 5)
die;
// the Newton-Raphson step
vec3 delta = normalize(∇F(x)) * F(x)/|∇F(x)|;
// we've converged if the change is small enough
if (|delta| < epsilon)
stabilize;
x -= delta;
steps += 1;
}
}
initially { RootFind(lerp(-2, 2, -0.5, ui, grid-0.5),
lerp(-2, 2, -0.5, vi, grid-0.5),
lerp(-2, 2, -0.5, wi, grid-0.5))
| wi in 0..(grid-1), vi in 0..(grid-1), ui in 0..(grid-1) };
``` | 646 | 1,580 | {"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-2021-17 | latest | en | 0.454395 |
https://edutimes.org/en/charles-law-definition-580951174/ | 1,679,683,105,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945288.47/warc/CC-MAIN-20230324180032-20230324210032-00290.warc.gz | 290,942,107 | 55,701 | HomeenDefinition of Charles's Law
Definition of Charles’s Law
Charles’ law is an empirical law, that is, based on experimental observations, which establishes the relationship between the volume and the temperature of a gas when the pressure and mass or number of moles are constant. The first to enunciate it was the French physicist Jacques Alexandre César Charles, at the end of the 18th century. According to this law, the volume of a fixed sample of a gas held at constant pressure is directly proportional to the absolute temperature . In other words:
This law states that if a gas is doubled in absolute temperature, its volume will also double. In fact, if the temperature is multiplied by any factor, the volume will also be multiplied by the same factor, as long as the amount of gas and its pressure are kept constant.
Charles’s law in equation form
Like any law of proportionality, the above relationship can be rewritten in the form of an equation simply by introducing a suitable constant of proportionality. That is to say:
where K is a constant of proportionality that depends on the amount of gas and its pressure.
As can be seen, this equation has the form of an increasing linear function with slope K. It is observed experimentally that this slope increases with the number of moles of the gas and decreases with the pressure. In addition, all the lines that are built at different values of P and n, when extrapolated to a volume of zero, intersect the temperature axis at -273.15 °C, which corresponds to absolute zero. This behavior is shown below:
Changes of state and Charles’s law
Charles’ law can be rearranged by dividing both sides of the equation by the temperature, in which case the right-hand side will just be the constant of proportionality:
In other words, Charles’s law predicts that if the pressure and the number of moles are held constant, the relationship between the volume and the absolute temperature will remain constant. This means that if we carry out a process in which a gas changes from an initial to a final state in an isobaric manner (at P = constant), the relationship between the initial volume and the temperature will be equal to the relationship between the volume and the final temperature, that is:
This equation can be used to determine both the volume and the initial or final temperature, when the other three variables are already known.
Examples of the application of Charles’s law
Below are two examples of typical gas-related problems that can be solved using Charles’s law.
Example 1: Doubling the volume
Determine the final temperature of an ideal gas that is initially at 25°C and that is heated until its volume increases to twice its initial value.
Solution
The data provided by the problem is:
T i = 25 °C
V f = 2. V i
The first thing we must do is transform the temperature to Kelvin, since Charles’s law relates volume to absolute temperature and the centigrade scale is a relative scale.
We can now apply Charles’s law to determine the final temperature. We don’t need to know the values of the volumes, just the relationship between them.
Therefore, the final temperature will be 596.30 K or 323.15 °C.
Example 2: Lowering the temperature by half
If a helium sample was originally at -130.15°C, cooled to -180.15°C at constant pressure, and its final volume turned out to be 10.0 L, what was the initial volume?
Solution
In this case, we have the following data:
T i = -130.15 °C
T f = -180.15 °C
V f = 10.0L
As before, we must start by determining the absolute temperatures, and then apply Charles’s law.
Now we can apply Charles’s law:
The helium sample must have started from an initial volume of 15.38 L.
Charles’ law constant of proportionality and the ideal gas law
The ideal gas law represents an equation of state that completely describes an ideal gas when we know three of four state functions, namely pressure, temperature, volume, or number of moles. The equation is given by:
where R is the universal ideal gas constant, P is the pressure of the gas, and all other variables are the same as in Charles’ law. This equation can be rewritten as:
This law applies to ideal gases under any set of conditions, including those in which Charles’s law applies. Therefore, in the case that the pressure and the number of moles are kept constant, the above expression must be equivalent to Charles’s law. By comparison, we can see that the Charles’ law constant of proportionality is then equal to the factor in parentheses:
As can be seen, this expression for the constant of proportionality agrees with the experimental observation that it remains constant when n and P are constant; increases as n increases and decreases as P increases.
References
Britannica, The Editors of Encyclopaedia. (2020, February 18). Charles’s law | Definition & Facts . Encyclopedia Britannica. https://www.britannica.com/science/Charless-law
Britannica, The Editors of Encyclopaedia. (2021, November 8). Jacques-Charles | French physicist . Encyclopedia Britannica. https://www.britannica.com/biography/Jacques-Charles
Chang, R. (2021). Chemistry ( 11th ed.). MCGRAW HILL EDUCATION.
Gas Laws . (nd). Chem.FSU. https://www.chem.fsu.edu/chemlab/chm1045/gas_laws.html
Libretexts. (2020, August 22). Gas Laws: Overview . Chemistry LibreTexts. https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Gas_Laws/Gas_Laws%3A_Overview
Libretexts. (2021, April 30). 14.4: Charles’s Law . Chemistry LibreTexts. https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Introductory_Chemistry_(CK-12)/14%3A_The_Behavior_of_Gases/14.04%3A_Charles’s_Law | 1,340 | 5,819 | {"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-2023-14 | latest | en | 0.946543 |
http://isabelle.in.tum.de/repos/isabelle/diff/3d7f74fd9fd9/src/HOL/Library/State_Monad.thy | 1,579,792,284,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250610919.33/warc/CC-MAIN-20200123131001-20200123160001-00182.warc.gz | 80,263,924 | 2,175 | changeset 24253 3d7f74fd9fd9 parent 24224 a5c95bbeb31d child 25595 6c48275f9c76
1.1 --- a/src/HOL/Library/State_Monad.thy Mon Aug 13 21:22:41 2007 +0200
1.2 +++ b/src/HOL/Library/State_Monad.thy Mon Aug 13 21:22:42 2007 +0200
1.3 @@ -226,16 +226,24 @@
1.5 print_translation {*
1.6 let
1.7 - fun unfold_monad (Const (@{const_syntax mbind}, _) \$ f \$ Abs (abs as (_, ty, _))) =
1.8 + fun dest_abs_eta (Abs (abs as (_, ty, _))) =
1.9 + let
1.10 + val (v, t) = Syntax.variant_abs abs;
1.11 + in ((v, ty), t) end
1.12 + | dest_abs_eta t =
1.13 let
1.14 - val (v', g') = Syntax.variant_abs abs;
1.15 - in Const ("_mbind", dummyT) \$ Free (v', ty) \$ f \$ unfold_monad g' end
1.16 + val (v, t) = Syntax.variant_abs ("", dummyT, t \$ Bound 0);
1.17 + in ((v, dummyT), t) end
1.18 + fun unfold_monad (Const (@{const_syntax mbind}, _) \$ f \$ g) =
1.19 + let
1.20 + val ((v, ty), g') = dest_abs_eta g;
1.21 + in Const ("_mbind", dummyT) \$ Free (v, ty) \$ f \$ unfold_monad g' end
1.22 | unfold_monad (Const (@{const_syntax fcomp}, _) \$ f \$ g) =
1.23 Const ("_fcomp", dummyT) \$ f \$ unfold_monad g
1.24 - | unfold_monad (Const (@{const_syntax Let}, _) \$ f \$ Abs (abs as (_, ty, _))) =
1.25 + | unfold_monad (Const (@{const_syntax Let}, _) \$ f \$ g) =
1.26 let
1.27 - val (v', g') = Syntax.variant_abs abs;
1.28 - in Const ("_mbind", dummyT) \$ Free (v', ty) \$ f \$ unfold_monad g' end
1.29 + val ((v, ty), g') = dest_abs_eta g;
1.30 + in Const ("_let", dummyT) \$ Free (v, ty) \$ f \$ unfold_monad g' end
1.31 | unfold_monad (Const (@{const_syntax Pair}, _) \$ f) =
1.32 Const ("return", dummyT) \$ f
1.33 | unfold_monad f = f; | 671 | 1,800 | {"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} | 2.578125 | 3 | CC-MAIN-2020-05 | latest | en | 0.374563 |
http://www.ridingthebeast.com/numbers/nu13.php | 1,519,531,238,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816094.78/warc/CC-MAIN-20180225031153-20180225051153-00649.warc.gz | 520,858,028 | 18,509 | New! Calculate the number of your name with the Gematria Calculator
# Properties of the number 13
## Symbolism
• Number that cleans and purifies.
• The number 13 brings the test, the suffering and the death. It symbolizes the death to the matter or to oneself and the birth to the spirit: the passage on a higher level of existence.
• For the superstitious, this number brings the bad luck or the misfortune.
• For the cabalist, the number 13 is the meaning of the Snake, the dragon, Satan and the murderer. But it is also for Christians the representative number of the Virgin Mary, she whose mission is to crush the head of Satan.
• Number in relation with the cross and also to the family, since by reduction we obtain four: 1 + 3 = 4.
• It is the element of too, that which makes pass from a cycle to another with what this change implies of anxieties by the arrival of a new unknown cycle.
• Represents the eternal love illustrated by Jacob and his twelve son, Jesus-Christ and his twelve apostles.
• If we consider 13 as a wheel to 12 rays, that is to say as 12 units around a center, it is beneficial. If we take it as prime number, it is maleficent. It is especially maleficent when we are at table and when one believes in his power. But, as observed judiciously Grimod: "The number 13 is to be feared only insofar as there would be to eat only for twelve."
• If we represent 12 under the form of the Zodiac, 13=12+1 is the number of the eternal return. The 13th hour is also the first, just like the 25th or the 37th.
• According to R. Allendy, this number "represents a principle of activity 3 exerting in the Unit of a whole 10 which contains it and which makes it produce only one cycle of perpetually identical renewal (1 + 3 = 4). Or it is the mechanism of its organization which submits the Universe to a permanent mode of oscillations and which leads it to specialize in nature."
• The image of the creature kneeling ahead the Throne of God, the Holy Trinity Father, Son and Holy Spirit, is symbolized by the number 13: the 1 ahead 3.
• It is "the manifestation of the good or bad generating power", according to R. Schwaller.
• Number representing the Son of God, according to Abellio.
• The thirteenth mystery of the Tarot does not have a name. It marks the uncertainty, the hesitation, the fickleness or again a transformation, the end of something (the death) and a renewal, a rupture, that is to say a very important change.
## Bible
• The thirteen guests at the Last supper of the Christ.
• To chapter thirteen of the Gospel of Saint John, during of its Last supper with his disciples, Jesus declares that one of then will betray him. And after the announcement of the treason of Judas, Jesus prophesies the Peter's denial.
• The thirteenth chapter of the Revelation is reserved to the Antichrist and to the Beast.
• In 13th psalm, it is written: "The fool has said in his heart, There is no God".
• The mystery of the Seven Churches of the Revelation presents the winner of 13 rewards on the whole.
• The thirteen sons of David which were born when he was to Jerusalem. (1 Ch 3,5-9)
• It is the thirteenth day of the twelfth month, named Adar, that the people celebrated the victory of Judas Maccabaeus on the army of Nicanor to Adasa. (1 M 7,43-49)
• A decree of Jew extermination was decided by Haman during of a convocation addressed to scribes the thirteenth day of the first month. (Est 3,12)
## General
• The apostle James says the Minor directed the Church of Jerusalem during thirteen years.
• The thirteen ecclesiastical titles of the sacerdotal hierarchy of the Roman Church.
• The father of Job had thirteen children, according to visions' of Ann-Catherine Emmerick
• .
• The celebration of the Epiphany takes place the thirteenth day after the nativity of the Lord. The number 13 is called theophanicus for this reason.
• We find often the number thirteen associated with the Blessed Virgin Mary. Her Assumption occurred a Friday 13, in August, at 3 o'clock of the evening, according to visions of Mary Agreda. However, according to revelations of Mary Jane Even in 1994, the Virgin would have died on August 13 and would have resuscitated two days later, that is to say on August 15 to be then received Body and Soul in the Sky. Also, the first and the last appearance of the Virgin Mary in Fatima occurred respectively on May 13 and on October 13, 1917 and it is on July 13, 1917 that the children of Fatima had their vision of the Hell, showing thus that the thirteen is also closely in relation to the suffering and to the death. Still today, in the end of time, the Virgin appears to some seers and clairvoyant only the 13 of each month. The 13th day of the month in the Christendom would be thus particularly dedicated to the Virgin Mary. Moreover several received particular messages tend to show it as it is the case of following messages. In one of messages given to a privileged soul of Quebec, Our-Lord recommended that the 13 of each month is in the honor of his Mother and established in each family. In another message given by the Virgin Mary to Sister Lucy of Fatima the 1st May 1987 for the celebration of the 70th anniversary of the day when she appeared her to Fatima on May 13, 1917, She asked her to celebrate the 13 of each month by songs and the praises in spirit of repair and of expiation. Remind also that it is on May 13, 1981 that occurred the attack of the Pope John-Paul II, in the Saint-Peter place. What saved him from the death, it is that he turned the head to look an engraving of Our-Lady of Fatima at the same moment where the ball of the gunner passed. The France, devoted to Mary since Louis XIII, celebrated her by processions.
• According to visions of Maria Valtorta, during the descent of the Holy Spirit on apostles in prayer in the upper room (Cenacle), the Fire of the Holy Spirit, then in form of sphere very shining above the head of Mary, was shared in 13 very shining melodious flames to go down on the 12 apostles and the Virgin Mary.
• The value of the letter M of Mary, center of the Latin alphabet, is 13. Likewise 13 words compose the invocation to the Virgin on the Miraculous Medal: O Mary, conceive without sin, pray for we who have recourse to you.
• For believer, Friday 13 (of the month of nisan) is the day where the Christ is dead on the Cross; it is also a Friday 13, day placed under the sign of Venus, that Eve, tempted by the demon, made eat an apple to Adam, what entailed their expulsion of the terrestrial Paradise. God being indeed rested the seventh day of the Creation, first Saturday (Sabbath of the Jews) was this day. The next week there had therefore a Friday 13, day of the original sin, since they was redeemed an other Friday 13, that the death of the Christ. Particularly the Jewish Passover was the 14 of the month of nisan, and the Crucifixion took place the day before the Sabbath of Passover, therefore a Friday 13.
• In the visions of Maria Valtorta, Jesus speaks about the thirteen veins of the humanity by which are distributed the divine graces, first by Himself, and then by His 12 apostles, choose by Him to represent the whole humanity and in which all the humanity is gathered in His 12 apostles.
• Some authors tell that Jesus would be born in a year counting thirteen months. When Hebrews celebrated their first Passover, they abandoned the solar Egyptian calendar and adopted the lunar calendar. Thus, to maintain the correspondence between the month of Pescha and the beginning of the spring, they had to introduce all the three years approximately a thirteenth month into the year.
• On the Miraculous Medal, the M of Mary surmounts the Holy Cross of Christ, this one being associated to the number 13. And the letter M is also 13th letter of the alphabet.
• The 13th glorious mystery of the Saint Rosary refers directly to Pentecost.
• According to the Rule in the Order of the Saint Saviour, given by the Christ to saint Brigitte of Sweden (who lived from 1303 to 1373), in the monastery, thirteen priests have to sing daily the mass and the office of the ecclesiastical year.
• The Jewish Faith states thirteen articles that are called fundamental dogmas of the Judaism. They were formulated by Maimonid.
• The Witnesses of Jehovah have 13 fundamental doctrines, without speaking their internal rules.
1. The bible is the inspired word of Jehovah
2. Jehovah is the only true God
3. Jesus-Christ is the fathered unique son of God
4. Satan is the "chief of this world"
5. The Kingdom of Jehovah and the Christ will replace all human governments and will be the only government of all the humanity
6. Since 1914 we live the "time of the end"
7. A only path leads to God, all the other religions are not approved by Jehovah
8. The death is a consequence of the sin of Adam
9. Only 144000 go to the sky
10. The others will live eternally on the earth under the Kingdom of Jehovah
11. Respect the authority of this world, if they don't hinder their works
12. Refuse the blood: transfusion and foodstuff
13. Do the will of Jehovah
• The 13 is the number of skies for the Aztecs, and the hair of the Ancient of Days had thirteen buckles and his beard thirteen wicks. The ancient Mexico divided also the time in cycles of 52 years divided themselves in four periods of thirteen years. They had also a week of thirteen days. Thirteen was also, for the Aztecs, a time number, which represented the completion of the temporal series.
• The sacred cord of Druids has thirteen segments.
• The thirteen evil spirits according to the Cabal.
• The thirteenth in a group appears in antiquity as the most powerful and the most sublime. For example, Ulysses, the thirteenth of his group, escapes the appetite devouring of the Cyclops.
• For the superstitious ones, Fridays 13 are real nightmares: if that day there are 13 guests to the table, that precedes a death in the year (to be 13 to table would carry misfortune); to see a black cat Friday 13 carries misfortune; it is preferable not to exit that day, but in the opposite case, if one leaves by a door, it is always necessary to enter by the same door. The superstition of Friday 13 was also revivified in this era of the computer by some viruses introduced into the computer systems appearing only the day of Fridays 13. The fortune teller on the other hand predicts better the future on Friday 13. In France, as soon as there is a Friday 13 to the calendar, the National Lottery organizes a special drawing because some choose that day to bet the money. But for some, less credulous and more prone to the optimism, the number 13 is a lucky number on which they hope to make a success of what they do or to try their chance. For example, one of these succeeds to convince the British navy to dissipate fears of superstitious sailors who refused to go up a ship a Friday 13. It was indeed decided that they would proceed to the launching of a new ship on a Friday 13. This ship was baptized the H.M.S. Friday. It was controlled by the Friday Captain and took finally the sea a Friday 13. They never see again the ship and the crew.
• For the Mayas, the time is divided into several cycles beginning with the birth of Venus. And the cycle in which we are would have begun August 13, 3114 before J.-C. and would end on December 22, 2012. According to them, this date corresponds to the fifth and last cycle of the Earth, what would end to the destruction of the world.
• One tells that Philip II, king of Macedonian from 356 to 336 before J.-C., had the misfortune, during of a procession, to add his statue to those of the twelve major gods of the Greek mythology. He was assassinated few time after while he was going to combat against Persia.
• Number in relation with the moon: it covers on the average thirteen degrees per day and there are thirteen lunations in the year.
• The Creation would be divided into thirteen dimensions and levels: in the first dimension, there are 13 levels; in the second dimension, there are 12 levels; in the third dimension, there are 10 levels. And so on until the thirteenth dimension, which is the dimension of the portal, which is the Dimension of the Christ. And there, there is one level.
• The Sumerian used a zodiac including 13 constellations and 26 main stars.
• The shield of the old Slav divinity Prono was decorated with thirteen white points.
• The thirteen cords of the harp in Japan.
• The thirteen gates of the human body of the woman: 2 eyes, 2 ears, 2 nostrils, the mouth, 2 breasts, the navel, the anus, the urethra and the vagina.
• Some advance that the Universe created is governed by thirteen fundamental constants of the physics which are amongst other the speed of the light, constants of Planck, Boltzmann and Eddington, the load of the proton, the mass of the proton and the electron to the rest, etc. But this is far to make the unanimity for all seekers and scientists.
• In none of the most modern American buildings there is a 13th floor et bedroom number 13. They has also taken the practice to proscribe the number 13 of the numbering in some streets. No more 13 also for some airlines and to the automotive race departure.
• It had been formed in Bordeaux, at the 19th century, a society of 13. These jolly fellows organized banquets Friday of each week and committed all their businesses a Friday. The feast of the society was celebrated 13th Friday of each year. Before sitting down at table, they never omitted to reverse the salt boxes.
• To the historical viewpoint, there had: 13 primitive cantons in Switzerland, 13 States primitive in the USA, 13 Länder of the Federal German Republic.
• The Apollo-13 capsule of the NASA is the only Apollo not to have succeeded to land on the Moon. This mission, which proceeded from the 11 to April 17, 1970, was already at midway of the Moon when an explosion occurred in an oxygen tank and paralyzed a part of instruments. The capsule had then to return on the Earth as soon as possible.
• The relationship between the volume of the Earth and that of the Sun is approximately 13 times a power of 10, that is to say 1 / (13.01 x 10E5) to be precise.
• The card deck includes 13 hearts, 13 spades, 13 squares, 13 clovers.
• Weight of the soul in ounce.
• Birthday of marriage: lace weddings.
## Gematria
• Fischer calls thirteen the Jehovah factor since some words of the Hebraic language have been inspired by the essence of Jehovah whose numerical value is 26, that is to say equal to 2 x 13. The following words have all a numerical value (gematria in "N") whose common factor is 13: Moses, 351 = 27 x 13; Joseph, 156 = 12 x 13; Isaac, 208 = 16 x 13; Abraham, 104 = 8 x 13; Torah, 611 = 47 x 13; Jacob, 182 = 14 x 13; Israel, 546 = 42 x 13; Sinai, 130 = 10 x 13; month, 312 = 12 x 26 = 12 x 2 x 13; the divinity written in Hebrew, aleph, he, beth, he, 1+5+2+5 = 13. By using the gematria in "n", the following words have also common the factor 13: Elohim written in Hebrew, aleph, lamed, he, yod, mem final, 1+12+5+10+24 = 52 = 4 x 13; Amen written in Hebrew, aleph, mem, nun final, 1+13+25 = 39 = 3 x 13. It is also the numerical value of word "one", aleph, heth, daleth, 1+8+4=13, and love, aleph, he, beth, he, 1+5+2+5=13, in Hebrew; and the numerical value of the name of Jesus in Hebrew, yod, he, waw, shin, ayin, is 10+5+6+300+70 = 391, given 13 by reduction. In the Ostervald Bible, the Old and the New Testament count 1040 chapters for the whole of 66 books. And 1040 = 80x13. More, always in Ostervald Bible, the Old Testament counts 780 chapters for the whole of 39 books. And 780 = 60 x 13.
• The number 13 would be in a certain manner in correlation with our Earth planet which is called in Hebrew Eretz and that is written aleph, resh and tzade. The letter tzade put at the end of a word has not 90 but 900 as value. The numeral value gives therefore 1+200+900=1101. This number, interpreted in the mathematical base two (binary) is equivalent to the number 13 in the decimal base.
• The numerical value of the Hebrew word BEV, meaning chaos, gives 13.
## Occurrence
• The number 13 is used 28 times in the OT and it is never used in the NT.
• Numbers 17, 22 and 120000 are used 13 times in the Bible and the number 120 is used 13 times in the OT in its cardinal form.
• In the Gospel of Saint John, Jesus uses on the whole 13 comparisons or titles to designate who he is really:
1. I am the Bread (Jn 6,35 and 6,48 and 6,51)
2. I am the Light (Jn 8,12 and 9,5)
3. I am the Gate (Jn 10,7 and 10,9)
4. I am the Good Shepherd (Jn 10,11 and 10,14)
5. I am the Resurrection (Jn 11,25)
6. I am the Way (Jn 14,6)
7. I am the Truth (Jn 14,6)
8. I am the Life (Jn 14,6)
9. I am the Vine (Jn 15,5)
10. I am the King (Jn 18,37 and 19,21)
11. I am Son of God (Jn 10,36)
12. I am in the Father (Jn 14,10 and 14,11 and 14,20 and 17,8)
13. I am (Jn 8,24 and 8,28 and 8,58 and 13,19)
• The word star (or star) is used 13 times in the Koran. The words sickness, tear, dragon and the term "Son of God" are used 13 times in the NT. The words carnal and treason are used 13 times in the Bible.
• In the Bible, 26 numbers (written in their cardinal form) are multiple of 13.
• In the New Testament, 8 chapters possess 13 verses on the whole. And always in the New Testament, only 13 different numbers are equal or higher than 2000.
### Return
Last modification: December 19, 1998
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## 446 Comments for Number 13 Symbolism, 13 Meaning and Numerology
### Comment on Number 13 Symbolism, 13 Meaning and Numerology
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### Friday the 13th
by Cindy - 2/19/18 8:21 PM
I was born on July 13, 1973 @ 13:08 GMT after 13 ours of labor. 13 has been a lucky number for me.
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### I was born on Friday the 13th
by Mahogany - 5/08/09 1:35 AM
I am thankful to the author of the post above! I was born on Friday 13th and can not fathom equating such negativity and bad luck to my existence. I will say this though: there seems to be a HIGH, MIGHTY CALL on my life that I not only asked for or desired. However, along with that call has come a great deal of tests and challenges. My father is a pastor and I grew up in a very spiritual home. I have made some choices, good and bad, that have affected my life negatively and positively (like everyone else). But for some reason, God has decided to share things with me in visions and dreams. Tears are streaming down my face as I type right now because often times I feel somewhat isolated and called to something that most won't understand. God has also given me an incredibly soft heart, which causes me to sympathise and empathise on a very large scale. As I prepare for the LSAT in June, so that i can hopefully start Law School in the fall, I am reminded that I already have 3 degrees and attempted a Ph.D. ... more...
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### RE:I was born on Friday the 13th
by Yash Patel - 11/18/17 12:45 PM
I was born on 13 june 1997 which was Friday..... I think for those who have born on Friday 13th.... This day seem to be lucky for the rest of our lives... Also the number 13 seems to be very lucky. I would advice you to love yourself and those who love you..... People will respect you a lot if you would do that and will be there for you all the time. Time may become worse but in the end.... You will achieve a lot to get ahead and more than anything else is that you will win hearts more than you will win achievements. I...not only celebrate my birthday on 13 june but also on every Friday 13th. Consider it at it is your own day.... Make the most of it.
Regards
Yash Patel from India 🇮🇳
### RE:I was born on Friday the 13th
by Lancelot - 12/01/17 9:01 PM
I was bor 07/13/2001 which july friday the 13th
### RE:I was born on Friday the 13th
by my initials WFW - 12/08/17 3:09 AM
I was born on a Friday 13 in December 1991 metal ram fire archer next Friday the 13th I will turn the double digit number of 33 wow I am so cursed I totally understand
### RE:I was born on Friday the 13th
by my initials WFW - 12/08/17 3:21 AM
the univers has one awful plan for me I love everybody but my sole was meant for something terrible and the torcher of my life over the years my bad luck man the univers really wants to make me angry and evil I will make sure im not around on this planet when the universe calls upon me to do what my sole was placed her to do I love man kind I will never turn evil or go with the plan
### RE: I was born on Friday the 13th
by Anonymous - 12/17/17 1:45 AM
I don't feel these numbers are being directed in a negative way towards you personally, but they are numbers of reminders that exist, due to the evil that is influencing death on this earth....the last supper for example, Adam and Eve etc. Use it to know truths and dont associate yourself with negative thoughts....we have the power of choice in Christ Jesus.
### RE:I was born on Friday the 13th
by Odunjo Adeola from Nigeria - 12/31/17 7:11 AM
I was born on Friday the 13th
### RE:I was born on Friday the 13th
by Odunjo Adeola from Nigeria - 12/31/17 7:14 AM
I was born on Friday the 13th
13/may/2005
### RE:I was born on Friday the 13th
by Donnie Schumacher - 1/13/18 10:23 PM
I was born on Friday the 13th as well. 01/13/1989 I can relate and I also am soft hearted person!
### RE:I was born on Friday the 13th
by Unknown - 1/20/18 9:47 AM
I was born 2006 April 13 and I have had exteream bad luck and I have been wondering why me and this person has cleared it up for me and I pray for him/her
### RE:I was born on Friday the 13th
by Fernando - 2/08/18 6:46 PM
I was born on Good Friday 13th April 1990. reading all these is scary and confusing but I believe life has much more in store for me.
Read 42 more replies
### All the time I look at my watch, car clock, vcr clock, microwave clock, at all times of the day and night. The time shows 13. 6:13, 12:13, 3:13, 8:13. So on and so forth.
by Gus - 3/18/16 2:44 AM
Anyone please help! What does that mean? Since this been occurring, ai been having the worst of luck.
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### RE:All the time I look at my watch, car clock, vcr clock, microwave clock, at all times of the day and night. The time shows 13. 6:13, 12:13, 3:13, 8:13. So on and so forth.
by Jeremiah - 9/06/17 1:57 PM
I was born on Jan 13 the mother of my children April 13 a love I had a deep a connection or possible past life soul mate passed away June 13 everyday I see 13 on clocks marking my time of arrival and departure from work and home it seems to have a significant impact on my awareness of what's going on around me
### RE:All the time I look at my watch, car clock, vcr clock, microwave clock, at all times of the day and night. The time shows 13. 6:13, 12:13, 3:13, 8:13. So on and so forth.
by Anonymous - 1/15/18 1:00 AM
I usually don't get involved with these threads but i am confused to recent comment on putting the mirror in front of the clock... Is this to make it stop why would you want it to stop... You should feel blesses these synchronicity's are happening it is a good thing. Many are to clouded by the illusion of this world to notice. It means the ascended masters the angels your angels are trying to communicate with you. This will most likely evolve into other repeating numbers and signs... 13 having the energy of 1 and 3 is literally the beginning the vibration of 1 directly relates to new beginning and the start of a new push forward into something great a spiritual gift. 3 relates to spiritual growth literally through communication and higher levels of intuition and preparing you for some type of transformation and growth if your willing to see what is in front of you. There may be some initial chaos in your life but these are for karmic reasons and know that something much bigger is there to help it is... more...
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### Number 13 very prevalant in my life.
by Christine - 1/14/18 8:42 AM
I am a leap year baby (1960), and I gave birth to a New Years baby. The number 13 has been very prevalent in my life since 2001 when I started a job (for a major one of a kind company in the world) on Jan. 3,2001 (1-3-01) 13! That's when I kept realizing it. To this very day it appears to me on the news, tv shows, conversations, articles, on vehicle tags in front of me, receipts (gas receipt was \$13.13 once), the clock every time I look at it it's 1:13, 12:13, etc. EVERY DAY MULTIPLE TIMES SINCE 2001. I'm not superstitious and I believe there truly is something to this number appearing to me so much...I just wish I knew what it meant to me and my life. I am a very kind, loving, sympathetic/empathetic person, but always seem to get into bad situations with me family. I've always been the black sheep of 4 kids and have been treated indifferent to them. I've always felt like I'm here for a very specific purpose, but I've never been able to find my way...but I have always felt A PRESENCE. There have been... more...
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### The Number Thirteen
by Theophrastus - 2/02/14 7:37 PM
Those born on the 13th are under its sign, but it is a predisposition not a fate that we have no control over. Thirteen marks a point of inflection or change. Coming to a 13, we determine whether that change will be good or bad. If we are aware of this predisposition, we can control the number for good in our lives. Surrendering to evil with a 13 predisposition is very dangerous, but using its power creatively can lead to great success.
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### RE: The Number Thirteen
by Anonymous - 12/24/17 4:14 PM
this number is very lucky indeed!
### Number 13
by Nancy - 11/19/17 8:57 AM
I was born on Wednesday August 13, 1969. I've heard of Wednesday's child full of "Woe" what does this mean? And on top of that I was born in a hospital called Elm St how conquiencendal.
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### I Was Born 13/04/64
by Caroline Roberts - 8/17/17 12:57 AM
I was born on the 13th and I consider myself very lucky my best friend who we met when I was sitxteen was born 13/04/64 in the same hospital He very lucky 2 🍀 Zodiac twins ..our paths in life are very much the same..
### RE: I Was Born 13/04/64
by Anonymous - 11/15/17 9:56 AM
theres only 12 months tho
### Number 13
by Edward A Stack Sr - 11/05/17 7:55 AM
Number 13 is not unlucky. Number 13 is an even number totaling 4. The number 4 is a lucky number.
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### I was Born The 13th of October 1974
by April - 10/28/17 11:37 PM
1,4,26,11,13,6,r all my lucky numbers,in A - blood type,left Handed green eyed Empath,I'm very honored 2 b a part of Gods Plan😄
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### The number 13
by Anonymous - 10/23/17 8:13 PM
I was born on the 13th July when time was 13:00 on point. I'm not lying just to get attention. And both my first name and last name starts with the letter M wich is the 13th letter, is it normal?
### RE: The number 13
by Anonymous - 10/23/17 8:19 PM
And I forgot to say that my first name is Matteo wich means gift from god
### The number 13
by Nate - 9/09/17 1:42 AM
Infamously as well, Hitler escaped death when he left a speaking engagement 13 minutes early. Considering the amount of jewish tie-in with this number I find this relevant.
### RE: The number 13
by Anonymous - 10/13/17 9:32 AM
Will pray for understanding that surpasses all underwings!!? Through GOD
### Im born on the 13 of august 1993
by Berlynne rose mckelvey - 8/30/17 8:22 AM
Dose my birthday mean any thing
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### I was born on 13th July
by won - 7/24/17 9:52 PM
my old mather was born on July 13th ,I was born on July 13th too.my
home number is 13,and may apartament number is 13 too.why ?
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### I was born on Wednesday (3) July 13th
by Urban - 7/12/17 9:12 AM
I'm that person who at least check my watch many times a day and at least 3 times I 'be the same same number
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### Friday the 13th
by Saul - 6/25/17 1:00 PM
Every 6 years my birthday falls on Friday the 13th pretty fascinating
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### I was born the 3rd month, 13th day at 3 p.m
by Sa - 6/10/17 10:11 PM
I also turned 3 on Friday the 13th and 13 on Friday the 13th. Strange things seem to happen quite often. I feel cursed an blessed at the same time my entire life
### RE: I was born the 3rd month, 13th day at 3 p.m
by Ryan J P Baker - 6/13/17 5:26 PM
### Jericho
by Evangelist Alfred Mensah - 6/12/17 1:01 AM
When the Israelites got to the wall of Jericho, the most high requested them to go round the wall once everyday but on the seventh day He asked them to go round seven times which in total makes it 13 and indeed the great wall collapsed. So i see it that the number 13 is a time when God shows his supremacy over the devil and proves His existence to His children by crushing the head of the devil. Thank you all. On Facebook i am Evangelist Alfred Mensah
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### 13
by Robert - 1/07/17 12:16 PM
13 has been showing up since January 3rd
Too many to say.What does this mean?
### RE: 13
by Anonymous - 5/21/17 6:29 PM
more than 20 yrs ago I was offered the chance to leave the Earth at the 13th hour ... 'they' were offering that same chance to others and said that if we did not accept the invitation to leave, we would only remember the encounter as a dream ... perhaps you declined the offer and since we are getting closer to the 13th hour you are now remembering it
### 13 Popping up
by Eric Dominguez - 5/16/17 11:22 PM
I see 13 everywhere now since I took off for college. Videos games cards, random events, like driving to go to study at friends house that lived on 13th street before I found out what street it was on, plus a car was driving 13 mph on a 30 to that house which was strange. So one day I bought a lottery ticket from box 13, it was called "break the bank" (I didn't win), but I got 6/30/17 as my lucky numbers on game 2 and the top number was 13. There was 2 games and opposite of 13 was 31 on the opposite game (game 1) on the bottom. I see it everywhere and don't what omen it's trying to telling me
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### Interpretation of my dream!
by Talitha - 4/20/17 11:50 PM
No.13 repeatedly appeared in my dream with month of May marked in circle.
### RE: Interpretation of my dream!
by Anonymous - 5/13/17 7:20 PM
Are you ok?
### I am here now do not worry your hearts
by C E 2017 - 5/04/17 10:43 AM
Born on MMOTHERS DAY May 13th 1956 Planets, stars, Moon, Sun were all aligned perfectly. Astrological number is 4 Persue Copernicus's model for validation and celebration, The Son, King, savior of heaven and earth has returned with love, peace, hope, and move of all love, but do not be fooled or be a fool, Evil will destroyed. Ya-weh ****++++
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### Keep seeing number 13 or 1 & 3 over the last 6 months or so
by Benjamin - 4/20/17 10:32 AM
Was out walking today in the countryside and decided to take my phone out and looked at it and it said 13:13. But as the subject says I have been seeing quite a lot of the numbers 1 and 3 over the last 6 months. 911 is also another one.
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### 13 follow me
by SANTOSH ANAND - 3/21/17 2:37 AM
Last few years 13 follow me. i see it about 10 times in a day. its caused me worried and not feeling alone but a strange feeling.What the hell it means ????
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### RE: 13 follow me
by Anonymous - 4/14/17 9:39 PM
I see it everywhere all the time. It hunts me down. Its so bad I put tape over the digital number displays of my cell phone and clock radio. I basically refuse to look at clocks . If its not an actual 1&3 ,it will be a series of numbers that add up to 13. 9:13 is a terrible time to see bc it contains 13 twice (9+1'3=13) it goes on and on and on. 13th verses in the bible are often bad
### My number 13
by Magma - 4/06/17 7:31 AM
Ciao! I want to know why i saw 13 everywhere. In all the love story. In all of my biggest decision. Or when i m going to die: was the 13 of (i don t remember the month) when i did an accident with car, but i m alive. My grandmother gift me a number 13 (il pendolo di una collana) and the other grandma, is called lucia. Santa lucia is 13 december. The last time that i have noted 13 is this morning when i saw that there was an extintor number 13, when i was passed another time, this extintor didn t was located where i saw that. And a lot of other stories. People that are my friend saw this, and, sometimes, coincidences, start to saw 13 too! Why? I m a little scaried by this because it s a big big responsability.
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### death follows me
by Adam - 3/22/17 6:41 PM
i never knew about the # 13 until i found out about the connection with me and it . i was born feb 13 . i see it everywhere but the worst time was when i moved into my 1st apartment where a baby had died and the mother was arrested i soon found out my mailbox # was 13. i also have a friend whos grandpa was a medal of honor recipient born on feb 13 and a accident occured by exit 13 the car was on fire and burning with people still in it it was very bad and a pistol i brought is numbered 13 because those parts are the best and also my friends grandpa his old place was right in front of my ex's house where very bad things were happening.than another ex has a daughter born on feb 13 she lives close to my other unfortunate ex when you open her front door you see the street of my unfortunate ex right in front of her apartment.i am pretty amazed i see it on the news associating with death and i killed my first big game animal when i was thirteen shot it right in the heart.im gonna join the marine corp its been a... more... | 8,960 | 33,815 | {"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} | 3.296875 | 3 | CC-MAIN-2018-09 | latest | en | 0.952629 |
https://www.statisticssolutions.com/sample-size-calculation-2/ | 1,726,489,933,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.32/warc/CC-MAIN-20240916112213-20240916142213-00894.warc.gz | 919,166,397 | 25,227 | Sample Size Calculation
Sample Size
Sample size calculation plays a very important role in statistical analysis. Sample size calculation refers to how much data we need for particular research to make a correct decision. If we have more data, then our decision will be more accurate, and there will be less error of the parameter estimate. Some of the factors that affect the sample size calculation are the type of data, including the power of the sample size, the Technique used for analysis, the marginal error, the level of significance, the standard deviation, the missing value, etc. First of all, we should consider the type of data level and the measurement of a specific sample size.There are four types of data levels:
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1. nominal data;
2. ordinal data;
3. interval data; and
4. ratio data.
Nominal data is simply categorical data. Ordinal data is the data in which ranks are assigned to the data. Interval data is when an interval is given between the cases. Ratio data is metric or continuous data, on which we can perform all analysis which cannot be performed on nominal, ordinal, or interval data. In determining the sample size calculation, we should consider the level of significance or the level of alpha. For instance, two tailed test alpha level is 5%, which is equal to 1.96. When the sample size calculation is done, we should consider the marginal error as well. Marginal error is simply the error that a researcher is willing to accept for a particular sample size. For example, in continuous data t value of alpha for 5% is 1.96 and SD in population is 1.167, and marginal error is .21. Then we can calculate the sample size by using the following formula:
N=
N= sample size
t= level of alpha
S= standard deviation
D= marginal error
When the data is categorical, then we can use the probability of method instead of the standard deviation. For example, when we have two categories for samples, then we can use .5 probability of the first category and .5 probability for the second category. We can use the following formula in the case of categorical data:
N=
t= level of alpha
P= probability of event happening
Q= probability of second event happening
D= marginal error
These are some basic formulas for sample size calculation. But sample size calculation differs from technique to technique. For example, when we are comparing the means of two populations, if the sample size is less than 30, then we will use the t-test. If the sample size is greater than 30, then we will use the Z-test for comparing the two populations’ sample means. As a rule of thumb, in regression analysis, there should be 10 cases for each independent variable. For example, if we have two independent variables, then the minimum sample size should be 20 in order to reach a correct decision about the regression parameter. If the sample size is less than the given criteria, then the decision will not be correct. When data are categorical and the level of alpha decreases, then the sample size should be bigger for the same technique. If the population size is smaller, then we need a bigger sample size, and if the population is large, then we need a smaller sample size as compared to the small population. Sample size will differ with different margin error. Missing value also affects the sample size. When data has missing value, we need bigger sample sizes as compared to the non-missing value sample. In an analysis of variance test, we need to determine how many covariates we can use for a particular treatment variable. We can use sample size to determine the covariate. For example, with a sample size of 50, and a number of groups in 3 treatment factor, we can use only 3 covariates in an analysis of variance study. Variance also affects the sample size. When variance is more for the variable taken into study, then the sample size needs to be bigger to reach a correct decision about the parameter estimate. When we do not know the population standard deviation, then we can use the range to know the standard deviation. Sample size also depends on the power. With more power, we need a bigger sample size. | 991 | 4,709 | {"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} | 4.09375 | 4 | CC-MAIN-2024-38 | latest | en | 0.855862 |
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18.3 Ex 26 - 32
# 18.3 Ex 26 - 32 - 1260 C H A P T E R 18 F U N D A M E N TA...
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1260 C H A P T E R 18 FUNDAMENTAL THEOREMS OF VECTOR ANALYSIS (ET CHAPTER 17) 25. Prove that div ( f × ∇ g ) = 0. SOLUTION We compute the cross product: f × ∇ g = f x , f y , f z × g x , g y , g z = i j k f x f y f z g x g y g z = f y g z f z g y , f z g x f x g z , f x g y f y g x We now compute the divergence of this vector. Using the Product Rule for scalar functions and the equality of the mixed partials, we obtain div ( f × ∇ g ) = x ( f y g z f z g y ) + y ( f z g x f x g z ) + z ( f x g y f y g x ) = f yx g z + f y g zx f zx g y f z g yx + f zy g x + f z g xy f xy g z f x g zy + f xz g y + f x g yz f yz g x f y g xz = ( f yx f xy ) g z + ( g zx g xz ) f y + ( f xz f zx ) g y + ( g xy g yx ) f z + ( f zy f yz ) g x + ( g yz g zy ) f x = 0 In Exercises 26–28, let denote the Laplace operator defined by ϕ = ϕ 2 x 2 + ϕ 2 y 2 + ϕ 2 z 2 26. Prove the identity curl ( curl ( F )) = ∇ ( div ( F )) F where F denotes F 1 , F 2 , F 3 . SOLUTION We compute the left-hand side of the identity. We have curl ( F ) = i j k x y z F 1 F 2 F 3 = F 3 y F 2 z , F 1 z F 3 x , F 2 x F 1 y Hence, curl ( curl ( F )) = y F 2 x F 1 y z F 1 z F 3 x , z F 3 y F 2 z x F 2 x F 1 y , x F 1 z F 3 x y F 3 y F 2 z = 2 F 2 y x 2 F 1 y 2 2 F 1 z 2 + 2 F 3 z x , 2 F 3 z y 2 F 2 z 2 2 F 2 x 2 + 2 F 1 x y , 2 F 1 x z 2 F 3 x 2 2 F 3 y 2 + 2 F 2 y z (1) We now compute the right-hand side of the given identity: ( div ( F )) = ∇ F 1 x + F 2 y + F 3 z = x F 1 x + F 2 y + F 3 z , y F 1 x + F 2 y + F 3 z , z F 1 x + F 2 y + F 3 z = 2 F 1 x 2 + 2 F 2 x y + 2 F 3 x z , 2 F 1 y x + 2 F 2 y 2 + 2 F 3 y z , 2 F 1 z x + 2 F 2 z y + 2 F 3 z 2 Therefore, ( div ( F )) F = ∇ ( div ( F )) 2 F 1 x 2 + 2 F 1 y 2 + 2 F 1 z 2 , 2 F 2 x 2 + 2 F 2 y 2 + 2 F 2 z 2 , 2 F 3 x 2 + 2 F 3 y 2 + 2 F 3 z 2
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S E C T I O N 18.3 Divergence Theorem (ET Section 17.3) 1261 = 2 F 2 x y + 2 F 3 x z 2 F 1 y 2 2 F 1 z 2 , 2 F 1 y x + 2 F 3 y z 2 F 2 x 2 2 F 2 z 2 , 2 F 1 z x + 2 F 2 z y 2 F 3 x 2 2 F 3 y 2 (2) Since the mixed partials are equal, the expressions obtained in (1) and (2) are the same. This proves the desired identity.
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18.3 Ex 26 - 32 - 1260 C H A P T E R 18 F U N D A M E N TA...
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Ask a homework question - tutors are online | 1,219 | 2,716 | {"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} | 3.296875 | 3 | CC-MAIN-2018-13 | latest | en | 0.346887 |
https://engineering.stackexchange.com/questions/52120/what-are-the-practical-solutions-to-manufacturing-this-mechanical-piece | 1,721,912,060,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763858305.84/warc/CC-MAIN-20240725114544-20240725144544-00208.warc.gz | 217,783,297 | 43,224 | What are the practical solutions to manufacturing this mechanical piece?
My Custom Piece (Measures are in mm)
I'm designing a custom mechanism that needs this piece to be able to function, but when i started thinking about how to manufacture it, I started getting confused about how to make it.
what I know is:
• how to make a material block into a cylinder using a "Lathe" (For Example).
• How to make the hole in the middle of the cylinder using the "Lathe" or the "Milling Machine" (For Example).
but when i get to making the saw shape every 60°, I don't know what to use.
• should I use a CNC Machine? But is it gonna be able to make that shape, since there isn't much space for the "Drill Bit" "Endmill/Cutter"?
I know that a similar piece is also used in some one way rotation mechanisms so there should be multiple ways to get this shape.
So if you know some ways to make a saw shaped cylinder,
if you have any advice or suggestion on practically making this piece, or if you even have have some corrections or other ways to get the cylinder or the hole.
it would be appreciated.
EDIT: some remarks:
• This drawing is measured in mm
• The purpose of this part is to stop the rotation of a similar part sitting on top of it
• So the sharpness of the edges is tolerable, as long as it blocks a similar part's rotation
• I like @Jonathan.R.Swift comment about separating the inner cylinder from the outer cylinder.
• yes i didn't know at first what Endmill is called in English.
• People who design things tend to need to understand how the parts can be made AND how to assemble them, otherwise they draw things that cannot be made. Check out Escher. Commented Aug 14, 2022 at 17:31
• is this drawing in mm o or inches? Commented Aug 15, 2022 at 14:04
• If the central 15mm dia cylinder were a separate part, this would be much easier to manufacture. There is no keying feature on that bore so a bushing would work? Commented Aug 16, 2022 at 6:13
• This is one of those things where engineer go nuts because you don't have rounded inner corners. Drawings are to be taken very literally. If not then specify a ISO 13715:2019 unspecified corner allowance. BTW you should make your summery axes obvious. Commented Aug 16, 2022 at 16:25
• @joojaa all the measures are in mm/deg. Commented Aug 22, 2022 at 14:18
should i use a CNC Machine? But is it gonna be able to make that shape, since there isn't much space for the "Drill Bit"?
On a mill, it is called an endmill, not a drill bit. I assume this is just a language issue though since I can't imagine anyone who knows how to use a mill doesn't know the proper name for an endmill.
But endmills aren't the only cutting tools you can run in a mill, and there are multiple types of endmills. Dovetail endmills for instance, and angular cutters which are not endmills at all.
If nothing need to actually sit in that sharp corner then why do you care if it's sharp or not? Just specify a fillet radius which will also spread out stresses. And if something does need to sit right in that corner then it can't be sharp anyways because you need to specify a relief which is round a cut into corner to to give room so debris and burrs on the other piece don't prevent it from being unable to sit flush against the corner. That's why you see a little circular cutout on the inside corner of machinist squares.
https://www.kinexmeasuring.com/en/products/squares
You can do it with the disc laying flat and have the end mill cut incrementally step up the Z-axis to cut the slope.
Or you could mount the entire disc to an angle plate of the same angle as the sloped face. Then the the mill can stay in the same Z-plane and just move around in the X-Y plane to cut the sloped faces which may make them smoother. In this case I suspect you might want to cut most of the face with a facemill if you could fit on in there. Then you go in with a dovetail end mill to get the undercut on that corner (that vertical face will be an undercut relative to the mill if the disc is sitting on an angle plate).
There exists a lot of metal working machines that most people don't know about. There exists more than just mills and lathes. Even amongst mills and lathes there are many different types. I bet you never even knew a horizontal mills or vertical lathes existed.
Now that I brought up horizontal mills, I believe you can also cut most of that sloped face by mounting the disc flat on a rotary table and use a horizontal mill with an single-angle angular cutter. You won't get the vertical edges around the shaft hole with that alone though it will be a radiused toward the center.
And finally, a 5-axis CNC mill does this easily: https://www.youtube.com/watch?v=2tankRyWaEs
• My appreciation of your answer surpassed my remarks about your style of giving advice, and i would thank you and appreciate your answer since you did give me multiple solutions i did not think of, and i did overall learn new things from you. so i would thank you again, and i would like to express my satisfaction of your answer and solution that solved my problem. Commented Aug 22, 2022 at 14:30
To add to DKNguyen's answer, it depends on the required strength of the part. If the part is highly-stressed, you would hot forge it out of steel alloy in a press.
If the stresses are significant but not excessively high, you would machine it from steel or aluminum in a CNC machine. All the slopes and angles shown in your drawing are easily cut on a 3-axis CNC milling machine.
If the stresses are relatively low and minimizing cost is important, you would die cast the part in copper-aluminum die casting compound in a closed-mold molding machine.
If the stresses are minimal and production volumes are very large and the part will not be subjected to hot conditions during use, you would injection mold the part in glass-filled polymer in an injection molding machine.
• I guess I was also thinking from the perspective of making a one-off. If you're making many you wouldn't use method I mentioned. Commented Aug 16, 2022 at 19:05
• Thank you a Lot, you gave me a lot of insight on how to go on, you're answer helped me a lot, and if i could make two answers correct i would definitely mark yours correct as well. Commented Aug 22, 2022 at 14:42 | 1,473 | 6,287 | {"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} | 2.609375 | 3 | CC-MAIN-2024-30 | latest | en | 0.9613 |
https://www.justintools.com/unit-conversion/time.php?k1=olympiads&k2=lustrums | 1,720,951,767,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514564.41/warc/CC-MAIN-20240714094004-20240714124004-00341.warc.gz | 735,355,037 | 26,793 | Please support this site by disabling or whitelisting the Adblock for "justintools.com". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :)
# TIME Units Conversionolympiads to lustrums
= 0.80054794520548 Lustrums
Category: time
The base unit for time is seconds (SI Unit)
[Lustrums] symbol/abbrevation: (L)
How to convert Olympiads to Lustrums (o to L)?
1 o = 0.80054794520548 L.
1 x 0.80054794520548 L = 0.80054794520548 Lustrums.
Always check the results; rounding errors may occur.
Definition:
In relation to the base unit of [time] => (seconds), 1 Olympiads (o) is equal to 126230400 seconds, while 1 Lustrums (L) = 157680000 seconds.
1 Olympiads to common time units
1 o = 126230400 seconds (s)
1 o = 2103840 minutes (min)
1 o = 35064 hours (hr)
1 o = 1461 days (day)
1 o = 208.71428571429 weeks (wk)
1 o = 4.0027397260274 years (yr)
1 o = 48.032876712329 months (mo)
1 o = 0.40022320862397 decades (dec)
1 o = 0.040022320862397 centuries (cent)
1 o = 0.0040022320862397 millenniums (mill)
1 o = 0.80054794520548 L
2 o = 1.601095890411 L
3 o = 2.4016438356164 L
4 o = 3.2021917808219 L
5 o = 4.0027397260274 L
6 o = 4.8032876712329 L
7 o = 5.6038356164384 L
8 o = 6.4043835616438 L
9 o = 7.2049315068493 L
10 o = 8.0054794520548 L
20 o = 16.01095890411 L
30 o = 24.016438356164 L
40 o = 32.021917808219 L
50 o = 40.027397260274 L
60 o = 48.032876712329 L
70 o = 56.038356164384 L
80 o = 64.043835616438 L
90 o = 72.049315068493 L
100 o = 80.054794520548 L
200 o = 160.1095890411 L
300 o = 240.16438356164 L
400 o = 320.21917808219 L
500 o = 400.27397260274 L
600 o = 480.32876712329 L
700 o = 560.38356164384 L
800 o = 640.43835616438 L
900 o = 720.49315068493 L
1000 o = 800.54794520548 L
2000 o = 1601.095890411 L
4000 o = 3202.1917808219 L
5000 o = 4002.7397260274 L
7500 o = 6004.1095890411 L
10000 o = 8005.4794520548 L
25000 o = 20013.698630137 L
50000 o = 40027.397260274 L
100000 o = 80054.794520548 L
1000000 o = 800547.94520548 L
1000000000 o = 800547945.20548 L | 869 | 2,196 | {"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-2024-30 | latest | en | 0.614897 |
https://physics.stackexchange.com/questions/518459/are-the-basins-of-attraction-of-two-static-gravitational-sources-two-half-planes | 1,725,723,789,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650883.10/warc/CC-MAIN-20240907131200-20240907161200-00698.warc.gz | 453,952,964 | 49,901 | # Are the basins of attraction of two static gravitational sources two half-planes?
Consider the following setup:
Two massive bodies of mass $$M$$ are fixed at the positions $$\vec{A}=(d, 0)$$ and $$\vec{B}=(-d, 0)$$.
Now imagine a test particle $$p$$ with initial position $$\vec{r}_0=(x_0,y_0)$$ and initial velocity $$\vec{v}_0=(0,0)$$. Its position $$\vec{r}(t)=\left(x(t),y(t)\right)$$ will change over time under the influence of the gravitational potential. If the test particles trajectory goes through $$A$$ (or $$B$$) the initial position $$\vec{r}_0$$ lies in the basin of attraction of the mass at point $$A$$ (or $$B$$).
## Or simpler yet: Does $$x_0 \gt 0$$ imply that $$x(t)\gt0$$ for all $$t$$?
NOTE: This problem is known as a special case of Euler's three-body problem. It is discussed in great detail in "Integrable Systems in Celestial Mechanics" by Diarmuid Ó Mathúna, where the equations of motion are given in closed form using Jacobi elliptic functions, by transforming to what Mathúna calls "Louiville coordinates". I am currently working through the derivation of this, but as it is very technical, it will take me some time. It is still possible, that the question may be answered without the explicit equations of motion.
Background:
I'm currently working on a (2 dimensional) simulation of gravity which visualizes gravitational basins of attraction.
The simulation works as follows:
$$n$$ massive bodies are given an intial position. They are treated as static and will never change position during the simulation. Each body is also given a unique color property (for visualization purposes later).
Next, the simulation space is filled with particles. Their initial velocity is set to zero. Their trajectories are calculated via numerical integration.
When a collision between a particle and one of the massive bodies is detected, the initial position of the particle is assumed to lie in the basin of attraction of that body (with which it collided). A collision occurs when the distance to a body is below a certain threshold. The initial position of the particle is then marked in the color of the planet it collided with.
In general the resulting pictures are very complex.
Here are some more examples: 1, 2, 3, 4*
As you can see the basins are generally fractals.
Yet, when I simulate the above described setup, where $$n=2$$ and the masses of the two bodies are the same, the following picture is produced.
Where blue pixels represent initial positions in the basin of attraction of the mass at point $$A$$, red pixels those in the basin of the mass at point $$B$$ and black pixels in neither. (A thin strip of black pixels lie on the line $$x=0$$, where they oscillate around the origin)
## The picture is clearly devided into two half planes, yet I can not explain why. Any help explaining this fact would be much appreciated.
I tried solving the problem analytically using Lagrange's equations of the second kind. I set: $$T=\frac{m}{2}(\dot{x}^2+\dot{y}^2)$$ $$V=-GM m\left[\frac{1}{\sqrt{(x-d)^2+y^2}}+\frac{1}{\sqrt{(x+d)^2+y^2}}\right]$$ And obtained the following equations of motion: $$\ddot{x} = -GM\left[\frac{x-d}{\left((x-d)^2+y^2\right)^{\frac{3}{2}}} + \frac{x+d}{\left((x+d)^2+y^2\right)^{\frac{3}{2}}}\right]$$ $$\ddot{y} = -GM\left[\frac{y}{\left((x-d)^2+y^2\right)^{\frac{3}{2}}} + \frac{y}{\left((x+d)^2+y^2\right)^{\frac{3}{2}}}\right]$$
Where $$x$$ and $$y$$ are the coordinates of the particle, $$m$$ the mass of the particle, $$M$$ the mass of the massive bodies and $$d$$ the distance (in the $$x$$-direction) from the origin of the massive bodies.
I converted this into a system of four first-order differential equations and tried solving it using Maple's dsolve, but it's been stuck Evaluating for two hours now, so I don't think it's going to finish...
Another thing I tried is showing that a particle on one side will never have enough potential Energy to ever cross the line $$x=0$$, but then I considered the starting location $$\vec{r}_0=(2d,y_0)$$ with arbitrary $$y_0$$. $$V((2d,y_0))=-GMm\left[\frac{1}{\sqrt{d^2 + y_0^2}} + \frac{1}{\sqrt{9d^2 + y_0^2}}\right]$$ Then I could easily show that there exists a point on the line $$x=0$$ that has less potential, therefore I imagined the particle could reach that point and still have "kinetic energy leftover" to cross the line. One such point would be $$\vec{r}_1=(0,y_0)$$, since: $$V((0,y_0))=-GMm\left[\frac{1}{\sqrt{d^2 + y_0^2}} + \frac{1}{\sqrt{d^2 + y_0^2}}\right]$$ From which it became evident that $$V((2d,y_0))\gt V((0,y_0))$$ A flaw in this argument could be, that the particle could never reach $$\vec{r}_1$$ (or any point on $$x=0$$ which has less potential). Another fact I can not prove.
• I would guess that the version of the problem where the sources orbit each other has been studied, but since this version is unphysical for astronomical bodies, maybe it hasn't been. Yours would be like the physical version in the rotating frame, but with the centrifugal and Coriolis forces turned off. This went against my intuition, as I assumed chaotic effects would not appear until I added a third body. You do have three bodies: the two static sources, and the test particle.
– user4552
Commented Dec 8, 2019 at 19:20
• @Ben Crowell: I assumed the velocity of the two bodies was negligible when compared to those of the test particles. After your rewording of the problem in your comment, I can see how that "simplification" was not one I can make. Thank you for clarifying. Although the initial question still interests me. Commented Dec 8, 2019 at 19:28
• Is it 3D or 2D and what is the size of the integration region? Commented Dec 8, 2019 at 19:29
• I'm thinking a symmetry argument could be sufficient here. Of course there is a mirror symmetry, but there is also a scale symmetry relating to velocities: if you scale the distance, of course the basin regions would need to scale as well; but this is equivalent to scaling down velocities; so that any velocity must proportionally match the eps-velocity basin picture in a way. The only figures that satisfy this invariance (plus some other constraints) ought to be planes (note how it doesn't change with scale). Particles (assuming 0 initial velocity o.c.) should never even leave their side ?
– Real
Commented Dec 13, 2019 at 1:33
• Interestingly, the paths of the particles seem to ergodically fill regions that in elliptical coordinates correspond to $\mu < \mu(0)$ and $\nu < \nu(0)$. However, the Lagrangian in elliptical coordinates is not very enlightening either. This might be of help: en.wikipedia.org/wiki/Euler%27s_three-body_problem Commented Dec 13, 2019 at 21:30
The good news is that your conjecture is true. The bad news is that my proof offers little insight. I'm not sure a small proof by some symmetry argument is possible, since it's not simply the symmetry of the setup that counts, the result is specific to inverse-square forces. Any perturbation to the exponent of the force leads to the particle skipping from side to side, generally behaving quite chaotically. This is reminiscent of the Laplace-Runge-Lenz vector, I wouldn't be surprised if there is some correspondence.
Anyway, first off, I'm assuming $$GM = m = 1$$. By symmetry considerations, we can assume the motion happens in a plane, so we'll just use two coordinates $$x$$ and $$y$$. Now let's use elliptical coordinates, of course placing the static masses at the foci $$x = -a$$ and $$x = +a$$
$$x = a \cosh \mu \cos \nu ,\\ y = a \sinh \mu \sin \nu .$$
Working through the algebra (and with the help of Mathematica) our Lagrangian turns out to be
\begin{align*} \mathcal{L} &= \frac{\dot{x}^2 + \dot{y}^2}{2} + \frac{1}{\sqrt{(x-a)^2+y^2}}+\frac{1}{\sqrt{(x+a)^2+y^2}} \\ & = \frac{a^2 (\cosh^2 \mu - \cos^2 \nu) (\dot{\mu}^2+\dot{\nu}^2)}{2}+\frac{1}{a} \frac{2 \cosh \mu}{\cosh^2 \mu - \cos^2 \nu} \\ &= a^2 u \frac{\dot{\mu}^2+\dot{\nu}^2}{2} + \frac{1}{a} \frac{2 \cosh \mu}{u}. \end{align*}
For convenience, I have defined $$u(\mu, \nu) = \cosh^2 \mu - \cos^2 \nu$$. To formulate the Hamiltonian, we need the conjugate momenta
$$p_\mu = \frac{\partial \mathcal{L}}{\partial \dot\mu} = a^2 u \dot{\mu} \\ p_\nu = \frac{\partial \mathcal{L}}{\partial \dot\nu} = a^2 u \dot{\nu}.$$
The Legendre transform yields
$$\mathcal{H} = p_\mu \dot{\mu} + p_\nu \dot{\nu} - \mathcal{L} = \frac{1}{u} \left( \frac{p_\mu^2 + p_\nu^2}{2a^2} + \frac{2 \cosh \mu}{a}\right).$$
As this Hamiltonian is time-independent, we can define a conserved energy $$E$$ (which will be negative in our case). At this point, I'm uncertain about how exactly the next step works, and how correct it is. I could be mistaken. Taking inspiration from this paper, we introduce a new fictitious time coordinate $$\tau$$ satisfying
$$d\tau = u dt ,$$
and a new Hamiltonian
$$\mathcal{H}' = u(\mathcal{H} - E).$$
Taking partial derivatives we find an equivalent of Hamilton's equations
$$-\frac{\partial \mathcal{H}'}{\partial q} = -\frac{\partial u}{\partial q}(\mathcal{H}-E) - u \frac{\partial \mathcal{H}}{\partial q} = u \frac{dp}{dt} = \frac{dp}{d\tau} \\ \frac{\partial \mathcal{H}'}{\partial p} = u \frac{\partial \mathcal{H}}{\partial q} = u \frac{dq}{dt} = \frac{dq}{d\tau},$$
where we used the fact that $$\mathcal{H} = E$$ on-shell in the second step of the first line. These equations allow us to solve the equations of motion in function of the fictitious time coordinate $$\tau$$. Given the implicit definition of $$\tau$$, based on the coordinates itself, this might not seem like it would help, but for this purpose qualitative arguments based on the general form of the Hamiltonian will suffice.
Substituting in everything, in our case the new Hamiltonian looks like
$$\mathcal{H}' = \frac{p_\mu^2 + p_\nu^2}{2 a^2} - \frac{2 \cosh \mu}{a} - E \cosh^2 \mu + E \cos^2 \nu .$$
It turns out $$\mu$$ and $$\nu$$ completely decouple, and basically look like two harmonic oscillators (up to a coordinate transformation)! If the two oscillation periods are incommensurate, as they will be for generic initial conditions, the particle will ergodically fill the region it is energetically allowed inside, coming arbitrarily close to any point inside of it, including the massive bodies themselves. This agrees with my simulations, an example of which is pictured. Here the blue line shows the trajectory of the particle, and the green and orange lines are curves of respectively constant $$\mu$$ and $$\nu$$.
The part corresponding to $$\nu$$,
$$\mathcal{H}_\nu' = \frac{p_\mu^2}{2a^2} - |E| \cos^2 \nu ,$$
implies that $$\cos^2 \nu > \cos^2 \nu_0$$. Since the half-plane $$x = 0$$ corresponds to $$\nu = \pm \frac{\pi}{2}$$, where the cosine achieves a minimum, this also proves that a particle starting with zero initial velocity will be confined to its own side.
• So the movement of the particle is confined to the intersection of two ellipses? Commented Dec 19, 2019 at 17:15
• @Eli I need $\mathcal{H}'$ to decouple the two coordinates. The equations of motion following from the normal Hamiltonian are very complex and coupled. As far as I can tell, the conclusion doesn't follow from a simple algebraic statement like $\mathcal{H} = E$. Or do I misunderstand your question? Commented Dec 19, 2019 at 17:58
• @NiveaNutella Lines of constant $\mu$ (like the green one) are ellipses, but lines of constant $\nu$ (like the orange one) are hyperbolae. Commented Dec 19, 2019 at 18:00
finally
In the following $$G=M=d=1$$.
The equipotentials look like
For a particle starting from rest, the contours also represent the level curves of the first constant of motion, total energy $$E$$. Each point on the contour represents a starting point for a trajectory. In addition, a trajectory can't leave the interior of the contour. Since the contours for $$V\le-2$$ (the "$$\infty$$" curve) don't cross over to the other plane, for trajectories in these regions with $$|x|>0$$, $$E$$ alone segregates the basins into different plains.
Now consider the second constant of motion,
$$R=r_1^2r_2^2\dot{\theta_1}\dot{\theta_2}-2(\cos\theta_1+\cos\theta_2)$$
where the symbols are as as in- As before we plot the level curves for this constant and see what they look like. Each curve indicates a starting point of the trajectory.Each pair$$(E,R)$$ uniquely identifies the entire trajectory.
We observe that,$$R$$ seems to divide the starting points into 3 classes:
$$R\gtrless0$$ corresponds to $$x\lessgtr0$$.
with $$R=0$$ implying $$x=0$$ or $$(y=0,0<|x|<1)$$. So trajectories starting in the right half-plane never enter the left as this requires a constant of motion to change sign. The $$R=0$$ with $$(y=0,0<|x|<1)$$ is explicitly checked and found also to satisfy the above.
Since the equipotential bounds the trajectory on the right, the particle is doomed to meet its corresponding source of field.
The formula used for proving the correspondence was
$$R(x_0,y_0,\dot{x}=0,\dot{y}=0)= -2\left(\frac{x_0+1}{(x_0+1)^2+y_0^2}+\frac{x_0-1}{(x_0-1)^2+y_0^2}\right)$$
# Earlier attempts
## 1
An object($$|x_0|>0$$) starting from rest, will only roam within the equipotential it was on at the beginning. So for points within $$\infty$$ contour($$V=-2$$), the answer is in the affirmative: no crossing over to the other plane. This is evident from the direction of forces too.
Now consider the field plot for the direction of the horizontal component of force
Trivially, the answer is affirmative for $$y_0=0$$ as y component of force vanishes.
The red dots(numerically calculated) indicate the boundary (excluding $$x=0$$), say $$B$$, where the horizontal component switches sign. Within this region, the particle always moves away from $$x=0$$.
If we could show that the particle never crosses B, then all such trajectories must terminate. Since the only point of singularity for these would be the mass location so answer be in the affirmative for these points too.
Alas! this isn't true.
## 2
If we can show that the particle never crosses $$y=\pm m x$$ where $$m=y_0/x_0$$ and $$0, the basins would indeed be half planes. This seems to be indicated by sims.
work in progress
To prove the this hypothesis, we''ll try the following program.
1. First we show that the force $$F$$ on $$y=\pm mx$$ always points into the region $$mx>|y|$$. As a result at least initially, the particle isn't going to cross.
2. Let energy $$E(x,y,\dot{x},\dot{y})$$ and something else, $$R(x,y,\dot{x},\dot{y})$$ be the 2 constants of motion.
3. We can eliminate $$\dot{y}$$ in $$R$$ using $$E=E(x_0,y_0,0,0)=E_0$$ to get $$R(x,y,\dot{x})$$.
4. For any trajectory, $$R=R(x_0,y_0,0)=R_0$$. Now comes the important part:
If the trajectory intersects $$y=m x$$ at another point $$x'$$, then $$R'=R(x',m x',\dot{x'})=R_0$$. We then analyze the behaviour of $$R'-R_0 \forall 0 hoping to find no real root.
5. A similar procedure for $$y=-m x$$ would bound the particle on the left while the equipotential already bounds it on the right.
The $$R$$ being used is
$$R=r_1^2r_2^2\dot{\theta_1}\dot{\theta_2}-2(\cos\theta_1+\cos\theta_2)$$
where the parameters are as indicated in the main answer above.
Currently, this procedure seems to fail. e.g. for $$m=1, x_0=y_0=3$$, here's what the plot is:
(the blue plane is $$R_0=R'$$) So for the following $$x,\dot{x}$$ values
there seem to exist values $$(x',\dot{x'})$$ for which, the trajectory swivels back to the line.
However, we should note that the existence of a negative $$\dot{x}$$ can't alone imply crossing as some large value of $${\dot{y}}$$ may still keep the trajectory inbound.
• I've read about the constant of motion($R$) in the literature, but I could not find the symmetry to which it corresponds. Any ideas, or did you get it from somewhere? (Maybe I'm wrong, but I thought every symmetry yields a Noether-Charge, which is a constant of motion. And every constant of motion corresponds to a symmetry..) Commented Dec 19, 2019 at 15:37
• emabarassed to say I got it from wiki but the references check out: 1) E.T. Whittaker, A treatise on the analytical dynamics of particles and rigid bodies , pg 283 2) Coulson's paper,doi.org/10.1002%2Fqua.560010405 Commented Dec 19, 2019 at 15:50
• Thanks for the reference Commented Dec 19, 2019 at 15:54
• as for the converse of Noether's theorem, given conserved charge what is the symmetry, it seems complicated. physics.stackexchange.com/questions/24596/… Commented Dec 19, 2019 at 16:01
• great question, by the way Commented Dec 19, 2019 at 16:01
As stated, the problem is to start a particle at rest at every point in the plane and track its trajectory until it hits one of the stationary masses.
The problem can be transformed to that of starting the particle near a stationary mass and looking for initial velocity vectors that result in the particle's trajectory reaching a point where the particle is momentarily stationary. One such counterexample is enough, if the stationary point is on the opposite half-plane from the starting point.
The stationary point could only be on an equipotential surface that corresponds to the particle's initial kinetic energy. And, I think the particle must necessarily approach its ultimate stationary point in a direction normal to the equipotential surface, which might provide a useful further constraint.
I have done analogous searches: for gravitational slingshot trajectories. The search was done using a genetic algorithm, and proceeded very rapidly. Because your problem is only 2D, you might be able to use a more direct technique like Newton's Method.
This approach might be able to disprove the two half-plane hypothesis, but it can't prove it. | 4,912 | 17,707 | {"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": 122, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-38 | latest | en | 0.913666 |
https://crushthecpaexam.com/cpa-riddles/ | 1,695,681,094,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510100.47/warc/CC-MAIN-20230925215547-20230926005547-00816.warc.gz | 205,655,520 | 35,299 | # CPA Riddles to Puzzle Your Finance Skills
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## CPA Riddles: The Missing Dollar
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Opt for the comprehensive Pro course at a generous 35% discount. | 695 | 2,988 | {"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-2023-40 | latest | en | 0.926426 |
https://hextobinary.com/unit/angle/from/minangle/to/cycle/193 | 1,718,635,102,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861719.30/warc/CC-MAIN-20240617122241-20240617152241-00385.warc.gz | 250,677,066 | 15,000 | # 193 Minutes in Cycles
Angle
Minute
Cycle
193 Minutes = 0.0089351851851852 Cycles
## How many Cycles are in 193 Minutes?
The answer is 193 Minutes is equal to 0.0089351851851852 Cycles and that means we can also write it as 193 Minutes = 0.0089351851851852 Cycles. Feel free to use our online unit conversion calculator to convert the unit from Minute to Cycle. Just simply enter value 193 in Minute and see the result in Cycle. You can also Convert 194 Minutes to Cycles
## How to Convert 193 Minutes to Cycles (193 min to cycle)
By using our Minute to Cycle conversion tool, you know that one Minute is equivalent to 0.000046296296296296 Cycle. Hence, to convert Minute to Cycle, we just need to multiply the number by 0.000046296296296296. We are going to use very simple Minute to Cycle conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Minute} = \text{0.000046296296296296 Cycles}$$
$$\text{193 Minutes} = 193 \times 0.000046296296296296 = \text{0.0089351851851852 Cycles}$$
## What is Minute Unit of Measure?
Minute of angle is a unit of angular measurement. It is also known as arcmin. A minute of angle is equal to 60 arcsec.
## What is the symbol of Minute?
The symbol of Minute is min. This means you can also write one Minute as 1 min.
## What is Cycle Unit of Measure?
The cycle is a unit of plane angle measurement. Like turn, cycle is also referred as full circle, complete rotation, revolution or turn. One cycle in general is equal to 360 degrees.
## What is the symbol of Cycle?
The symbol of Cycle is cycle. This means you can also write one Cycle as 1 cycle.
## Minute to Cycle Conversion Table (193-202)
Minute [min]Cycle [cycle]
1930.0089351851851852
1940.0089814814814815
1950.0090277777777778
1960.0090740740740741
1970.0091203703703704
1980.0091666666666667
1990.009212962962963
2000.0092592592592593
2010.0093055555555556
2020.0093518518518519
## Minute to Other Units Conversion Table
Minute [min]Output
193 minutes in arcmin is equal to193
193 minutes in arcsecond is equal to11580
193 minutes in cycle is equal to0.0089351851851852
193 minutes in degree is equal to3.22
193 minutes in gradian is equal to3.57
193 minutes in gon is equal to3.57
193 minutes in octant is equal to0.071481481481481
193 minutes in quadrant is equal to0.035740740740741
193 minutes in radian is equal to0.056141424272484
193 minutes in sextant is equal to0.053611111111111
193 minutes in sign is equal to0.10722222222222
193 minutes in turn is equal to0.0089351851851852
193 minutes in circle 1/10 is equal to0.089351851851852
193 minutes in circle 1/16 is equal to0.14296296296296
193 minutes in circle 1/2 is equal to0.01787037037037
193 minutes in circle 1/4 is equal to0.035740740740741
193 minutes in circle 1/6 is equal to0.053611111111111
193 minutes in circle 1/8 is equal to0.071481481481481
193 minutes in full circle is equal to0.0089351851851852
193 minutes in mil is equal to57.19
193 minutes in point is equal to0.28592592592593
193 minutes in second is equal to11580
Disclaimer:We make a great effort in making sure that conversion is as accurate as possible, but we cannot guarantee that. Before using any of the conversion tools or data, you must validate its correctness with an authority. | 934 | 3,270 | {"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} | 3.359375 | 3 | CC-MAIN-2024-26 | latest | en | 0.848118 |
https://math.icalculator.com/3x3-matrix-inverse-calculator.html | 1,696,167,073,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510888.64/warc/CC-MAIN-20231001105617-20231001135617-00720.warc.gz | 422,061,946 | 6,161 | # 3x3 Matrix Inverse Calculator
The inverse of a matrix is something which can be very difficult to calculate and becomes more difficult when the order of the given matrix is 3 x 3.
Matrix (A)
3x3 Matrix Inverse Calculator Results
Determinant (|A|)
Inverse of Matrix = (adj A)/|A|
Please provide a rating, it takes seconds and helps us to keep this resource free for all to use
## About the 3 x 3 matrix inverse calculator
The inverse of a matrix can only be found in the case if the matrix is a square matrix and the determinant of that matrix is a non-zero number. After that, you have to go through numerous lengthy steps, which are more time consuming in order to find the inverse of a matrix. Who would want to work so hard just to find out the inverse of a 3 x 3 matrix? This is the reason why iCalculator made this good online calculator that will save your manual calculations and save you a lot of time.
You can find out the inverse of a matrix (say A) by finding out the value of 'I' in the above equation: A = IA.
The use of this calculator is very easy. You just have to enter the values of the respective 3 x 3 order matrix in the required fields and hit the enter button. You will get the desired results immediately.
Calculating the inverse of a 3x3 matrix can be a daunting task, but with the help of our 3x3 Matrix Inverse Calculator, it's as easy as 1-2-3! In this tutorial, we'll guide you through the process of using our calculator step-by-step.
## Step 1: Enter the Matrix
The first step is to enter your 3x3 matrix into the calculator. You can do this by typing the values of the matrix into the corresponding boxes. Make sure to enter the values in the correct order, starting with the top-left element and ending with the bottom-right element.
## Step 2: Calculate the Determinant
The next step is to calculate the determinant of the matrix. You can do this by clicking the "Calculate" button next to the "Determinant" field. The determinant of a 3x3 matrix can be calculated using the following formula:
det(A) = a11(a22a33 - a32a23) - a12(a21a33 - a31a23) + a13(a21a32 - a31a22)
where A is the matrix, and aij represents the element in the ith row and jth column.
## Step 3: Check for Invertibility
Before we calculate the inverse, we need to check if the matrix is invertible. A matrix is invertible if and only if its determinant is nonzero. If the determinant is zero, then the matrix is singular, and its inverse does not exist.
## Step 4: Calculate the Adjoint
Assuming the matrix is invertible, we can now calculate the adjoint of the matrix. The adjoint of a matrix is obtained by taking the transpose of the matrix of cofactors. The cofactor of an element is obtained by multiplying its minor by (-1)^(i+j), where i and j are the row and column indices of the element. The minor of an element is obtained by deleting its row and column from the matrix and calculating the determinant of the resulting 2x2 matrix.
## Step 5: Calculate the Inverse
Finally, we can calculate the inverse of the matrix using the formula: | 734 | 3,072 | {"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} | 4.0625 | 4 | CC-MAIN-2023-40 | longest | en | 0.903174 |
https://forums.paddling.com/t/paddle-power-curves/42927?page=2 | 1,712,978,653,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816535.76/warc/CC-MAIN-20240413021024-20240413051024-00227.warc.gz | 238,004,831 | 11,020 | Wing curve
– Last Updated: Sep-21-08 9:59 AM EST –
I am not sure that the wing curve falls off below the others. I think with proper rotation and angle it can maintain higher power peak over the entire range.
Why would a GP or a Euro have more power at the middle or end of stroke compared to Wing? Have not paddled GP enough to make conclusinos there but to me the wing has plenty of power in the second half of the stroke if one rotates well.
Again, my thoughts are based non-numerical data -;) And I'm an engineer so I can't trust that with my life, but my "engineering gut feeling" tells me so -;)))
One more clarification is needed. The horizontal axis may be representing time in the water, of which the GP may just as well spend more during a stroke cycle compared to a Wing. But the wing spends a shorter amount of time, all of it at higher power IMO, assumint early take out.
So, to me the charts as they are directly superimposed, do not yeld a meaningful comparison, beyond the relative measure within each curve (not compared to other curves) of how fast the power during the catch and release cycle raises and falls for each paddle individually.
More meaningful would be some sort of measure about the energy transfered to forward motion during the stroke and I am willing to bet that the more efficient paddles in the water (wing) will be on top in all segments of the curve.
Lastly, I think the point is well taken that "in-water" efficiency is one thing and "muscle and body efficiency" is another. And a paddle may as well have better "body efficiency" with a low "water efficiency" and thus make it a better choice for some people, while others would be better served with the opposite. I whish there was a way to measrure the two factros (water & body efficiency) and come-up with a compund measure in relation to paddling distance or energy output by the body over time or something like that at different speeds (e.g vs. different boat resistance in the water, to eliminate the type of boat as a factor).
There, now someone in physics or physical education with a decent math background or help can go collect and analyze data and write a thesis on this -;)
Engineers in the crowd?
Sorry I’m not an engineer but I started my career as an engineer and have a background in molecular physics, which I do not use any more.
I think the curves are actually his guess of “Force vs Time” not “Power Application”, so the integrated area would be the total power and the first derivative would be more representative of “Power” application.
It’s all a guess. I think my Onno surf paddle would have a much different profile from, my Mike Johnson surf paddle, or even my Onno full tour which would be hugely different from my Hungarian no name plastic/aluminum europaddle I bought with my first boat. They are all europaddles and have hugely different blade shapes.
No data. It’s fun to speculate but don’t believe any of it.
OK, count me in
– Last Updated: Sep-21-08 12:05 PM EST –
I'm designing the modification of a tow tank into a water channel at the moment. It will be used for teaching ship model studies in a naval architecture program. However, I am designing it for two ulterior motives: 1/4 scale kayak model and full-size paddle studies. I plan to start with flow visualization of paddle vortex dynamics. Depending on what that tells me, I will plan a hard-data experimental program.
Greyak's curves I find very useful - I consider them to be his hypothesis of power distribution during the strokes of different paddles, using an idealized technique with each. To those who bash the graphs, that's fine, but he never said they were hard data. A hypothesis is there to be proved or disproved by those with research funding (I'm planning to be that nerd). But research without a starting point is just noodling around.
One of the better points made above is that as shown, the curves are normalized to be the same width in the horizontal axis, which is clearly not ideal. I think it would be easy to alter them to make an indication of average paddle stroke duration, but that's just quibbling.
By the way, I'm keeping the wave generator that was part of the tow tank and hope to incorporate it into the water channel. I think it will be interesting to make time-dependent drag measurements of kayaks in a wavy sea, a test that has not been done as far as I know.
Not until you logged on here! Ha.
Greyak
– Last Updated: Sep-21-08 1:30 PM EST –
No offense intended. With an off hand guess, your charts are probably right on the money. But if you took two people, and gave them each one of those paddles, your results would vary. I was just agreeing with Charlie as to the method.
Last year at Raystown, I asked David Yost what his method was in designing canoes. IE testing, mathamatical equations, etc... He essentially said to me that his experience has shown him what hull aspects would best fit the type of canoe he was designing. Intuitive would probably the the best description. But that intuition comes from many years of experience and success.
Your original post says simply "Paddle Power curves" as if that were gospel. Perhaps you should have said a bit more.
Andy
horizontal axis
They are very useful visualization curves. I would label the vertical axis as force, not power. The horizontal axis should be labeled time. But then the wing stroke, which is shorter, should end before the GP stroke.
The wing stroke develops max force just after the catch, then exits the water quickly. The GP develops force much more gradually and holds it longer because the stroke is longer.
Efficiency is only slightly related to the force time history. Most losses occur in the human body not the paddle/water interaction. Whichever paddle enables you body to work most efficiently is the most efficient paddle. For instance, “slippage” or “grabbing the water” just are different ways of saying there are different drag coefficients. And a higher blade drag coefficient is not necessarily more efficient. Each paddler will have a paddle drag coefficient and paddle length that are most efficient for him, at the speed he intends to paddle. At a different speed, another paddle might be more efficient.
Curious how you envision…
… setting up and doing these paddle tests.
Anything I’ve seen has the paddle being drawn back through still water, and and looking at flow across/around the blades. That’s just not how they work when paddling (despite what people think they feel) unless you’ve tied your stern to something solid.
Since you’ll have a tow tank, I’m assuming you’re planning something more interesting - with paddle stroke simulation at speed (seems very difficult to rig - unless you just use real paddlers for the mechanics (you did say full size). I also assume that you’ve seen stop motion photos of this, and you look at the paddle primarily as a couple (lever), and not as something slipping back through the watter like a paddle wheel as most do (very secondary and in arcs vs. linear to what extent it does exist).
I totally agree technically
Just not looking at it that technically.
Depending on the axes, same paddle at different cadence, or slight technique variation, would have a somewhat different curve too. Interesting, but gets into more than I’m looking to share. Same for comparing 5 different euros (though there are better things to look at on that level than this).
Even these curves should probably not be same lengths, but that gets into really subjecting things about long vs short strokes (which can be changed with any paddled).
These only compare general type/technique combos - for same paddler - with both force and time as relative things. So force is the force the paddler applies (however much that is), through the duration of the stroke (however long that is).
– Last Updated: Sep-22-08 12:19 AM EST –
How to study paddle dynamics - a good question without a clear answer. My facility is being modified from a tow tank (still water, moving model) to a recirculating water channel (moving water, still model). It's about 4 ft wide by 2ft deep, although we may make it deeper.
For initial flow viz studies, I envision a kayak cockpit set up next to the channel, with the water moving by at cruising speed. A paddler will sit in the cockpit and do his best imitation of someone paddling at cruising speed. This may require another body of water on the other side of the 'kayak', the cockpit may need to be on gimbals, etc. If done properly, it should be possible to reproduce the relative (vector) velocities between the paddle and the moving water. High speed video from multiple directions (side, top, end) and release of dye trails from ports in the paddle should allow visualization of the vortex flow around the paddles.
All of that would be qualitative, but it is usually better to do visualization first to see what's going on. Quantitative data could be got by connecting the cockpit mounts to force transducers, but quantifying the flow downstream of the paddle would be hideously difficult. I'm a flow viz guy, so I tend to concentrate on that, which I think gets you 90% of the way to understanding the dynamics.
No study will ever account for the huge range in paddles and paddling styles - there's just too much variation. But hey, that's no reason not to try - we can start with the basics and see what there is to learn. If nothing else, I should be able to say whether there is a fundamental difference in the vortex dynamics of different style paddles, e.g. GP and Euro - that would be an interesting result in itself.
PS for the record, if the graphs show force vs. time, then the area under the graph is called the Impulse. Impulse measures the change in momentum from before the stroke to after; momentum is mass times velocity:
(mV)state 1 + Impulse = (mV)state 2
My favourite quote
42.5% of statistics are invented on spot.
Things that are not right with the graphs:
1. I heard that wing paddles are tunable - that is can tuned for a particular cadence, power delivery, arm span. You graph sort of forgets that. Which paddle was used?
2. Euro paddle - the recreational special is shown, right?
3. Which hybrid?
7. Is it really a figment of your imagination?
My favorite stat
NM
NM
Feel free to start a “good” one NM
This is a Rorschach test
interesting graphs
– Last Updated: Sep-23-08 10:45 AM EST –
however without a real biometric powermeter flat water test tank and a stop watch it is speculation.
Cycling sort of has paddling beat by being able to measure scientifically how many watts you generate with the drivetrain. But it is an expensive piece of equipment.
I tend to agree with the curves to an extent. But they don't mean anything. This is actually more speculative than the pull-test shawna and Leon had us do at Symposium.
Greyak never posts stuff without
good reason.
paul
Well as a fellow hornet nest
kicker I applaud him.
boat curves?
Greyak - fun stuff. As an engineer I was also curious around how the curves were generated but knowing that they represent your (subjective) experience and are to be used to generate dialogue…they are great thought-starters.
Sometimes I ponder similar characteristics in boats (canoes) themselves…Hemlock Peregrine requires some muscle at beginning of stroke but then delivers big speed/glide at low effort for remainder of stroke, Bell Fire boats take virtually no muscle at beginning but start to take muscle at end of stroke if you go fast enough…Merlin II is easier at beginning of stroke than Peregrine but does not deliver same speed…
fun stuff! | 2,536 | 11,647 | {"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-2024-18 | latest | en | 0.963911 |
https://drewvid.pythonanywhere.com/tedtestsAPI2?radio=titles&text=A%20performance%20of%20Mathemagic | 1,568,542,091,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514571027.62/warc/CC-MAIN-20190915093509-20190915115509-00223.warc.gz | 468,823,267 | 2,111 | full transcript
"From the Ted Talk by Arthur Benjamin: A performance of "Mathemagic""
#### Unscramble the Blue Letters
Make sure everybody knows that I got the answer right, because this is my big fisinh, OK. So, without any more stalling, here we go. I'll start the problem in the middle, with 57 times 683. 57 times 68 is 3,400, plus 476 is 3876, that's 38,760 plus 171, 38,760 plus 171 is 38,931. 38,931; double that to get 77,862. 77,862 becomes coioke fission, cookie fission is 77,822. That seems right, I'll go on. Cookie fission, OK. Next, I do 57 sarequd, which is 3,249, so I can say, three billion. Take the 249, add that to cookie, 249, oops, but I see a carry coming — 249 — add that to cookie, 250 plus 77, is 327 million — fiosisn, fission, OK, fallniy, we do 683 squared, that's 700 times 666, plus 17 squared is 466,489, rev up if I need it, rev up, take the 466, add that to fission, to get, oh gee — 328,489.
#### Open Cloze
Make sure everybody knows that I got the answer right, because this is my big ______, OK. So, without any more stalling, here we go. I'll start the problem in the middle, with 57 times 683. 57 times 68 is 3,400, plus 476 is 3876, that's 38,760 plus 171, 38,760 plus 171 is 38,931. 38,931; double that to get 77,862. 77,862 becomes ______ fission, cookie fission is 77,822. That seems right, I'll go on. Cookie fission, OK. Next, I do 57 _______, which is 3,249, so I can say, three billion. Take the 249, add that to cookie, 249, oops, but I see a carry coming — 249 — add that to cookie, 250 plus 77, is 327 million — _______, fission, OK, _______, we do 683 squared, that's 700 times 666, plus 17 squared is 466,489, rev up if I need it, rev up, take the 466, add that to fission, to get, oh gee — 328,489.
1. fission
2. squared
3. finally
4. finish
#### Original Text
Make sure everybody knows that I got the answer right, because this is my big finish, OK. So, without any more stalling, here we go. I'll start the problem in the middle, with 57 times 683. 57 times 68 is 3,400, plus 476 is 3876, that's 38,760 plus 171, 38,760 plus 171 is 38,931. 38,931; double that to get 77,862. 77,862 becomes cookie fission, cookie fission is 77,822. That seems right, I'll go on. Cookie fission, OK. Next, I do 57 squared, which is 3,249, so I can say, three billion. Take the 249, add that to cookie, 249, oops, but I see a carry coming — 249 — add that to cookie, 250 plus 77, is 327 million — fission, fission, OK, finally, we do 683 squared, that's 700 times 666, plus 17 squared is 466,489, rev up if I need it, rev up, take the 466, add that to fission, to get, oh gee — 328,489.
#### ngrams of length 2
collocation frequency
digit number 22
applause ends 4
number sir 4
digit numbers 3
digit calculator 3
month audience 3
sir audience 3
#### ngrams of length 3
collocation frequency
digit number sir 4
number sir audience 3
#### ngrams of length 4
collocation frequency
digit number sir audience 3
3. big
4. billion
5. carry
6. coming
8. double
9. finally
10. finish
11. fission
12. gee
13. middle
14. million
15. oops
16. problem
17. rev
18. squared
19. stalling
20. start
21. times | 1,031 | 3,142 | {"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-2019-39 | longest | en | 0.91246 |
https://p2p.wrox.com/139706-post2.html | 1,606,721,724,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141211510.56/warc/CC-MAIN-20201130065516-20201130095516-00278.warc.gz | 417,243,701 | 7,798 | View Single Post
April 23rd, 2005, 04:01 PM
planoie
Friend of Wrox
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Thanked 16 Times in 16 Posts
This definately calls for some recursion. I don't normally provide complete solutions (unless I'm getting paid :D), but this particular challenge was interesting. Plus I tried writing out what I thought you should try, but I simply could not word it well enough. So I figured I'd just give it a go and see if I could make it work:
Dim aryJaggedData(2)() As String
aryJaggedData(0) = New String() {"Red", "Yellow", "Blue"}
aryJaggedData(1) = New String() {"8", "9"}
aryJaggedData(2) = New String() {"Covered"}
strResult = RecurseData(aryJaggedData, 0, String.Empty, String.Empty))
Private Function RecurseData(ByVal aryData As String()(), ByVal intRecurseLevel As Integer, ByVal strResult As String, ByVal strLine As String) As String
Dim strPositionVal As String
Dim strMyLine As String
For i As Integer = 0 To aryData(intRecurseLevel).GetUpperBound(0)
strPositionVal = aryData(intRecurseLevel)(i)
If strLine.Length > 0 Then
strMyLine = String.Format("{0}/{1}", strLine, strPositionVal)
Else
strMyLine = strPositionVal
End If
If intRecurseLevel < aryData.GetUpperBound(0) Then
'Need to traverse down the y axis
strResult = RecurseData(aryData, intRecurseLevel + 1, strResult, strMyLine)
Else
strResult = String.Format("{0}{1}{2}", strResult, Environment.NewLine, strMyLine)
End If
Next
Return strResult
End Function
-Peter | 460 | 1,548 | {"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} | 2.890625 | 3 | CC-MAIN-2020-50 | latest | en | 0.662455 |
https://dice.mpi-inf.mpg.de/fact/cash-in/causes-gain-money | 1,680,441,558,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950528.96/warc/CC-MAIN-20230402105054-20230402135054-00539.warc.gz | 247,578,868 | 4,529 | # cash in: causes gain money
from ConceptNet
## Related concepts
Parents - Siblings -
## Related properties
Similarity Property causes gain money 1.00 causes lose of money 0.86 causes increase in money 0.85 causes have money 0.84 causes make money 0.83 causes you get money 0.83 be used to get money 0.82 causes receive money 0.80 requires earn lot of money 0.80 be used to get your money 0.80
## Clauses
### Salient implies Plausible
0.28
Rule weight: 0.28
Evidence weight: 0.99
Similarity weight: 1.00
Evidence: 0.99
Plausible(cash in, causes gain money)
Evidence: 0.91
¬ Salient(cash in, causes gain money)
### Similarity expansion
0.73
Rule weight: 0.85
Evidence weight: 0.99
Similarity weight: 0.86
Evidence: 0.99
Plausible(cash in, causes gain money)
Evidence: 0.90
¬ Plausible(cash in, causes lose of money)
0.72
Rule weight: 0.85
Evidence weight: 0.99
Similarity weight: 0.85
Evidence: 0.99
Plausible(cash in, causes gain money)
Evidence: 0.98
¬ Plausible(cash in, causes increase in money)
0.71
Rule weight: 0.85
Evidence weight: 0.99
Similarity weight: 0.84
Evidence: 0.99
Plausible(cash in, causes gain money)
Evidence: 0.99
¬ Plausible(cash in, causes have money)
0.70
Rule weight: 0.85
Evidence weight: 0.99
Similarity weight: 0.83
Evidence: 0.99
Plausible(cash in, causes gain money)
Evidence: 0.99
¬ Plausible(cash in, causes make money)
0.70
Rule weight: 0.85
Evidence weight: 0.99
Similarity weight: 0.83
Evidence: 0.99
Plausible(cash in, causes gain money)
Evidence: 1.00
¬ Plausible(cash in, causes you get money)
0.69
Rule weight: 0.85
Evidence weight: 0.99
Similarity weight: 0.82
Evidence: 0.99
Plausible(cash in, causes gain money)
Evidence: 0.99
¬ Plausible(cash in, be used to get money)
0.69
Rule weight: 0.85
Evidence weight: 0.94
Similarity weight: 0.86
Evidence: 0.91
Salient(cash in, causes gain money)
Evidence: 0.67
¬ Salient(cash in, causes lose of money)
0.69
Rule weight: 0.85
Evidence weight: 0.96
Similarity weight: 0.84
Evidence: 0.06
Typical(cash in, causes gain money)
Evidence: 0.04
¬ Typical(cash in, causes have money)
0.68
Rule weight: 0.85
Evidence weight: 0.97
Similarity weight: 0.83
Evidence: 0.06
Typical(cash in, causes gain money)
Evidence: 0.03
¬ Typical(cash in, causes you get money)
0.68
Rule weight: 0.85
Evidence weight: 0.96
Similarity weight: 0.83
Evidence: 0.06
Typical(cash in, causes gain money)
Evidence: 0.04
¬ Typical(cash in, causes make money)
0.68
Rule weight: 0.85
Evidence weight: 0.99
Similarity weight: 0.80
Evidence: 0.99
Plausible(cash in, causes gain money)
Evidence: 0.99
¬ Plausible(cash in, causes receive money)
0.68
Rule weight: 0.85
Evidence weight: 0.99
Similarity weight: 0.80
Evidence: 0.99
Plausible(cash in, causes gain money)
Evidence: 0.91
¬ Plausible(cash in, requires earn lot of money)
0.67
Rule weight: 0.85
Evidence weight: 0.99
Similarity weight: 0.80
Evidence: 0.99
Plausible(cash in, causes gain money)
Evidence: 0.99
¬ Plausible(cash in, be used to get your money)
0.67
Rule weight: 0.85
Evidence weight: 0.96
Similarity weight: 0.82
Evidence: 0.06
Typical(cash in, causes gain money)
Evidence: 0.04
¬ Typical(cash in, be used to get money)
0.67
Rule weight: 0.85
Evidence weight: 0.92
Similarity weight: 0.85
Evidence: 0.06
Typical(cash in, causes gain money)
Evidence: 0.08
¬ Typical(cash in, causes increase in money)
0.67
Rule weight: 0.85
Evidence weight: 0.92
Similarity weight: 0.85
Evidence: 0.91
Salient(cash in, causes gain money)
Evidence: 0.87
¬ Salient(cash in, causes increase in money)
0.66
Rule weight: 0.85
Evidence weight: 0.97
Similarity weight: 0.80
Evidence: 0.06
Typical(cash in, causes gain money)
Evidence: 0.04
¬ Typical(cash in, be used to get your money)
0.66
Rule weight: 0.85
Evidence weight: 0.92
Similarity weight: 0.84
Evidence: 0.91
Salient(cash in, causes gain money)
Evidence: 0.92
¬ Salient(cash in, causes have money)
0.65
Rule weight: 0.85
Evidence weight: 0.92
Similarity weight: 0.83
Evidence: 0.91
Salient(cash in, causes gain money)
Evidence: 0.92
¬ Salient(cash in, causes make money)
0.65
Rule weight: 0.85
Evidence weight: 0.95
Similarity weight: 0.80
Evidence: 0.06
Typical(cash in, causes gain money)
Evidence: 0.06
¬ Typical(cash in, causes receive money)
0.65
Rule weight: 0.85
Evidence weight: 0.92
Similarity weight: 0.83
Evidence: 0.91
Salient(cash in, causes gain money)
Evidence: 0.93
¬ Salient(cash in, causes you get money)
0.64
Rule weight: 0.85
Evidence weight: 0.94
Similarity weight: 0.80
Evidence: 0.91
Salient(cash in, causes gain money)
Evidence: 0.67
¬ Salient(cash in, requires earn lot of money)
0.64
Rule weight: 0.85
Evidence weight: 0.92
Similarity weight: 0.82
Evidence: 0.91
Salient(cash in, causes gain money)
Evidence: 0.91
¬ Salient(cash in, be used to get money)
0.63
Rule weight: 0.85
Evidence weight: 0.92
Similarity weight: 0.80
Evidence: 0.91
Salient(cash in, causes gain money)
Evidence: 0.91
¬ Salient(cash in, causes receive money)
0.63
Rule weight: 0.85
Evidence weight: 0.86
Similarity weight: 0.85
Evidence: 0.83
Remarkable(cash in, causes gain money)
Evidence: 0.81
¬ Remarkable(cash in, causes increase in money)
0.63
Rule weight: 0.85
Evidence weight: 0.85
Similarity weight: 0.86
Evidence: 0.83
Remarkable(cash in, causes gain money)
Evidence: 0.87
¬ Remarkable(cash in, causes lose of money)
0.63
Rule weight: 0.85
Evidence weight: 0.92
Similarity weight: 0.80
Evidence: 0.91
Salient(cash in, causes gain money)
Evidence: 0.90
¬ Salient(cash in, be used to get your money)
0.61
Rule weight: 0.85
Evidence weight: 0.83
Similarity weight: 0.86
Evidence: 0.06
Typical(cash in, causes gain money)
Evidence: 0.18
¬ Typical(cash in, causes lose of money)
0.61
Rule weight: 0.85
Evidence weight: 0.85
Similarity weight: 0.84
Evidence: 0.83
Remarkable(cash in, causes gain money)
Evidence: 0.88
¬ Remarkable(cash in, causes have money)
0.60
Rule weight: 0.85
Evidence weight: 0.85
Similarity weight: 0.83
Evidence: 0.83
Remarkable(cash in, causes gain money)
Evidence: 0.89
¬ Remarkable(cash in, causes make money)
0.59
Rule weight: 0.85
Evidence weight: 0.84
Similarity weight: 0.83
Evidence: 0.83
Remarkable(cash in, causes gain money)
Evidence: 0.93
¬ Remarkable(cash in, causes you get money)
0.59
Rule weight: 0.85
Evidence weight: 0.84
Similarity weight: 0.82
Evidence: 0.83
Remarkable(cash in, causes gain money)
Evidence: 0.91
¬ Remarkable(cash in, be used to get money)
0.59
Rule weight: 0.85
Evidence weight: 0.86
Similarity weight: 0.80
Evidence: 0.06
Typical(cash in, causes gain money)
Evidence: 0.15
¬ Typical(cash in, requires earn lot of money)
0.59
Rule weight: 0.85
Evidence weight: 0.85
Similarity weight: 0.80
Evidence: 0.83
Remarkable(cash in, causes gain money)
Evidence: 0.85
¬ Remarkable(cash in, causes receive money)
0.58
Rule weight: 0.85
Evidence weight: 0.84
Similarity weight: 0.80
Evidence: 0.83
Remarkable(cash in, causes gain money)
Evidence: 0.91
¬ Remarkable(cash in, requires earn lot of money)
0.57
Rule weight: 0.85
Evidence weight: 0.84
Similarity weight: 0.80
Evidence: 0.83
Remarkable(cash in, causes gain money)
Evidence: 0.94
¬ Remarkable(cash in, be used to get your money)
### Typical and Remarkable implies Salient
0.14
Rule weight: 0.14
Evidence weight: 1.00
Similarity weight: 1.00
Evidence: 0.91
Salient(cash in, causes gain money)
Evidence: 0.06
¬ Typical(cash in, causes gain money)
Evidence: 0.83
¬ Remarkable(cash in, causes gain money)
### Typical implies Plausible
0.48
Rule weight: 0.48
Evidence weight: 1.00
Similarity weight: 1.00
Evidence: 0.99
Plausible(cash in, causes gain money)
Evidence: 0.06
¬ Typical(cash in, causes gain money) | 2,684 | 7,589 | {"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} | 2.5625 | 3 | CC-MAIN-2023-14 | latest | en | 0.689085 |
https://discuss.codecademy.com/t/help-please/191526 | 1,537,863,261,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267161214.92/warc/CC-MAIN-20180925063826-20180925084226-00202.warc.gz | 489,017,004 | 5,064 | #1
I thought my code was producing the desired result...all numbers in start_list squared then sorted. The result I get back is correct but it asks me not to modify start_list. I'm not sure how to move forward. Any help would be appreciated.
``````start_list = [5, 3, 1, 2, 4]
square_list = []
for start_list in start_list:
square_list.append(start_list**2)
square_list.sort()
print square_list``````
More with 'for'
#2
Oh, I just noticed I've helped you before . So, when looping, we need to create a new variable that will iterate through our list.
Example:
``````for new_variable in original_variable:
#etc...``````
We dont need to define a variable like this in out loop : `new_variable = #whatever`
we can just do in in our loop .
#3
I'm sorry, but I just don't understand. I'm a visual learner. Can you please show me in context how that might look so I can figure this out?
#4
So currently, we're looping through `start_list`. Where is that? it's the `start_list` after `in`. The `start_list` after `for` is the one that will loop through `start_list = [5, 3, 1, 2, 4]`.
But this is bad because Codecademy doesn't want us to modify our list (`start_list`).
What should we do? Use something else like a new iterator to loop through `start_list`. Exapmle:
``````for new_iterator in my_list:
#anything happens here...``````
`new_iterator` will hold the same values as `start_list`, the only difference is: `new_iterator` is not a list!
Now... whats next in your code:
Since we don't want to modify our list (like I mentioned) we can use our new variable to square (`** 2`) and append to `square_list`.
#5
Thank you for your help! I finally have understanding and closure.
#6
This topic was automatically closed 7 days after the last reply. New replies are no longer allowed. | 468 | 1,799 | {"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-2018-39 | latest | en | 0.894858 |
https://www.netlib.org/lapack/explore-html-3.6.1/d0/d62/dsbtrd_8f_source.html | 1,713,695,372,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817765.59/warc/CC-MAIN-20240421101951-20240421131951-00613.warc.gz | 845,724,942 | 13,491 | LAPACK 3.6.1 LAPACK: Linear Algebra PACKage
dsbtrd.f
Go to the documentation of this file.
1 *> \brief \b DSBTRD
2 *
3 * =========== DOCUMENTATION ===========
4 *
5 * Online html documentation available at
6 * http://www.netlib.org/lapack/explore-html/
7 *
8 *> \htmlonly
10 *> <a href="http://www.netlib.org/cgi-bin/netlibfiles.tgz?format=tgz&filename=/lapack/lapack_routine/dsbtrd.f">
11 *> [TGZ]</a>
12 *> <a href="http://www.netlib.org/cgi-bin/netlibfiles.zip?format=zip&filename=/lapack/lapack_routine/dsbtrd.f">
13 *> [ZIP]</a>
14 *> <a href="http://www.netlib.org/cgi-bin/netlibfiles.txt?format=txt&filename=/lapack/lapack_routine/dsbtrd.f">
15 *> [TXT]</a>
16 *> \endhtmlonly
17 *
18 * Definition:
19 * ===========
20 *
21 * SUBROUTINE DSBTRD( VECT, UPLO, N, KD, AB, LDAB, D, E, Q, LDQ,
22 * WORK, INFO )
23 *
24 * .. Scalar Arguments ..
25 * CHARACTER UPLO, VECT
26 * INTEGER INFO, KD, LDAB, LDQ, N
27 * ..
28 * .. Array Arguments ..
29 * DOUBLE PRECISION AB( LDAB, * ), D( * ), E( * ), Q( LDQ, * ),
30 * \$ WORK( * )
31 * ..
32 *
33 *
34 *> \par Purpose:
35 * =============
36 *>
37 *> \verbatim
38 *>
39 *> DSBTRD reduces a real symmetric band matrix A to symmetric
40 *> tridiagonal form T by an orthogonal similarity transformation:
41 *> Q**T * A * Q = T.
42 *> \endverbatim
43 *
44 * Arguments:
45 * ==========
46 *
47 *> \param[in] VECT
48 *> \verbatim
49 *> VECT is CHARACTER*1
50 *> = 'N': do not form Q;
51 *> = 'V': form Q;
52 *> = 'U': update a matrix X, by forming X*Q.
53 *> \endverbatim
54 *>
55 *> \param[in] UPLO
56 *> \verbatim
57 *> UPLO is CHARACTER*1
58 *> = 'U': Upper triangle of A is stored;
59 *> = 'L': Lower triangle of A is stored.
60 *> \endverbatim
61 *>
62 *> \param[in] N
63 *> \verbatim
64 *> N is INTEGER
65 *> The order of the matrix A. N >= 0.
66 *> \endverbatim
67 *>
68 *> \param[in] KD
69 *> \verbatim
70 *> KD is INTEGER
71 *> The number of superdiagonals of the matrix A if UPLO = 'U',
72 *> or the number of subdiagonals if UPLO = 'L'. KD >= 0.
73 *> \endverbatim
74 *>
75 *> \param[in,out] AB
76 *> \verbatim
77 *> AB is DOUBLE PRECISION array, dimension (LDAB,N)
78 *> On entry, the upper or lower triangle of the symmetric band
79 *> matrix A, stored in the first KD+1 rows of the array. The
80 *> j-th column of A is stored in the j-th column of the array AB
81 *> as follows:
82 *> if UPLO = 'U', AB(kd+1+i-j,j) = A(i,j) for max(1,j-kd)<=i<=j;
83 *> if UPLO = 'L', AB(1+i-j,j) = A(i,j) for j<=i<=min(n,j+kd).
84 *> On exit, the diagonal elements of AB are overwritten by the
85 *> diagonal elements of the tridiagonal matrix T; if KD > 0, the
86 *> elements on the first superdiagonal (if UPLO = 'U') or the
87 *> first subdiagonal (if UPLO = 'L') are overwritten by the
88 *> off-diagonal elements of T; the rest of AB is overwritten by
89 *> values generated during the reduction.
90 *> \endverbatim
91 *>
92 *> \param[in] LDAB
93 *> \verbatim
94 *> LDAB is INTEGER
95 *> The leading dimension of the array AB. LDAB >= KD+1.
96 *> \endverbatim
97 *>
98 *> \param[out] D
99 *> \verbatim
100 *> D is DOUBLE PRECISION array, dimension (N)
101 *> The diagonal elements of the tridiagonal matrix T.
102 *> \endverbatim
103 *>
104 *> \param[out] E
105 *> \verbatim
106 *> E is DOUBLE PRECISION array, dimension (N-1)
107 *> The off-diagonal elements of the tridiagonal matrix T:
108 *> E(i) = T(i,i+1) if UPLO = 'U'; E(i) = T(i+1,i) if UPLO = 'L'.
109 *> \endverbatim
110 *>
111 *> \param[in,out] Q
112 *> \verbatim
113 *> Q is DOUBLE PRECISION array, dimension (LDQ,N)
114 *> On entry, if VECT = 'U', then Q must contain an N-by-N
115 *> matrix X; if VECT = 'N' or 'V', then Q need not be set.
116 *>
117 *> On exit:
118 *> if VECT = 'V', Q contains the N-by-N orthogonal matrix Q;
119 *> if VECT = 'U', Q contains the product X*Q;
120 *> if VECT = 'N', the array Q is not referenced.
121 *> \endverbatim
122 *>
123 *> \param[in] LDQ
124 *> \verbatim
125 *> LDQ is INTEGER
126 *> The leading dimension of the array Q.
127 *> LDQ >= 1, and LDQ >= N if VECT = 'V' or 'U'.
128 *> \endverbatim
129 *>
130 *> \param[out] WORK
131 *> \verbatim
132 *> WORK is DOUBLE PRECISION array, dimension (N)
133 *> \endverbatim
134 *>
135 *> \param[out] INFO
136 *> \verbatim
137 *> INFO is INTEGER
138 *> = 0: successful exit
139 *> < 0: if INFO = -i, the i-th argument had an illegal value
140 *> \endverbatim
141 *
142 * Authors:
143 * ========
144 *
145 *> \author Univ. of Tennessee
146 *> \author Univ. of California Berkeley
147 *> \author Univ. of Colorado Denver
148 *> \author NAG Ltd.
149 *
150 *> \date November 2011
151 *
152 *> \ingroup doubleOTHERcomputational
153 *
154 *> \par Further Details:
155 * =====================
156 *>
157 *> \verbatim
158 *>
159 *> Modified by Linda Kaufman, Bell Labs.
160 *> \endverbatim
161 *>
162 * =====================================================================
163 SUBROUTINE dsbtrd( VECT, UPLO, N, KD, AB, LDAB, D, E, Q, LDQ,
164 \$ work, info )
165 *
166 * -- LAPACK computational routine (version 3.4.0) --
167 * -- LAPACK is a software package provided by Univ. of Tennessee, --
168 * -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
169 * November 2011
170 *
171 * .. Scalar Arguments ..
172 CHARACTER UPLO, VECT
173 INTEGER INFO, KD, LDAB, LDQ, N
174 * ..
175 * .. Array Arguments ..
176 DOUBLE PRECISION AB( ldab, * ), D( * ), E( * ), Q( ldq, * ),
177 \$ work( * )
178 * ..
179 *
180 * =====================================================================
181 *
182 * .. Parameters ..
183 DOUBLE PRECISION ZERO, ONE
184 parameter ( zero = 0.0d+0, one = 1.0d+0 )
185 * ..
186 * .. Local Scalars ..
187 LOGICAL INITQ, UPPER, WANTQ
188 INTEGER I, I2, IBL, INCA, INCX, IQAEND, IQB, IQEND, J,
189 \$ j1, j1end, j1inc, j2, jend, jin, jinc, k, kd1,
190 \$ kdm1, kdn, l, last, lend, nq, nr, nrt
191 DOUBLE PRECISION TEMP
192 * ..
193 * .. External Subroutines ..
194 EXTERNAL dlar2v, dlargv, dlartg, dlartv, dlaset, drot,
195 \$ xerbla
196 * ..
197 * .. Intrinsic Functions ..
198 INTRINSIC max, min
199 * ..
200 * .. External Functions ..
201 LOGICAL LSAME
202 EXTERNAL lsame
203 * ..
204 * .. Executable Statements ..
205 *
206 * Test the input parameters
207 *
208 initq = lsame( vect, 'V' )
209 wantq = initq .OR. lsame( vect, 'U' )
210 upper = lsame( uplo, 'U' )
211 kd1 = kd + 1
212 kdm1 = kd - 1
213 incx = ldab - 1
214 iqend = 1
215 *
216 info = 0
217 IF( .NOT.wantq .AND. .NOT.lsame( vect, 'N' ) ) THEN
218 info = -1
219 ELSE IF( .NOT.upper .AND. .NOT.lsame( uplo, 'L' ) ) THEN
220 info = -2
221 ELSE IF( n.LT.0 ) THEN
222 info = -3
223 ELSE IF( kd.LT.0 ) THEN
224 info = -4
225 ELSE IF( ldab.LT.kd1 ) THEN
226 info = -6
227 ELSE IF( ldq.LT.max( 1, n ) .AND. wantq ) THEN
228 info = -10
229 END IF
230 IF( info.NE.0 ) THEN
231 CALL xerbla( 'DSBTRD', -info )
232 RETURN
233 END IF
234 *
235 * Quick return if possible
236 *
237 IF( n.EQ.0 )
238 \$ RETURN
239 *
240 * Initialize Q to the unit matrix, if needed
241 *
242 IF( initq )
243 \$ CALL dlaset( 'Full', n, n, zero, one, q, ldq )
244 *
245 * Wherever possible, plane rotations are generated and applied in
246 * vector operations of length NR over the index set J1:J2:KD1.
247 *
248 * The cosines and sines of the plane rotations are stored in the
249 * arrays D and WORK.
250 *
251 inca = kd1*ldab
252 kdn = min( n-1, kd )
253 IF( upper ) THEN
254 *
255 IF( kd.GT.1 ) THEN
256 *
257 * Reduce to tridiagonal form, working with upper triangle
258 *
259 nr = 0
260 j1 = kdn + 2
261 j2 = 1
262 *
263 DO 90 i = 1, n - 2
264 *
265 * Reduce i-th row of matrix to tridiagonal form
266 *
267 DO 80 k = kdn + 1, 2, -1
268 j1 = j1 + kdn
269 j2 = j2 + kdn
270 *
271 IF( nr.GT.0 ) THEN
272 *
273 * generate plane rotations to annihilate nonzero
274 * elements which have been created outside the band
275 *
276 CALL dlargv( nr, ab( 1, j1-1 ), inca, work( j1 ),
277 \$ kd1, d( j1 ), kd1 )
278 *
279 * apply rotations from the right
280 *
281 *
282 * Dependent on the the number of diagonals either
283 * DLARTV or DROT is used
284 *
285 IF( nr.GE.2*kd-1 ) THEN
286 DO 10 l = 1, kd - 1
287 CALL dlartv( nr, ab( l+1, j1-1 ), inca,
288 \$ ab( l, j1 ), inca, d( j1 ),
289 \$ work( j1 ), kd1 )
290 10 CONTINUE
291 *
292 ELSE
293 jend = j1 + ( nr-1 )*kd1
294 DO 20 jinc = j1, jend, kd1
295 CALL drot( kdm1, ab( 2, jinc-1 ), 1,
296 \$ ab( 1, jinc ), 1, d( jinc ),
297 \$ work( jinc ) )
298 20 CONTINUE
299 END IF
300 END IF
301 *
302 *
303 IF( k.GT.2 ) THEN
304 IF( k.LE.n-i+1 ) THEN
305 *
306 * generate plane rotation to annihilate a(i,i+k-1)
307 * within the band
308 *
309 CALL dlartg( ab( kd-k+3, i+k-2 ),
310 \$ ab( kd-k+2, i+k-1 ), d( i+k-1 ),
311 \$ work( i+k-1 ), temp )
312 ab( kd-k+3, i+k-2 ) = temp
313 *
314 * apply rotation from the right
315 *
316 CALL drot( k-3, ab( kd-k+4, i+k-2 ), 1,
317 \$ ab( kd-k+3, i+k-1 ), 1, d( i+k-1 ),
318 \$ work( i+k-1 ) )
319 END IF
320 nr = nr + 1
321 j1 = j1 - kdn - 1
322 END IF
323 *
324 * apply plane rotations from both sides to diagonal
325 * blocks
326 *
327 IF( nr.GT.0 )
328 \$ CALL dlar2v( nr, ab( kd1, j1-1 ), ab( kd1, j1 ),
329 \$ ab( kd, j1 ), inca, d( j1 ),
330 \$ work( j1 ), kd1 )
331 *
332 * apply plane rotations from the left
333 *
334 IF( nr.GT.0 ) THEN
335 IF( 2*kd-1.LT.nr ) THEN
336 *
337 * Dependent on the the number of diagonals either
338 * DLARTV or DROT is used
339 *
340 DO 30 l = 1, kd - 1
341 IF( j2+l.GT.n ) THEN
342 nrt = nr - 1
343 ELSE
344 nrt = nr
345 END IF
346 IF( nrt.GT.0 )
347 \$ CALL dlartv( nrt, ab( kd-l, j1+l ), inca,
348 \$ ab( kd-l+1, j1+l ), inca,
349 \$ d( j1 ), work( j1 ), kd1 )
350 30 CONTINUE
351 ELSE
352 j1end = j1 + kd1*( nr-2 )
353 IF( j1end.GE.j1 ) THEN
354 DO 40 jin = j1, j1end, kd1
355 CALL drot( kd-1, ab( kd-1, jin+1 ), incx,
356 \$ ab( kd, jin+1 ), incx,
357 \$ d( jin ), work( jin ) )
358 40 CONTINUE
359 END IF
360 lend = min( kdm1, n-j2 )
361 last = j1end + kd1
362 IF( lend.GT.0 )
363 \$ CALL drot( lend, ab( kd-1, last+1 ), incx,
364 \$ ab( kd, last+1 ), incx, d( last ),
365 \$ work( last ) )
366 END IF
367 END IF
368 *
369 IF( wantq ) THEN
370 *
371 * accumulate product of plane rotations in Q
372 *
373 IF( initq ) THEN
374 *
375 * take advantage of the fact that Q was
376 * initially the Identity matrix
377 *
378 iqend = max( iqend, j2 )
379 i2 = max( 0, k-3 )
380 iqaend = 1 + i*kd
381 IF( k.EQ.2 )
382 \$ iqaend = iqaend + kd
383 iqaend = min( iqaend, iqend )
384 DO 50 j = j1, j2, kd1
385 ibl = i - i2 / kdm1
386 i2 = i2 + 1
387 iqb = max( 1, j-ibl )
388 nq = 1 + iqaend - iqb
389 iqaend = min( iqaend+kd, iqend )
390 CALL drot( nq, q( iqb, j-1 ), 1, q( iqb, j ),
391 \$ 1, d( j ), work( j ) )
392 50 CONTINUE
393 ELSE
394 *
395 DO 60 j = j1, j2, kd1
396 CALL drot( n, q( 1, j-1 ), 1, q( 1, j ), 1,
397 \$ d( j ), work( j ) )
398 60 CONTINUE
399 END IF
400 *
401 END IF
402 *
403 IF( j2+kdn.GT.n ) THEN
404 *
405 * adjust J2 to keep within the bounds of the matrix
406 *
407 nr = nr - 1
408 j2 = j2 - kdn - 1
409 END IF
410 *
411 DO 70 j = j1, j2, kd1
412 *
413 * create nonzero element a(j-1,j+kd) outside the band
414 * and store it in WORK
415 *
416 work( j+kd ) = work( j )*ab( 1, j+kd )
417 ab( 1, j+kd ) = d( j )*ab( 1, j+kd )
418 70 CONTINUE
419 80 CONTINUE
420 90 CONTINUE
421 END IF
422 *
423 IF( kd.GT.0 ) THEN
424 *
425 * copy off-diagonal elements to E
426 *
427 DO 100 i = 1, n - 1
428 e( i ) = ab( kd, i+1 )
429 100 CONTINUE
430 ELSE
431 *
432 * set E to zero if original matrix was diagonal
433 *
434 DO 110 i = 1, n - 1
435 e( i ) = zero
436 110 CONTINUE
437 END IF
438 *
439 * copy diagonal elements to D
440 *
441 DO 120 i = 1, n
442 d( i ) = ab( kd1, i )
443 120 CONTINUE
444 *
445 ELSE
446 *
447 IF( kd.GT.1 ) THEN
448 *
449 * Reduce to tridiagonal form, working with lower triangle
450 *
451 nr = 0
452 j1 = kdn + 2
453 j2 = 1
454 *
455 DO 210 i = 1, n - 2
456 *
457 * Reduce i-th column of matrix to tridiagonal form
458 *
459 DO 200 k = kdn + 1, 2, -1
460 j1 = j1 + kdn
461 j2 = j2 + kdn
462 *
463 IF( nr.GT.0 ) THEN
464 *
465 * generate plane rotations to annihilate nonzero
466 * elements which have been created outside the band
467 *
468 CALL dlargv( nr, ab( kd1, j1-kd1 ), inca,
469 \$ work( j1 ), kd1, d( j1 ), kd1 )
470 *
471 * apply plane rotations from one side
472 *
473 *
474 * Dependent on the the number of diagonals either
475 * DLARTV or DROT is used
476 *
477 IF( nr.GT.2*kd-1 ) THEN
478 DO 130 l = 1, kd - 1
479 CALL dlartv( nr, ab( kd1-l, j1-kd1+l ), inca,
480 \$ ab( kd1-l+1, j1-kd1+l ), inca,
481 \$ d( j1 ), work( j1 ), kd1 )
482 130 CONTINUE
483 ELSE
484 jend = j1 + kd1*( nr-1 )
485 DO 140 jinc = j1, jend, kd1
486 CALL drot( kdm1, ab( kd, jinc-kd ), incx,
487 \$ ab( kd1, jinc-kd ), incx,
488 \$ d( jinc ), work( jinc ) )
489 140 CONTINUE
490 END IF
491 *
492 END IF
493 *
494 IF( k.GT.2 ) THEN
495 IF( k.LE.n-i+1 ) THEN
496 *
497 * generate plane rotation to annihilate a(i+k-1,i)
498 * within the band
499 *
500 CALL dlartg( ab( k-1, i ), ab( k, i ),
501 \$ d( i+k-1 ), work( i+k-1 ), temp )
502 ab( k-1, i ) = temp
503 *
504 * apply rotation from the left
505 *
506 CALL drot( k-3, ab( k-2, i+1 ), ldab-1,
507 \$ ab( k-1, i+1 ), ldab-1, d( i+k-1 ),
508 \$ work( i+k-1 ) )
509 END IF
510 nr = nr + 1
511 j1 = j1 - kdn - 1
512 END IF
513 *
514 * apply plane rotations from both sides to diagonal
515 * blocks
516 *
517 IF( nr.GT.0 )
518 \$ CALL dlar2v( nr, ab( 1, j1-1 ), ab( 1, j1 ),
519 \$ ab( 2, j1-1 ), inca, d( j1 ),
520 \$ work( j1 ), kd1 )
521 *
522 * apply plane rotations from the right
523 *
524 *
525 * Dependent on the the number of diagonals either
526 * DLARTV or DROT is used
527 *
528 IF( nr.GT.0 ) THEN
529 IF( nr.GT.2*kd-1 ) THEN
530 DO 150 l = 1, kd - 1
531 IF( j2+l.GT.n ) THEN
532 nrt = nr - 1
533 ELSE
534 nrt = nr
535 END IF
536 IF( nrt.GT.0 )
537 \$ CALL dlartv( nrt, ab( l+2, j1-1 ), inca,
538 \$ ab( l+1, j1 ), inca, d( j1 ),
539 \$ work( j1 ), kd1 )
540 150 CONTINUE
541 ELSE
542 j1end = j1 + kd1*( nr-2 )
543 IF( j1end.GE.j1 ) THEN
544 DO 160 j1inc = j1, j1end, kd1
545 CALL drot( kdm1, ab( 3, j1inc-1 ), 1,
546 \$ ab( 2, j1inc ), 1, d( j1inc ),
547 \$ work( j1inc ) )
548 160 CONTINUE
549 END IF
550 lend = min( kdm1, n-j2 )
551 last = j1end + kd1
552 IF( lend.GT.0 )
553 \$ CALL drot( lend, ab( 3, last-1 ), 1,
554 \$ ab( 2, last ), 1, d( last ),
555 \$ work( last ) )
556 END IF
557 END IF
558 *
559 *
560 *
561 IF( wantq ) THEN
562 *
563 * accumulate product of plane rotations in Q
564 *
565 IF( initq ) THEN
566 *
567 * take advantage of the fact that Q was
568 * initially the Identity matrix
569 *
570 iqend = max( iqend, j2 )
571 i2 = max( 0, k-3 )
572 iqaend = 1 + i*kd
573 IF( k.EQ.2 )
574 \$ iqaend = iqaend + kd
575 iqaend = min( iqaend, iqend )
576 DO 170 j = j1, j2, kd1
577 ibl = i - i2 / kdm1
578 i2 = i2 + 1
579 iqb = max( 1, j-ibl )
580 nq = 1 + iqaend - iqb
581 iqaend = min( iqaend+kd, iqend )
582 CALL drot( nq, q( iqb, j-1 ), 1, q( iqb, j ),
583 \$ 1, d( j ), work( j ) )
584 170 CONTINUE
585 ELSE
586 *
587 DO 180 j = j1, j2, kd1
588 CALL drot( n, q( 1, j-1 ), 1, q( 1, j ), 1,
589 \$ d( j ), work( j ) )
590 180 CONTINUE
591 END IF
592 END IF
593 *
594 IF( j2+kdn.GT.n ) THEN
595 *
596 * adjust J2 to keep within the bounds of the matrix
597 *
598 nr = nr - 1
599 j2 = j2 - kdn - 1
600 END IF
601 *
602 DO 190 j = j1, j2, kd1
603 *
604 * create nonzero element a(j+kd,j-1) outside the
605 * band and store it in WORK
606 *
607 work( j+kd ) = work( j )*ab( kd1, j )
608 ab( kd1, j ) = d( j )*ab( kd1, j )
609 190 CONTINUE
610 200 CONTINUE
611 210 CONTINUE
612 END IF
613 *
614 IF( kd.GT.0 ) THEN
615 *
616 * copy off-diagonal elements to E
617 *
618 DO 220 i = 1, n - 1
619 e( i ) = ab( 2, i )
620 220 CONTINUE
621 ELSE
622 *
623 * set E to zero if original matrix was diagonal
624 *
625 DO 230 i = 1, n - 1
626 e( i ) = zero
627 230 CONTINUE
628 END IF
629 *
630 * copy diagonal elements to D
631 *
632 DO 240 i = 1, n
633 d( i ) = ab( 1, i )
634 240 CONTINUE
635 END IF
636 *
637 RETURN
638 *
639 * End of DSBTRD
640 *
641 END
subroutine dlaset(UPLO, M, N, ALPHA, BETA, A, LDA)
DLASET initializes the off-diagonal elements and the diagonal elements of a matrix to given values...
Definition: dlaset.f:112
subroutine dsbtrd(VECT, UPLO, N, KD, AB, LDAB, D, E, Q, LDQ, WORK, INFO)
DSBTRD
Definition: dsbtrd.f:165
subroutine drot(N, DX, INCX, DY, INCY, C, S)
DROT
Definition: drot.f:53
subroutine dlargv(N, X, INCX, Y, INCY, C, INCC)
DLARGV generates a vector of plane rotations with real cosines and real sines.
Definition: dlargv.f:106
subroutine dlartv(N, X, INCX, Y, INCY, C, S, INCC)
DLARTV applies a vector of plane rotations with real cosines and real sines to the elements of a pair...
Definition: dlartv.f:110
subroutine xerbla(SRNAME, INFO)
XERBLA
Definition: xerbla.f:62
subroutine dlartg(F, G, CS, SN, R)
DLARTG generates a plane rotation with real cosine and real sine.
Definition: dlartg.f:99
subroutine dlar2v(N, X, Y, Z, INCX, C, S, INCC)
DLAR2V applies a vector of plane rotations with real cosines and real sines from both sides to a sequ...
Definition: dlar2v.f:112 | 6,931 | 17,266 | {"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} | 2.5625 | 3 | CC-MAIN-2024-18 | latest | en | 0.26834 |
https://www.ademcetinkaya.com/2023/03/mwy-midway-limited.html | 1,680,400,039,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950373.88/warc/CC-MAIN-20230402012805-20230402042805-00050.warc.gz | 693,094,768 | 61,315 | Outlook: MIDWAY LIMITED is assigned short-term Ba1 & long-term Ba1 estimated rating.
Time series to forecast n: 11 Mar 2023 for (n+3 month)
Methodology : Modular Neural Network (Market Direction Analysis)
## Abstract
MIDWAY LIMITED prediction model is evaluated with Modular Neural Network (Market Direction Analysis) and Statistical Hypothesis Testing1,2,3,4 and it is concluded that the MWY stock is predictable in the short/long term. According to price forecasts for (n+3 month) period, the dominant strategy among neural network is: Buy
## Key Points
1. Stock Forecast Based On a Predictive Algorithm
2. Can machine learning predict?
3. Fundemental Analysis with Algorithmic Trading
## MWY Target Price Prediction Modeling Methodology
We consider MIDWAY LIMITED Decision Process with Modular Neural Network (Market Direction Analysis) where A is the set of discrete actions of MWY stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
F(Statistical Hypothesis Testing)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Modular Neural Network (Market Direction Analysis)) X S(n):→ (n+3 month) $\stackrel{\to }{R}=\left({r}_{1},{r}_{2},{r}_{3}\right)$
n:Time series to forecast
p:Price signals of MWY stock
j:Nash equilibria (Neural Network)
k:Dominated move
a:Best response for target price
For further technical information as per how our model work we invite you to visit the article below:
How do AC Investment Research machine learning (predictive) algorithms actually work?
## MWY Stock Forecast (Buy or Sell) for (n+3 month)
Sample Set: Neural Network
Stock/Index: MWY MIDWAY LIMITED
Time series to forecast n: 11 Mar 2023 for (n+3 month)
According to price forecasts for (n+3 month) period, the dominant strategy among neural network is: Buy
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Grey to Black): *Technical Analysis%
## IFRS Reconciliation Adjustments for MIDWAY LIMITED
1. A similar example of a non-financial item is a specific type of crude oil from a particular oil field that is priced off the relevant benchmark crude oil. If an entity sells that crude oil under a contract using a contractual pricing formula that sets the price per barrel at the benchmark crude oil price minus CU10 with a floor of CU15, the entity can designate as the hedged item the entire cash flow variability under the sales contract that is attributable to the change in the benchmark crude oil price. However, the entity cannot designate a component that is equal to the full change in the benchmark crude oil price. Hence, as long as the forward price (for each delivery) does not fall below CU25, the hedged item has the same cash flow variability as a crude oil sale at the benchmark crude oil price (or with a positive spread). However, if the forward price for any delivery falls below CU25, the hedged item has a lower cash flow variability than a crude oil sale at the benchmark crude oil price (or with a positive spread).
2. When designating a risk component as a hedged item, the hedge accounting requirements apply to that risk component in the same way as they apply to other hedged items that are not risk components. For example, the qualifying criteria apply, including that the hedging relationship must meet the hedge effectiveness requirements, and any hedge ineffectiveness must be measured and recognised.
3. When assessing a modified time value of money element, an entity must consider factors that could affect future contractual cash flows. For example, if an entity is assessing a bond with a five-year term and the variable interest rate is reset every six months to a five-year rate, the entity cannot conclude that the contractual cash flows are solely payments of principal and interest on the principal amount outstanding simply because the interest rate curve at the time of the assessment is such that the difference between a five-year interest rate and a six-month interest rate is not significant. Instead, the entity must also consider whether the relationship between the five-year interest rate and the six-month interest rate could change over the life of the instrument such that the contractual (undiscounted) cash flows over the life of the instrument could be significantly different from the (undiscounted) benchmark cash flows. However, an entity must consider only reasonably possible scenarios instead of every possible scenario. If an entity concludes that the contractual (undiscounted) cash flows could be significantly different from the (undiscounted) benchmark cash flows, the financial asset does not meet the condition in paragraphs 4.1.2(b) and 4.1.2A(b) and therefore cannot be measured at amortised cost or fair value through other comprehensive income.
4. The rebuttable presumption in paragraph 5.5.11 is not an absolute indicator that lifetime expected credit losses should be recognised, but is presumed to be the latest point at which lifetime expected credit losses should be recognised even when using forward-looking information (including macroeconomic factors on a portfolio level).
*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.
## Conclusions
MIDWAY LIMITED is assigned short-term Ba1 & long-term Ba1 estimated rating. MIDWAY LIMITED prediction model is evaluated with Modular Neural Network (Market Direction Analysis) and Statistical Hypothesis Testing1,2,3,4 and it is concluded that the MWY stock is predictable in the short/long term. According to price forecasts for (n+3 month) period, the dominant strategy among neural network is: Buy
### MWY MIDWAY LIMITED Financial Analysis*
Rating Short-Term Long-Term Senior
Outlook*Ba1Ba1
Income StatementB1B3
Balance SheetCaa2B2
Leverage RatiosCBaa2
Cash FlowB1Caa2
Rates of Return and ProfitabilityB2Caa2
*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?
### Prediction Confidence Score
Trust metric by Neural Network: 73 out of 100 with 533 signals.
## References
1. Li L, Chen S, Kleban J, Gupta A. 2014. Counterfactual estimation and optimization of click metrics for search engines: a case study. In Proceedings of the 24th International Conference on the World Wide Web, pp. 929–34. New York: ACM
2. Hartigan JA, Wong MA. 1979. Algorithm as 136: a k-means clustering algorithm. J. R. Stat. Soc. Ser. C 28:100–8
3. Clements, M. P. D. F. Hendry (1995), "Forecasting in cointegrated systems," Journal of Applied Econometrics, 10, 127–146.
4. Pennington J, Socher R, Manning CD. 2014. GloVe: global vectors for word representation. In Proceedings of the 2014 Conference on Empirical Methods on Natural Language Processing, pp. 1532–43. New York: Assoc. Comput. Linguist.
5. Imai K, Ratkovic M. 2013. Estimating treatment effect heterogeneity in randomized program evaluation. Ann. Appl. Stat. 7:443–70
6. Bierens HJ. 1987. Kernel estimators of regression functions. In Advances in Econometrics: Fifth World Congress, Vol. 1, ed. TF Bewley, pp. 99–144. Cambridge, UK: Cambridge Univ. Press
7. L. Prashanth and M. Ghavamzadeh. Actor-critic algorithms for risk-sensitive MDPs. In Proceedings of Advances in Neural Information Processing Systems 26, pages 252–260, 2013.
Frequently Asked QuestionsQ: What is the prediction methodology for MWY stock?
A: MWY stock prediction methodology: We evaluate the prediction models Modular Neural Network (Market Direction Analysis) and Statistical Hypothesis Testing
Q: Is MWY stock a buy or sell?
A: The dominant strategy among neural network is to Buy MWY Stock.
Q: Is MIDWAY LIMITED stock a good investment?
A: The consensus rating for MIDWAY LIMITED is Buy and is assigned short-term Ba1 & long-term Ba1 estimated rating.
Q: What is the consensus rating of MWY stock?
A: The consensus rating for MWY is Buy.
Q: What is the prediction period for MWY stock?
A: The prediction period for MWY is (n+3 month) | 2,085 | 8,946 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "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} | 2.609375 | 3 | CC-MAIN-2023-14 | latest | en | 0.796947 |
https://en.wikipedia.org/wiki/Color_histogram | 1,721,234,938,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514789.19/warc/CC-MAIN-20240717151625-20240717181625-00684.warc.gz | 204,987,924 | 22,760 | # Color histogram
In image processing and photography, a color histogram is a representation of the distribution of colors in an image. For digital images, a color histogram represents the number of pixels that have colors in each of a fixed list of color ranges, that span the image's color space, the set of all possible colors.
The color histogram can be built for any kind of color space, although the term is more often used for three-dimensional spaces such as RGB or HSV. For monochromatic images, the term intensity histogram may be used instead. For multi-spectral images, where each pixel is represented by an arbitrary number of measurements (for example, beyond the three measurements in RGB), the color histogram is N-dimensional, with N being the number of measurements taken. Each measurement has its own wavelength range of the light spectrum, some of which may be outside the visible spectrum.
If the set of possible color values is sufficiently small, each of those colors may be placed on a range by itself; then the histogram is merely the count of pixels that have each possible color. Most often, the space is divided into an appropriate number of ranges, often arranged as a regular grid, each containing many similar color values. The color histogram may also be represented and displayed as a smooth function defined over the color space that approximates the pixel counts.
Like other kinds of histograms, the color histogram is a statistic that can be viewed as an approximation of an underlying continuous distribution of color values.
## Overview
Color histograms are flexible constructs that can be built from images in various color spaces, whether RGB, rg chromaticity or any other color space of any dimension. A histogram of an image is produced first by discretization of the colors in the image into a number of bins, and counting the number of image pixels in each bin. For example, a Red–Blue chromaticity histogram can be formed by first normalizing color pixel values by dividing RGB values by R+G+B, then quantizing the normalized R and B coordinates into N bins each. A two-dimensional histogram of Red-Blue chromaticity divided into four bins (N=4) might yield a histogram that looks like this table:
red 0-63 64-127 128-191 192-255 blue 0-63 43 78 18 0 64-127 45 67 33 2 128-191 127 58 25 8 192-255 140 47 47 13
A histogram can be N-dimensional. Although harder to display, a three-dimensional color histogram for the above example could be thought of as four separate Red-Blue histograms, where each of the four histograms contains the Red-Blue values for a bin of green (0-63, 64-127, 128-191, and 192-255).
The histogram provides a compact summarization of the distribution of data in an image. The color histogram of an image is relatively invariant with translation and rotation about the viewing axis, and varies only slowly with the angle of view.[1] By comparing histogram signatures of two images and matching the color content of one image with the other, the color histogram is particularly well suited for the problem of recognizing an object of unknown position and rotation within a scene. Importantly, translation of an RGB image into the illumination invariant rg-chromaticity space allows the histogram to operate well in varying light levels.
1. What is a histogram?
A histogram is a graphical representation of the number of pixels in an image. In a more simple way to explain, a histogram is a bar graph, whose X-axis represents the tonal scale(black at the left and white at the right), and Y-axis represents the number of pixels in an image in a certain area of the tonal scale. For example, the graph of a luminance histogram shows the number of pixels for each brightness level(from black to white), and when there are more pixels, the peak at the certain luminance level is higher.
2. What is a color histogram?
A color histogram of an image represents the distribution of the composition of colors in the image. It shows different types of colors appeared and the number of pixels in each type of the colors appeared. The relation between a color histogram and a luminance histogram is that a color histogram can be also expressed as “Three Luminance Histograms”, each of which shows the brightness distribution of each individual Red/Green/Blue color channel.
## Characteristics of a color histogram
A color histogram focuses only on the proportion of the number of different types of colors, regardless of the spatial location of the colors. The values of a color histogram are from statistics. They show the statistical distribution of colors and the essential tone of an image.
In general, as the color distributions of the foreground and background in an image are different, there might be a bimodal distribution in the histogram.
For the luminance histogram alone, there is no perfect histogram and in general, the histogram can tell whether it is over-exposure or not, but there are times when you might think the image is over exposed by viewing the histogram; however, in reality it is not.
## Principles of the formation of a color histogram
The formation of a color histogram is rather simple. From the definition above, we can simply count the number of pixels for each 256 scales in each of the 3 RGB channel, and plot them on 3 individual bar graphs.
In general, a color histogram is based on a certain color space, such as RGB or HSV. When we compute the pixels of different colors in an image, if the color space is large, then we can first divide the color space into certain numbers of small intervals. Each of the intervals is called a bin. This process is called color quantization. Then, by counting the number of pixels in each of the bins, we get the color histogram of the image.
The concrete steps of the principles can be viewed in Example 1.
## Examples
### Example 1
Given the following image of a cat (an original version and a version that has been reduced to 256 colors for easy histogram purposes), the following data represents a color histogram in the RGB color space, using four bins.
Bin 0 corresponds to intensities 0-63
Bin 1 is 64-127
Bin 2 is 128-191 and Bin 3 is 192-255.
Red Green Blue Pixel Count
Bin 0 Bin 0 Bin 0 7414
Bin 0 Bin 0 Bin 1 230
Bin 0 Bin 0 Bin 2 0
Bin 0 Bin 0 Bin 3 0
Bin 0 Bin 1 Bin 0 8
Bin 0 Bin 1 Bin 1 372
Bin 0 Bin 1 Bin 2 88
Bin 0 Bin 1 Bin 3 0
Bin 0 Bin 2 Bin 0 0
Bin 0 Bin 2 Bin 1 0
Bin 0 Bin 2 Bin 2 10
Bin 0 Bin 2 Bin 3 1
Bin 0 Bin 3 Bin 0 0
Bin 0 Bin 3 Bin 1 0
Bin 0 Bin 3 Bin 2 0
Bin 0 Bin 3 Bin 3 0
Bin 1 Bin 0 Bin 0 891
Bin 1 Bin 0 Bin 1 13
Bin 1 Bin 0 Bin 2 0
Bin 1 Bin 0 Bin 3 0
Bin 1 Bin 1 Bin 0 592
Bin 1 Bin 1 Bin 1 3462
Bin 1 Bin 1 Bin 2 355
Bin 1 Bin 1 Bin 3 0
Bin 1 Bin 2 Bin 0 0
Bin 1 Bin 2 Bin 1 101
Bin 1 Bin 2 Bin 2 882
Bin 1 Bin 2 Bin 3 16
Bin 1 Bin 3 Bin 0 0
Bin 1 Bin 3 Bin 1 0
Bin 1 Bin 3 Bin 2 0
Bin 1 Bin 3 Bin 3 0
Bin 2 Bin 0 Bin 0 1146
Bin 2 Bin 0 Bin 1 0
Bin 2 Bin 0 Bin 2 0
Bin 2 Bin 0 Bin 3 0
Bin 2 Bin 1 Bin 0 2552
Bin 2 Bin 1 Bin 1 9040
Bin 2 Bin 1 Bin 2 47
Bin 2 Bin 1 Bin 3 0
Bin 2 Bin 2 Bin 0 0
Bin 2 Bin 2 Bin 1 8808
Bin 2 Bin 2 Bin 2 53110
Bin 2 Bin 2 Bin 3 11053
Bin 2 Bin 3 Bin 0 0
Bin 2 Bin 3 Bin 1 0
Bin 2 Bin 3 Bin 2 170
Bin 2 Bin 3 Bin 3 17533
Bin 3 Bin 0 Bin 0 11
Bin 3 Bin 0 Bin 1 0
Bin 3 Bin 0 Bin 2 0
Bin 3 Bin 0 Bin 3 0
Bin 3 Bin 1 Bin 0 856
Bin 3 Bin 1 Bin 1 1376
Bin 3 Bin 1 Bin 2 0
Bin 3 Bin 1 Bin 3 0
Bin 3 Bin 2 Bin 0 0
Bin 3 Bin 2 Bin 1 3650
Bin 3 Bin 2 Bin 2 6260
Bin 3 Bin 2 Bin 3 109
Bin 3 Bin 3 Bin 0 0
Bin 3 Bin 3 Bin 1 0
Bin 3 Bin 3 Bin 2 3415
Bin 3 Bin 3 Bin 3 53929
### Example 2
Application in camera:
Nowadays, some cameras have the ability to show the 3 color histograms when we take photos.
We can examine clips (spikes on either the black or white side of the scale) in each of the 3 RGB color histograms. If we find one or more clipping on a channel of the 3 RGB channels, then this would result in a loss of detail for that color.
To illustrate this, consider this example:
1. We know that each of the three R, G, B channels has a range of values from 0-255 (8 bit). So consider a photo that has a luminance range of 0-255.
2. Assume the photo we take is made of 4 blocks that are adjacent to each other and we set the luminance scale for each of the 4 blocks of original photo to be 10, 100, 205, 245. Thus, the image looks like the topmost figure on the right.
3. Then, we overexpose the photo a little, say, the luminance scale of each block is increased by 10. Thus, the luminance scale for each of the 4 blocks of new photo is 20, 110, 215, 255. Then, the image looks like the second figure on the right.
There is not much difference between both figures, all we can see is that the whole image becomes brighter (the contrast for each of the blocks remain the same).
4. Now, we overexpose the original photo again, this time the luminance scale of each block is increased by 50. Thus, the luminance scale for each of the 4 blocks of the new photo is 60, 150, 255, 255. The new image now looks like the third figure on the right.
Note that the scale for the last block is 255 instead of 295, for 255 is the top scale and thus the last block has clipped! When this happens, we lose the contrast of the last 2 blocks, and thus, we cannot recover the image no matter how we adjust it.
To conclude, when taking photos with a camera that displays histograms, always keep the brightest tone in the image below the largest scale 255 on the histogram in order to avoid losing details.
## Drawbacks and other approaches
The main drawback of histograms for classification is that the representation is dependent on the color of the object being studied, ignoring its shape and texture. Color histograms can potentially be identical for two images with different object content which happens to share color information. Conversely, without spatial or shape information, similar objects of different color may be indistinguishable based solely on color histogram comparisons. There is no way to distinguish a red and white cup from a red and white plate. Put it another way: histogram-based algorithms have no concept of a generic 'cup', and a model of a red and white cup is no use when given an otherwise identical blue and white cup. Another problem is that color histograms have high sensitivity to noisy interference such as lighting intensity changes and quantization errors. High dimensionality (bins) color histograms are also another issue. Some color histogram feature spaces often occupy more than one hundred dimensions.[2]
Some of the proposed solutions have been color histogram intersection, color constant indexing, cumulative color histogram, quadratic distance, and color correlograms. Although there are drawbacks of using histograms for indexing and classification, using color in a real-time system has several advantages. One is that color information is faster to compute compared to other invariants. It has been shown in some cases that color can be an efficient method for identifying objects of known location and appearance.
Further research into the relationship between color histogram data to the physical properties of the objects in an image has shown they can represent not only object color and illumination but relate to surface roughness and image geometry and provide an improved estimate of illumination and object color.[3]
Usually, Euclidean distance, histogram intersection, or cosine or quadratic distances are used for the calculation of image similarity ratings.[4] Any of these values do not reflect the similarity rate of two images in itself; it is useful only when used in comparison to other similar values. This is the reason that all the practical implementations of content-based image retrieval must complete computation of all images from the database, and is the main disadvantage of these implementations.
Another approach to representative color image content is two-dimensional color histogram. A two-dimensional color histogram considers the relation between the pixel pair colors (not only the lighting component).[5] A two-dimensional color histogram is a two-dimensional array. The size of each dimension is the number of colors that were used in the phase of color quantization. These arrays are treated as matrices, each element of which stores a normalized count of pixel pairs, with each color corresponding to the index of an element in each pixel neighborhood. For comparison of two-dimensional color histograms it is suggested calculating their correlation, because constructed as described above, is a random vector (in other words, a multi-dimensional random value). While creating a set of final images, the images should be arranged in decreasing order of the correlation coefficient.
The correlation coefficient may also be used for color histogram comparison. Retrieval results with correlation coefficient are better than with other metrics.[6]
## Intensity histogram of continuous data
The idea of an intensity histogram can be generalized to continuous data, say audio signals represented by real functions or images represented by functions with two-dimensional domain.
Let ${\displaystyle f\in L^{1}(\mathbb {R} ^{n})}$ (see Lebesgue space), then the cumulative histogram operator ${\displaystyle H}$ can be defined by:
${\displaystyle H(f)(y)=\mu \{x:f(x)\leq y\}}$.
${\displaystyle \mu }$ is the Lebesgue measure of sets. ${\displaystyle H(f)}$ in turn is a real function. The (non-cumulative) histogram is defined as its derivative.
${\displaystyle h(f)=H(f)'}$.
## References
1. ^ Shapiro, Linda G. and Stockman, George C. "Computer Vision" Prentice Hall, 2003 ISBN 0-13-030796-3
2. ^ Xiang-Yang Wang, Jun-Feng Wu, and Hong-Ying Yang "Robust image retrieval based on color histogram of local feature regions" Springer Netherlands, 2009 ISSN 1573-7721
3. ^ Anatomy of a color histogram; Novak, C.L.; Shafer, S.A.; Computer Vision and Pattern Recognition, 1992. Proceedings CVPR '92., 1992 IEEE Computer Society Conference on 15–18 June 1992 Page(s):599 - 605 doi:10.1109/CVPR.1992.223129
4. ^ Integrated Spatial and Feature Image Systems: Retrieval, Analysis and Compression; Smith, J.R.; Graduate School of Arts and Sciences, Columbia University, 1997
5. ^ Effectiveness estimation of image retrieval by 2D color histogram; Bashkov, E.A.; Kostyukova, N.S.; Journal of Automation and Information Sciences, 2006 (6) Page(s): 84-89
6. ^ Content-Based Image Retrieval Using Color Histogram Correlation; Bashkov, E.A.; Shozda, N.S.; Graphicon proceedings, 2002 Page(s): [1] Archived 2012-07-07 at the Wayback Machine | 3,639 | 14,647 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 6, "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} | 2.703125 | 3 | CC-MAIN-2024-30 | latest | en | 0.912076 |
http://www.worksheetlibrary.com/subjects/math/measurement/measureoperations/ | 1,553,080,579,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202326.46/warc/CC-MAIN-20190320105319-20190320131319-00462.warc.gz | 379,025,158 | 4,593 | Worksheets By : Grade Levels Subjects Standards
Operations With Measurement Worksheets
1000s K-8 Math Worksheets for Members ........................................................ 100 Free Math Worksheets
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Adding Time and Weight Units of Measurement - Intermediate Skills Intermediate level worksheets (3) on adding times and weights. Standard: MATH 3 Grade(s): 6-8 Page(s): 4 Download Worksheet
Adding and Subtracting Measurements - Basic Skills Basic worksheets (2) on converting times and weights. Standard: MATH 3 Grade(s): 6-8 Page(s): 3 Download Worksheet
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Adding Measurements of Mass, Length, and Capacity - Basic Skills Basic worksheets (2) on adding mass, length, and capacity. Standard: MATH 3 Grade(s): 6-8 Page(s): 3 Download Worksheet
Adding Measurements of Mass, Length, and Capacity - Intermediate Skills Intermediate level worksheets (3) on adding mass, length, and capacity. Standard: MATH 3 Grade(s): 6-8 Page(s): 4 Download Worksheet
Adding and Subtracting Measurements with Fractions - Basic Skills Basic worksheets (2) on adding and subtracting fractional units of measurement. Standard: MATH 3 Grade(s): 6-8 Page(s): 3 Download Worksheet
Adding and Subtracting Measurements with Fractions - Intermediate Skills Intermediate level worksheets (3) on adding and subtracting fractional units of measurement. Standard: MATH 3 Grade(s): 6-8 Page(s): 4 Download Worksheet
Multiplying English Units of Measurement - Basic Skills Basic worksheets (2) on multiplying units of measurement. Standard: MATH 3 Grade(s): 6-8 Page(s): 3 Download Worksheet
Multiplying English Units of Measurement - Intermediate Skills Intermediate level worksheets (3) on multiplying units of measurement. Standard: MATH 3 Grade(s): 6-8 Page(s): 3 Download Worksheet | 458 | 3,348 | {"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-2019-13 | latest | en | 0.365988 |
https://www.epj.org/index.php?option=com_content&view=article&id=774%3Aepjb-highlight-probing-the-edge-of-chaos&catid=110%3Aepj-b&Itemid=429&lang=en | 1,716,073,810,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057516.1/warc/CC-MAIN-20240518214304-20240519004304-00842.warc.gz | 684,621,384 | 7,933 | # EPJ B Highlight - Probing the edge of chaos
To understand natural phenomena with a chaotic nature, the key is to find out how their variable physical characteristics behave at the very point preceding the onset of chaos
The edge of chaos—right before chaos sets in—is a unique place. It is found in many dynamical systems that cross the boundary between a well-behaved dynamics and a chaotic one. Now, physicists have shown that the distribution—or frequency of occurrence—of the variables constituting the physical characteristics of such systems at the edge of chaos has a very different shape than previously reported distributions. The results have been published in EPJ B by Miguel Angel Fuentes from the Santa Fe Institute in New Mexico, USA, and Universidad del Desarrollo, Chile, and Alberto Robledo from the National Autonomous University of Mexico, Mexico City. This could help us better understand natural phenomena with a chaotic nature.
In probability theory, the central limit theorem was first developed by an 18th century French mathematician called Abraham de Moivre. It applies to independent random physical quantities or variables, each with a well-defined expected value and well-defined way of varying. This theorem states that once iterated a sufficiently large number of times, these variable physical quantities will be approximately distributed along a central limit—also referred to as the attractor. In chaotic and standard random systems, such distribution is in the shape of a bell curve.
Now, new central limit theorems are emerging for more complex physical processes, such as natural phenomena. In this study, the authors took existing knowledge of the specific position of the attractor at the edge of chaos. To do so, they employed a mathematical formula called the logistic map as a particular example of the dynamic system under study. They found that the distribution of physical properties of such dynamic systems at this specific point at the edge of chaos has a fractal structure not previously known.
M. A. Fuentes and A. Robledo (2014), Sums of variables at the onset of chaos, European Physical Journal B, DOI 10.1140/epjb/e2014-40882-1
(EPJ A)
(EPJ B)
(EPJ E)
(EPJ H)
(EPJ N)
(EPJ PLUS)
(EPJ QT)
(EPJ ST)
(EPJ TI) | 479 | 2,274 | {"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-2024-22 | latest | en | 0.938027 |
https://www.ademcetinkaya.com/2022/09/how-accurate-is-machine-learning-in_74.html | 1,714,015,734,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297284704.94/warc/CC-MAIN-20240425032156-20240425062156-00542.warc.gz | 543,136,330 | 59,295 | ## Abstract
We evaluate Hang Seng Index prediction models with Ensemble Learning (ML) and Statistical Hypothesis Testing1,2,3,4 and conclude that the Hang Seng Index stock is predictable in the short/long term. According to price forecasts for (n+16 weeks) period: The dominant strategy among neural network is to Hold Hang Seng Index stock.
Keywords: Hang Seng Index, Hang Seng Index, stock forecast, machine learning based prediction, risk rating, buy-sell behaviour, stock analysis, target price analysis, options and futures.
## Key Points
1. How do predictive algorithms actually work?
2. How can neural networks improve predictions?
3. Game Theory
## Hang Seng Index Target Price Prediction Modeling Methodology
We consider Hang Seng Index Stock Decision Process with Statistical Hypothesis Testing where A is the set of discrete actions of Hang Seng Index stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
F(Statistical Hypothesis Testing)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Ensemble Learning (ML)) X S(n):→ (n+16 weeks) $\stackrel{\to }{R}=\left({r}_{1},{r}_{2},{r}_{3}\right)$
n:Time series to forecast
p:Price signals of Hang Seng Index stock
j:Nash equilibria
k:Dominated move
a:Best response for target price
For further technical information as per how our model work we invite you to visit the article below:
How do AC Investment Research machine learning (predictive) algorithms actually work?
## Hang Seng Index Stock Forecast (Buy or Sell) for (n+16 weeks)
Sample Set: Neural Network
Stock/Index: Hang Seng Index Hang Seng Index
Time series to forecast n: 02 Sep 2022 for (n+16 weeks)
According to price forecasts for (n+16 weeks) period: The dominant strategy among neural network is to Hold Hang Seng Index stock.
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Yellow to Green): *Technical Analysis%
## Conclusions
Hang Seng Index assigned short-term Ba2 & long-term Ba3 forecasted stock rating. We evaluate the prediction models Ensemble Learning (ML) with Statistical Hypothesis Testing1,2,3,4 and conclude that the Hang Seng Index stock is predictable in the short/long term. According to price forecasts for (n+16 weeks) period: The dominant strategy among neural network is to Hold Hang Seng Index stock.
### Financial State Forecast for Hang Seng Index Stock Options & Futures
Rating Short-Term Long-Term Senior
Outlook*Ba2Ba3
Operational Risk 7562
Market Risk6380
Technical Analysis7834
Fundamental Analysis8188
Risk Unsystematic5049
### Prediction Confidence Score
Trust metric by Neural Network: 86 out of 100 with 573 signals.
## References
1. Imbens GW, Rubin DB. 2015. Causal Inference in Statistics, Social, and Biomedical Sciences. Cambridge, UK: Cambridge Univ. Press
2. L. Busoniu, R. Babuska, and B. D. Schutter. A comprehensive survey of multiagent reinforcement learning. IEEE Transactions of Systems, Man, and Cybernetics Part C: Applications and Reviews, 38(2), 2008.
3. Artis, M. J. W. Zhang (1990), "BVAR forecasts for the G-7," International Journal of Forecasting, 6, 349–362.
4. E. Altman. Constrained Markov decision processes, volume 7. CRC Press, 1999
5. G. J. Laurent, L. Matignon, and N. L. Fort-Piat. The world of independent learners is not Markovian. Int. J. Know.-Based Intell. Eng. Syst., 15(1):55–64, 2011
6. Burgess, D. F. (1975), "Duality theory and pitfalls in the specification of technologies," Journal of Econometrics, 3, 105–121.
7. Blei DM, Lafferty JD. 2009. Topic models. In Text Mining: Classification, Clustering, and Applications, ed. A Srivastava, M Sahami, pp. 101–24. Boca Raton, FL: CRC Press
Frequently Asked QuestionsQ: What is the prediction methodology for Hang Seng Index stock?
A: Hang Seng Index stock prediction methodology: We evaluate the prediction models Ensemble Learning (ML) and Statistical Hypothesis Testing
Q: Is Hang Seng Index stock a buy or sell?
A: The dominant strategy among neural network is to Hold Hang Seng Index Stock.
Q: Is Hang Seng Index stock a good investment?
A: The consensus rating for Hang Seng Index is Hold and assigned short-term Ba2 & long-term Ba3 forecasted stock rating.
Q: What is the consensus rating of Hang Seng Index stock?
A: The consensus rating for Hang Seng Index is Hold.
Q: What is the prediction period for Hang Seng Index stock?
A: The prediction period for Hang Seng Index is (n+16 weeks) | 1,291 | 4,840 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "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 | 0.817566 |
https://www.airmilescalculator.com/distance/svl-to-tmp/ | 1,675,817,967,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500664.85/warc/CC-MAIN-20230207233330-20230208023330-00075.warc.gz | 636,893,348 | 21,328 | # How far is Tampere from Savonlinna?
The distance between Savonlinna (Savonlinna Airport) and Tampere (Tampere–Pirkkala Airport) is 179 miles / 289 kilometers / 156 nautical miles.
The driving distance from Savonlinna (SVL) to Tampere (TMP) is 242 miles / 390 kilometers, and travel time by car is about 5 hours 25 minutes.
179
Miles
289
Kilometers
156
Nautical miles
## Distance from Savonlinna to Tampere
There are several ways to calculate the distance from Savonlinna to Tampere. Here are two standard methods:
Vincenty's formula (applied above)
• 179.438 miles
• 288.777 kilometers
• 155.927 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 178.783 miles
• 287.724 kilometers
• 155.358 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Savonlinna to Tampere?
The estimated flight time from Savonlinna Airport to Tampere–Pirkkala Airport is 50 minutes.
## What is the time difference between Savonlinna and Tampere?
There is no time difference between Savonlinna and Tampere.
## Flight carbon footprint between Savonlinna Airport (SVL) and Tampere–Pirkkala Airport (TMP)
On average, flying from Savonlinna to Tampere generates about 51 kg of CO2 per passenger, and 51 kilograms equals 113 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Savonlinna to Tampere
See the map of the shortest flight path between Savonlinna Airport (SVL) and Tampere–Pirkkala Airport (TMP).
## Airport information
Origin Savonlinna Airport
City: Savonlinna
Country: Finland
IATA Code: SVL
ICAO Code: EFSA
Coordinates: 61°56′35″N, 28°56′42″E
Destination Tampere–Pirkkala Airport
City: Tampere
Country: Finland
IATA Code: TMP
ICAO Code: EFTP
Coordinates: 61°24′50″N, 23°36′15″E | 578 | 2,079 | {"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} | 2.796875 | 3 | CC-MAIN-2023-06 | latest | en | 0.827535 |
http://mizar.uwb.edu.pl/version/current/html/proofs/partit1/31 | 1,568,729,913,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-39/segments/1568514573080.8/warc/CC-MAIN-20190917141045-20190917163045-00000.warc.gz | 127,673,750 | 13,359 | let Y be non empty set ; :: thesis: for PA, PB being a_partition of Y holds ERl (PA '\/' PB) = (ERl PA) "\/" (ERl PB)
let PA, PB be a_partition of Y; :: thesis: ERl (PA '\/' PB) = (ERl PA) "\/" (ERl PB)
A1: PA is_finer_than PA '\/' PB by Th16;
A2: PB is_finer_than PA '\/' PB by Th16;
A3: union PA = Y by EQREL_1:def 4;
A4: union PB = Y by EQREL_1:def 4;
A5: ERl (PA '\/' PB) c= (ERl PA) "\/" (ERl PB)
proof
let x, y be object ; :: according to RELAT_1:def 3 :: thesis: ( not [x,y] in ERl (PA '\/' PB) or [x,y] in (ERl PA) "\/" (ERl PB) )
assume [x,y] in ERl (PA '\/' PB) ; :: thesis: [x,y] in (ERl PA) "\/" (ERl PB)
then consider p0 being Subset of Y such that
A6: p0 in PA '\/' PB and
A7: x in p0 and
A8: y in p0 by Def6;
A9: p0 is_min_depend PA,PB by ;
then A10: p0 is_a_dependent_set_of PA ;
A11: p0 is_a_dependent_set_of PB by A9;
consider A1 being set such that
A12: A1 c= PA and
A1 <> {} and
A13: p0 = union A1 by A10;
consider a being set such that
A14: x in a and
A15: a in A1 by ;
reconsider Ca = { p where p is Element of PA : ex f being FinSequence of Y st
( 1 <= len f & f . 1 = x & f . (len f) in p & ( for i being Nat st 1 <= i & i < len f holds
ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) ) )
}
as set ;
reconsider pb = union Ca as set ;
reconsider Cb = { p where p is Element of PB : ex q being set st
( q in Ca & p /\ q <> {} )
}
as set ;
reconsider x9 = x as Element of Y by A7;
reconsider fx = <*x9*> as FinSequence of Y ;
A16: fx . 1 = x by FINSEQ_1:def 8;
then A17: fx . (len fx) in a by ;
( 1 <= len fx & ( for i being Nat st 1 <= i & i < len fx holds
ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & fx . i in p1 & x0 in p1 /\ p2 & fx . (i + 1) in p2 ) ) ) by FINSEQ_1:40;
then A18: a in Ca by A12, A15, A16, A17;
then consider y5 being set such that
A19: x in y5 and
A20: y5 in Ca by A14;
consider z5 being set such that
A21: x9 in z5 and
A22: z5 in PB by ;
y5 /\ z5 <> {} by ;
then A23: z5 in Cb by ;
Ca c= PA
proof
let z be object ; :: according to TARSKI:def 3 :: thesis: ( not z in Ca or z in PA )
assume z in Ca ; :: thesis: z in PA
then ex p being Element of PA st
( z = p & ex f being FinSequence of Y st
( 1 <= len f & f . 1 = x & f . (len f) in p & ( for i being Nat st 1 <= i & i < len f holds
ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) ) ) ) ;
hence z in PA ; :: thesis: verum
end;
then A24: pb is_a_dependent_set_of PA by A18;
A25: pb c= union Cb
proof
let x1 be object ; :: according to TARSKI:def 3 :: thesis: ( not x1 in pb or x1 in union Cb )
assume x1 in pb ; :: thesis: x1 in union Cb
then consider y being set such that
A26: x1 in y and
A27: y in Ca by TARSKI:def 4;
ex p being Element of PA st
( y = p & ex f being FinSequence of Y st
( 1 <= len f & f . 1 = x & f . (len f) in p & ( for i being Nat st 1 <= i & i < len f holds
ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) ) ) ) by A27;
then consider z being set such that
A28: x1 in z and
A29: z in PB by ;
y /\ z <> {} by ;
then z in Cb by ;
hence x1 in union Cb by ; :: thesis: verum
end;
union Cb c= pb
proof
let x1 be object ; :: according to TARSKI:def 3 :: thesis: ( not x1 in union Cb or x1 in pb )
assume x1 in union Cb ; :: thesis: x1 in pb
then consider y1 being set such that
A30: x1 in y1 and
A31: y1 in Cb by TARSKI:def 4;
A32: ex p being Element of PB st
( y1 = p & ex q being set st
( q in Ca & p /\ q <> {} ) ) by A31;
then consider q being set such that
A33: q in Ca and
A34: y1 /\ q <> {} ;
consider pd being set such that
A35: x1 in pd and
A36: pd in PA by ;
A37: ex pp being Element of PA st
( q = pp & ex f being FinSequence of Y st
( 1 <= len f & f . 1 = x & f . (len f) in pp & ( for i being Nat st 1 <= i & i < len f holds
ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) ) ) ) by A33;
then consider fd being FinSequence of Y such that
A38: 1 <= len fd and
A39: fd . 1 = x and
A40: fd . (len fd) in q and
A41: for i being Nat st 1 <= i & i < len fd holds
ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & fd . i in p1 & x0 in p1 /\ p2 & fd . (i + 1) in p2 ) ;
reconsider x1 = x1 as Element of Y by ;
reconsider f = fd ^ <*x1*> as FinSequence of Y ;
len f = (len fd) + (len <*x1*>) by FINSEQ_1:22;
then A42: len f = (len fd) + 1 by FINSEQ_1:40;
1 + 1 <= (len fd) + 1 by ;
then A43: 1 <= len f by ;
A44: f . ((len fd) + 1) in y1 by ;
y1 meets q by ;
then consider z0 being object such that
A45: ( z0 in y1 & z0 in q ) by XBOOLE_0:3;
A46: z0 in y1 /\ q by ;
A47: dom fd = Seg (len fd) by FINSEQ_1:def 3;
A48: for k being Nat st 1 <= k & k <= len fd holds
f . k = fd . k
proof
let k be Nat; :: thesis: ( 1 <= k & k <= len fd implies f . k = fd . k )
assume ( 1 <= k & k <= len fd ) ; :: thesis: f . k = fd . k
then k in dom fd by ;
hence f . k = fd . k by FINSEQ_1:def 7; :: thesis: verum
end;
then A49: ( (fd ^ <*x1*>) . ((len fd) + 1) = x1 & f . 1 = x ) by ;
A50: f . (len fd) in q by ;
A51: for i being Nat st 1 <= i & i < len fd holds
ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 )
proof
let i be Nat; :: thesis: ( 1 <= i & i < len fd implies ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) )
assume A52: ( 1 <= i & i < len fd ) ; :: thesis: ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 )
then A53: f . i = fd . i by A48;
( 1 <= i + 1 & i + 1 <= len fd ) by ;
then f . (i + 1) = fd . (i + 1) by A48;
hence ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) by ; :: thesis: verum
end;
for i being Nat st 1 <= i & i < len f holds
ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 )
proof
let i be Nat; :: thesis: ( 1 <= i & i < len f implies ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) )
assume that
A54: 1 <= i and
A55: i < len f ; :: thesis: ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 )
A56: i <= len fd by ;
now :: thesis: ( ( 1 <= i & i < len fd & ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) ) or ( 1 <= i & i = len fd & ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) ) )
per cases ( ( 1 <= i & i < len fd ) or ( 1 <= i & i = len fd ) ) by ;
case ( 1 <= i & i < len fd ) ; :: thesis: ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 )
hence ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) by A51; :: thesis: verum
end;
case ( 1 <= i & i = len fd ) ; :: thesis: ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 )
hence ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) by A32, A37, A44, A46, A50; :: thesis: verum
end;
end;
end;
hence ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) ; :: thesis: verum
end;
then pd in Ca by A35, A36, A42, A43, A49;
hence x1 in pb by ; :: thesis: verum
end;
then A57: pb = union Cb by ;
Cb c= PB
proof
let z be object ; :: according to TARSKI:def 3 :: thesis: ( not z in Cb or z in PB )
assume z in Cb ; :: thesis: z in PB
then ex p being Element of PB st
( z = p & ex q being set st
( q in Ca & p /\ q <> {} ) ) ;
hence z in PB ; :: thesis: verum
end;
then A58: pb is_a_dependent_set_of PB by ;
now :: thesis: pb c= p0
assume not pb c= p0 ; :: thesis: contradiction
then pb \ p0 <> {} by XBOOLE_1:37;
then consider x1 being object such that
A59: x1 in pb \ p0 by XBOOLE_0:def 1;
A60: not x1 in p0 by ;
consider y1 being set such that
A61: x1 in y1 and
A62: y1 in Cb by ;
A63: ex p being Element of PB st
( y1 = p & ex q being set st
( q in Ca & p /\ q <> {} ) ) by A62;
then consider q being set such that
A64: q in Ca and
A65: y1 /\ q <> {} ;
A66: ex pp being Element of PA st
( q = pp & ex f being FinSequence of Y st
( 1 <= len f & f . 1 = x & f . (len f) in pp & ( for i being Nat st 1 <= i & i < len f holds
ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) ) ) ) by A64;
then consider fd being FinSequence of Y such that
A67: 1 <= len fd and
A68: fd . 1 = x and
A69: fd . (len fd) in q and
A70: for i being Nat st 1 <= i & i < len fd holds
ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & fd . i in p1 & x0 in p1 /\ p2 & fd . (i + 1) in p2 ) ;
reconsider x1 = x1 as Element of Y by ;
reconsider f = fd ^ <*x1*> as FinSequence of Y ;
len f = (len fd) + (len <*x1*>) by FINSEQ_1:22;
then A71: len f = (len fd) + 1 by FINSEQ_1:40;
1 + 1 <= (len fd) + 1 by ;
then A72: 1 <= len f by ;
A73: f . ((len fd) + 1) in y1 by ;
y1 meets q by ;
then consider z0 being object such that
A74: ( z0 in y1 & z0 in q ) by XBOOLE_0:3;
A75: z0 in y1 /\ q by ;
A76: dom fd = Seg (len fd) by FINSEQ_1:def 3;
A77: for k being Nat st 1 <= k & k <= len fd holds
f . k = fd . k
proof
let k be Nat; :: thesis: ( 1 <= k & k <= len fd implies f . k = fd . k )
assume ( 1 <= k & k <= len fd ) ; :: thesis: f . k = fd . k
then k in dom fd by ;
hence f . k = fd . k by FINSEQ_1:def 7; :: thesis: verum
end;
then A78: ( (fd ^ <*x1*>) . ((len fd) + 1) = x1 & f . 1 = x ) by ;
A79: f . (len fd) in q by ;
A80: for i being Nat st 1 <= i & i < len fd holds
ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 )
proof
let i be Nat; :: thesis: ( 1 <= i & i < len fd implies ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) )
assume A81: ( 1 <= i & i < len fd ) ; :: thesis: ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 )
then A82: f . i = fd . i by A77;
( 1 <= i + 1 & i + 1 <= len fd ) by ;
then f . (i + 1) = fd . (i + 1) by A77;
hence ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) by ; :: thesis: verum
end;
A83: for i being Nat st 1 <= i & i < len f holds
ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 )
proof
let i be Nat; :: thesis: ( 1 <= i & i < len f implies ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) )
assume that
A84: 1 <= i and
A85: i < len f ; :: thesis: ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 )
A86: i <= len fd by ;
now :: thesis: ( ( 1 <= i & i < len fd & ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) ) or ( 1 <= i & i = len fd & ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) ) )
per cases ( ( 1 <= i & i < len fd ) or ( 1 <= i & i = len fd ) ) by ;
case ( 1 <= i & i < len fd ) ; :: thesis: ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 )
hence ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) by A80; :: thesis: verum
end;
case ( 1 <= i & i = len fd ) ; :: thesis: ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 )
hence ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) by A63, A66, A73, A75, A79; :: thesis: verum
end;
end;
end;
hence ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) ; :: thesis: verum
end;
for i being Nat st 1 <= i & i < len f holds
ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 & x0 in p2 & f . (i + 1) in p2 )
proof
let i be Nat; :: thesis: ( 1 <= i & i < len f implies ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 & x0 in p2 & f . (i + 1) in p2 ) )
assume ( 1 <= i & i < len f ) ; :: thesis: ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 & x0 in p2 & f . (i + 1) in p2 )
then consider p1, p2, x0 being set such that
A87: ( p1 in PA & p2 in PB & f . i in p1 ) and
A88: x0 in p1 /\ p2 and
A89: f . (i + 1) in p2 by A83;
( x0 in p1 & x0 in p2 ) by ;
hence ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 & x0 in p2 & f . (i + 1) in p2 ) by ; :: thesis: verum
end;
hence contradiction by A7, A10, A11, A60, A71, A72, A78, Th21; :: thesis: verum
end;
then y in pb by A8, A9, A24, A58;
then consider y1 being set such that
A90: y in y1 and
A91: y1 in Cb by ;
A92: ex p being Element of PB st
( y1 = p & ex q being set st
( q in Ca & p /\ q <> {} ) ) by A91;
then consider q being set such that
A93: q in Ca and
A94: y1 /\ q <> {} ;
A95: ex pp being Element of PA st
( q = pp & ex f being FinSequence of Y st
( 1 <= len f & f . 1 = x & f . (len f) in pp & ( for i being Nat st 1 <= i & i < len f holds
ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) ) ) ) by A93;
then consider fd being FinSequence of Y such that
A96: 1 <= len fd and
A97: fd . 1 = x and
A98: fd . (len fd) in q and
A99: for i being Nat st 1 <= i & i < len fd holds
ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & fd . i in p1 & x0 in p1 /\ p2 & fd . (i + 1) in p2 ) ;
reconsider y9 = y as Element of Y by A8;
reconsider f = fd ^ <*y9*> as FinSequence of Y ;
len f = (len fd) + (len <*y9*>) by FINSEQ_1:22;
then A100: len f = (len fd) + 1 by FINSEQ_1:40;
then A101: 1 + 1 <= len f by ;
A102: f . ((len fd) + 1) in y1 by ;
y1 meets q by ;
then consider z0 being object such that
A103: ( z0 in y1 & z0 in q ) by XBOOLE_0:3;
A104: z0 in y1 /\ q by ;
A105: dom fd = Seg (len fd) by FINSEQ_1:def 3;
A106: for k being Nat st 1 <= k & k <= len fd holds
f . k = fd . k
proof
let k be Nat; :: thesis: ( 1 <= k & k <= len fd implies f . k = fd . k )
assume ( 1 <= k & k <= len fd ) ; :: thesis: f . k = fd . k
then k in dom fd by ;
hence f . k = fd . k by FINSEQ_1:def 7; :: thesis: verum
end;
then A107: f . (len fd) in q by ;
A108: for i being Nat st 1 <= i & i < len fd holds
ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 )
proof
let i be Nat; :: thesis: ( 1 <= i & i < len fd implies ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) )
assume A109: ( 1 <= i & i < len fd ) ; :: thesis: ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 )
then A110: f . i = fd . i by A106;
( 1 <= i + 1 & i + 1 <= len fd ) by ;
then f . (i + 1) = fd . (i + 1) by A106;
hence ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) by ; :: thesis: verum
end;
A111: for i being Nat st 1 <= i & i < len f holds
ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 )
proof
let i be Nat; :: thesis: ( 1 <= i & i < len f implies ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) )
assume that
A112: 1 <= i and
A113: i < len f ; :: thesis: ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 )
A114: i <= len fd by ;
now :: thesis: ( ( 1 <= i & i < len fd & ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) ) or ( 1 <= i & i = len fd & ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) ) )
per cases ( ( 1 <= i & i < len fd ) or ( 1 <= i & i = len fd ) ) by ;
case ( 1 <= i & i < len fd ) ; :: thesis: ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 )
hence ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) by A108; :: thesis: verum
end;
case ( 1 <= i & i = len fd ) ; :: thesis: ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 )
hence ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) by ; :: thesis: verum
end;
end;
end;
hence ex p1, p2, x0 being set st
( p1 in PA & p2 in PB & f . i in p1 & x0 in p1 /\ p2 & f . (i + 1) in p2 ) ; :: thesis: verum
end;
A115: ( (fd ^ <*y9*>) . ((len fd) + 1) = y & 1 <= len f ) by ;
A116: f . 1 = x by ;
for i being Nat st 1 <= i & i < len f holds
ex u being set st
( u in Y & [(f . i),u] in (ERl PA) \/ (ERl PB) & [u,(f . (i + 1))] in (ERl PA) \/ (ERl PB) )
proof
let i be Nat; :: thesis: ( 1 <= i & i < len f implies ex u being set st
( u in Y & [(f . i),u] in (ERl PA) \/ (ERl PB) & [u,(f . (i + 1))] in (ERl PA) \/ (ERl PB) ) )
assume ( 1 <= i & i < len f ) ; :: thesis: ex u being set st
( u in Y & [(f . i),u] in (ERl PA) \/ (ERl PB) & [u,(f . (i + 1))] in (ERl PA) \/ (ERl PB) )
then consider p1, p2, u being set such that
A117: p1 in PA and
A118: p2 in PB and
A119: f . i in p1 and
A120: u in p1 /\ p2 and
A121: f . (i + 1) in p2 by A111;
A122: u in p1 by ;
A123: u in p2 by ;
reconsider x2 = f . i as set ;
reconsider y2 = f . (i + 1) as set ;
A124: [x2,u] in ERl PA by ;
A125: [u,y2] in ERl PB by ;
( ERl PA c= (ERl PA) \/ (ERl PB) & ERl PB c= (ERl PA) \/ (ERl PB) ) by XBOOLE_1:7;
hence ex u being set st
( u in Y & [(f . i),u] in (ERl PA) \/ (ERl PB) & [u,(f . (i + 1))] in (ERl PA) \/ (ERl PB) ) by ; :: thesis: verum
end;
then [x9,y9] in (ERl PA) "\/" (ERl PB) by ;
hence [x,y] in (ERl PA) "\/" (ERl PB) ; :: thesis: verum
end;
for x1, x2 being object st [x1,x2] in (ERl PA) \/ (ERl PB) holds
[x1,x2] in ERl (PA '\/' PB)
proof
let x1, x2 be object ; :: thesis: ( [x1,x2] in (ERl PA) \/ (ERl PB) implies [x1,x2] in ERl (PA '\/' PB) )
assume [x1,x2] in (ERl PA) \/ (ERl PB) ; :: thesis: [x1,x2] in ERl (PA '\/' PB)
then ( [x1,x2] in ERl PA or [x1,x2] in ERl PB ) by XBOOLE_0:def 3;
then A126: ( ex A being Subset of Y st
( A in PA & x1 in A & x2 in A ) or ex B being Subset of Y st
( B in PB & x1 in B & x2 in B ) ) by Def6;
now :: thesis: ( ( x1 in Y & x2 in Y & ex A being Subset of Y st
( A in PA & x1 in A & x2 in A ) & [x1,x2] in ERl (PA '\/' PB) ) or ( x1 in Y & x2 in Y & ex B being Subset of Y st
( B in PB & x1 in B & x2 in B ) & [x1,x2] in ERl (PA '\/' PB) ) )
per cases ( ( x1 in Y & x2 in Y & ex A being Subset of Y st
( A in PA & x1 in A & x2 in A ) ) or ( x1 in Y & x2 in Y & ex B being Subset of Y st
( B in PB & x1 in B & x2 in B ) ) )
by A126;
case ( x1 in Y & x2 in Y & ex A being Subset of Y st
( A in PA & x1 in A & x2 in A ) ) ; :: thesis: [x1,x2] in ERl (PA '\/' PB)
then consider A being Subset of Y such that
A127: A in PA and
A128: ( x1 in A & x2 in A ) ;
ex y being set st
( y in PA '\/' PB & A c= y ) by ;
hence [x1,x2] in ERl (PA '\/' PB) by ; :: thesis: verum
end;
case ( x1 in Y & x2 in Y & ex B being Subset of Y st
( B in PB & x1 in B & x2 in B ) ) ; :: thesis: [x1,x2] in ERl (PA '\/' PB)
then consider B being Subset of Y such that
A129: B in PB and
A130: ( x1 in B & x2 in B ) ;
ex y being set st
( y in PA '\/' PB & B c= y ) by ;
hence [x1,x2] in ERl (PA '\/' PB) by ; :: thesis: verum
end;
end;
end;
hence [x1,x2] in ERl (PA '\/' PB) ; :: thesis: verum
end;
then (ERl PA) \/ (ERl PB) c= ERl (PA '\/' PB) ;
then (ERl PA) "\/" (ERl PB) c= ERl (PA '\/' PB) by EQREL_1:def 2;
hence ERl (PA '\/' PB) = (ERl PA) "\/" (ERl PB) by A5; :: thesis: verum | 8,592 | 19,675 | {"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} | 3.234375 | 3 | CC-MAIN-2019-39 | latest | en | 0.889774 |
https://www.wepapers.com/samples/project-1-essay/ | 1,632,420,262,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057427.71/warc/CC-MAIN-20210923165408-20210923195408-00472.warc.gz | 1,079,243,616 | 20,786 | # Project 1 Essay
Type of paper: Essay
Topic: Force, Energy, Stress, Pressure, Water, Vehicles, Law, Cooking
Pages: 6
Words: 1650
Published: 2020/12/21
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As an example of Physics in our daily life activities we can consider the interaction between our foot and the road when we walk. According to Newton`s third law for every action force there is an equal and opposite reaction force. (Harris, W., n.d.). Force is the pull or push needed to do a work. A force exists as a result of interaction, whenever there is an interaction between the two objects there is a force acting upon each object. When one object exerts a force on the second object, the second object also exerts an equal and opposite force on the first object. So, when we push on something, then it is also pushes us back. When we walk, we push backward on the road, and the road also pushes us forward. These two pushing forces are the result of friction force between our feet and the road. The friction force enables us to walk normally, walking would be impossible without this friction force between the surface of the road and our feet. In our daily activities, when we play football, baseball etc. we use Physics in these activities. In an interaction between the baseball bat and a baseball the bat strikes a baseball, the baseball and the bat exert equal forces on each other in opposite direction. But, compared to the mass of the bat, mass of the ball is very small, and the baseball- bat interaction accelerates the ball on getting struck.
In our daily activities we use lots of machines, because machines are the devices that require less force to accomplish an amount of work. Levers are such an important tool that allows us to do a work with less effort. A lever is a rigid object that is free to turn about fulcrum or the part of lever that supports the length of bar, and force is exerted upon the lever arm at a specific point to lift or move heavy objects (load). The closer the fulcrum to the load, the less force required to lift the load. The closer the fulcrum to the force, the greater the force required to lift the load. Levers are divided into three classes. A first class lever has fulcrum at the centre between the force and load. In a second class lever the load arm is in between the fulcrum and the force arm, and the fulcrum is usually closer to the load, which reduces to force required to accomplish the work. In class three levers the force arm lies between the fulcrum and load arm. Due to this arrangement large force is required for the displacement of load. (Tucker, K., n.d.). In our daily activities we use varieties of levers to accomplish different types of work with less effort. Scissors, pliers, water pump, car jack, are the example of class one levers. Bottle openers are the example of second class levers, and example of class three levers are tweezers, an arm lifting a weight, spade, and shovel.
When we drive a vehicle with hydraulic brake system, we apply small amount of pressure on the brake pedal to stop it. Since the actual brakes require much greater force to stop the car the braking system multiply the small pressure applied by the foot, and the system transmit the force to the tires using friction, and the tires transmit the frictional force to the road to stop the car. When we press the brake pedal the pressure applied by the foot is transmitted through the enclosed fluid of the braking system. It is an application of Pascal`s law in our daily activities to achieve a greater force by an externally applied pressure. Pascal`s law states that when there is an increase in pressure in an confined fluid, the pressure is transmitted at every other point in the container. The hydraulic brakes use incompressible fluids to transmit the forces from one point to another, within the fluid. Most aircraft use hydraulic brake system, and Pascal`s law is applied in hydraulic car lift. (Hodanbosi, C., 2014)
The boiling point of water (or any other liquid) depends upon the pressure on its surface. Under the normal atmospheric pressure, boiling point of water is 100°C, and the boiling point varies with different atmospheric pressure. Applying more heat to an open pot of water, the liquid water convert into water vapour without the increase in temperature. If the pressure is reduced the boiling point of water will also be decreased, because the water molecules will escape the surface of water easily. If we increase the pressure on the liquid then it will be difficult for the water molecules to escape from the surface, and boiling point of the water will be increased. In a closed container the water and vapour are at same temperature with no space for the water molecules to escape. When heat is applied to the closed container it will increase the pressure inside the surface of the water with an increase in temperature of the system. The observation that water boils at higher temperature, when the external pressure is increased is applied in pressure cooker for faster cooking, with less consumption of fuel. A pressure cooker consists of a metal pot and lid. The lid had a rubber ring to seal off the space between the lid and the pot. A pressure regulator, safety valve of low melting point alloy, a vent to allow the steam to escape generated inside the pot is fitted on top of the pressure cooker. (Chen, L., Anderson, J., & Wang, D.,2009) In the cooking process the pressure cooker generates high pressure, which allows the water inside the pot to boil at a greater temperature higher than its normal boiling point. At the beginning of boiling, water absorbs heat, and converts into steam. The increased pressure inside the pressure cooker raises the boiling temperature of water. Once the temperature reaches the melting point of the safety valve, the valve melts, and releases the steam and pressure inside the cooker to prevent an explosion.
In our common experience, if we try to jump out of a speeding vehicle in the opposite direction of the vehicle or suddenly stop our movement we fall on the ground. So, we move forward in the direction of motion of the vehicle to avoid a fall. This activity is associated with Newton`s first law of motion. According to this law a body continues in its state of rest or uniform motion in a straight line unless it is compelled by some external force to change the state. (Bibel, G., 2013) When we try to jump out of the speeding vehicle we are also in motion and share the velocity of the vehicle, and the body continues in its state of motion unless it exercises some force to change the state. In other words, if somehow we could eliminate gravity and air resistance a ball thrown straight upward would continue to move upward forever.
During our regular physical activities of walking, playing, running swimming, etc., our body derive energy from the ingested food, and for powering car energy is derived from liquid fuel source. Energy is the ability to do work, and the tendency of energy is that it flows from higher concentration to lower concentration. Foods like carbohydrate, fat etc, and liquid fuels like gasoline have the highly concentrated potential energy stored in their chemical bonds. According to the law of thermodynamics, energy exists in different forms, and it can be converted from one form to another. The thermodynamic variable or entropy of an isolated system remain constant or increases with time, and entropy is the unavailable amount of energy to do work. In the process of our regular physical activities like walking, playing, running etc., or for powering car, energy is converted into different form, and high concentration of energy gets transformed into useful energy for the bodies and machines for normal function. Every time as the energy is converted, some of the energy becomes less useful. In our daily activities, food energy is converted into work and less useful heat energy, and the vehicle engines convert the fuel energy into work and less useful heat energy. In the process of energy conversion although some parts of the energy gets converted into low grade energy, but, there is no change in the actual amount of energy. In our daily activities, we use Physics according to the law of thermodynamics, and fuel our human body and the machines, and convert the energy into work. In our daily activities we use several electrical appliances like electric heater, bulb, washing, television, where the conversion of energy take place, and electrical energy is finally converted into low grade heat energy that heat up the atmosphere around us.
## Reference:
Bibel, G. (2013, August 2). The Physics of Disaster: An Exploration of Train Derailments [Excerpt]. Retrieved March 16, 2015, from http://www.scientificamerican.com/article/the-physics-of-disaster/.
Chen, L., Anderson, J., & Wang, D. (2009, June 19). Cooking Under Pressure: Applying the Ideal Gas Law in the Kitchen. Retrieved March 16, 2015, from http://www.cameron.edu/~garyb/CHEM1004Online/CaseStudy/pressure_cooking.pdf.
Harris, W. (n.d.). How Newton's Laws of Motion Work. Retrieved March 15, 2015, from http://science.howstuffworks.com/innovation/scientific-experiments/newton-law-of-motion4.htm.
Hodanbosi, C. (2014, June 12). Pascal's Principle and Hydraulics. Retrieved March 15, 2015, from http://www.grc.nasa.gov/WWW/k-12/WindTunnel/Activities/Pascals_principle.html.
Tucker, K. (n.d.). Levers Used in Everyday Life. Retrieved March 15, 2015, from http://www.ehow.com/info_8435160_levers-used-everyday-life.html.
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Chat now | 2,564 | 11,387 | {"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} | 3.859375 | 4 | CC-MAIN-2021-39 | longest | en | 0.933998 |
https://deepai.org/publication/dynamics-of-cycles-in-polyhedra-i-the-isolation-lemma | 1,643,026,643,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304528.78/warc/CC-MAIN-20220124094120-20220124124120-00649.warc.gz | 260,280,423 | 60,618 | Dynamics of Cycles in Polyhedra I: The Isolation Lemma
A cycle C of a graph G is isolating if every component of G-V(C) is a single vertex. We show that isolating cycles in polyhedral graphs can be extended to larger ones: every isolating cycle C of length 8 ≤ |E(C)| < 2/3(|V(G)|+3) implies an isolating cycle C' of larger length that contains V(C). By “hopping” iteratively to such larger cycles, we obtain a powerful and very general inductive motor for proving and computing long cycles (we will give an algorithm with running time O(n^2)). This provides a method to prove lower bounds on Tutte cycles, as C' will be a Tutte cycle of G if C is. We also prove that E(C') ≤ E(C)+3 if G does not contain faces of size five, which gives a new tool for proving results about cycle spectra and evidence that these face sizes obstruct long cycles. As a sample application, we test our motor on a conjecture on essentially 4-connected graphs. A planar graph is essentially 4-connected if it is 3-connected and every of its 3-separators is the neighborhood of a single vertex. Essentially 4-connected graphs have been thoroughly investigated throughout literature as the subject of Hamiltonicity studies. Jackson and Wormald proved that every essentially 4-connected planar graph G on n vertices contains a cycle of length at least 2/5(n+2), and this result has recently been improved multiple times, culminating in the lower bound 5/8(n+2). However, the best known upper bound is given by an infinite family of such graphs in which every graph G on n vertices has no cycle longer than 2/3(n+4); this upper bound is still unmatched. Using isolating cycles, we improve the lower bound to match the upper (up to a summand +1). This settles the long-standing open problem of determining the circumference of essentially 4-connected planar graphs.
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1 Introduction
One of the unchallenged milestones in planar graph theory is the result by Tutte [11] that every 4-connected planar graph on vertices is Hamiltonian, i.e. has circumference , where the circumference of a graph is the length of a longest cycle of . However, decreasing the connectivity assumption from 4 to 3 reveals infinitely many planar graphs that do not have long cycles: Moon and Moser [9] showed that there are infinitely many 3-connected planar (i.e., polyhedral) graphs that have circumference at most for , and this upper bound is best possible up to constant factors, as there is a constant such that every polyhedral graph contains a cycle of length at least .
One of the biggest remaining open problems in this area ever since is to characterize the connectivity properties between connectivity 3 and 4 that imply long cycles. Essential -connectivity is such a property and will be a focus of this paper.
Indeed, essentially -connected graphs have been thoroughly investigated throughout literature for this purpose. An upper bound for the circumference of essentially 4-connected planar graphs was given by an infinite family of such graphs on vertices in which every graph satisfies [2]; the graphs in this family are in addition maximal planar. Regarding lower bounds, Jackson and Wormald [8] proved that for every essentially 4-connected planar graph on vertices. Fabrici, Harant and Jendroľ [2] improved this lower bound to ; this result in turn was recently strengthened to [4], and then further to [5]. For the restricted case of maximal planar essentially 4-connected graphs, the matching lower bound was proven in [3]; however, the method used there is specific to maximal planar graphs. For the general polyhedral case, it is still an open conjecture that every essentially -connected planar graph on vertices satisfies and thus the upper bound; while this conjecture has been an active research topic at workshops for over a decade111personal communication with Jochen Harant, it was only recently explicitly stated [3, Conjecture 2].
Here, we show that for every essentially 4-connected planar graph; this matches the upper bound up to the summand . Moreover, this is only an implication of a much more general result about polyhedral graphs, which we present here. While one part of the proof scheme follows the established approach of using Tutte cycles in combination with the discharging method, we contribute an intricate intersection argument on the weight distribution between groups of neighboring faces, which we call tunnels. This methods differs substantially from the result in [3] (which exploits the inherent structure of maximal planar graphs) and is, unlike the previous results, able to harness the dynamics of extending cycles in polyhedral graphs. In particular, we will discharge weights along an unbounded number of faces, which was an obstacle that was not needed to solve for the weaker bounds.
In fact, we will prove this tool in the following variant that provides also a method to prove many different cycle lengths. Let a cycle of a graph be extendable if contains a larger isolating cycle such that and , where is the number of faces of size five in .
Lemma 1 (Isolation Lemma).
Every isolating cycle of length in a polyhedral graph on vertices is extendable.
The assumption is redundant if and only if . The Isolation Lemma may be seen as a polyhedral relative of Woodall’s Hopping Lemma [13] that allows cycle extensions through common neighbors of cycle vertex pairs even when none of these pairs have distance two in . Despite this correlation, the Isolation Lemma makes inherently use of planarity; in fact, it fails hard for non-planar graphs, as the graphs for any and any (isolating) cycle of length in these show. We state some immediate corollaries.
Corollary 2.
If a polyhedral graph contains an isolating cycle of length , contains cycles of at least different lengths in .
Corollary 3.
If a bipartite polyhedral graph contains an isolating cycle of length , contains a cycle of length for every even .
The conclusion of Corollary 3 holds even when the polyhedral graph does not have faces of size 5. In view of the sheer number of results in Hamiltonicity studies that use subgraphs involving faces of size five (see e.g. the Tutte Fragment or the one of Faulkner and Younger), this provides evidence that these faces are indeed key to a small circumference. Another corollary is that polyhedral graphs on vertices, in which all cycles have length less than , do not contain any isolating cycle; for example, this holds for the sufficiently large Moon-Moser graphs [9] and the 18 graph classes of [7, Theorem 1] that have shortness exponent less than 1.
Finally, the Isolation Lemma implies also the following theorem.
Theorem 4.
Every essentially 4-connected planar graph on vertices contains an isolating Tutte cycle of length at least and such a cycle can be computed in time .
Proof.
We only consider existence and will give an algorithm at the end of this paper. It is well-known that every 3-connected plane graph on at most 10 vertices is Hamiltonian [1]. Since these graphs contain in particular the essentially 4-connected planar ones, this implies the theorem if ; we therefore assume . For , it was shown in [2, Lemma 4(i)+(ii)] that contains an isolating Tutte cycle of length at least . If , is already long enough, since ; otherwise, applying iteratively the Isolation Lemma to gives the claim and preserves a Tutte cycle, as no vertices of are deleted. ∎
Theorem 4 encompasses and strengthens most of the results known for the circumference of essentially 4-connected planar graphs, some of which can be found in [2, 6, 14].
2 Preliminaries
We use standard graph-theoretic terminology and consider only graphs that are finite, simple and undirected. For a vertex of a graph , denote by the degree of in . We omit subscripts if the graph is clear from the context. Two edges and are adjacent if they share at least one end vertex. The distance of two edges in a connected graph is the length of a shortest path that contains both. We denote a path of that visits the vertices in the given order by .
A separator of a graph is a subset of such that is disconnected; we call a -separator if . Let a cycle of a graph be isolating if every component of is a single vertex (see Figure 0(a) for a example). We do not require that these single vertices have degree three (this differs e.g. from [2, 4, 5]). A chord of a cycle is an edge for which and are in . According to Whitney [12], every 3-connected planar graph has a unique embedding into the plane (up to flipping and the choice of the outer face). Hence, we assume in the following that such graphs are equipped with a fixed planar embedding, i.e., are plane. Let be the set of faces of a plane graph .
3 Proof of the Isolation Lemma
Let be a 3-connected plane graph on vertices, and let be an isolating cycle of of length . We assume to the contrary that is not extendable.
Let be the subset of that is contained in the maximal path-connected open set (i.e. region) of that is bounded (hence, strictly inside ), and . Without loss of generality, we assume . Since and , we have . Let be the plane graph obtained from by deleting either all chords of if or otherwise all chords of whose interior point set is contained in the bounded region of (see Figure 0(a)). Let and .
For a face of or of , the edges of that are incident with are called -edges of and their number is denoted by . A -vertex of is a vertex that is incident to a -edge of ; a -vertex of is extremal if it is incident to at most one -edge of and non-extremal otherwise. A face is called -face if . If
has an odd number of
-vertices or -edges, we call their unique vertex or edge in the middle the middle -vertex or -edge of .
Let two faces and of be opposite if and have a common -edge. If has an -edge , let the -opposite face of be the face of that is different from and incident to . Let a face of be thin if and is in ; otherwise, let be thick. A face of is called minor if it is either thick and incident to exactly one vertex of or thin and incident to exactly one edge that is not in ; otherwise, is called major. Let and be the sets of minor faces of and , respectively. For a minor thick face of , let be the unique vertex of incident to . A 2-sandwiched vertex is the middle -vertex of a thick minor 2-face of .
Construct a graph (see Figure 0(b)) obtained from with vertex set and the following edge-set. First, for every face , add the edge to . Second, for every major face of (in arbitrary order), fix any vertex that is incident to and add the edge to for every vertex that is incident to . We first prove that is a tree.
Lemma 5.
is a tree with inner vertex set , leaf set and no vertex of degree two.
Proof.
Consider two faces and of that are incident to a common edge . Since does not contain any chord of , is not a chord of , so that and are incident to a common vertex . By construction of , all inner vertices of that are incident to or (in particular, ) are connected in . Hence, is connected. As is isolating, every two faces of are incident to at most one common vertex of . Hence, the union of the acyclic graphs that are constructed for every major face of , and thus itself, is acyclic. We conclude that is a tree with inner vertex set and leaf set .
Note that may contain vertices of unbounded degree even when every vertex of has degree three in (for example, in Figure 0(b)). If , Lemma 5 holds by symmetry also for the tree that is constructed from in the same way as from .
We have . If , .
Proof.
Since , has at least four vertices by Lemma 5. It is well-known that every tree on at least two vertices has exactly leaves, where is the number of vertices of degree at least 3 in . Since has no vertex of degree two, this gives the first claim by Lemma 5. The second claim follows from by the same argument applied to . ∎
In both equalities of Lemma 6, the last summand is non-negative but may be zero, as a longest cycle of the graph obtained from the octahedron by inserting a vertex of degree three in every face shows (see also the graph obtained from Figure 0(b) by deleting ).
Consider a minor 1-face of with -edge ; since there are no thin minor 1-faces, is thick. Then the cycle obtained from by replacing with the path shows that is extendable, which contradicts our assumption. We conclude that has no minor 1-face. Since is isolating and has no chords in , has no minor 0-face. To summarize our assumptions so far, we know that is not extendable, , and every minor face of satisfies .
For the final contradiction, we aim to prove
2c ≥4(|M−|+|M+|) if V+≠∅ and (1) 2c ≥4|M−|+|M+| if V+=∅ (2)
Assume . By Lemma 6, and . Since , Inequality 1 implies and thus the claim . In the special case , we have , so that Inequality 2 implies . Since every minor face of has a non-extremal -vertex (since has no minor 1-faces), minimum degree 3 in implies that we have at least one chord of in , so that . This gives (hence, we are only off by in the case ).
In order to prove Inequalities (1) and (2), we will charge every -face of with weight ; hence, the total charge has weight . Then we discharge (i.e., move) these weights to minor faces such that no face has negative weight. We will prove that after the discharging every minor face of has sufficiently high weight to satisfy the inequalities.
3.1 Arches and Tunnels
For a face of , a path of is an arch of if is minor and is either the maximal path in all of whose edges are incident to (in this case we say that is proper; then has length one or two depending on whether is thin or thick) or a chord of whose inner point set is contained in and that does not join the two extremal -vertices of (see Figure 0(a)). Hence, an arch is proper if and only if , so that every minor face has exactly one proper arch. The face of an arch is the minor face of that contains the inner point set of .
Let the archway of an arch be the path in between the extremal vertices of whose edges are incident to in . Since and its archway close a face in the graph , we define , thickness, and the -vertices and -edges of just as the corresponding terms for the face ; in particular, denotes the number of edges of the archway of , and is thick if is thick. Then every arch has exactly two extremal -edges, as has no minor 1-face and, by the last condition of the definition of arches, two arches of have never the same pair of extremal -edges (this prevents that two arches of the same face “overlap”). We call an arch a -arch if . If and are arches or faces of , let be the number of -edges that and have in common; then and are opposite if . An arch of an arch is an arch of such that every -edge of is a -edge of ; we also say that has arch .
Consider the 3-arches in Figure 3.1 and assume that every is thick and proper, so that every is a minor 3-face. Since every receives only initial weight 3 and is unbounded, every local method of transferring weights to reach weight 4 per minor face is bound to fail. Unfortunately, 3-faces are not the only example where non-local methods are needed: in fact, there is an unbounded number of faces in which weights must be transferred non-locally. We will therefore design the upcoming discharging rule in such a way that weight transfers are not dependent on minor face but instead on arches; this will reduce all structures that have to be handled non-locally to one common structure (called tunnel), which is the one in Figure 3.1.
Let two 3-arches and be consecutive if . The reflexive and transitive closure of this symmetric relation partitions the set of 3-arches; we call the sets of this partition tunnels (see Figure 3.1). Since is plane, imposes a notion of clockwise and counterclockwise on ; in the following, both directions always refer to . The counterclockwise track of a tunnel (which will transfer weights counterclockwise around ) is the sequence of all 3-arches of such that is the clockwise consecutive successor of for every . We call a tunnel track and its tunnel cyclic if and and are consecutive, and acyclic otherwise.
The exit pair of a counterclockwise track consists of the counterclockwise extremal -edge of and the -opposite face of . Clockwise tracks and exit pairs are defined analogously. The exit pairs and of a tunnel are then the two exit pairs of the counterclockwise and clockwise tracks of ; we call and exit faces of ). Clearly, we have if and only if is cyclic, and if so, and are opposite faces, so that . Hence, the exit pairs of every acyclic tunnel are different, while the exit faces may be identical.
In order to describe the weight transfers through tunnels, we define the following reflexive and symmetric relation for faces and of and extremal -edges and of (not necessarily different) 3-arches of an acyclic tunnel such that and are incident to and , respectively. Let be on-track with if the following statements are equivalent (see Figure 3.1).
• and are contained in the same region of
• the distance between and in the union of the -edges of (measured by the length of a path that does not exceed ) is a multiple of .
Clearly, this relation is an equivalence relation. Moreover, if is an extremal -edge of a 3-arch of a tunnel , is on-track with exactly one exit pair of . Tunnels will serve as objects through which we can pull weight over long distances. We will later prove that tunnels transfer weights only one-way, i.e. use (the on-track pairs of) at most one of their tracks. Based on the structure of , weight may not be transferred through the whole tunnel track; the following definition restricts the parts where weight transfers may occur.
Let be an acyclic tunnel track, an extremal -edge of an arch of such that is on-track with the exit pair of , the extremal -edge of different from , and the -opposite face of (see Figure 3.1; informally, is the on-track pair in that precedes ). Recursively, we define that is a transfer pair of if is either the exit pair of or a transfer pair, and
• the -opposite face of is minor and ,
• is either an extremal -edge of or adjacent to that edge, and in the latter case the middle -edge of is incident to or a major face, and
• the middle -edge of is not incident to (in particular, ).
For a tunnel track , an arch of that has an extremal -edge such that is a transfer pair of is called transfer arch of . Note that a transfer arch of is not necessarily a transfer arch of the other tunnel track of .
3.2 Discharging Rule
By saying that a face pulls weight over its -edge for a positive weight , we mean that is added to and subtracted from the -opposite face of ; we sometimes omit if the precise value is not important, but positive.
Definition 7 (Discharging Rule).
For every minor face of and every -edge of (both in arbitrary order), pulls weight 1 over from the -opposite face of for every of the following conditions that is satisfied (see Figure 2).
1. is major
2. is minor, and is a thick 2-face
3. is minor and , is the middle -edge of a 3-arch of and not an extremal -edge of a 3-arch of , either is a 3-face or an extremal -edge of is an extremal -edge of and the other extremal -edge of is incident to an extremal -vertex of and not incident to a major face such that, if is incident to , has a 4-arch that has
4. is minor, is a non-extremal -edge of a 4-arch of such that the extremal -edge of that is adjacent to is an extremal -edge of and the other extremal -edge of is incident to a thick minor 2-face , is not the middle -edge of a 3-arch of , and
5. is minor, is a non-extremal -edge of a 4-arch of such that the extremal -edge of that is adjacent to is an extremal -edge of and the other extremal -edge of is an extremal -edge of a 3-arch of a face , is a transfer pair, no 2-arch has extremal -vertex , there is no 3-arch of , is not an extremal -edge of a 3-arch of , and
6. is minor, is a non-extremal -edge of a thick minor 4-face such that the extremal -edge of that is not adjacent to is the middle -edge of a 3-arch of a face , and is not the middle -edge of a 3-arch of .
7. is minor, is a transfer pair of an acyclic tunnel track , and the exit pair of satisfies (in the notation and ) at least one of the conditions
Note that the weight transfers of this rule are solely dependent on and (and not on the current weight transfers). This holds in particular for the ones caused by , as these do not depend on (but instead of on the exit pair). Since tunnels partition the set of 3-arches, it suffices to evaluate once for the transfer pairs of each tunnel track.
After the discharging rule has been applied, effectively routes weight 1 through an extremal part of a tunnel track towards its exit face if this exit face pulls weight from by any other condition. By definition of , the only faces that do not have an arch of a tunnel (i.e. reside “outside” ) and pull over a -edge of such an arch are the exit faces of ; in this sense, weight may leave only through an exit face of .
3.3 Structure of Tunnels and Transfers
We give further insights into the structure of tunnels and the location of edges over which our discharging rule pulls (positive) weight.
Lemma 8.
Every tunnel track with satisfies . In particular, every tunnel is acyclic.
We remark that it is possible, but slightly more involved, to prove Lemma 8 solely by using the discharging rule of Definition 7. Using Lemma 8, we assume from now on that every tunnel is acyclic. We now show that does not contain the dotted edges of Figure 2 for the respective conditions; this sheds first light on the implications that are triggered by the assumption that is not extendable.
Lemma 9.
For any satisfied condition , none of the red dotted arches in the respective Figure 1(a)1(e) exist in the depicted face of . If , .
Proof.
We use the notation of Figure 2. Assume . If (or, by symmetry, ) in Figure 1(a) is a -edge of a 2-arch of , is an extremal -vertex of , since is polyhedral. Then is extendable by the path replacement , which adds one or two new vertices to (depending on whether is thick). If (or, by symmetry, ) is the middle -edge of a 3-arch, we have , as , since is polyhedral; hence, neither nor is the middle -edge of a 3-arch. Using the same argument, has no 4-arch with extremal -edges and .
Assume . Then is not the middle -edge of a 3-arch of , as is polyhedral. By definition of C3, is not an extremal -edge of a 3-arch of .
Assume . Then the first argument for implies . In addition, , as otherwise is extendable by the path replacement . Hence, is not an extremal -edge of a 3-arch of . If is an extremal -edge of a 3-arch of , is extendable by the path replacement , as this adds at most three new vertices to . Note that we have in this replacement if is proper ( is thick because of ), as has no minor 1-face (here, with -edge ). In addition, is not the middle -edge of a 3-arch of , as otherwise would be a 2-separator of by the previous claims.
Assume . By definition of , it only remains to prove that is not the middle -edge of a 3-arch of . If it is, is a 2-separator of , since and are not contained in . This contradicts that is polyhedral.
Assume . Then , as otherwise is extendable by the replacement . Since is polyhedral, has degree at least three in , which implies as only remaining option. Then , as otherwise is extendable by the replacement . Sinec is polyhedral, this implies that there is no 2-arch of with extremal -vertices and . Assume that has a 2-arch with extremal -vertices and . Then is thick, as has a 2-arch, but non-proper, as otherwise would be a minor 1-face of . Hence, , so that is extendable by the replacement . This implies in addition that is not an extremal -edge of a 3-arch of , as would have degree two in . ∎
For a -edge of a face of and a condition , let denote that is satisfied for and in Definition 7. For notational convenience throughout this paper, whenever pulls weight from a face , we denote by the extremal -vertex of whose clockwise neighbor in is -vertex of , and denote by the th vertex modulo in a clockwise traversal of starting at .
So far, a tunnel might transfer weights through both of its tracks simultaneously. The next lemma shows that this never happens.
Lemma 10.
Let and be the exit pairs of a tunnel such that pulls weight over . Then
1. is minor, and no other condition is satisfied for ,
2. every 2-arch of an arch of has a -edge such that is on-track with ,
3. and there is no 2-arch of that has -edge ,
4. does not pull any weight over .
5. for every 4-arch that has an arch of , the common extremal -edge of and satisfies that is on-track with ,
6. every arch of that is consecutive to two transfer arches of satisfies
By Lemma 104, all weight transfers that are caused within a tunnel by Condition C7 go one-way, i.e., use only one track of . As an immediate implication, the following lemma shows that all weight transfers strictly within a tunnel are solely dependent on the weight transfers on its exit pairs.
Lemma 11.
Let be a transfer pair of a tunnel track with exit pair such that the -opposite face of is minor. Then pulls weight over if and only if pulls weight over (and if so, and ).
Proof.
Assume that pulls weight over . By Lemma 8, is acyclic. By Lemma 101, . Hence, is satisfied for , so that pulls weight over .
Assume to the contrary that for some and does not pull any weight over . The latter implies . Since is minor, . Since is a -edge of a 3-arch of with , . By planarity, . By Lemma 9, , which is a contradiction. ∎
An immediate implication of the discharging rule in Definition 7 is that every face pulls a non-negative integer weight over every edge, as every satisfied condition adds 1 to that weight. We next prove that no two of the conditions C1–C7 are satisfied simultaneously for the same face and edge ; hence, pulls either weight 0 or 1 over . This is crucial for keeping the amount of upcoming arguments on a maintainable level; in fact, our conditions were designed that way.
Lemma 12.
The total weight pulled by a face of over its -edge is either 0 or 1. If it is 1, the -opposite face does not pull any weight over .
Proof.
Assume to the contrary that is incident to two faces and of such that and for conditions and and ; without loss of generality, we assume that is not stated before in Definition 7. In general, implies . If , is major, which implies and thus ; then is major, which contradicts . Hence, , so that both and are minor. If , Lemmas 104 and 11 imply that and is an extremal -edge of two consecutive 3-arches; then by Lemma 9 and by planarity, which is a contradiction. Hence , which implies .
We distinguish the remaining options for and in .
Assume ; by our notational convention, is the 2-sandwiched -vertex of . Then , which implies , as the remaining options for require . Since is thick, exists. By Lemma 9 (for C2), . Assume . If is not incident to any extremal -edge of (see Figure 1(c)), contains neither nor by Lemma 9 (for C2), which contradicts . Otherwise, consider Figure 2(a). Then implies , which contradicts Lemma 9 (for C4). If , contradicts the assumption of that case. If , Lemma 9 implies that neither nor is contained in , which contradicts .
Assume . Then is the middle -edge of a 3-arch of , which does not satisfy and by definition of these conditions. We conclude . Then contradicts Lemma 9, and contradicts that is plane.
Assume . If , , as 2-faces do not have 3-arches, so let . Assume (see Figure 2(b) for ). By Lemma 9 (applied for and ), neither nor is adjacent to a vertex of in , so that is a 2-separator of . If , we get a contradiction to planarity.
Assume . If , by planarity, so let . Assume . By Lemma 9 (applied for and ), neither nor is adjacent to a vertex of in , so that is a 2-separator of . If , we get a contradiction to planarity.
Assume . Then and thus , which contradicts that is plane. ∎
By Lemma 12, we know that whenever weight 1 is pulled over some edge by some condition , no other condition is satisfied on and 1 is the final amount of weight transferred over .
3.4 The Proof
Throughout this section, let denote the weight function on the set of faces of after our discharging rule has been applied. Clearly, still holds. For and the set of -edges of a face of , let the (weight) contribution of to be (i.e. the initial weight these edges give to ) plus the sum of weights pulled by over edges in minus the sum of weights pulled by opposite faces of over edges in . The contribution of an arch to is the contribution of the -edges of to ; in particular, every proper arch contributes weight to . Note that we have if a set contributes weight to , as may loose weight at most 1 for every of its -edges that is not in by Lemma 12, which cancels the initial weight 1 given by this -edge.
Lemma 13.
For two -edges and of a minor face , let and such that and and are not contained in . Then
1. (and thus ) is either an extremal -edge of or adjacent to that, and
2. and have distance at least three in .
Proof.
For Claim 1, assume to the contrary that is neither an extremal -edge of nor adjacent to that. Then and . Since is minor, . Since the definition of requires an edge that is not incident to , . For , the edge in Figures 1(c)1(e) is not incident to , which contradicts the choice of .
For Claim 2, assume to the contrary that and have distance at most two in . Since is minor, . Let and let be the 3-arch of that has -edge . Then the existence of and planarity imply , and Lemma 9 implies . Hence, . Then by planarity and the fact that, in the notation of the conditions , and , we have and no 3-arch with -edge . This is a contradiction. ∎
Lemma 14.
Let , be the -opposite face of , be the arch of shown in Figure 2 (for , let be the proper arch of ), and be the set of common -edges of and .
• If and , contributes weight at least to .
• If , contributes weight at least to .
• If , contributes weight at least to .
Lemma 15.
For a minor face , let be an arch of with minimal such that a face pulls weight over a -edge of by Condition . Then .
In particular, we may choose in Lemma 15 as the proper arch of . This implies the following helpful corollary.
Corollary 16.
Every minor face that has a -edge over which an opposite face of pulls weight by or satisfies .
We now show that Inequalities 1 and 2 hold, which proves the Isolation Lemma.
Lemma 17.
Let be a face of . Then , if is thick and minor, if is a thin minor 2-face, and if is thin and minor such that .
Proof.
By Lemma 12, an opposite face of pulls weight at most weight one over any -edge of . Since the initial weight of such an edge for is one, we have . In the remaining proof, let be minor. Assume that has a -edge such that the -opposite face of pulls weight over . Then by Lemma 15. We therefore assume that no opposite face of pulls weight over a -edge of by Condition or .
Let . If is thick, Condition and Lemma 12 imply . If is thin, assume to the contrary that . Then has a -edge such that for the -opposite face of . Since , Lemma 13 implies that the -edge of different from contributes weight one to . This contradicts .
Let . Assume to the contrary that has an -edge such that for the -opposite face of . Then, for every , Lemma 9 contradicts that is a -edge of . Hence, . If is thin, this gives the claim, so let be thick. Let be the middle -edge of . Then is not an extremal -edge of a 3-arch, as then is extendable by the replacement consisting of the two 3-arches and their two common -edges. Hence, , which gives the claim .
Let . Assume that has an -edge such that for the -opposite face of , as otherwise . Let . If (this is not possible for by Lemma 9), is extendable by Figure 3(a) for | 7,917 | 32,875 | {"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-2022-05 | longest | en | 0.941324 |
http://www.thestudentroom.co.uk/wiki/Revision:Introduction_to_Groups | 1,477,338,525,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719677.59/warc/CC-MAIN-20161020183839-00414-ip-10-171-6-4.ec2.internal.warc.gz | 734,515,220 | 30,716 | • # Revision:Introduction to Groups
TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > Introduction to Groups
Many mathematical objects have some degree of symmetry - by this, we mean that we can do something to them which doesn't change what they look like. As an example, consider a square. We can rotate it by 90 degrees (a quarter turn), 180 degrees (a half turn) or 270 degrees (a three quarter turn) and it will look the same afterwards. We can also reflect it in its horizontal and vertical axes of symmetry, or either of its diagonal axes of symmetry, and it will look the same afterwards.
More importantly, we can perform any of these rotations or reflections in a row, and the square will still look the same. If we call any rotation or reflection of the square a transformation then we can talk of composing transformations, i.e. doing them one after the other. We say that any composition of transformations gives us another transformation.
We gave seven examples of transformations here (three rotations and four reflections). Are there any more symmetries? In fact there is just one more, which is the most obvious one of all. We could simply do nothing to the square, which will obviously leave it looking the same! This gives us eight symmetries in total. It might seem weird to think of "doing nothing" as a transformation, but it is important that we do so -- you will see why later.
As a last point in this introduction, we should point out that you can 'undo' any of the transformations. If you rotate the square, you can 'undo' that rotation by rotating the square in the opposite direction, which is the same as doing nothing. If you reflect the square in one of its axes of symmetry, you can 'undo' that reflection by reflecting again in the same axis of symmetry. This is why it's important to count doing nothing as a transformation -- we want the composition of any two transformations to be another transformation.
This all looks very interesting, and it would be good to see if we can find something like a general mathematical theory for talking about symmetry in this way. In fact we can, and that is what we call group theory.
# Definitions
We define a group formally by saying that a group G is a set (whose elements are our transformations) with a binary operation , which we use to talk about the composition of transformations. Formally, is a function from G G G. To be a group, our set G and its binary operation have to satisfy four axioms:
A1. If a and b are elements of G then a b must also be an element of G. If this is true then we say that G is closed.
A2. The operation must be associative, i.e. for any a, b, cG we must have a (b c) = (a b) c.
A3. We must have an identity element eG, which represents doing nothing. The identity element satisfies e g = g, and g e = g.
A4. We must be able to undo any of our transformations. We say that for every gG we must have an inverse element g-1, which satisfies g-1 g = e, and also g g-1 = e.
These four properties of closure, associativity, identity and inverse define a group.
# Examples of groups
### Symmetry group of the square
We already saw that the transformations of a square which leave it unchanged form a group. We can express this in a more mathematical language. Let's write ρ as a shorthand for the transformation "rotate by 90 degrees". If we want to do a 180 degree rotation then we simply do two 90 degree rotations one after the other. We write this using our composition operation, as ρ ρ, which we read as "perform ρ, then perform ρ again". In fact we can compress this notation even more -- we think of composition as something like a multiplication, and we just write ρ2 to mean ρ ρ.
Similarly, if we want to do a rotation by 270 degrees then we simply rotate by 90 degrees three times, and we write this as ρ3. Now if we rotate once more, then we are back to where we started -- that is, rotating by 360 degrees is the same as not rotating at all! If we use the notation e for our identity element "do nothing" then we can write ρ4 = e.
The reflections are a little more complicated. We will use the notation m to mean "reflect in the vertical axis of symmetry", i.e. reflecting the square left-to-right. Now we could come up with new symbols for the other three reflections, but it turns out that each of the other reflections are just combinations of rotations with the left-right reflection! It's easiest to see this with the help of a picture:
### The integers
The integers Z form a group under the operation of addition. To check this, we just need to check that each of our four properties of closure, associativity, identity and inverse are satisfied. We can do these one by one:
A1. (Closure). We need to know that whenever a, bZ then a + bZ as well. But this is obvious, because adding together two whole numbers always gives another whole number!
A2. (Associativity). We want to show that whenever a, b, cZ we have (a + b) + c = a + (b + c). But again this is obvious, because brackets don't matter when we're adding numbers together.
A3. (Identity). We want to find an integer which leaves every other number unchanged when you add it -- but this is exactly what zero does. We have a + 0 = a and 0 + a = a, so our identity is 0.
A4. (Inverse). For every integer aZ we want to find an integer b such that a + b = b + a = 0. But if we choose b = -a then this is satisfied for every a, and so the inverse of a is just -a.
Since all four of our properties are satisfied, Z must be a group under addition.
Notice that Z is not a group under the operation of multiplication. The identity would have to be 1, since a x 1 = 1 x a = a. But then 0 x a = 0 for every integer a, so we could never find an integer a which is the inverse of 0. Since 0 doesn't have an inverse, Z is not a group under multiplication. This is an important point to notice -- the operation that we choose is just as important in defining the group as the set which we apply that operation to.
In our introduction, we said that groups were associated with symmetries of objects. What object are the integers the symmetry group of? The answer is curious -- the integers are the symmetry group of themselves! An intuitive picture of the integers is as points on a number line, stretching off to infinity in both directions. If we add an integer a to every number on this number line then we will just get another copy of the number line. What used to be the point -a will now be the point 0, and what used to be the point 0 will now be the point a. However, looked at as a whole, the number line will look exactly the same as it did before you added a to every number.
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Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE | 1,696 | 7,261 | {"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} | 4.34375 | 4 | CC-MAIN-2016-44 | latest | en | 0.945829 |
https://www.codecogs.com/library/engineering/fluid_mechanics/orifice/geometry/drowned-orifice.php | 1,601,032,010,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400223922.43/warc/CC-MAIN-20200925084428-20200925114428-00507.warc.gz | 773,196,911 | 7,903 | I have forgotten
• https://me.yahoo.com
# Drowned Orifice
Discharge through a drowned orifice
View version details
### Key Facts
Gyroscopic Couple: The rate of change of angular momentum () = (In the limit).
• = Moment of Inertia.
• = Angular velocity
• = Angular velocity of precession.
Blaise Pascal (1623-1662) was a French mathematician, physicist, inventor, writer and Catholic philosopher.
Leonhard Euler (1707-1783) was a pioneering Swiss mathematician and physicist.
## Discharge Through A Partially Drowned Orifice
If the outlet side of the orifice is partly under water, it is known as a partly drowned or partly submerged orifice as shown in fig.2.
The discharge through a partially drowned orifice is obtained by treating the lower portion as a drowned orifice and the upper portion as an orifice running free, and then by adding the two discharges thus obtained.
We know the discharge through the free portion,
and discharge through the drowned orifice,
Now total discharge,
Example:
[metric]
##### Example - Discharge through a partially drowned orifice
Problem
An orifice in one side of a large tank is rectangular in shape, 2 meters broad and 1 meter deep. The water level on one side of the orifice is 4 meters above its top edge. The water level on the other side of the orifice is 0.5 meter below its top edge as shown in fig.
Calculate the discharge through the orifice per second if = 0.63
Workings
Given,
• = 2m
• = 1m
• = 4m
• = 4+1 = 5m
• = 4+0.5 = 4.5m
• = 0.63
Since the orifice is partially drowned, therefore let us split up the orifice into two portion will be treated as a free orifice and the lower portion as a drowned orifice.
The discharge through the free portion of the orifice,
The discharge through the drowned portion of the orifice,
Total discharge = = 2.88 + 5.92 = 8.8m3 /s
Solution
Total discharge = 8.8m3 /s | 491 | 1,886 | {"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": 16, "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-2020-40 | longest | en | 0.922636 |
https://svesterauholzappel.de/how/much/is/the/boiler/consume/fuel/monthly/16376.html | 1,627,081,969,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150067.51/warc/CC-MAIN-20210723210216-20210724000216-00353.warc.gz | 560,138,438 | 4,501 | how much is the boiler consume fuel monthly
### How To calculate fuel consumption in boiler ?? - YouTube
Use our free and easy-to-use AFUE Savings Calculator to instantly see how much you can save by upgrading to a more energy efficient heating system. Compare furnaces or boilers with different AFUE ratings and fuel types (natural gas, oil, propane, or electricity).
### How Much Gas Does a Boiler Use Per Hour? | Viessmann
How Much Gas Does a Boiler Use Per Hour? | Viessmann
### Boiler Fuel Consumption Calculation--ZBG
For example, The calculation formula of fuel demand for steam boiler(saturated steam boilers) is as follows: Bj=B (1-q4/100) Bj- calculation of fuel consumption, kg/h B- fuel consumption, kg/h q4- boiler mechanical incomplete combustion loss,% B=100Q/ (Qdw*) B- fuel consumption, kg/h Q- effective heat utilization of boiler, kj/h Qdw- Low calorific value of fuel, kj/kg η- boiler thermal efficiency,%. Q = D (Ibh-Igs-rω / 100) + Dpw (Ibs-Igs) Q- boilers use
### How Much Home Heating Oil Will I Use This - COD Fuel Oil
Using kW and kWh, you can work out how much gas your boiler is using and therefore how much it costs. A 24 kW boiler will use 24 kWh of energy per hour. According to Choose, the cost is approximately 3.8 pence per kWh, which means it would cost around 91 pence to run a 24 kW boiler for one hour. This will change depending on the cost of gas and the size and age of your boiler.
### How to Calculate Gas Consumption in a Gas Burner | Hunker
Mar 24, 2020 · Let's understand more about boiler efficiency and common methods used for calculations of gas consumption. The initial cost of a boiler is usually less than its time to time maintenance and other associated expenses. Quality boilers tend to consume less energy, demand less maintenance, and cost less than traditional boilers in the long run.
### How to Check Gas Consumption of the Boiler
Estimating the Fuel Consumption of Boilers and Furnaces
### How Much Gas Does Your Boiler Consume? | 128 Plumbing
Oct 22, 2019 · Gas boiler fuel consumption depends on a number of factors, including the age and condition of the boiler, its type, and how much cold air seeps into your home. If you want to find out exactly how much gas your boiler consumes, calculate it from your gas bill, taking the gas consumption of other appliances into account. | 549 | 2,358 | {"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} | 3.171875 | 3 | CC-MAIN-2021-31 | latest | en | 0.895775 |
https://socratic.org/questions/59b8dedf11ef6b08ef85ae10 | 1,576,332,518,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541157498.50/warc/CC-MAIN-20191214122253-20191214150253-00038.warc.gz | 542,163,321 | 5,968 | # Question 5ae10
Sep 13, 2017
I tried this:
#### Explanation:
The two scales are related but in a slight complicated way (because the difference in intervals between fixed points of the two). Anyway, we have:
"Temp. in Celsius"=5/9xx("Temp. in Farhenheit"-32)#
you need to insert $98.6$ into the formula and evaluate the Temperature in Celsius.
[You should get ${37}^{\circ} C$] | 107 | 385 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2019-51 | latest | en | 0.816783 |
https://zbmath.org/0862.33003 | 1,657,039,424,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104585887.84/warc/CC-MAIN-20220705144321-20220705174321-00162.warc.gz | 1,165,640,747 | 11,189 | ## Pfaff’s method. II: Diverse applications.(English)Zbl 0862.33003
The paper discusses Pfaff’s proof of the Saalschütz summation, which actually preceded Saalschütz’ work by hundred years. Set $$S_n(a,b,c) ={}_3F_2{-n,a,b;1 \choose c,1+a +b-c-n}$$, and $$\sigma_n (a,b,c) = {(c-a)_n(c-b)_n \over (c)_n (c-a-b)_n}$$. The Saalschütz summation states $$S_n(a,b,c) = \sigma_n (a,b,c)$$, and Pfaff proved it in the simplest possible way: showed that $$S_n(a,b,c) - S_{n-1} (a,b,c)$$ and $$\sigma_n (a,b,c)- \sigma_{n-1} (a,b,c)$$ admit the same recurrence. This Pfaffian approach is shown to be effective for Bailey’s, Dougall’s, Lakin’s and Kummer’s summation identities. It is noted that the Pfaffian approach seems most effective for balanced and well-poised hypergeometric series. It is often the case that the Pfaffian approach has to prove a cluster of related identities, and not just one of them.
### MSC:
33C20 Generalized hypergeometric series, $${}_pF_q$$ 05A19 Combinatorial identities, bijective combinatorics
Full Text:
### References:
[1] Andrews, G.E., On the q-analog of Kummer’s theorem and applications, Duke math. J., 40, 525-528, (1973) · Zbl 0266.33003 [2] Andrews, G.E., Connection coefficient problems and partitions, (), 1-24 · Zbl 0186.30203 [3] Andrews, G.E., Plane partitions III: the weak Macdonald conjecture, Invent. math., 53, 193-225, (1979) · Zbl 0421.10011 [4] G.E. Andrews, Pfaff’s method I: the Mills-Robbins-Rumsey determinant, Discrete Math., to appear. · Zbl 1069.15009 [5] Andrews, G.E.; Burge, W.H., Determinant identities, Pacific J. math., 158, 1-14, (1993) · Zbl 0793.15001 [6] G.E. Andrews and D.W. Stanton, Determinants in plane partitions enumeration, to appear. · Zbl 0908.05007 [7] Askey, R., Variants of Clausen’s formula for the square of special _{2}F1, (), 1-12, Bombay · Zbl 0756.33002 [8] Bailey, W.N., Some identities involving generalized hypergeometric series, (), 503-516, (2) · JFM 55.0219.05 [9] Bailey, W.N., Generalized hypergeometric series, (1935), Cambridge Univ. Press London and New York, (Reprinted: Hafner, New York, 1964) · Zbl 0011.02303 [10] Burchnall, J.L.; Lakin, A., The theorems of saalschutz and dougall, Quart. J. math. Oxford, 1, 2, 161-164, (1950) · Zbl 0040.03401 [11] Daum, J.A., The basic analog of Kummer’s theorem, Bull. amer. math. soc., 48, 711-713, (1942) · Zbl 0060.19808 [12] Dougall, J., On Vandermonde’s theorem and some more general expansions, (), 114-132 · JFM 38.0313.01 [13] Gessel, I.; Stanton, D., Strange evaluations of hypergeometric series, SIAM J. math. anal., 13, 295-308, (1982) · Zbl 0486.33003 [14] Jackson, F.H., Transformations of q-series, Messenger math., 39, 145-153, (1910) [15] Jackson, F.H., Summation of q-hypergeometric series, Messenger math., 50, 101-112, (1921) [16] Lakin, A., A hypergeometric identity related to Dougall’s theorem, J. London math. soc., 27, 229-234, (1952) · Zbl 0046.07401 [17] Mills, W.H.; Robbins, D.P.; Rumsey, H., Enumeration of a symmetry class of plane partitions, Discrete math., 67, 43-55, (1987) · Zbl 0656.05006 [18] Pfaff, J.F., Observationes analyticae ad L. Euler institutiones calculi integralis, vol. IV, supplem. II et IV, historia de 1793, Nova acta acad. sci. petropolitanae, 11, 38-57, (1797) [19] Saalschultz, L., Eine summationsformel, Z. math. phys., 35, 186-188, (1890) · JFM 22.0262.03 [20] Wilf, H.S.; Zeilberger, D., Rational functions certify combinatorial identities, J. amer. math. soc., 3, 147-158, (1990) · Zbl 0695.05004
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 1,279 | 3,869 | {"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} | 2.578125 | 3 | CC-MAIN-2022-27 | longest | en | 0.743936 |
https://jacobbradjohnson.com/faculty-of-education-centre-for-security-studies-to/ | 1,556,254,410,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578759182.92/warc/CC-MAIN-20190426033614-20190426055614-00540.warc.gz | 433,192,451 | 9,215 | ## FACULTY OF EDUCATION CENTRE FOR SECURITY STUDIES TO
FACULTY OF EDUCATION
CENTRE FOR SECURITY STUDIES
TO : MR PATRICK JAMBO
FROM : MASTON CHITSONGA
REG. NO : CSS/06/1
COURSE TITTLE: INTRODUCTION TO GEOSPATIAL
TECHNOLOGIES
COURSE CODE : SSGE 1201
TASK : FUNDAMEMTALS OF A GOOD MAP
DUE DATE : 30th JUNE, 2017
FUNDAMENTALS OF A GOOD MAP
A map is a diagrammatic representation of an area of land or sea showing physical features, cities, roads etc. (https: // www google. com.)
For a map to qualify a good map, it must be drawn in line with the required basics. These basics are also referred to fundamentals of a map.
The first fundamental of a good map is the scale. A scale of a map is defined as the ratio of a distance on a map to the corresponding distance on the ground. Martin, (1995), as cited by Heywood Cornelius Carver, p: 260. Scale gives an indication of how much smaller than reality a map is. It is also important when using spatial entities like: points, lines and areas to represent generalised two dimensional versions of real world features. There are three ways of presenting a scale on a map. The three ways are as follows: verbal scale, graphical scale and ratio scale. (Heywood Cornelius Carver, 2011 p.260).
The second fundamental of a good map is the legend. Legend is a key of the map. Information needed to read a map is found in a map legend. It provides colour and symbol. It is used to look up details for the map element. A legend is a standard element on most lay outs. Wider audience maps may exclude certain given feature types drawn with standard symbology such as blue water bodies or green land masses, however, these are left to the discretion of the map author. Complicated legends with many items necessitate the usage of grouping levels. The two forms of grouping levels most commonly seen are: the categorical group and the shape type group. When categorical separations are not needed, shape type groupings are often displayed in points, lines and polygons. (Gretchen N. Peterson, p.34).
The third fundamental of a good map is the tittle. A good map should have a tittle which its purpose is to pronounce the intent of the map. It also identifies the geographical location of the map as well as the authorising agency. The tittle is either the primary or secondary layout element and if secondary, it is only second to the map element. Tittles should be spelled out, and avoid acronym at all cost. Avoid any redundant terms such as “map of ……” or “analysis of……” Avoid using jargon such as “framework,” or “model,” as these result in useless mind confusion for the reader. Cartographers may list the tittle simply or artistically. Tittles of maps typically appear at the top of the map, but not always. (Paul A. Longly, Michael F. Good Child, David J. Magwire and David W. Rhind.)
The fourth fundamental of a good map is a directional indicator. A good map should have a directional indicator which illustrate the orientation of a map to the viewer. Some cartographers place an arrow that points to the North Pole on the map. This is a “North arrow.” The North Arrow enables your map audience to be familiar about where the North is. It also shows at least four, and sometimes more directions. Other maps indicate direction by using a “compass rose,” with arrows pointing to all four cardinal directions. (Daniel Dorling & David Fairbain, 2014.)
The fifth fundamental of a good map is projection. Map projection is a systematic transformation of the latitudes and longitudes of locations from the surface of the sphere or an ellipsoid into locations on a plane. Maps cannot be created without map projections. Map projections transforms the earth on to a two dimensional surface. In doing so, they approximate the true shape of the earth. All map projections necessarily distort the surface in some fashion. Transverse Mercator projection and Mollweide projection are two forms of map projection. There are many types of map projections but the common used types are: cylindrical projection, azimuthal projection, and conic projection. (Anson, R.W. & Ormerling, F.J., 1993)
The sixth fundamental of a good map is a date. A date and year on which the lay out was printed must be included on a map. The dates for your data sources are addressed in the data citation section. It is important to include a date on most lay outs that are intended to be stand-alone prints. The date gives the audience an idea of the maps vintage for maps that endure. (Gretchen N. Peterson, p.40).
The seventh fundamental of a good map is inset maps. Some maps feature inset maps. Inset maps are smaller maps on the same sheet of paper. They provide additional information not shown on the larger map. Inset maps are drawn at a larger more readable scale. They usually feature areas of interest related to the large map. Inset maps are commonly used on tourist and travel maps. (https: www. bsu. edu/library/ collections/ gcmc/ feedback form/)
To sum up, scale, legend, tittle, directional indicator, projection, date, and inset maps are fundamentals of a good map. Learning more about common fundamentals of maps, will enhance the understanding of the world. Maps really are a geographer’s most important tool.
References
Anson, R.W., ; Omerling, F.J. (1993) Basic Cartography. Oxford: Butterworth Heinemann.
Carver H.C. (2011), Introduction to Geographical Information Systems. Harlow: Pearson
Education.
Dorling D. ; Fairbain D. (2014), Ways of Representing the World. London and New York:
Rout ledge Taylor and Francis Group.
https: www. Bsu. edu/library/collections/gcmc/feedback form/ Accessed on 23rd June, 2017.
https://www. Google. Com/ Accessed on 20th June, 2017.
Longley P.A., Good child M.F., Magwire D.J. ; Rhind D.W. (2005), Geographical
Information Systems and Science. Barcelona: Grafos SA.
Peterson G.N., GIS Cartography, a Guide to Effective Map Design. London: CRS Press
Tylor ; Francis Group. | 1,390 | 5,946 | {"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} | 2.859375 | 3 | CC-MAIN-2019-18 | latest | en | 0.859725 |
https://se.mathworks.com/matlabcentral/cody/problems/63-encode-roman-numerals/solutions/1404806 | 1,604,076,704,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107911027.72/warc/CC-MAIN-20201030153002-20201030183002-00529.warc.gz | 508,123,711 | 17,019 | Cody
# Problem 63. Encode Roman Numerals
Solution 1404806
Submitted on 4 Jan 2018 by Daniel Rubin
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
x = 1990; y_correct = 'MCMXC'; assert(isequal(dec2rom(x),y_correct))
2 Pass
x = 2008; y_correct = 'MMVIII'; assert(isequal(dec2rom(x),y_correct))
3 Pass
x = 1666; y_correct = 'MDCLXVI'; assert(isequal(dec2rom(x),y_correct))
4 Pass
x = 49; y_correct = 'XLIX'; assert(isequal(dec2rom(x),y_correct))
5 Pass
x = 45; y_correct = 'XLV'; assert(isequal(dec2rom(x),y_correct))
6 Pass
x = 0; y_correct = ''; assert(isempty(dec2rom(x)))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 257 | 838 | {"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} | 2.546875 | 3 | CC-MAIN-2020-45 | latest | en | 0.53525 |
https://chess.stackexchange.com/questions/21417/can-you-mate-with-each-of-your-16-pieces-on-its-original-square | 1,701,846,038,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100583.31/warc/CC-MAIN-20231206063543-20231206093543-00007.warc.gz | 202,216,646 | 47,272 | # Can you mate with each of your 16 pieces on its original square?
An opinion poll conducted by problem composer Roberto Osorio over 10 master-level chess players in the Argentine Chess Club produced 100% wrong answers! :-)
When the right answer is posted here, I will post a follow-up problem.
EDIT: OK here is the follow-up:
``````[title "R.Osorio - StrateGems 2007 "]
[fen "rnb2rqb/pppp4/6p1/7p/6n1/7k/PPPPPPNP/RNBQKBNR w - - 0 1"]
``````
This is the position after White's 16th move. What was the exact sequence of moves?
• Can you mate with each of your 16 pieces on its original square? Yes. Let's say, you have all your pieces, he has just the king, then you can mate him on c2. Drive him to c8 with your pieces, put your queen to e7. Next, place your remaining pieces on their original squares, take care that you don't stalemate him. Then it is a mate in 8 with your queen on d1. A player already did this in his games and posted it to the forum on Lichess, but I didn't bookmark it. Mar 2, 2022 at 23:23
Obviously the last white move was 16. Nf3g1+.
I note that black needs at least 15 moves to place the pieces as they are:
• 4 moves to swap queen with rook as they are (various movements possible),
• 2 moves for the bishop (Bf8g7, Bg7h8),
• 2 moves for the knight (Ng8f6, Nf6g4 or Ng8h6, Nh6g4)
• 1 move each for the two pawns (g7g6 and h7h5)
• 5 moves for the king (any direct route)
Together with the information that it is black's 16th move, this means that all of black's pieces must have taken a direct route to their destination.
Note that black did not capture any white pieces, because white still has 16 pieces on the board. This means in particular that the pawns on g6 and h5 are the pawns from g7 and h7 respectively.
White has promoted the g pawn for a knight. Since black has still the original g pawn, and all pieces except for two pawns, this is only possible, if the white g pawn moved to g6, captured the black pawn on f7 and promoted on f8 for a knight.
The white pawn could only be promoted after the square on f8 was emptied, i.e. the order of moves is roughly: 1) white pawn moves to g6, 2) white pawn captures black pawn on f7, 3) black plays g7g6, and Bf8g7
Before white captured g6xf7, the black e pawn must have been gone as otherwise (with e7, Qd8, Bf8 and Ke8 present), black would have been forced to recapture the white pawn on f7 since it is check (and the black king would have no squares to escape).
Counting the minimum number of white moves now until the position after blacks's 15th move (i.e. with white knights on f3 and g2)
• white's g pawn: 5 moves (g2g4, g4g5, g5g6, g6xf7 f7f8N)
• return the promoted knight from f8 to either f3 or g2: 3 moves
• capturing the e7 pawn and returning the piece: If this was done by the rook, it would take at least 8 moves. This would make the total (5+3+8>15), so is not possible. So e7 was captured by a white knight, which returned. Both options (a) white knight from g1 capturing e7 and returning to g2 or f3 or (b) white knight from b1 capturing e7 and returning to b1 plus the move Ng1f3; would take 7 moves.
5+3+7=15, meaning that also white moved its pieces on shortest possible routes without any time for waiting moves.
As for the rest, basically it comes down to check how to achieve the main plans in order, i.e.:
1. capture the pawn on e7 with a knight and return the knight at least partially (so that the black king is able to get to e7); as mentioned this needs to be done before white captures gxf7
2. capture gxf7 and promote the pawn
3. return all white pieces to their final destination
During phase "1." black is very limited in moves since the g pawn cannot move yet (has to wait for the white pawn to play gxf7). This leaves as the only possible first moves for black 1....h5 2...Rh6 ... 3...Rf6 4....Nh6.
These four moves are just enough time for white to capture the e7 pawn with the knight on b1 and to return the knight to c3 (making space for the king on e7, i.e. giving black additional "waiting" moves (waiting for white to push the pawn to g6 and play gxf7). Note that white has to capture the e7 pawn with the knight on b1, because trying to capture it with the knight on g1 would take one move longer and black would run out of waiting moves. So the first four moves are: 1. Nc3 h5 2. Nd5 Rh6 3. Nxe7 Rf6 4. Nd5 Nh6 5. Nc3 Ke7.
Phase two starts and white has to push the g pawn forward, because playing for instance 6. Nb1 black would run out of "waiting" moves and would not manage to regroup queen/rook/bishop in time. So, the next three and a half moves (6. g4 Ke6 7. g5 Kf5 8. g6 Kg4 9. gxf7 ) are basically black pushing the king on the shortest available route towards h3 and white pushing the g pawn forward. As mentioned black has to wait with playing g6 until white captures gxf7, so black can only push the king during this phase.
The final phase (9....g6 10. Nf3 Bg7 11. f8=N Bh8 12. Ne6 Qg8 13. Nf4 Rf8 14. Ng2 Kh3 15. Nb1 Ng4 16. Ng1#) is basically just moving the pieces to their final squares. There are no alternative move orders possible. For instance white cannot play Nb1 earlier, because with 14. Ng2 he is just in time to block the bishop from attacking h3, allowing 14. ... Kh3.
• White has enough moves left to capture the e7 pawn with the queen's knight and return. However, I can find only three waiting moves (h5, Nh6, Rh7) which Black can play before the King must move out. Apr 18, 2018 at 18:05
• @Glorfindel: Black could move the rook via h6-f6 to f8, giving him an extra waiting move. Apr 18, 2018 at 18:11
The thoughts mentioned in user1583209's answer more or less summed up mine, but I couldn't find a way to get the king out in time without requiring extra moves. The final trick is
to have the b1 knight capture the e7 pawn; this opens the way for the black king. Black has just enough moves (thanks again @user1583209) he can make with his kingside:
``````[FEN ""]
1. Nc3 h5 2. Nd5 Rh6 3. Nxe7 Rf6 4. Nd5 Nh6 5. Nc3 Ke7 6. g4 Ke6 7. g5 Kf5 8. g6 Kg4 9. gxf7 g6 10. Nf3 Bg7 11. f8=N Bh8 12. Ne6 Qg8 13. Nf4 Rf8 14. Ng2 Kh3 15. Nb1 Ng4 16. Ng1#
``````
• Why is it not possible to capture e7 with the g1 knight? Apr 18, 2018 at 18:15
• Takes one more move, so one more 'waiting' move from Black. Apr 18, 2018 at 18:16
• Ah, I see, the e7 pawn needs to be captured quickly and only the b1 knight can do that. Amazing problem and all of it forced as far as I can see. Apr 18, 2018 at 18:17
• @user1583209 Great team work, glor & user! Roberto's theme is 3 white knight circuits. I will give the tick to user because the detail of his response is educational for anyone trying to learn about proof games. (This is a retro area of chess problem composition where we are definitely in the "golden age" right now, thanks to composers like Roberto, Nicolas Dupont, Silvio Baier, Andrei Frolkin, Michel Caillaud etc.) Apr 19, 2018 at 5:57
I'd say
yes, but you need to allow at least one pawn to promote (and return to its original square). E.g. promote the white a pawn to a queen, move it back to a2 and put a knight on a3. Then, move the black king to a4, black pawns to a5 and b4 and a black rook to b5. In the meantime, White can just play Ng1-h3-g1 etc. The move Na3-b1 will then be checkmate, and the white pieces will all be on their original square.
• The question should be "Can you mate with 15 out of your 16 pieces on their original squares?" Apr 17, 2018 at 19:39
• @Glorfindel yes that's the solution. Clearly one might ask "but is a promoted pawn the same piece?" This is a question of semantics, and we don't want to spoil Roberto's excellent joke. Now I will post one of his problems. If you solve it, you definitely get the "tick mark" for correct answer. If someone else solves it, I'll have to decide if you get the tick mark or they do :-) Apr 18, 2018 at 2:10
• Great! It's better to post the follow-up as a new question. With the diagram, you're already spoiling the solution. Apr 18, 2018 at 5:40
• @Glorfindel: thanks for your comment. If I had posted as 2 questions then someone might come along later and say that Q1 is just a subset of Q2 and vote to close it. I don't think your solution is "spoiled" at all. It remains good and you have got a decent amount of reputation from it. I did point out in my initial post that there would be a second part. Bottom line: from a long term stackexchange perspective, it is better to have one post about Roberto's idea rather than two. Apr 18, 2018 at 17:46
After some thinking, as I did not look at the other answers, I concluded that it is indeed possible to checkmate with all 16 pieces on their original squared. Doing this requires a promoted piece to travel back to its “original square.”
Then I looked at Roberto’s excellent problem and thought “there must be a faster way to do that.” Of course, his problem is special in its own right because there is only one way to reach the position. But the effect that it shows is beatable without the uniqueness
There are two variants to consider. Type #1 focuses on the least possible disturbance, meaning that Black has 13 pieces (think about it) at home when mated. Type #2 is without regard to that. @Rebecca J. Stones did much-appreciated work on this in the comments. Here are the results.
``````[Title "Rebecca J. Stones, chess.stackexchange.com 5/10/2019, Type #1, Non-Unique Proof Game In 11.5 Moves"]
[FEN ""]
1. g4 g5 2. Nf3 f5 3. gxf5 Bh6 4. f6 Kf8 5. f7 Kg7 6. f8=N Kf6 7. Ne6 Kf5 8. Nf4 Kg4 9. Ng2 Kh3 10. Ne5 Bf8 11. Nf3 g4 12. Ng1#
``````
``````[Title "Rebecca J. Stones, chess.stackexchange.co 5/11/2019, Type #2, Non-Unique Proof Game In 7.5 Moves"]
[FEN ""]
1. a4 f6 2. a5 Kf7 3. a6 e5 4. axb7 Ne75. bxc8=Q Kg8 6. Qxd7 Ng6 7. Qa4 Nd7 8. Qa2#
``````
Secondly, it is also of interest what can be done with discovered checks.
``````[Title "me, chess.stackexchange.com 5/10/2019, Type #1, Non-Unique Proof Game In 15.5 Moves"]
[FEN ""]
1. h4 g5 2. hxg5 f6 3. g6 Nh6 4. g7 Kf7 5. g8=R Ke6 6. Rg3 Kf5 7. Rgh3 Kg4 8. R3h2 Ng8 9. Nh3 Kh5 10. g4+ Kh4 11. g5 Nh6 12. g6 Ng8 13. g7 Nh6 14. g8=R Kh5 15. Rgg2 Ng8 16. Ng1#
``````
``````[Title "Rebecca J. Stones, chess.stackexchange.co 5/11/2019, Type #2, Non-Unique Proof Game In 10.5 Moves"]
[FEN ""]
1. Nf3 h5 2. e4 h4 3. e5 f6 4. exf6 Kf7 5. fxe7 Kg6 6. exd8=Q Kh5 7. Qe7 g6 8. Qee2 Rh6 9. Nc3 Be7 10. Nb1 Bg5 11. Ng1#
``````
• Here's a 12-mover: `1. g4 g5 2. Nf3 f5 3. gxf5 Bh6 4. f6 Kf8 5. f7 Kg7 6. f8=N Kf6 7. Ne6 Kf5 8. Nf4 Kg4 9. Ng2 Kh3 10. Ne5 Bf8 11. Nf3 g4 12. Ng1#` It also achieves the minimum number of black pieces not on their original squares. May 10, 2019 at 22:30
• I found this 8-mover if we don't care about black's pieces, where the a pawn promotes to a queen: `1. a4 f6 2. a5 Kf7 3. a6 e5 4. axb7 Ne7 5. bxc8=Q Kg8 6. Qxd7 Ng6 7. Qa4 Nd7 8. Qa2#`. May 11, 2019 at 1:06
• Good job on getting it faster. Do note that the problem by Osorio is a proof game. What it lacks in speed, it makes up for with uniqueness. There is no other way to reach the diagrammed position after white's 16th move with any other move sequence than the solution. May 11, 2019 at 1:19
• I found one in 11 moves: `1. Nf3 h5 2. e4 h4 3. e5 f6 4. exf6 Kf7 5. fxe7 Kg6 6. exd8=Q Kh5 7. Qe7 g6 8. Qee2 Rh6 9. Nc3 Be7 10. Nb1 Bg5 11. Ng1#`. I'm not sure how good this is. May 11, 2019 at 14:55
• The Osorio problem is not only a proof game (perhaps 1 in 1000 positions admit a unique proof game?) but also includes the artistic feature that three white knights leave their starting squares and come back (of course one was not a knight when it set out on that journey) Jun 8, 2020 at 6:01 | 3,695 | 11,643 | {"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} | 3.203125 | 3 | CC-MAIN-2023-50 | longest | en | 0.972845 |
https://explore.openaire.eu/search/publication?articleId=altaiap_____::e16b1e0f0505315c39dce162d0bc9964 | 1,539,950,171,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512395.23/warc/CC-MAIN-20181019103957-20181019125457-00500.warc.gz | 687,647,864 | 10,348 | ## THE PARAMETERS OF FIRE DANGER FLAME RETARDANT FIBREBOARD
Article Russian OPEN
Антонов (Antonov), Александр (Aleksandr) Викторович (Viktorovich) ; Петрушева (Petrusheva), Надежда (Nadezhda) Александровна (Aleksandrovna) ; Алашкевич (Alashkevich), Юрий (Iurii) Давыдович (Davydovich) (2016)
• Publisher: Altai State University
• Journal: Khimiia rastitel'nogo syr'ia (Chemistry of plant raw material) (issn: 1029-5143, eissn: 1029-5151)
• Related identifiers:
• Subject: fiberboard; fire danger; loss of the sample mass; the maximum temperature gaseous products of combustion; mathematical description, vermiculite | древесно-волокнистая плита; пожарная опасность; потеря массы образца; время достижения максимальной температуры газообразных продуктов горения; математическое описание; вермикулит
The paper presents the results of research of dependence of indicators of fire danger of wood-fiber plates with reduced fire risk from technological and design parameters of the milling process. The work is aimed at obtaining hardboard wet process production with indicators of fire danger, reduced to the level of flame-resistant according to GOST 12.1.044-89 «Fire and explosion hazards of substances and materials. Nomenclature of indices and methods of their determination». For the task solution in the work there were used methods of mathematical planning with the aim of obtaining the mathematical description of grinding process of wood-fiber mass during the manufacture of hardboard panels with a reduced fire hazard. Construction of mathematical models of object is an important part of scientific research. Developed mathematical model of the object are a good research tool. The equations describing the studied processes of preparation of wood materials, in our view, allow for obtaining high-quality wood mass depending on the set modes of the milling process; under certain values of the constructive and technological parameters of the refining installations and by changing the values of the mass fraction of expanded vermiculite in wood-fiber composition, to lower the parameters and indicators of fire danger plates made of it.
• Similar Research Results (2)
ФИЗИКО-МЕХАНИЧЕСКИЕ СВОЙСТВА ТРУДНОВОСПЛАМЕНЯЕМЫХ ДРЕВЕСНОВОЛОКНИСТЫХ ПЛИТ (2016) 92% ВОПРОСЫ СНИЖЕНИЯ ПОЖАРНОЙ ОПАСНОСТИ ДРЕВЕСНОВОЛОКНИСТЫХ ПЛИТ (2016) 70%
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# chapter3q - Chapter 3 Stoichiometry of Formulas and...
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Chapter 3 Stoichiometry of Formulas and Equations Created: 9:27:51 AM MST 1. Calcium fluoride, CaF 2 , is a source of fluorine and is used to fluoridate drinking water. Calculate its molar mass. A. 118.15 g/mol B. 99.15 g/mol C. 78.07 g/mol D. 59.08 g/mol E. 50.01 g/mol 5. Aluminum sulfate, Al 2 (SO 4 ) 3 , is used in tanning leather, purifying water, and manufacture of antiperspirants. Calculate its molar mass. A. 450.06 g/mol B. 342.15 g/mol C. 315.15 g/mol D. 278.02 g/mol E. 74.98 g/mol 9. What is the mass in grams of 0.250 mol of the common antacid calcium carbonate? A. 4.00 × 10 2 g B. 25.0 g C. 17.0 g D. 4.00 × 10 -2 g E. 2.50 × 10 -3 g 13. Calculate the number of oxygen atoms in 29.34 g of sodium sulfate, Na 2 SO 4 . A. 1.244 × 10 23 O atoms B. 4.976 × 10 23 O atoms C. 2.409 × 10 24 O atoms D. 2.915 × 10 24 O atoms E. 1.166 × 10 25 O atoms 23. Hydroxylamine nitrate contains 29.17 mass % N, 4.20 mass % H, and 66.63 mass O. If its molar mass is between 94 and 98 g/mol, what is its molecular formula? A. NH 2 O 5 B. N 2 H 4 O 4 C. N 3 H 3 O 3 D. N 4 H 8 O 2 E. N 2 H 2 O 4 25. A compound containing chromium and silicon contains 73.52 mass percent chromium. Determine its empirical formula. A. CrSi 3 B. Cr 2 Si 3 C. Cr 3 Si D. Cr 3 Si 2 E. Cr 2 S 27. Terephthalic acid, used in the production of polyester fibers and films, is composed of carbon, hydrogen, and oxygen. When 0.6943 g of terephthalic acid was subjected to combustion analysis it produced 1.471 g CO 2 and 0.226 g H 2 O. What is its empirical formula? A.
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## This note was uploaded on 03/18/2010 for the course CHEM 210 taught by Professor Mcomber during the Spring '10 term at Skyline College.
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Building an interest payment projection forcast
Hi all,
I was wondering if someone might know how to build out a interest projection piece for an app?
For example if I had a loan which had a life of 3 years with 4 payments a year with a 6% interest rate. How would I go about projecting this out so that as the days draw closer and each interest period passes the current date the projections update dynamically to show the next 4 periods.
Below is roughly what I am trying to build towards
Many thanks,
Conor
1 ACCEPTED SOLUTION
Accepted Solutions
Super User
hi there @cnr
Please find the following, the scenario is a loan amount, interest based on the term and the total repayment is divided by the term. This is them shown to the user base on date in reference to the payment term.
If this is not your exact requirement you can take some bits from it.
Result
Steps
``````ClearCollect(
colPayments,
ForAll(
Sequence(12), //3 years with 4 payments
{
payment: yourPaymentValue,
yourStartDatePicker.SelectedDate,
Value * 4, //you wish to have 4 payments per month
Months
)
}
)
)``````
• Add a horizontal gallery and set its Item = colPayments
• Add two labels with Text values
• lblDate = ThisItem.month
• lblPayment = ThisItem.payment
The above will give you all the value in your collection.
To make it time relevant filter the collection
• Filter(colPayments, month >= Today())
The above will onlydisplaythe dates which newer than the current date.
Hope this helps
R
hey there if you liked the post give it a thumbs up, and if it solved your question please accept it as a solution.
Super User
hi there @cnr
Please find the following, the scenario is a loan amount, interest based on the term and the total repayment is divided by the term. This is them shown to the user base on date in reference to the payment term.
If this is not your exact requirement you can take some bits from it.
Result
Steps
``````ClearCollect(
colPayments,
ForAll(
Sequence(12), //3 years with 4 payments
{
payment: yourPaymentValue,
yourStartDatePicker.SelectedDate,
Value * 4, //you wish to have 4 payments per month
Months
)
}
)
)``````
• Add a horizontal gallery and set its Item = colPayments
• Add two labels with Text values
• lblDate = ThisItem.month
• lblPayment = ThisItem.payment
The above will give you all the value in your collection.
To make it time relevant filter the collection
• Filter(colPayments, month >= Today())
The above will onlydisplaythe dates which newer than the current date.
Hope this helps
R
hey there if you liked the post give it a thumbs up, and if it solved your question please accept it as a solution.
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Author Mollie Sherman
Posted May 13, 2022
Reads 151
There are various ways to answer this question, depending on what information is known.
If the question is asking how many ounces are in 16 grams, the answer is 0.563. This can be determined by using a conversion factor or by using dimensional analysis.
If the question is asking how many ounces of a substance weigh 16 grams, then the answer will depend on the specific gravity of the substance. For example, if the substance is water, then 16 grams of water would be equal to 1.04 ounces.
## How many ounces are in 16 grams?
There are 16 ounces in 1 pound.
## How do you convert 16 grams to ounces?
In the metric system, a gram is a unit of mass and an ounce is a unit of weight. The two units are not equivalent, so it is not possible to simply convert 16 grams to ounces.
To convert grams to ounces, you first need to determine the weight of the object in grams. You can do this by using a scale or by calculating the density of the object. Once you have the weight in grams, you can divide it by 28.35 to get the equivalent weight in ounces.
For example, let's say you have a book that weighs 16 grams. The weight of the book in ounces would be 16 divided by 28.35, which is equal to 0.566 ounces.
If you need to convert ounces to grams, the process is the reverse. You would multiply the weight in ounces by 28.35 to get the weight in grams.
## How many ounces are in a gram?
Grams are a unit of weight and ounces are a unit of volume, so you can't directly convert one to the other. However, you can convert grams to ounces by using the density of the material you're interested in. For example, the density of water is 1 gram per milliliter, so 1 gram of water would be the same as 1 milliliter, or about 0.035 ounces. To convert ounces to grams, you would need to know the density of the material you're interested in.
## How do you convert grams to ounces?
To convert grams to ounces, divide the number of grams by 28.35.
## What is the conversion factor for grams to ounces?
There are many factors that contribute to changing a grams to ounces. The first thing you need to do is take the weight in grams and divide it by the number of grams in an ounce, which is 28.35. This will give you the number of ounces. However, there are different types of ounces so you will need to specify which one you want. The most common types of ounces used are the avoirdupois ounce, which is the type used in the United States, and the troy ounce, which is used for measurements of precious metals. There are also fluid ounces, but they are not commonly used to measure weight. To convert grams to avoirdupois ounces, you would divide the weight in grams by 28.35. For troy ounces, you would divide the weight in grams by 31.1.
## Frequently Asked Questions
### How many ounces is in 16 gram?
16 grams is equal to 3.54 ounces.
### How many grams Makes 1 ounces?
There are 3.14 grams within one ounce.
### How do you convert grams into ounces?
There is no one definitive answer to this question. Typically, you would divide an ingredient's gram measurement by 28 to convert it to ounces.
### What weight is 1 oz in grams?
1 ounce is equal to 28.4 grams.
### Which is more 1 oz or 1g?
1 oz is more 1 oz.
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https://en.m.wikipedia.org/wiki/Elliptic_paraboloid | 1,566,783,816,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330913.72/warc/CC-MAIN-20190826000512-20190826022512-00401.warc.gz | 450,965,393 | 20,476 | Paraboloid
(Redirected from Elliptic paraboloid)
Paraboloid of revolution
In geometry, a paraboloid is a quadric surface that has exactly one axis of symmetry and no center of symmetry. The term "paraboloid" is derived from parabola, which refers to a conic section that has a similar property of symmetry.
Every plane section of a paraboloid by a plane parallel to the axis of symmetry is a parabola. The paraboloid is hyperbolic if every other plane section is either a hyperbola, or two crossing lines (in the case of a section by a tangent plane). The paraboloid is elliptic if every other nonempty plane section is either an ellipse, or a single point (in the case of a section by a tangent plane). A paraboloid is either elliptic or hyperbolic.
Equivalently, a paraboloid may be defined as a quadric surface that is not a cylinder, and has an implicit equation whose part of degree two may be factored over the complex numbers into two different linear factors. The paraboloid is hyperbolic if the factors are real; elliptic if the factors are complex conjugate.
An elliptic paraboloid is shaped like an oval cup and has a maximum or minimum point when its axis is vertical. In a suitable coordinate system with three axes x, y, and z, it can be represented by the equation[1]:892
${\displaystyle z={\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}.}$
where a and b are constants that dictate the level of curvature in the xz and yz planes respectively. In this position, the elliptic paraboloid opens upward.
Hyperbolic paraboloid
A hyperbolic paraboloid (not to be confused with a hyperboloid) is a doubly ruled surface shaped like a saddle. In a suitable coordinate system, a hyperbolic paraboloid can be represented by the equation[2][3]:896
${\displaystyle z={\frac {y^{2}}{b^{2}}}-{\frac {x^{2}}{a^{2}}}.}$
In this position, the hyperbolic paraboloid opens downward along the x-axis and upward along the y-axis (that is, the parabola in the plane x = 0 opens upward and the parabola in the plane y = 0 opens downward).
Any paraboloid (elliptic or hyperbolic) is a translation surface, as it can be generated by a moving parabola directed by a second parabola.
Properties and applications
Elliptic paraboloid
With a = b an elliptic paraboloid is a paraboloid of revolution: a surface obtained by revolving a parabola around its axis. It is the shape of the parabolic reflectors used in mirrors, antenna dishes, and the like; and is also the shape of the surface of a rotating liquid, a principle used in liquid mirror telescopes and in making solid telescope mirrors (see rotating furnace). This shape is also called a circular paraboloid.
Mesh Grid of Paraboloid surface
Paraboloid Surface (0.5 transparency)
There is a point called the focus (or focal point) on the axis of a circular paraboloid such that, if the paraboloid is a mirror, light from a point source at the focus is reflected into a parallel beam, parallel to the axis of the paraboloid. This also works the other way around: a parallel beam of light incident on the paraboloid parallel to its axis is concentrated at the focal point. This applies also for other waves, hence parabolic antennas. For a geometrical proof, click here.
Plane sections
As plane sections of an elliptic paraboloid with equation
${\displaystyle z={\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}}$
one gets the following cases:
• a parabola, if the plane is parallel to the z-axis,
• an ellipse or a point or empty, if the plane is not parallel to the z-axis.
• a point, if the plane is a tangent plane.
Obviously, any elliptic paraboloid of revolution contains circles. This is also true, but less obvious, in the general case (see circular section).
Remark: an elliptic paraboloid is projectively equivalent to a sphere.
Hyperbolic paraboloid
The hyperbolic paraboloid is a doubly ruled surface: it contains two families of mutually skew lines. The lines in each family are parallel to a common plane, but not to each other. Hence the hyperbolic paraboloid is a conoid.
These properties characterize hyperbolic paraboloids and are used in one of the oldest definitions of hyperbolic paraboloids: a hyperbolic paraboloid is a surface that may be generated by a moving line that is parallel to a fixed plane and crosses two fixed skew lines. This property makes it simple to manufacture a hyperbolic paraboloid from a variety of materials and for a variety of purposes, from concrete roofs to snack foods.
Pringles potato chips resemble a truncated hyperbolic paraboloid.[4] Their uniform shape allows them to be stacked in sturdy tubular containers, and the strength of the hyperbolic paraboloid shape helps prevent them from breaking while stacked.[5]
Examples in architecture
Plane sections
hyperbolic paraboloid with hyperbolas and parabolas
As plane sections of a hyperbolic paraboloid with equation
${\displaystyle z={\frac {x^{2}}{a^{2}}}-{\frac {y^{2}}{b^{2}}}}$
one gets the following cases:
• a parabola, if the plane is parallel to the z-axis with equation ${\displaystyle ux+vy+w=0,\ au\neq \pm bv}$ ,
• a line, if the plane is parallel to the z-axis with equation ${\displaystyle ux+vy+w=0,\ au=\pm bv}$ ,
• a pair of intersecting lines, if the plane is a tangent plane,
• a hyperbola, if the plane is not parallel to the z-axis and not a tangent plane.
Remarks:
1. A hyperbolic paraboloid is a ruled surface (contains lines), but not developable (in this case it is unlike a cylinder or cone).
2. The Gauss curvature at any point is negative. Hence it is a saddle surface.
3. The unit hyperbolic paraboloid with equation ${\displaystyle z=x^{2}-y^{2}}$ can be represented by ${\displaystyle z=2xy}$ after a rotation around the z-axis with an angle of 45° degrees.
4. A hyperbolic paraboloid is projectively equivalent to a hyperboloid of one sheet.
Cylinder between pencils of elliptic and hyperbolic paraboloids
elliptic paraboloid, parabolic cylinder, hyperbolic paraboloid
The pencil
${\displaystyle z=x^{2}+{\frac {y^{2}}{b^{2}}},\ b>0,}$
of elliptic paraboloids and the pencil
${\displaystyle z=x^{2}-{\frac {y^{2}}{b^{2}}},\ b>0,}$
of hyperbolic paraboloids approach the same surface
${\displaystyle z=x^{2}}$
for ${\displaystyle b\rightarrow \infty }$ , which is a parabolic cylinder (see image).
Curvature
The elliptic paraboloid, parametrized simply as
${\displaystyle {\vec {\sigma }}(u,v)=\left(u,v,{\frac {u^{2}}{a^{2}}}+{\frac {v^{2}}{b^{2}}}\right)}$
${\displaystyle K(u,v)={\frac {4}{a^{2}b^{2}\left(1+{\frac {4u^{2}}{a^{4}}}+{\frac {4v^{2}}{b^{4}}}\right)^{2}}}}$
${\displaystyle H(u,v)={\frac {a^{2}+b^{2}+{\frac {4u^{2}}{a^{2}}}+{\frac {4v^{2}}{b^{2}}}}{a^{2}b^{2}{\sqrt {\left(1+{\frac {4u^{2}}{a^{4}}}+{\frac {4v^{2}}{b^{4}}}\right)^{3}}}}}}$
which are both always positive, have their maximum at the origin, become smaller as a point on the surface moves further away from the origin, and tend asymptotically to zero as the said point moves infinitely away from the origin.
The hyperbolic paraboloid,[2] when parametrized as
${\displaystyle {\vec {\sigma }}(u,v)=\left(u,v,{\frac {u^{2}}{a^{2}}}-{\frac {v^{2}}{b^{2}}}\right)}$
has Gaussian curvature
${\displaystyle K(u,v)={\frac {-4}{a^{2}b^{2}\left(1+{\frac {4u^{2}}{a^{4}}}+{\frac {4v^{2}}{b^{4}}}\right)^{2}}}}$
and mean curvature
${\displaystyle H(u,v)={\frac {-a^{2}+b^{2}-{\frac {4u^{2}}{a^{2}}}+{\frac {4v^{2}}{b^{2}}}}{a^{2}b^{2}{\sqrt {\left(1+{\frac {4u^{2}}{a^{4}}}+{\frac {4v^{2}}{b^{4}}}\right)^{3}}}}}.}$
Geometric representation of multiplication table
If the hyperbolic paraboloid
${\displaystyle z={\frac {x^{2}}{a^{2}}}-{\frac {y^{2}}{b^{2}}}}$
is rotated by an angle of π/4 in the +z direction (according to the right hand rule), the result is the surface
${\displaystyle z=\left({\frac {x^{2}+y^{2}}{2}}\right)\left({\frac {1}{a^{2}}}-{\frac {1}{b^{2}}}\right)+xy\left({\frac {1}{a^{2}}}+{\frac {1}{b^{2}}}\right)}$
and if a = b then this simplifies to
${\displaystyle z={\frac {2xy}{a^{2}}}}$ .
Finally, letting a = 2, we see that the hyperbolic paraboloid
${\displaystyle z={\frac {x^{2}-y^{2}}{2}}.}$
is congruent to the surface
${\displaystyle z=xy}$
which can be thought of as the geometric representation (a three-dimensional nomograph, as it were) of a multiplication table.
The two paraboloidal 2 → ℝ functions
${\displaystyle z_{1}(x,y)={\frac {x^{2}-y^{2}}{2}}}$
and
${\displaystyle z_{2}(x,y)=xy}$
are harmonic conjugates, and together form the analytic function
${\displaystyle f(z)={\frac {z^{2}}{2}}=f(x+yi)=z_{1}(x,y)+iz_{2}(x,y)}$
which is the analytic continuation of the ℝ → ℝ parabolic function f(x) = x2/2.
Dimensions of a paraboloidal dish
The dimensions of a symmetrical paraboloidal dish are related by the equation
${\displaystyle 4FD=R^{2},}$
where F is the focal length, D is the depth of the dish (measured along the axis of symmetry from the vertex to the plane of the rim), and R is the radius of the rim. They must all be in the same unit of length. If two of these three lengths are known, this equation can be used to calculate the third.
A more complex calculation is needed to find the diameter of the dish measured along its surface. This is sometimes called the "linear diameter", and equals the diameter of a flat, circular sheet of material, usually metal, which is the right size to be cut and bent to make the dish. Two intermediate results are useful in the calculation: P = 2F (or the equivalent: P = R2/2D) and Q = P2 + R2, where F, D, and R are defined as above. The diameter of the dish, measured along the surface, is then given by
${\displaystyle {\frac {RQ}{P}}+P\ln \left({\frac {R+Q}{P}}\right),}$
where ln x means the natural logarithm of x, i.e. its logarithm to base e.
The volume of the dish, the amount of liquid it could hold if the rim were horizontal and the vertex at the bottom (e.g. the capacity of a paraboloidal wok), is given by
${\displaystyle {\frac {\pi }{2}}R^{2}D,}$
where the symbols are defined as above. This can be compared with the formulae for the volumes of a cylinder (πR2D), a hemisphere (/3R2D, where D = R), and a cone (π/3R2D). πR2 is the aperture area of the dish, the area enclosed by the rim, which is proportional to the amount of sunlight a reflector dish can intercept. The surface area of a parabolic dish can be found using the area formula for a surface of revolution which gives
${\displaystyle A={\frac {\pi R\left({\sqrt {(R^{2}+4D^{2})^{3}}}-R^{3}\right)}{6D^{2}}}.}$ | 3,016 | 10,504 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 30, "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} | 4.25 | 4 | CC-MAIN-2019-35 | latest | en | 0.923368 |
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# 10,000 is deposited in a certain account that pays r percent
Author Message
Manager
Joined: 02 Sep 2008
Posts: 102
10,000 is deposited in a certain account that pays r percent [#permalink]
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17 Mar 2009, 21:17
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\$10,000 is deposited in a certain account that pays r percent annual interest compounded annually,
the amount D(t), in dollars, that the deposit will grow to in t years is given by D(t) = 10,000
{1+(r/100)}t. What amount will the deposit grow to in 3 years?
(1) D(t) = 11,000
(2) r =10
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is
sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
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Manager
Joined: 02 Mar 2009
Posts: 125
Re: Compound Interest + DS [#permalink]
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17 Mar 2009, 23:17
I think it should be B.
1. You will have 2 unknowns. r and t. You do not know that 11,000 has grown after how many years.
2. You can replace t with 3 and replace r with 10% to get the answer.
Intern
Joined: 16 Mar 2009
Posts: 25
Re: Compound Interest + DS [#permalink]
### Show Tags
17 Mar 2009, 23:44
it should be B
what is the OA ?
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Re: Compound Interest + DS [#permalink] 17 Mar 2009, 23:44
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# Are Sensor Pixels Square?
Started Oct 9, 2012 | Discussions thread
I Was Wrong
The tentative conclusions in my previous posting are incorrect. The macro magnification close to the lens-barrel is extremely sensitive to distance and focusing. In my photos of the clear plastic ruler, the horizontal measurements were made on different photos. The following photo (Macro with object touching the lens) is used to determine horizontal and vertical image size in a single photo:
Calibration of SX30 Macro Magnification with Object Touching the Lens
(Small size for posting)
The "object" diameter at the outer edge of the radiating lines is 9.90 +/- 0.03 mm both vertically and horizontally. I used the Canon Image Browser with the "Image Viewer" at its maximum size, which was a frame of 265 mm x 199 mm on my monitor, in order to calculate image size and magnification.
The Screen-Image Diameter at the outer edge of the radiating lines was 99.5 mm +/- 0.3 mm in both the horizontal and vertical directions, so the horizontal and vertical magnifications are the same within better than 1%.
Since the image is not distorted, it appears that the sensor of the 6.17 mm x 4.55 mm is not completely used because it does not have a 4:3 ratio. The Canon specs,
http://www.canon.ca/inetCA/arcproducts?m=gp&pid=4812#_030
say there are 14.5 total Megapixels and 14.1 Effective Megapixels on the sensor.
Assuming that the total area of 6.17 mm x 4.55 mm = 28.0735 mm^2 corresponds to 14.5 Total Megapixels and that the "Effective Area" corresponds to 14.1 Effective Megapixels:
Effective Area = (14.1 /14.5) x 28.0735 mm^2 = 27.299 mm^2
= 6.033 mm x 4.525 mm effective width and height
actually in use on the sensor contributing to the final image.
If we use 4320 x 3240 pixels = 14.0 Megapixels, the effective area becomes
Effective Area = (14.0 /14.5) x 28.0735 mm^2 = 27.1054 mm^2
= 6.01 mm x 4.51 mm (Effective width & height with 4:3 ratio)
Using these numbers, the vertical and horizontal macro magnifications for the photo above are:
Horizontal Magnification = (99.5 / 265) x 6.01 mm / 9.9 mm = 0.228 (for object touching the lens)
Vertical Magnification = (99.5 / 199) x 4.51 mm / 9.9 mm = 0.228
(The numbers in brackets are the on-screen measurements of image size and frame size, made on my computer monitor.)
So, using "Effective Sensor Size" of 6.01 mm x 4.51 mm, the image magnification on the sensor is less than when the sensor dimensions from the specs are used, and the discrepancy between horizontal and vertical magnifications has been removed.
Does this look reasonable?
Stephen Barrett's gear list:Stephen Barrett's gear list
Canon PowerShot SX30 IS +3 more
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View Full Version : depth buffer problem
deans
11-14-2002, 10:41 PM
I've reduced my problem to the simplest program I can. The more distant quad is drawn in front of the nearer one. They draw correctly if I reverse their draw order but that shouldn't matter. The near/far ratio looks good. It happens on two different machines. I'm sure it's something dumb. Please help! http://www.opengl.org/discussion_boards/ubb/smile.gif
# include "glutinit.h"
# include <stdio.h>
# include <GL/glut.h>
void RenderAll()
{
glClear (GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);
gluLookAt( 204.59, 255.48, 5.37, // cam position x,y,z
50, 50, 0, // cam target x,y,z
0, 0, 1 ); // up vector x,y,z
// draw 2 squares
{
glColor3f( 0.f, 0.f, 1.f );
glVertex3f( -50, 50, 50 );
glVertex3f( -50, 50, -50 );
glVertex3f( 50, 50, -50 );
glVertex3f( 50, 50, 50 );
glColor3f( 1.f, 0.f, 0.f );
glVertex3f( -50, -50, 50 );
glVertex3f( -50, -50, -50 );
glVertex3f( 50, -50, -50 );
glVertex3f( 50, -50, 50 );
glEnd();
}
glFlush ();
glutSwapBuffers();
}
void ReshapeWindow( int width, int height )
{
glViewport (0, 0, (GLsizei) width, (GLsizei) height);
glMatrixMode (GL_PROJECTION);
float aspect = (float)height / (float)width;
glFrustum (-0.7f, 0.7f, -0.7f * aspect, 0.7f * aspect, 2.f, 500.0f);
glMatrixMode (GL_MODELVIEW);
}
void HandleInput(unsigned char key, int x, int y)
{
if( key == 27 ) exit( 0 );
}
void main( int argc, char** argv )
{
glutInit(&argc, argv);
glClearColor (0.0, 0.0, 0.0, 0.f );
glutInitDisplayMode( GLUT_DEPTH |
GLUT_DOUBLE |
GLUT_RGB );
glEnable( GL_DEPTH_TEST );
glutInitWindowSize (800, 800);
glutInitWindowPosition (100, 50);
glutCreateWindow (argv[0]);
glutDisplayFunc( RenderAll );
glutReshapeFunc( ReshapeWindow );
glutKeyboardFunc( HandleInput );
glutIdleFunc( RenderAll );
glutMainLoop();
}
satan
11-15-2002, 01:13 AM
The winding of your quads is clockwise while default for OpenGL is counterclockwise.
Just change the order of your glVertex calls
and it should work.
deans
11-15-2002, 01:51 AM
hey, thanks for looking at this. http://www.opengl.org/discussion_boards/ubb/smile.gif that didn't fix it though, nor did it invert the problem; the wrong one was still in front if drawn in the same order. This makes sense to me because i'm not backface culling based on the winding (=normal) of a polygon. so, if the winding is the wrong way in camera space, then we're just looking at the back of the quad, which we're still drawing. it should still occlude poly's in the distance, but alas it doesn't. still puzzled. any other ideas?
http://www.opengl.org/discussion_boards/ubb/smile.gif
bakery2k
11-15-2002, 02:58 AM
Gotcha! I sat and played with this for a few minutes, and realised what the problem is.
You must call glEnable(GL_DEPTH_TEST) AFTER creating your window. i.e. you should have:
void main( int argc, char** argv )
{
glutInit(&argc, argv);
glClearColor (0.0, 0.0, 0.0, 0.f );
glutInitDisplayMode( GLUT_DEPTH |
GLUT_DOUBLE |
GLUT_RGB );
glutInitWindowSize (800, 800);
glutInitWindowPosition (100, 50);
glutCreateWindow (argv[0]);
glEnable( GL_DEPTH_TEST );
glutDisplayFunc( RenderAll );
glutReshapeFunc( ReshapeWindow );
glutKeyboardFunc( HandleInput );
glutIdleFunc( RenderAll );
glutMainLoop();
}
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matched data and agreement
# matched data and agreement - MATCHED PAIR DATA AND...
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MATCHED PAIR DATA AND AGREEMENT Sometimes the best way to control confounding is to match on the confounding factor - Age distribution unexposed is young - Age distribution of exposed is older - Among 0-5 years, we will have enough unexposed and 5-15 have exposed, So we don’t have same age to compare - So, can’t make comparable comparisons - Can be matched for case, or other exposures Example matched case control study Case control study of acute prolapsed lumbar intervertebral disk. Kelsey et al. (1975) For each case, a matched control of the same age and gender was obtained from the same hospital or radiology service 217 matched pairs One exposure of interest: whether subject was a driver Each pari is a separate stratum - Analysis adjusted for matched factor - Cases and control are patched pair - Here, m=no of pair - Each pair creates its own 2*2 table - So, H 0 : All OR is same in all table=1 - H A : OR in those 2*2 table ≠1 Summary - There are four possible table for each pair Stratum specific Odds Ratio In this case Type of combination No of tables Both case and control exposed 144 Case exposed, control not 41 Control exposed, case not 19 Both case and control unexposed 13 Total pair 217 M-H gives common OR by aggregating the data from all tables Another summary Counts are pairs, not individuals M-H OR is ratio of number of discordant pairs: 41/19 = 2.16 Without matching - 185 = 144+41 63 = 144+19 - 32 = 13+19 54 = 13+41 Stratum- specific MH-Test Contributions χ 2 MH
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{[ snackBarMessage ]} | 462 | 1,904 | {"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} | 2.859375 | 3 | CC-MAIN-2018-17 | latest | en | 0.838094 |
https://numbermatics.com/n/15368239360395829/ | 1,544,395,301,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823183.3/warc/CC-MAIN-20181209210843-20181209232843-00324.warc.gz | 796,585,408 | 9,054 | # 15368239360395829
## 15,368,239,360,395,829 is an odd composite number composed of two prime numbers multiplied together.
15368239360395829 is an odd composite number. It is composed of two distinct prime numbers multiplied together. It has a total of four divisors.
## Prime factorization of 15368239360395829:
### 409 × 37575157360381
See below for interesting mathematical facts about the number 15368239360395829 from the Numbermatics database.
### Names of 15368239360395829
• Cardinal: 15368239360395829 can be written as Fifteen quadrillion, three hundred sixty-eight trillion, two hundred thirty-nine billion, three hundred sixty million, three hundred ninety-five thousand, eight hundred twenty-nine.
### Scientific notation
• Scientific notation: 1.5368239360395829 × 1016
### Factors of 15368239360395829
• Number of distinct prime factors ω(n): 2
• Total number of prime factors Ω(n): 2
• Sum of prime factors: 37575157360790
### Divisors of 15368239360395829
• Number of divisors d(n): 4
• Complete list of divisors:
• Sum of all divisors σ(n): 15405814517756620
• Sum of proper divisors (its aliquot sum) s(n): 37575157360791
• 15368239360395829 is a deficient number, because the sum of its proper divisors (37575157360791) is less than itself. Its deficiency is 15330664203035038
### Bases of 15368239360395829
• Binary: 110110100110010101010100001111111001100100101000110101 2
### Squares and roots of 15368239360395829
• 15368239360395829 squared (153682393603958292) is 236182781038419599235415560597241
• 15368239360395829 cubed (153682393603958293) is 3629713511802389751200795915872739549541405307789
• The square root of 15368239360395829 is 123968703.1488021537
• The cube root of 15368239360395829 is 248623.0397681557
### Scales and comparisons
How big is 15368239360395829?
• 15,368,239,360,395,829 seconds is equal to 488,662,474 years, 29 weeks, 6 days, 7 hours, 43 minutes, 49 seconds.
• To count from 1 to 15,368,239,360,395,829 would take you about one trillion, five million, eighty-seven thousand, four hundred twenty-three year!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000.
Note: we do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!
• A cube with a volume of 15368239360395829 cubic inches would be around 20718.6 feet tall.
### Recreational maths with 15368239360395829
• 15368239360395829 backwards is 92859306393286351
• The number of decimal digits it has is: 17
• The sum of 15368239360395829's digits is 82
• More coming soon!
The information we have on file for 15368239360395829 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
Keywords: Divisors of 15368239360395829, math, Factors of 15368239360395829, curriculum, school, college, exams, university, STEM, science, technology, engineering, physics, economics, calculator. | 915 | 3,346 | {"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} | 2.90625 | 3 | CC-MAIN-2018-51 | latest | en | 0.737134 |
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# How do you write the logarithmic expression as the sum, difference or multiple of logarithms and simplify as much as possible for ${\log _5}\left( {\dfrac{{25}}{x}} \right)$?
Last updated date: 09th Aug 2024
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Hint: The above question is based on the concept of logarithm. The main approach towards solving the question is by applying various logarithmic properties on the above given expression. Properties like quotient rule and power rule are applied to the expression and further it is simplified.
Complete step by step solution:
Logarithms, like exponents, have many useful properties that can be used to reduce by simplifying logarithmic expressions and solve logarithmic equations. Logarithm is the exponent or power to which base must be raised to yield a given number. When expressed mathematically x is the logarithm of base n to the base b if ${b^x} = n$ ,then we can write it as $x = {\log _b}n$.
The above given expression is ${\log _5}\left( {\dfrac{{25}}{x}} \right)$.So first we will apply quotient rule which is given below:
${\log _b}\left( {\dfrac{M}{N}} \right) = {\log _b}M - {\log _b}N$
Now this states that the log of quotients is the difference of the log of dividend and divisor. So, applying this on the expression,
$\Rightarrow \log \left( {\dfrac{{25}}{x}} \right) = {\log _5}\left( {25} \right) - {\log _5}x \\ \Rightarrow {\log _5}\left( {\dfrac{{25}}{x}} \right) = {\log _5}\left( {{5^2}} \right) - {\log _5}x \\$
Now by applying the power rule which is given below:
${\log _b}\left( {{M^p}} \right) = p{\log _b}\left( M \right)$
Since we have the first term in power therefore will apply the other property.
${\log _5}\left( {\dfrac{{25}}{x}} \right) = {\log _5}\left( {{5^2}} \right) - \log x \\ \Rightarrow{\log _5}\left( {\dfrac{{25}}{x}} \right) = 2{\log _5}\left( 5 \right) - \log x \\ \therefore{\log _5}\left( {\dfrac{{25}}{x}} \right) = (2 \times 1) - {\log _5}x = 2 - {\log _5}x$
Therefore, we get the above solution after simplifying.
Note: An important thing to note is that these properties can have any values for M,N and b where $M,N > 0$and $0 < b \ne 1$.The reason for this is the argument of the logarithm must be positive and the base of the logarithm must also be positive and not equal to 1. | 711 | 2,382 | {"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} | 4.78125 | 5 | CC-MAIN-2024-33 | latest | en | 0.853013 |
https://studylib.net/doc/5627707/%E6%94%AF%E6%8C%81%E5%90%91%E9%87%8F%E6%9C%BAsvm | 1,555,721,024,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578528433.41/warc/CC-MAIN-20190420000959-20190420022959-00199.warc.gz | 576,778,674 | 54,350 | 支持向量机SVM
Support Vector machine
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1
SVM寻优算法
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Chap8 SL Zhongzhi Shi
2
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Chap8 SL Zhongzhi Shi
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1900-1920 数据描述
1920-1940 统计模型的曙光
1940-1960 数理统计时代
1990-1999 建模复杂的数据结构
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Chap8 SL Zhongzhi Shi
4
Tikhonov, Ivanov 和 Philips 发现的关于解决不适定
Parzen, Rosenblatt 和Chentsov 发现的非参数统计学；
Vapnik 和Chervonenkis 发现的在泛函数空间的大数
Kolmogorov, Solomonoff 和Chaitin 发现的算法复杂
2015/4/8
Chap8 SL Zhongzhi Shi
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Chap8 SVM Zhongzhi Shi
6
（1）搜集数据：采样、实验设计
（2）分析数据：建模、知识发现、可视化
（3）进行推理：预测、分类
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SVM是一种基于统计学习理论的机器
Vapnik在COLT-92上首次提出，从此
Vapnik V N. 1995. The Nature of
Statistical Learning Theory. SpringerVerlag, New York
Vapnik V N. 1998. Statistical Learning
Theory. Wiley-Interscience Publication,
John Wiley&Sons, Inc
Chap8 SVM Zhongzhi Shi
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P(X,Y)，P(X,Y)反映了某种知识。学习问
( independently drawn and identically
distributed )的观测样本train set，
(x1,y1),(x2,y2),…,(xn,yn)
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Chap8 SVM Zhongzhi Shi
9
(typically in Rn), independently drawn
from some fixed distribution F(x)
训练器Supervisor (S) labels each input
x with an output value y according to
some fixed distribution F(y|x)
学习机Learning Machine (LM) “learns”
from an i.i.d. l-sample of (x,y)-pairs
output from G and S, by choosing a
function that best approximates S from
a parameterised function class f(x,),
where is in the parameter set
functions f(x,) and the equivalent
representation of each f using its index
G
x
S
LM
y
y^
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R( w) L( y, f ( x, w))dF ( x, y )
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1 l
Remp ( w) L( yi , f ( xi , w))
l i 1
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R(w) Remp (w) (h / l )
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Chap8 SVM Zhongzhi Shi
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VC维
VC维(Vapnik-Chervonenkis Dimension)。模式识别方法
(h / l )
h(ln(2l / h 1) ln( / 4)
l
h是函数H=f(x, w)的VC维, l是样本数.
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Problem: how rich class of classifications q(x;θ) to use.
underfitting
good fit
overfitting
Problem of generalization: a small emprical risk Remp does not
imply small true expected risk R.
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1. 学习过程的一致性理论
What are (necessary and sufficient) conditions for consistency (convergence
of Remp to R) of a learning process based on the ERM Principle?
2.学习过程收敛速度的非渐近理论
How fast is the rate of convergence of a learning process?
3. 控制学习过程的泛化能力理论
How can one control the rate of convergence (the generalization
ability) of a learning process?
4. 构造学习算法的理论
How can one construct algorithms that can control the
generalization ability?
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ERM is intended for relatively large
samples (large l/h)
Large l/h induces a small which
decreases the the upper bound on risk
Small samples? Small empirical risk
doesn’t guarantee anything!
…we need to minimise both terms of the
RHS of the risk bounds
1.
2.
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The empirical risk of the chosen
An expression depending on the VC dimension of
Chap8 SVM Zhongzhi Shi
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The Structural Risk Minimisation (SRM)
Principle
Let S = {Q(z,),}. An admissible
structure S1S2…Sn…S:
For each k, the VC dimension hk of Sk is finite and
h1≤h2≤…≤hn≤…≤hS
Every Sk is either is non-negative bounded, or
satisfies for some (p,k)
Q
sup
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k
p
1
p
z,dFz
R
k, p 2
Chap8 SVM Zhongzhi Shi
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S1
S2
Sn
h1
h*
hn
h
The SRM Principle continued
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For given z1,…,zl and an admissible structure
S1S2…Sn… S, SRM chooses function
Q(z,lk) minimising Remp in Sk for which the
guaranteed risk (risk upper-bound) is minimal
Thus manages the unavoidable trade-off of
quality of approximation vs. complexity of
approximation
Chap8 SVM Zhongzhi Shi
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on the risk
Confidence interval
h1
Sn
hn
h
S*
S1
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h*
risk
S*
Sn
Chap8 SVM Zhongzhi Shi
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SVMs are learning systems that
use a hyperplane of linear functions
in a high dimensional feature space — Kernel function
trained with a learning algorithm from optimization
theory — Lagrange
Implements a learning bias derived from statistical
learning theory — Generalisation SVM is a classifier
derived from statistical learning theory by Vapnik and
Chervonenkis
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x
f
yest
f(x,w,b) = sign(w. x - b)
denotes +1
denotes -1
How would you
classify this data?
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denotes +1
x
f
yest
f(x,w,b) = sign(w. x - b)
denotes -1
How would you
classify this data?
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denotes +1
x
f
yest
f(x,w,b) = sign(w. x - b)
denotes -1
How would you
classify this data?
© 2001, 2003, Andrew W. Moore
Chap8 SVM Zhongzhi Shi
25
denotes +1
x
f
yest
f(x,w,b) = sign(w. x - b)
denotes -1
How would you
classify this data?
© 2001, 2003, Andrew W. Moore
Chap8 SVM Zhongzhi Shi
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denotes +1
x
f
yest
f(x,w,b) = sign(w. x - b)
denotes -1
How would you
classify this data?
© 2001, 2003, Andrew W. Moore
Chap8 SVM Zhongzhi Shi
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denotes +1
x
f
f(x,w,b) = sign(w. x - b)
The maximum
margin linear
classifier is the
linear classifier
with the maximum
margin.
denotes -1
© 2001, 2003, Andrew W. Moore
yest
Linear SVM
Chap8 SVM Zhongzhi Shi
This is the
simplest kind of
SVM (Called an
LSVM)
28
Training set: (xi, yi), i=1,2,…N; yi{+1,-1}
Hyperplane: wx+b=0
This is fully determined by (w,b)
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According to a theorem
from Learning Theory,
from all possible linear
decision functions the one
that maximises the margin
of the training set will
minimise the
generalisation error.
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Note1: decision functions (w,b) and
(cw, cb) are the same
Note2: but margins as measured by the
outputs of the function xwx+b are
not the same if we take (cw, cb).
Definition: geometric margin: the margin
w
given by the canonical decision
function, which is when c=1/||w||
Strategy:
1) we need to maximise the geometric
wx+b>0
margin! (cf result from learning
wx+b<0
theory)
wx+b=0 2) subject to the constraint that
training examples are classified
correctly
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According to Note1, we can demand the function output for
the nearest points to be +1 and –1 on the two sides of the
decision function. This removes the scaling freedom.
Denoting a nearest positive example x+ and a nearest negative
example x-, this is wx b 1 and wx b 1
Computing the geometric margin (that has to be maximised):
1 w
b
w
b
1
1
(
x
x
)
(wx b wx b)
2 || w ||
|| w || || w ||
|| w ||
2 || w ||
|| w ||
And here are the constraints:
wxi b 1 for yi 1
wxi b 1 for yi 1
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yi (wxi b) 1 0 for all i
Chap8 SVM Zhongzhi Shi
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Maximum margin – summing up
Given a linearly separable
training set (xi, yi),
i=1,2,…N; yi{+1,-1}
Minimise ||w||2
Subject to
yi (wxi b) 1 0, i 1,..N
programming problem with
linear inequality constraints.
wx+b=1 There are well known
wx+b=0 procedures for solving it
wx+b=-1
wx+b>1
wx+b<1
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The training points
that are nearest to the
separating function
are called support
vectors.
What is the output of
our decision function
for these points?
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T {( x1, y1 ), ,( xl , yl )} ( x y)l
yi y {1, 1} 是输出指标，或输出.
2维空间上的分类问题)
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n维空间上的分类问题.
Chap8 SVM Zhongzhi Shi
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,( xl , yl )} ( x y)l
x
f ( x) sgn( g ( x))
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SVM分类问题大致有三种：线性可分问题、近似线性可分
2015/4/8
Chap8 SVM Zhongzhi Shi
37
l3
2
)
l2和 l3 之间的距
w
。
2015/4/8
Chap8 SVM Zhongzhi Shi
38
l2 , l3 可表示为
(w x) b k1，(w x) b k 2
（w x) b k , (w x) b -k
w
b
, b ，则两式可以等价写为
k
k
（w x) b 1, (w x) b -1
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Chap8 SVM Zhongzhi Shi
39
1- b 1 b
2
D
w2
w1
w2
D
w2
2
[x]1
（w x ) b 1
（w x) b 1
1 b
w2
D
[x]2
1 b
D
w2
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Chap8 SVM Zhongzhi Shi
2
w12 w2 2
2
w
40
“间隔”
2
|| w ||
w b和 的最优化问题
min
w ,b
s.t.
1
|| w ||2 ,
2
yi (( w xi ) b) 1, i 1, , l
(1.2.1)
(1.2.2)
n
R
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Margin =
2
W
H1平面：
W X1 b 1
H2平面：
W X 2 b 1
yi [(W X i ) b] 1 0
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…..(1)
Chap8 SVM Zhongzhi Shi
…..(2)
42
l
1 l l
min
yiyj i j xi xj j
2 i 1 j 1
j 1
l
s.t.
y
i
i
0,
i 1
i 0, i 1 l
i 为原始问题中与每个约束条件对应的Lagrange乘子。这是
2015/4/8
Chap8 SVM Zhongzhi Shi
*
43
* (a1* ,
, al* )T
l
*
*
*
*
i 1
l
b yj yi i* xi xj
*
i 1
*
*
w
x
b
0 ,决策函数 f x sgn((w* x) b* )
*
*
*
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Chap8 SVM Zhongzhi Shi
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（即“软化” 约束条件）
1 ,
l
T
l
i 作
i 1
2
w
l
i 1
i
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Chap8 SVM Zhongzhi Shi
45
min
w,b ,
s.t
l
1 2
w C i
2
i 1
yi ((w xi ) b) 1 i , i 1,
l
i 0, i 1, l
l
体现了经验风险，而
i 1
i
w 则体现了表达能力。所以
SVC的原始问题。
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Chap8 SVM Zhongzhi Shi
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(广义)线性支持向量分类机算法
1. 设已知训练集 T {( x1 , y1 ),
l
,( xl , yl )} ( x y)，其中
xi x Rn , yi y {1, 1}, i 1, , l
2. 选择适当的惩罚参数 C 0 ，构造并求解最优化问题
l
1 l l
min
yi y j i j xi x j j
2 i 1 j 1
j 1
l
y
s.t.
i 1
i
i
0
0 i C , i 1,
l
l
al* )T
*
*
3. 计算 w yii xi ，选择 * 的一个分量 0 j C，并据此
*
i 1
l
*
i 1
*
*
4. 构造分划超平面 (w* x) b* 0 ,决策函数 f ( x) sgn((w x) b )
2015/4/8
Chap8 SVM Zhongzhi Shi
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2015/4/8
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Non-linear Classification
What can we do if the boundary is nonlinear ?
Idea:
2015/4/8
transform the data vectors to a
space where the separator is
linear
Chap8 SVM Zhongzhi Shi
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Non-linear Classification
The transformation many times is made to an infinite
dimensional space, usually a function space.
Example: x cos(uTx)
2015/4/8
Chap8 SVM Zhongzhi Shi
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Non-linear SVMs
Transform x (x)
The linear algorithm depends only on xxi, hence
transformed algorithm depends only on (x)(xi)
Use kernel function K(xi,xj) such that K(xi,xj)= (x)(xi)
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T
[w]1 2[w]2[ x]1 2[w]3[ x]2 2[w]4[ x]1[ x]2 [w]5[ x]12 [w]6[ x]22 b 0
T
( x) (1, 2[ x]1, 2[ x]2 , 2[ x]1[ x]2 ,[ x]12 ,[ x]22 )T
(1)
[w]1[ X ]1 2[w]2[ X ]2 2[w]3[ X ]3 2[w]4[ X ]4 [w]5[ X ]5 [w]6[ X ]6 b 0
2015/4/8
Chap8 SVM Zhongzhi Shi
52
(w* x) b* 0，其中w* ([w* ]1, [w* ]6 )T
[w* ]1 2[w* ]2[ x]1 2[w* ]3[ x]2 2[w* ]4[ x]1[ x]2 [w* ]5[ x]12 [w* ]6[ x]22 b 0
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Chap8 SVM Zhongzhi Shi
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l
1 l l
min
yi y j i j ( xi ) ( x j ) j
2 i 1 j 1
j 1
l
s.t.
y
i 1
i
i
0
0 i C , i 1,
l
( xi ) (1, 2[ xi ]1, 2[ xi ]2 , 2[ xi ]1[ xi ]2 ,[ xi ]12 ,[ xi ]22 )T
( x j ) (1, 2[ x j ]1, 2[ x j ]2 , 2[ x j ]1[ x j ]2 ,[ x j ]12 ,[ x j ]22 )T
( ( xi ) ( x j )) 1 2[ xi ]1[ x j ]1 2[ xi ]2 [ x j ]2 2[ xi ]1[ xi ]2[ x j ]1[ x j ]2
[ xi ]12 [ x j ]12 [ xi ]22 [ x j ]22
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(2)
54
*
*
* T
(
,
1
l ) 后，得到分划超平面
(w* x) b* 0
l
w* yii* ( xi ), j { j | 0 *j C}
i 1
l
b y j yi i ( ( xi ) ( x j ))
*
i 1
2015/4/8
f ( x) sgn(( w* ( x)) b* )
l
f ( x) sgn( yii ( ( xi ) ( x)) b* )
i 1
Chap8 SVM Zhongzhi Shi
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(( xi x j ) 1)2
[ xi ]12 [ x j ]12 [ xi ]22 [ x j ]22 1 2[ xi ]1[ x j ]1[ xi ]2 [ x j ]2
2[ xi ]1[ x j ]1 2[ xi ]2 [ x j ]2
(3)
( ( xi ) ( x j )) K ( xi , x j ) (( xi x j ) 1)2
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Chap8 SVM Zhongzhi Shi
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(( xi x j ) 1)2
l
1 l l
min
yi y j i j K xi , x j j
2 i 1 j 1
j 1
l
s.t.
y
i 1
i
i
0
0 i C , i 1,
*
*
(
1,
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l
l* )T，而决策函数
Chap8 SVM Zhongzhi Shi
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l
f ( x) sgn( yii K ( xi , x)) b* )
i 1
l
b* y j yii K ( xi , x j ) j { j | 0 *j C}
i 1
K ( xi , x) － 核函数
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:
x
( x)
K ( x, x) ( ( x) ( x))
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K ( x, xi ) [( x xi ) c]q 得到q 阶多项式分类器
K ( x, xi ) exp{
| x xi |2
2
}
Sigmoind内核
K ( x, xi ) tanh( ( x xi ) c)
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K ( x, y ) x y
The kind of kernel represents the inner
product of two vector(point) in a feature
space of n d 1 dimension.
d
d
For example
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SVM寻优算法
需要计算存储核函数矩阵。当样本点数较大时，需要很
SVM在二次型寻优过程中要进行大量的矩阵运算，通常
－
Edgar Osuna(Cambridge,MA)等人在IEEE NNSP’97发表
Machines ,提出了SVM的分解算法，即将原问题分解为若
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SVM寻优算法
1. 块算法(Chunking Algorithm)：
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SVM寻优算法
SMO使用了块与分解技术，而SMO算法则将分解算法思想推
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Chap8 SVM Zhongzhi Shi
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SVM寻优算法
SMO算法每次迭代时，在可行的区域内选择两点，
2015/4/8
Chap8 SVM Zhongzhi Shi
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SVM寻优算法
146
4
97.26
83
0
100
137
3
97.81
32
2
93.75
106
2
98.11
90
1
98.89
34
1
97.06
87
2
97.70
111
2
98.20
40
1
97.50
91
1
98.90
3
0
100
24
2
91.67
984
21
97.87
2015/4/8
Chap8 SVM Zhongzhi Shi
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SVM寻优算法
SMO算法核缓存算法
SMO算法在每次迭代只选择两个样本向量优化目标
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SVM寻优算法
SMO算法核缓存算法
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SVM寻优算法
1
5624
5624*23
40726
7:06
10
5624
5624*233
40726
3:50
20
5624
5624*466
40726
2:41
30
5624
5624*699
40726
1:56
40
5624
5624*932
40726
1:29
50
5624
5624*1165
40726
1:23
60
5624
5624*1398
40726
1:08
70
5624
5624*1631
40726
1:05
80
5624
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40726
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90
5624
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40726
1:07
100
5624
5624*2330
40726
1:37
250
5624
5624*5624
40726
1:12
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SVM寻优算法
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SVM寻优算法
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SVM寻优算法
alpha=C）是否以后不再被启发式选择，或者不再被判定为
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SVM寻优算法
•这样就可以在SMO算法收敛前，提前将边界上的点从训
•训练开始前，训练活动集样本初始化为全部训练样本。每
•检查完当前活动向量集中所有样本后，产生了新的活动向
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SVM寻优算法
2. 固定工作样本集 (Osuna et al.)：
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SVM applications
Pattern recognition
o
DNA array expression data analysis
o
Features: words counts
Features: expr. levels in diff. conditions
Protein classification
o
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Features: AA composition
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Handwritten Digits Recognition
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Applying SVMs to Face Detection
The SVM face-detection system
1. Rescale the
input image
several times
2. Cut 19x19
window patterns
out of the scaled
image
5. If the class corresponds
to a face, draw a rectangle
around the face in the
output image.
3. Preprocess the
light correction and
histogram equalization
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4. Classify the
pattern using
the SVM
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Applying SVMs to Face Detection
Experimental results on static images
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Set A: 313 high-quality, same number of faces
Set B: 23 mixed quality, total of 155 faces
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Applying SVMs to Face Detection
Extension to a real-time
system
An example
of the skin
detection
module
implemented
using SVMs
Face
Detection
on the PCbased
Color Real
Time
System
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References
Vladimir Vapnik. The Nature of Statistical Learning Theory, Springer, 1995
Andrew W. Moore. cmsc726: SVMs.
http://www.cs.cmu.edu/~awm/tutorials
C. Burges. A tutorial on support vector machines for pattern
recognition. Data Mining and Knowledge Discovery, 2(2):955-974,
1998. http://citeseer.nj.nec.com/burges98tutorial.html
Vladimir Vapnik. Statistical Learning Theory. Wiley-Interscience;
1998
Thorsten Joachims (joachims_01a): A Statistical Learning Model of Text
Classification for Support Vector Machines
Ben Rubinstein. Statistical Learning Theory. Dept. Computer Science &
Software Engineering, University of Melbourne; and Division of Genetics &
Bioinformatics, Walter & Eliza Hall Institute
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www.intsci.ac.cn/shizz/
Questions?!
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Arab people
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Biology 1105 Exam 1
# Biology 1105 Exam 1 - Biology 1105 Exam#1 Exercise 2...
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Biology 1105 Exam #1 Exercise 2: Measurements in Biology Metric system – used by scientists throughout the world to make measurements Also used everywhere in everyday life except in the US -where most of the measurements are the antiquated English system of pounds, inches, feet, etc- Based on units of 10, unlike the English system Conversions 1 inch = 2.5 centimeters 1 foot = 30 centimeters 1 yard = 0.9 meter 1 mile = 1.6 kilometers 1 ounce = 28 grams 1 pound = 0.45 kilograms 1 fluid ounce = 30 milliliters 1 pint = 0.47 liter 1 quart = 0.95 liter 1 gallon = 3.8 liters Metric Units: meter (m) – basic unit of length liter (L) – the basic unit of volume kilogram (kg) – the basic unit of mass degree Celsius ( 0 C) – the basic unit of temperature Units of are indicated by Latin/Greek prefixes before the base units: Prefix (Latin) Division of Metric Unit Deci (d) 0.1 10 ^-1 Centi (c) 0.01 10^-2 Milli (m) 0.001 10^-3 Micro (u) 0.000001 10^-6 Nano (n) 0.000000001 10^-9 Pico (p) 0.000000000001 10^-12 Prefix (Greek) Multiple of Metric Unit Deka (da) 10 10^1 Hecto (h) 100 10^2 Kilo (k) 1000 10^3 Mega (M) 1000000 10^6 Giga (G) 1000000000 10^9 K | H | D | *(m,l,g) | D | C | M
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Meter (m) – basic unit of length Units of area are squared units of length: o 1m = 100 cm = 1000 mm = 0.001 km = 1 X 10 -3 km o I cm 2 = 100 mm 2 Most biologists measure lengths with metric rulers or metersticks Volume – the space occupied by an object Units of volume are cubed (3-D) units of length The liter (L) is the basic unit of volume o 1 L = 1000 cm 3 = 1000 mL a PIPET is used to extract and dispense volumes of liquid used to measure small volumes (25 mL or less) liquid is drawn into a pipet using a bulb o pipet pump GRADUATED CYLINDERS are used to measure larger volumes meniscus – the interface i.e. between water and air; curved because of surface tension and the adhesion of water to the sides of the cylinder o when measuring liquid, position our eyes level w/ meniscus and read volume at lowest level Kilogram (kg) – the basic unit of mass a kg = the mass of one thousand cubic centimeters (cm 3 ) of water at 4 0 C Biologists often measure mass with a TRIPLE BEAM BALANCE 3 beams o closest beam has 0.1-g graduations o middle beam has 100-g graduations o farthest beam has 10-g graduations the mass of the object is the sum of the masses indicated by the weights on the 3 beams Density – mass per unit volume density of water = mass/volume = about 1 Temperature – measure of how kinetic energy of molecules; the amount of heat in a system measured with a thermometer calibrated in degrees Celsius Celsius scale is based on water freezing at 0 0 C and boiling at 100 0 C Can interconvert o C and degrees Fahrenheit ( o F) using the formula o 5( o F) = 9( o C) + 160
Numerical Data: Statistics - offer a way to organize, summarize, and describe data the data are usually samples of information from a much larger population of values allow us to analyze the sample and draw inferences about the entire population make decisions without complete data about a population
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Biology 1105 Exam 1 - Biology 1105 Exam#1 Exercise 2...
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Ask a homework question - tutors are online | 1,051 | 3,788 | {"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} | 3.609375 | 4 | CC-MAIN-2018-13 | latest | en | 0.777181 |
https://aprove.informatik.rwth-aachen.de/eval/lowerbounds/html/proofs/104015.master.html | 1,723,324,912,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640822309.61/warc/CC-MAIN-20240810190707-20240810220707-00887.warc.gz | 69,295,887 | 3,841 | ### (0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
revconsapp(C(x1, x2), r) → revconsapp(x2, C(x1, r))
deeprevapp(C(x1, x2), rest) → deeprevapp(x2, C(x1, rest))
deeprevapp(V(n), rest) → revconsapp(rest, V(n))
deeprevapp(N, rest) → rest
revconsapp(V(n), r) → r
revconsapp(N, r) → r
deeprev(C(x1, x2)) → deeprevapp(C(x1, x2), N)
deeprev(V(n)) → V(n)
deeprev(N) → N
second(V(n)) → N
second(C(x1, x2)) → x2
isVal(C(x1, x2)) → False
isVal(V(n)) → True
isVal(N) → False
isNotEmptyT(C(x1, x2)) → True
isNotEmptyT(V(n)) → False
isNotEmptyT(N) → False
isEmptyT(C(x1, x2)) → False
isEmptyT(V(n)) → False
isEmptyT(N) → True
first(V(n)) → N
first(C(x1, x2)) → x1
goal(x) → deeprev(x)
Rewrite Strategy: INNERMOST
### (1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
### (2) Obligation:
Complexity Dependency Tuples Problem
Rules:
revconsapp(C(z0, z1), z2) → revconsapp(z1, C(z0, z2))
revconsapp(V(z0), z1) → z1
revconsapp(N, z0) → z0
deeprevapp(C(z0, z1), z2) → deeprevapp(z1, C(z0, z2))
deeprevapp(V(z0), z1) → revconsapp(z1, V(z0))
deeprevapp(N, z0) → z0
deeprev(C(z0, z1)) → deeprevapp(C(z0, z1), N)
deeprev(V(z0)) → V(z0)
deeprev(N) → N
second(V(z0)) → N
second(C(z0, z1)) → z1
isVal(C(z0, z1)) → False
isVal(V(z0)) → True
isVal(N) → False
isNotEmptyT(C(z0, z1)) → True
isNotEmptyT(V(z0)) → False
isNotEmptyT(N) → False
isEmptyT(C(z0, z1)) → False
isEmptyT(V(z0)) → False
isEmptyT(N) → True
first(V(z0)) → N
first(C(z0, z1)) → z0
goal(z0) → deeprev(z0)
Tuples:
REVCONSAPP(C(z0, z1), z2) → c(REVCONSAPP(z1, C(z0, z2)))
REVCONSAPP(V(z0), z1) → c1
REVCONSAPP(N, z0) → c2
DEEPREVAPP(C(z0, z1), z2) → c3(DEEPREVAPP(z1, C(z0, z2)))
DEEPREVAPP(V(z0), z1) → c4(REVCONSAPP(z1, V(z0)))
DEEPREVAPP(N, z0) → c5
DEEPREV(C(z0, z1)) → c6(DEEPREVAPP(C(z0, z1), N))
DEEPREV(V(z0)) → c7
DEEPREV(N) → c8
SECOND(V(z0)) → c9
SECOND(C(z0, z1)) → c10
ISVAL(C(z0, z1)) → c11
ISVAL(V(z0)) → c12
ISVAL(N) → c13
ISNOTEMPTYT(C(z0, z1)) → c14
ISNOTEMPTYT(V(z0)) → c15
ISNOTEMPTYT(N) → c16
ISEMPTYT(C(z0, z1)) → c17
ISEMPTYT(V(z0)) → c18
ISEMPTYT(N) → c19
FIRST(V(z0)) → c20
FIRST(C(z0, z1)) → c21
GOAL(z0) → c22(DEEPREV(z0))
S tuples:
REVCONSAPP(C(z0, z1), z2) → c(REVCONSAPP(z1, C(z0, z2)))
REVCONSAPP(V(z0), z1) → c1
REVCONSAPP(N, z0) → c2
DEEPREVAPP(C(z0, z1), z2) → c3(DEEPREVAPP(z1, C(z0, z2)))
DEEPREVAPP(V(z0), z1) → c4(REVCONSAPP(z1, V(z0)))
DEEPREVAPP(N, z0) → c5
DEEPREV(C(z0, z1)) → c6(DEEPREVAPP(C(z0, z1), N))
DEEPREV(V(z0)) → c7
DEEPREV(N) → c8
SECOND(V(z0)) → c9
SECOND(C(z0, z1)) → c10
ISVAL(C(z0, z1)) → c11
ISVAL(V(z0)) → c12
ISVAL(N) → c13
ISNOTEMPTYT(C(z0, z1)) → c14
ISNOTEMPTYT(V(z0)) → c15
ISNOTEMPTYT(N) → c16
ISEMPTYT(C(z0, z1)) → c17
ISEMPTYT(V(z0)) → c18
ISEMPTYT(N) → c19
FIRST(V(z0)) → c20
FIRST(C(z0, z1)) → c21
GOAL(z0) → c22(DEEPREV(z0))
K tuples:none
Defined Rule Symbols:
revconsapp, deeprevapp, deeprev, second, isVal, isNotEmptyT, isEmptyT, first, goal
Defined Pair Symbols:
REVCONSAPP, DEEPREVAPP, DEEPREV, SECOND, ISVAL, ISNOTEMPTYT, ISEMPTYT, FIRST, GOAL
Compound Symbols:
c, c1, c2, c3, c4, c5, c6, c7, c8, c9, c10, c11, c12, c13, c14, c15, c16, c17, c18, c19, c20, c21, c22
### (3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)
Removed 2 leading nodes:
GOAL(z0) → c22(DEEPREV(z0))
DEEPREV(C(z0, z1)) → c6(DEEPREVAPP(C(z0, z1), N))
Removed 18 trailing nodes:
ISEMPTYT(C(z0, z1)) → c17
ISNOTEMPTYT(C(z0, z1)) → c14
ISVAL(N) → c13
ISEMPTYT(V(z0)) → c18
REVCONSAPP(V(z0), z1) → c1
DEEPREV(N) → c8
DEEPREVAPP(N, z0) → c5
ISEMPTYT(N) → c19
SECOND(V(z0)) → c9
ISNOTEMPTYT(N) → c16
FIRST(V(z0)) → c20
REVCONSAPP(N, z0) → c2
ISVAL(C(z0, z1)) → c11
DEEPREV(V(z0)) → c7
FIRST(C(z0, z1)) → c21
SECOND(C(z0, z1)) → c10
ISVAL(V(z0)) → c12
ISNOTEMPTYT(V(z0)) → c15
### (4) Obligation:
Complexity Dependency Tuples Problem
Rules:
revconsapp(C(z0, z1), z2) → revconsapp(z1, C(z0, z2))
revconsapp(V(z0), z1) → z1
revconsapp(N, z0) → z0
deeprevapp(C(z0, z1), z2) → deeprevapp(z1, C(z0, z2))
deeprevapp(V(z0), z1) → revconsapp(z1, V(z0))
deeprevapp(N, z0) → z0
deeprev(C(z0, z1)) → deeprevapp(C(z0, z1), N)
deeprev(V(z0)) → V(z0)
deeprev(N) → N
second(V(z0)) → N
second(C(z0, z1)) → z1
isVal(C(z0, z1)) → False
isVal(V(z0)) → True
isVal(N) → False
isNotEmptyT(C(z0, z1)) → True
isNotEmptyT(V(z0)) → False
isNotEmptyT(N) → False
isEmptyT(C(z0, z1)) → False
isEmptyT(V(z0)) → False
isEmptyT(N) → True
first(V(z0)) → N
first(C(z0, z1)) → z0
goal(z0) → deeprev(z0)
Tuples:
REVCONSAPP(C(z0, z1), z2) → c(REVCONSAPP(z1, C(z0, z2)))
DEEPREVAPP(C(z0, z1), z2) → c3(DEEPREVAPP(z1, C(z0, z2)))
DEEPREVAPP(V(z0), z1) → c4(REVCONSAPP(z1, V(z0)))
S tuples:
REVCONSAPP(C(z0, z1), z2) → c(REVCONSAPP(z1, C(z0, z2)))
DEEPREVAPP(C(z0, z1), z2) → c3(DEEPREVAPP(z1, C(z0, z2)))
DEEPREVAPP(V(z0), z1) → c4(REVCONSAPP(z1, V(z0)))
K tuples:none
Defined Rule Symbols:
revconsapp, deeprevapp, deeprev, second, isVal, isNotEmptyT, isEmptyT, first, goal
Defined Pair Symbols:
REVCONSAPP, DEEPREVAPP
Compound Symbols:
c, c3, c4
### (5) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
revconsapp(C(z0, z1), z2) → revconsapp(z1, C(z0, z2))
revconsapp(V(z0), z1) → z1
revconsapp(N, z0) → z0
deeprevapp(C(z0, z1), z2) → deeprevapp(z1, C(z0, z2))
deeprevapp(V(z0), z1) → revconsapp(z1, V(z0))
deeprevapp(N, z0) → z0
deeprev(C(z0, z1)) → deeprevapp(C(z0, z1), N)
deeprev(V(z0)) → V(z0)
deeprev(N) → N
second(V(z0)) → N
second(C(z0, z1)) → z1
isVal(C(z0, z1)) → False
isVal(V(z0)) → True
isVal(N) → False
isNotEmptyT(C(z0, z1)) → True
isNotEmptyT(V(z0)) → False
isNotEmptyT(N) → False
isEmptyT(C(z0, z1)) → False
isEmptyT(V(z0)) → False
isEmptyT(N) → True
first(V(z0)) → N
first(C(z0, z1)) → z0
goal(z0) → deeprev(z0)
### (6) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
REVCONSAPP(C(z0, z1), z2) → c(REVCONSAPP(z1, C(z0, z2)))
DEEPREVAPP(C(z0, z1), z2) → c3(DEEPREVAPP(z1, C(z0, z2)))
DEEPREVAPP(V(z0), z1) → c4(REVCONSAPP(z1, V(z0)))
S tuples:
REVCONSAPP(C(z0, z1), z2) → c(REVCONSAPP(z1, C(z0, z2)))
DEEPREVAPP(C(z0, z1), z2) → c3(DEEPREVAPP(z1, C(z0, z2)))
DEEPREVAPP(V(z0), z1) → c4(REVCONSAPP(z1, V(z0)))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
REVCONSAPP, DEEPREVAPP
Compound Symbols:
c, c3, c4
### (7) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
REVCONSAPP(C(z0, z1), z2) → c(REVCONSAPP(z1, C(z0, z2)))
DEEPREVAPP(C(z0, z1), z2) → c3(DEEPREVAPP(z1, C(z0, z2)))
DEEPREVAPP(V(z0), z1) → c4(REVCONSAPP(z1, V(z0)))
We considered the (Usable) Rules:none
And the Tuples:
REVCONSAPP(C(z0, z1), z2) → c(REVCONSAPP(z1, C(z0, z2)))
DEEPREVAPP(C(z0, z1), z2) → c3(DEEPREVAPP(z1, C(z0, z2)))
DEEPREVAPP(V(z0), z1) → c4(REVCONSAPP(z1, V(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(C(x1, x2)) = [2] + x2
POL(DEEPREVAPP(x1, x2)) = [4] + [4]x1 + [3]x2
POL(REVCONSAPP(x1, x2)) = [5] + [2]x1
POL(V(x1)) = [3]
POL(c(x1)) = x1
POL(c3(x1)) = x1
POL(c4(x1)) = x1
### (8) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
REVCONSAPP(C(z0, z1), z2) → c(REVCONSAPP(z1, C(z0, z2)))
DEEPREVAPP(C(z0, z1), z2) → c3(DEEPREVAPP(z1, C(z0, z2)))
DEEPREVAPP(V(z0), z1) → c4(REVCONSAPP(z1, V(z0)))
S tuples:none
K tuples:
REVCONSAPP(C(z0, z1), z2) → c(REVCONSAPP(z1, C(z0, z2)))
DEEPREVAPP(C(z0, z1), z2) → c3(DEEPREVAPP(z1, C(z0, z2)))
DEEPREVAPP(V(z0), z1) → c4(REVCONSAPP(z1, V(z0)))
Defined Rule Symbols:none
Defined Pair Symbols:
REVCONSAPP, DEEPREVAPP
Compound Symbols:
c, c3, c4
### (9) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty | 3,406 | 7,734 | {"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-2024-33 | latest | en | 0.484781 |
https://www.csc.ncsu.edu/news/1243 | 1,726,296,816,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651559.58/warc/CC-MAIN-20240914061427-20240914091427-00490.warc.gz | 669,006,377 | 7,027 | CSC News
November 09, 2011
How Will Players Respond To Game Changes? Do The Math
Matt Shipman | News Services | 919.515.6386
Dr. David Roberts | 919.513.7182
The Abstract Blog
When it comes to video games, imitation can be dangerous. If one game makes changes that players love, there’s no guarantee that players of a different game would welcome the same changes. So, do you make the change anyway? Or do you sit tight and ride out the status quo? Time to do some math.
Because players behave differently in different gaming environments, it’s not as simple as comparing apples to apples. It’s more like comparing apples to oranges. Or possibly apples to manatees. Here’s where the math comes in.
When developers – or researchers – make a change to a gaming environment, they can quantify how players respond to that change. And researchers have come up with a math-based technique to extrapolate how a similar change would play out in a different gaming environment.
“This is the first principled approach that allows us to quantify information about a new environment based on information from an old environment,” says Dr. David Roberts, an assistant professor of computer science at NC State and co-author of a paper on the research. “To an extent, developers will always rely on their intuition, but this could help them take advantage of previous mistakes or successes by themselves or other developers.” Roberts also notes that this research looks exclusively at interactive storytelling environments (like RPGs) – so its usefulness for arcade style games is not yet known.
Roberts did the research with Fred Roberts, a professor of mathematics at Rutgers University, and essentially put a new spin on mathematical psychology concepts that have been around since the Hoover administration: models of utility and preference. These models use concise mathematical equations to describe human behavior. The paper is available here.
In these models, a person’s preference for a given alternative is a function of its “utility” – or value. In other words, the more value a person places on something, the more likely he/she is to select it.
In this work, the researchers developed a technique where they evaluated baseline behaviors in one game (Game A), as well as alterations in behavior after a specific change was made to the game. This behavioral data was then used to quantify the utility (or value) of the various behaviors exhibited by players in Game A. Essentially, they analyzed how and why players behaved differently after the change.
The researchers then evaluated baseline behaviors in a second game (Game B). Researchers used Game B’s baseline data and the utility data from Game A to estimate what the utility of behaviors would be in Game B if it made the same changes that had been made to Game A. With me so far? Basically, you need to understand the essential value of doing specific things in Game B if you want to predict how its players will respond to change.
While it may sound complicated (and you should really see the equations), it works pretty well – with error rates ranging from +/-0.38 percent to +/-27 percent, depending on the quality of input data.
Let’s explain what I mean by error rates. Let’s say researchers use the new technique, and predict that 72 percent of players will respond to a game-change in the same way. If the error rate is +/-2 percent, that means that between 70 and 74 percent of the players actually responded that way (they were within 2 percent of the correct outcome).
“We don’t think this is going to make a huge difference in the way games are designed,” David Roberts says, “but it could help take some of the guesswork out of the process.”
Roberts will be presenting the paper at the 4th International Conference on Interactive Digital Storytelling, being held Nov. 28 through Dec. 1 in Vancouver, British Columbia.
~shipman~ | 796 | 3,913 | {"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} | 2.578125 | 3 | CC-MAIN-2024-38 | latest | en | 0.941189 |
http://cr4.globalspec.com/thread/112402/Nitrogen-Flow-Calculation | 1,513,149,053,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948522205.7/warc/CC-MAIN-20171213065419-20171213085419-00115.warc.gz | 61,379,461 | 16,815 | CR4® - The Engineer's Place for News and Discussion®
Previous in Forum: Rubber and Heat Next in Forum: Nitroglycerin Question
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Posts: 2
Nitrogen Flow Calculation
05/10/2017 2:30 AM
Can anybody explain how to calculate the nitrogen flow (m3/Hr) in 1" line at 1 Kg/Cm2(g) and if On/Off valve opens it will exposed to atm(0kg/cm2 g)
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Anonymous Poster #1
#1
Re: Nitrozen flow calculation
05/10/2017 3:09 AM
For nitrozen flow, just go with the Tao.
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#2
Re: Nitrozen flow calculation
05/10/2017 3:43 AM
What is TAO
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Anonymous Poster #1
#3
Re: Nitrozen flow calculation
05/10/2017 3:59 AM
I am being a bit facetious, but not totally so. The Tao is a Chinese philosophy associated with Lao Tse and Chuang Tse (various spellings for each). It is commonly translated as "the way", and it tends to be open-minded, non-resistant, go-with-the-flow. The spelling "nitrozen" suggested Zen Buddhism, which might be loosely related. There is some Zen (s)wordplay in all of this.
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#4
Re: Nitrozen flow calculation
05/10/2017 4:13 AM
If this is nitrogen gas rather than liquid, you won't go too far wrong by simply assuming it is air. There are lots of Web-accessible calculators for air flow versus pressure difference. Pick a pipe size the next (or second next) larger size than you need, and then throttle with a valve.
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#5
Re: Nitrozen flow calculation
05/10/2017 4:19 AM
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#6
Re: Nitrozen flow calculation
05/10/2017 4:39 AM
That can probably be found by an Easter-egg hunt or wild-goose chase....
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#7
Re: Nitrozen flow calculation
05/10/2017 4:53 AM
My post was mainly for OP's benefit. I replied to yours as you'd mentioned air vs N2 and websites
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#10
Re: Nitrozen flow calculation
05/10/2017 3:35 PM
I think he wanting to make a nitrogen powered 'tater launcher. Unfortunately, the pressure is all wrong, the valve and line size is wrong, except for "new" potatoes.
Line length not given, I suppose he does not need a muzzle reaction suppressor.
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#8
Re: Nitrogen Flow Calculation
05/10/2017 11:40 AM
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#9
Re: Nitrogen Flow Calculation
05/10/2017 3:33 PM
In a word, NO! No line length given.
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#11
Re: Nitrogen Flow Calculation
05/11/2017 6:07 AM
Nitrogen flowing out of the end of a piece of pipe isn't doing anything useful. Depending on the length of pipe, the velocity out of the end could approach the speed of sound at [pressure, temperature], though it certainly cannot exceed it.
In order for this nitrogen to do something useful, and the only thing it is doing at present is making a big load of unwanted noise (in which case air would be a better choice), it has to be connected to something that presents back-pressure. So the premise in the original posting <...exposed to atm(0kg/cm2 g)...> is absurd.
So, nobody <...Can...explain...> until the purpose of this currently-abstruse contrivance is explained to the forum. Until that happens, any attempt at calculation is futile.
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https://www.gradesaver.com/textbooks/math/calculus/calculus-8th-edition/chapter-12-vectors-and-the-geometry-of-space-12-4-the-cross-product-12-4-exercises-page-862/32 | 1,575,988,031,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540527620.19/warc/CC-MAIN-20191210123634-20191210151634-00158.warc.gz | 722,271,292 | 12,471 | Calculus 8th Edition
(a) $\lt -4,7,-10 \gt$ (b) $\frac{ \sqrt {165}}{2}$
(a) Given: $P(-1,3,1),Q(0,5,2)$ and $R(4,3,-1)$ $PQ ^\to=\lt 0-(-1),5-3,2-1\gt=\lt 1,2,1 \gt$ $PR ^\to=\lt 4-(-1),3-3, -1-1\gt=\lt 5,0,-2 \gt$ $\lt 1,2,1 \gt \times \lt 5,0,-2 \gt=\lt -4,7,-10 \gt$ (b) Area of a vector with vertices at P,Q, and R is $Area=\frac{1}{2}|PQ ^\to \times PR ^ \to|$ $PQ ^\to \times PR ^ \to=\lt 1,2,1 \gt \times \lt 5,0,-2 \gt=\lt -4,7,-10 \gt$ $|PQ ^\to \times PR ^ \to|=\sqrt {(-4)^2+(7)^2+(-10)^2}= \sqrt {165}$ $Area=\frac{1}{2}|PQ ^\to \times PR ^ \to|=\frac{ \sqrt {165}}{2}$ | 318 | 583 | {"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} | 3.96875 | 4 | CC-MAIN-2019-51 | latest | en | 0.217552 |
https://diydrones.com/profiles/blogs/servo-redundancy?xg_source=activity | 1,571,234,722,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986668994.39/warc/CC-MAIN-20191016135759-20191016163259-00361.warc.gz | 464,041,720 | 22,015 | # Servo Redundancy
I continue to develop my "full scale" gas powered, variable pitch quad-copter.
After watching the video below at
Amazing DIY Projects
I realized (more vividly) how vulnerable a quad is to any fault in the control system.
Then I saw his video here --> Redundant flight controls where at around 23:30 in the video he informs that he will use quintuple redundancy. That got me thinking about how I could follow the same approach using multiple independent flight controls to provide redundancy for the control system aspect of the project. His project is different in that he has many outputs (~76 electric motors), but I only have 4 collective "swash plates" that I have to move, so the challenge was to come up with a linkage to connect multiple servos in a way that they won't fight each other, but will also be able to retain control if one of the servo's fails.
Below is an image of the non-redundant setup I'm using for collective pitch.
I struggled quite a bit with many different ideas, but finally arrived at this which I think is elegantly simple.
The servo arm is connected to a rod-end which is screwed into the bottom of the lower (non-rotating) portion of the swash plate. If the angle of the servo sweep is kept reasonable (60 to 90 degrees total) this arrangement does not create too much binding or bending of the servo arm.
Looking back at the first image, you can see that I've connected the servo arms together in an equilateral triangle arrangement, with the rod-end at the center of the triangle. The motion path of the rod-end is exactly the same as it would be with a single servo. Of course, this still needs to work if one of the servos fails. If that happens, lets assume the servo just stops moving, and stays in place (gear friction is greater than the linkage torque so servo doesn't "freewheel").
The issue here is that the rod-end only moves one-third the distance of any given single servo movement. So with one failed servo, the rod-end will only move 2/3 the distance of the remaining two active servos. For this reason, the servos should be adjusted so that they have 50% extra travel (only use 2/3 of their range when all 3 servos working).
Another issue is tuning: The system will have lower gain when one servo fails, so I will need to test the flight stability in both cases (all servo's working, and again with a failed servo) to ensure the PID loops will be able to safely control in both situations.
This design works best if the failed servo doesn't freewheel. I'm don't think it will work if the failed servo doesn't freeze in it's failed position.
The idea here, is that each of these servo's would be controlled by a separate flight controller, so the whole control system would be tripley (word?) redundant. Actually, this linkage concept could be extended to make it 4,5 or even 6x redundant, but only 3 servo's can be directly connected to each other by the "Y" linkage. Beyond 3, I will need to join multiple linkages together.
Below is a photo of my non-redundant collective linkage on my most recent test stand. Getting ready to test the vibration of a 2 cycle engine driving a variable pitch 2 blade rotor.
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Comment by Armin Strobel on March 14, 2018 at 7:54am
Nice project! I would consider some feedback from the servos that you know which one failed that the other ones can compensate for the lost travel. You may want to consider a module in front of every servo which is connected to all PWM inputs and to the feedback of every single servo. That module can do two things:
1.Identify if one of the flight controllers disagrees with the other two ones and the use the average of the two other inputs.
2. It can identify that one of the other servos failed and compensate for that
Or in short having smart servos which observes the other servos and identifies disagreements of the flight controllers
Comment by Randy Sonnicksen on March 14, 2018 at 8:28am
Armin, I hear what you are saying, and....this project has already gotten more complex than I had originally hoped.
So, my approach at this point is to let the healthy systems compensate for the unhealthy ones. I acknowledge that during a failure, there will be a loss in performance and responsiveness, but enough to safely land.
But you bring up a good question, of how do I even know there has been a failure? My concern with introducing a bunch of feedback signals, is that from my experience, the feedback is just as likely to fail as the actuator, so I am introducing a whole new set of failure modes (servo working OK, but feedback signal error).
You've got me thinking though, and I will need to implement some method to tell me when I've got a problem in one of my control systems so that I know to terminate the flight. Thanks for your input!
Comment by Armin Strobel on March 14, 2018 at 9:08am
I understand. Also additional complexity may not reduce the risk if not done very carefully.
I agree. Your system needs to be tuned towards robustness that it can handle safely one servo / flight controller failure.
You are also right you really need a health monitoring system. Otherwise you could have a failure without even noticing (if you did a great job with the tuning) and you flying around without failsafe without knowing it. The servo OK, but feedback signal error is fine, since the feedback is only for the health monitoring and should force a safety landing. That is unfortunately but you would probably also stop your car if the red warning light for low oil pressure is coming up, eve though it is much more likely that you have a failed sensor or connection to the sensor.
Another thing is to think that theoretically with integrators involved in your flight controllers they could theoretically work against each other. Sorry for bringing all that up.
Comment by Randy Sonnicksen on March 14, 2018 at 10:24am
Armin,
The algorithm I'm playing with has a roll/pitch angle proportional controller cascaded to a roll/pitch rate PID controller.
If I'm thinking correctly, the separate controllers shouldn't "wind up" their integral term, because the PID loop is working on a "roll rate" signal. I will need to think about this some more, but yes, if they were straight roll angle to PID loop, they would certainly diverge with one PID hitting a high limit, while another hits a low limit. Not good.
Comment by Armin Strobel on March 14, 2018 at 2:05pm
I was assuming that it would be a cascaded PID. Straight roll/pitch I highly would doubt that it would work reliable, if at all.
The integral part might be actually a port of the solution. In normal operation it might could happen that one integral parts winds up to the higher end and the other one on the lower end. As long if the limits are set somewhat decent set it will not matter. But if one servo fails then the integral part of the others will compensate for that. You will not have the same gain as you mentioned before, but the maybe resulting offset (operating point) can be compensated by the integral parts
Comment by Randy Sonnicksen on March 15, 2018 at 2:20pm
Armin, Yes this is a bit tricky. I was wrong before thinking that PID on the roll/pitch rate control would avoid divergence between controllers. This will be a problem if I use integral control. I wonder how the guy in Amazing DIY Projects handled this. Is it possible his FC's only use P and D control? That would make Him, the pilot, the I controller.
If I went with the P only algorithm, I would probably sense when a servo failed, because I would likely have to hold the stick off-center to level the craft.
So let me think this through. Lets say my quad is a "+" configuration to make it simple.
Lets say a servo on my right rotor fails, and just holds its current position. So far no problem.
Now a wind pushes me to the right. So I add left cyclic. This increases the pitch in the right rotor, and decreases the pitch in the left rotor. But, because only 2 of 3 servos are moving on the right, and 3/3 on the left, I get a smaller increase in lift on the right than the decrease on the left. So my total amount of lift decreases, and I begin to descend. Also, if these rotors are spinning CW, the CCW counter-torque from the right rotor increases less than the decrease in the left rotor. So a net decrease in the CCW counter-torque is experienced, and the craft begins to yaw clockwise.
So if a servo fails (stationary) I will see lift and yaw effects with cyclic stick inputs.
Conversely, I will also likely see pitch/roll and yaw effects with throttle stick inputs.
NOW - What if a servo on the right rotor fails, and say it goes rapidly to minimum position.
The craft will immediately roll right, yaw right, and descend. The roll and yaw axis controllers will respond to the sudden rate change from the gyro, and send a correcting signal, and the roll angle controller will aid this as the roll angle deviates from level. (assuming you were hovering). The craft would be drifting right from the right roll, and this will need to be corrected by the pilot with left cyclic in addition to more throttle, and left rudder.
If the P gains are high, the roll/pitch and yaw controls will immediately help to stabilize the craft in the event of a servo failure. Lift (throttle) will need to be compensated for by the pilot unless the craft is operating in some form of altitude hold mode. With a variable pitch quad, there is an actual throttle output which literally controls the gas engine throttle. This will likely be from a speed controller, or will be linked to a mechanical governor to regulate engine speed.
Comment by Armin Strobel on March 15, 2018 at 3:40pm
I just would make sure that you limit the maximum servo position of every servo. I also agree the integral parts could run in different direction, but if you make sure that you have some kind of anti windup for your integrator in a range where it makes sense, you should be fine.
I might shouln't have raised those concerns, just build it and test all possible cases. If you discover than problems then ask here for advice. I am an others are more than happy to help you.
Comment by Randy Sonnicksen on March 16, 2018 at 10:16am
I posted a comment on Axel Borg's (AmazingDIYProjects) youtube video wondering how he handled integral control with 5 independent controllers, and....HE ANSWERED! Which is very uncharacteristic for him. He indicated that the flight controllers only use P control. I gain is set to zero. He, the pilot, is the I controller. There are of course some negative consequences (like having to hold the cyclic off center if your Cg and Cl don't line up) but that's a minor nuisance. Thanks Axel for your response!
You are absolutely correct; the controllers are not connected to each other, so I had to turn down the I-value to zero. I, the pilot, handle diversion that occur over time, its quite a natural thing to do when flying it from the pilot seat; Im not breaking any sweat doing the I-component work.
Comment by Chris Card on March 21, 2018 at 5:18pm
Interesting project.
Regarding the coupling of the servos as depicted in the image (top of page)... What happens if one of the three servos breaks in way that allows it to "free-wheel" , no resistance ?
If a servo fails and is immobilized , and just happens to be centered such that there is no offset caused by its' failure , the pitch control range for that rotor, (as you have already mentioned ) , would be reduced. This might be noticeable to the operator and hopefully prompt a timely investigation. I don't know if such a failure would be revealed in the data flash logs, but it would be visible to the human eye during a pre-flight or post-flight mechanical inspection.
If you used the Pixhawks' current monitoring (Amperes) to log the current flow for each of the three servo groups (three Pixhawks) independently . There may be a deviation in the current consumption that might show up in the logged flight data , especially if all three data logs were compared with each other.... That would require splitting the servo power distribution for each of the three groups.
..additionally, during the bench testing , one could establish base values for the expected current (amps). Such logging (with analysis) might also help to detect if a servo is binding or becoming troubled or even an engine/governor is having issues.
Does the motor to propeller drive coupling use a spar-clutch (one-way clutch) for auto-gyration like helicopters (R/C & full-scale) utilize?.
Regards
Chris
Comment by Randy Sonnicksen on March 21, 2018 at 6:20pm
Chris, Yes, if the broken servo free-wheels, I'm SOL.
Regarding pitch control range, I'm planning to only use the middle 2/3 of the servo's range during normal operation so that extra range on the servo is available to compensate for a non-responsive servo.
Good idea on monitoring the servo current, but I must confess, I haven't selected a flight controller yet. In any case, this redundancy is intended to keep the bird in the air in the event of a malfunction during flight. Each servo will function will need to be verified during pre-flight.
Yes, the motor to propeller drive includes a sprag clutch. I wrestled with the best way to do this. I eventually decided to put them as close to the rotor as possible so that the rotor could free-wheel even if the drive belt became entangled in the pulley or it's axle. With all 4 rotors running at different RPM during autorotation, I will need some help from the controller to keep all the rotors in an acceptable speed range, and keep the bird level.
The good news is I have quite a bit of headroom on rotor RPM. I should be able to easily go 50% or even 100% over nominal RPM without a structural issue. I think the tolerance is much tighter for a full scale helo. (10%).
Someone has also suggested a global monitoring system, that doesn't perform any control function, but simply monitors the health of all the components, and parameters and alarms if a fault is detected.
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Season Two of the Trust Time Trial (T3) Contest
A list of all T3 contests is here. The current round, the Vertical Horizontal one, is here | 3,255 | 14,414 | {"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-2019-43 | latest | en | 0.948164 |
https://www.urbanpro.com/bank-clerical-exam/syllogism-tricks | 1,725,885,593,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651098.19/warc/CC-MAIN-20240909103148-20240909133148-00109.warc.gz | 986,585,298 | 56,460 | true
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# Syllogism Tricks
Ashish Vastri
06/02/2017 0 0
Syllogism Tricks
Syllogism is a major part of reasoning in various competetive exams. It's simple but brain teaser too. Most of the students find it difficult to learn the concepts of venn diagram and apply accurately to questions.
I would try to help you all with some useful concepts and tricks that you may apply in competitive exams.
All of these tricks will surely help you in Banking, Railways, SSC, NDA, MBA and many other exams.
All + All = All
All + No = No
All + Some = No conclusion
Some + All = Some
Some + No = Some not
Some + Some = No conclusion
No + All = Some not (Reversed)
No + Some = Some not (Reversed )
No + No = No conclusion
Note:
Some not/ Some Not Reversed + Anything = No conclusion
To solve syllogism questions you may take help of two types of cancellation :
• Cross Cancellation
• Vertical Cancellation
Cross Cancellation
STATEMENTS
1. All pens are Erasers
2. All Erasers are Books
3. No book is Table
CONCLUSIONS
1. No Eraser is table
All + No = No
All Erasers are books
No book is a Table
If we will cross cancel books, we get
CONCLUSION: No Eraser is table .
Hence conclusion follows
Vertical Cancellation
STATEMENTS
1. Some apples are fruits
2. All bananas are fruits
CONCLUSION
1. No banana is apple
All + Some = No conclusion
Some apples are fruits
All bananas are fruits
Here we can cancel fruits vertically but the direction of adding will get Reversed , i.e. Conclusion will be ALL + Some = No conclusion
Hence, Conclusion 1 doesn't follow.
There are more key points regarding Syllogism which I will post in next lesson.
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The best Tutors for Bank Clerical Exam Coaching Classes are on UrbanPro | 811 | 3,405 | {"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} | 4.0625 | 4 | CC-MAIN-2024-38 | latest | en | 0.898549 |
https://wcipeg.com/problem/mockccc14j2#comment9158 | 1,643,389,674,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306301.52/warc/CC-MAIN-20220128152530-20220128182530-00484.warc.gz | 648,667,261 | 5,711 | ## Problem J2: Rope Unfolding
Alice's love potion failed because Bob overdosed. Now, her new tactic is to impress Bob. Bob is a computer scientist who specializes in ropes. In computer science, a rope is a data structure used for efficiently manipulating strings. Bob has a rope of characters. However, it's too long to fit in his pocket, so he folds it into N segments. Each fold is at a letter which is not the last. The fold takes the rest of the rope after that letter, reverses it, and puts it above the current letter. For example, the rope "`ABRACADABRA`", when folded "`AB|RACA|DAB|RA`", yields:
``` AR
DAB
ACAR
AB
```
Alice notices that Bob often gets a headache pulling the rope out of his pocket and spending a long time untangling it. She concludes that if she finds a way to unfold it for Bob, Bob will be so impressed that he can't help but fall madly in love with her. You must write a program to help Alice!
### Input Format
The first line of input will contain the positive integer N (1 ≤ N ≤ 100), the number of lines to follow. The next N lines each contain a segment of the folded rope.
The entire, unfolded rope will be no greater than 100 characters, and will only consist of uppercase letters from 'A' to 'Z'.
### Output Format
The output should contain one line — the unfolded rope.
```4
AR
DAB
ACAR
AB
```
### Sample Output 1
```ABRACADABRA
```
```5
D
LRO
W
OL
HEL
```
### Sample Output 2
```HELLOWORLD
```
Point Value: 3
Time Limit: 2.00s
Memory Limit: 16M
Added: Feb 27, 2014
Author: Alex
Languages Allowed:
C++03, PAS, C, HASK, ASM, RUBY, PYTH2, JAVA, PHP, SCM, CAML, PERL, C#, C++11, PYTH3
## Comments (Search)
• (0/0)
2
CD
BA
Would both ABCD and BADC be accepted?
• (0/0)
no, only BADC is accepted
• (0/0)
Why isn't ABCD accepted? It doesn't say anywhere in the problem statement, that the last line of input goes from left to right.
The "help" says "You should always assume that any test case that is not explicitly forbidden by the problem statement can and will appear in the test data". So, I can't see why this can't be a valid test case.
RA
BADA
_RAC
_BA
• (0/1)
Input:
4
---GH
--EF
-CD
AB
Output: ABCDEFGH
I think this relaxation would break a lot of solutions.
• (4/0)
Isn't this illegal by the conditions of the problem statement?
• (0/0)
Yes it is illegal
• (0/0)
the output should be ABDCEFHG | 673 | 2,361 | {"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} | 2.609375 | 3 | CC-MAIN-2022-05 | latest | en | 0.91491 |
http://www.jiskha.com/display.cgi?id=1394414007 | 1,462,396,270,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860124045.24/warc/CC-MAIN-20160428161524-00135-ip-10-239-7-51.ec2.internal.warc.gz | 618,273,034 | 3,564 | Wednesday
May 4, 2016
# Homework Help: Physics
Posted by help on Sunday, March 9, 2014 at 9:13pm.
A 2 kg block has an initial velocity of 16 m/s at the bottom of the ramp (see
diagram below). The coefficient of kinetic friction between the block and the
ramp is 0.1. The spring has length 3 m and has a spring constant of 100
N/m. What maximum compression will the spring experience when the block
strikes it? | 117 | 412 | {"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-2016-18 | longest | en | 0.935918 |
https://www.cheenta.com/circumference-of-a-semicircle-amc-8-2014-problem-25/ | 1,632,037,456,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056752.16/warc/CC-MAIN-20210919065755-20210919095755-00037.warc.gz | 725,112,806 | 42,776 | How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?
# Circumference of a Semicircle | AMC 8, 2014 | Problem 25
Try this beautiful problem from AMC-8-2014 (Geometry) based on Circumference of a Semicircle
## Circumference of a Semicircle- AMC 8, 2014 - Problem 25
On A straight one-mile stretch of highway, 40 feet wide, is closed. Robert rides his bike on a path composed of semicircles as shown. If he rides at 5miles per hour, how many hours will it take to cover the one-mile stretch?
• $\frac{\pi}{11}$
• $\frac{\pi}{10}$
• $\frac{\pi}{5}$
### Key Concepts
Geometry
Semicircle
Distance
Answer:$\frac{\pi}{10}$
AMC-8, 2014 problem 25
Challenges and Thrills of Pre College Mathematics
## Try with Hints
Find the circumference of a semi-circle
Can you now finish the problem ..........
If Robert rides in a straight line, it will take him $\frac {1}{5}$ hours
can you finish the problem........
If Robert rides in a straight line, it will take him $\frac {1}{5}$ hours. When riding in semicircles, let the radius of the semicircle r, then the circumference of a semicircle is ${\pi r}$. The ratio of the circumference of the semicircle to its diameter is $\frac {\pi}{2}$. so the time Robert takes is $\frac{1}{5} \times \frac{\pi}{2}$. which is equal to $\frac{\pi}{10}$ | 379 | 1,306 | {"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} | 4.25 | 4 | CC-MAIN-2021-39 | latest | en | 0.824574 |
https://byjus.com/question-answer/a-car-is-moving-along-a-straight-line-it-s-displacement-x-time-t-graph-7/ | 1,713,634,419,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817670.11/warc/CC-MAIN-20240420153103-20240420183103-00201.warc.gz | 126,130,521 | 25,109 | 1
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Question
# A car is moving along a straight line. It's displacement (x) - time (t) graph is shown. Match the entries in column I with points on graph as in Column-II. Column I (p) x → negative,v → positive,a → positive (q) x → positive, v → negative,a → negative (r) x → negative,v → negative,a → positive (s) x → positive, v → positive,a → negative. Column II
A
(p - d) (q - b) (r - c) (s - a)
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B
(p - b) (q - d) (r - c) (s - a)
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C
(p - a) (q - c) (r - b) (s - d)
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D
(p - c) (q - a) (r - d) (s - b)
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Solution
## The correct option is A (p - d) (q - b) (r - c) (s - a) At,A,x is positive, v is positive as the the slope is positive a is negative as the slope is decreasing At,B,x is positive, v is positive as the the slope is negative a is negative as the slope is still decrease At C,x is negative, At D, x is negative v is negative v is positive a is positive a is positive
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Join BYJU'S Learning Program | 427 | 1,386 | {"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} | 3.265625 | 3 | CC-MAIN-2024-18 | latest | en | 0.897099 |
http://www.shmoop.com/quadratic-formula-function/imaginary-complex-number-exercises-4.html | 1,472,206,299,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982295424.4/warc/CC-MAIN-20160823195815-00039-ip-10-153-172-175.ec2.internal.warc.gz | 682,562,851 | 9,079 | Imaginary and Complex Numbers Exercises
# Imaginary and Complex Numbers Exercises
### Example 1
Simplify.
3i3 – 4i(6i16) + 2
Simplify.
i2346
### Example 3
Simplify.
5 + i – (i7 – 4)(2i + 1)
Simplify.
(2 – i)4
### Example 5
Simplify.
-1 + 3i2 + 6(4 – i) | 106 | 266 | {"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} | 3.390625 | 3 | CC-MAIN-2016-36 | latest | en | 0.727257 |
https://www.convertunits.com/from/yoctojoule/to/foot+poundal | 1,674,863,266,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499468.22/warc/CC-MAIN-20230127231443-20230128021443-00508.warc.gz | 745,265,210 | 12,635 | ## ››Convert yoctojoule to foot poundal
yoctojoule foot poundal
How many yoctojoule in 1 foot poundal? The answer is 4.214011E+22.
We assume you are converting between yoctojoule and foot poundal.
You can view more details on each measurement unit:
yoctojoule or foot poundal
The SI derived unit for energy is the joule.
1 joule is equal to 1.0E+24 yoctojoule, or 23.730360457056 foot poundal.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between yoctojoules and foot poundal.
Type in your own numbers in the form to convert the units!
## ››Want other units?
You can do the reverse unit conversion from foot poundal to yoctojoule, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Yoctojoule
The SI prefix "yocto" represents a factor of 10-24, or in exponential notation, 1E-24.
So 1 yoctojoule = 10-24 joules.
The definition of a joule is as follows:
The joule (symbol J, also called newton meter, watt second, or coulomb volt) is the SI unit of energy and work. The unit is pronounced to rhyme with "tool", and is named in honor of the physicist James Prescott Joule (1818-1889).
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 456 | 1,675 | {"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} | 2.609375 | 3 | CC-MAIN-2023-06 | latest | en | 0.828941 |
http://www.fas.org/man/dod-101/navy/docs/fun/part08.htm | 1,419,129,210,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802770616.6/warc/CC-MAIN-20141217075250-00024-ip-10-231-17-201.ec2.internal.warc.gz | 500,665,985 | 43,653 | # PRINCIPLES OF UNDERWATER SOUND Chapter 8
PRINCIPLES OF UNDERWATER SOUND
8.1 OBJECTIVES AND INTRODUCTION
Objectives
1. Understand why sound energy is employed for underwater surveillance and detection.
2. Understand the following terms associated with the fundamentals of sound production: wave motion & propagation, acoustic pressure, acoustic intensity, characteristic impedance.
3. Know the SI units employed in acoustic measurements and be able to convert to other measurement standards.
4. Understand the decibel method of measuring sound levels, including decibel calculations.
5. Be able to calculate intensity levels and sound pressure levels.
6. Be able to obtain an overall intensity level from individual sources.
7. Know how the following factors affect transmission loss: spreading, absorption, scattering, and bottom loss.
8. Be able to calculate appropriate values of transmission loss.
9. Understand the sources and effects of self-noise and ambient noise.
10. Understand the sources and effects of cavitation.
11. Be able to use Wenz curves to estimate values of ambient noise.
12. Be able to apply the concept of signal-to-noise ratio to underwater sound.
13. Know the basic sonar equation and its passive and active derivations.
14. Understand the significance of the terms in the sonar equations and their relationship to the figure-of-merit.
15. Be able to calculate figure-of-merit.
16. Be able to calculate values of sound speed based upon an understanding of temperature, pressure, and salinity effects.
17. Know the basic thermal and sound-speed structure of the ocean and how field observations of sound speed are made.
18. Be acquainted with the Ray Theory solution to the Wave Equation.
19. Understand the use of Snell's Law in determining ray path structure by calculating simple straight-line and curved ray paths.
20. Know the three basic types of sound-speed gradients and how they affect sound propagation to produce the following types of paths: surface duct, shadow zone, sound channel, convergence zone, and bottom bounce.
21. Demonstrate understanding of the theoretical and operational principles, applications capabilities and limitations of underwater sound through class discussion and problem solving.
Introduction
The effectiveness of the present-day submarine depends upon its ability to remain undetected for long periods of time while it searches, tracks, or attacks from beneath the sea surface. This medium of concealment, however, is advantageous to the submarine only so long as it is not detected or deprived of its ability to detect. Before a submarine can be attacked, it must be detected and its subsequent positions determined within the requirements of the available weapons system. Detection and position fixing can take place in two ways. There may either be some radiation or reflection of energy from the submarine to the searcher, or else the submarine may disturb one of the natural, static, spatial fields, such as the earth's magnetic field, thereby betraying its presence.
The choice of energy to be used for underwater detection is determined by three factors:
1. Range of penetration in the medium.
2. Ability to differentiate between various objects in the medium.
3. Speed of propagation.
Of all the known physical phenomena, light has excellent differentiation ability and high speed of transmission, but its range in water is very limited, on the order of tens of meters, thereby restricting its operational usefulness. This is not to say that light will never be used in ASW, for continuing experimentation and perfection of laser light detectors might yet add this method to the arsenal. Radio frequency waves also are propagated with extreme rapidity and to great distances through certain mediums, but sea water is essentially impervious to them for most frequencies. VLF signals will penetrate only about 10 meters, whereas higher frequency penetration depths can be measured in millimeters. Magnetic and gravitational field distortions are detectable only at very short ranges because the anomaly diminishes proportionally with the inverse of the range cubed. While their detection range is greater than either light or radio frequency, it is only of the magnitude of several hundred meters and therefore is insufficient for normal surveillance.
Acoustic energy, while lacking the propagation speed of electromagnetic waves, is capable of being transmitted through the sea to distances that are operationally significant. Because of this, sound is the physical phenomenon used for antisubmarine warfare, underwater communications, and underwater navigation. It must not be inferred, however, that sound is a panacea. It too has significant limitations to its effective employment, all of which must be thoroughly understood by the operators of underwater sound equipment. The optimum use of sound requires a thorough understanding of its limitations so that these effects can be minimized. For example, sea water is not uniform in pressure, temperature, or salinity, and all these characteristics have important effects on sound propagation through the sea. The requirement for predicting these effects on sonar performance has become a necessity, and a difficult one at that.
8.2 FUNDAMENTAL CONCEPTS
All sound, whether produced by a cowbell or a complicated electronic device, behaves in much the same manner. Sound originates as a wave motion by a vibrating source and requires for its transmission an elastic medium such as air or water. For example, consider a piston suspended in one of these mediums. As the piston is forced to move forward and backward, the medium is compressed on the forward stroke and decompressed or rarefied on the return stroke. Thus, a wave motion or series of compressions and rarefactions is caused to move from the source out through the medium. In the fluid medium the molecular motion is back and forth, parallel to the direction of the piston's movement. Because the fluid is compressible, this motion results in a series of detectable pressure changes. This series of compressions and rarefactions, such as is produced by the piston, constitutes a compressional wave train. Another way of explaining the phenomenon of acoustic wave propagation is to consider the medium of transmission as a loosely packed collection of mass elements connected by springy bumpers. A disturbance of the elements at some point (e.g., piston motion) moves along in the fluid by the successive extension and compression of the springs as the elements swing back and forth, each communicating its motion to its neighbor through the connecting bumpers. In this way, the agitation of a cluster of elements is propagated through the medium even though the individual elements do no more than move about their equilibrium positions without actually migrating. The sound wave propagates parallel to the source resulting in a longitudinal wave. Recall from radar principles that the electromagnetic wave propagated perpendicular to the source, resulting in a transverse wave.
FFigure 8-1. Pictorial representation of simple longitudinal wave.
8.2.1 The Sound Wave
The sine wave of figure 8-1 is graphical representation of the compressional wave train. As the wave passes a given point, the fluid elements are compressed and extended in the manner depicted by the sine wave's oscillations above and below the static pressure. The compressions and rarefactions are so labeled on the curve. There are two important things to note on this curve. The first is the maximum amplitude of the sine wave, labeled P, which represents the maximum pressure excursion above and below the static or hydrostatic pressure that exists in the fluid at the location of the wave train. The second thing to note is that the wave train would be passing, or propagating at the speed of sound c in the medium. The units for speed of sound are meters per second. The relationship between the three acoustic quantities that can be derived from figure 8-1 is:
Frequency (f) = Speed of sound (c) (8-1)
Wavelength ()
This is exactly the same as described for electromagnetic energy in chapter 1 and thus will not be elaborated upon.
One final point should be made about the sine wave representation of figure 8-1. Though somewhat difficult to imagine and more difficult to describe pictorially, the displaced parallel lines in figure 8-1a represent the motion of the elements within the field as the wave train passes. As the elements are compressed and extended, their motion can also be mathematically described by a sine wave; however, the elements would be oscillating to and fro about their static position. The maximum amplitude would then be the maximum displacement from the static position. To provide an example of the order of magnitude of these displacements, consider that the faintest 1,000 Hz tone that can just be heard in air has pressure variations of only 2/10,000,000,000 of one atmosphere of pressure. The corresponding particle displacement is about 10-9 cm. By the way of comparison, the diameter of an atom is about
10-8 cm.
As this pressure disturbance propagates through the medium, the pressure at any point in the medium can be expressed as a function of the distance, r, from the source and time, t, since a particular wave departed the source.
P(r,t) = P(r)sin | 2 (r - ct) | (8-2)
| |
Note how the maximum amplitude, P(r), is dependent on the distance from the source. Neglecting wave interference effects, the amplitude of a pressure disturbance will always diminish, and equation 8-2 therefore represents a decreasing amplitude sinusoid wave. If, however, the distance is fixed, then pressure is solely a function of time, and equation 8-2 simplifies to
P(t) = PAsin [2(ft)] (8-3)
where PA is now the maximum amplitude at the range r.
8.2.2 Intensity
A propagating sound wave carries mechanical energy with it in the form of kinetic energy of the particles in motion plus the potential energy of the stresses set up in the elastic medium. Because the wave is propagating, a certain amount of energy per second, or power, is crossing a unit area and this power per unit area, or power density is called the intensity, I, of the wave. The intensity is proportional to the square of the acoustic pressure. Before giving the defining relationship for intensity, however, two variables must be explained. The value of peak pressure, P, as shown in figure 8-1 is not the "effective" or root-mean-square pressure. An analogy exists between acoustic pressure and the voltages measured in AC circuits. Most voltmeters read the rms voltage. The rms value of a sinusoidal voltage is simply the peak voltage divided by the square root of two. For example, the common 115-volt line voltage has a peak value of about 162 volts. In order to obtain the rms pressure, P must likewise be divided by the square root of two.
Pe = Prms = P (8-4)
/2
In this text, the effective or rms pressure as measured by a pressure-sensitive hydrophone will be labeled Pe. The explanation of units for pressure will be fully discussed at a later time.
8.2.3 Characteristic Impedance
The second variable that must be explained is the proportionality factor that equates intensity to effective pressure squared. It consists of two terms multiplied together - fluid density, and the propagation speed of the wave, c. The quantity, c, is called the characteristic impedance; it is that property of a sound medium that is analogous to resistance or impedance in electrical circuit theory, where power equals voltage squared divided by resistance. Additionally, it can be illustrated by a simple example: When two dissimilar mediums, such as air and water, are adjacent to each other, the boundary between the two is called a discontinuity. When sound energy is traveling through one medium and encounters a discontinuity, part of the energy will be transferred across the boundary and part will be reflected back into the original medium. The greater the difference between the characteristic impedances, the greater will be the percentage of energy reflected. (The difference between the c values for air and water in SI units is approximately 1.5 x 106.) Thus, when sound is traveling through water and it reaches the surface, only a small amount is transmitted into the air. Most of the energy is reflected by the air/ocean boundary back into the water. Obviously, it is important to maintain a consistent set of units when comparing characteristic impedances, and care must be exercised when dealing with different sources of acoustic information.
With the concepts of rms pressure and characteristic impedance understood, it is now possible to formulate an expression for acoustic intensity, the average power per unit area normal to the direction of wave propagation.
I = pe2 (pcsea water 1.5 x 105 dyne-sec/cm3) (8-5)
pc
The units of acoustic intensity are normally watts/m2. The importance of equation 8-5 is that it clearly shows the dependence of the power transmitting capacity of a train of acoustic waves on the pressure. If the rms pressure in the water can be measured, then the sound intensity can be determined. One way to do this is by means of a hydrophone, an electrical-acoustic device, much like a microphone, that transforms variations in water pressure into a variable electric voltage. Thus, after appropriate calibration, Pe can be read directly from a voltmeter attached to the output of a hydrophone.
8.3 MEASUREMENT OF ACOUSTIC PARAMETERS
8.3.1 A convenient system is needed in order to measure and discuss acoustic parameters. Pressure is defined as a force per unit area. Although many people are familiar with the British units of pounds per square inch (psi), it has long been the convention in acoustics to use metric units, namely newtons per square meter (N/m2), or dynes per square centimeter (dynes/cm2). Of the two metric units, the dynes/cm2 has been the most commonly used. It has an alternate name, microbar (bar), and is equivalent to approximately 1/1,000,000 of a standard atmosphere. For underwater sounds, a reference pressure of 1 bar was established from which all others were measured. The corresponding reference pressure for airborne sounds was 0.0002 bar, because this was the approximate intensity of a 1,000-Hz tone that was barely audible to human ears. The previously less commonly used N/m2 also has an alternate name, a Pascal (Pa), and the reference standard derived from this was the micropascal (Pa), which is equivalent to 10-6N/m2.
With such a profusion of reference standards and measurement systems, there were ample opportunities for misunderstandings as an operator or planner consulted different sources of acoustic information. In 1971 the Naval Sea Systems Command directed that thereafter all sound pressure levels should be expressed in the Systeme Internationale (SI) units of micropascals. Although all new publications contain the updated standards, older references will not until they are revised. To assist in making conversions until all publications are revised, table (8-1) summarizes some conversion values.
Throughout this text, the acoustic pressure reference standard, Po, is 1 Pa unless otherwise noted.
Table 8-1. Acoustic Reference Conversion Factors
__________________________________________________________________
1 bar = 1 dyne/cm2 = 0.1 N/m2
= 105 Pa = 10-6 atmospheres
1 Pa = 10-6 N/m2 = 10-5 bar
= 10-5 dyne/cm2 = 10-11 atmospheres
8.3.2 Sound Pressure Level
In theoretical investigations of acoustic phenomena, it is often convenient to express sound pressures in newtons/m2 and sound intensities in watts/m2. However, in practical engineering work it is customary to describe these same quantities through the use of logarithmic scales known as sound pressure levels. The reason is related, in part, to the subjective response of the ear. The human ear can hear sounds having pressure disturbances as great as
100,000,000 micropascals and as small as 10 micropascals. A problem is encountered when discussing pressures that vary over so great a range, in that the minimum audible disturbance is one ten-millionth that of the maximum. In underwater acoustics, useful pressures having even greater variations in magnitude are commonly encountered. In order to make the numbers more manageable, both in magnitude and for actual manipulation, logarithms are used rather than the numbers themselves. Suppose two acoustic signals are to be compared, one having a Pe, of 100,000,000 Pa and the other a Pe of 10 Pa. Their ratio would be
P1 = 100,000,000 Pa = 10,000,000 = 107
P2 10 Pa
In underwater acoustics, however, the attribute of primary interest is sound intensity, or power, rather than pressure. As with pressure, acoustic intensities are referenced to some standard intensity, designated Io, and the logarithm of the ratio taken. Intensity level is therefore defined as
IL = 10 log(I/Io) (8-6)
where IL is measured in dB. However, as there is only one intensity reference (10-12 watt/m2 in air) and many pressure references, IL must be able to be expressed in terms of pressure. By inserting equation (8-5) into equation (8-6), a new expression of IL can be obtained, which is based on pressure rather than intensity per sec.
Pe2
IL = 10 log pc (8-7)
Po2
pc
or
IL = 10 log Pe2 = 20 log Pe
Po2 Po
Under the assumption that the reference intensity and the reference pressure are measured in the same acoustic wave, then a new sound level can be defined called sound pressure level.
SPL = 20log Pe (8-8)
Po
Since the voltage outputs of the microphones and hydrophones commonly used in acoustic measurements are proportional to pressure, acoustic pressure is the most readily measured variable in a sound field. For this reason, sound pressure level is more widely used in specifying sound levels, and this is also why only pressure references are used in underwater acoustics. Note that IL and SPL are numerically equivalent.
IL = 10logI = 20logPe = SPL (8-9)
Io Po
where
Io = P2
o
pc
8.3.3 Decibels
SPL has the dimensionless units of decibels (dB). The decibel system was selected by acousticians for a number of logical reasons. First, it is a logarithmic system, which is convenient for dealing with large changes in quantities. It also simplifies computations since multiplication and division are reduced to addition or subtraction, respectively. Second, human senses have an approximate logarithmic response to stimuli such as light, sound, and heat. For example, the human ear perceives about the same change in loudness between 1 and 10 units of pressure as it perceives between 10 and 100 units of pressure. And finally, in the area of underwater acoustics, the primary interest is in ratios of power levels and signal levels rather than absolute numerical values. In the decibel system, the bel is the fundamental division of a logarithmic scale for expressing the ratio of two amounts of power. The number of bels to express such a ratio is the logarithm to the base 10 of the ratio. Acousticians decided the bel was a unit too large for application in their field, and subsequently adopted the decibel (1/10 bel) as their basic logarithmic unit. The conversion factors in table 8-1 can in themselves be cumbersome to use, but when expressed in dB, only addition or subtraction is required. When converting from a pressure referenced to 1 bar to one referenced to 1 Pa, simply add 100dB. When converting from 0.0002 bar to 1 Pa, simply add 26dB. If converting from 1 Pa to the others, merely subtract the appropriate values. Table 8-2 shows some representative values of conversions between different reference levels. Note that the new micropascal reference standard is small enough that negative values of decibels are rarely encountered.
As an aid in interpreting and understanding the decibel scale and its relation to intensity and pressure, it is useful to remember that
a factor 2 in intensity is + 3 dB
a factor of 0.5 in intensity is -3 dB
a factor of 10 in intensity is + 10 dB
a factor of 0.1 in intensity is -10 dB
a factor of 2 in pressure is + 6 dB
a factor of 0.5 in pressure is -6 dB
a factor of 10 in pressure is + 20 dB
a factor of 0.1 in pressure is -20dB
In understanding intensity levels and sound pressure levels, it is important to note that the decibel scale is a ratio of power or energy, no matter what quantities are being ratioed. A problem commonly arising in acoustic calculations is that of obtaining the overall intensity level after the individual intensities within the applicable bandwidths have been calculated. Such a situation is encountered in calculating a term in the sonar equations (to be discussed later) called noise level, which is actually a combination of ambient noise and self-noise. Because we are dealing with decibels, it is not possible to merely add similar intensity levels together and work with their sum. For example, two 30 dB signals combine to give a total intensity level of 33 dB, not 60 dB as might be expected. The reason for this is that, as shown above, doubling the intensity is represented in decibels by a + 3 dB change. The process is more complicated when dealing with levels of unequal value. Figure 8-2 can be used to determine the dB increase above a level of IL1, in terms of the difference IL1 - IL2, to be expected when IL1 and IL2 are combined. This process can be expanded to include any number of intensity levels. However, when dealing with more than two intensities, it is often easier to use anti-logs, to convert each IL back to its intensity units, add all the intensities together, then reconvert to dB levels. Either method may be used and should result in the same numerical value of noise level.
Although standardization has been reached for measuring intensities, such is not the case for other quantities. Ranges are expressed in yards, kilometers, and nautical miles. Depths are given in feet, meters, or fathoms. Sound speed is stated in feet per second or meters per second and ship speed in knots.
Figure 8-2. Nomogram for combining dB levels.
Temperatures are commonly specified in degrees Fahrenheit or degrees Celsius. These diverse units should warn the user to exercise due caution when discussing the various facets of underwater sound to ensure that misunderstandings do not occur. In this text, the SI units of measure will be used wherever possible.
8.4 SPEED OF SOUND IN THE SEA
From physics it will be remembered that when gas is the transmitting medium, the denser the gas, the slower the speed of sound, and yet the speed of sound in water is about four times greater than that in air. Although this seems contradictory, it is not, because there is another more important factor that influences the speed of sound. In truth, the speed of sound is determined primarily by the elasticity of the medium and only secondarily by the density.
8.4.1 Bulk Modulus
Elasticity is defined as that property of a body that causes it to resist deformation and to recover its original shape and size when the deforming forces are removed. Of specific concern is volume elasticity or bulk modulus - that is, the ratio of force per unit area (stress) to the change in volume per unit volume (strain).
Thus,
Bulk Modulus = Stress
Strain
In order to bring about a change in the volume of a liquid, it is necessary to exert a force of much greater magnitude than is required to bring about an equivalent change in the same volume of air. Therefore, the value of bulk modulus is much greater for a liquid than for a gas. This bit of information, however, is meaningless until it is applied in the formula for the speed of sound.
The speed of sound, c, in a fluid is equal to the square root of the ratio of bulk modulus to density. Thus,
c = Bulk Modulus (8-10)
Density
Although seawater is almost a thousand times denser than air, the enormous bulk modulus of water is the more important factor determining sound speed. Of concern, however, are not the differences of the two mediums but the conditions in water that cause changes in sound speed. Contrary to the assumptions made up to this point, the ocean is not a homogeneous medium, and the speed of sound varies from point to point in the ocean. This variation in sound speed is one of the most important characteristics affecting the transmission of sound. The three main environmental factors affecting the speed of sound in the ocean are salinity, pressure, and temperature.
8.4.2 Salinity
Salinity, which on the average ranges from 32 to 38 parts per thousand (ppt), is fairly constant in the open ocean. A change of salinity will cause a small corresponding change in density with a resulting change in bulk modulus, causing variation of sound speed. The greatest variation in salinity in the open ocean exists in the vicinity of "oceanic fronts," which are narrow zones separating water masses of different physical characteristics, usually exhibiting very large horizontal gradients of temperature and salinity (figure 8-3). Even greater variation in salinity can be expected around the mouths of rivers, heavy ice, and in areas of extraordinary rainfall (e.g., the monsoon) where a layer of fresh water overrides a layer of salt water. A change in salinity of one part per thousand will result in a change in sound speed of approximately 1.3 meters per second.
8.4.3 Pressure
Pressure in most circumstances is more important than salinity, but in the sea its change is constant and thus predictable. It also causes a change in bulk modulus and density, and the result is an increase in sound speed of 0.017 m/sec for every meter of depth increase. This slight change, which is important when temperature remains constant, causes a sound beam to bend upward at great depths as will be discussed later.
8.4.4 Temperature
Temperature, the foremost factor affecting sound speed, usually decreases with depth, and this leads to an accompanying decrease in sound speed at the rate of approximately 3 m/sec per degree Celsius. Below a depth of about 1,000 m, however, temperature is
fairly constant, and the predominant factor affecting sound speed becomes pressure. At first glance it would seem that a temperature decrease would increase sound speed due to the increased water density, but not so. As the temperature of a medium decreases, bulk modulus decreases while density increases. Considering these effects in terms of the sound speed formula in equation (8-10), it is clear that a decrease in temperature brings an attendant decrease in sound speed. It also should be noted that temperature differs bulk modulus and density at a variable rate. A change in temperature at one point on the scale, therefore, affects sound speed differently than an equal change at another point on the scale. It should be noted that the effect of temperature is relatively large compared to the other factors. It takes a depth change of about 165 meters to cause the same change in sound speed as a one-degree temperature change. As will be discussed, temperature is therefore the only factor normally measured and evaluated under operational conditions.
8.4.5 Sound Speed Equation
Dealing with these three factors to arrive at values for bulk modulus and density, and thence sound speed, is very cumbersome. To overcome this, numerous empirical relationships have been developed for converting the three factors directly to sound speed. A simplified version of such sound speed equations developed by Wilson in 1960 is present below.
c = 1449 + 4.6T + 0.055T2 + 0.003T3
+ (1.39 - 0.012T)(S - 35) + 0.017d (8-11)
where
T = temperature in degrees Celsius
S = salinity in parts per thousand
d = depth in meters
Given accurate temperature, salinity, and depth data, this equation is accurate within 0.6 meters/sec, 96 percent of the time. By way of contrast, the equation for the speed of sound in air is approximately
c = 331.6 + 0.6T
In making calculations involving the transmission of sound through the sea, it frequently is adequate to use a standard speed rather than the more accurate value given by equation 8-11. Although in seawater c can vary from a low of about 1,420 m/s to over 1,560 m/s depending on environmental conditions, a standard speed of 1,500 m/s may be assumed for computation purposes unless otherwise noted.
8.4.6 Field Observations of Sound Speed
Knowledge of sound velocity is important to the ASW tactician and physical oceanographer because of the effect that variations in sound velocity have upon acoustic absorption and refraction. Two different devices are in use today for finding the speed of sound in the sea.
8.4.6.1 Bathythermograph. The first device is called a bathy-thermograph. As previously stated, temperature is the predominant ocean variable affecting sound speed. Not only is it relatively easy to measure, but when applied to empirical relationships such as equation (8-11), sound speed can be computed. Older BT systems employed a mechanical device that was lowered on a cable and the temperature was scribed on a smoked piece of glass. This had a number of inherent disadvantages that have been overcome through the development of the expendable bathythermograph (XBT), which does not require retrieval of the sensing unit. A diagram view of an XBT is shown in figure 8-4. It consists of a thermistor probe that is ejected from the launching platform and sinks at a known non-linear rate. The XBT is connected to a special recorder on board the launching platform by a fine wire. As it sinks, ther thermistor changes its electrical resistance with changing temperature, and as a result a temperature vs. depth trace is obtained. Because the wire uncoils from both the probe and its launcher, there is no tension on the wire until the probe has sunk to its full depth. At this point, the wire snaps and the recording stops. Variants of the basic XBT have been developed for use aboard submarines and from aircraft through incorporation into sonobuoys.
When the XBT temperature vs. depth trace is converted to sound speed vs. depth, it produces a sound speed profile very similar to that obtainable from a sound velocimeter, and is of sufficient accuracy for operational requirements.
8.4.6.2 Sound Velocimeter
The second and most accurate method is the sound velocimeter. Its principle advantage is that it can measure sound speed directly, without need for conversions, by transmitting a pulse of sound over a very short path on the order of 1/2 meter or less. When the pulse arrives at the receiver, another pulse is then triggered from the transmitter; this is known as the "sing-around" or "howler" principle. The faster the speed of sound in the water in which the velocimeter is submerged, the faster the pulse will travel and the sooner it will arrive at the receiver to trigger the succeeding pulse. Since nearly all the time delay between pulses occurs as acoustic delay in the water, the PRF of the pulses is determined by the local sound speed and is directly proportional to it. Thus, knowing the path length and observing the PRF can lead directly to computation of sound speed. Until recently sound velocimeters were expensive and awkward to use, thus eliminating their use tactical-ly. The recent development of the expendable sound velocimeter (XSV) has made it possible to reduce sound velocity measurement er-rors to less than .25 meters per second at reasonable expense with-out reduction of the mobility of combatant units. Today's sophis-ticated sonars and acoustic navigation systems can provide improved information in many oceanic regions when actual sound-velocity pro-files are used rather than extrapolated sound velocity values based on temperature profiles and assumed salinity data. Based on the variability in the sea with time, a policy of regular periodic mea-surement of the velocity profile is required during an ASW opera-tion. Normally, one or two ships in the force are assigned the bathythermograph guard duty. These ships periodically measure the temperature or velocity profile, and promulgate it to all ASW units in company. These sound velocity profiles are essential in deter-mining the sound propagation paths available.
8.4.7 Typical Sound Speed Profiles
It is important to remember that while temperature is the dominant factor, the sound-speed profile is really a composite of the pressure, salinity, and temperature profiles as shown in
figure 8-6. In the area of ocean fronts, where salinity may vary up to 3 ppt. from assumed values, the use of temperature data alone may result in an error of up to 4.2 meters per second in the calculation of sound speed.
Figure 8-6. Graphical relationship of sound speed
to pressure, salinity, and temperature
A typical composite deep-sea sound-speed profile is shown in greater detail in figure 8-7. The profile may be divided into four major layers each having different thermal characteristics. Just below the sea surface is the surface layer, in which the speed of sound is susceptible to daily and local changes of heating, cooling, and wind action. The surface layer may contain isothermal water that is formed due to mixing by the action of wind as it blows across the water. Below the surface layer lies the seasonal thermocline - the word "thermocline" denoting a layer in which the temperature changes rapidly with depth. The seasonal thermocline is characterized by a negative sound-speed gradient that varies with the seasons. During the summer and fall, when the near-surface waters of the sea are warm, the seasonal thermocline is strong and well defined; during the winter and spring, and in the Arctic, it tends to merge with, and be indistinguishable from, the surface layer. Underlying the seasonal thermocline is the permanent thermocline, which is affected only slightly by seasonal changes. Below the permanent thermocline and extending to the sea bottom is the deep isothermal layer, having a nearly constant temperature of about 4oC, in which the speed of sound has a
positive gradient because of the effect of pressure on sound speed.
Between the negative speed gradient of the permanent thermocline and the positive gradient of the deep isothermal layer, there is a speed minimum toward which sound traveling at great depths tends to be bent or focused by refraction. This is the deep sound channel and will be discussed later. The refraction of sound, however, is much more complex than this simple four-layer ocean model would indicate. There are ocean eddies, fronts, interfaces between currents, underwater mountains and ridges. For instance, a submarine detected in the Labrador Current but crossing into the Gulf Stream has been compared to a person going out of an open field and disappearing into the nearby woods.
8.5 RAY THEORY
The propagation of sound in an elastic medium can be described mathematically by solutions of the wave equation using the ap-propriate boundary and medium conditions for a particular problem. The wave equation is a partial differential equation relating the acoustic pressure P to the coordinate x, y, z, and the time t, and may be written as
2P = c2 (2P + 2P + 2P) (8-12)
t2 (x2 y2 z2)
8.5.1 Normal-Mode Theory
There are two theoretical approaches to a solution of the wave equation. One is called normal-mode theory, in which the prop-agation is described in terms of characteristic functions called normal modes, each of which is a solution of the equation. The normal modes are combined additively to satisfy the boundary and source conditions of interest. The result is a complicated math-ematical function which, though adequate for computations on a digital computer, gives little insight, compared to ray theory, on the distribution of the energy of the source in space and time. Normal-mode theory is well suited for a description of sound prop-agation in shallow water, but will not be discussed in this text.
8.5.2 Ray Acoustics
The other form of solution of the wave equation is ray theory, and the body of results and conclusions therefrom is called ray acous-tics. The essence of ray theory is (1) the postulate of wave
fronts, along which the phase or time function of the solution is constant, and (2) the existence of rays that describe where in space the sound emanating from the source is being sent. Like its analog in optics, ray acoustics has considerable intuitive appeal and presents a picture of the propagation in the form of the ray diagram.
For almost all operational problems, the sound-speed gradient, with respect to horizontal changes of location, can be assumed to be zero. The major gradient of interest is the vertical gradient, dc/dz, where dz is the amount of depth change. If a source of sound at the surface of the sea radiates omnidirectionally, a wave front expanding from this source in all directions transfers energy from one particle in the water to another, and by this means the wave is propagated. If some point on this wave front is selected, and from it a line is drawn in the direction of energy propagation, then connecting these points as the wave expands in space will result in a line called a ray, as illustrated in figure 8-8.
A sound wave, or ray, which enters another medium or layer of the same medium having a different characteristic impedance, will undergo an abrupt change in direction and speed. Depending upon the angle of incidence and the abruptness of change in c, a portion of the impinging acoustic energy will be reflected off the medium boundary, and a portion will be refracted or bent passing
through the boundary. A sound ray will always bend toward the region of slower sound speed.
One of the most important practical results of ray theory is Snell's Law, which describes the refraction of waves in mediums of variable speeds. Snell's Law states that the angle of incidence, 1, at a boundary is related to the angle of refraction 2, by the following expression:
sin 1 = c1 (8-13)
sin 2 c2
where
c1 = sound speed in medium 1
c2 = sound speed in medium 2
If the wave is considered to be passing through three horizontal layers or strata, in each of which the sound speed is considered to be constant, then Snell's Law can be rewritten as
c1 = c2 = c3 = cn (8-14)
cos 1 cos 2 cos 3 cos n
where
cn = speed of sound at any point in the medium
n = angle made with horizontal at that point
Note that the angle in equation 8-14 is the complement of the angle usually expressed in Snell's basic law. It is commonly referred to as the grazing angle or angle of inclination. This expression is the basis of ray computation used by most computers, since it enables a particular ray to be "traced out" by following it through the successive layers into which the speed profile may have been divided. In a layered medium having layers of constant speed, the rays consist of a series of straight-line segments joined together, in effect, by Snell's Law.
In practice, however, temperature does not change abruptly, but rather the gradient will normally decrease or increase at a measurable rate. For such a situation, the sound speed at any depth z would be given by
c(z) = co + gz (8-15)
where
co = speed at the surface or transducer depth
g = speed gradient dc/dz between the surface and depth z
The net result is that, in reality, ray traces appear as curves rather than straight lines. By combining equations (8-14) and (8-15), an expression can be developed for the radius of curvature R of any ray at any point along the ray path, as shown by equation (8-16) and figure 8-11.
R = -co = c (8-16)
g gcos
Under operational conditions, values of R are very large, approaching several tens of kilometers.
8.6 PROPAGATION PATHS
8.6.1 Thermal Structure
The thermal structure of the ocean governs the refractive con-ditions for a given water mass. Despite infinite vertical temperature variations in the ocean, the temperature structure normally can be related to three basic types: (1) isothermal,
(2) negative temperature gradient, and (3) positive temperature gradient. In discussing sound propagation, it is customary to use the temperature profile as an indicator of sound speed conditions at various depths, because it has the greatest effect. It must be remembered, however, that changes in sound-beam direction result from changes in the sound-speed profile, which is influenced not only by temperature but pressure and salinity as well.
8.6.2 Direct Path
In an isothermal condition, the water's temperature is almost constant. If there is a slight decrease in temperature, and it is just balanced out by the pressure increase, the result is an iso-sound-speed condition. This causes a straight-line ray, leaving the source in lines that continue with little or no change in angle. Long ranges are possible when this type of structure is present.
When there is a negative temperature gradient, sound speed decreases with depth, and sound rays bend sharply downward. This condition is common near the surface of the sea. At some horizon-tal distance from the sound source, beyond where the rays bend downward, is a region in which sound intensity is negligible (fig-ure 8-22); it is called a shadow zone. The magnitude of the tem-perature gradient determines the amount of bending of the sound beam and thus the range of the shadow zone. For example, if the decrease in temperature to a depth of 10 meters totals 2oC or more, the shadow zone would begin beyond a horizontal range of 1,000 met-ers due to the sharp curvature of the sound beam.
When the temperature of the water has a positive gradient, sound speed increases with depth, and sound rays are refracted upward.
Longer ranges are attained with this temperature structure than with a negative gradient because the rays are refracted upward and then reflect off the surface. Unless the surface of the sea is very rough, most of the rays are repeatedly reflected at the surface to longer ranges.
Circumstances usually produce conditions where combinations of temperatures occur. One of these combinations includes a layer of isothermal water over water with a negative gradient. Approximately 90 per cent of all the bathythermograph records from all over the world show this type of thermal structure. One ray, labeled "the critical ray," becomes horizontal at the boundary or division between the isothermal layer and the negative gradient. The speed of sound is a maximum at this boundary point.
Consequently, we define the layer depth (z) as that depth of greatest sound speed (c) above the seasonal thermocline (see figure 8-7). One half of the critical beam bends toward the upper region at a reduced speed, and the other half bends toward the lower region at a reduced speed. The angle that the critical ray makes with the horizontal at the point of projection is called the critical angle.
All rays in the sound beam directed at an angle less than the critical angle will follow paths entirely within the isothermal layer and will be bent upward to the surface. All rays directed at an angle greater than the critical angle follow paths that penetrate the boundary and are subsequently refracted downward. No rays enter the region bounded by the two branches of the split critical ray, and for this reason it is also called a shadow zone. Sharp shadow zones are not fully developed because of diffraction and other effects, though the sound intensity in this area is quite low. Submarine commanders deliberately use this phenomenon, when it exists, to attempt to escape detection when approaching a target. The optimum depth for close approach to a target with minimum probability of counter-detection is approximately
Best depth = 17 Z (8-33)
where z is the layer depth in meters.
This is accurate down to a layer depth of 60 meters. Below that, the best depth for approach is a constant 60 meters below layer depth.
8.6.3 Convergence Zone
In the deep ocean, temperature usually decreases with depth to approximately 1,000 meters. Deeper than this, temperature is a constant 4oC and sound speed increases as a result of pressure. A negative speed gradient overlays a positive speed gradient, allowing equal speeds at two different depths with slower speed conditions in between. Sound originating in the thermocline, traveling nearly parallel to the surface initially, thus bends toward greater depths. But as the detected sound penetrates into deep water, it passes the region of minimum sound speed and enters the deep isothermal layer. Now the gradient in sound speed operates in the other direction; the sound path bends upward rather than downward, and the sound returns to the surface. This produces a convergence zone, where the sound waves concentrate as if they had been focused. It typically lies at a distance of about fifty kilometers from the source. Beyond this convergence zone is a second zone of silence, where again the acoustic waves diffract downward; then another convergence zone, fifty kilometers out, and so forth. The mapping of these zones is a routine part of submarine operations; by measuring deep-water temperatures witha bathythermograph, one obtains data that readily allow a computer to calculate the appropriate sound-wave paths.
When water with a negative speed gradient overlays a positive speed gradient, a sound channel is produced. Under these circumstances, any sound signal traveling in this area is refracted back and forth so that it becomes horizontally channeled. Sound rays originating with an initial upward inclination are refracted upward. Rays from a sound source in this layer that make a small angle with the horizontal are roughly sinusoidal, crossing and
recrossing the layer of minimum speed. This reinforcement of rays within the sound channel can continue until the sound is absorbed, scattered, or intercepted by some obstacle. Sounds traveling in this manner sometimes are received at extremely great distances from the source. These long ranges occur primarily as a result of two factors: absorption is small for low-frequency sound and most of the sound energy from a sound source at the axis is confined to the channel.
Under certain circumstances, a sound channel can exist near the surface of the sea. In a surface layer with a strong positive
temperature gradient the upward bending of sound rays combined with reflections from the surface will form such a channel. Sonar ranges many times greater than normal have been observed where sound channels exist. However, the conditions that produce such sound channels near the surface are rare and not very stable.
The region of minimum sound velocity, at a depth exceeding a kilometer, is the deep sound channel. It acts as an acoustic waveguide; sound propagating either upward or downward encounters sound-velocity gradients and bends back into this channel. Within the deep sound channel, sound undergoes only a cylindrical spreading loss, in which intensity drops off only as the first power of the distance. If sound spread uniformly in all directions, known as spherical spreading, this falloff would follow the square of the distance. The concept of cylindrical and spherical spreading and their importance to sonar is presented in the next section. Figure (8-18) shows the ray paths existing for a deep sound channel. Rays A and B are bounded by the sound channel. Other rays may follow propagation paths that are reflected at the sea surface and bottom (ray C), or refracted and reflected from either the sea bottom or the sea surface (rays D and E). Ray D is commonly called the refracted surface reflected (RSR) path, while ray E is called the refracted bottom reflected (RBR) path.
Figure 8-18. Sound Velocity Profile and Typical Ray
Path Diagram for a Sound Source (or Receiver) on the Axis
of the Deep Sound Channel
8.6.4 Bottom Bounce
In addition to being refracted by varying conditions in the medium, sound can be reflected in the manner of a light beam striking a mirrorlike surface and lose little of its intensity. The two surfaces that can produce this type of reflection are the surface of the water and the bottom of the ocean. Rarely if ever are these surfaces smooth enough to give a mirror reflection, but in many instances the majority of the sound is reflected as a beam. Some sonars make use of this phenomenon which is called bottom bounce. A beam is directed against the bottom from which it is reflected to the surface of the water. From the surface it is reflected back to the bottom again. Thus, the ray bounces from one to the other until its energy is dissipated or until it strikes a target and returns to the sonar. As with reflected light, the angle of reflection is equal to the angle of incidence. Obviously, ranging over a fairly flat bottom is more effective than ranging over a rough or sloping bottom. This path is highly dependent upon depth and absorption of sound by the ocean bottom.
It is obvious, then, that many sound paths are available, as indicated in the composite drawing in figure 8-20. For simplicity in the figure, the bottom-bounce path is depicted as a single beam, whereas in the real world, a wider beam corresponding to a multidirectional sound source would be the normal case.
Figure 8-20 Three Typical Sound Paths Between
a Source and a Receiver in Deep Water in Mid-North Atlantic Ocean
and Nominal Values of One-Way Propagation Loss
The convergence zone path is shown as having depth based on the concept of a wide sound beam or "bundle" of sound rays emanating from a source. By definition, the convergence zone is formed when the upper ray of the bundle becomes horizontal. This is the depth at which the sound velocity is equal to the high of the velocities at either the surface or at the bottom of the surface layer. The difference in depth between the ocean bottom and the depth at which the upper ray becomes horizontal is called the depth excess. The depth excess available defines the depth of the bundle of rays or "amount of sound" that forms the convergence zone path. As the sound energy travels upward and approaches the surface, the path narrows, tending to focus the sound energy resulting in convergence gains. A maximum of 1200 ft depth excess is required to produce an operational useful convergence zone with reliable convergence gains. (As velocity increases with depth, the corresponding amount by which the velocity of the bottom ray exceeds that of the top ray in the convergence zone path bundle is called velocity excess.) These concepts are depicted in Figure (8-21)
Although many paths are available, only by observing the environment carefully and paying close attention to his equipment will the operator be able to use them to best advantage and not find them a liability.
Figure 8.21 Depth Excess and Velocity Excess Concepts
for the Convergence Zone Sound Path
8.7 SOUND PROPAGATION THROUGH THE SEA
The sea, together with its boundaries, forms a remarkably complex medium for the propagation of sound. It possesses an internal structure and a peculiar upper and lower surface that create many diverse effects upon the sound emitted from an underwater source. In traveling through the sea, an underwater sound signal becomes delayed, distorted, and weakened. The transmission loss term in the sonar equations expresses the magnitude of these effects.
8.7.1 Transmission Loss
Consider a source of sound located in the sea. The intensity of the sound can be measured at any point in the sea, near to or far from the source. For purposes of measuring intensity at the source, the intensity measurement is generally taken at one unit distance from the source and labeled Io. The intensity can then be measured at any distant point where a hydrophone is located and denoted I. It is operationally significant to compare the two values. One way to do this is to form the ration Io/I. Note that if the ratio, denoted n, is greater than unity, the intensity at the source is greater than at the receiver, as would be expected. If n = Io/I, then
10 log n = 10 log Io - 10 log I
= sound intensity level at the source minus sound
The value 10 log n is called the transmission loss and, of course, is measured in decibels. Most of the factors that influence transmission loss have been accounted for by scientific research, and can be grouped into two major categories: spreading and attenuation.
To understand spreading loss, it is convenient to imagine a theoretical ocean that has no boundaries and in which every point has the same physical properties as any other point - i.e., an infinite, homogeneous medium. In such a medium, sound energy would propagate from a point source in all directions along straight paths and would have a spherical wave front.
Under these conditions the change in power density with distance from the point source would be due only to the spherical divergence of energy. Note that there is no true loss of energy as might be implied, but rather the energy is simply spread over a progressively larger surface area, thus reducing its density. For this model, the amount of power spread over the surface of a sphere of radius r, centered at a point source, is expressed by
Pt(watts) (8-17)
Power density (watts/m2) at r = 4r2
where P, is the acoustic power level immediately adjacent to the source. This concept of power density (watts/m2) was used to develop the radar equation in chapter 2. As stated before, the units of acoustic energy are watts/m2. Therefore, equation 8-17 can be written
It = Pt (watts/m2 = P2e
4r2 pc
where Pe, is measured at range r, and Pt is the acoustic power level immediately adjacent to the source.
The intensity of the sound immediately adjacent to the source is measured, by convention, at a unit distance (1 meter) from the source. It will be labeled I1 and is given by
I1 = Pt
4(1)2
The acoustic intensity (Ir) at some distance, r, from the source will be less than the acoustic intensity (I1) at 1 meter from the source. This is the result of spreading a fixed amount of power over a geometrically increasing surface area (a sphere, figure 8-22). The reduction of the acoustic intensity as a function of distance, r, is shown in the ratio
Pt
Ir = 4(r)2 = 1
I1 Pt r2
4(1)2
However, the ratio Ir to I1 at any significant range is typically so small that these values are best examined on a logarithmic scale using the decibel.
10 log Ir = 10 log 1 = 10 log(1) - 10 log r2 = -20 log r
I1 r2
or
10 log Ir = 10 log I1 - 20 log r
The reduction in acoustic intensity (I1) due to range (spreading)
is called the transmission loss (TL), and for spherical spreading
TL = 20 log r (r in meters) (8-18)
and
10 log Ir = 10 log I1 - TL
The ocean is not an unbounded medium, however, and all sources are not omnidirectional point sources. For sources that radiate energy only in a horizontal direction, sound energy diverges more like the surface of an expanding cylinder. Also, since the ocean is bounded at the surface and bottom, cylindrical divergence is usually assumed for ranges that are large compared to the depth of the water or for when sound energy is trapped within a thermal layer or sound channel. For this model, the acoustic intensity of energy at the surface of a cylinder of radius r is expressed by
Ir = Pt watts/m2 (8-19)
2rh
where h is the vertical distance between upper and lower boundaries
(see figure 8-22). The transmission loss is therefore
10 log Ir = 10 log 1 = 10 log (1) - 10 log r = - 10 log r
I1 r
or
l0 log Ir = 10 log I1 - 10 log r
or
TL = 10 log r (8-20)
and
10 log Ir = 10 log I1 - TL
Equation (8-20) represents cylindrical divergence of sound energy, sometimes referred to as inverse first power spreading. Except at short ranges, it is the most commonly encountered type of spreading loss. It should be noted that the loss of intensity of a sound wave due to spreading is a geometrical phenomenon and is independent of frequency. As range increases, the percentage of intensity lost for a given distance traveled becomes increasingly less.
8.7.3 Attenuation Loss
Attenuation of sound energy in seawater arises principally through the action of two independent factors, absorption and scattering, with an additional contribution from bottom loss.
8.7.3.1 Absorption. The primary causes of absorption have been attributed to several processes, including viscosity, thermal con-ductivity, and chemical reactions involving ions in the seawater. (1) The viscosity of the medium causes sound energy to be convert-ed into heat by internal friction. (2) Some sound energy is con-verted into heat because sound waves alternately raise and lower the temperatures. (3) Suspended particles are set to oscillating by the sound waves and in this process some of the sound energy is dissipated in the form of heat. This is especially the case if the particles are air bubbles. While each of these factors offers its own unique contribution to the total absorption loss, all of them are caused by the repeated pressure fluctuations in the medium as the sound waves are propagated. They involve a process of conver-sion of acoustic energy into heat and thereby represent a true loss of acoustic energy to the environment.
Experimentation has produced a factor , called the absorption coefficient, which when multiplied by the range gives the total loss in dB due to absorption. Water temperature and the amount of magnesium sulphate (MgSO4) are important factors influencing the magnitude of , because the colder the average water temperature and the greater the amount of MgSO4 present, the greater will be the losses due to absorption. However, it is the frequency of the sound wave that causes the most significant variation in the absorption coefficient. While the formula will change slightly with temperature and geographical location, an equation for the value of in decibels per meter for seawater at 5oC is
= 0.036f2 + 3.2 x 10-7f2 (8-21)
f2 + 3600
where
f = frequency in kHz
While this formula is rather cumbersome, the important thing to observe is that increases roughly as the square of the frequency. This relationship is of major importance to the naval tactician. It tells him that if higher frequencies are chosen for sonar operation in order to achieve greater target definition, the price he must pay is greater attenuation. The higher the frequency, the greater the attenuation and the less the range of detection. For this reason, where long-range operation of sonar equipment is desired, the lower the frequency used the better. Figure 8-23 depicts typical values of the absorption coefficient of seawater at 5oC for varying frequencies.
To obtain the transmission loss due to absorption, is merely multiplied by the range in meters. Thus,
TL = r (8-22)
8.7.3.2 Scattering
Another form of attenuation is scattering, which results when sound strikes foreign bodies in the water, and the sound energy is re-flected. Some reflectors are boundaries (surface, bottom, and shores), bubbles, suspended solid and organic particles, marine
life, and minor inhomogeneities in the thermal structure of the ocean. The amount of energy scattered is a function of the size, density, and concentration of foreign bodies present in the sound path, as well as the frequency of the sound wave. The larger the area of the reflector compared to the sound wavelength, the more effective it is as a scatterer. Part of the reflected sound is re-turned to the source as an echo, i.e, is backscattered, and the re-mainder is reflected off in another direction and is lost energy. Back-scattered energy is known as reverberation and is divided into three types: volume, surface and bottom.
Volume reverberation is caused by various reflectors, but fish and other marine organisms are the major contributors. Additional causes are suspended solids, bubbles, and water masses of markedly different temperatures. Volume reverberation is always present during active sonar operations, but is not normally a serious factor in masking target echoes. The one exception involves the deep scattering layer (DSL), which is a relatively dense layer of marine life present in most areas of the ocean. During daylight hours, the layer is generally located at depths of about 600 meters and does not pose a serious problem. At night, however, the layer migrates toward the surface and becomes a major source of reverberation. It is rarely opaque to sound when detected with a sonar looking down on it from directly above, as with a fathometer, but this is not the case with a search sonar transmitting in a more or less horizontal direction. By pinging horizontally, the sound waves encounter many more organisms, and the effect can vary from partial transmission of sound to total reflection and scattering, thereby hiding a submarine.
Surface reverberation is generated when transmitted sound rays strike the surface of the ocean, i.e., the underside of the waves. It is always a factor in active sonar operations, and is directly related to wind speed because it controls wave size and the angle of incidence.
Bottom reverberation occurs whenever a sound pulse strikes the ocean bottom. In deep water this condition normally does not cause serious problems, but in shallow water, bottom reverberation can dominate the background and completely mask a close target. The amount of energy lost through scattering will vary with the roughness of the bottom and the frequency of the incident sound.
Sound reflected from the ocean floor usually suffers a significant loss in intensity. Part of this loss is caused by the scattering effects just described, but most of it results from the fact that a portion of sound energy will enter the bottom and travel within it as a new wave, as illustrated in figure 8-24. The net result is that the strength of the reflected wave is greatly reduced. The amount of energy lost into the bottom varies with the bottom composition, sound frequency, and the striking angle of the sound wave. The total of these losses can vary from as low as 2 dB/bounce to greater than 30 dB/bounce. In general, bottom loss will tend to increase with frequency and with the angle of incidence. Soft bottoms such as mud are usually associated with high bottom losses (10 to 30 dB/bounce); hard bottoms such as smooth rock or sand produce lower losses.
While it is possible to derive equations that will compute precise values of TL, associated with each of these additional scattering and bottom loss factors, the ocean characteristics are so variable that there is little utility in doing so. It is customary, therefore, in operational situations, to make an educated guess as to their values and lump them together into one term "A," known as the transmission loss anomaly, which is included in the transmission loss equation.
8.7.4 Total Propagation Loss
It would be useful to have a simple mathematical relationship that would describe all the effects of the various factors influencing transmission loss as they occur in the ocean. But the state of the physical conditions encountered in the ocean are very complex and not at all easy to represent. A few mathematical models do exist that provide close approximations for some sets of conditions, but at present, no single model accounts for all the conditions encountered.
A simplified model used to obtain approximate values of transmission loss for the spherical spreading case is
TL = 20 log r + r + A (8-23)
and for the cylindrical spreading case
TL = 10 log r + r + A (8-24)
It is important to realize that sound transmission in the ocean is three-dimensional and that transmission loss versus horizontal range alone is not sufficient information for most operational situations. Areas of no sonar coverage occur at various intervals of range because of refraction, reflection, and interference between waves traveling via different paths. Therefore, while the TL equations are interesting and somewhat useful, they are not always totally accurate.
8.8 SOUND SOURCES AND NOISE
Background noise, like reverberation, interferes with the reception of desired echoes. Unlike reverberation, however, it does not result from unwanted echoes of the transmitted pulse but from active noise-makers located in the ship or in the water.
Noise produced by these sources is classified as self-noise and ambient noise. Self-noise is associated with the electronic and mechanical operation of the sonar and the ship. Ambient noise encompasses all of the noises in the sea.
8.8.1 Self-Noise
Self-noise is produced by noisy tubes and components in the sonar circuitry, water turbulence around the housing of the transducer, loose structural parts of the hull, machinery, cavitation, and hydrodynamic noises caused by the motion of the ship through the water.
8.8.1.1 Machinery Noise. The dominant source of machinery noise in a ship is its power plant and the power distribution system that supplies power to the other machinery on the vehicle, such as compressors, generators, propellers, etc. Machinery noise is normally always present and is kept to a minimum by acoustically isolating the various moving mechanical components.
The gearing connecting the propellers is an important source of machinery noise. If the engine runs at a relatively low speed, as is the case with a reciprocating heat engine, gears may not be required between the engine and the propeller. High-speed power sources, however, such as steam or gas turbines, usually require reduction gears to the propeller. The frequency of the explosions in the cylinders of a reciprocating engine is not likely to be a source of ultrasonic (above 15 kHz) noise but might be an important source of low-frequency sonic noise. A more crucial source of noise in the reciprocating engine is the clatter of the valves opening and closing. In gas turbines, the noise generated is in the ultrasonic region at a radian frequency equal to the angular velocity of the turbine buckets.
Noise in the ultrasonic region is extremely important in sonar and acoustic torpedo performance. The noise produced by auxiliary units, such as pumps, generators, servos, and even relays, is often more significant than the power-plant noise. The large masses involved in the power plant usually keep noise frequencies relatively low. For this reason, a small relay may interfere more with the operation of a torpedo than an electric motor that generates several horsepower. Small, high-speed servomotors, however, may be serious sources of ultrasonic noise.
8.8.1.2 Flow Noise. Flow noise results when there is relative motion between an object and the water around it. This flow is easiest to understand by assuming that the object is stationary and that the water is moving past it. Under ideal conditions such that the object is perfectly streamlined and smooth, water movement will be even and regular from the surface outward as shown by the flow lines in figure 8-25A. This idealized condition is called laminar flow and produces no self-noise. Irregular objects can achieve nearly laminar flow conditions only at very low speeds (i.e., 1 or 2 knots or below).
As flow speed increases, friction between the object and the water increases, resulting in turbulence (figure 8-25B) and progressively increasing noise due to fluctuating static pressure in the water. Thus we have, in effect, a noise field. If a hydrophone is placed in such a region, fluctuations of pressure will occur on its face, and it will measure the result in flow noise in the system.
As pressures fluctuate violently at any one point within the eddy, they also fluctuate violently from point to point inside the eddy. Moreover, at any given instant the average pressure of the eddy as a whole differs but slightly from the static pressure. Thus, very little noise is radiated outside the area of turbulence, and although a ship-mounted hydrophone may be in an intense flow-noise field, another hydrophone at some distance from the ship may be unable to detect the noise at all. Flow noise, then, is almost exclusiverly a self-noise problem.
Actually, not much information is known about flow noise, but these general statements may be made about its effect on a shipborne sonar:
1. It is a function of speed with a sharp threshold. At very low speeds there is no observable flow noise. A slight increase in speed changes the flow pattern from laminar to turbulent, and strong flow noise is observed immediately. Further increases in speed step up the intensity of the noise.
2. It is essentially a low-frequency noise.
3. It has very high levels within the area of turbulence, but low levels in the radiated field. In general, the noise field is strongest at the surface of the moving body, decreasing rapidly with distance from the surface.
4. The amount of flow noise can be related to the degree of marine fouling on the ship's bottom and sonar dome. Fouling is the attachment and growth of marine animals and plants upon submerged objects. More than 2,000 species have been recorded, but only about 50 to 100 species are the major troublemakers. These nuis-ances can be divided into two groups - those with shells and those without. Particularly important of the shell forms are tube-worms, barnacles, and mollusks because they attach firmly and resist being torn loose by water action. Nonshelled forms such as algae, hy-droids, and tunicates commonly attach to relatively stationary ob-jects and therefore do not cuase much trouble on ships that fre-quently get underway.
Most fouling organisms reproduce by means of the fertilization of eggs by sperm, and the resulting larvae drift with the currents. Depending on the species, the larval stage may vary from a few hours to several weeks. Unless the larvae attach themselves to a suitable surface during this period, they will not grow into adulthood.
Geography governs fouling by isolating species with natural barriers. Temperature is the most important factor governing distribution of individual species, limiting reproduction or killing adults.
On a local scale, salinity, pollution, light, and water movement affect the composition and development of fouling communities. Most fouling organisms are sensitive to variations in salinity, developing best at a normal seawater concentrations of 30 to 35 parts per thousand. Pollution, depending on type, may promote or inhibit fouling. Fouling algae and microscopic plants, the food supply of many fouling animals, are dependent upon sufficient light for proper growth.
To forestall fouling on ship bottoms and sonar domes, the Navy uses special paints with antifouling ingredients. The most common active ingredient of antifouling paint is cuprous oxide, and the effective life of the paint is from 2 to 2 1/2 years.
8.8.1.3 Cavitation. As the speed of the ship or object is increased, the local pressure drops low enough at some points behind the object to allow the formation of steam. This decrease in pressure and the resulting bubbles of vapor represent the onset of cavitation. As the ship moves away from the bubbles, however, the pressure increases, causing the bubbles to collapse and produce a continuous, sharp, hissing noise signal that is very audible.
Because the onset of cavitation is related to the speed of the object, it is logical that cavitation first appears at the tips of the propeller blades, inasmuch as the speed of the blade tips is considerably greater than the propeller hub. This phenomenon, known as blade-tip cavitation, is illustrated in figure 8-26.
As the propeller speed increases, a greater portion of the propeller's surface is moving fast enough to cause cavitation, and the cavitating area begins to move down the trailing edge of the blade. As the speed increases further, the entire back face of the blade commences cavitating, producing what is known as sheet cavitation, as shown in figure 8-27.
The amplitude and frequency of cavitation noise are affected considerably by changing speeds, and changing depth in the case of a submarine. As speed increases, so does cavitation noise. As depth increases, the cavitation noise decreases and moves to the higher frequency end of the spectrum in much the same manner as though speed had been decreased. This decrease in noise is caused by the increased pressure with depth, which inhibits the formation of cavitation bubbles.
A given propeller at a given depth of submergence will produce no bubbles unless its speed exceeds a certain critical value. When the speed (RPM) exceeds this critical value the number of bubbles formed increases rapidly, but not according to any known law. The critical speed itself, however, depends in a simple manner on the depth of the propeller beneath the sea surface. This dependence is given by
No2 = constant No - critical speed
d d - depth
Thus, if a given propeller begins to cavitate at 50 rpm when at a depth of 15 ft, it will begin to cavitate at 100 rpm when at a depth of 60 ft.
Since all torpedo homing systems and many sonar systems operate in the ultrasonic region, cavitation noise is a serious problem. Torpedoes generally home on the cavitation noise produced by ships, and any cavitation noise produced by the torpedo interferes with the target noise it receives. Because the speed at which a vehicle can operate without cavitating increases as the ambient pressure is increased, some acoustic torpedoes are designed to search and attack from depths known to be below the cavitating depth. In the same way, a submarine commander, when under attack, will attempt to dive to depths at which he can attain a relatively high speed without producing cavitation.
8.8.1.4 Quieting
It is important for an ASW vessel to operate with a very low noise level. Noise reduction methods can be applied to other military vessels, and to that matter to commercial vessels. For protection against acoustic mines and to reduce the risk of being detected early by sonar listening devices, noise minimizing is necessary. A low level of noise also permits own ship to receive a lower level of signal from the enemy. Strict levels on the limits of structureborne noise and of airborne noise levels improve the habitability for the ship's crew.
8.8.1.4.1 Surface Quieting. Considerations must be given to noise emission planning in design of the ship. Attention must be paid to the layout of the propellers and the hull's appendages with regard to waterborne noise generation. Model tests on propellers are necessary to check cavitation and to shift the cavitation point to a relatively high speed. Above this speed a propeller air emission system can be used to reduce cavitation-generated waterborne noise. In the U.S. this bubble injection system is called Prairie Masker. Prairie is an acronym for propeller air-induced emission. A system on some U.S. vessels for injecting air through hull girths is called Masker. This reduces somewhat the flow noise. See figure 8-28.
Other measures can be taken to reduce noise:
- Reducation in structure borne noise transmission through the provision of low tuned flexible mounts for critical items combined with high impedance foundations.
- Encapsulation of all diesel prime movers, compressors and the gas turbines. Provision of high attenuation silencers for the exhaust ducts of all diesel prime movers and the gas turbines as well as for the combustion air inlets of the cruising diesels and the gas turbines.
Special attention must be paid to diesel units, the compressors, and converters positioned near the ship's sonar. As those are considered to be the most critical structureborne noise sources, they have been double elastic-mounted. In addition, the intermediate foundation frame structure of the diesel units is damped with reinforced concrete blocks at the foundation springs.
The design of the main reduction gear units is important to noise reduction. The rotating parts and meshes were machined to the highest accuracy to minimize gear contact noise. The reduction gear assemblies are hard elastic-mounted. In the design of elastic mounts attention must be paid to tuning with regard to the potential vibration exciting sources of the ship and its equipment. The behavior of the mounts under shock and rough sea conditions has to be examined.
8.8.1.4.2 Submarine Quieting. Self noise is a limiting factor in sonar performance, both from a counter detection point of view, the point at which an ememy is able to detect you, and with respect to target detection and tracking. Thus, there has been a strong trend toward building quieter submarines.
Significant improvement was realized in using a "bed plate" or acoustically-isolated foundation for the gears, turbines, condens-ers and turbo generators of the propulsion plant. Its noise-iso-lating mountings, however, added a good deal of volume and weight when compared with the conventional alternative of mounting the machinery directly to the hull.
The reduction gearing, which converts the horsepower of the rapidly-rotating turbine shaft to that of the slowly-rotating propeller, is a particularly strong noise source. An attempt was made to replace this gearing with a heavy but quiet turbine-electric drive system, featuring a turbo generator that drives a low-noise electric motor of low rotational speed and high torque. This arrangement, however, proved to be less efficient than the standard geared-turbine system.
Significant headway has been made in reducing propeller cavit-ation. However, recent advances in submarine quieting, due to im-provements in propeller fabrication, have been neutralized as this technology has been more widely disseminated than hoped.
Further silencing gains have been found in the use of a natur-al-circulation nuclear reactor, pump-jet propulsors, and hull-mounted sound-absorbing acoustic tiles.
8.8.2 Ambient Noise
Ambient noise is background noise in the sea due to either natural or man-made causes, and may be divided into four general categories: hydrodynamic, seismic, ocean traffic, and biological.
8.8.2.1 Hydrodynamic Noise. Hydrodynamic noise is caused by the movement of the water itself as a result of tides, winds, currents, and storms. The level of hydrodynamic noise present in the sea is directly related to the condition of the sea surface. As the surface becomes agitated by wind or storm, the noise level rises, reducing detection capability. Very high hydrodynamic noise levels caused by severe storms in the general area of the ship can result in the complete loss of all signal reception.
8.8.2.2 Seismic Noise. Seismic noises are caused by land movements under or near the sea - as, for example, during an earthquake. They are rare and of short duration, and hence will not be elaborated upon.
8.8.2.3 Ocean Traffic. Ocean traffic's effect on ambient noise level is determined by the area's sound propagation characteristics, the number of ships, and the distance of the shipping from the area. Noises caused by shipping are similar to those discussed under the heading of self-noise, with the frequencies depending on the ranges to the ships causing the noise. Noises from nearby shipping can be heard over a wide spectrum of frequencies, but as the distance becomes greater, the range of frequencies becomes smaller, with only the lower frequencies reaching the receiver because the high frequencies are attenuated. In deep water, the low frequencies may be heard for thousands of kilometers.
8.8.2.4 Biological Noise. Biological noises produced by marine life are part of ambient background noise and at times are an important factor in ASW. Plants and animals that foul the ships are passive and contribute to self-noise by increasing water turbulence. Crustaceans, fish, and marine mammals are active producers of sounds, which are picked up readily by sonar equipment.
During and since World War II, a great deal of research on sound-producing marine animals has been carried out. The object was to learn all the species of animals that produce sound, their methods of production, and the physical characteristics of the sounds (frequencies, intensities, etc.). Sounds produced by many species have been analyzed electronically, and considerable physical data have been obtained.
All important sonic marine animals are members of one of three groups crustaceans, fish, and mammals.
Crustaceans, particularly snapping shrimp, are one of the most important groups of sonic marine animals. Snapping shrimp, about 2 centimeters long, bear a general resemblance to the commercial species, but are distinguishable from them by one long, large claw with a hard movable finger at the end. They produce sound by snapping the finger against the end of the claw. Distribution of snapping shrimp appears to be governed by temperature, and they are found near land in a worldwide belt lying between latitudes 35oN. and 40oS. In some places, such as along the coast of Europe, they range as far north and south as 52o. The largest colonies or beds of snapping shrimp occur at depths of less than 30 fathoms on bottoms of coral, rock, and shell. There are exceptions, however. They may, for example, occur as deep as 250 fathoms, and have been found on mud and sand bottoms covered with vegetation.
A shrimp bed is capable of producing an uninterrupted crackle resembling the sound of frying fat or burning underbrush. Frequencies range from less than 1 to 50 kHz. Noise is constant, but there is a diurnal cycle, with the maximum level at sunset. Over beds, a pressure level of 86 db re 1 Pa has been noted. Intensity drops off rapidly as the range from the bed increases. Lobsters, crabs, and other crustaceans may make minor contributions to background noise.
Fish produce a variety of sounds that may be placed in three categories, depending upon how the sounds are caused. The first category includes sound produced by the air bladder, a membranous sac of atmospheric gases lying in the abdomen. The sound is caused by the movement of muscles inside or outside the bladder or by the general movement of the body. The second division includes sounds produced by various parts of the body such as fins, teeth, and the like rubbing together. This noise is called stridulatory sound. The third class includes sounds that are incidental to normal activities, such as colliding with other fish or the bottom while swimming, biting and chewing while feeding, and so on.
The majority of the sonic fish inhabit coastal waters, mostly in temperate and tropical climates. Although fish are the most pre-valent, and therefore the most important, sound producers, their activity is not as continuous in intensity as that of snapping shrimp. The level of sound produced by them increases daily when they feed (usually at dawn and dusk), and annually when they breed. Fish sounds range in frequency from about 50 to 8000 Hz. Sounds of air bladder origin have most of their energy concentrated at the lower end of this spectrum, 75 to 150 Hz, whereas stridulatory sounds characteristically are concentrated at the higher end of the spectrum.
Marine mammals, in addition to returning echoes from sonar equipment, produce sound vocally and stridulously. Seals, sea lions, and similar animals posses vocal cords, and bark or whistle or expel air through their mouths and nostrils to produce hisses and snorts. Whales, porpoises, and dolphins force air through soft-walled nasal sacs, and the sounds produced in this way have been described as echo-ranging pings, squeals, and long, drawn-out moans similar to a foghorn blast. Other sounds probably produced by stridulation and attributed principally to whales, porpoises, and dolphins are described as clicking and snapping.
8.8.3 Wenz Curve.
While difficult to determine accurately on an operational basis, ambient noise levels are nonetheless important factors to be considered in determining sonar performance. Figure 8-29is an example of a set of Wenz curves that can be used to estimate noise levels from a variety of sources within the frequency range of interest. Figure 8-30 is an expansion of the shipping noise/wind noise portion of the Wenz curves.
SAMPLE PROBLEM USING WENZ CURVES
A SOSUS station is searching for an enemy submarine known to be producing sound at a frequency of 300 Hz. It is suspected that the sub is patrolling in shipping lanes that currently have 6 ft. seas. Use Wenz curves to determine an approximate value of ambient noise.
From figure 8-30 the ambient noise level due to shipping at 300 Hz is 65 dB and that due to 6 ft seas is 66 dB.
Using the nomogram, figure 8-2 combine the signals of 65 dB and 66 dB
IL1 - IL2 = 66-65 = 1
From the nomogram add 2.4 dB to 66 dB ... 68.4 dB.
8.9 THE SONAR EQUATIONS
The key to success in antisubmarine warfare is initial detection. For this work the major sensor in use today is sonar, both active and passive, and the present state of knowledge of the physical world suggests no change in this situation for many years. An understanding of sonar can only be achieved through a comprehension of the sonar equations and the concept called figure of merit. Many of the phenomena and effects associated with underwater sound may conveniently and logically be related in a quantitative manner by the sonar equations. For many problems in ASW, the sonar equa-tions are the working relationships that tie together the effects of the medium, the target, and the equipment, so that the operator can effectively use and understand the information received and provide prediction tools for additional information. Therefore, the purpose of this section is to spell out the sonar equations and figure of merit, to state the specifics of their usefulness, and to indicate how the various parameters in the sonar equations, in-cluding the figure of merit, can be measured.
8.9.1 Signal to Noise Ratio.
The sonar equations are based on a relationship or ratio that must exist between the desired and undesired portion of the received energy when some function of the sonar set, such as detection or classification, is performed. These functions all involve the re-ception of acoustic energy occurring in a natural acoustic back-ground. Of the total acoustic energy at the receiver, a portion is from the target and is called signal. The remainder is from the environment and is called noise. The oceans are filled with noise sources, such as breaking waves, marine organisms, surf, and dis-tant shipping, which combine to produce what is known as ambient noise. Self-noise, in contrast, is produced by machinery within the receiving platform and by motion of the receiving platform through the water. Further, in active systems, scatterers such as fish, bubbles, and the sea surface and bottom produce an unwanted return called reverberation, which contributes to the masking of the desired signal.
The function of the design engineer is to optimize the signal-to-noise (S/N) ratio for all conditions as detailed in the original design specifications of the sonar set. The operator, using his knowledge of the design specifications, his known ability in certain circumstances, the predicted conditions extrapolated from previously determined measurements, and actual on-board measurements, can then predict the detection probability.
In order to predict performance, the operator's interaction with the sonar set must be defined or quantified in a manner that provides a measure of predictability for varying signal and noise levels. This quantity, known as Detection Threshold (DT), attempts to describe in a single number everything that happens once the signal and its accompanying noise are received at the sonar. Detection threshold is defined as the signal minus noise level required inboard of the hydrophone array in order that an operator can detect a target. Actually, the business of detecting a sonar signal is a chance process for several reasons, one of which is that a human being is involved. The decision to call a target may be either right or wrong if a target is really present, then there is a detection; if a target is not present, then there is a false alarm. Hence, the definition of DT is normally qualified by adding the requirement that an operator "can detect a target on 50 percent of those occasions for which a target presents itself." Thus, if the average value of provided signal-to-noise equals the average of required signal-to-noise, a detection occurs in 50 percent of the times that a detection could occur. To summarize:
If average provided = average required, then detection probability is 50%
If average provided > average required, then detection probability is 50% to 100%
If average provided < average required, then detection probability is 50% to 0%
Note that the instantaneous value of the provided or required signal-to-noise can vary over a wide range due to the variability of operators, and an individual operator's moods, as well as time fluctuations in propagation loss, target radiated signal, and own ship noise. Hence, while the average value of provided signal-to-noise may be less than the average value of required signal-to-noise, at times the instantaneous value may be greater than the required value, and a detection may occur. Thus, a probability of detection greater than zero exists.
Putting this all together, it can be seen that if detection is to occur with a specified degree of probability, then the signal, expressed in decibels, minus the noise, expressed in decibels, must be equal to or greater than a number, the Detection Threshold, which also is expressed in decibels.
S - N > DT (8-25)
This equation is the foundation upon which all the versions of the sonar equations are based, and is simply a specialized statement of the law of conservation of energy.
The next step is to expand the basic sonar equation in terms of the sonar parameters determined by the equipment, the environment, and the target.
Two pairs of the parameters are given the same symbol (Own Sonar Source Level/Target Source Level, and Self-noise Level/Ambient-noise Level) because they are accounted for similarly in the sonar equations. This set of parameters is not unique, nor is the symbolism the same in all publications, but they are the ones conventionally used in technical literature. It should be noted as the discussion progresses how each of these parameters will fit into the mold of the basic sonar equation.
Parameter Active/Passive Symbol Determined by
Own sonar source level Active SL Equipment
Self noise level Active/Passive NL Equipment
Receiving Directivity Active (Noise Limited
Index Passive DI Equipment
Detection Threshold Active/Passive DT Equipment
Transmission Loss Active (2TL) TL Environment
Passive TL
Reverberation level Active RL Environment
Ambient Noise Level Active NL Environment
Passive
Target Strength Active TS Target
Target Source Level Passive SL Target
_________________________________________________________________
The transmission loss is frequency dependent as shown in Section 8.7. When calculating the acoustic signal resulting from a signal of interest superimposed on the ambient background noise of the ocean (signal to noise ratio), this calculation must be made using signals of the same frequency to be valid. Since sonar systems are designed to operate in specific frequency bands, calculations for these systems must relate to the design frequency band of the specific sonar to be valid.
8.9.2 Passive Sonar Equation
A passive sonar depends on receiving a signal that is radiated by a target. The target signal can be caused by operating machinery, propeller noise, hull flow noise, etc., but the same fundamental signal-to-noise ratio requirement must be satisfied. At the receiver, the passive equation begins as S - N > DT. If the target radiates an acoustic signal of SL (Target Source Level), the sound intensity is diminished while en route to the receiver because of any one or more of the following: spreading, ray path bending, absorption, reflection, and scattering. The decrease in intensity level due to this is called Transmission Loss (TL) and is also measured in decibels. Hence the intensity level of the signal arriving at the ship is
S = SL - TL (8-26)
Noise, N, acts to mask the signal and is not wanted. Therefore, the receiver is composed of many elements, sensitive primarily in the direction of the target so that it can discrimminate against noise coming from other directions. This discrimination against noise can be referred to as a spatial processing gain and is called the Receiving Directivity Index, DI. DI gives the reduction in noise level obtained by the directional properties of the transducer array. Therefore, in the basic equation, noise is now reduced and becomes
N = NL - DI (8-27)
There are two things to note in this simple equation: the first is that DI is always a positive quantity, so that NL - DI is always less than or equal to NL; the second is that the parameter NL represents both Self-noise Level and Ambient-noise Level, for by its definition it is the noise at the hydrophone location and can come from any, or all, sources.
The passive sonar equation can now be constructed in terms of signal and noise. When S and N are substituted from equations (8-26) and (8-27) into (8-25) the result is
SL - TL - NL + DI > DT (8-28)
which is the simplest form of the passive sonar equation. In words, equation 8-28 says that the source level of the target minus the loss due to propagation through the medium, minus the sum of all interfering noises plus improvement by the spatial processing gain of the receiver, must be equal to or greater than the detection threshold for a target to be detected with the specified probability of detection. However, the greater-than or equal-to condition is normally written as an equality. It is then understood to mean that if the left-hand side's algebraic sum is greater than DT, detection is possible with a greater probability than that specified by DT. If the sum is less than DT, detection probability decreases. Generally speaking, these two conditions imply that either detection is highly probable or seldom occurs. As a further aid to understanding the passive sonar equation, figure 8-31 illustrates the sonar parameters and indicates where
each term involved interacts to produce the desired results.
8.9.3 Active Sonar Equation
In an active sonar, acoustic energy is transmitted, and the received signal is the echo from the target. Two different, but related, equations are needed to describe the active sonar - one for an ambient-noise-limited situation and the other for the reverberation-limited situation. As developed previously, sonar performance is governed by the requirement that signal minus noise must be equal to or greater than detection threshold. The differ-ence in the two active sonar equations that satisfy this require-ment depends upon the characteristics of the noise that is actually present at the receiver when the signal is detected. The ambient noise may be described as either isotropic-i.e., as much noise power arrives from one direction as from any other - or as reverb-eration, in which noise returns primarily from the direction in which the sonar has transmitted.
Before developing the active sonar equations, the two types of noise should be briefly explained. Ambient noise consists of those noises present even when no sound is being radiated by a sonar. These include such noises as sea animals, machinery, propulsion noises generated by the echo-ranging platform, and the turbulence generated in the vicinity of the sonar. This type is the same as the noise level term discussed in the passive sonar equation. The second type, reverberation, consists of a multiplicity of echoes returned from small scatterers located in the sound beam and near the target when they reflect the transmitted energy. The combined effect of echoes from all of these scatterers produces a level of noise at the receiver that will tend to mask the returning echo from any wanted target.
8.9.3.1 Noise-Limited. The development of the active sonar equation is similar to that for the passive equation. In other words, the formal sonar parameters will be fitted to the signal and noise terms of equation (8-25). If a sonar transmits an acoustic pulse with an initial source level of SL dB, the transmitted pulse will suffer a transmission loss in traveling to the target. The target will scatter acoustic energy, some of which will return to the sonar. The back-scattered intensity is called target strength and is related to the scattering cross section of the target. The returning echo will again undergo a propagation loss, and thus the signal at the sonar will be
S = SL - 2TL + TS (8-29)
As long as the source of the radiated energy and the receiver for the echo are located together, the transmission loss experienced is equal to twice the one-way transmission loss.
When the echo returns, under some conditions the reverberation background due to the initial transmission will have disappeared, and only ambient noise will be present. This noise will be identical to that described in the passive sonar equation, modified by the receiving directivity index. The fundamental relationship can then be expressed as
SL - 2TL + TS - NL + DI > DT (8-30)
which is the basic active sonar equation used when the sonar is operating in a noise-limited situation.
8.9.3.2 Reverberation-Limited. If, on the other hand, the echo returns when the reverberation background has not decayed to a level below the ambient noise level, the background noise is given by RL. In this case, the parameter DI, defined in terms of an isotropic background, is inappropriate, inasmuch as reverberation is by no means isotropic. For a reverberation background the terms NL - DI are replaced by an equivalent reverberation level observed at the hydrophone terminals, and the sonar equation takes the form.
SL - 2TL + TS - RL > DT (8-31)
which is known as the reverberation-limited active sonar equation. Detailed quantification for the new term, RL, is difficult at best, for it is a time-varying function resulting from the inhomogeneties in the medium. One thing to note is that in the normal conversion from the basic equation to the active equations, the inequality again becomes an equality. As discussed under the passive sonar equation, it is understood that when the terms on the left-hand side exceed the detection threshold by a significant amount, detection is highly probable, and when it is significantly less than the detection threshold, detection seldom occurs. Figure 8-31 pictorially depicts the active sonar equations.
Of special interest in the active sonar equations is the term TS, and the fact that it usually is on the order of 15 to 25 dB. The variability of the value of TS is a function of the target aspect presented to the incoming signal. A beam target presents a greater reflective area then the bow, hence more energy is reflected from a beam target then from a bow target. Because TS is 10 times the logarithm of the reflected intensity divided by the inbound intensity, this statement apparently says that more energy is reflected than is incident, a condition clearly not possible.
The key lies in the definition of terms:
I reflected is the intensity, I, of the reflected signal measured one meter from the target, assuming the target is a point source.
I inbound is the intensity, I, of the signal inbound from transmitting ship to the target measured at a point on the target.
Intensity is actually power per unit area striking the target at some point, and thus the total sound power striking the target is I inbound times an effective area. If one assumes that the major portion of this power is reflected not from the original effective area (which is almost the same as the profile area of the target), but instead from a point source, it necessarily follows that the reflected energy computed in this way must be greater because of the reduced area from which the energy emanates. Thus I reflected is greater than I inbound if both are defined as indicated above.
In this case, there is no such wave as the one that is defined as originating from the point source. This construct is merely a convenient way of duplicating the actual measured value of I re-flected when the wave is 1,000 meters or more away from the point source enroute back to the transmitting ship. Thus, if one were to measure I reflected and I inbound both at 1,000 meters from the target, then I inbound would definitely be greater than I reflect-ed. Therefore, I reflected would have been computed to suffer a greater attenuation in traveling 1,000 meters from the constructed point source than I inbound will suffer in going 1,000 meters to the target. The explanation for this is the rapid attenuation due to spreading from the point source as compared to that undergone by the inbound wave, which is very near a plane wave when it is within 1,000 meters of the target.
8.10 FIGURE OF MERIT
The possible detection range of particular equipment should be known so that a tactician will then have a measure of the sonar's capability and a feel for what the sonar can do in a given tactical situation. Unfortunately, with no change in basic sonar configuration its detection capability measured in terms of range can increase or decrease severalfold simply because the ocean itself has changed. To state this another way, sonar equipment can only be designed to detect the arrival of a certain sound energy intensity. The range from which that sound intensity arrives is highly dependent on how much energy was lost en route, and therefore detection range alone is a poor measure of sonar capability. A better measure is the ability of the sonar to detect a certain level of sound energy intensity just outboard of its receiver. The key to this better measure of performance is to separate the sonar from the ocean in which it must operate. Only then can sonar capability be discussed in terms of the unchanging sonar hardware as distinguished from the ever-changing ocean.
The better measure for sonar capability is called figure of merit (FOM), and it equals the maximum allowable one-way transmission loss in passive sonars, or the maximum allowable two-way transmission loss in active sonars for a detection probability of 50 percent. Therefore, solving equations (8-28) and (8-30) for transmission loss, we get
Passive FOM = SL - NL + DI - DT (8-32)
Active FOM = SL + TS - NL + DI - DT (8-33)
This combination of terms is probably the most used performance parameter for sonars, and it is important to understand just what it means. The FOM of a sonar system is the maximum transmission loss that it can tolerate and still provide the necessary detection probability as specified by DT. FOM is improved by raising the source level (this can be accomplished by increasing the transmitted power in the active case or finding a noisier target in the passive case), decreasing the ambient noise level, increasing the absolute value of the receiving directivity index, and decreasing the detection threshold. The value of figure of merit is that, with no knowledge of the intervening propagation path between a ship and a target, a quantitative comparison of two different sonars can be made. The figure of merit may be utilized for comparing the relative performance of two passive sonars provided the calculations of the comparative figures of merit are made for the same frequency. The difference in FOM's represent additional allowable propagation loss that can be sustained by the sonar having the higher FOM and still make a detection; thus a longer detection range results.
8.10.1 Prop Loss Curves
Naturally, to the tactician, detection ranges are of prime importance, and although FOM can be interpreted in terms of range, it can be done only if the propagation losses involved are known. The measure of propagation loss (total transmission loss) in many parts of the ocean has proven that not only does sound propagation change with frequency and location but with season as well. Thus, the sound velocity profile will determine the propagation path(s) available, which along with frequency and location will determine transmission loss versus range. Therefore, in order to convert FOM to range, one must have a propagation loss curve for the frequency or the sonar (s) concerned and for the area and season of the year for which range prediction is desired.
Figure 8-33 is a typical propagation loss curve for the waters off Iceland in the summer for a frequency of 2 Khz. Note that the propagation losses for the three sound paths previously described are plotted. Multiple bottom bounce losses were measured and mul-tiple convergence zones were estimated.
Assume three different passive sonars with the following values of FOM at 2 Khz: 80, 95, 105. Based on Figure 8-32, the range in Kyds at which 50% probability of detection is predicted for each sonar for each sound transmission path in the Iceland area is as follows:
Sonar (FOM) Direct Bottom Bounce Convergence Zone
A 80 7.5 None None
B 95 21 50 81
C 105 36 90 160
Note that the sonar with the higher figure of merit permits the use of sound paths (bottom bounce and convergence zone) not available to the sonar with relatively low figure of merit. The higher the FOM the longer the detection range for a given path. Propagation loss curves can be made from a ray tracing program or actual measurements and smoothed data.
In summary, sonar performance is the key to ASW success, and figure of merit is the key to sonar performance. With knowledge of his sonar's FOM, a commanding officer can ensure that his equipment is peaked, and also predict detection ranges against possible enemy targets. The war planner can do likewise for either a real or hypothetical enemy. Because the changing ocean results in dramatic changes in propagation loss versus range, to state a sonar's capability in terms of range is only half the story, and may even be misleading. Using figure of merit, however, the sonar with the higher FOM will always be the better sonar when comparing sonars in the same mode.
8.10.2 Figure of Merit Sample Problem
Your sonar is capable of either passive or active operation. You are operating in the shipping lanes with a sea state of 2. Water depth is 200 fathoms. Using the following information, you must decide which mode to use. Intelligence information indicates that the threat will be a Zebra-class submarine (all dB are reference 1Pa).
Target parameters:
radiated noise source level 100 dB
target strength 15 dB
target detection range l0,000 m
Sonar parameters: Active Passive
Source level 110 dB -
Frequency 1.5 kHz -
Self-noise at 15 kts 50 dB 50 dB
Directivity index 10 dB 8 dB
Detection threshold -2 dB 3 dB
In order to determine which sonar to use, it is necessary to calculate the FOM and total transmission loss for each mode.
First, calculate the total transmission loss for each mode. Since the desired target detection range of 10 km is much greater than the water depth of 200 fathoms, we will use equation (8-24) for cylindrical spreading:
TL = 10 log r + r + A
The quantity A is assumed to be zero since no information is available. The absorption coefficient () is calculated by substituting signal frequencies for each mode into equation (8-21):
= .036f2 + 3.2 x 10-7 f2 (where f is in kilohertz)
f2 + 3600
Active mode Passive mode
= .036(1.5)2 = .036(.5)2
(1.5)2 + 3600 (.5)2 + 3600 + (3.2 x 10-7)(1.5)2 +(3.2 x 10-7)(.5)2
= 2.32 x 10-5 = 2.6 x 10-6
TL = 10 log(10,000) + 2.32 TL = 10 log(10,000) + 2.6 x
x 10-5(10,000) 10-6(10,000)
TL = 40 + .232 TL = 40 + .026
TL = 40.232 (one way) TL = 40.026dB (Total TL)
2 x TL = 80.464dB (Total TL)
Note that TL depends only upon the detection range and the frequency of the signal.
Next calculate the FOM using equation (8-33) for the active mode and equation (8-32) for the passive case. Note that the only values that must be determined are the noise levels using the Wenz curves and nomograms for the different frequencies.
Active Mode Passive Mode
f = 1500 Hz f = 500 Hz
AN(shipping) = negligible AN(shipping) = 57dB
AN(sea state) = 58dB AN(sea state) = 61dB
Self-Noise = 50dB Self-Noise = 50dB
Using the nomogram, combine the signals for each mode:
58-50 = 8dB 61 - 57 = 4 dB
NL = 58 + .65 (from nomogram) AN = 61 + 1.5 (from nomogram)
NL = 58.65 dB AN = 62.5 dB
62.5 - 50 = 12.5 dB
NL = 62.5 + .25 = 62.75 dB
Note that when three noise signals are involved, a two-step signal-combining process is required. The resultant is always added to the higher signal level. Determining the FOM is now a simple matter of substituting the calculated and given values into the appropriate equations:
Active Mode Passive Mode
FOM = SL + TS - NL + DI FOM = SL - NL + DI - DT
- DT FOM = 100 - 62.75 + 8 - 3
FOM = 110 + 15 - 58.65 + FOM = 42.25 dB
10 -(-2)
FOM = 78.35 dB
Compare the FOM of each mode with the total TL for each mode to determine which mode is optimum for this target. The FOM for the active case is less than the total TL. Therefore, the active mode will give your ship less than 50% probability of detection. The FOM for the passive case is greater than the total TL. Therefore, the passive mode will give a greater than 50% probability of detection, which means that the passive mode should be used.
8.11 SUMMARY
Of all the energy forms available, sound, even with its inherent disadvantages, is the most useful for underwater detection of submarines. It travels as a series of compressions and rarefactions at a speed equal to the product of its frequency and wavelength. The pressure of the wave can be expressed as a function of both time and its distance from the source. Acoustic power, called its intensity, is a measure of the amount of energy per unit area and has the units watts/m2. Acoustic pressure is expressed in micro-pascals. To make comparisons easier, both are normally converted to the logarithmic decibel system. The speed of sound in the sea is related to the bulk modulus and density of the water, which are affected by the temperature, pressure, and salinity. Temperature is the most important of these environmental factors, and therefore the thermal structure of the ocean is of significant tactical importance. The tracing out of sound paths in water is known as ray theory and is governed by Snell's Law. Various unique propagation paths can be identified according to the thermal structure of the water, but in practice such paths are a complex combination of simpler structures. Actual sound propagation through the sea is subject to geometric spreading and attenuation, both of which decrease the acoustic intensity at the receiver. The active and passive sonar equations are an expression of various factors determined by the equipment, the medium, and the target, which lead to an overall measure of sonar performance called figure of merit.
8.14 REFERENCES
Cheney, R. E., and D. E. Winfrey, "Distribution and Classification
of Ocean Fronts," NAVOCEANO Technical Note 3700-56-76. Washington, D.C.: GPO, 1976.
Chramiec, Mark A. Unpublished lecture notes on Figure of Merit and
Transmission Loss, Raytheon Company, 1983.
Cohen, Philip M., "Bathymetric Navigation and Charting", United
States Naval Institute, Annapolis, Maryland, 1970.
Commander, Naval Ordnance Systems Command. Elements of Weapons
Systems. NAVORD OP 3000, vol. 1, 1st Rev. Washington, D.C.:
GPO, 1971.
Corse, Carl D. Introduction to Shipboard Weapons, Annapolis, MD:
Naval Institute Press, 1975.
Cox, Albert W., "Sonar and Underwater Sound", Lexington Books,
Lexington, MA, 1974.
Duxbury, Alyn C. The Earth and its Oceans. Reading, Massachusetts:
Heppenheimer, T. A., "Anti-Submarine Warfare: The Threat, The
Strategy, The Solution", Pasna Publications Inc., Arlington, Va.
1989.
Honhart, D. C. Unpublished class notes on Acoustic Forecasting,
King, L. F., and D. A. Swift. Development of a Primer on Underwater for ASW. Master's thesis, Naval Postgraduate
School, 1975.
Kinsler, L. E., and A. R. Frey, Fundamentals of Acoustics, 2nd ed.
New York: John Wiley and Sons, 1962.
Myers, J. J., C. H. Holm, and R. F. McAllister, eds. Handbook of
Ocean and Underwater Engineering, New York: McGraw-Hill Book
Company, 1969.
National Defense Research Committee, "Principles and Applications
of Underwater Sound", Washington, D.C., 1976.
Naval Education and Training Command. Sonar Technical G3 2(u),
NAVEDTRA 10131-D, Washington, D.C.: GPO, 1976.
Naval Operations Department, U.S. Naval War College, Technological
Factors and Constraints in System Performance Study-Sonar
Fundamentals. Vol I-1, 1975.
Naval Training Command. The Antisubmarine Warfare Officer (U).
NAVTRA 10778-C. Washington, D.C.: GPO, 1973.
Operations Committee, Naval Science Department, U.S. Naval Academy,
Naval Operations Analysis, Annapolis, MD: U.S. Naval Institute,
1968.
Sollenberger, R.T., and T. R. Decker, Environmental Factors Affecting Antisubmarine Warfare Operations. Master's thesis,
Sverdrup, H. U., M. W. Johnson, and R. H. Fleming. The Oceans.
Englewood Cliffs, N.J.: Prentice-Hall, Inc. 1942.
Urick, R. J., Principles of Underwater Sound, 2nd ed. New York:
McGraw-Hill Book Company, 1975.
U.S. Naval Oceanographic Office. Ocean Thermal Structure Fore-
casting. SP 105 ASWEPS Manual Series, vol. 5, 1st ed., by R. W.
James. Washington, D.C.: GPO, 1966. | 24,204 | 113,885 | {"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-2014-52 | latest | en | 0.844012 |
https://www.physicsforums.com/threads/why-protons-and-neutrons-have-spins.70409/ | 1,718,901,142,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861957.99/warc/CC-MAIN-20240620141245-20240620171245-00443.warc.gz | 823,256,310 | 19,406 | Why protons and neutrons have spins?
• Winga
In summary: GraberIn summary, the spin quantum numbers of 1/2 and -1/2 can be explained in terms of quarks as the result of two up quarks and one down quark (uud) making up the proton, which is a baryon. The spin value of each quark is 1/2, but due to the complex nature of quantum mechanics, the overall spin of the proton is not simply the sum of the individual spin values. Instead, it involves Clebsch-Gordan coefficients and other intricate algebraic calculations. This phenomenon is still being studied and fully understood in the field of quantum mechanics.
Winga
Their spin quantum numbers are 1/2 and -1/2.
How to explain these in terms of quarks?
Because they are made of two up quarks and one down quark (uud), protons are baryons. So are neutrons (udd). The magnitude of the spin-value of each quark is 1/2
marlon
Composing 3 spins of 1/2,according to Clebsch-Gordan's theorem,doesn't necessarily yield a spin 1/2 state...In fact,it yields a linear combination of 1/2 & 3/2 states.Something more is needed to explain the spin of those 2 barions.
Daniel.
Angular momentum doesn't add linearly. That is, a proton contains three quarks, but its spin is not 3/2, but 1/2. When you add angular momentum, you have a range of possible results, which goes from the maximum (the arithmetic sum) down to the minimum (the larger minus the smaller) with possible values at unit steps in between. If you add two spin-1/2 fermions, you get either spin 0 or spin 1 (a boson). If you add 5 units of angular momentum to 2 units, you get 3, 4, 5, 6, or 7 units of angular momentum. If you add 3/2 units and 2 units, you get 1/2, 3/2, 5/2, or 7/2 units. A proton has spin 1/2. As you should expect, there's another particle with the same quark content as the proton, but spin 3/2, the Δ+.
marlon
Are quarks also have spins, +1/2 & -1/2?
Eigenvalues of $$\hat{S}_{z}$$...?Of course.
Daniel.
Let me just say that the truth is far more complex than the simple constituent quark model could let one think
see for instance : Present understanding of the nucleon spin structure
there are several contribution to the spin of the nucleon, and it is a superbe illustration of the strength of QFT that they precisely add up to exactly one-half. The experimetal resolution of this puzzle is currently being performed.
Is there any easier way to explain this phenomenon?
Winga said:
Is there any easier way to explain this phenomenon?
No, sorry, this is QM. Though the explanation by Humanino is entirely correct, i suggest you make sure you understand how addition of the J-operator and the associated Clebsch-Gordan coefficients work. This is basic QM-stuff.
Question is : did you study any QM yet ?
regards
marlon
Mmmm...not yet!
Winga said:
Mmmm...not yet!
In that case, Houston we have a problem...
Within what context do you need to answer this question of proton-spin ?
regards
marlon
No offense,but "Their spin quantum numbers are 1/2 and -1/2.
How to explain these in terms of quarks?" and "Not yet" when asked about studying QM,do not really match.There's an order to everything...You can ask about everything,it's a forum,but the explanations you asked for "...in terms of quarks" are not simple to understand,if you have no theoretical preparation...
Daniel.
The first site given by astronuc indeed gives a splendid explanation as to why there are six quarks. However, it does not explain the problem at hand here. one cannot bypass the QM-nature of it's solution.
regards
marlon
marlon said:
Because they are made of two up quarks and one down quark (uud), protons are baryons. So are neutrons (udd). The magnitude of the spin-value of each quark is 1/2
marlon
For protons, the spin of one up quark cancels out the spin of one down quark, and one up quark left with spin value 1/2?
Unfortunately its not that simple. Spins don't add like regular numbers in quantum mechanics much less field theory. You have to play an intricate game with things called Clebsch Gordon Coefficients and the like. A first course in QM should suffice to explain the correct algebra.
Would the correct algebra be able to be taught here? If not, where can I find an easy explination of what to do?
The CRC Handbook for 2002 says:
Winga said:
Their spin quantum numbers are 1/2 and -1/2.
How to explain these in terms of quarks?
Hi Winga,
The free neutron undergoes, with half-life of 614 seconds, a convulsive transformation to a proton + a 782 KeV beta- and its spin is +1/2 and its magnetic dipole moment is -1.913 nuclear magneton units.
The residual Proton's spin is also +1/2 but its magnetic dipole moment is
+2.793 nuclear magniton units. Cheers, Jim
1. Why do protons and neutrons have spins?
The spin of a particle is an intrinsic property that cannot be explained by classical physics. Protons and neutrons, as fundamental particles, have a spin of 1/2, which is a quantum mechanical property. This spin is necessary to explain the magnetic properties of these particles.
2. How is the spin of a proton or neutron determined?
The spin of a particle is determined experimentally through scattering experiments. These experiments involve firing particles at high speeds and measuring their interactions. The results of these experiments have shown that protons and neutrons have a spin of 1/2.
3. Is the spin of a proton or neutron always the same?
Yes, the spin of a proton or neutron is a fundamental property that does not change. However, the direction of the spin can vary, resulting in different spin states for a particle. This is known as spin angular momentum and is also a quantum mechanical property.
4. How does the spin of a proton or neutron relate to its magnetic moment?
The spin of a particle is directly related to its magnetic moment. The magnetic moment is a measure of the strength of a particle's magnetic field. For protons and neutrons, their spin is the primary contributor to their magnetic moment, making them behave like tiny magnets.
5. Are there any other particles with a spin of 1/2?
Yes, the spin of 1/2 is a common property among fundamental particles. In addition to protons and neutrons, electrons, muons, and neutrinos also have a spin of 1/2. This spin is a fundamental aspect of the Standard Model of particle physics and is essential in understanding the interactions between particles.
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1K | 1,757 | 7,002 | {"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": 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} | 3.09375 | 3 | CC-MAIN-2024-26 | latest | en | 0.877388 |
http://www.thecodingforums.com/threads/help-needed-is-quick-union-find-the-right-solution-to-this-problem.437683/ | 1,469,791,195,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257830064.24/warc/CC-MAIN-20160723071030-00055-ip-10-185-27-174.ec2.internal.warc.gz | 727,710,345 | 11,446 | # Help needed: Is Quick-Union-Find the right solution to this problem ?
Discussion in 'C Programming' started by aredo3604gif@yahoo.com, Apr 11, 2005.
1. ### Guest
The user can dynamically enter and change the rule connection between
objects. The rule is a "<" and so given two objects:
a < b simply means that b < a can't be set, also it must be a != b.
And with three objects a < b , b < c means a < c
I studied Quick Union Find algorithms a bit and if I understood them
correctly, once the user gives the input setting the rules/connection
between objects, the algorithm will give as output the result of the
solved connections between objects.
As far as I check on the array or given list of objects from user
input that it's never added anything like a = b nor b<a when a<b it's
already in the list, would it work as expected so that I know which
objects in the list are "the strongest" ones as per user given rules ?
I tried thinking about using simple logic like AND,OR,XOR of compare
results values but I couldn't find a way to ensure that the
transitivity rule could be mantained.
So, is a DAG or Quick Union Find the way to go ?
My only concern about Quick Union Find is that it constructs a graph
with a single highest level root like a tree, right ? But if I am not
wrong the user could set things in such a way that the graph would
have more than one "root" .. ?! | 336 | 1,384 | {"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-2016-30 | latest | en | 0.958018 |
https://www.teacherspayteachers.com/Product/Two-Digit-Addition-NO-regrouping-task-cards-Summer-theme-3103527 | 1,506,233,875,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689874.50/warc/CC-MAIN-20170924044206-20170924064206-00057.warc.gz | 869,191,952 | 24,327 | ## Main Categories
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This resource is also available in a BUNDLE. Here's the link: BUNDLE of Two Digit Addition and Subtraction ( with and without regrouping) task cards (Summer theme)
This is a colorful set of 24 task cards to practice two-digit addition with NO regrouping with summer theme.This set is a wonderful addition to your lessons!
I've included a recording sheet and answer key, too!
◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈
These activities would work for grades 1-3!
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Here are some possible uses for these in your classroom:
✿ early finishers
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✎"Love the task cards - nice themes...." Keri S.
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Here are some other math resources you might to check out:
Fidget Spinner Subtraction WITH regrouping
BUNDLE of Two Digit Addition and Subtraction ( with and without regrouping) task cards (Pirate theme)
Addition and Subtraction Facts to 20 Bundle
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\$3.75 | 1,073 | 2,445 | {"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-2017-39 | latest | en | 0.739234 |
https://ccm.net/forum/affich-271477-excel-forumla-help-if-formula | 1,723,549,683,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641076695.81/warc/CC-MAIN-20240813110333-20240813140333-00663.warc.gz | 121,584,120 | 25,423 | # Excel Forumla help (IF formula)
Solved/Closed
pleasehelp1101 - Feb 22, 2010 at 01:54 AM
pleasehelp1101 - Feb 22, 2010 at 04:07 PM
Hello,
I need some help, I just can't figure the IF formula out. On my Excel 2007 spreadsheet, I am trying to figure out if I put the letter grade of a B in one column and then in another separate column it automatically produces a 3. Also, I need each column to be flexible so that if I get an A in on course, then the corresponding column next to it will produce a 4 in it. I am trying to make a spreadsheet to keep track of my ongoing grades and I am stuck on this step. Thanks for the help in advance.
Related:
## 1 response
rizvisa1 Posts 4478 Registration date Thursday January 28, 2010 Status Contributor Last seen May 5, 2022 766
Feb 22, 2010 at 06:11 AM
in that column where you wants the digits to appear, you need to use this sort of formula
=IF(A1="", "", IF(A1="A", 4, IF(A1="B", 3, IF(A1="C", 2, IF(A1="D", 1, IF(A1="F",0,"Invalid Entry")))))) | 299 | 994 | {"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-2024-33 | latest | en | 0.907588 |
https://ask.sagemath.org/question/35354/sage-seems-to-be-improperly-computing-an-infinite-sum-and-giving-an-incorrect-answer/?answer=37683 | 1,568,527,949,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514570740.10/warc/CC-MAIN-20190915052433-20190915074433-00326.warc.gz | 381,985,370 | 15,827 | # Sage seems to be improperly computing an infinite sum, and giving an incorrect answer
Reference this question: https://ask.sagemath.org/question/353...
Here is the evaluation of an infinite sum in sage:
var('n')
f(n) = (-1)^(n+1)/(3*n+6*(-1)^n)
sum(f(2*n)+f(2*n+1),n,0,oo)
1/3*log(2) - 7/9
Evaluating the same sum in Mathematica:
f[n_] := (-1)^(n + 1)/(3*n + 6*(-1)^n)
Sum[f[2*n] + f[2*n + 1], {n, 0, Infinity}]
1/6 (-2 + Log[4])
Sage seems to be giving an incorrect solution. Am I missing something?
edit retag close merge delete
Thanks for reporting this bug!
( 2016-11-02 22:33:40 -0500 )edit
Sort by ยป oldest newest most voted
I don't think you're missing anything: Sage is not computing the infinite sum correctly. The summand is half of 1/((n+1)*(2n-1)), and that's easier to type:
sage: sum(1/((n+1)*(2*n-1)), n, 0, 1000).n()
-0.205068171626375
sage: sum(1/((n+1)*(2*n-1)), n, 0, 10000).n()
-0.204618542543703
sage: sum(1/((n+1)*(2*n-1)), n, 0, 100000).n() # seems to be converging
-0.204573546255870
sage: sum(1/((n+1)*(2*n-1)), n, 0, oo).n() # but not to this number
-1.09345743518226
sage: sum(1/((n+1)*(2*n-1)), n, 0, oo)
2/3*log(2) - 14/9
sage: sum(1/((n+1)*(2*n-1)), n, 0, oo).n() - sum(1/((n+1)*(2*n-1)), n, 0, 10000).n()
-0.888838892638556
The infinite sum differs from the partial sum by about 8/9, which is consistent with what Mathematica says.
more
I've opened https://trac.sagemath.org/ticket/21801 though I don't have time to report upstream now (meeting in a few minutes)
( 2016-11-02 15:07:26 -0500 )edit
@krisman thanks!
( 2016-11-02 15:29:13 -0500 )edit
It turns out that this is a bug in Maxima, which is what Sage uses for this kind of symbolic manipulation. The Maxima bug is being tracked at https://sourceforge.net/p/maxima/bugs....
( 2016-11-02 20:00:04 -0500 )edit
If you use simplify over the expression for the sum you get some kind of "double fraction" like 1/2/... what is a bit strange. See here. Maybe this is unrelated to the problem and is just some kind of non very standard notation.
( 2016-11-02 23:11:35 -0500 )edit
2
That term is just (1/2) multiplied by 1/(n+1).
( 2016-11-03 09:24:04 -0500 )edit
This is now fixed, so i am retagging this question from confirmed_bug to solved_bug:
sage: f(n) = (-1)^(n+1)/(3*n+6*(-1)^n)
sage: sum(f(2*n)+f(2*n+1),n,0,oo)
1/3*log(2) - 1/3
Thanks to all for making this happen !
more | 890 | 2,405 | {"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} | 3.703125 | 4 | CC-MAIN-2019-39 | longest | en | 0.84293 |
https://visualfractions.com/calculator/subtract-fractions/what-is-1-1-minus-27-54/ | 1,638,795,142,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363292.82/warc/CC-MAIN-20211206103243-20211206133243-00252.warc.gz | 661,111,668 | 9,840 | # How to Calculate 1/1 Minus 27/54
Are you looking to work out and calculate how to subtract 1/1 from 27/54? In this really simple guide, we'll teach you exactly what 1/1 - 27/54 is and walk you through the step-by-process of how to subtract one fraction from another.
Want to quickly learn or show students how to calculate 1/1 minus 27/54? Play this very quick and fun video now!
To start with, the number above the line in a fraction is called a numerator and the number below the line is called the denominator.
Why do you need to know this? Well, because to subtract a fractions from another we need to first make sure both fractions have the same denominator.
Let's set up 1/1 and 27/54 side by side so they are easier to see:
1 / 1 - 27 / 54
Our denominators are 1 and 54. What we need to do is find the lowest common denominator of the two numbers, which is 54 in this case.
If we multiply the first denominator (1) by 54 we will get 54. If we we multiply the second denominator (54) by 1 we will also get 54. We also need to multiply the numerators above the line by the same amounts so that the fraction values are correct:
1 x 54 / 1 x 54 - 27 x 1 / 54 x 1
This is what 1/1 minus 27/54 looks like with the same denominator:
54 / 54 - 27 / 54
Now that these fractions have been converted to have the same denominator, we can subtract one numerator from the other to make one fraction:
54 - 27 / 54 = 27 / 54
You're done! You now know exactly how to calculate 1/1 - 27/54. Hopefully you understood the process and can use the same techniques to add other fractions together. The complete answer is below (simplified to the lowest form):
1/2
## Convert 1/1 minus 27/54 to Decimal
Here's a little bonus calculation for you to easily work out the decimal format of the fraction we calculated. All you need to do is divide the numerator by the denominator and you can convert any fraction to decimal:
27 / 54 = 0.5
If you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support!
• "How to Calculate 1/1 minus 27/54". VisualFractions.com. Accessed on December 6, 2021. http://visualfractions.com/calculator/subtract-fractions/what-is-1-1-minus-27-54/.
• "How to Calculate 1/1 minus 27/54". VisualFractions.com, http://visualfractions.com/calculator/subtract-fractions/what-is-1-1-minus-27-54/. Accessed 6 December, 2021.
### Preset List of Fraction Subtraction Examples
Below are links to some preset calculations that are commonly searched for: | 678 | 2,604 | {"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} | 4.71875 | 5 | CC-MAIN-2021-49 | latest | en | 0.917072 |
https://studylib.net/doc/10406843/math-5320-exam-ii-november-5--2004-in-class-make-up | 1,713,384,913,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817171.53/warc/CC-MAIN-20240417173445-20240417203445-00539.warc.gz | 487,813,256 | 12,290 | # MATH 5320 Exam II November 5, 2004 In-Class Make-Up
```MATH 5320
Exam II
November 5, 2004
In-Class Make-Up
Answer the problems on separate paper. You do not need to rewrite the problem statements on your
answer sheets. Work carefully. Do your own work. Show all relevant supporting steps!
1.
Determine the radius of convergence of each of the following series:
a.
n
n + ( − 1) n
(2 z − i ) n
∑
2
n
n = 2 n + ( − 1)
∞
b.
n
4 + ( − 1) n
( z + 1) n
∑
n +1
n = 2 9 + ( −1)
∞
2.
Let G be a region in £ and let f ∈A (G ) . Prove that if Re f ( z ) + Im f ( z ) ≡ 0 on G, then
f is constant on G.
3.
Show that for all complex z the following hold:
a.
|cos z |2 = cos 2 x + sinh 2 y for z = x + iy
b.
cos3 z = cos 3 z − 3cos z sin 2 z
4.
Let f ( z) = (1 − z )1+i . Identify and sketch the image of the line segment (0, i) under f .
5.
Let M be the Möbius transformation which maps
i - 1, 2i, i + 1 to 1 , 1 , 1 , resp. Find a
i + 1 2i i − 1
formula for M and identify images of the unit quarter
discs under M , i.e., the images of D1 , D2 , D3 , D4 ,
where the unit quarter disc D j is given by
D j = Qj ∩ B(0,1) . See figure.
``` | 480 | 1,160 | {"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-2024-18 | latest | en | 0.741008 |
https://calculatorguru.net/converter/furlong-to-exameter/ | 1,716,966,308,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059206.29/warc/CC-MAIN-20240529041915-20240529071915-00552.warc.gz | 119,678,805 | 33,041 | Furlong to Exameter converter | fur to Em conversion
Furlong to Exameter converter| fur to Em conversion
Are you struggling with converting Furlong to Exameter? Don’t worry! Our online “Furlong to Exameter Converter” is here to simplify the conversion process for you.
Here’s how it works: simply input the value in Furlong. The converter instantly gives you the value in Exameter. No more manual calculations or headaches – it’s all about smooth and effortless conversions!
Think of this Furlong (fur ) to Exameter (Em) converter as your best friend who helps you to do the conversion between these length units. Say goodbye to calculating manually over how many Exameter are in a certain number of Furlong – this converter does it all for you automatically!
What are Furlong and Exameter?
In simple words, Furlong and Exameter are units of length used to measure the size or distance of something. It helps us understand the length of objects, spaces, or dimensions. The short form of Furlong is “fur” and the short form for Exameter is “Em”
In everyday life, we use length units to express the size of anything in various contexts, such as measuring furniture, determining the length of a room, or specifying the dimensions of an object. Furlong and Exameter are also two common units of length.
How to convert from Furlong to Exameter?
If you want to convert between these two units, you can do it manually too. To convert from Furlong to Exameter just use the given formula:
Em = Value in fur * 2.011679999E-16
here are some examples of conversion,
• 2 fur = 2 * 2.011679999E-16 Em = 4.023359998E-16 Em
• 5 fur = 5 * 2.011679999E-16 Em = 1.0058399995E-15 Em
• 10 fur = 10 * 2.011679999E-16 Em = 2.011679999E-15 Em
Furlong to Exameter converter: conclusion
Here we have learn what are the length units Furlong (fur ) and Exameter (Em)? How to convert from Furlong to Exameter manually and also we have created an online tool for conversion between these units.
Furlong to Exameter converter” or simply fur to Em converter is a valuable tool for simplifying length unit conversions. By using this tool you don’t have to do manual calculations for conversion which saves you time. | 545 | 2,197 | {"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-2024-22 | latest | en | 0.895758 |
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Showing results 1 to 6 of 6 Search took 0.01 seconds. Search: Posts Made By: LaurV
Forum: Puzzles 2020-04-08, 04:55 Replies: 44 Views: 6,382 Posted By LaurV Well, if you look for the exact number**, then... Well, if you look for the exact number**, then here it is (https://oeis.org/A126100): 72489. But generating all those is tedious, a lot of symmetry tricks to check... I would still prefer to go...
Forum: Puzzles 2020-04-08, 04:01 Replies: 44 Views: 6,382 Posted By LaurV See my post #13 (the edit for virgo) - you forget... See my post #13 (the edit for virgo) - you forget the back-spread. edit: whoops, didn't read all the thread (second page), this reply was not for the last post, but for tgan's post (last on the...
Forum: Puzzles 2020-04-05, 11:10 Replies: 44 Views: 6,382 Posted By LaurV Maybe for you... :rofl: :bow: Maybe for you... :rofl: :bow:
Forum: Puzzles 2020-04-03, 10:39 Replies: 44 Views: 6,382 Posted By LaurV This is not what you initially said. I learned... This is not what you initially said. I learned from you, to read it as it is, and not to interpret it. The two examples are the same, if I can label arbitrary the A node. Yes, the labeling is...
Forum: Puzzles 2020-04-02, 03:31 Replies: 44 Views: 6,382 Posted By LaurV You re totally right about connected graphs.... You re totally right about connected graphs. However I was talking about "brute force" and I didn't actually think to a way to enumerate connected graphs without generating all of them and check...
Forum: Puzzles 2020-04-02, 02:43 Replies: 44 Views: 6,382 Posted By LaurV Right. So? (how does this change the puzzle? It... Right. So? (how does this change the puzzle? It does not) The puzzle is well formulated. I also get their number for the example, with a little Excel calculation. (edit: @virgo: you forgot the...
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May 21, 2013
# Posts by Miley
Total # Posts: 324
math/calc
The hypotenuse of a right triangle has one end at the origin and one end on the curve y+x^2e^(-3x), with x > or = 0. One of the other two sides is on the x-asis, the other side is parallel to the y-axis. Find the maximum area of such a triangle. At what x-value does it occur?
MUSIC
Tell me about maurice ravel his childhood ( fam and friends) his love life his passion his accomplishments his death
Math
yep that is correct.
Math
yep that is correct.
math
100 sided polygon has = n(n-3)/2 diagonals 100(100-3)/2 4850 diagonals
Math
Math
its realy .34 An example: .336633663 3 is the tenths place (0.1) 3 is the hundredths place (0.01) 6 is in the thousandths place (0.001) roud to the nearest tenth would be .3 the answer is .3 * 100 = 30%
math/calc
The hypotenuse of a right triangle has one end at the origin and one end on the curve y+x^2e^(-3x), with x > or = 0. One of the other two sides is on the x-asis, the other side is parallel to the y-axis. Find the maximum area of such a triangle. At what x-value does it occur?
Math
perimeter = 2*(w+h) 2* (20 +17)= 2* (37)= 74
calculus
a closed box has a fixed surface area A and a suqare base wide side X a) find the formula for its volume, v, as a function x b) sketch the graph of v against x c) find the max value of V
Pages: <<Prev | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | Next>> | 469 | 1,442 | {"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} | 3.59375 | 4 | CC-MAIN-2013-20 | latest | en | 0.937463 |
https://stackoverflow.com/questions/16941330/c-programming-xor-bitwise-operation/16941997 | 1,544,902,762,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826968.71/warc/CC-MAIN-20181215174802-20181215200802-00614.warc.gz | 756,512,639 | 29,022 | # C Programming - XOR Bitwise Operation
What operation does the following ‘C’ statement perform?
star = star ^ 0b00100100;
(A) Toggles bits 2 and 5 of the variable star.
(B) Clears all bits except bits 2 and 5 of the variable star.
(C) Sets all bits except bits 2 and 5 of the variable star.
(D) Multiplies value in the variable star with 0b00100100.
• Grumpy response: none, since that won't compile. C doesn't support a `0b` prefix for binary numbers, as standard. – unwind Jun 5 '13 at 13:33
• Really? But this question came out on my past exam paper so I assume it will work though. – D3FTY Jun 5 '13 at 13:34
• What's the original value of `star`? – SubSevn Jun 5 '13 at 13:36
• – NightWhisper Jun 5 '13 at 13:36
• @D3FTY By the way, if you know this is the XOR operator, why don't you 1. already know what it does (it's expected from someone doing CS and stuff), 2. if you don't, why don't you google it either? – user529758 Jun 5 '13 at 13:38
XOR operator (also called "logical addition") is defined like this:
``````a b a^b
-----------
0 0 0
0 1 1
1 0 1
1 1 0
``````
So `a^0` leaves `a` intact while `a^1` toggles it.
For multiple-bit values, the operation is performed bitwise, i.e. between corresponding bits of the operands.
• Alright, cheers for that. – D3FTY Jun 5 '13 at 13:44
If you know how XOR works, and you know that `^` is XOR in C, then this should be pretty simple. You should know that XOR will flip bits where 1 is set, bits 2 and 5 of 0b00100100 are set, therefore it will flip those bits.
From an "during the test" standpoint, let's say you need to prove this to yourself, you really don't need to know the initial value of `star` to answer the question, If you know how `^` works then just throw anything in there:
`````` 00100100
---------
10001110 (star's new value)
bit position: | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0
|---|---|---|---|---|---|---|---
star's new v: | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0
|---|---|---|---|---|---|---|---
star's old v: | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0
``````
(A) Toggles bits 2 and 5 of the variable star. (Yes)
(B) Clears all bits except bits 2 and 5 of the variable star. (Nope)
(C) Sets all bits except bits 2 and 5 of the variable star. (Nope)
(D) Multiplies value in the variable star with 0b00100100. (36x170 = 142? Nope)
It is (A) toggles bits 2 and 5.
The following is the truth table for the XOR operation:
``````x y x^y
0 0 0
1 0 1
0 1 1
1 1 0
``````
You can see from the table that `x XOR 0 = x` and `x XOR 1 = !x`.
XOR is a bitwise operation, so it operates on individual bits. Therefore if you XOR `star` with some constant, it will toggle the `1` bits in the constant.
You can find some explanation e.g. here.
• Any specific explanation to it? I still don't understand it. – D3FTY Jun 5 '13 at 13:38
• @H2CO3 Well, I'm sorry if I'm dumb. But you can't expect everyone to have the same level of knowledge like you, smart alex. No offense. – D3FTY Jun 5 '13 at 13:44
The `exclusive OR` has this truth table:
``````A B A^B
-----------
1 1 0
1 0 1
0 1 1
0 0 0
``````
We can see that if `B` is `true` (`1`) then `A` is flipped (toggled), and if it's `false` (`0`) `A` is left alone. So the answer is (A).
• How would that make the answer C? – interjay Jun 5 '13 at 14:18
• Whoopsie! Thanks. – John Jun 5 '13 at 14:19
XOR operator returns 0 if both inputs are same otherwise returns 1 if both inputs are different.For Example the Given Truth Table :-
• a=1 b=1 => a^b=0,
• a=0 b=0 => a^b=0,
• a=0 b=1 => a^b=1,
• a=1 b=0 => a^b=1.
well xor is binary operator that work on bits of 2 nos.
rule of xoring:for same bit ans is 0 and for different bit ans is 1 let
```a= 1 0 1 0 1 1
b= 0 1 1 0 1 0
--------------
c= 1 1 0 0 0 1
--------------```
compare bit of a and b bit by bit if same put 0 else put 1 xor is basically used to find the unique in given set of duplicate no. just xor all nos. and u will get the unique one(if only single unique is present) | 1,375 | 4,003 | {"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} | 2.84375 | 3 | CC-MAIN-2018-51 | latest | en | 0.891767 |
https://www.daniweb.com/tags/octal | 1,586,501,204,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371886991.92/warc/CC-MAIN-20200410043735-20200410074235-00524.warc.gz | 872,294,701 | 9,850 | 6 Years Ago 6 Years Ago I want to convert some numbers to binary, octal, and hexadecimal.. for example: . DECIMAL......BINARY......OCTAL.....HEXADECIMAL ................................................................... 1.................00000001.....001.......1 2.................00000010.....002.......2 And so on.. 0 19 7 Years Ago 7 Years Ago Hello all. I'm writing a program that converts from any cominbation of base 2, 8, 10, and 16. So far I have this code and I want to convert from hexadecimal to octal but it doesn't work. I would appreciate any help. Thank you! Here is the code: #include "stdafx.h" #include #include #include #include #include #include #include #include #include #include #include #include #define ULONG unsigned long #define FALSE 0 #define TRUE 1 #define BOOL int #define MAX 1000 using namespace std; void B2D(); void B2O(); void B2H(); void … 0 26 7 Years Ago Hi everyone I'm stuck on a school project. The Following is the prompt: > Write an assembly language program that prompts the user to enter a string to be interpreted as an 8-digit hexadecimal number. Your program must convert the ASCII string representation of this hexadecimal number into a 32-bit unsigned binary number. Then, your program must prepare and store an ASCII string for the octal (base 8) representation of the same number and print it out.Your program should generate an error if the input string contains a character that is not 0 through 9 or A through F (except … 0 1 7 Years Ago 7 Years Ago Hi Guys, I have written the below program to convert binary to octal, it is giving me erroneous results. On entering 1111 as the input number I should get 17 as the output, whereas I am getting 16707000337 as the output. Kindly help me out, below is my program: /*program to convert binary to octal*/ #include #include #include int main() { int bin,dec[100]={0},i=0,j=0,k=0,num=0,output[100]={0},number=0,rem1,rem,output1; printf("enter the binary number to be converted into octal"); scanf("%d",&bin); rem1=bin; /*counts the number of digits in the input binary number*/ while(rem1!=0) { rem1=rem1/10; num+=1; } /*converts binary number to decimal */ for(i=0;i
The End. | 560 | 2,178 | {"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} | 2.703125 | 3 | CC-MAIN-2020-16 | latest | en | 0.723344 |
https://www.jiskha.com/display.cgi?id=1351130393 | 1,516,397,652,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888135.38/warc/CC-MAIN-20180119204427-20180119224427-00564.warc.gz | 935,628,784 | 3,719 | # Science
posted by .
Question: The greatest expansion of water occurs when: isn't it when it turns to ice?? The other answers are at 4C it gains temperature or at 4C it loses temperature. In the winter the pipes can break because the water in them freezes that's why I thought it was that. Thanks
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More Similar Questions | 557 | 2,266 | {"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} | 3.234375 | 3 | CC-MAIN-2018-05 | latest | en | 0.918091 |
https://codegolf.stackexchange.com/questions/63127/an-infinite-ftw | 1,721,659,236,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517878.80/warc/CC-MAIN-20240722125447-20240722155447-00544.warc.gz | 152,254,950 | 64,387 | # An infinite FTW
The infinite Fibonacci word is a specific, infinite sequence of binary digits, which are calculated by repeated concatenation of finite binary words.
Let us define that a Fibonacci-type word sequence (or FTW sequence) is any sequence ⟨Wn that is formed as follows.
• Commence with two arbitrary arrays of binary digits. Let us call these arrays W-1 and W0.
• For each n > 0, let Wn ≔ Wn-1 ∥ Wn-2, where denotes concatenation.
A consequence of the recursive definition is that Wn is always a prefix of Wn+1 and, therefore, of all Wk such that k > n. In a sense, this means the sequence ⟨Wn converges to an infinite word.
Formally, let W be the only infinite array such that Wn is a prefix of W for all n ≥ 0.
We'll call any infinite word formed by the above process an infinite FTW.
Write a program or function that accepts two binary words W-1 and W0 as input and prints W, abiding by the following, additional, rules:
• You may accept the words in any order; as two arrays, an array of arrays, two strings, an array of strings, or a single string with a delimiter of your choice.
• You may print the digits of the infinite word either without a delimiter or with a consistent delimiter between each pair of adjacent digits.
• For all purposes, assume that your code will never run out of memory, and that its data types do not overflow.
In particular, this means that any output to STDOUT or STDERR that is the result of a crash will be ignored.
• If I run your code on my machine (Intel i7-3770, 16 GiB RAM, Fedora 21) for one minute and pipe its output to wc -c, it must print at least one million digits of W for (W-1, W0) = (1, 0).
• Standard rules apply.
### Example
Let W-1 = 1 and W0 = 0.
Then W1 = 01, W2 = 010, W3 = 01001, W4 = 01001010 … and W = 010010100100101001010….
This is the infinite Fibonacci word.
### Test cases
All test cases contain the first 1,000 digits of the infinite FTW.
Input: 1 0
Output: 0100101001001010010100100101001001010010100100101001010010010100100101001010010010100100101001010010010100101001001010010010100101001001010010100100101001001010010100100101001001010010100100101001010010010100100101001010010010100100101001010010010100101001001010010010100101001001010010100100101001001010010100100101001001010010100100101001010010010100100101001010010010100101001001010010010100101001001010010010100101001001010010100100101001001010010100100101001001010010100100101001010010010100100101001010010010100101001001010010010100101001001010010010100101001001010010100100101001001010010100100101001001010010100100101001010010010100100101001010010010100101001001010010010100101001001010010010100101001001010010100100101001001010010100100101001010010010100100101001010010010100100101001010010010100101001001010010010100101001001010010010100101001001010010100100101001001010010100100101001010010010100100101001010010010100100101001010010010100101001001010010010100101001001010010100100101001001
Input: 0 01
Output: 0100101001001010010100100101001001010010100100101001010010010100100101001010010010100100101001010010010100101001001010010010100101001001010010100100101001001010010100100101001001010010100100101001010010010100100101001010010010100100101001010010010100101001001010010010100101001001010010100100101001001010010100100101001001010010100100101001010010010100100101001010010010100101001001010010010100101001001010010010100101001001010010100100101001001010010100100101001001010010100100101001010010010100100101001010010010100101001001010010010100101001001010010010100101001001010010100100101001001010010100100101001001010010100100101001010010010100100101001010010010100101001001010010010100101001001010010010100101001001010010100100101001001010010100100101001010010010100100101001010010010100100101001010010010100101001001010010010100101001001010010010100101001001010010100100101001001010010100100101001010010010100100101001010010010100100101001010010010100101001001010010010100101001001010010100100101001001
Input: 11 000
Output: 0001100000011000110000001100000011000110000001100011000000110000001100011000000110000001100011000000110001100000011000000110001100000011000110000001100000011000110000001100000011000110000001100011000000110000001100011000000110000001100011000000110001100000011000000110001100000011000110000001100000011000110000001100000011000110000001100011000000110000001100011000000110001100000011000000110001100000011000000110001100000011000110000001100000011000110000001100000011000110000001100011000000110000001100011000000110001100000011000000110001100000011000000110001100000011000110000001100000011000110000001100000011000110000001100011000000110000001100011000000110001100000011000000110001100000011000000110001100000011000110000001100000011000110000001100011000000110000001100011000000110000001100011000000110001100000011000000110001100000011000000110001100000011000110000001100000011000110000001100011000000110000001100011000000110000001100011000000110001100000011000000110001100000011000110000001100000011
Input: 10 010
Output: 0101001001010010100100101001001010010100100101001010010010100100101001010010010100100101001010010010100101001001010010010100101001001010010100100101001001010010100100101001001010010100100101001010010010100100101001010010010100100101001010010010100101001001010010010100101001001010010100100101001001010010100100101001001010010100100101001010010010100100101001010010010100101001001010010010100101001001010010010100101001001010010100100101001001010010100100101001001010010100100101001010010010100100101001010010010100101001001010010010100101001001010010010100101001001010010100100101001001010010100100101001001010010100100101001010010010100100101001010010010100101001001010010010100101001001010010010100101001001010010100100101001001010010100100101001010010010100100101001010010010100100101001010010010100101001001010010010100101001001010010010100101001001010010100100101001001010010100100101001010010010100100101001010010010100100101001010010010100101001001010010010100101001001010010100100101001001010
Input: 101 110
Output: 1101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101101011101011101101011101011101101011101
• Fibonacci-type words FTW! Commented Nov 7, 2015 at 3:47
v%w=w++w%(v++w)
The infix function % produces an infinite string, which Haskell prints forever because Haskell is cool like that.
>> "1"%"0"
"010010100100101001010010010100100101001010010010100101001001010010010100101001001010010010100101001001010010100100101001001010010100100101001010010010100100101001010010010100100101001010010010100101001001010010010100101001001010010010100101001001010010100100101001001010010100100101001010010010100100101001010010010100100101001010010010100101001001010010010100101001001010010100100101001001010010100100101001001010010100100101001010010010100100101001010010010100100101001010010010100101001001010010010100101001001010010100100101001001010010100100101001001010010100100101001010010010100100101001010010010100100101001010010010100101001001010010010100101001001010010100100101001001010010100100101001001010010100100101001010010010100100101001010010010100101001001010010010100101001001010010010100101001001010010100100101001001010010100100101001001010010100100101001010010010100100101001010010010100101001001010010010100101001001010010010100101001001010010100100101001001010010100100101001010010010100100101001010010010100100101
The recursive idea is similar to Zgarb's solution. Writing f for the function %, and + for string concatenation, it implements:
f(v,w) = w + f(w,v+w)
The infinite output string starts with w, and the remainder of it is the result for the Fibonacci-shifted strings w and v+w.
This has no problem generating a million characters in a minute.
# Pyth, 8 bytes
u,peGsGQ
Input in the form "W-1", "W0".
Proof of completion:
$time pyth -cd 'u,peGsGQ' <<< '"1", "0"' 2>/dev/null | head -c1000000 > /dev/null real 0m0.177s user 0m0.140s sys 0m0.038s Proof of correctness: Here is the series as internally generated. It is printed in concatenation by the program. [0, 10, 010, 10010, 01010010, 1001001010010,...] Compare to the following, printed in concatenation, which is simply the string added to the previous string on each step: [0, 1, 0, 01, 010, 01001, 01001010, 0100101001001, ...] We want to prove these are equivalent. Clearly, they are the same through the first few steps. Let's compare them after a bit: 010 010 10010 01001 01010010 01001010 1001001010010 0100101001001 We see that the pairs of strings are alternately of the forms: 01a 10b a10 b01 Where a and b are arbitrary sequences on 0s and 1s. Let's continue the sequence for a bit, to prove is continues forever by induction: 01a 10b 10b01a 10b01a10b a10 b01 a10b01 b01a10b01 2 steps later, it is of the correct form. 10b 01a 01a10b 01a10b01a b01 a10 b01a10 a10b01a10 2 steps later, it is of the correct form. So by induction, the strings always match once concatenated. • +1 for writing working code that you don't understand. Commented Nov 6, 2015 at 16:20 • I believe your 8 byte solution works because it prints starting at W0 but instead of printing W1 it prints W-1 || W0, which is the "wrong" concatenation order. I think this works out to be equivalent, but I haven't come up with a proof... Commented Nov 6, 2015 at 16:25 ## Haskell, 31 bytes w#v=v++drop(length v)(v#(v++w)) This defines an infix function # that takes two strings and returns an infinite string. Usage: > let w#v=v++drop(length v)(v#(v++w)) in "11"#"000" "000110000001100011000000110000... If I query the millionth element of the sequence defined by "1" and "0", even the online interpreter prints the result in less than a second: > let w#v=v++drop(length v)(v#(v++w)) in ("1"#"0") !! 1000000 '0' ## Explanation w#v= -- The result of w#v is v++ -- v concatenated to v#(v++w) -- the infinite word v#(v++w), drop(length v) -- with the first v characters dropped. Basically, we know that w#v == v#(v++w) and w#v begins with v, and use these facts to define the result. Since Haskell is lazy, this "magically" just works. ## CJam, 12 11 bytes llL{@_o+_}h This takes the two words on separate lines, in the opposite order, e.g. 0 1 gives 0100101001001010010100100101001... ### Explanation The idea is to build up the word naively (by remembering the current word and appending the previous one to it) and while we're doing that, we print whatever we just appended (in order not to repeat the prefix that was already printed). To avoid having to handle the starting point separately, we start from an empty word, such that W0 is the first thing we append (and print). ll e# Read the two lines separately. L e# Push an empty string. { e# Infinite loop... @ e# Pull up the previous FTW. _o e# Print it. +_ e# Append it to the current FTW and duplicate it. }h # Pip, 8 bytes Hey, tied with Pyth! (fOba.b) Straightforward recursive definition borrowed from xnor's Haskell answer. With spaces added for clarity: (f Ob a.b) Every program in Pip is an implicit function that takes the command-line args as its arguments (assigned to variables a through e) and prints its return value. O is an operator that outputs and then returns its operand, so the first thing that happens here is the second argument is displayed (sans trailing newline). Now, the Lisp-inspired syntax (f x y) in Pip is a function call, equivalent to f(x,y) in C-like languages. The f variable refers to the current function--in this case, the top-level program. Thus, the program recursively calls itself with b and a.b as the new arguments. I was pleasantly surprised to find that this approach is plenty fast: dlosc@dlosc:~$ time pip -e '(fOba.b)' 1 0 2>/dev/null | head -c1000000 > /dev/null
real 0m0.217s
user 0m0.189s
sys 0m0.028s
It takes about 30 seconds on my Ubuntu machine for the program to hit the max recursion depth, at which point it has printed out somewhere over a billion digits.
This iterative solution is slightly faster and hogs less memory, at the cost of one byte:
W1Sba.:Ob
# APL, 24 18
{(,/⌽⍵),⊂⍞←↑⍵}⍣≡⍞⍞
Using xnor's formulation allowed to shave off few characters.
⍞ ⍝ Read W₋₁ as a character string.
{ }⍣≡ ⍝ Apply the following function repeatedly until fixpoint:
⍝ It takes a pair of strings (A B),
⍞←↑⍵ ⍝ prints A
(,/⌽⍵),⊂ ↑⍵ ⍝ and returns (BA A).
On an infinite machine in infinite time it would actually print W thrice—first incrementally while running the loop, and then twice as a result of the whole expression when the ⍣≡ fixpoint operator finally returns.
It's not very fast but fast enough. In GNU APL:
$printf '%s\n' '{(,/⌽⍵),⊂⍞←↑⍵}⍣≡⍞⍞' 1 0 | apl -s | head -c 100 0100101001001010010100100101001001010010100100101001010010010100100101001010010010100100101001010010$
$time printf '%s\n' '{(,/⌽⍵),⊂⍞←↑⍵}⍣≡⍞⍞' 1 0 | apl -s | head -c 1000000 >/dev/null 0m3.37s real 0m2.29s user 0m1.98s system • Two infinite numbers. O.O +1 Commented Nov 7, 2015 at 19:34 • I didn't know about ⍣≡; it sounds very useful. Commented Nov 8, 2015 at 2:05 ## PowerShell, 97 76 Bytes param($a,$b)write-host -n$b;while(1){write-host -n $a;$e=$b+$a;$a=$b;$b=$e}
Edit - Umm, writing out $e.substring($b.length) after we just concatenated $a and $b together is equivalent to writing out just $a ... derp. Wow, verbose. PowerShell will, by default, spit out a newline every time you output something. Really the only way to get around that is with write-host -n (short for -NoNewLine), and that absolutely kills the length here. Essentially, this iterates through the sequence, building $e as the "current" Wn as we go. However, since we're wanting to build the infinite word instead of the sequence, we leverage our previous variables to print out the suffix $a that was populated in our previous loop. Then we setup our variables for the next go-round and repeat the loop. Do note that this expects the input to be explicitly delimited as strings, else the + operator gets used for arithmetic instead of concatenation. ### Example: PS C:\Tools\Scripts\golfing> .\infinite-ftw.ps1 "111" "000" 0001110000001110001110000001110000001110001110000001110001110000001110000001110001110000001110000001110001110000001110001110000001110000001110001110000001110001110000001110000001110001110000001110000001110001110000001110001110000001110000001110001110000001110000 ... # Pure bash, 58 a=$1
b=$2 printf$b
for((;;)){
printf $a t=$b
b+=$a a=$t
}
I run out of memory before 1 minute, but have plenty of digits by then - after 10 seconds I have 100 million digits:
$./ftw.sh 1 0 | head -c100 0100101001001010010100100101001001010010100100101001010010010100100101001010010010100100101001010010ubuntu@ubuntu:~$
${ ./ftw.sh 1 0 & sleep 10 ; kill$! ; } | wc -c
102334155
$ # Mathematica, 56 bytes $IterationLimit=∞;#0[$Output~WriteString~#2;#2,#<>#2]& # Dyalog APL, 9 {⍵∇⍺,⍞←⍵} This one uses ∇ to define a recursive function. It's a direct translation of this xnor's Python 3 answer. It takes W0 as right, and W−1 as its left argument, both should be character vectors. # C, 76 (gcc) main(c,v)char**v;{int p(n){n>2?p(n-1),p(n-2):puts(v[n]);}for(p(4);;p(c++));} This is a fairly straightforward recursive printer, implemented as a nested function (a GNU C extension not supported by clang) to avoid having to pass v around. p(n) prints Wn-2, where W-1 and W0 must be provided in v[1] and v[2]. This initially calls p(4) to print W2. It then loops: it calls p(3) to print W1, making the complete output W2W1, which is W3. It then calls p(4) to print W2, making the complete output W4, etc. Performance is slightly better than my earlier answer: I'm seeing 1875034112 values in a minute. # C, 81 (clang) This is a completely different approach from the above that I feel is worth keeping up, even though it scores worse. s[],*p=s;main(c,v)char**v;{for(++v;;)for(puts(v[*++p=*++p!=1]);*p+1==*--p;++*p);} This has all kinds of undefined behaviour, mainly for fun. It works with clang 3.6.2 on Linux and with clang 3.5.2 on Cygwin for the test cases in the question, with or without special command-line options. It doesn't break when optimisations are enabled. It doesn't work with other compilers. You may accept the words in any order; as two arrays, an array of arrays, two strings, an array of strings, or a single string with a delimiter of your choice. I accept the words as command-line arguments in string format. You may print the digits of the infinite word either without a delimiter or with a consistent delimiter between each pair of adjacent digits. I use newline as the consistent delimiter. For all purposes, assume that your code will never run out of memory, and that its data types do not overflow. In particular, this means that any output to STDOUT or STDERR that is the result of a crash will be ignored. I access s out of bounds. This must surely end with a segfault or an access violation at some point. s happens to get placed at the end of the data segment, so it shouldn't clobber other variables and give wrong output before that segfault. Hopefully. If I run your code on my machine (Intel i7-3770, 16 GiB RAM, Fedora 21) for one minute and pipe its output to wc -c, it must print at least one million digits of W for (W-1, W0) = (1, 0). Testing using { ./program 1 0 | tr -d '\n' & sleep 60; kill$!; } | wc -c, I get 1816784896 digits in one minute on my machine when the program was compiled with -O3, and 1596678144 when it was compiled with optimisations diabled.
Ungolfed, no UB, with explanation:
#include <stdio.h>
// Instead of starting with -1 and 0, start with 0 and 1. I will use lowercase w for that,
// so that wx = W(x-1).
// Declare a variable length array of numbers indicating what has been printed.
// The length is indicated through a pointer just past the end of the values.
// The first element of the array is a special marker.
// [0 3 1] means that w3 w1 has been printed.
// The array is initialised to [0] meaning nothing has been printed yet.
int stack[99999];
int *ptr = stack + 1;
int main(int argc, char *argv[]) {
++argv; // Let argv[0] and argv[1] refer to w0 and w1.
for(;;) {
// Determine the word to print next, either 0 or 1.
// If we just printed 1 that wasn't the end of a different word, we need to print 0 now.
// Otherwise, we need to print 1.
int word = ptr[-1] != 1;
// Print the word, and mark the word as printed.
puts(argv[word]);
*ptr++ = word;
// If we just printed w(x) w(x-1) for any x, we've printed w(x+1).
while (ptr[-1] + 1 == ptr[-2]) {
--ptr;
++ptr[-1];
}
}
}
• Your evil s[] trick works well with clang (just installed it). I'm quite surprised this actually works. For all purposes, assume that your code will never run out of memory, and that its data types do not overflow. Unfortunately, that means simply printing W97 is not considered valid. Could you call p recursively? That would eliminate the need for a main. Commented Nov 7, 2015 at 16:43
• @Dennis To be fair, at the current rate, the version that cheats by printing W97 would do the right thing in printing W∞ for >3000 years. I'll see if I can improve on that. :)
– hvd
Commented Nov 7, 2015 at 20:52
• @Dennis I managed to get it working with the same number of bytes for the program, but making it specific to GCC, and no longer having a clean function. I don't see how to put the logic of repeatedly calling p into p itself without adding more code, but if I do find a way I'll edit again.
– hvd
Commented Nov 7, 2015 at 21:18
# Javascript (53 bytes)
(a,c)=>{for(;;process.stdout.write(a))b=a,a=c,c=b+a;}
Input should be string and not number ('0' and not just 0).
• Welcome to Programming Puzzles & Code Golf! Our rules for code golf challenges state that, by default, submissions must be full programs or functions. As such, they have to accept some sort of user input; hardcoding the input is not allowed. Furthermore, your code prints the sequence Wn, not its limit. Commented Nov 6, 2015 at 18:40
## Perl 5, 4555 49 bytes
44 bytes, plus 1 for -E instead of -e
sub{$i=1;{say$_[++$i]=$_[$i].$_[$i-1];redo}} Use as e.g. perl -E'sub{$i=1;{say$_[++$i]=$_[$i].$_[$i-1];redo}}->(1,0)'
This prints successive approximations to W and thus, if you wait long enough, prints, on its last line of output, W to any desired length, as required. The digits of the word are concatenated without a delimiter.
Since I'm on Windows, I tested it for the "at least one million digits of W" requirement by running it with output redirected to a file and killing it after about 59 seconds, then running GnuWin32's wc -L, which printed 701408733.
## Update:
The OP clarified in a comment on this answer (and probably I should have realized anyway) that the extra output preceding W disqualifies the above. So instead here's a 55-byte solution that prints only W:
sub{print$_[0];{print$_[1];unshift@_,$_[0].$_[1];redo}}
Used the same way, but with the arguments in reverse order, and doesn't require -E:
perl -e'sub{print$_[0];{print$_[1];unshift@_,$_[0].$_[1];redo}}->(0,1)'
Doubtless it can be golfed further, but I don't see how to do so right now.
## Further update:
Dennis shaved five bytes by using -a (thus reading <> to remove sub) and reassigning the parameter passed to print at the end of the redo block:
With -ane and reading from <> (both inputs on one line, space-separated, in reverse order); 48 + 2 bytes:
$_=$F[0];{print;unshift@F,$F[0].($_=$F[1]);redo} And, based on that, I shaved one more byte (same as above, but now the inputs are in the correct order); 47+2 bytes: $_=$F[1];{print;push@F,$F[-1].($_=$F[-2]);redo}
## REXX, 48
arg a b
do forever
b=a||b
say b
a=b||a
say a
end
ftw.rex
exec ftw.rex 0 1
Currently can't test performance because I used an online compiler to write it. The "forever" can be replaced with any number where as the printed ftw numbers are (number + 2).
I also wrote a small (messy) solution in Prolog. Didn't figure how to test performance with this one but its probably atrocious anyway.
f(A,B,C):-atom_concat(B,A,D),(C=D;f(B,D,C)).
:- C='0';C='1';(f(1,0,C)).
## Python 2, 67 bytes
a,b=input()
print'\n'.join(b+a)
while 1:a,b=b,b+a;print'\n'.join(a)
Accepts input as a comma-separated pair of strings: "1","0", for the example in the question.
No online interpreter because infinite loops are bad. Buffered output made me gain a lot of bytes. :( Thanks Dennis for pointing out that 1 digit per line is valid.
Timing on my (significantly-weaker) machine:
$time python golf.py <<< '"1","0"' 2>/dev/null | head -c2000000 > /dev/null real 0m1.348s user 0m0.031s sys 0m0.108s • The question allows a consistent delimiter between digits. You can save at least 28 bytes by printing each digit on a separate line. Commented Nov 6, 2015 at 18:16 ## Minkolang 0.11, 62 bytes (od" "=,6&xI0G2@dO$I)I1-0G($d2[I2:g]Xx0c2*1c-0g0g-d[icO]0G0G1) Try it here. Expects input in the order W0, W-1 with a space in between. ### Explanation ( Open while loop (for input-reading) od Read in character from input and duplicate " "=, 0 if equal to " ", 1 otherwise 6& Jump 6 characters if this is non-zero xI0G2@ Dump, push length of stack, move to front, and jump next two dO Duplicate and output as character if 1$I) Close while loop when input is empty
I1-0G Push length of stack - 1 and move it to the front
Meta-explanation for the following is that at this point in time, we have two numbers followed by a string of "0"s and "1"s with no separation. If the lengths of W0 and W-1 are a and b, respectively, then the two numbers at the front of the stack are <a+b> and <a>, in that order. The word formed by concatenating Wi+1 and Wi, i.e. Wi+1 + Wi, is equal to 2 * Wi+1 - Wi. So the following code duplicates the stack (2 * Wi+1), pops off the top <a> elements (- Wi), and then replaces <a+b> and <a> with their successors, <a+2b> and <b>.
( Open while loop (for calculation and output)
$d Duplicate the whole stack 2[I2:g] Pull <a+b> and <a> from the middle of the stack Xx Dump the top <a> elements (and <a+b>) 0c2*1c- Get <a+b> and <a>, then calculate 2*<a+b> - <a> = <a+2b> = <a+b> + <b> 0g0g- Get <a+b> and <a>, then subtract d[icO] Output as character the first <b> elements 0G0G Move both to front 1) Infinite loop (basically, "while 1:") • (Note: this doesn't produce 1 million digits in a minute...only 0.5 million. Given that this is naturally a relatively slow language, I think I can be cut a little slack. :P) Commented Nov 10, 2015 at 7:23 ## Python 3, 32 def f(a,b):print(end=b);f(b,a+b) The same recursive idea as my Haskell answer, except the prefix is printed because Python can't handle infinite strings. Used a trick from Sp3000 to print without spaces by putting the string as the end argument in Python 3 # Perl, 32 bytes #!perl -pa$#F=3}for(@F){push@F,$F[-1].$_
Counting the shebang as two, input is taken from stdin, space separated as W0, W-1. Output for 1MB times at ~15ms, most of which can be attributed to interpreter launch.
Sample Usage
$echo 0 1 | perl inf-ftw.pl 0100101001001010010100100101001001010010100100101001010010010100100101001010010010100100101001010010010100101001001010010010100101001001010010100100101001001010010100100101001001010010100100101001010010010100100101001010010010100100101001010010010100101001001010010010100101001001010010100100101001001010010100100101001001010010100100101001010010010100100101001010010010100101001001010010010100101001001010010010100101001001010010100100101001001010010100100101001001010010100100101001010010010100100101001010010010100101001001010010010100101001001010010010100101001001010010100100101001001010010100100101001001010010100100101001010010010100100101001010010010100101001001010010010100101001001010010010100101001001010010100100101001001010010100100101001010010010100100101001010010010100100101001010010010100101001001010010010100101001001010010010100101001001010010100100101001001010010100100101001010010010100100101001010010010100100101001010010010100101001001010010010100101001001010010100100101001001...$ echo 110 101 | perl inf-ftw.pl
1101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101101011101011101101011101011101101011101101011101011101101011101101011101011101101011101011101101011101...
# Prolog, 69 bytes
q(A,B):-atom_concat(B,A,S),write(A),q(B,S).
p(A,B):-write(B),q(A,B).
Example input: p('1','0')
Haven't found a way to remove the extra write.
Should be able to improve on this if I figure out how to do that.
# J, 22 bytes including 6 for input chars
similar to Dyalog APL solution
'1'((]stdout@])\$:,)'0'
Try it online! | 8,737 | 28,334 | {"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-30 | latest | en | 0.880025 |
https://www.codewars.com/users/pawptart/published_kumite | 1,638,446,892,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362219.5/warc/CC-MAIN-20211202114856-20211202144856-00570.warc.gz | 770,968,757 | 26,051 | Club's riddle
Fundamentals
Games
Code
Diff
• ``````def riddle(word):
return word.rindex("?")``````
• 1 1 ```def riddle(word): ``` 2 − ``` return len(word) - 1 ``` 2 + ``` return word.rindex("?") ```
One Away (Levenshtein)
A bit faster:
``````Rehearsal ------------------------------------
3.207147 0.008056 3.215203 ( 3.215506)
0.760552 0.000013 0.760565 ( 0.760609)
1.287006 0.000000 1.287006 ( 1.287230)
0.466994 0.000000 0.466994 ( 0.467006)
0.105617 0.000005 0.105622 ( 0.105649)
0.065559 0.000017 0.065576 ( 0.065609)
--------------------------- total: 5.900966sec
user system total real
1.493939 0.000000 1.493939 ( 1.494093) # Eileenandrea
0.785715 0.000000 0.785715 ( 0.785772) # Eileenandrea, a==b check
1.238191 0.000018 1.238209 ( 1.238325) # 深紅心
0.484410 0.000000 0.484410 ( 0.484429) # 深紅心, a==b check
0.076156 0.000000 0.076156 ( 0.076182) # pawptart
0.065472 0.000000 0.065472 ( 0.065472) # pawptart (refactored)
``````
Code
Diff
• ``````def one_away(a, b)
return true if a == b
case a.size - b.size
when -1 then compare(a, b)
when 0 then compare(b, a, 1)
when 1 then compare(b, a)
else false
end
end
def compare(a, b, o = 0)
(0..b.size).each do |i|
return a[0, i] == b[0, i] && a[i + o, a.size] == b[i + 1, b.size] if a[i] != b[i]
end
true
end``````
• 1 1 ```def one_away(a, b) ``` 2 2 ``` return true if a == b ``` 3 3 ``` ``` 4 4 ``` case a.size - b.size ``` 5 − ``` when -1..1 then compare(a, b) ``` 5 + ``` when -1 then compare(a, b) ``` 6 + ``` when 0 then compare(b, a, 1) ``` 7 + ``` when 1 then compare(b, a) ``` 6 6 ``` else false ``` 7 7 ``` end ``` 8 8 ```end ``` 9 9 10 − ```def compare(a, b) ``` 11 − ``` c = 0 ``` 12 − ``` i = 0 ``` 13 − ``` x, y = [a, b].sort_by(&:size) # Don't have to do this, but it makes the code easier ``` 14 − ``` x_off = x.size < y.size ? 0 : 1 # Used in case the string sizes are mismatched ``` 15 − ``` ``` 16 − ``` # Find the first difference ``` 17 − ``` # Use the index to split the strings into substrings ``` 18 − ``` # Compare the substrings ``` 19 − ``` while i < y.size ``` 20 − ``` if x[i] != y[i] ``` 21 − ``` return x[0, i] == y[0, i] && x[i + x_off, x.size] == y[i + 1, y.size] ``` 22 − ``` end ``` 23 − ``` i += 1 ``` 12 + ```def compare(a, b, o = 0) ``` 13 + ``` (0..b.size).each do |i| ``` 14 + ``` return a[0, i] == b[0, i] && a[i + o, a.size] == b[i + 1, b.size] if a[i] != b[i] ``` 24 24 ``` end ``` 16 + ``` ``` 17 + ``` true ``` 25 25 ```end ```
One Away (Levenshtein)
Added some benchmarks, improved some existing solutions
Substring comparison is orders of magnitude faster than iterating over each letter of the string in Ruby.
`==` drops down into C to perform the comparison instead of relying on Ruby's relatively slow iteration.
Sample benchmark:
``````Rehearsal ------------------------------------
2.953821 0.028788 2.982609 ( 2.983072) # Eileenandrea
1.069810 0.000000 1.069810 ( 1.069971) # 深紅心
0.414153 0.000074 0.414227 ( 0.414269) # 深紅心 (modified)
0.087044 0.003973 0.091017 ( 0.091033) # pawptart
--------------------------- total: 4.557663sec
user system total real
1.521839 0.000000 1.521839 ( 1.522046)
1.044602 0.000028 1.044630 ( 1.044804)
0.407469 0.000000 0.407469 ( 0.407515)
0.065739 0.000000 0.065739 ( 0.065745)
``````
Code
Diff
• ``````def one_away(a, b)
return true if a == b
case a.size - b.size
when -1..1 then compare(a, b)
else false
end
end
def compare(a, b)
c = 0
i = 0
x, y = [a, b].sort_by(&:size) # Don't have to do this, but it makes the code easier
x_off = x.size < y.size ? 0 : 1 # Used in case the string sizes are mismatched
# Find the first difference
# Use the index to split the strings into substrings
# Compare the substrings
while i < y.size
if x[i] != y[i]
return x[0, i] == y[0, i] && x[i + x_off, x.size] == y[i + 1, y.size]
end
i += 1
end
end``````
• 1 − ```def one_away(a,b) ``` 2 − ``` case a.length - b.length ``` 3 − ``` when 0 ``` 4 − ``` i = 0 ``` 5 − ``` while i < a.length && a[i] == b[i] ``` 6 − ``` i += 1 ``` 7 − ``` end ``` 8 − ``` i += 1 ``` 9 − ``` # probably slower than just using substring equality, ``` 10 − ``` while i < a.length && a[i] == b[i] ``` 11 − ``` i += 1 ``` 12 − ``` end ``` 13 − ``` # but the allure of pretty complexities is too much ``` 14 − ``` return i >= a.length ``` 15 − ``` when 1 ``` 16 − ``` return helper(a,b) ``` 17 − ``` when -1 ``` 18 − ``` return helper(b, a) ``` 19 − ``` else ``` 20 − ``` return false ``` 1 + ```def one_away(a, b) ``` 2 + ``` return true if a == b ``` 3 + ``` ``` 4 + ``` case a.size - b.size ``` 5 + ``` when -1..1 then compare(a, b) ``` 6 + ``` else false ``` 21 21 ``` end ``` 22 22 ```end ``` 23 23 24 − ```def helper(longer, shorter) ``` 25 − ``` i = 0 ``` 26 − ``` while i < shorter.length && longer[i] == shorter[i] ``` 27 − ``` i += 1 ``` 28 − ``` end ``` 29 − ``` while i < shorter.length && longer[i + 1] == shorter[i] ``` 30 − ``` i += 1 ``` 10 + ```def compare(a, b) ``` 11 + ``` c = 0 ``` 12 + ``` i = 0 ``` 13 + ``` x, y = [a, b].sort_by(&:size) # Don't have to do this, but it makes the code easier ``` 14 + ``` x_off = x.size < y.size ? 0 : 1 # Used in case the string sizes are mismatched ``` 15 + ``` ``` 16 + ``` # Find the first difference ``` 17 + ``` # Use the index to split the strings into substrings ``` 18 + ``` # Compare the substrings ``` 19 + ``` while i < y.size ``` 20 + ``` if x[i] != y[i] ``` 21 + ``` return x[0, i] == y[0, i] && x[i + x_off, x.size] == y[i + 1, y.size] ``` 22 + ``` end ``` 23 + ``` i += 1 ``` 31 31 ``` end ``` 32 − ``` return i >= shorter.length ``` 33 33 ```end ```
Array#flip method
Recursion
Algorithms
Computability Theory
Theoretical Computer Science
Arrays
Methods
Functions
Object-oriented Programming
Control Flow
Basic Language Features
Fundamentals
Classes
Code
Diff
• ``````class Array
def flip
reverse.map{|e|e.flip rescue e}
end
end``````
• 1 − ```class Array # Opens up the Array class for method creation ``` 1 + ```class Array ``` 2 2 ``` def flip ``` 3 − ``` self.reverse.map{|e| e.is_a?(Array) ? e.flip : e} ``` 3 + ``` reverse.map{|e|e.flip rescue e} ``` 4 4 ``` end ``` 5 5 ```end ```
Code
Diff
• ``````from functools import reduce
def prod(numbers):
return reduce(lambda x, y: x * y, numbers or [0], 1)
``````
• 1 + ```from functools import reduce ``` 2 + 1 1 ```def prod(numbers): ``` 2 − ``` return numbers[0] * (prod(numbers[1:]) or 1) if numbers else 0 ``` 4 + ``` return reduce(lambda x, y: x * y, numbers or [0], 1) ``` 5 + ``` ```
Minimal Square
Fundamentals
Puzzles
Games
Code
Diff
• ``````# It doesn't get any simpler, but we can definitely make it harder!
def minimal_square(a, b):
# This solution abuses the fact that we can multiply by True/False:
# For instance, 1 * True = 1 and 1 * False = 0
#
# A step further:
# (a * (a > b ) + b * ( b >= a )) is equivalent to max(a, b)
#
# And:
# (a * ( a < b ) + b * ( b <= a )) is equivalent to min(a, b)
#
# Putting it all together, the following is equivalent to:
# max(min(a,b)*2,max(a,b))**2
side = (((a*(a<b)+b*(b<=a))*2)*(((a*(a<b)+b*(b<=a))*2)>(a*(a>b)+b*(b>=a))))+\
((a*(a>b)+b*(b>=a))*(((a*(a<b)+b*(b<=a))*2)<=(a*(a>b)+b*(b>=a))))
return side**2
``````
• 1 + ```# It doesn't get any simpler, but we can definitely make it harder! ``` 2 + 1 1 ```def minimal_square(a, b): ``` 2 − ``` return max(min(a, b)*2, max(a, b))**2 ``` 3 − ```# This is my first kumite (∩_∩) ``` 4 − ```# I think this is the best solution there's not much to improve on ``` 5 − ```# The problem isn't too hard ``` 4 + ``` # This solution abuses the fact that we can multiply by True/False: ``` 5 + ``` # For instance, 1 * True = 1 and 1 * False = 0 ``` 6 + ``` # ``` 7 + ``` # A step further: ``` 8 + ``` # (a * (a > b ) + b * ( b >= a )) is equivalent to max(a, b) ``` 9 + ``` # ``` 10 + ``` # And: ``` 11 + ``` # (a * ( a < b ) + b * ( b <= a )) is equivalent to min(a, b) ``` 12 + ``` # ``` 13 + ``` # Putting it all together, the following is equivalent to: ``` 14 + ``` # max(min(a,b)*2,max(a,b))**2 ``` 15 + 16 + ``` side = (((a*(a(a*(a>b)+b*(b>=a))))+\ ``` 17 + ``` ((a*(a>b)+b*(b>=a))*(((a*(ab)+b*(b>=a)))) ``` 18 + ``` return side**2 ```
Mathematics
Algorithms
Numbers
Code
Diff
• ``````def average(g)
g==[] ? nil : g.sum.to_f/g.size
end ``````
• 1 1 ```def average(g) ``` 2 − ``` g.size < 1 ? nil : g.reduce(:+).to_f / g.size ``` 2 + ``` g==[] ? nil : g.sum.to_f/g.size ``` 3 3 ```end ```
Series of Calculators - Easy
Code
Diff
• ``````def calculator(a, operator, b):
return "Not a valid operation" if operator not in ['+', '-', '/', '*'] else eval(str(a) + operator + str(b))
``````
• 1 1 ```def calculator(a, operator, b): ``` 2 − ``` return eval(str(a) + operator + str(b)) ``` 2 + ``` return "Not a valid operation" if operator not in ['+', '-', '/', '*'] else eval(str(a) + operator + str(b)) ```
Series of Calculators - Easy
Code
Diff
• ``````def calculator(a, operator, b):
return eval(str(a) + operator + str(b))
``````
• 1 1 ```def calculator(a, operator, b): ``` 2 − ``` operations = {"+": lambda x, y: x + y, "-": lambda x, y: x - y, ``` 3 − ``` "*": lambda x, y: x * y, "/": lambda x, y: x / y} ``` 4 − ``` return operations[operator](a, b) ``` 2 + ``` return eval(str(a) + operator + str(b)) ```
find highest and lowest number in a string array
Code
Diff
• ``````def high_and_low(str_in)
str_in.split.map(&:to_i).minmax.reverse.join(' ')
end``````
• 1 − ```def high_and_low(str_in): ``` 2 − ``` arr = sorted(int(n) for n in str_in.split(" ")) ``` 3 − ``` return "{} {}".format(arr[-1], arr[0]) ``` 1 + ```def high_and_low(str_in) ``` 2 + ``` str_in.split.map(&:to_i).minmax.reverse.join(' ') ``` 3 + ```end ```
Is Odd ?
Numbers
Algorithms
One character less
Code
Diff
• ``````def isOdd(n):
return str(n % 2 > 0)``````
• 1 1 ```def isOdd(n): ``` 2 − ``` return str(n % 2 != 0) ``` 2 + ``` return str(n % 2 > 0) ```
Is Odd ?
Numbers
Algorithms
Code
Diff
• ``````def isOdd(n):
return str(n % 2 != 0)``````
• 1 − ```def isOdd(n) : ``` 2 − ``` return "False" if n % 2 == 0 else "True" ``` 1 + ```def isOdd(n): ``` 2 + ``` return str(n % 2 != 0) ```
gavrashenko's Kumite #5
Code
Diff
• ``````function findBiggestSubArray(arr) {
let maxSum = null;
let startIndex, lastIndex = 0;
for (let i = 0; i < arr.length; i++) {
let sequence = [];
for (let j = i; j < arr.length; j++) {
sequence.push(arr[j]);
let sum = sequence.reduce((c,a)=>c+a);
if (maxSum == null || sum > maxSum) {
startIndex = i;
lastIndex = j;
maxSum = sum;
}
}
}
return [maxSum, startIndex, lastIndex];
}``````
• 1 1 ```function findBiggestSubArray(arr) { ``` 2 − ``` return [] ``` 2 + ``` let maxSum = null; ``` 3 + ``` let startIndex, lastIndex = 0; ``` 4 + ``` ``` 5 + ``` for (let i = 0; i < arr.length; i++) { ``` 6 + ``` let sequence = []; ``` 7 + ``` for (let j = i; j < arr.length; j++) { ``` 8 + ``` sequence.push(arr[j]); ``` 9 + ``` let sum = sequence.reduce((c,a)=>c+a); ``` 10 + ``` if (maxSum == null || sum > maxSum) { ``` 11 + ``` startIndex = i; ``` 12 + ``` lastIndex = j; ``` 13 + ``` maxSum = sum; ``` 14 + ``` } ``` 15 + ``` } ``` 16 + ``` } ``` 17 + ``` ``` 18 + ``` return [maxSum, startIndex, lastIndex]; ``` 3 3 ```} ```
Get iterated digit sum
Tiny, tiny performance bump under certain circumstances.
Code
Diff
• ``````require 'benchmark'
strings = []
10000.times do
string = Array.new(rand(500..1000)) do rand(10) end.join
strings << string
end
def solution1(s) s.chars.sum(&:to_i) end
def solution2(s) s.sum - s.size * 48 end
def solution3(s) s.sum do |x| x - 48 end end
def solution4(s) s.sum do |x| x - '0' end end
def solution5(s) s.sum{|x|x-'0'} end
Benchmark.bmbm do |x|
x.report do strings.each(&method(:solution1)) end # farekkusu
x.report do strings.each(&method(:solution2)) end # ttakuru88
x.report do strings.each(&method(:solution3)) end # steffan153 (test code author)
x.report do strings.each(&method(:solution4)) end # rowcased
x.report do strings.each(&method(:solution5)) end # pawptart
end
puts "<PASSED::>" # prevent from saying failed tests``````
• 88
```end
```
99
1010
```def solution1(s) s.chars.sum(&:to_i) end
```
1111
```def solution2(s) s.sum - s.size * 48 end
```
1212
```def solution3(s) s.sum do |x| x - 48 end end
```
1313
```def solution4(s) s.sum do |x| x - '0' end end
```
14+
```def solution5(s) s.sum{|x|x-'0'} end
```
1414
1515
```Benchmark.bmbm do |x|
```
1616
``` x.report do strings.each(&method(:solution1)) end # farekkusu
```
1717
``` x.report do strings.each(&method(:solution2)) end # ttakuru88
```
1818
``` x.report do strings.each(&method(:solution3)) end # steffan153 (test code author)
```
1919
``` x.report do strings.each(&method(:solution4)) end # rowcased
```
21+
``` x.report do strings.each(&method(:solution5)) end # pawptart
```
2020
```end
```
2121
2222
```puts "<PASSED::>" # prevent from saying failed tests
```
Return Hello World
• ``````def hello_world
• 1 − ```char* Hi (void) ``` 2 − ```{char* ans; ``` 3 − ```asprintf(&ans, "Hello World."); ``` 4 − ```return ans;} ``` 1 + ```def hello_world ``` 2 + ``` print(HELLO_WORLD) ``` 3 + ```end ``` | 4,682 | 13,135 | {"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-2021-49 | latest | en | 0.363553 |
http://www.slideserve.com/aviva/electrical-circuits | 1,490,804,216,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218190753.92/warc/CC-MAIN-20170322212950-00207-ip-10-233-31-227.ec2.internal.warc.gz | 677,571,820 | 16,233 | This presentation is the property of its rightful owner.
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Electrical Circuits PowerPoint PPT Presentation
Electrical Circuits. EQ: How can we illustrate the transformation of electrical energy to sound, light, heat, and mechanical motion?
Electrical Circuits
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Electrical Circuits
EQ: How can we illustrate the transformation of electrical energy to sound, light, heat, and mechanical motion?
**To answer this question, as we work through the presentation write down at LEAST one example of electrical energy to sound, light, heat and mechanical motion as presented in the illustrations in the EQ blank.
• What is an energy transformation?
• Describe at least one example you know.
Electric Circuits
• An electric circuit contains a source of electrical energy, a conductor of the electrical energy (wire) connected to the energy source, and a device that uses and transforms the electrical energy.
• All these components must be connected in a complete, unbroken path in order for energy transformations to occur.
Sources of electrical energy
• The electrical energy in a battery comes from stored chemical energy.
• The electrical energy in a solar cell comes from light energy from the sun.
• The electrical energy in outlets may come from chemical energy (burning fuels) which powers a generator in a power plant.
Light
• Electrical energy can be transformed into light energy in an electric circuit if a light bulb is added to the circuit.
• The transformation in this case might be that chemical energy in a battery is transformed into electrical energy in the circuit which is transformed into light and heat energy in the light bulb.
Sound
• Electrical energy can be transformed into sound energy in an electric circuit if a bell, buzzer, radio, or TV is added to the circuit.
• The transformation in this case might be that chemical energy in a battery is transformed into electrical energy in the circuit which is transformed into sound energy by the buzzer.
Heat
• Electrical energy can be transformed into heat energy in an electric circuit if a toaster, stove, or heater is added to the circuit.
• How is the nuclear power plant an example of chemical energy transformed to mechanical energy?
• The transformation in this case might be that chemical energy from the fuel at the power plant is transformed into heat energy which is transformed into mechanical energy to turn a generator.
• The generator transforms the mechanical energy into electrical energy.
• Then the electrical energy in the circuit is transformed into heat energy in the heater.
Mechanical motion
• Electrical energycan be transformed into the energy of mechanical motion if a fan or motor is added to the circuit.
• Transformation in this case might be that chemical energy in a battery is transformed into electrical energy in the circuit which is transformed into the energy of mechanical motion by the fan or motor. | 679 | 3,393 | {"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-2017-13 | longest | en | 0.8852 |
http://60.83.publicone.net/multiplying-decimals-worksheet-grade-6/ | 1,606,922,425,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141711306.69/warc/CC-MAIN-20201202144450-20201202174450-00663.warc.gz | 1,663,986 | 30,878 | In Free Printable Worksheets186 views
4.34 / 5 ( 123votes )
Top Suggestions Multiplying Decimals Worksheet Grade 6 :
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Top | 1,898 | 9,321 | {"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} | 3.171875 | 3 | CC-MAIN-2020-50 | latest | en | 0.761497 |
https://www.crazy-numbers.com/en/92955 | 1,713,118,971,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816893.19/warc/CC-MAIN-20240414161724-20240414191724-00663.warc.gz | 670,171,975 | 3,408 | Warning: Undefined array key "numbers__url_substractions" in /home/clients/df8caba959271e8e753c9e287ae1296d/websites/crazy-numbers.com/includes/fcts.php on line 156
Number 92955: mathematical and symbolic properties | Crazy Numbers
Discover a lot of information on the number 92955: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 92955
Is 92955 a prime number? No
Is 92955 a perfect number? No
Number of divisors 8
List of dividers
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1, 3, 5, 15, 6197, 18591, 30985, 92955
Sum of divisors 148752
Prime factorization 3 x 5 x 6197
Prime factors
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3, 5, 6197
## How to write / spell 92955 in letters?
In letters, the number 92955 is written as: Ninety-two thousand nine hundred and fifty-five. And in other languages? how does it spell?
92955 in other languages
Write 92955 in english Ninety-two thousand nine hundred and fifty-five
Write 92955 in french Quatre-vingt-douze mille neuf cent cinquante-cinq
Write 92955 in spanish Noventa y dos mil novecientos cincuenta y cinco
Write 92955 in portuguese Noventa e dois mil novecentos cinqüenta e cinco
## Decomposition of the number 92955
The number 92955 is composed of:
2 iterations of the number 9 : The number 9 (nine) represents humanity, altruism. It symbolizes generosity, idealism and humanitarian vocations.... Find out more about the number 9
1 iteration of the number 2 : The number 2 (two) represents double, association, cooperation, union, complementarity. It is the symbol of duality.... Find out more about the number 2
2 iterations of the number 5 : The number 5 (five) is the symbol of freedom. It represents change, evolution, mobility.... Find out more about the number 5 | 570 | 2,039 | {"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} | 2.78125 | 3 | CC-MAIN-2024-18 | latest | en | 0.692724 |
https://oeis.org/A180498 | 1,686,077,259,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653071.58/warc/CC-MAIN-20230606182640-20230606212640-00162.warc.gz | 476,313,332 | 3,727 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A180498 a(n) = n^2 - 5*floor(n/sqrt(5))^2. 1
1, 4, 4, 11, 5, 16, 4, 19, 1, 20, 41, 19, 44, 16, 45, 11, 44, 4, 41, 80, 36, 79, 29, 76, 20, 71, 9, 64, 121, 55, 116, 44, 109, 31, 100, 16, 89, 164, 76, 155, 61, 144, 44, 131, 25, 116, 4, 99, 196, 80, 181, 59, 164, 36, 145, 11, 124, 239, 101, 220, 76, 199, 49, 176, 20, 151 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS a(n)=1 for n=A023039. LINKS Zak Seidov, Table of n, a(n) for n = 1..1000 Zak Seidov, Graphic of A180498. EXAMPLE a(4)=11 since 4^2-5*floor(4/sqrt(5))^2=16-5=11. CROSSREFS Cf. A023039, A082532. Sequence in context: A161719 A343090 A161433 * A107856 A212102 A168373 Adjacent sequences: A180495 A180496 A180497 * A180499 A180500 A180501 KEYWORD nonn AUTHOR Carmine Suriano, Sep 08 2010 STATUS approved
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Last modified June 6 14:05 EDT 2023. Contains 363147 sequences. (Running on oeis4.) | 517 | 1,276 | {"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-2023-23 | latest | en | 0.628224 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=19&t=34036&p=111726 | 1,571,274,286,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986672431.45/warc/CC-MAIN-20191016235542-20191017023042-00066.warc.gz | 573,433,060 | 11,815 | ## H bar
$\Delta p \Delta x\geq \frac{h}{4\pi }$
Daniel Bowen 3I
Posts: 32
Joined: Fri Sep 28, 2018 12:24 am
### H bar
Does anyone know if h bar will be a given constant or if we will have to find its value mathematically in each problem
Alicia Gibbons 1B
Posts: 31
Joined: Fri Sep 28, 2018 12:17 am
### Re: H bar
What do you mean? H bar is a constant that equals h/2pi. I think it's written this way so it can theoretically be simplified using the values of delta p and delta x.
Chloe Qiao 4C
Posts: 65
Joined: Fri Sep 28, 2018 12:27 am
### Re: H bar
I do not think h bar will be given as a constant during the test, but the equation of delta x * delta p >=h/4pi will probably on the equation sheet. So don't worry about h bar(unless you really like to use it and if so just remember h bar equals to h/2pi).
Megan Gianna Uy 3L
Posts: 30
Joined: Fri Sep 28, 2018 12:26 am
### Re: H bar
He has the equation sheet posted on his website with all the equations he's giving us for the tests. In this case the equation on the sheet is (delta p)(delta x)>=h/(4pi), so you dont need to know the value of hbar since hbar =h/2pi.
Hannah Yates 1K
Posts: 59
Joined: Fri Sep 28, 2018 12:27 am
### Re: H bar
H bar will not be given, which just means you need to do one more step when using Heisenberg indeterminacy equation. But, h will definitely be given.
Dong Hyun Lee 4E
Posts: 68
Joined: Fri Sep 28, 2018 12:23 am
### Re: H bar
H bar just stands for h/2pi. He does have the equation given on the first page of the exams but I would memmorize the equation he has taught us in the class where (indeterminancy in momentum) times (indeterminancy in position) = h/(4pi). As this is just easier to memmorize and use. It is useful to know the concepts behind it but at least for me, when solving questions this format is just easier to use. | 560 | 1,843 | {"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": 0, "codecogs_latex": 1, "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-2019-43 | latest | en | 0.899278 |
https://blog.hmns.org/2009/06/how-much-is-oil-worth/ | 1,632,832,481,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060803.2/warc/CC-MAIN-20210928122846-20210928152846-00480.warc.gz | 186,401,663 | 16,997 | How much is oil worth? | BEYONDbones
# How much is oil worth?
June 3, 2009
4035 Views
photo credit: tvol
Ever wonder how common items such as coffee, beer, or detergent compare to the price of oil? Don’t worry, I have already done the math for you. By breaking down the measurements into smaller units, you can compare the price of a glass of milk to a cup of oil, but I don’t recommend drinking it.
1 barrel of oil (bbl) contains 42 gallons.
1 gallon contains 16 cups.
Therefore, 1 barrel contains (42 * 16) = 672 cups.
If a barrel of oil today costs \$68, then a cup of oil would come out to \$0.10
That’s right; a nice cold glass of oil would only cost you 10 cents. Some of you coffee drinkers may notice that a Grande Latte from Starbucks, which contains about 2 cups of liquid, costs\$3.10. The same amount of oil would be about 20 cents.
Lets take that a bit further. How much would a barrel of Grande Latte cost?
Remember that there are 672 cups of liquid in a barrel, and that a cup of Grande Latte is \$1.55 a cup.
The cost of a barrel of Grande Latte (is it still a Grande if it is in a barrel?) = \$1041.60. But think how long you could stay awake!
Grande Lattes not quite your cup of tea? Here is a list of the prices at Starbucks in case you don’t like lattes.
Now let’s move on to something a little harder, and a little more alcoholic:
Beer provides more of a challenge because it does not come in gallons or cups, but in bottles measured in fluid ounces (fl.oz.).
In a six-pack there are 6 bottles of 12 fl.oz. each. If a six pack costs \$8, then each bottle would cost \$1.33 (cost of the six pack) /6 = per bottle price.
8 fl.oz. = 1 cup
If each bottle in a six pack has 12 fl.oz., then there are 72 fl.oz. of liquid. This comes out to 9 cups. At 8\$ a sixpack, this comes out to just \$0.11 per fl.oz. That means each cup of beer would cost 88 cents. That seems cheap compared to a Grande Latte at \$1.55 a cup, but a bit pricy compared to oil at \$0.10 a cup.
At \$0.88 a cup, a barrel of beer would cost \$591. For those of you interested:
photo credit: G & A Sattler
(672 * 8 = number of fl.oz in a barrel) / 12 fl.oz / bottle = the number of bottles in the barrel.
This means there are 448 bottles of beer in your \$591 barrel of beer.
You can continue on with any number of liquids.
Milk is an easy one to work with because it already comes in gallons.
If 1 gallon of milk is \$2, then to get the price per barrel you simply multiply by 42, so milk comes out to be \$84 per barrel.
You can go the other way and figure out that milk (16 cups in a gallon) costs \$0.13 per cup.
I have included a few household items here for you to compare:
Cup Barrel Oil \$0.10 \$68 Grand Latte \$1.55 \$1041.60 Beer \$0.88 \$591 Milk \$0.13 \$84 Coke \$0.24 \$161.28 Nyquil \$6.00 \$4032 Tide Detergent \$2.40 \$1612.80 Listerine \$2.00 \$1344
Try these conversions with your other favorite liquids such as water, juice, shampoo and sun screen!
Authored By Daniel Burch
An inveterate punster, amateur chef, and fencer, Daniel B has a double degree in History and Museum Science from Baylor. He currently serves as the Assistant Program Coordinator for the Wiess Energy Hall and Adult Education at HMNS.
### 2 responses to “How much is oil worth?”
1. Lyndell says:
How much per serving or use size?
2. Daniel B says:
A per serving size can vary depending on item.
A Grande Latté is one serving with 188 calories. So that’s 1 serving (2 cups or 16 fl.oz.) for \$3.10
A bottle of Beer is 1 serving of 12 fl.oz. with 150 calories for \$1.33
In a gallon of milk there are 16 serving of 8 fl.oz. each with 91 calories for \$0.13
In Coca cola 1 serving can is 12 fl.oz. with 140 calories for \$.36
Nyquil dose is 1 fl.oz. for \$0.75
Tide Detergent is 1.5 fl.oz. for each load for \$0.30
Listerine is 1 fl.oz. for \$0.25
And crude oil?
You can’t do much with unrefined crude oil, but if you turn into other products, like gasoline.
You get 19.15 gallons of gasoline from 1 barrel of crude oil or about 45% of a barrel of crude oil
(See the DoE’s EIA )
1 gallon of gasoline is about \$2.40
So if you drive about 7 miles to work or school (14 mile round trip)and your car gets about 25 miles per gallon, then you use .56 of a gallon (or 17360 calories). Then it would cost you \$1.34 to get to and from school and you would have used 3% of the gasoline that comes out of 1 barrel of crude oil or 1% of a barrel of crude oil
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(281) 242-3055 | 1,397 | 4,915 | {"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} | 4.0625 | 4 | CC-MAIN-2021-39 | latest | en | 0.942347 |
https://www.nagwa.com/en/videos/367138103521/ | 1,582,464,365,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145774.75/warc/CC-MAIN-20200223123852-20200223153852-00455.warc.gz | 817,523,211 | 6,308 | # Video: Verifying Whether the Mean is a Good Measure of Central Tendency for a Data Set Represented on a Line Plot
A group of girl scouts is selling boxes of cookies to raise money for charity. The line plot shows how many boxes of cookies they sold. State whether the mean would be a good measure of central tendency for the data or not. Explain your answer. [A] Yes, the mean closely represents the middle of the data. (B) No, the mean is skewed too high because of the data point 36.
03:58
### Video Transcript
A group of girl scouts is selling boxes of cookies to raise money for charity. The line plot shows how many boxes of cookies they sold. State whether the mean would be a good measure of central tendency for the data or not. Explain your answer. Is it (A) yes, the mean closely represents the middle of the data or (B) no, the mean is skewed too high because of the data point 36?
The mean is a good measure of central tendency when there are no outliers or extreme values in our data set. In order to answer this question, we need to calculate the mean with and without the possible outlier of 36, the highest number of boxes of cookies sold. If the mean of the data alters significantly when this point is removed, then it will not be a good measure of central tendency.
We recall that in order to calculate the mean, we need to divide the sum of the values by the number of values. In this case, we need to add two 20s, four 22s, two 24s, and so on. An easier way to calculate the sum would be to multiply the number of data points at each value. We have two data points at 20. And two multiplied by 20 is 40. There are four data points at 22. Four multiplied by 22 is 88. Two multiplied by 24 is 48. Three multiplied by 26 is 78. Repeating this process, we have 28, 90, 32, and 36.
There are no data points at 34, 38, and 40. This means that none of the girl scouts sold this number of boxes of cookies. The sum of these numbers is 440. We need to divide this by 17 as there are 17 data points. This means that the group contained 17 girl scouts. Dividing 440 by 17 gives us 25.882 and so on. Rounding this to one decimal place, we have 25.9. The mean number of boxes of cookies sold by the 17 girls was 25.9.
When we remove the girl who sold 36 boxes, we have 404 boxes in total as 440 minus 36 is 404. We need to divide this by 16 as this is the number of other girl scouts. 404 divided by 16 is 25.25. After we have removed the value of 36, the mean has decreased from 25.9 to 25.25. Whilst the mean has decreased slightly, it is still between two of the data points, 24 and 26. This means that the value of 36 has had very little impact on the mean. We can, therefore, conclude that 36 is not an outlier. And this means that the correct answer is: yes, the mean closely represents the middle of the data.
Had the value of 36 been significantly higher, then it would’ve impacted the mean. In that situation, the mean wouldn’t have been a good measure of central tendency. | 745 | 3,001 | {"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} | 4.53125 | 5 | CC-MAIN-2020-10 | latest | en | 0.929459 |
http://s3.us-east-2.amazonaws.com/eijgetboattrip/maths-10th-ncert-71-zip.html | 1,638,350,611,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359976.94/warc/CC-MAIN-20211201083001-20211201113001-00405.warc.gz | 78,201,553 | 5,569 | ## Aluminum Bass Boats For Sale In Texas
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## Maths 10th Ncert 7.1 Zip,Boat Excursion West Palm Beach 001,Hakvoort Yacht Builders 75 - How to DIY
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Find the ratio in which y-axis divides the Ncert Solutions Of Class 10th Maths Chapter 7 Exercise 7.3 line segment joining the points A 5,�6 and B �1, �4. From 1 and 2we can say that ABCD is not a rectangle therefore it is a parallelogram. Name the type of quadrilateral formed, if any, by the following 10th Ncert Maths Deleted Portion Id points, and give reasons for your answer. Find maths 10th ncert 7.1 zip point on the x-axis which is equidistant from Maths 10th Ncert 8.4 Zip 2, -5 and -2, 9. The midpoints maths 10th ncert 7.1 zip the sides of a triangle are 3,44,1 and 2,0. The sign of x and y-coordinates in Lorem lpsum 296 boatplans/online/ranger-vs-tracker-aluminum-boats-online here of the quadrant is shown below:. The x and y taken together in order is called coordinte of point denoted by x, y. | 397 | 1,427 | {"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-2021-49 | latest | en | 0.928917 |
https://en.m.wikipedia.org/wiki/Euler%27s_theorem | 1,606,663,297,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141198409.43/warc/CC-MAIN-20201129123729-20201129153729-00360.warc.gz | 283,860,029 | 19,229 | # Euler's theorem
In number theory, Euler's theorem (also known as the Fermat–Euler theorem or Euler's totient theorem) states that if n and a are coprime positive integers, then a raised to the power of the totient of n is congruent to one, modulo n, or:
${\displaystyle a^{\varphi (n)}\equiv 1{\pmod {n}}}$
where ${\displaystyle \varphi (n)}$ is Euler's totient function. In 1736, Leonhard Euler published his proof of Fermat's little theorem,[1] which Fermat had presented without proof. Subsequently, Euler presented other proofs of the theorem, culminating with "Euler's theorem" in his paper of 1763, in which he attempted to find the smallest exponent for which Fermat's little theorem was always true.[2]
The converse of Euler's theorem is also true: if the above congruence is true, then ${\displaystyle a}$ and ${\displaystyle n}$ must be coprime.
The theorem is a generalization of Fermat's little theorem, and is further generalized by Carmichael's theorem.
The theorem may be used to easily reduce large powers modulo ${\displaystyle n}$. For example, consider finding the ones place decimal digit of ${\displaystyle 7^{222}}$, i.e. ${\displaystyle 7^{222}{\pmod {10}}}$. The integers 7 and 10 are coprime, and ${\displaystyle \varphi (10)=4}$. So Euler's theorem yields ${\displaystyle 7^{4}\equiv 1{\pmod {10}}}$, and we get ${\displaystyle 7^{222}\equiv 7^{4\times 55+2}\equiv (7^{4})^{55}\times 7^{2}\equiv 1^{55}\times 7^{2}\equiv 49\equiv 9{\pmod {10}}}$.
In general, when reducing a power of ${\displaystyle a}$ modulo ${\displaystyle n}$ (where ${\displaystyle a}$ and ${\displaystyle n}$ are coprime), one needs to work modulo ${\displaystyle \varphi (n)}$ in the exponent of ${\displaystyle a}$:
if ${\displaystyle x\equiv y{\pmod {\varphi (n)}}}$, then ${\displaystyle a^{x}\equiv a^{y}{\pmod {n}}}$.
Euler's theorem underlies RSA cryptosystem, which is widely used in Internet communications. In this cryptosystem, Euler's theorem is used with n being a product of two large prime numbers, and the security of the system is based on the difficulty of factoring such an integer.
## Proofs
1. Euler's theorem can be proven using concepts from the theory of groups:[3] The residue classes modulo n that are coprime to n form a group under multiplication (see the article Multiplicative group of integers modulo n for details). The order of that group is φ(n). Lagrange's theorem states that the order of any subgroup of a finite group divides the order of the entire group, in this case φ(n). If a is any number coprime to n then a is in one of these residue classes, and its powers a, a2, ... , ak modulo n form a subgroup of the group of residue classes, with ak ≡ 1 (mod n). Lagrange's theorem says k must divide φ(n), i.e. there is an integer M such that kM = φ(n). This then implies,
${\displaystyle a^{\varphi (n)}=a^{kM}=(a^{k})^{M}\equiv 1^{M}=1\equiv 1{\pmod {n}}.}$
2. There is also a direct proof:[4][5] Let R = {x1, x2, ... , xφ(n)} be a reduced residue system (mod n) and let a be any integer coprime to n. The proof hinges on the fundamental fact that multiplication by a permutes the xi: in other words if axjaxk (mod n) then j = k. (This law of cancellation is proved in the article Multiplicative group of integers modulo n.[6]) That is, the sets R and aR = {ax1, ax2, ... , axφ(n)}, considered as sets of congruence classes (mod n), are identical (as sets—they may be listed in different orders), so the product of all the numbers in R is congruent (mod n) to the product of all the numbers in aR:
${\displaystyle \prod _{i=1}^{\varphi (n)}x_{i}\equiv \prod _{i=1}^{\varphi (n)}ax_{i}=a^{\varphi (n)}\prod _{i=1}^{\varphi (n)}x_{i}{\pmod {n}},}$ and using the cancellation law to cancel each xi gives Euler's theorem:
${\displaystyle a^{\varphi (n)}\equiv 1{\pmod {n}}.}$
## Euler quotient
The Euler quotient of an integer a with respect to n is defined as:
${\displaystyle q_{n}(a)={\frac {a^{\varphi (n)}-1}{n}}}$
The special case of an Euler quotient when n is prime is called a Fermat quotient.
Any odd number n that divides ${\displaystyle q_{n}(2)}$ is called a Wieferich number. This is equivalent to saying that 2φ(n) ≡ 1 (mod n2). As a generalization, any number n that is coprime to a positive integer a, and such that n divides ${\displaystyle q_{n}(a)}$ , is called a (generalized) Wieferich number to base a. This is equivalent to saying that aφ(n) ≡ 1 (mod n2).
a numbers n coprime to a that divide ${\displaystyle q_{n}(a)}$ (searched up to 1048576) OEIS sequence 1 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, ... (all natural numbers) A000027 2 1, 1093, 3279, 3511, 7651, 10533, 14209, 17555, 22953, 31599, 42627, 45643, 52665, 68859, 94797, 99463, 127881, 136929, 157995, 228215, 298389, 410787, 473985, 684645, 895167, 1232361, 2053935, 2685501, 3697083, 3837523, 6161805, 11512569, ... A077816 3 1, 11, 22, 44, 55, 110, 220, 440, 880, 1006003, 2012006, 4024012, 11066033, 22132066, 44264132, 55330165, 88528264, 110660330, 221320660, 442641320, 885282640, 1770565280, 56224501667, 112449003334, ... A242958 4 1, 1093, 3279, 3511, 7651, 10533, 14209, 17555, 22953, 31599, 42627, 45643, 52665, 68859, 94797, 99463, 127881, 136929, 157995, 228215, 298389, 410787, 473985, 684645, 895167, ... 5 1, 2, 20771, 40487, 41542, 80974, 83084, 161948, 643901, 1255097, 1287802, 1391657, 1931703, 2510194, 2575604, 2783314, 3765291, 3863406, 4174971, 5020388, 5151208, 5566628, 7530582, 7726812, 8349942, 10040776, 11133256, 15061164, 15308227, 15453624, 16699884, ... A242959 6 1, 66161, 330805, 534851, 2674255, 3152573, 10162169, 13371275, 50810845, 54715147, 129255493, 148170931, 254054225, 273575735, 301121113, 383006029, 646277465, ... A241978 7 1, 4, 5, 10, 20, 40, 80, 491531, 983062, 1966124, 2457655, 3932248, 4915310, 6389903, 9339089, 9830620, 12288275, 12779806, 18678178, 19169709, 19661240, 24576550, 25559612, ... A242960 8 1, 3, 1093, 3279, 3511, 7651, 9837, 10533, 14209, 17555, 22953, 31599, 42627, 45643, 52665, 68859, 94797, 99463, 127881, 136929, 157995, 206577, 228215, 284391, 298389, 383643, 410787, 473985, 684645, 895167, ... 9 1, 2, 4, 11, 22, 44, 55, 88, 110, 220, 440, 880, 1760, 1006003, ... 10 1, 3, 487, 1461, 4383, 13149, 39447, 118341, 355023, 56598313, 169794939, 509384817, ... A241977 11 1, 71, 142, 284, 355, 497, 710, 994, 1420, 1491, 1988, 2485, 2840, 2982, 3976, 4970, 5680, 5964, 7455, 9940, 11928, 14910, 19880, 23856, 29820, 39760, 59640, 79520, 119280, 238560, 477120, ... A253016 12 1, 2693, 123653, 1812389, 2349407, 12686723, 201183431, 332997529, ... A245529 13 1, 2, 863, 1726, 3452, 371953, 743906, 1487812, 1747591, 1859765, 2975624, 3495182, 3719530, 5242773, 6990364, 7439060, 8737955, 10485546, 14878120, 15993979, 17475910, 20971092, 26213865, 29756240, 31987958, 34951820, 41942184, 47981937, 52427730, 59512480, ... A257660 14 1, 29, 353, 3883, 10237, 19415, 112607, 563035, ... 15 1, 4, 8, 29131, 58262, 116524, 233048, 466096, ... 16 1, 1093, 3279, 3511, 7651, 10533, 14209, 17555, 22953, 31599, 42627, 45643, 52665, 68859, 94797, 99463, 127881, 136929, 157995, 228215, 298389, 410787, 473985, 684645, 895167, ... 17 1, 2, 3, 4, 6, 8, 12, 24, 48, 46021, 48947, 92042, 97894, 138063, 146841, 184084, 195788, 230105, 276126, 293682, 368168, 391576, 414189, 460210, 552252, 587364, 598273, 690315, 736336, 783152, 828378, 920420, ... 18 1, 5, 7, 35, 37, 49, 185, 245, 259, 331, 1295, 1655, 1813, 2317, 3641, 8275, 9065, 11585, 12247, 16219, 18205, 25487, 33923, 57925, 61235, 81095, 85729, 91025, 127435, 134717, 169615, 178409, 237461, 306175, 405475, 428645, 455125, 600103, 637175, 673585, 892045, 943019, ... 19 1, 3, 6, 7, 12, 13, 14, 21, 26, 28, 39, 42, 43, 49, 52, 63, 78, 84, 86, 91, 98, 104, 117, 126, 129, 137, 147, 156, 168, 172, 182, 196, 234, 252, 258, 273, 274, 294, 301, 312, 364, 387, 411, 441, 468, 504, 516, 546, 548, 559, 588, 602, 624, 637, 728, 774, 819, 822, 882, 903, 936, 959, 1032, 1092, 1096, 1118, 1176, 1204, 1274, 1456, 1548, 1638, 1644, 1677, 1764, 1781, 1806, 1872, 1911, 1918, 2107, 2184, 2192, 2236, 2329, 2408, 2457, 2548, 2709, 2877, 3096, 3276, 3288, 3354, 3528, 3562, 3612, 3822, 3836, 3913, 4214, 4368, 4472, 4658, 4914, 5031, 5096, 5343, 5418, 5733, 5754, 5891, 6321, 6552, 6576, 6708, 6713, 6987, 7124, 7224, 7644, 7672, 7826, 8127, 8428, 8631, 8736, 8944, 9316, 9828, 10062, 10192, 10686, 10836, 11466, 11508, 11739, 11782, 12467, 12642, 13104, 13152, 13416, 13426, 13974, 14248, 14448, 14749, 15093, 15288, 15344, 15652, 16029, 16254, 16303, 16856, 17199, 17262, 17673, 18632, 18963, 19656, 20124, 20139, 21372, 21672, 22932, 23016, 23478, 23564, 24934, 25284, 26208, 26832, 26852, 27391, 27948, 28496, 29498, 30186, 30277, 30576, 30688, 31304, 32058, 32508, 32606, 34398, 34524, 35217, 35346, 37264, 37401, 37926, 39312, 40248, 40278, 41237, 42744, 43344, 44247, 45864, 46032, 46956, 47128, 48909, 49868, 50568, 53019, 53664, 53704, 54782, 55896, 56889, 56992, 58996, 60372, 60417, 60554, 61152, 62608, 64116, 65016, 65212, 68796, 69048, 70434, 70692, 74528, 74802, 75852, 76583, 78624, 80496, 80556, 82173, 82474, 85488, 87269, 88494, 90831, 91728, 92064, 93912, 94256, 97818, 99736, 100147, 101136, 105651, 106038, 107408, 109564, 111792, 112203, 113778, 113984, 114121, 117992, 120744, 120834, 121108, 123711, 125216, 128232, 130032, 130424, 132741, 137592, 138096, 140868, 141384, 146727, 149056, 149604, 151704, 153166, 160992, 161112, 164346, 164948, 170976, 174538, 176988, 181662, 183456, 184128, 187824, 188512, 191737, 195636, 199472, 200294, 211302, 211939, 212076, 214816, 219128, 223584, 224406, 227556, 228242, 229749, 241488, 241668, 242216, 246519, 247422, 256464, 260848, 261807, 265482, 272493, 275184, 276192, 281736, 282768, 288659, 293454, 298112, 299208, 300441, 303408, 306332, 316953, 322224, 328692, 329896, 336609, 341952, 342363, 349076, 353976, 363324, 371133, 375648, 383474, 391272, 398223, 398944, 400588, 422604, 423878, 424152, 438256, 447168, 448812, 455112, 456484, 459498, 482976, 483336, 484432, 493038, 494844, 512928, 521696, 523614, 530964, 536081, 544986, 550368, 552384, 563472, 565536, 575211, 577318, 586908, 596224, 598416, 600882, 612664, 633906, 635817, 644448, 657384, 659792, 673218, 683904, 684726, 689247, 698152, 701029, 707952, 726648, 739557, 742266, 751296, 766948, 782544, 785421, 796446, 797888, 801176, 845208, 847756, 848304, 865977, 876512, 894336, 897624, 901323, 910224, 912968, 918996, 966672, 968864, 986076, 989688, 1025856, 1027089, 1043392, 1047228, ... 20 1, 281, 1967, 5901, 46457, ... 21 1, 2, ... 22 1, 13, 39, 673, 2019, 4711, 8749, 14133, 26247, 42399, 61243, 78741, 183729, 551187, ... 23 1, 4, 13, 26, 39, 52, 78, 104, 156, 208, 312, 624, 1248, ... 24 1, 5, 25633, 128165, ... 25 1, 2, 4, 20771, 40487, 41542, 80974, 83084, 161948, 166168, 323896, 643901, ... 26 1, 3, 5, 9, 15, 45, 71, 213, 355, 497, 639, 1065, 1491, 1775, 2485, 3195, 4473, 5325, 7455, 12425, 13419, 15975, 22365, 37275, 67095, 111825, 335475, ... 27 1, 11, 22, 44, 55, 110, 220, 440, 880, 1006003, ... 28 1, 3, 9, 19, 23, 57, 69, 171, 207, 253, 437, 513, 759, 1265, 1311, 1539, 2277, 3795, 3933, 4807, 11385, 11799, 14421, 24035, 35397, 43263, 72105, 129789, 216315, 389367, 648945, ... 29 1, 2, ... 30 1, 7, 160541, ...
The least base b > 1 which n is a Wieferich number are
2, 5, 8, 7, 7, 17, 18, 15, 26, 7, 3, 17, 19, 19, 26, 31, 38, 53, 28, 7, 19, 3, 28, 17, 57, 19, 80, 19, 14, 107, 115, 63, 118, 65, 18, 53, 18, 69, 19, 7, 51, 19, 19, 3, 26, 63, 53, 17, 18, 57, ... (sequence A250206 in the OEIS)
## Notes
1. ^ See:
2. ^ See:
• L. Euler (published: 1763) "Theoremata arithmetica nova methodo demonstrata" (Proof of a new method in the theory of arithmetic), Novi Commentarii academiae scientiarum Petropolitanae, 8 : 74–104. Euler's theorem appears as "Theorema 11" on page 102. This paper was first presented to the Berlin Academy on June 8, 1758 and to the St. Petersburg Academy on October 15, 1759. In this paper, Euler's totient function, ${\displaystyle \varphi (n)}$ , is not named but referred to as "numerus partium ad N primarum" (the number of parts prime to N; that is, the number of natural numbers that are smaller than N and relatively prime to N).
• For further details on this paper, see: The Euler Archive.
• For a review of Euler's work over the years leading to Euler's theorem, see: Ed Sandifer (2005) "Euler's proof of Fermat's little theorem"
3. ^ Ireland & Rosen, corr. 1 to prop 3.3.2
4. ^ Hardy & Wright, thm. 72
5. ^ Landau, thm. 75
6. ^
## References
The Disquisitiones Arithmeticae has been translated from Gauss's Ciceronian Latin into English and German. The German edition includes all of his papers on number theory: all the proofs of quadratic reciprocity, the determination of the sign of the Gauss sum, the investigations into biquadratic reciprocity, and unpublished notes.
• Gauss, Carl Friedrich; Clarke, Arthur A. (translated into English) (1986), Disquisitiones Arithemeticae (Second, corrected edition), New York: Springer, ISBN 0-387-96254-9
• Gauss, Carl Friedrich; Maser, H. (translated into German) (1965), Untersuchungen uber hohere Arithmetik (Disquisitiones Arithemeticae & other papers on number theory) (Second edition), New York: Chelsea, ISBN 0-8284-0191-8
• Hardy, G. H.; Wright, E. M. (1980), An Introduction to the Theory of Numbers (Fifth edition), Oxford: Oxford University Press, ISBN 978-0-19-853171-5
• Ireland, Kenneth; Rosen, Michael (1990), A Classical Introduction to Modern Number Theory (Second edition), New York: Springer, ISBN 0-387-97329-X
• Landau, Edmund (1966), Elementary Number Theory, New York: Chelsea | 5,925 | 13,677 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 26, "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} | 4.125 | 4 | CC-MAIN-2020-50 | latest | en | 0.874611 |
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# LINEAR EQUATIONS AND THEIR SOLUTIONS
Mathematics
Study Guide
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1
LINEAR EQUATIONS AND THEIR SOLUTION
1. X-3Y=-6………….1
5X+3Y=42…………2
ANS: By adding the equations, 3y and -3y will cancel out, then
6x = 36
Now dividing by 6 on both sides, we get
X = 6
To find the value of y we put x = 6 in equation 1, then
6 -3y = -6
Adding -6 on both sides, we get
-3y = -12
Dividing by – 3 on both sides, we get
y = 4
Hence (x, y) = (6, 4)
2. 2X-6Y=7 …………3
5X-2Y=10 ………..4
ANS: Multiply equation 3 by 2 and multiply equation 4 by -6, we get
4x - 12y = 14
-30x + 12y = -60
Now adding the above equations, we get
-26x = -46
Now dividing by-26 on both sides, we get
X = 46/26 or
2
LINEAR EQUATIONS AND THEIR SOLUTION
X = 23/13
Now put the value of x in equation 4x - 12y = 14, we get
4(23/13) - 12y = 14
92/13 - 12y = 14
Now multiply by 13 on both sides, we get
92 - 156y = 182
Adding -92 on both sides, we get
-156y = 182
Dividing by -156 on both sides, we get
Y = -91/78
HENCE (x, y) = (23/13, -91/78)
3. 2X+16Y=9………….5
1/4X+4Y=8…………..6
ANS: Multiplying equation 6 by -4, we get
2x + 16y = 9
-x - 16y = 8
X = 17
Putting the value of x in equation 5, we get
2(17) + 16y = 9
34 + 16y = 9
Now adding -34 on both sides, we get
16y = -25
3
LINEAR EQUATIONS AND THEIR SOLUTION
Dividing by 16 on both sides, we get
y = -25/16
HENCE (x, y) = (17, -25/16)
4. X-2Y=7………..7
9X+6Y=15……….8
ANS: Multiplying equation 7 by 3, we get
3x - 6y = 21
9x + 6y = 15
Adding the above equations, we get
12x = 36
Dividing by 12 on both sides, we get
x = 3
Now putting the values in equation 7, we get
3 - 2y = 7
Adding -3 on both sides, we get
-2y = 4
Dividing by -2 on both sides, we get
y = -2
HENCE (x, y) = (3, -2)
5. 7/4X-5/2Y=2………..9
1/4X+7/2Y=8……….10
ANS: Multiply both equations with 4, we get
7x - 10y = 8………..A
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